93
Warm-UP: What do you think is happening to this cell? Homework: 10 Key Ideas Section 12.1 AP Test Money DUE: NEXT FRIDAY! 3/6 Remember, the 2 nd semester Final Exam will be an AP test (the week before the official AP test); you might as well take the official test too!

Warm-UP: What do you think is happening to this cell? Homework: 10 Key Ideas Section 12.1 AP Test Money DUE: NEXT FRIDAY! 3/6 Remember, the 2 nd semester

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Warm-UP

What do you think is happening to this cell

Homework10 Key IdeasSection 121

AP Test Money DUE NEXT FRIDAY 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

>

Topicsbull Concept 131 What are the advantages to sexualasexual reproduction

What is the relationship between a gene a chromosome and DNA

bull Concept 132 What happens to a cell during meiosis Compare mitosis and meiosis Why bother doing meiosis

bull Concept 134 When is genetic diversity increased during meiosis How What is the advantage to this

bull Concept 141 How can we predict the offspring of parents using punnett squares (monohybrid dihybrid) Explain why some traits are dominant to others Explain how variation is increased due to independent assortment

bull Concept 142 Use probability laws to solve punnett squares predicting the outcomes of crosses where many traits are involved

bull Concept 143 Inheritance patterns are more complex that simple dominancerecessive incomplete dominance blood types codominance

bull Concept 144 Use a pedigree analysis of a family to make predictions about future offspring

bull Concept 151 What did Morgan tell us that Mendel couldnrsquot How do chromosomes explain Mendelrsquos 31 ratio

bull Concept 152 Explain how sex is determined Why do some diseases show in boys more often than girls

bull Concept 153 How do linked genes complicate Mendelrsquos findings

Essential Knowledge

bull 3a2 In eukaryotes heritable info is passed to the next generation via mitosis and meiosis

bull 3a3 The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from parent to offspring

bull 3a4 The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

bull 3c2 Biological systems have multiple processes that increase genetic variation

Unit 7 Inheritance

Big Idea In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis which has multiple processes that increases genetic variation The inheritance pattern of many traits cannot be explained by simple Mendelian genetics The chromosomal basis of inheritance explains pattern of transmission of genes from parent to offspring

Time (seconds) Drawing Description025 ndash

030 ndash

035ndash

040 ndash

045 ndash

050-

In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tapebull string

Procedure1 add 3-5 beads to a pipe cleaner to

model genes2 twist a 2nd pipe cleaner to the 1st to

model the double helix3 tie string to the beads to model

mRNA4 wrap the whole model around a

cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Chromosome Model

Procedure1 add 3-5 beads to a pipe cleaner

to model genes2 twist a 2nd pipe cleaner to the

1st to model the double helix3 tie string to the beads to model

mRNA transcribing DNA4 wrap the whole model around

a cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA supercoils

around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected (DNA Replication not quite finished)

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Topicsbull Concept 131 What are the advantages to sexualasexual reproduction

What is the relationship between a gene a chromosome and DNA

bull Concept 132 What happens to a cell during meiosis Compare mitosis and meiosis Why bother doing meiosis

bull Concept 134 When is genetic diversity increased during meiosis How What is the advantage to this

bull Concept 141 How can we predict the offspring of parents using punnett squares (monohybrid dihybrid) Explain why some traits are dominant to others Explain how variation is increased due to independent assortment

bull Concept 142 Use probability laws to solve punnett squares predicting the outcomes of crosses where many traits are involved

bull Concept 143 Inheritance patterns are more complex that simple dominancerecessive incomplete dominance blood types codominance

bull Concept 144 Use a pedigree analysis of a family to make predictions about future offspring

bull Concept 151 What did Morgan tell us that Mendel couldnrsquot How do chromosomes explain Mendelrsquos 31 ratio

bull Concept 152 Explain how sex is determined Why do some diseases show in boys more often than girls

bull Concept 153 How do linked genes complicate Mendelrsquos findings

Essential Knowledge

bull 3a2 In eukaryotes heritable info is passed to the next generation via mitosis and meiosis

bull 3a3 The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from parent to offspring

bull 3a4 The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

bull 3c2 Biological systems have multiple processes that increase genetic variation

Unit 7 Inheritance

Big Idea In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis which has multiple processes that increases genetic variation The inheritance pattern of many traits cannot be explained by simple Mendelian genetics The chromosomal basis of inheritance explains pattern of transmission of genes from parent to offspring

Time (seconds) Drawing Description025 ndash

030 ndash

035ndash

040 ndash

045 ndash

050-

In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tapebull string

Procedure1 add 3-5 beads to a pipe cleaner to

model genes2 twist a 2nd pipe cleaner to the 1st to

model the double helix3 tie string to the beads to model

mRNA4 wrap the whole model around a

cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Chromosome Model

Procedure1 add 3-5 beads to a pipe cleaner

to model genes2 twist a 2nd pipe cleaner to the

1st to model the double helix3 tie string to the beads to model

mRNA transcribing DNA4 wrap the whole model around

a cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA supercoils

around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected (DNA Replication not quite finished)

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Unit 7 Inheritance

Big Idea In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis which has multiple processes that increases genetic variation The inheritance pattern of many traits cannot be explained by simple Mendelian genetics The chromosomal basis of inheritance explains pattern of transmission of genes from parent to offspring

Time (seconds) Drawing Description025 ndash

030 ndash

035ndash

040 ndash

045 ndash

050-

In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tapebull string

Procedure1 add 3-5 beads to a pipe cleaner to

model genes2 twist a 2nd pipe cleaner to the 1st to

model the double helix3 tie string to the beads to model

mRNA4 wrap the whole model around a

cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Chromosome Model

Procedure1 add 3-5 beads to a pipe cleaner

to model genes2 twist a 2nd pipe cleaner to the

1st to model the double helix3 tie string to the beads to model

mRNA transcribing DNA4 wrap the whole model around

a cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA supercoils

around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected (DNA Replication not quite finished)

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Time (seconds) Drawing Description025 ndash

030 ndash

035ndash

040 ndash

045 ndash

050-

In eukaryotic organisms heritable information is passed to the next generation via mitosis and meiosis

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tapebull string

Procedure1 add 3-5 beads to a pipe cleaner to

model genes2 twist a 2nd pipe cleaner to the 1st to

model the double helix3 tie string to the beads to model

mRNA4 wrap the whole model around a

cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Chromosome Model

Procedure1 add 3-5 beads to a pipe cleaner

to model genes2 twist a 2nd pipe cleaner to the

1st to model the double helix3 tie string to the beads to model

mRNA transcribing DNA4 wrap the whole model around

a cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA supercoils

around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected (DNA Replication not quite finished)

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tapebull string

Procedure1 add 3-5 beads to a pipe cleaner to

model genes2 twist a 2nd pipe cleaner to the 1st to

model the double helix3 tie string to the beads to model

mRNA4 wrap the whole model around a

cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Chromosome Model

Procedure1 add 3-5 beads to a pipe cleaner

to model genes2 twist a 2nd pipe cleaner to the

1st to model the double helix3 tie string to the beads to model

mRNA transcribing DNA4 wrap the whole model around

a cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA supercoils

around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected (DNA Replication not quite finished)

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Chromosome Model

Procedure1 add 3-5 beads to a pipe cleaner

to model genes2 twist a 2nd pipe cleaner to the

1st to model the double helix3 tie string to the beads to model

mRNA transcribing DNA4 wrap the whole model around

a cotton ball to represent how DNA supercoils around histones in order to get smaller

5 replicate the 1st to model a sister chromatid

6 tape it to the 1st at about the middle to model the centromere

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA supercoils

around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected (DNA Replication not quite finished)

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Chromosome Model

Materialsbull pipe cleanersbull beadsbull cotton ballsbull tape

Partsbull chromosome tightly coiled DNAbull double helix DNA is 2 strands of

nucleotides twisted togetherbull gene small sections of DNA that

encode traitsbull histone proteins that DNA

supercoils around makes the DNA smaller so that it can fit in our cells

bull sister chromatid a duplicated copy of the original DNA still connected to the 1st

bull centromere the place where the 2 sister chromatids are connected

bull mRNA copies of DNA that read genes

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
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  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP Can you ldquounwindrdquo your model like in the picture to show the parts Sketch and label your model (or the picture of the model) Labels histone mRNA DNA sister chromatid centromere

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight Mitosis Worksheet

UNIT 78 TEST March 19th

Inheritance and Regulation

Due Now Stamp Sheet

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight 122

UNIT 78 TEST March 19th

Inheritance and Regulation

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework for Tonight

Warm-UP

1 Is blowing up a balloon a good model for how cells or organisms grow Explain

2 In what ways are cells different than each other the same

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosis

Cell Division 2 Types Mitosis and Meiosis

Mitosisbull Purpose

growth repair asexual reproduction (actually

binary fission in prokaryotes)bull occurs after DNA replicationbull is followed by cytokinesisbull 2 genetically identical daughter cells

(clones)bull Steps

1 Replication2 Alignment3 Separation

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

1 Draw a big cell with a nucleus inside2 Add homologous pairs of chromosomes

ndash 1 big yellow 1 big whitendash 1 medium yellow 1 medium whitendash 1 small yellow 1 small white

3 Clone each chromosome (DNA Replication) (making sister chromatids)

4 Nucleus breaks down (erase it)5 Clones separate one to each side of cell6 Cytokinesis cell membrane grows between the 2

halves7 Nucleus grows around chromosomes

Mitosis Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
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  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

bull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 127 Stages of Mitosis p232At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid

Mitosis Modeling

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Analysis

1 What is the outcome of mitosis2 What steps of mitosis ensure a perfect

outcome3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have

4 Why do cells go through mitosis

Mitosis Modeling

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Words to Knowbull Nucleusbull DNAbull Histonesbull Chromatinbull Chromosomebull Chromatidbull Centromere

Copyright copy 2002 Pearson Education Inc publishing as Benjamin Cummings

Interphase ldquoSrdquo phase

Mitosis

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP

1Advantages of Sex

2Disadvantages of Sex

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Thurs Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Fig 13-2

Bacteria

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environmentPlants

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Fig 13-2

Parent

Bud

05 mmAsexual Reproduction

Advantages1 Can be done in isolation2 If you are perfect why change

Best when an organism is ldquoperfectlyrdquo suited for its environment

3 Less ldquomachineryrdquo needed no waste on flowers ldquonewrdquo types of cells or other secondary sexual characteristics

Disadvantages1 Less variety Poor option in a

changing environment

Aspen

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Fig 13-1

Sexual ReproductionAdvantages1 Increased variety Best when

environment is changing

Increased Variety is DUE TObull random fertilizationbull meiosis

bull independent assortmentbull crossing-over

Disadvantages1 Resource investment

bull specialized cells gametesbull secondary sexual

characteristics mate attraction2 Requires a partner (except when

plants self-fertilize)

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
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  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 81
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  • Slide 83
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  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

In eukaryotic organisms heritable information is passed to the next generation via mitosis and

meiosisCell Division 2 Types

Meiosis sex cells (gametes) (spermeggs)bull Purpose

ensures that each gamete receives one complete haploid (1n) set of chromosomes

Ensures chromosome number stays the same each offspring receives frac12 (1n) of the full set for the species (2n)

bull Steps1 DNA Replication2 homologous pairs form a tetrad

(independent assortment) (genetic variation increases)

3 cross over between homologous chromosomes may occur (genetic variation increases)

4 Meiosis I homologous pairs separate5 Meiosis II sister chromatids separate

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb Add homologous pairs of chromosomes

bull 1 big white 1 big yellowbull 1 medium white 1 medium yellowbull 1 small white 1 small yellow

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model crossing

over for now)5 Homologous pairs separate one to each side of cell all paternal (or all maternal) do NOT

have to go together (independent assortment)6 Cytokinesis cell membrane grows between the 2 halves7 Clones (sister chromatids) separate (centromere ldquobreaksrdquo as DNA polymerization

completes) one to each side of cell8 Cytokinesis cell membrane grows between the 2 halves9 Nucleus grows around chromosomes 4 cells made (gametes which have half the

number of chromosomes (haploids)

Meiosis Modeling

When yoursquore confident on the process show Mr Jones

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
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  • Slide 86
  • Slide 87
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Meiosis Modelingbull When yoursquore confident on the process show Mr Jones

bull In your notebookTable 138 Stages of Meiosis p255At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Meiosis Modeling

Analysis

1 What is the outcome of meiosis2 What steps of meiosis cause variation3 If the first cell has 46 chromosomes (before

cloning) how many does each daughter cell have4 How many unique gametes did you form How

many could have formed5 Why do cells go through meiosis

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP 1 Why do meiosis2 How does meiosis increase variation in organisms

AP Test Money DUE 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow Meiosis Wksht

UNIT 78 TEST March 19th

Inheritance and Regulation

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Meiosis has multiple processes that increase genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Meiosis has multiple processes that increases genetic variation

Genetic Variation is due to

Mutations (last unit)

Meiosisbull Independent assortment of chromosomes paternal (red)

chromosomes maymay not move together during Meiosis Indash causes an organism with n homologous pairs to be able to

produce 2n varieties of gametes instead of only 2ndash ex humans 223=8388608

bull Crossing over swapping of sections of homologous pairs during Meiosis I produces recombinant chromosomesndash causes ldquolinkedrdquo genes to become ldquounlinkedrdquo 2 alleles on

the same chromosome maymay not move together during Meiosis I

ndash recombinant chromosomes have unique combinations of alleles

Random fertilization the particular combination of egg and sperm that results in the offspring is not predictable

ndash ex humans dadrsquos varieites of sperm (8388608) and momrsquos varieites of eggs (8388608) can produce varieties 70368744177664 offpring

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP Watch the video What is happening Why does this happen How does this increase variation

AP Test Money DUE TOMORROW 36Remember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow

UNIT 78 TEST March 19th

Inheritance and Regulation

>

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Fig 13-9a

MITOSIS MEIOSIS

MEIOSIS I

Prophase I

Chiasma

Chromosomereplication

Homologouschromosomepair

Chromosomereplication

2n = 6

Parent cell

Prophase

Replicated chromosome

Metaphase Metaphase I

Anaphase ITelophase I

Haploid n = 3

Daughter cells ofmeiosis I

MEIOSIS II

Daughter cells of meiosis II

nnnn

2n2n

Daughter cellsof mitosis

AnaphaseTelophase

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Compare

Contrast

Independent Assortment

Crossing-over Sexual Fertilization Mutation

Types of Changes to Cells Think About what where when how how much variation created

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 92
  • Slide 93

Warm-UP Now we need to keep track of the genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons Barbie has Huntingtons What is the chance her child will get Huntingtons

AP Test Money DUE 36 TOMORROWRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

Homework DUE Tomorrow 141 5 Key Ideas AND Concept Check 1-3

H=dominant (remember Huntingtonrsquos is dominant)

h=recessive (no Huntingtonrsquos)

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 83
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

AP Test Money DUE 36 TODAYRemember the 2nd semester Final Exam will be an AP test (the week before the official AP test) you might as well take the official test too

DUE next Thursday Monohybrid Crosses

Homework DUE Monday Test Corrections for 10 Make-UP

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP Now we need to keep track of TWO UNLINKED genes on the chromosomes that are inherited Notice the letters on the chromosomes They represent different alleles for Huntingtons and red hair color Barbie has Huntingtons and is a blond (but a carrier) Ken does not have Huntingtonrsquos and is also a carrier for red hair What is the chance their child will get Huntingtons and red hair

Homework DUE Tomorrow 142 5 Key Ideas AND Concept Check 1-3

H= Huntingtonrsquosh=no HuntingtonrsquosR= no red hairr= red hair

DUE Now Test Corrections for 10 Make-UP

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
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  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
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  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
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The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
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  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
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  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Mendel 1857bull P Crossed purebredsbull F1 all one phenotypebull F2 31 ratio

Conclusion1 ldquoAllelesrdquo segregate (in meiosis)

independently2 Some alleles (dominant) hide the

expression of others (recessive)3 Recessive alleles can appear again

IF the dominant are no longer present

Termsbull allele variety of a genebull homozygous both alleles are the

same (purebred)bull heterozygous alleles are different

(carriers) (hybrids)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
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  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
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  • Slide 45
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  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
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  • Slide 43
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  • Slide 50
  • Slide 51
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  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 81
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
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  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 83
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  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Inheritance Patterns

bull Simple DominanceRecessive Mendelbull monohybrid cross one traitbull di or tri hybrid cross two or three traits

assuming independent assortmentbull laws of probability vs punnett squares

bull incomplete dominance phenotype of hybrids is between the phenotypes of the purebreds

bull codominance two dominant alleles affect the phenotype in separate distinguishable ways

bull polygenic inheritance an additive effect of two or more genes on a single phenotype

bull x-linked genes genes on the x chromosome are expressed by boys more often than girls bc boys only have one x chromosome (they cannot ldquomaskrdquo recessive alleles)

bull linked genes when 2 alleles are on the same chromosome they show up together more often

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow 143 5 Key Ideas AND Concept Check 1-3

DUE Now 142DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

UNIT 78 TEST March 19th

Inheritance and Regulation

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
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  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 83
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoods

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
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  • Slide 55
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with RrYy

1 Chance of baby with Rr 052 Chance of baby with Yy 053 Chance of baby with RrYy 025

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
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  • Slide 50
  • Slide 51
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  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 0752 Chance of baby with at least one Y 0753 Chance of baby with RrYy 05625 (916)

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 93

The inheritance pattern of many traits cannot be explained by simple Mendelian genetics

Laws of Probabilitybull skip Punnett squaresbull instead multiply likelyhoodsbull ex

Chance of baby with both dominant phenotypes

1 Chance of baby with at least one R 10

2 Chance of baby with at least one Y 10

3 Chance of baby with both dominant phenotypes 10

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

Homework DUE Tomorrow Linked vs Unlinked Genes AND Monohybrid Crosses to tunritincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with

vestigial wings and cinnabar eyes Predict the phenotypic ratio of the F2 individuals

Homework DUE Tomorrow 153 5 Key Ideas and Concept Check 1-3DUE Thursday Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 143

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 89
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  • Slide 92
  • Slide 93

Warm-UP

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn) The recessive vg allele causes small malformed wings called vestigial wings The recessive cn allele causes bright-red eyes called cinnabar eyes In the first cross a purebred female with wild-type wings and red eyes is mated with a purebred male with vestigial wings and cinnabar eyes 1 Predict the phenotypic ratio of the F1 individuals 2 In the second cross female F1 flies are mated with males with vestigial wings and cinnabar eyes Predict the

phenotypic ratio of the F2 individuals

3 The results of the cross from 2 show the following outcome Individuals of each phenotype are shown in the table Which of the following is the most likely explanation of the results a The two genes are located on two different chromosomes b The two genes are sex-linked c The two genes are located on mitochondrial DNA d The two genes are linked on an autosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Morganbull experiments with fruit flies bull Interested in how linkage (when

characters are on the same chromosome) affects inheritance of two characters

bull crossed flies that differed in traits of body color and wing size

Conclusionsbull body color and wing size are usually

inherited together in specific combinations (parental phenotypes)

bull these genes sometimes assort independentlybull they must linked on the same

chromosomebull but the linkage is incomplete

cross-over sometimes creates recombinant chromosomes

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
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  • Slide 41
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  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
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  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 87
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

1 Build your diploid cell (an homologous pair for each chromosome)a Draw a big cell with a nucleus insideb DRAW homologous pairs of chromosomes

bull 1 big blue 1 big redbull 1 small blue 1 small red

ndash Add genes to your chromosomebull Write an H and an R on big blue and an h and an r on big redbull Write a T on small blue and a t on small red

2 Clone each chromosome (DNA Replication) (making sister chromatids)3 Nucleus breaks down (erase it)4 Homologous pairs form a tetrad crossing over may occur (note do not model

crossing over for now)5 Finish meiosis (see notes for help)6 Nucleus grows around chromosomes 4 cells made (gametes which have half

the number of chromosomes (haploids)7 Repeat Steps 1-6 this time at Step 4 include ONE crossover event

Recombination Modeling

When yoursquore confident on the process show Mr Jones

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

bull When yoursquore confident on the process show Mr Jones

bull In your notebookFigure 1510 Recombination p295At each stage describe what is happening in 1-2 sentences DO NOT COPY

DRAW with LABELS cell membrane nucleus chromosome DNA replication cytokinesis sister chromatid crossing over recombinant chromosome independent assortment Meiosis I Meiosis II

Recombination Modeling

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Analysis

1 How does recombination increase the variety of gametes formed

2 Which types of gametes were more likely Gametes with recombinant chromosomes or ones without

3 Recombination is a random process How might it benefithurt the survival of an individualspecies

Recombination Modeling

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
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  • Slide 44
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  • Slide 49
  • Slide 50
  • Slide 51
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  • Slide 53
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  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP

1 How could you calculate recombination frequency2 What would the recombination frequency tell you about the relative position of

genes on a chromosome3 Genes A B and C are on the same chromosome Testcrosses show that the

recombination frequency between A and B is 28 and between A and C is 12 Can you determine the linear order of these genes

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Homework DUE Tomorrow Linked and Unlinked GenesDUE Tonight Monohybrid Crosses to turnitincomDUE Friday Dihybrid Crosses to turnitincom

DUE Now 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP Talk about your Concept Check with your table team Do you have agreement Do you have questions

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Sturtevant Genetic Mapping

bull linkage map an ordered list of the genetic loci along a particular chromosome based on recombination frequencies

bull map unitbull one map unit represents a 1

recombination frequencybull indicate relative distance and order

not precise locations of genesbull 50 frequency of recombination is

observed for any two genes on different chromosomes

bull the farther apart two genes are the higher the probability a crossover will occur between them and therefore the higher the recombination frequency

bull genes that are far apart on the same chromosome can have a recombination frequency near 50 Such genes are physically linked but genetically unlinked and behave as if found on different chromosomes

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 92
  • Slide 93

The chromosomal basis of inheritance provides an understanding of the pattern of transmission of genes from

parent to offspring Abnormal Chromosome Numberbull In nondisjunction pairs of homologous

chromosomes do not separate normally during meiosis

bull As a result one gamete receives two of the same type of chromosome and another gamete receives no copy

bull Aneuploidy results from the fertilization of gametes in which nondisjunction occurred

bull Offspring with this condition have an abnormal number of a particular chromosome

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
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  • Slide 49
  • Slide 50
  • Slide 51
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  • Slide 53
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  • Slide 55
  • Slide 56
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Warm-UP What can we learn about inheritance from looking at a pedigree (use example of the Royal Family of Europe from the early 1900s)

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 83
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Warm-UP In the following human pedigree squares represent males circles represent females and shaded symbols indicate individuals affected with a disorder One of the affected males from the third generation has a child with a female who is a carrier For the pedigree shown which of the following best expresses the probability that the couplersquos first son will be affected with the disorder

(A) 25 (B) 50 (C) 75 (D) 100

DUE Tonight Dihybrid Crosses to turnitincom

DUE Linked vs Unlinked Genes 153

UNIT 78 TEST March 19th

Inheritance and Regulation

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
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  • Slide 45
  • Slide 46
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  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
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  • Slide 83
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  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Is a widowrsquos peak a dominant or recessive trait

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Is a widowrsquos peak a dominant or recessive trait

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

Fig 14-UN5

George

Sandra Tom Sam

Arlene

Wilma Ann Michael

Carla

Daniel Alan Tina

Christopher

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

A heterozygous tall and a homozygous purple flowered plant is crossed with a short and white flowered plant Predict the offspring Assume independent assortment

1

P=purplep=whiteT=tallt=short

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
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  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
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  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

For a plant with the genotype WwXxYyZz what is the probability that the plant will produce a gamete with a haploid genotype of Wxyz Give your answer as a fraction or as a value between 0 and 1 to four decimal places

2

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

A completely-plated stickleback from a marine population was mated to a low-plated stickleback from a freshwater population The resulting F1 hybrids all displayed a completely plated phenotype When the F1 hybrids were allowed to interbreed the resulting F2 generation included completely plated offspring and low-plated offspring in an approximate 31 ratio Which of the following conclusions is best supported by the results of the breeding experiments

a Phenotypic variation in the F2 generation suggests that armor morphology is controlled by many alleles of a single gene

b The completely-plated phenotype is controlled by a dominant allele of a single gene

c Armor loss is an acquired characteristic that is affected by one or more environmental factors

d Patterns of armor plating in stickleback populations are regulated by sex-specific signals

3

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

If a man with type AB blood marries a woman with type O what blood types would you expect in their children What fraction would you expect of each type

4

IA= Type AIB= Type Bi= Type O

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

If a King marries a Queen that is a carrier of hemophilia (sex-linked recessive disease that is historically fatal before adulthood) what is the chance of the Kingdom having an heir to the kingrsquos throne assuming only one child is born

5

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
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  • Slide 89
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  • Slide 91
  • Slide 92
  • Slide 93

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

6Is an attached earlobe a dominant or recessive trait Draw the pedigree and fill in the genotypes for each offspring

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
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  • Slide 43
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  • Slide 49
  • Slide 50
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  • Slide 54
  • Slide 55
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  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
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  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

7 Gregor Mendelrsquos pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome

a Pea color and pea shape display independent inheritance patterns b Offspring of a given cross show all possible combinations of traits in

equal proportions c Most offspring of a given cross have a combination of traits that is

identical to that of either one parent or the other d Recessive phenotypes can skip a generation showing up only in the

parental and F 2 generations

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
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  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

8

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Based on the data in Table I which of the following best explains why there are no individuals with constricted pods in the F 1 generation a Inflated pod shape is dominant to constricted pod shape b The inflated-pod offspring in the F sub one generation are

homozygous c Constricted pod shape typically arises from a new mutation in the F 1

generation d The constricted-pod offspring are carriers for the inflated pod shape

allele

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Sexual Reproduction
  • Slide 25
  • Slide 26
  • Slide 27
  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
  • Slide 31
  • Slide 32
  • Slide 33
  • Slide 34
  • Slide 35
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • Slide 42
  • Slide 43
  • Slide 44
  • Slide 45
  • Slide 46
  • Slide 47
  • Slide 48
  • Slide 49
  • Slide 50
  • Slide 51
  • Slide 52
  • Slide 53
  • Slide 54
  • Slide 55
  • Slide 56
  • Slide 57
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
  • Slide 70
  • Slide 71
  • Slide 72
  • Slide 73
  • Slide 74
  • Slide 75
  • Slide 76
  • Slide 77
  • Slide 78
  • Slide 79
  • Slide 80
  • Slide 81
  • Slide 82
  • Slide 83
  • Slide 84
  • Slide 85
  • Slide 86
  • Slide 87
  • Slide 88
  • Slide 89
  • Slide 90
  • Slide 91
  • Slide 92
  • Slide 93

9

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table I the ratio of phenotypes in the offspring from the testcross with F 1 plants that had purple flowers and inflated pods suggests that the genes for flower color and pod shape are located a close together on the same autosome b on the X chromosome c on different chromosomes d on a mitochondrial chromosome

10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

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  • Topics
  • Unit 7 Inheritance
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  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
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  • Sexual Reproduction
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  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
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  • Recombination Modeling
  • Recombination Modeling (2)
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10

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

Which of the following provides the best justification for an assumption that might have been used in the computer simulation (Table II) a The broad allele is recessive to the narrow allele because broad

leaves appear in every generation b The purple allele is dominant to the white allele because all the

offspring from the cross of purple-flowered and white-flowered plants had purple flowers

c The narrow allele is codominant with the purple allele because the purple-flower trait and the narrow-leaf trait segregate together

d The white allele is dominant to both the broad and narrow alleles because plants with either type of leaf shape can have white flowers

11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
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  • Slide 8
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  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
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  • Sexual Reproduction
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  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
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  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
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11

Table I shows the results of breeding experiments to examine the inheritance of flower color (purple versus white) and pod shape (inflated versus constricted) For the crosses recorded in Table I true-breeding parents were crossed to produce F1 offspring which were then testcrossed to homozygous recessive individuals Table II shows the results of computer-simulated crosses to model the inheritance of leaf shape (broad versus narrow) and flower color (purple versus white)

In Table II the F 1 offspring of the cross between broad-leaved white-flowered plants with narrow-leaved purple-flowered plants have a phenotype that differs from that of either parent However many testcross offspring have the same phenotype as one of the two plants in the parental cross but relatively few testcross offspring have the same phenotype as the F 1 offspring Which of the following best explains the observation a Recombination between the leaf-shape and flower-color genes resulted in

chromosomes carrying a dominant allele of both genes b Recombination between the broad and narrow alleles of the leaf-shape gene resulted in

chromosomes carrying three different alleles at the same genetic locus c Independent assortment of homologous chromosomes resulted in the combinations of

alleles present in the parental generation d The computer model cannot capture the possible assortments of gametes when

multiple genes are considered

  • Slide 1
  • Topics
  • Unit 7 Inheritance
  • Slide 4
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  • Slide 7
  • Slide 8
  • Slide 9
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  • Mitosis Modeling
  • Mitosis Modeling (2)
  • Mitosis Modeling (3)
  • Slide 19
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  • Sexual Reproduction
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  • Meiosis Modeling
  • Meiosis Modeling (2)
  • Meiosis Modeling (3)
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  • Recombination Modeling
  • Recombination Modeling (2)
  • Recombination Modeling (3)
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