23-1 Honors Biology chapter 23 Genetics John Regan Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display

23-1

Honors Biology

chapter 23Genetics

John Regan

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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23.1 Mendel’s laws

• Heredity - Biological inheritance– Passing of traits from parents to offspring

• Genetics- Study of heredity

• Gregor Mendel– Published a paper in 1866 stating that parents pass discrete

heritable factors on to their offspring• Factors retain individuality generation after generation• Identified that each trait is inherited by a pair of factors, one

from each parent

– Law of Dominance– One form of a factor may be dominant over an alternative form

– Dominant (capital letter)

– Prevents the expression of the other form –recessive (small letter)

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Mendel’s laws cont’d.• Alleles- alternate forms of a gene for the same trait

– Capital letter- represents the dominant allele – Small letter- represents the recessive allele

• Dominant-a certain trait will result if the individual has at least 1 dominant allele

• Recessive- for a recessive trait to result the individual must have 2 copies of the recessive allele

– Genotype- genetic composition of an individual with regard to a specific trait

• 2 copies of the dominant allele- homozygous dominant

• 1 copy of the dominant allele and 1 of the recessive- heterozygous

• 2 copies of the recessive allele- homozygous recessive

– Phenotype- physical characteristics of a trait - outward appearance of a trait

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Mendel’s laws cont’d.

• Mendel’s law of segregation– Each individual has two factors (genes) for each trait– The factors segregate (separate) during the formation

of gametes– Each gamete contains only one factor from each pair

of factors– Fertilization gives each new individual 2 factors for

each trait

The Inheritance of Traits The parent

generation is also known as the P generation.

The offspring of this P cross are called the first filial (F1) generation.

The second filial (F2) generation is the offspring from the F1 cross.

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• What we know now– Genes are sections of chromosomes– Chromosomes come in pairs called homologous pairs

• Homologous pairs have genes controlling the same traits

• Genes are located at the same point or locus, on each member of the pair

• Information contained within the homologous genes is not necessarily the same (ex: blues vs. brown eyes)

– Alternative forms of a gene for a trait are called alleles

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Gene locus

• Fig. 23.2

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• The inheritance of a single trait cont’d.– Phenotype- physical appearance of the individual with regard to

a trait• Homozygous dominant individual and heterozygous individual will

have the same genotype in this instance

• Homozygous recessive individual will have a different phenotype

– Example- inheritance of a widow’s peak vs. a straight hairline in people

• Alternative forms of alleles for hairline shape

• Widow’s peak is dominant to straight

– W=allele for widow’s peak

– w= allele for straight hairline

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Widow’s peak

• Fig. 23.3

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• Inheritance of a single trait cont’d.– Gamete formation

• During meiosis, homologous chromosomes separate so there is only 1 member of each pair in a gamete

• There is one allele for each trait, such as hairline, in each gamete

• No two letters in a gamete can be the same letter of the alphabet

– If genotype is Ww, then gametes from this individual will contain either a W or a w

– If the genotype is WwLl (looking at 2 traits), gametes can contain any of the following combinations

» WL, Wl, wL, or wl

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• Practice problems– For each of the following genotypes, give all possible gametes

• WW

• WWSs

• Tt

• Ttgg

• AaBb

– For each of the following, state whether the genotype or a gamete is represented

• D

• Ll

• Pw

• LlGg

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Genotype related to phenotype

• Table 23.1

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• One trait crosses-monohybrid cross– Let’s consider a specific cross

• If a homozygous woman with a widow’s peak reproduces with a man with a straight hairline, what kind of hairline will their children have?

– Use W= widow’s peak, w=straight hairline

– figure out the genotype of each parent

» Woman has a widow’s peak and we are told she is homozygous so she is WW

» Man has a straight hairline so he must be ww

– Determine their gametes: her eggs will all have W and his sperm will all have w

– Put together all possible combinations: all offspring will receive a W from her and a w from him so all Ww

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• One-trait cross cont’d.– In this example there was only 1 combination of eggs and sperm

possible– What if the parents were each Ww?

• Then sperm could have either W or w, and eggs also could have W or w

– One way to figure out the possible combinations of eggs and sperm that could occur is to use a Punnett square

– This is illustrated on the following slide

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A punnett square is an easy way to figure out all possible combinations of

eggs and sperm

• Fig 23.4

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• One-trait cross cont’d.– After the genotypes and phenotypes of offspring are determined,

we can determine the ratios• In our previous example, in the Punnett square we had the following

offspring: WW, Ww, Ww, and ww

– The genotypic ratio is 1 WW:2 Ww:1 ww

– The phenotypic ratio is 3 individuals with a widow’s peak to 1 individual with a straight hairline

• Another way to phrase the phenotypic ratio is in terms of probability

– This couple has a 75% chance of producing a child with a widow’s peak and a 25% chance of producing a child with a straight hairline

• The probability will be the same for each pregnancy between this couple

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• One-trait crosses and probability– Product rule of probability

• The chance of 2 or more independent events occurring together is the product of their chance of occurring separately

– In the cross Ww X Ww, what is the chance of obtaining either a W or a w from a parent?

» Chance of W = ½ and the chance of w = ½

» Therefore the probability of having these genotypes is as follows

» Chance of WW= ½ X ½ = ¼

» Chance of Ww = ½ X ½ = ¼

» Chance of wW= ½ X ½ = ¼

» Chance of ww = ½ X ½ = ¼

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• One-trait crosses and probability cont’d.– Sum rule of probability-the chance of an event that can occur in

more than one way is the sum of the individual chances• To calculate the chance of an offspring having a widow’s peak, add

the chances of WW, Ww, or wW from the preceding slide

– ¼ + ¼ + ¼ = ¾ or 75%

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• The one-trait test cross– You cannot distinguish between a homozygous dominant

individual and a heterozygous individual just by looking• They are phenotypically the same

– Breeders of plants and animals may do a test cross to determine the likely genotype of an individual with the dominant phenotype

• Cross with a recessive individual-has a known genotype

• If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous

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One-trait testcross

• Fig. 23.5

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• Practice problems– Both a man and a woman are heterozygous for freckles.

Freckles are dominant over no freckles. What is the chance that their child will have freckles?

– Both you and your sibling have attached ear lobes, but your parents have unattached lobes. Unattached earlobes (E) are dominant over attached (e). What are the genotypes of your parents?

– A father has dimples, the mother of his children does not, and all 5 of their children have dimples. Dimples (D) are dominant over no dimples (d). Give the probable genotypes of all persons concerned.

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• The inheritance of two traits– Mendel reasoned from the results of his pea plant crosses that

each pair of factors assorts independently into gametes• If one pair of homologous chromosomes has a W on one and

w on the other, and a second pair has an R on one and an r on the other, when gametes form all possible combinations of these letters are possible: WR, Wr, wR, and wr

• Called the Law of Independent Assortment

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Segregation and independent assortment

• Fig. 23.6

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• The law of independent assortment states :– Each pairs of factors assorts independently (without

regard to how the others separate)– All possible combinations of factors can occur in the

gametes

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• Two-trait crosses-dihybrid cross– Let’s consider two traits, hairline and finger length

• We’ll use the W and w as before, and now S=short fingers and s=long fingers

– A person who is WwSs widow’s peak and short fingers and a person who is also WwSs have children

• Figure the gametes for each parent

– WS, Ws, wS, and ws for both

• Make a big Punnett square as illustrated on the following slide

• Figure the genotypes of the offspring

– Phenotypic ratio: 9 widow’s peak, short fingers: 3 widow’s peak, long fingers; 3 straight hairline, short fingers; 1 straight hairline, long fingers

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Dihybrid cross

• Fig. 23.7

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• Two-trait crosses and probability– Probability laws

• Probability of widow’s peak = ¾

Probability of short fingers= ¾

• Probability of straight hairline= ¼

Probability of long fingers= ¼

– Using the product rule• Probability of widow’s peak and short fingers = ¾ X ¾ = 9/16

• Probability of widow’s peak and long fingers = ¾ X ¼ = 3/16

• Probability of straight hairline and short fingers = ¼ X ¾ = 3/16

• Probability of straight hairline and long fingers = ¼ X ¼ = 1/16

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• The two-trait testcross– Cross an individual with the dominant phenotype for each trait

with an individual with the recessive phenotype of both traits• If the dominant individual is homozygous for both traits, all offspring

will show the dominant phenotypes

• If the dominant individual is heterozygous, then a Punnett square shows the following phenotypic ratio of offspring

– 1 dominant for both: 1 dominant for first trait recessive for second: 1 recessive for first trait dominant for second: and 1 recessive for both

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Two-trait testcross

• Fig. 23.8

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Mendel’s laws

• Practice problems– Attached earlobes are recessive, What genotype do children

have if one parent is homozygous for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline?

– If an individual from this cross reproduces with another of the same genotype, what are the chances that they will have a child with a straight hairline and attached earlobes?

– A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant) and a woman who does not. What are the genotypes of all persons concerned?

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23.2 Beyond simple inheritance patterns

• Polygenic inheritance– Controlled by 2 or more sets of alleles– Each dominant allele codes for a product and effects are

additive– Result is a continuous range of phenotypes

• Distribution resembles a bell curve

• The more gene pairs involved, the more continuous the pattern of variation

• Ex: human height, skin pigmentation

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Polygenic inheritance

• Fig. 23.9

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Beyond simple inheritance patterns cont’d.

• Polygenic inheritance cont’d.– Skin color

• Controlled by many gene pairs and many alleles

• Assuming the simplest model, 2 alleles at loci

– A and B

– When an AaBb person has children with another AaBb person, children can range from very light to very dark

AABB Very dark

AABb or AaBB Dark

AaBb, AAbb, aaBB

Medium brown

Aabb or aaBb Light

aabb Very light

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• Environmental influences– Environment can influence gene expression and therefore

phenotype• Ex: sunlight exposure on skin-suntan;

• coat color in Himalayan rabbits only make black pigment where body temp. is colder- extremities

– Human twin studies• Polygenic traits are most influenced

• “nature vs. nurture”

• Identical twins separated at birth are studied

– If they share a trait in common even though raised in different environments, it is likely genetic

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Coat color in Himalayan rabbits

• Fig. 23.10

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Beyond simple inheritance patterns cont’d.• Incomplete dominance and codominance

– Incomplete dominance• Heterozygous individual has a phenotype intermediate to the two

homozygous individuals

• Ex: 4 o’clock flowers

red incompletely dominant with white; heterozygote is pink

• Ex: curly-haired Caucasian woman and a straight-haired Caucasian man produce wavy-haired children

– When 2 wavy-haired people have children, the phenotypic ratio is 1 curly: 2 wavy: 1 straight

– Codominance • Occurs when both alleles are equally expressed

• Ex: horses:chestnut codominant w/cream; heterozygote is palomino

• Ex: cattle: red codominant w/ white; heterozygote is roan

• Ex: type AB blood has both A antigens and B antigens on red blood cells

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Incomplete dominance

• Fig. 23.11

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• Multiple allele inheritance– The gene exists in several allelic forms, but each person still has

only 2 of the possible alleles– ABO blood types

• IA = A antigens on RBCs• IB = B antigens on RBCs• i = has neither A nor B antigens on RBCs

– Both IA and IB are dominant over I, IA and IB are codominantPhenotype GenotypeA IAIA or IAiB IBIB or IBiAB IAIB

O ii

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• ABO blood types cont’d.– Paternity testing- ABO blood groups often used

• Can disprove paternity but not prove it

– Rh factor- another antigen on RBCs• Rh positive people have the antigen

• Rh negative people lack it

– There are multiple alleles for Rh negative, but all are recessive to Rh positive

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Inheritance of blood type

• Fig. 23.12

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• Practice problems– A polygenic trait is controlled by three pairs of alleles. What are

the two extreme genotypes for this trait?– What is the genotype of the lightest child that could result from a

mating between two medium-brown individuals?– A child with type O blood is born to a mother with type A blood.

What is the genotype of the child? The mother? ‘what are the possible genotypes of the father?

– From the following blood types determine which baby belongs to which parents:

• Baby 1 type O Mrs. Doe type A Mrs. Jones type A

• Baby 2 type B Mr. Doe type A Mr. Jones type AB

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23.3 Sex-linked inheritance

• Sex chromosomes– 22 pairs of autosomes, 1 pair of sex chromosomes

• X and y

– In females, the sex chromosomes are XX

– In males, the sex chromosomes are XY

» Note that in males the sex chromosomes are not homologous

– Traits controlled by genes in the sex chromosomes are called sex-linked traits

• X chromosome has many genes, the Y chromosome does not

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Sex-linked inheritance cont’d.

• X-linked traits– Red-green colorblindness is X-linked

• The X chromosome has genes for normal color vision

– XB = normal vision

– Xb – colorblindness

– Genotypes Phenotypes

XBXB female with normal color vision

XBXb carrier female with normal color vision

XbXb colorblind female

XBY male with normal color vision

XbY colorblind male

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Cross involving an X-linked allele

• Fig. 23.13

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Sex-linked inheritance-practice problems

• Both the mother and the father of a colorblind male appear to be normal. From whom did the son inherit the allele for colorblindness? What are the genotypes of the mother, father, and the son?

• A woman is colorblind. What are the chances that her son will be colorblind? If she is married to a man with normal vision, what are the chances that her daughters will be colorblind? Will be carriers?

• Both the husband and the wife have normal vision. The wife gives birth to a colorblind daughter. Is it more likely the father had normal vision or was colorblind? What does this lead you to deduce about the girl’s parentage?

• What is the genotype of a colorblind male with long fingers is s=long fingers? If all his children have normal vision and short fingers, what is the likely genotype of the mother?

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23.4 Inheritance of linked genes

• A single chromosome has many genes – The sequence is fixed because each allele has a specific locus

• All genes on a single chromosome form a linkage group

• When linkage is complete, a dihybrid produces only 2 types of gametes

• Any time traits are inherited together, a linkage group is suspected

• Or, if very few recombined phenotypes appear in offspring, linkage is also suspected

– Recall that during meiosis crossing over sometimes occurs• If crossing over occurs between 2 alleles of interest, then 4 types of

gametes are formed instead of 2

– Fig. 23.14 on the following slide illustrates linkage

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Linkage group

• Fig. 23.14

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Inheritance of linked genes cont’d.

• Ocurrence of crossing over can indicate the sequence of genes on a chromosome – Crossing over occurs more frequently between distant genes

than between those close together on a chromosome

• Practice problems– When AaBb individuals reproduce, the phenotypic ratio is about

3:1. What ratio was expected? What may have caused the observed ratio?

– The genes for ABO blood type and for fingernails are on the same homologous pair of chromosomes. In an actual family, 45% of offspring have type B blood and no fingernails, and 45% have type O blood and fingernails; 5% have type B blood and fingernails, and 5% have type O blood and no fingernails. What process accounts for the recombinant phenotypes?

Chapter 23 Objectives

1. Know vocabulary terms like allele, homozygous, heterozygous, genotype, phenotype.

2. Know Mendel’s laws.

3. Be able to solve a 1-trait genetics problem for simple dominance using a Punnett square and use it to determine chances of a trait appearing in the offspring.

4. Be able to determine genotype of members of a family from phenotypic information.

5. Be able to solve incomplete dominance and codominace problems.

6. Be able to solve a 2-trait genetics problem.

7. Know that the phenotypic ratio for a dihybrid cross is 9:3:3:1.

8. Be able to solve ABO blood type problems.

9. Know how multiple alleles and polygenic traits affect inheritance.

10. Know that environmental factors can affect phenotypes.

11. Be able to solve problems for sex-linked traits.

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23-1 Honors Biology chapter 23 Genetics John Regan Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display