GENETICSgoodman.pdf

8/18/2019 GENETICSgoodman.pdf

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GENETICSGENETICS


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GeneticsGenetics

• The study of heredityheredity.

•• Gregor Mendel (1860Gregor Mendel (1860’’s)s) discovered the

fundamental principlesprinciples of geneticsgenetics by breedingbreedinggarden peasgarden peas.


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GeneticsGenetics•• Alleles Alleles

1. Alternative forms of genes.genes.2. Units that determine heritable traits.

3.Dominant allelesDominant alleles (

TTTT -

tall pea plantstall pea plants)

a. homozygous dominanta. homozygous dominant

4. Recessive allelesRecessive alleles (tttt - dwarf pea plantsdwarf pea plants)

a. homozygous recessivea. homozygous recessive

5. HeterozygousHeterozygous (TtTt - tall pea plantstall pea plants)


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PhenotypePhenotype

•• Outward appearanceOutward appearance•• Physical characteristicsPhysical characteristics

•• Examples:Examples:

1.1. tall pea planttall pea plant

2.2.

dwarf pea plantdwarf pea plant


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GenotypeGenotype

•• Arrangement of genes that produces the Arrangement of genes that produces the

phenotypephenotype•• Example:Example:

1.1. tall pea planttall pea plant

TT = tall (homozygous dominant)(homozygous dominant)

2.2. dwarf pea plantdwarf pea plant

tt = dwarf (homozygous recessive)(homozygous recessive)3.3. tall pea planttall pea plant

Tt = tall (heterozygous)(heterozygous)


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Punnett squarePunnett square

• A Punnett squarePunnett square is used to show thepossible combinationscombinations of gametesgametes.


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Breed the P generationP generation

•• tall (TT) vs. dwarf (tt) pea plantstall (TT) vs. dwarf (tt) pea plants

t

t

T T


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tall (TT) vs. dwarf (tt) pea plantstall (TT) vs. dwarf (tt) pea plants

t

t

T T

Tt

Tt

Tt

Tt All Tt = tall(heterozygous tall)

produces theFF11 generationgeneration


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Breed the FF11 generationgeneration

•• tall (Tt) vs. tall (Tt) pea plantstall (Tt) vs. tall (Tt) pea plants

T

t

T t


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tall (Tt) vs. tall (Tt) pea plantstall (Tt) vs. tall (Tt) pea plants

TT

Tt

Tt

tt

T

t

T t

produces the

FF22 generationgeneration

1/4 (25%) = TT

1/2 (50%) = Tt

1/4 (25%) = tt

1:2:1 genotype1:2:1 genotype

3:1 phenotype3:1 phenotype


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Monohybrid CrossMonohybrid Cross

• A breeding experiment that tracks the inheritanceof a single trait.single trait.

•• MendelMendel ’’ss ““ principle of segregationprinciple of segregation””

a. pairs of genes separate during gametegamete

formation (meiosis).(meiosis).

b. the fusion of gametesgametes at fertilization pairsgenes once again.


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Homologous ChromosomesHomologous Chromosomes

eye color locus

b = blue eyes

eye color locus

B = brown eyes

Paternal Maternal

This person would

have brown eyes (Bb)


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MeiosisMeiosis -- eye color eye color

Bb

diploid (2n)

B

b

meiosis I

B

B

b

b

sperm

haploid (n)

meiosis II


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•• ExampleExample: Cross between two heterozygotesheterozygotesfor brown eyes (Bb)

BB = brown eyes

Bb = brown eyesbb = blue eyes

B

b

B b

Bb x Bb

male

gametes

female gametes


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BB

Bb

Bb

bb

B

b

B b

Bb x Bb

1/4 = BB - brown eyed1/2 = Bb - brown eyed

1/4 = bb - blue eyed

1:2:1 genotype3:1 phenotype


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Dihybrid CrossDihybrid Cross

• A breeding experiment that tracks the inheritanceof two traits.two traits.

•• MendelMendel ’’ss ““ principle of independent assortmentprinciple of independent assortment””

a. each pair of alleles segregates independently

during gamete formation (metaphase I)(metaphase I)

b. formula: 22nn (n = # of heterozygotes)(n = # of heterozygotes)


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Independent AssortmentIndependent Assortment

•• Question:Question: How many gametes wil l be producedfor the following allele arrangements?

• Remember: 22nn

(n = # of heterozygotes)(n = # of heterozygotes)

1.1. RrYyRrYy

2.2. AaBbCCDd AaBbCCDd

3.3. MmNnOoPPQQRrssTtQqMmNnOoPPQQRrssTtQq


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Answer: Answer:

1. RrYy: 21. RrYy: 2nn = 2= 222 == 4 gametes4 gametes

RY Ry rY ryRY Ry rY ry

2. AaBbCCDd: 22. AaBbCCDd: 2nn = 2= 233 == 8 gametes8 gametes

ABCD ABCd AbCD AbCd ABCD ABCd AbCD AbCd

aBCD aBCd abCD abCDaBCD aBCd abCD abCD

3. MmNnOoPPQQRrssTtQq: 23. MmNnOoPPQQRrssTtQq: 2nn = 2= 266 == 64 gametes64 gametes


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•• Example:Example: cross between roundround and yellowyellowheterozygous pea seeds.

RR = round= roundr r = wrinkled= wrinkled

YY = yellow= yellow

yy = green= green RY Ry rY ryRY Ry rY ry x RY Ry rY ryRY Ry rY rypossible gametes produced

RrYyRrYy x RrYyRrYy


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RYRY RyRy rYrY ryry

RYRY

RyRy

rYrY

ryry


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RRYY

RRYy

RrYY

RrYy

RRYy

RRyy

RrYy

Rryy

RrYY

RrYy

rrYY

rrYy

RrYy

Rryy

rrYy

rryy

Round/Yellow: 9

Round/green: 3

wrinkled/Yellow: 3

wrinkled/green: 1

9:3:3:1 phenotypic ratio

RYRY RyRy rYrY ryry

RYRY

RyRy

rYrY

ryry


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Test CrossTest Cross

• A mating between an individual of unknown genotypeunknown genotypeand a homozygous recessivehomozygous recessive individual.

•• Example:Example: bbC__bbC__ x bbccbbcc

BB = brown eyesBb = brown eyes

bb = blue eyes

CC = curly hair

Cc = curly hair

cc = straight hair

bCbC b___ b___

bcbc


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Test CrossTest Cross

•• Possible results:Possible results:

bCbC b___ b___

bcbc bbCc bbCc

C bCbC b___ b___

bcbc bbCc bbccor

c


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Incomplete DominanceIncomplete Dominance

•• F1 hybridsF1 hybrids have an appearance somewhat inin

betweenbetween the phenotypesphenotypes of the two parentalvarieties.

•• Example:Example: snapdragons (flower)snapdragons (flower)

• red (RR) x white (rr)

RR = red flower RR = red flower rr = white flower

r

r

R R


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Incomplete DominanceIncomplete Dominance

Rr

Rr

Rr

Rr

r

r

R R

All Rr = pink(heterozygous pink)

produces theFF11 generationgeneration


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CodominanceCodominance

•• Two allelesTwo alleles are expressed (multiple allelesmultiple alleles)in heterozygous individualsheterozygous individuals.

•• Example:Example: bloodblood

1. type A = I AI A or I Ai

2. type B = IBIB or IBi

3. type AB = I AIB

4. type O = ii


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•• Example:Example: homozygous male B (IB

IB

)x

heterozygous female A (I Ai)

I AIB I AIB

IBi IBi

1/2 = I AIB1/2 = IBi

I A

IB IB

i


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•• Example:Example: male O (ii) x female AB (I AIB)

I Ai IBi

I Ai IBi

1/2 = I Ai

1/2 = IB

i

i

I A IB

i


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•• QuestionQuestion: If a boy has a blood type O andhis sister has blood type AB,

what are the genotypes and

phenotypes of their parents.

• boy - type O (ii) X girl - type AB (I AIB)


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•• Answer: Answer:

I AIB

ii

Parents:Parents:

genotypesgenotypes = I Ai and IBi

phenotypesphenotypes = A and B

IB

I A i

i


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SexSex--linked Traitslinked Traits

• Traits (genes) located on the sexsexchromosomeschromosomes

•• Example:Example: fruit fliesfruit flies

(redred-eyed male) X (whitewhite-eyed female)


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Sex ChromosomesSex Chromosomes

XX chromosome - female Xy chromosome - male

fruit fly

eye color


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•• Example:Example: fruit fliesfruit flies

(red-eyed male) X (white-eyed female)

•• Remember:Remember: the Y chromosomeY chromosome in malesdoes not carry traits.

RR = red eyed

Rr = red eyed

rr = white eyed

Xy = male

XX = female

Xr

XR y

Xr


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XR Xr

XR Xr

Xr y

Xr y

1/2 red eyed and female

1/2 white eyed and male

Xr

XR y

Xr


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Population GeneticsPopulation Genetics

• The study of genetic changesgenetic changes in populationspopulations.• The science of microevolutionary changesmicroevolutionary changes in

populationspopulations.

•• HardyHardy--Weinberg equilibrium:Weinberg equilibrium:

the principle that shuffling of genes that occurs

during sexual reproduction, by itself, cannot

change the overall genetic makeup of a population.

•• HardyHardy--Wienberg equation:Wienberg equation: 1 = p1 = p22 + 2pq + q+ 2pq + q22


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Question:Question:

•• How do we get this equation?How do we get this equation?

Answer: Answer: ““ SquareSquare”” 1 = p + q1 = p + q

1122 = (p + q)= (p + q)22

1 = p1 = p22 + 2pq + q+ 2pq + q22


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HardyHardy--Wienberg equationWienberg equation

•• Five conditionsFive conditions are required for Hardy-Wienbergequilibrium.

1.1. large populationlarge population

2.2. isolated populationisolated population3.3. no net mutationsno net mutations

4.4. random matingrandom mating

5.5. no natural selectionno natural selection


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•• Need to remember the following:Need to remember the following:

pp22 = homozygous dominant= homozygous dominant

2pq = heterozygous2pq = heterozygous

qq22 = homozygous recessive= homozygous recessive

ImportantImportant


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• Iguanas with webbed feet (recessive trait)(recessive trait) makeup 4% of the population. What in the population

is heterozygousheterozygous and homozygoushomozygous dominantdominant.

Question:Question:


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Answer: Answer:

1. q1. q22 = 4% or .04= 4% or .04 qq22 = .04= .04 q = .2q = .2

2. then use 1 = p + q2. then use 1 = p + q1 = p + .21 = p + .2 11 -- .2 = p.2 = p .8 = p.8 = p

3. for heterozygous use 2pq3. for heterozygous use 2pq

2(.8)(.2) = .32 or 32%2(.8)(.2) = .32 or 32%

4. For homozygous dominant use p4. For homozygous dominant use p22

.8.822 = .64 or 64%= .64 or 64%


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HardyHardy--Wienberg equationWienberg equation

1 = p1 = p22 + 2pq + q+ 2pq + q22

•• 64% = p64% = p22 = homozygous dominant= homozygous dominant

•• 32% = 2pq32% = 2pq = heterozygous= heterozygous

•• 04%04% = q= q22 = homozygous recessive= homozygous recessive

•• 100%100%

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GENETICSgoodman.pdf