hardy-weinberg equilibrium - washington state universitydybdahl/evolect6b-hwe.pdf · hardy-weinberg...
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Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium
• Rules of probability• A simple population model• Mechanisms of evolutionarychange
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Evolution & GeneticsEvolution & Genetics
???
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The Modern SynthesisThe Modern Synthesis
Sewell Wright
R.A. Fisher
J.B.S.Haldane
T. Dobzhansky
““Population GeneticsPopulation Genetics””
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Population GeneticsPopulation Genetics
Study of how genes behave in populations
Description: Fr(red) = 7/20 = 0.35
Involves description and also prediction
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Alleles & GenotypesAlleles & Genotypes
SF
Heterozygotes(FS)
Homozygotes(FF or SS)
Fr(F allele) = 11/28 = 0.393 Fr(S) = 0.607Fr(FS genotype) = 7/14 = 0.500
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Alleles & GenotypesAlleles & Genotypes
SF
Population Genetics is the study of howgenes behave in populations
Evolution = change in allele frequencies
Theory: Can we predict changes in alleleand genotype frequencies?
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p’ = [2(p2) + 1(2pq)]
2[p2 + q2 + 2pq]
Mathematical ModelsMathematical Models(don(don’’t be frightened)t be frightened)
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Rules of ProbabilityRules of ProbabilityThe probability of randomlyencountering an item of a certaintype is equal to the frequency of thattype in the population.
The probability of rolling a 6 with asingle die is 1/6.
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Rules of ProbabilityRules of ProbabilityAddition Rule: The probability thateither of two mutually exclusiveevents will occur is equal to the sumof their independent probabilities ofoccurrence
The probability of rolling a 6 or a 1with a single die is 1/6 + 1/6 = 1/3.
The sum of all possible outcomes = 1
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Rules of ProbabilityRules of ProbabilityMultiplication Rule: The probabilitythat two independent events will bothoccur is equal to the product of theirindependent probabilities ofoccurrence.
The probability of rolling a 6 andanother 6 with two dice is (1/6)*(1/6)= 1/36.
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Rules of ProbabilityRules of ProbabilityWhat is the probability ofrolling an 11 with two dice?
Possibility 1: Roll a 5 and a 6Probability = (1/6)*(1/6) = 1/36
Possibility 2: Roll a 6 and a 5Probability = (1/6)*(1/6) = 1/36
Total Prob. = (1/36) + (1/36) = 1/18
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Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium
What happens to allele andgenotype frequencies over time?
(simple null model)
Evolution = change in allele frequencies
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Hardy-Weinberg EquilibriumAssumptions about the population
Large Population No immigration or emigration No mutation Random mating Random reproductive success(i.e., no selection)
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Hardy-Weinberg EquilibriumFigure 5.1: Basicpopulation cycle
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A
aA a
A a
Aa
Aa
Aa
A
a
A
a
A
a
A a
Aa
A
a
Aa
A
a AaA
a
A
a A
a
A
a
Aa
A
a
a
gene pool
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
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Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
Create a new generation by randomlycombining gametes (random mating)
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Probability that 1st allele is an A =
Fr(AA) = p2 Fr(aa) = q2
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
p
Probability that 2nd allele is an A = p
Probability that both alleles are A = p2
Probability of creating an AA individual?
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What is probability of an Aa heterozygote?Prob. of A from dad and a from mom = pqProb. of a from dad and A from mom = pq
Fr(Aa) = 2pq
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
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If assumptions of Hardy-Weinberg hold, thenwe can predict the genotype freq’s in next gen:
Fr(AA) = p2
Fr(aa) = q2
Fr(Aa) = 2pqp2 + q2 + 2pq = 1
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
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What are allele freq.’s in next generation?
Fr(A)’ = p’ = [2(Fr(AA)) + 1(Fr(Aa))]
Total
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
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What are allele freq.’s in next generation?
Fr(A)’ = p’ = [2(p2) + 1(2pq)]
Total
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
AA,aa and Aa individuals,each with 2 alleles
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What are allele freq.’s in next generation?
Fr(A)’ = p’ = [2(p2) + 1(2pq)]
=[2p(p + q)]
2 = p
2[p2 + q2 + 2pq]
=2p
2
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumSingle locus with two alleles (A and a)Fr(A) = p Fr(a) = q p + q = 1
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Genotypes occur in predictablefrequencies
Allele frequencies do not change overtime (i.e., evolution does not occur)
Hardy-Weinberg EquilibriumHardy-Weinberg EquilibriumConclusions of the null model
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Hardy-Weinberg EquilibriumViolations of the Assumptions
Small Population --> Genetic Drift Immigration/emigration --> Gene Flow Mutation --> Mutation Pressure Non-random mating --> Pop. Structure Differential RS --> Natural Selection
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Hardy-Weinberg EquilibriumFig. 5.10:Mechanisms ofevolutionarychange
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Testing for HWETesting for HWE
Apple maggot fly (Rhagoletis pomonella)McPheron et al. 1988. Nature 336:64-66
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Testing for HWETesting for HWELocus: b-hydroxyacid dehydrogenase (Had)
Genotype
100/100
125/125
100/125
No.
260
180
360
Step 1: Calculate observed allele frequencies
800
Pop’n in HWE?Box 5.5
Fr(100) = (2*260 + 1*360)/(2*800) = 0.55 (= p)
Fr(125) = (2*180 + 1*360)/(2*800) = 0.45 (= q)
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Testing for HWETesting for HWELocus: b-hydroxyacid dehydrogenase (Had)
Genotype
100/100
125/125
100/125
No.
260
180
360
Step 2: Calculate expected genotype numbers
Exp.
242
162
396
p2*N
q2*N
2*p*q*N
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Testing for HWETesting for HWELocus: b-hydroxyacid dehydrogenase (Had)
Genotype
100/100
125/125
100/125
No.
260
180
360
Step 3: Compare observe to expected
Exp.
242
162
396
Stat.
1.34
3.27
2.00
= 6.61 χ2 = (O-E)2 E Σ df = 1
crit = 3.84
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Testing for HWETesting for HWELocus: b-hydroxyacid dehydrogenase (Had)
Genotype
100/100
125/125
100/125
No.
260
180
360
Exp.
242
162
396
Conclude: Population deviates from HWE
Fewer heterozygotes than expected
Why?
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Testing for HWETesting for HWERhagoletis pomonella Fruit type
preference:
Apples vs.hawthorn
Non-randommating
Deficiency ofheterozygotes