Hardy-Weinberg Equilibrium - Washington State Universitydybdahl/EvoLect6b-HWE.pdf · Hardy-Weinberg...

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium

• Rules of probability• A simple population model• Mechanisms of evolutionarychange

Evolution & GeneticsEvolution & Genetics

???

The Modern SynthesisThe Modern Synthesis

Sewell Wright

R.A. Fisher

J.B.S.Haldane

T. Dobzhansky

““Population GeneticsPopulation Genetics””

Population GeneticsPopulation Genetics

Study of how genes behave in populations

Description: Fr(red) = 7/20 = 0.35

Involves description and also prediction

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

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?

p’ = [2(p2) + 1(2pq)]

2[p2 + q2 + 2pq]

Mathematical ModelsMathematical Models(don(don’’t be frightened)t be frightened)

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.

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

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.

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

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium

What happens to allele andgenotype frequencies over time?

(simple null model)

Evolution = change in allele frequencies

Hardy-Weinberg EquilibriumAssumptions about the population

Large Population No immigration or emigration No mutation Random mating Random reproductive success(i.e., no selection)

Hardy-Weinberg EquilibriumFigure 5.1: Basicpopulation cycle

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


Create a new generation by randomlycombining gametes (random mating)

Probability that 1st allele is an A =

Fr(AA) = p2 Fr(aa) = q2


p

Probability that 2nd allele is an A = p

Probability that both alleles are A = p2

Probability of creating an AA individual?

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


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


What are allele freq.’s in next generation?

Fr(A)’ = p’ = [2(Fr(AA)) + 1(Fr(Aa))]

Total



Fr(A)’ = p’ = [2(p2) + 1(2pq)]

Total


AA,aa and Aa individuals,each with 2 alleles


Fr(A)’ = p’ = [2(p2) + 1(2pq)]

=[2p(p + q)]

2 = p

2[p2 + q2 + 2pq]

=2p

2


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

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

Hardy-Weinberg EquilibriumFig. 5.10:Mechanisms ofevolutionarychange

Testing for HWETesting for HWE

Apple maggot fly (Rhagoletis pomonella)McPheron et al. 1988. Nature 336:64-66

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)


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


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


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?

Testing for HWETesting for HWERhagoletis pomonella Fruit type

preference:

Apples vs.hawthorn

Non-randommating

Deficiency ofheterozygotes

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