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### Transcript of Hardy-Weinberg Equilibrium dybdahl/EvoLect6b-HWE.pdf · PDF file 2006-09-18 ·...

• Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium

• Rules of probability • A simple population model • Mechanisms of evolutionary change

• 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

S F

Heterozygotes (FS)

Homozygotes (FF or SS)

Fr(F allele) = 11/28 = 0.393 Fr(S) = 0.607 Fr(FS genotype) = 7/14 = 0.500

• Alleles & GenotypesAlleles & Genotypes

S F

Population Genetics is the study of how genes behave in populations

Evolution = change in allele frequencies

Theory: Can we predict changes in allele and 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 Probability The probability of randomly encountering an item of a certain type is equal to the frequency of that type in the population.

The probability of rolling a 6 with a single die is 1/6.

• Rules of ProbabilityRules of Probability Addition Rule: The probability that either of two mutually exclusive events will occur is equal to the sum of their independent probabilities of occurrence

The probability of rolling a 6 or a 1 with a single die is 1/6 + 1/6 = 1/3.

The sum of all possible outcomes = 1

• Rules of ProbabilityRules of Probability Multiplication Rule: The probability that two independent events will both occur is equal to the product of their independent probabilities of occurrence.

The probability of rolling a 6 and another 6 with two dice is (1/6)*(1/6) = 1/36.

• Rules of ProbabilityRules of Probability What is the probability of rolling an 11 with two dice?

Possibility 1: Roll a 5 and a 6 Probability = (1/6)*(1/6) = 1/36

Possibility 2: Roll a 6 and a 5 Probability = (1/6)*(1/6) = 1/36

Total Prob. = (1/36) + (1/36) = 1/18

• Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium

What happens to allele and genotype frequencies over time?

(simple null model)

Evolution = change in allele frequencies

• Hardy-Weinberg Equilibrium Assumptions about the population

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

• Hardy-Weinberg Equilibrium Figure 5.1: Basic population cycle

• A

aA a

A a

A a

A a

A a

A

a

A

a

A

a

A a

A a

A

a

A a

A

a A aA

a

A

a A

a

A

a

A a

A

a

a

gene pool

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single locus with two alleles (A and a) Fr(A) = p Fr(a) = q p + q = 1

• Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single locus with two alleles (A and a) Fr(A) = p Fr(a) = q p + q = 1

Create a new generation by randomly combining gametes (random mating)

• Probability that 1st allele is an A =

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

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single 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?

• What is probability of an Aa heterozygote? Prob. of A from dad and a from mom = pq Prob. of a from dad and A from mom = pq

Fr(Aa) = 2pq

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single locus with two alleles (A and a) Fr(A) = p Fr(a) = q p + q = 1

• If assumptions of Hardy-Weinberg hold, then we can predict the genotype freq’s in next gen:

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

Fr(Aa) = 2pq p2 + q2 + 2pq = 1

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single locus with two alleles (A and a) Fr(A) = p Fr(a) = q p + q = 1

• What are allele freq.’s in next generation?

Fr(A)’ = p’ = [2(Fr(AA)) + 1(Fr(Aa))]

Total

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single locus with two alleles (A and a) Fr(A) = p Fr(a) = q p + q = 1

• What are allele freq.’s in next generation?

Fr(A)’ = p’ = [2(p2) + 1(2pq)]

Total

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Single 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

• 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 Equilibrium Single locus with two alleles (A and a) Fr(A) = p Fr(a) = q p + q = 1

•  Genotypes occur in predictable frequencies

 Allele frequencies do not change over time (i.e., evolution does not occur)

Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium Conclusions of the null model

• Hardy-Weinberg Equilibrium Violations 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 Equilibrium Fig. 5.10: Mechanisms of evolutionary change

• Testing for HWETesting for HWE

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

• Testing for HWETesting for HWE Locus: 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)

• Testing for HWETesting for HWE Locus: 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

• Testing for HWETesting for HWE Locus: 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

• Testing for HWETesting for HWE Locus: 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?

• Testing for HWETesting for HWE Rhagoletis pomonella Fruit type

preference:

Apples vs. hawthorn

Non-random mating

Deficiency of heterozygotes