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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