Hardy-Weinberg EquilibriumHardy-Weinberg Equilibrium
• Rules of probability• A simple population model• Mechanisms of evolutionarychange
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)
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
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
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?
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
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
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
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
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
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
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)
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
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
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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