lec_4_cond_13

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Mapping Zygote pool to Adult pool Mapping Adult pool to Gamete pool (or gene pool) p = Prop.(A1 allele) = {2(PN) + 1(HN)}/2N = P + H/2 q = Prop.(A2 allele) = {2(QN) + 1(HN)}/2N = Q + H/2 Haploid Gamete Pool Diploid Adult Pool p = P + H/2 P A1A1 Q A2A2 H A1A2 q = Q + H/2 p A1 q A2 SUMMARY Mapping Gamete pool to Zygote pool 16 (with respect to most alleles in genotype) Adult Pool Gamete Pool Zygote Pool Syngamy Survival Gametogenesis Haploid Gamete Pool Diploid Zygote Pool p A1 q A2 P A1A1 Q A2A2 H A1A2 ? H A1A2 Zygote Pool P A1A1 Q A2A2 H A1A2 Adult Pool NO CHANGE since no evolutionary forces are operating Q A2A2 P A1A1

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Probability of 2 independent events (A & B) is the product of the probability of each Start with a simple result from probability theory: Probability of A and B = PROB(A) X PROB(B) Example: PROB(H & H) of 2 coin flips = 1/2 X 1/2 = 1/4 To make this mapping we need to systematically consider all possible matings in the population It is not important that you calculate every entry in the table -- but that you understand how the table was created and what it represents Random mating table Mating PROB. Offspring (Zygotes) A1A1 A1A2 A2A2 A1A1 A1A1 P 2 P 2 A1A1 A1A2 P x H PH / 2 PH / 2 A1A1 A2A2 Q 2 Q 2 A1A2 A1A1 A1A2 A1A2 A1A2 A2A2 H x P HP / 2 HP / 2 H 2 H 2 / 2 H 2 / 4 H 2 / 4 H x Q HQ / 2 HQ / 2 A2A2 A2A2 A2A2 A1A1 A1A2 A2A2 Q x P QP Q x H QH / 2 QH / 2 P x Q PQ Σ = 1.0 Σ = p 2 Σ = 2 p q Σ = q 2 Proportion of all offspring that are A2A2 Proportion of all offspring that are A1A2 Proportion of all offspring that are A1A1 17 Diploid Zygote or Adult Pool p = (P + H / 2) Haploid Gamete Pool p A1 q A2 P A1A1 Q A2A2 H A1A2 p 2 2 p q q 2 q = (Q + H / 2) (Quantitative solution) --because there is no selection Adult pool = Zygote pool Haploid Gamete Pool p A1 q A2 Diploid Zygote Pool P A1A1 Q A2A2 H A1A2 P = p 2 Q = q 2 H = 2 p q Putting all the pieces together We can summarize our table of all possible matings to map the transition from gene pool to zygote pool
Called the Binomial Square Principle since Lack of change in gene and genotype frequencies called HARDY-WEINBERG EQUILIBRIUM IF random mating & no evolutionary forces are operating THEN P, H, Q, p , & q remain constant over time, P = p 2 H = 2 p q Q = q 2 ( p + q) = p + 2 p q + q 2 2 2 Returning to the Δ CCR5/HIV data: We start with the antibody (-) data since these people probably have not been exposed to HIV 1st What is our estimate of p = proportion of Δ CCR5?

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