Mapping zygote pool to adult pool mapping adult pool

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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
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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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