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Unformatted text preview: Aa = 0.8 w aa = 0.6: Genotype frequencies are p 2 = 0.16, 2pq = 0.48, q 2 =0.36, thus: p t+1 = ((.16 x 1.0) + (.24 x .8))/((.16 x 1.0) + (.48 x .8) + (.36 x .6)) = .463; so q = .537 and thus f(AA) t+1 = .215, f(Aa) t+1 = .497 and f(aa) t+1 = .288. Note both allele frequencies and genotype frequencies have changed (compare to what we saw with inbreeding). This can be continued with the new allele frequencies and so on. When will the selection process stop? when p = 0, i.e., when p t+1 = p t . In some situations this will stop only when one allele is selected out of the population (p = 1.0). Now we can consider various regimes of selection (s = selection coefficient , (1s) is fitness ): AA Aa aa I 1 1 1  s selection against recessive II 1  s 1  s 1 selection against dominant III 1 1  hs 1  s incomplete dominance (0<h<1) IV 1  s 1 1  t selection for heterozygotes...
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 Fall '10
 JessicaDigirolamo
 Microbiology

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