26. In the peppered month (B ist on be ml aria), black individuals may be either homozygous (A₁A₁) or heterozygous (A₁A₂), whereas pale gray months are homozygous (A₂A₂). Suppose that in a sample of 250 months from one locality, 108 are black and 142 are gray, (a) Which allele is dominant? (b) Assuming that the locus is in Hardy-Weinberg equilibrium, what are the allele irequencies? (c) Under this assumption, please describe Ho (Observed heterozygosity), He (Expect heterozygosity) and F_is. (6 分）

詳解 (共 1 筆)

詳解提供者：Iry Kuo

(a)黑蛾基因是顯性。 (b)白蛾 a^2=142/250 a=0.75 A=1-0.75 = 0.25 ( A^2 + 2Aa + a^2 = 1 ) (c) Ho = 2Aa = 2 * 0.75 * 0.25 = 0.375 He = 0.375 Fis =