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教甄◆生物專業題庫

【非選題】
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 Fis. (6 分)
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Iry Kuo 小一上 (2016/04/27)
(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 =