Hardy - Weinberg Questions.

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Hardy - Weinberg Questions

Hardy-Weinberg In a population of 600 members, the numbers of different individuals of 3 different genotypes are AA=350 Aa=100 aa=150 What are the genotypic frequencies in this population? AA + Aa + aa = 350 + 100 + 150 = 600 individuals So ….. AA/Total = 350 / 600 = .58 (58%) AA = 0.58 Aa/Total = 100 / 600 = .17 (17%) Aa = 0.17 aa/ Total = 150 / 600 = .25 (25%) aa = 0.25

Hardy-Weinberg

Hardy-Weinberg In a population of 600 members, the numbers of different individuals of 3 different genotypes are AA=350 Aa=100 aa=150 What would be the expected genotypic frequencies? Since p = 0.5 and q = 0.5 Then p2 = (0.5)2 = 0.25 What this means is 2pq = 2 x (0.5) x (0.5) = .5 p2 + 2pq + q2 = 1 And q2 = (0.5)2 = 0.25 .25 + .5 + .25 = 1

Hardy-Weinberg In a population of 600 members, the numbers of different individuals of 3 different genotypes are AA=350 Aa=100 aa=150 Is this population in genetic equilibrium? Frequencies in the Population Expected Frequencies in Hardy-Weinberg equilibrium No 0.58 AA 0.25 0.17 Aa 0.50 0.25 aa 0.25