In a Hardy-Weinberg population the recessive phenotype (genotype aa) o…

Question

In a Hardy-Weinberg population the recessive phenotype (genotype aa) occurs in 16 percent of individuals. Assuming equilibrium, the frequency of heterozygous carriers (Aa) is closest to:

Accepted Answer

A. {'text': '0.48, computed as 2pq with p = 0.6 and q = 0.4', 'label': 'A'}

If q squared equals 0.16, then q equals 0.4 and p equals 0.6. Heterozygote frequency is 2pq, which is 2 times 0.6 times 0.4, equal to 0.48. The dominant homozygote frequency p squared is 0.36, and the total 0.36 plus 0.48 plus 0.16 equals 1.

Answer

{'text': '0.16, the same as the homozygous recessive frequency', 'label': 'B'}

Answer

{'text': '0.32, half the heterozygote-plus-recessive total', 'label': 'C'}

Answer

{'text': '0.64, dominated by the homozygous dominant share', 'label': 'D'}