In a population satisfying the Hardy-Weinberg principle for a single g…

Question

In a population satisfying the Hardy-Weinberg principle for a single gene with two alleles of frequencies p and q, the frequency of heterozygotes is:

Accepted Answer

D. {'text': '2pq (twice the product of the two frequencies)', 'label': 'D'}

Under Hardy-Weinberg equilibrium, genotype frequencies follow the binomial expansion of (p+q) squared, giving p squared plus 2pq plus q squared equal to 1. The heterozygote (Aa) frequency is therefore 2pq.

Answer

{'text': 'p squared (the homozygous dominant frequency)', 'label': 'A'}

Answer

{'text': 'q squared (the homozygous recessive frequency)', 'label': 'B'}

Answer

{'text': 'p plus q (which must equal one by definition)', 'label': 'C'}