If a population has p = 0.7 and q = 0.3 for a gene, which genotype frequencies are expected under Hardy-Weinberg equilibrium?

If a population has p = 0.7 and q = 0.3 for a gene, which genotype frequencies are expected under Hardy-Weinberg equilibrium?

• A. AA 0.45, Aa 0.40, aa 0.15
• B. AA 0.49, Aa 0.42, aa 0.09
• C. AA 0.51, Aa 0.29, aa 0.20
• D. AA 0.70, Aa 0.30, aa 0.00

Answer: (B)

Under Hardy-Weinberg equilibrium, genotype frequencies come from squaring the allele frequencies: p^2 for AA, 2pq for Aa, and q^2 for aa. With p = 0.7 and q = 0.3, compute p^2 = 0.49, 2pq = 0.42, and q^2 = 0.09. So the expected genotype frequencies are AA 0.49, Aa 0.42, aa 0.09. This matches the option that lists those exact values. The other options would require different allele frequencies or violate the p + q = 1 and p^2 + 2pq + q^2 = 1 relationships (for example, an aa frequency of 0.15 implies q ≈ 0.387, not 0.3; an aa of 0.00 would imply q = 0).

Under Hardy-Weinberg equilibrium, genotype frequencies come from squaring the allele frequencies: p^2 for AA, 2pq for Aa, and q^2 for aa. With p = 0.7 and q = 0.3, compute p^2 = 0.49, 2pq = 0.42, and q^2 = 0.09. So the expected genotype frequencies are AA 0.49, Aa 0.42, aa 0.09. This matches the option that lists those exact values. The other options would require different allele frequencies or violate the p + q = 1 and p^2 + 2pq + q^2 = 1 relationships (for example, an aa frequency of 0.15 implies q ≈ 0.387, not 0.3; an aa of 0.00 would imply q = 0).