Hardy-Weinberg: Heterozygote Frequency

If the gene frequency of A = 0.7 and a = 0.3, what is the genotypic frequency of the heterozygote under Hardy-Weinberg equilibrium?

Correct Answer: 0.42

From the notes (Table 8 — H-W Equation): Genotypic frequency of heterozygote = 2pq

Calculation

• p (frequency of A) = 0.7; q (frequency of a) = 0.3
• Heterozygote (Aa) = 2pq = 2 × 0.7 × 0.3 = 0.42

Full Genotypic Frequency Check

• AA (dominant homozygote) = p² = (0.7)² = 0.49
• Aa (heterozygote) = 2pq = 0.42 ✔
• aa (recessive homozygote) = q² = (0.3)² = 0.09
• Total: 0.49 + 0.42 + 0.09 = 1.00 ✔

Key Note (from notes)

Frequency of heterozygotes (2pq) cannot exceed 50% and is maximum when p = q = 0.5 — at which point 2pq = 2 × 0.5 × 0.5 = 0.50.