In a population of 5000 individuals, the allele frequency of the dominant allele A is 0.7 and that of recessive allele a is 0.3. Under Hardy–Weinberg equilibrium, how many individuals are expected to be heterozygous (Aa)?

Step-by-step solution

1. Apply Hardy–Weinberg law: "frequency of heterozygotes = 2pq".
2. Compute "2pq = 2 * 0.7 * 0.3 = 0.42".
3. Multiply by total population: "0.42 * 5000 = 2100" individuals are expected to be heterozygous.

Concept tested: Hardy–Weinberg equilibrium

Published 22 September 2026 · How this solution was prepared and checked