In a population of 3000 individuals, allele A has frequency 0.55. Assuming Hardy–Weinberg equilibrium, how many individuals are expected to be heterozygous (Aa)?

Step-by-step solution

1. Compute frequency of allele a as $q=1-p=0.45$.
2. Calculate heterozygote frequency $2pq=2\times 0.55\times 0.45=0.495$.
3. Expected number of heterozygotes = $3000\times 0.495=1485$.

Concept tested: Hardy–Weinberg heterozygote calculation

Published 24 September 2026 · How this solution was prepared and checked