In a village of 2500 people, the frequency of the recessive allele ‘r’ responsible for a genetic disorder is 0.15. Assuming Hardy–Weinberg equilibrium, how many individuals are expected to be homozygous recessive (rr)?

Step-by-step solution

1. Step 1: Let $p$ be the frequency of the dominant allele $R$, so $p=1-q=1-0.15=0.85$.
2. Step 2: Hardy–Weinberg predicts the frequency of homozygous recessive genotype as $q^2=0.15^2=0.0225$.
3. Step 3: Multiply this frequency by the total population: $0.0225\times 2500=56.25$.
4. Step 4: Since the number of individuals must be whole, round to the nearest integer: 56 individuals are expected to be rr.

Concept tested: Hardy–Weinberg population genetics

Published 1 October 2026 · How this solution was prepared and checked