A recessive X‑linked disease has a prevalence of 1 in 2000 among males. Assuming random mating and Hardy–Weinberg equilibrium, what is the carrier frequency among females?

Step-by-step solution

1. For males, disease frequency equals allele frequency $q=1/2000=0.0005$.
2. Female carrier frequency $=2pq\approx 2q$ (since $p\approx 1$) $=2\times 0.0005=0.001$.
3. Convert to per 1 000 individuals: $0.001\times 1000=1$, hence 1 in 1000.

Concept tested: X‑linked carrier frequency from male prevalence

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