In a population, an X‑linked recessive disease has a prevalence of 1 in 1000 among males. Assuming random mating and Hardy–Weinberg equilibrium, estimate the carrier frequency among females.
Answer
0.2% of females
The correct answer is 0.2% of females — Thus, about 0.2% of females are carriers.
Step-by-step solution
- Let q be the frequency of the disease allele. For X‑linked recessive, prevalence in males equals q, so "q = 1/1000 = 0.001".
- Dominant allele frequency p = 1 – q = 0.999.
- Carrier frequency in females = 2pq = 2 * 0.999 * 0.001 = 0.001998. This equals approximately 0.2% of females.
- Thus, about 0.2% of females are carriers.
Common mistake: Treating carrier frequency as q instead of 2pq
Concept tested: X‑linked recessive – prevalence and carrier frequency
Published · How this solution was prepared and checked
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