In a human population in Hardy-Weinberg equilibrium, the frequency of an autosomal recessive allele (q) is 0.20. What percentage of the population is expected to be heterozygous carriers (2pq)?
Answer and explanation
Correct answer: A. 32 percent
- Step 1: Allele frequencies satisfy p + q = 1. Given q = 0.20, p = 1 - 0.20 = 0.80. Step 2. The frequency of heterozygotes is given by 2pq = 2 (0.80) (0.20) = 0.32. Step 3. Converting to percentage: 0.32 100 = 32%. Answer: 32 percent. Why not the others:* 64% is p^2 (homozygous dominant), 4% is q^2 (homozygous recessive), and 16% is half the carrier frequency.
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