A diploid population of 100 individuals has allele A at frequency p = 0.4. Assuming random mating and Hardy–Weinberg equilibrium, what is the expected heterozygosity in this population?
Answer
0.48
The correct answer is 0.48 — Simplify: 2 × 0.24 = 0.48.
Step-by-step solution
- Compute frequency of allele a as q = 1 – p = 0.6.
- Use Hardy–Weinberg formula for heterozygosity: H = 2pq.
- Substitute p = 0.4 and q = 0.6: H = 2 × 0.4 × 0.6.
- Simplify: 2 × 0.24 = 0.48.
Common mistake: Using p or q alone instead of 2pq
Concept tested: Hardy–Weinberg heterozygosity
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