In a population of 2000 individuals, the allele frequency of the dominant allele A (p) is 0.6. Assuming Hardy–Weinberg equilibrium, how many individuals are expected to be heterozygous (Aa)?
Answer
960
The correct answer is 960 — Number of heterozygotes = 0.48 × 2000 = 960.
Step-by-step solution
- Allele frequency of recessive allele a (q) is 1 – p = 0.4.
- Under Hardy–Weinberg, frequency of heterozygotes is 2pq = 2 × 0.6 × 0.4 = 0.48.
- Number of heterozygotes = 0.48 × 2000 = 960.
Common mistake: Using p^2 for heterozygotes instead of 2pq
Concept tested: Hardy–Weinberg equilibrium
Published · How this solution was prepared and checked
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