In a population of 2000 individuals, the frequency of the dominant allele A (p) is 0.65. Assuming Hardy–Weinberg equilibrium, calculate the expected number of heterozygous individuals (Aa).
Answer
910 individuals are expected to be heterozygous (Aa).
The correct answer is 910 individuals are expected to be heterozygous (Aa). — Thus 910 individuals are expected to be Aa.
Step-by-step solution
- Determine the recessive allele frequency: q = 1 – p = 1 – 0.65 = 0.35.
- Calculate heterozygote frequency using 2pq: 2 × 0.65 × 0.35 = 0.455.
- Multiply the heterozygote frequency by the total population: 0.455 × 2000 = 910.
- Thus 910 individuals are expected to be Aa.
Common mistake: Using p² instead of 2pq for heterozygotes
Concept tested: Hardy–Weinberg heterozygote frequency
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