In a population of 500 at Hardy-Weinberg equilibrium, the frequency of allele t is 0.2. How many heterozygous Tt individuals are expected?
Correct answer A. 160
Explanation
p = 1 − 0.2 = 0.8. Heterozygote frequency 2pq = 2 × 0.8 × 0.2 = 0.32. Expected number = 0.32 × 500 = 160.
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