In a population in Hardy-Weinberg equilibrium, the frequency of the recessive allele q is 0.2. What is the frequency of heterozygotes?
Answer and explanation
Correct answer: A. 0.32
- Step 1: p = 1 - 0.2 = 0.8. Step 2. 2pq = 2 x 0.8 x 0.2 = 0.32. Answer: 0.32. Why not the others: 0.04 is q squared, 0.64 is p squared.
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