In the Hardy-Weinberg equilibrium with q = 0.3, what is the frequency of heterozygotes?
Answer and explanation
Correct answer: C. 0.42
- Step 1: p = 1 - 0.3 = 0.7. Step 2. Heterozygotes = 2pq = 2 x 0.7 x 0.3 = 0.42. Answer: 0.42. Why not the others: 0.21: That is pq, not 2pq. 0.09: That is q squared. 0.7: That is the allele frequency p.
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