In a Hardy-Weinberg population, if the frequency of the recessive allele $a$ is $q = 0.3$, the percentage of heterozygotes (Aa) in the population is:
A9%
B42%
C49%
D30%
Answer & Solution
Correct answer: B. 42%
$p + q = 1 \Rightarrow p = 0.7$. Heterozygotes: $2pq = 2(0.7)(0.3) = 0.42 = 42\%$. (Homozygous dominant AA = $p^2$ = 49%; homozygous recessive aa = $q^2$ = 9%.)
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