Hardy-Weinberg Equilibrium

AP Biology· difficulty 3/5

A population has 64 AA, 32 Aa, and 4 aa individuals. The frequency of allele A is

A
$0.60$
B
$0.40$
C
$0.96$
D
$0.80$
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Explanation

Total = 100. A alleles: 2(64) + 1(32) = 160. Total alleles = 200. p = 160/200 = 0.80.

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