AP Biology· difficulty 2/5
A population of 1000 individuals is at Hardy-Weinberg equilibrium with p² = 0.49 and q² = 0.09.
What is the expected number of heterozygotes (Bb)?
210
840
490
420
Explanation
p² + 2pq + q² = 1, so 2pq = 1 - 0.49 - 0.09 = 0.42. With N = 1000, expected Bb = 0.42 x 1000 = 420.
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