Hardy–Weinberg: heterozygote frequency
In a population at Hardy–Weinberg equilibrium, 16% of individuals show a recessive phenotype. What percentage are heterozygous?
Answer D: explanation for every choice
A
16% is q², the homozygous recessive frequency.
B
This is pq, forgetting the factor of 2.
C
36% is p², the homozygous dominant frequency.
D
Correct. q² = 0.16, so q = 0.4 and p = 0.6. Then 2pq = 2(0.6)(0.4) = 0.48.
TakeawayStart from q², take the square root, then compute 2pq.
Equation usedHardy–Weinberg
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