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If p is 0.6 in a Hardy-Weinberg population, the frequency of heterozygotes is:
A0.36 of the population
B0.16 of the population
C0.60 of the population
D0.48 of the population
Answer & Solution
Correct answer: D. 0.48 of the population
1. The equilibrium equation gives each genotype its own term.
2. The frequency of pq individuals is 2pq.
3. If p is 0.6 then q is 0.4, since the two allele frequencies sum to one.
4. So 2pq is 2 multiplied by 0.6 multiplied by 0.4.
5. That gives 0.48 of the population as heterozygotes.
_Source: OpenStax Biology 2e (CC BY 4.0), Ch 19 'The Evolution of Populations', sections 19.1-19.3_
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