Hardy-Weinberg Equilibrium

AP Biology· difficulty 3/5

In a randomly mating population at Hardy-Weinberg equilibrium, the dominant allele frequency p = 0.7 and the recessive allele frequency q = 0.3.

Allele frequencies in a population p = 0.7 q = 0.3 p^2 = 0.49 (AA) 2pq = 0.42 (Aa) q^2 = 0.09 (aa) Sum = 1

Assuming Hardy-Weinberg equilibrium, what fraction of the population is heterozygous?

  • A

    0.42

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  • B

    0.09

  • C

    0.49

  • D

    0.30

Explanation

Heterozygote frequency = 2pq = 2(0.7)(0.3) = 0.42, illustrating that intermediate allele frequencies maximize heterozygosity.

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