The dominant allele (A) occurs with a frequency of 0.6 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the frequency of heterozygous individuals?

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To determine the frequency of heterozygous individuals under Hardy-Weinberg equilibrium, we first identify the allele frequencies: the dominant allele A has frequency p = 0.6, so the recessive allele a has frequency q = 1 - p = 0.4.

Under Hardy-Weinberg, the genotype frequencies are:
- p^2 for AA (homozygous dominant)
- 2pq for Aa (heterozygous)
- q^2 for aa (homozygous recessive)

Calculating the heterozygote frequency: 2pq = 2 × 0.6 × 0.4 = 0.48.

Now let's examine each option:
- 0.24: This value does not match any standard Hardy-Weinberg proportion in this scenario. It might be mistaken for a half of 0.48 or a wrong product, but it does not align with p^2, 2pq, or q^2 here.
- 0.48: This is the correct heterozygote frequency, since it equals 2pq with p = 0.6 and q = 0.4.
- 0.36: This equals p^2 when p = 0.6, representing homozygous dominant individuals, not heterozygotes.
- 0.16: This equals q^2 when q = 0.4, representing homozygous recessive individuals, not heterozygotes.

In summary, the heterozygote class frequency is computed from the product 2pq, and plugging in the given allele frequencies yields 0.48. The other values correspond to the homozygous genotypes instead of heterozygotes.

2025SP-BZ-111-L02 Post-Lab 13: Natural Selection and Population Genetics

The dominant allele (A) occurs with a frequency of 0.6 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the frequency of heterozygous individuals?

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