A population is in Hardy-Weinberg equilibrium with two alleles, A and a; the frequency of allele a is 0.2.

What is the percentage of the population that is heterozygous for this allele?

Question

A population is in Hardy-Weinberg equilibrium with two alleles, A and a; the frequency of allele a is 0.2.

What is the percentage of the population that is heterozygous for this allele?

BlackTom AI · Accepted Answer

In Hardy–Weinberg the heterozygote frequency is given by 2pq. Here q (frequency of allele a) = 0.20, so first find p.

1) Determine allele frequencies:
- q = 0.20
- p = 1 − q = 1 − 0.20 = 0.80

2) Calculate genotype frequencies using Hardy–Weinberg formulas:
- Homozygous dominant (AA): p^2 = (0.80)^2 = 0.64 → 64.0%
  Calculation: 0.80 × 0.80 = 0.6400
- Homozygous recessive (aa): q^2 = (0.20)^2 = 0.04 → 4.0%
  Calculation: 0.20 × 0.20 = 0.0400
- Heterozygous (Aa): 2pq = 2 × 0.80 × 0.20 = 0.32 → 32.0%
  Calculation: 2 × 0.80 = 1.60; 1.60 × 0.20 = 0.3200

3) Why the correct answer is 32.0%:
The heterozygote fraction under Hardy–Weinberg equilibrium is explicitly 2pq. Substituting p = 0.80 and q = 0.20 yields 0.32, which as a percentage is 32.0%. This matches the provided correct answer.

4) Why the other options are incorrect:
- A) 0.80% (0.008) is far too small and does not correspond to any H–W genotype frequency here (likely a misplaced decimal).
- B) 5.12% (0.0512) and C) 7.68% (0.0768) do not equal p^2, q^2, or 2pq for p = 0.8 and q = 0.2.
- E) 44.7% is not any of the calculated genotype frequencies (p^2 = 64.0%, q^2 = 4.0%, 2pq = 32.0%).
- F) “None of these” is false because one option (32.0%) is correct.

Conclusion: The percentage of the population that is heterozygous (Aa) is 32.0%, confirming the provided answer.

A population is in Hardy-Weinberg equilibrium with two alleles, A and a; the frequency of allele a is 0.2. What is the percentage of the population that is heterozygous for this allele?单项选择题

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