Assuming the population is in Hardy-Weinberg equilibrium, if 16% of the individuals in a population show a recessive trait (q2=16%), what is the allelic frequency for the dominant allele?

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BlackTom AI · Accepted Answer

To analyze this question, start with the Hardy-Weinberg relationship where q^2 represents the frequency of the recessive phenotype, and q is the recessive allele frequency. Given q^2 = 0.16, take the square root to find q: q = sqrt(0.16) = 0.4.

Since p + q = 1, the dominant allele frequency p equals 1 − q = 1 − 0.4 = 0.6, or 60%.

Now evaluate each option:
- Option A: 91% — This is inconsistent with p = 0.6 derived from q = 0.4; 91% would require q ≈ 0.09, whose q^2 would be about 0.0081, not 0.16. Hence this option is incorrect.
- Option B: 60% — This matches the calculated p = 0.6 exactly, reflecting the correct application of p = 1 − q when q = 0.4. This option aligns with the Hardy-Weinberg calculation and is consistent with the given q^2 value.
- Option C: 84% — An allelic frequency of 0.84 would imply q ≈ 0.16, giving q^2 ≈ 0.026, which does not equal 0.16. Therefore this option is incorrect.
- Option D: 70% — This would correspond to p = 0.7, implying q = 0.3 and q^2 = 0.09, not 0.16. Thus this option is also incorrect.

In summary, only the option reflecting a dominant allele frequency of 60% is consistent with the given recessive phenotype frequency under Hardy-Weinberg equilibrium, since q = 0.4 and p = 0.6 satisfies p + q = 1 and q^2 = 0.16.

AGRI10051_2025_SM2 Supplementary or Special Exam: Genetics for Agriculture (AGRI10051_2025_SM2)- Requires Respondus LockDown Browser

Assuming the population is in Hardy-Weinberg equilibrium, if 16% of the individuals in a population show a recessive trait (q2=16%), what is the allelic frequency for the dominant allele?

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