numerical: hardy-weinberg

Question

In a population, frequency of recessive phenotype (aa) is 0.16. Assuming H-W equilibrium, what is frequency of heterozygotes?

RankGuru · Accepted Answer

1. **Identify Given Value**: Recessive phenotype frequency = $q^2 = 0.16$.
2. **Find Recessive Allele Frequency ($q$)**: 
   - $q = \sqrt{0.16} = 0.4$.
3. **Find Dominant Allele Frequency ($p$)**: 
   - Since $p + q = 1$, $p = 1 - 0.4 = 0.6$.
4. **Calculate Heterozygote Frequency ($2pq$)**: 
   - $Frequency = 2 	imes p 	imes q = 2 	imes 0.6 	imes 0.4$
   - $Frequency = 2 	imes 0.24 = \mathbf{0.48}$.
5. **Check**: $p^2 (0.36) + 2pq (0.48) + q^2 (0.16) = 1.00$.

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numerical: hardy-weinberg

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