Question:
If a natural population with 50 individuals is in Hardy-Weinberg equilibrium for a gene with two alleles A and a, with the gene frequency of allele A of 0.6, the genotype frequency of Aa will be:
If a natural population with 50 individuals is in Hardy-Weinberg equilibrium for a gene with two alleles A and a, with the gene frequency of allele A of 0.6, the genotype frequency of Aa will be:
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Use \( 2pq \) for heterozygous (Aa) frequency; \( p^2 + 2pq + q^2 = 1 \).
Updated On: Jun 19, 2025
- 0.16
- 0.36
- 0.24
- 0.48
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The Correct Option is D
Solution and Explanation
According to Hardy-Weinberg equilibrium, the genotype frequency of Aa is given by \( 2pq \), where \( p = 0.6 \) (frequency of A) and \( q = 1 - p = 0.4 \).
\[ 2pq = 2 \times 0.6 \times 0.4 = 0.48 \] The correct genotype frequency of Aa is actually \( 0.48 \)
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