Question

If a natural population with 200 individuals is in Hardy-Weinberg equilibrium for a gene with two alleles A and a, and with the gene frequency of allele A of 0.8, the genotype frequency of Aa will be :

• A. \( 0.8 \)
• B. \( 0.16 \)
• C. \( 0.32 \)
• D. \( 0.64 \)

Hint:

Remember the Hardy-Weinberg equations: \(p + q = 1\) (for allele frequencies) and \(p^2 + 2pq + q^2 = 1\) (for genotype frequencies). Always identify the given allele frequency first to find the other.

Answer 1

The Correct Option is C

According to the Hardy-Weinberg equilibrium, if there are two alleles A and a with frequencies \(p\) and \(q\) respectively, then the genotype frequencies are given by the equation: $$p^2 (\text{AA}) + 2pq (\text{Aa}) + q^2 (\text{aa}) = 1$$ Given that the frequency of allele A (\(p\)) is 0.8. Since there are only two alleles, the frequency of allele a (\(q\)) can be calculated as: $$p + q = 1$$ $$0.8 + q = 1$$ $$q = 1 - 0.8 = 0.2$$ The genotype frequency of heterozygotes (Aa) is given by \(2pq\). $$2pq = 2 \times 0.8 \times 0.2$$ $$2pq = 1.6 \times 0.2$$ $$2pq = 0.32$$ Therefore, the genotype frequency of Aa will be 0.32.