Question

In a population at Hardy–Weinberg equilibrium, for gene-X with two alleles, A and a, if the frequency of allele A is 0.2 and that of allele a is 0.8, the frequency of heterozygote genotype Aa in the population will be ______ (correct to 2 decimal places).

Hint:

In Hardy–Weinberg equilibrium: Homozygous dominant = $p^2$, Heterozygote = $2pq$, Homozygous recessive = $q^2$.

Answer 1

Correct Answer: 0.32

Step 1: Recall Hardy–Weinberg equation.
\[ p^2 + 2pq + q^2 = 1 \] where
$p$ = frequency of allele A,
$q$ = frequency of allele a.
Step 2: Given.
$p = 0.2, \quad q = 0.8$ Step 3: Compute heterozygote frequency.
\[ 2pq = 2(0.2)(0.8) = 0.32 \] Step 4: Conclusion.
\[ \boxed{\text{Frequency of Aa} = 0.32} \]