Question

In a diploid population at Hardy-Weinberg equilibrium, the locus A has two alleles \( A_1 \) and \( A_2 \). If the frequency of \( A_1 A_1 \) genotype is 0.01, then the frequency of the allele \( A_2 \) is ............ (round off to 2 decimal places)

Hint:

In Hardy-Weinberg equilibrium, if the genotype frequencies are given, use the square root of the homozygous genotype frequency to find the allele frequency.

Answer 1

Correct Answer: 0.89

Step 1: Hardy-Weinberg equation.
Let \( p \) be the frequency of allele \( A_1 \) and \( q \) be the frequency of allele \( A_2 \). According to Hardy-Weinberg equilibrium, we know: \[ p^2 + 2pq + q^2 = 1 \] The frequency of the genotype \( A_1 A_1 \) is \( p^2 = 0.01 \), so: \[ p = \sqrt{0.01} = 0.1 \] Step 2: Frequency of allele \( A_2 \).
Since \( p + q = 1 \), we have: \[ q = 1 - p = 1 - 0.1 = 0.9 \] Step 3: Conclusion.
Thus, the frequency of allele \( A_2 \) is \( \boxed{0.1} \).