Question

Imagine a population of diploid species in Hardy-Weinberg equilibrium. The population has two alleles for a gene which are ‘a’ and ‘A’. The number of individuals with ‘aa’ genotype in this population is 1 in 10000. The frequency of the allele ‘A’ in the population is ............ (up to two decimal places)

Hint:

In Hardy-Weinberg equilibrium, remember that the sum of the frequencies of both alleles (p and q) is always 1. Use the genotype frequency to find allele frequencies.

Answer 1

In Hardy-Weinberg equilibrium, the frequency of alleles in a population can be calculated using the Hardy-Weinberg principle. Let the frequency of allele 'A' be represented by \( p \) and the frequency of allele 'a' be represented by \( q \). According to Hardy-Weinberg equilibrium: \[ p + q = 1 \] Additionally, the frequency of the 'aa' genotype is \( q^2 \), and it is given that the frequency of the 'aa' genotype is \( \frac{1}{10000} \). Therefore: \[ q^2 = \frac{1}{10000} \] Taking the square root of both sides: \[ q = \frac{1}{100} \] Since \( p + q = 1 \), we can substitute the value of \( q \) to find \( p \): \[ p = 1 - q = 1 - \frac{1}{100} = \frac{99}{100} = 0.99 \] Thus, the frequency of allele 'A' in the population is 0.99.