Question

1. in a population in hardy - weinberg equilibrium, the frequency of homozygous recessive individuals is 0.2. what is the frequency of the dominant and recessive alleles in the population?

Explanation:

Step1: Recall Hardy - Weinberg equation

The frequency of homozygous recessive individuals ($q^{2}$) is given. The Hardy - Weinberg equation for allele frequencies is $p + q=1$, and for genotype frequencies is $p^{2}+2pq + q^{2}=1$, where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.

Step2: Calculate the frequency of the recessive allele ($q$)

Given $q^{2}=0.2$. Taking the square - root of both sides, we get $q=\sqrt{0.2}\approx0.447$.

Step3: Calculate the frequency of the dominant allele ($p$)

Since $p + q = 1$, then $p=1 - q$. Substituting $q = 0.447$, we have $p=1 - 0.447 = 0.553$.

Answer:

The frequency of the dominant allele ($p$) is approximately $0.553$ and the frequency of the recessive allele ($q$) is approximately $0.447$.