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 by the Hardy - Weinberg equation $p^{2}+2pq + q^{2}=1$, where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele, and $p + q=1$. We are given that $q^{2}=0.2$.

Step2: Calculate the frequency of the recessive allele

To find $q$, take the square - root of $q^{2}$. So, $q=\sqrt{q^{2}}=\sqrt{0.2}\approx0.447$.

Step3: Calculate the frequency of the dominant allele

Since $p + q = 1$, then $p=1 - q$. Substitute $q\approx0.447$ into the equation, so $p = 1-0.447 = 0.553$.

Answer:

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