Evolution - Hardy-Weinberg Principle

In a population of $1000$ individuals, $360$ belong to the genotype $AA$, $480$ to $Aa$, and the remaining $160$ to $aa$. Based on this data, find the frequency of allele $A$ in the population.

Total individuals $= 1000$. Frequency of $AA$ ($p^2$) $= \frac{360}{1000} = 0.36$. Frequency of $aa$ ($q^2$) $= \frac{160}{1000} = 0.16$. From $p^2 = 0.36$, we get $p = \sqrt{0.36} = 0.6$. Alternatively, from $q^2 = 0.16$, we get $q = \sqrt{0.16} = 0.4$. Since $p + q = 1$, $p = 1 - 0.4 = 0.6$.

The frequency of allele $A$ is denoted by $p$. It can be calculated by taking the square root of the frequency of homozygous dominant individuals or by subtracting the frequency of the recessive allele ($q$) from $1$.

If the frequency of a recessive phenotype in a stable population is $0.09$, calculate the frequency of the heterozygous genotype.

The recessive phenotype frequency corresponds to $q^2$. Given $q^2 = 0.09$, therefore $q = \sqrt{0.09} = 0.3$. Using $p + q = 1$, we find $p = 1 - 0.3 = 0.7$. The frequency of the heterozygous genotype is $2pq = 2 \times 0.7 \times 0.3 = 0.42$.

To find the heterozygotes ($2pq$), we must first determine $q$ from the recessive phenotype frequency ($q^2$), then find $p$, and finally multiply $2 \times p \times q$.