Hardy-Weinberg Calculation: Heterozygote Frequency

This solution explains how to calculate the frequency of heterozygotes ($A_1A_2$) in a population that is at Hardy-Weinberg equilibrium. We are given the frequency of the $A_1A_1$ genotype.

Allele Frequency ($p$) from $A_1A_1$ Genotype

The Hardy-Weinberg principle states that genotype frequencies are related to allele frequencies. For a population at equilibrium, the frequency of the homozygous genotype $A_1A_1$ is equal to $p^2$, where $p$ is the frequency of the $A_1$ allele.

We are given:

$ \text{Frequency}(A_1A_1) = p^2 = 0.01 $

To find the frequency of the $A_1$ allele ($p$), we take the square root:

$ p = \sqrt{0.01} $

$ p = 0.1 $

Allele Frequency ($q$) Calculation

In a population with two alleles ($A_1$ and $A_2$), the sum of their frequencies must equal 1 ($p + q = 1$). Using the calculated frequency of $A_1$ ($p$), we can find the frequency of the $A_2$ allele ($q$):

$ q = 1 - p $

$ q = 1 - 0.1 $

$ q = 0.9 $

Heterozygote Frequency ($A_1A_2$) Calculation

The frequency of the heterozygote genotype $A_1A_2$ in a Hardy-Weinberg equilibrium population is given by the expression $2pq$. Substituting the calculated allele frequencies $p=0.1$ and $q=0.9$:

$ \text{Frequency}(A_1A_2) = 2pq $

$ \text{Frequency}(A_1A_2) = 2 \times 0.1 \times 0.9 $

$ \text{Frequency}(A_1A_2) = 2 \times 0.09 $

$ \text{Frequency}(A_1A_2) = 0.18 $

Final Result: Heterozygote Frequency

The frequency of the heterozygotes $A_1A_2$ is calculated to be 0.18. This value is consistent with the provided answer range (0.179 to 0.181) and meets the requirement of being expressed up to 2 decimal places.