Question

The frequency of two alleles in a gene pool is 0.17 (A) and 0.83(a). Assume that the population

is in Hardy-Weinberg equilibrium.

Calculate the percentage of heterozygous individuals in the population.

Answer 1

You want to use the Hardy-Weinberg equilibrium equation.
\(p^2+2pq+q^2 = 1\)

\(p^2 =\) the frequency of homozygous dominant genotype \((A)\)
\(2pq =\) the frequency of heterozygous genotype
\(q^2 =\) the frequency of homozygous recessive genotype \((a)\)

Answer 2

Given that your question doesn't state the population, we'll assume 100. As well, given that A is considered the dominant allele.
\(Het = 2 (0.17 \times 0.83)\)

So the full equation would be:
\(0.17^2 \times 2(0.17 \times 0.83) \times 0.83^2 = 1\)

Though this is not required, as all you need here is \(2pq\).

Once you have \(2pq\), multiply it by your population amount (assuming 100 at this moment), and you will have the amount of people per your population that are heterozygous.

Although here it is percentage, so you'd multiply \(2pq\) by a hundred, which would give you your percentage of the population that is heterozygous.