In a certain population which is in Hardy Weinberg equilibrium, the frequency of allele A is 0.8 and that of allele a is 0.2. What would be the frequency of heterozygotes in this population at equilibrium?

The correct option is D 0.32
Hardy-Weinberg law states that the gene pool of a population remains constant under certain conditions. This is called genetic equilibrium or Hardy-Weinberg equilibrium.

The Hardy-Weinberg equation is an algebraic equation which is an expression of the Hardy-Weinberg principle. The frequencies of alleles and the genotypic ratios in a population can be calculated with the help of this equation.

According to Hardy Weinberg equation, if there are two alleles for a particular gene where A represents the dominant allele and a represents the recessive allele, then the frequency of A is p and a is q.
Combined frequencies of all alleles for a gene in a population is equal 1.
Therefore, p+q=1.
The frequency of homozygous dominant individuals (AA) in the population would be equal to pxp or $p^2$
The frequency of homozygous recessive individuals (aa) in the population would be equal to qxq or $q^2$
Since there are two ways of forming the heterozygote Aa, (allele A from the father and a from mother and vice versa) the frequency of heterozygous dominant individuals (Aa) in the population would be 2pq.
The sum of all three genotypic frequencies is $p^2+2pq+q^2=1$ which is a binomial expansion of ${(p+q)}^2$

p = frequency of the dominant allele (A)
q = frequency of the recessive allele(a)
$p^2$ = frequency of homozygous dominant individuals (AA)
$q^2$ = frequency of homozygous recessive individuals (aa)
2pq = frequency of heterozygous dominant individuals (Aa)

As per the question given, the frequency of the dominant allele, p is given as 0.8 and that of the recessive allele, q is 0.2.

Hence, the frequency of heterozygotes (Aa) in this population would be
2pq = 2(0.8)(0.2) = 0.32