The Hardy-Weinberg principle states that for a gene with two alleles, A and a, with frequencies $p$ and $q$ respectively, the genotype frequencies in a population at equilibrium are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Given that the frequency of allele A is $p = 0.1$. The frequency of the homozygous dominant genotype (AA) is represented by $p^2$.
Substitute the value of $p$ into the expression for the AA genotype frequency: $$p^2 = (0.1)^2$$ $$p^2 = 0.01$$
Final Answer: 0.01
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