The Hardy- Weinberg Principle That States If $p$ Is the Frequency of the Dominant Allele

The Hardy- Weinberg principle that states if $p$ is the frequency of the dominant allele $A$ in a gene pool and $q$ is the frequency of the recessive allele $a$ in the gene pool, then $p+q=1$ and $\mathrm{p}^{2}+2 \mathrm{pq}+\mathrm{q}^{2}=1$ .
*Portion of the population that is homozygous dominant (AA) : $\mathrm{p}^{2}$
*Portion of the population that is heterozygous dominant (Aa) : 2pq
*Portion of the population that is homozygous recessive (aa) : $\mathrm{q}^{2}$
Sickle-cell anemia is a recessive genetic condition, meaning that a person only has the disease if both alleles are recessive. If $0.35 \%$ of a population has sickle- cell anemia, what percentage of the population is homozygous dominant and what percentage is heterozygous dominant? Round $\mathrm{q}$ and $\mathrm{p}$ to the nearest thousandth, and round percentages to the nearest tenth of a perœent.

A) homozygous dominant: $86.2 \%$ , heterozygous dominant: $12.8 \%$
B) homozygous dominant: $94.1 \%$ , heterozygous dominant: $5.9 \%$
C) homozygous dominant: 94.5\%, heterozygous dominant: $5.5 \%$
D) homozygous dominant: 88.5\%, heterozygous dominant: $11.1 \%$