This is all based on the Hardy-Weinberg equilibrium: p^{2} + 2pq + q^{2} = 1(100%) and where p+q=1. p^{2} is frequency of homozygotes for "S" and q^{2} is frequency of homozygotes for "s" and 2pq is frequency of heterozygotes.

The only way to answer this question would be if you were only interested in the frequency of alleles in the African American (AA) population. The frequency of "ss" carriers in the AA population is 11% as you mentioned. 11%=.11. The square root of .11 would be .33 or ~33% for the recessive allele "s". So our p=.33 because p^{2} is .11 or 11%.

.33^{2} is .11 or 11%. So we get back to the 11% of "ss" carriers in the AA population.

1-.33 would be the frequency of the dominant allele "S" which would be .66 or ~66%. In order to get the frequency of heterozygotes you would multiply s*S*2 or 2*p*q which would be .43 or 43%.

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