Hardy-weinberg equilibrium

Question

If a population has 8 AA indivduals, 4 Aa individuals, and 3 aa individuals. how come when i do the problem the way counting up total (a) allele then divide by total alleles in population (for example 10/30=.33) i get a different answer than when i take the number of homozygous recessive individuals and divide by the toal # of individuals (3/15 = 0.2) and square root it (=.44) i thought both of these ways solved for p (frequency of the a allele)?

thanks in advance!

Answer 1

To apply Hardy-Weinberg empirically, you must have a large population, not just 15.

Suppose you had a large population with the same proportion of homozygous recessive individuals. Since f(a) = 0.44, f(A) = 0.56, and the proportion of AA individuals is (0.56)² = .31. So, 31% would be AA, 20% aa, and the remaining 49% heterozygous. As you can see the distribution of the small sample of 15 is not close to what it would have to be for the population. 5 AA, 7 Aa, 3 aa would have been much more reasonable as a typical sample.