BIO HELP! Hardy-Weinberg Equilibrium!?

> In a population that is 10 percent AA, 20 percent Aa, and 70 percent aa, what is the frequency of allele A? of a?

Let's say we have 100 individuals. Then we have 200 alleles total:

10 AA => 20 A alleles

20 Aa => 20 A + 20 a alleles

70 aa => 140 a alleles

freq(A) = (20 + 20) / 200 = 0.2 (answer)

freq(a) = (20 + 140) / 200 = 0.8 (answer)

> If A and a are the only alleles at the gene A locus, does the population seem to be a Hardy-Weinberg equilibrium?

Let's say we have 100 individuals. If in Hardy-Weinberg equilibrium we expect

number of homozygote A individuals = freq(AA) * population size = (freq(A))^2 * population size = 0.2^2 * 100 = 4

number of heterozygote individuals = 2*freq(A)*freq(a) * population size = 32

number of homozygote a individuals = freq(a)^2 * population size = 64

Those numbers are nowhere near the 10. 20. and 70 that we observed. This population is NOT in Hardy-Weinberg equilibrium. Something is stomping the heterozygotes.

It depends on what "seems to be" means. The H-W calculated values are fr(aa) = 0.7, fr(Aa) = 0.2733, and fr(AA) = 0.0267. The observed values are about the calculated values.

The calculated values: fr(A) = 0.1633 and fr(a) = 0.8367.