There is a population not at HWE where red eye = $a^+$ (dominant) and white eyes = $a^-$.
$a^+/a^+ = 0.6$
$a^-/a^+ = 0.1$
$a^-/a^- = 0.3$
what are the frequencies of the $a^+$ and $a^-$ alleles?
My attempt: So if it's not at HWE then I should divide the heterozygotes equally to get $a^+ = 0.65$ and $a^- = 0.35$.
We are now at HWE. What are the frequencies in the next generation?
My attempt: so if $a^+ (p) = 0.65$ and $a^- (q) = 0.35$ then $p^2 = 0.42$ and $q^2 = 0.12$. Then using $p^2 + 2pq + q^2 = 1$, I can figure out that the heterozygotes make up $0.46$. So the frequency of the $a^+$ allele is $0.42 + 0.46 = 0.88$, and then $a^-$ should be $0.12$ (because they should add to 1)
What is the frequency 5 generations later in HWE?
My attempt: Wouldn't it be the same answer as #2. I'm not sure how more generations in HWE would change the allele frequencies.