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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.

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2 Answers 2

You are almost correct, but with a few modifications/comments.

  • The calculations in Q1 are independent of HWE, and are just calculating the current allele frequencies in the population based on genotype frequencies (i.e. would be the same irrespectively of if the population is in HWE or not).
  • In Q2 it is unclear if the questions is asking for genotype frequencies or allele frequencies. Your answer of genotype frequencies of $p^2=0.42$ ($a^+a^+$), $q^2 =0.12$ ($a^-a^-$) and $2pq =0.46$ ($a^+a^-$) is correct if you assume the allele frequencies from Q1 and HWE. However, your allele frequencies are incorrect and should be $p = 0.42 + 0.46/2 = 0.65$ and $q = 0.12 + 0.46/2 = 0.35$, i.e. the same as in Q1. Remember that the allele frequencies shouldn't change under HWE.
  • Yes frequencies should not change in Q3. This is the textbook definition of HWE, i.e. that allele and genotype frequencies remain constant between generations in the absence of evolutionary forces (i.e. under the HWE).
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thanks. for #1 I thought that i calculated the correct allele frequencies (im given genotype frequencies). –  Kirby Sep 24 '13 at 14:02
@Kirby Yes the numbers in Q1 are correct, and I was only commenting on that you wrote "So if it's not at HWE then I should ...". The calculation of frequencies (as $p = f(a^+a^+)+f(a^+a^-)/2$ and $q = f(a^-a^-)+f(a^+a^-)/2$ ) is valid with or without HWE. –  fileunderwater Sep 24 '13 at 14:05
ok. for #2 you said that p^2 = 0.35, but isn't that just p and p^2 would be 0.42? –  Kirby Sep 24 '13 at 14:31
@Kirby Yes correct. That was a typo on my part. Allele freq $p = 0.65$ and genotype freq $p^2 = 0.42$ ($f(a^+a^+)$). –  fileunderwater Sep 24 '13 at 14:38
@Kirby Feel free to mark an answer as 'answered' if you find it satisfying. –  fileunderwater Oct 22 '13 at 10:37

I am not sure I fully understood your question. I did not really understand why your divided the heterozygotes frequencies by 2 and add this value to the homozygotes to get the allele frequencies when Hardy-Weinberg (HW) is not respected. I think you missunderstand the concept of Hardy-Weinberg principle.

The frequency of the the two (a- an a+) alleles depends upon selection or genetic drift. If both selection and drift are absent, then the allele frequencies do not change.

Under some assumptions, one can know the mathematic relation between genotype frequencies and allele frequencies. A population is at Hardy-Weinberg Equilibrium if :

organisms are diploid (Although the model can be extended to other cases)
only sexual reproduction occurs
generations are non overlapping
mating is random
population size is infinitely large
allele frequencies are equal in the sexes
there is no migration, mutation or selection


A population that is a hardy-Weinberg equilibrium (and if the conditions for this equilibrium is still respected at the next geenration) then no change in either allele or genotypes frequencies will occur in the following generations.

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thanks. but how can i find the allele frequencies in #1? –  Kirby Sep 24 '13 at 13:08

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