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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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3 Answers 3

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

phenylketonuria is an autosomal recessive disorder.in a population of 1800 people,648 members are suffering with PKU.find out the number of heterozygous individuals in that population.

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This appears to be a question. It is certainly not an answer. –  HDE 226868 1 hour ago
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  Bez 43 mins ago

At generation 1, the frequencies of $a^-$ is $p=0.35$ and the frequency of $a^+$ if $q=0.65$.

Assuming random mating (and no change in allele frequency) after any number of generation is given by $p^2$ (=0.1225), $2pq$ (=0.455) and $q^2$ (=0.4225).

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