# Possible Genotypes of 4 Alleles of Adh

There are four alleles of Adh which are Adh-1, Adh-2, Adh-3 and Adh-4. Their respective frequencies are 0.11, 0.84, 0.01 and 0.04. What are the Hardy-Weinberg frequencies of the possible 10 genotypes?

My Attempt: First I thought that there were more than 10 genotypes (4*3*2*1) = 24. I've never done HWE with 4 genotypes so im not sure how to use $p+q=1$ or $p^2+2pq+q^2=1$.

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I may be wrong but to get the number of possible genotypes isnt it 4+3+2+1=10?. You can have A1A1,A1A2,A1A3,A1A4,A2A2,A2A3,A2A4 and so on. When you multiply you are counting some genotypes twice (for example A1A2 and A2A1. They are the same) – von Mises Sep 25 '13 at 15:42

## 1 Answer

Those Hardy-Weinberg equations are the general case, used for only two alleles. This question is basically answered here, for three alleles; you've got a situation of four alleles. That means you need to have:

$(p+q+r+s)^2=1$

Where $p$, $q$, $r$, and $s$ are the frequencies of your respective alleles. This expands out to the rather unwieldy:

$p^2+2pq+2pr+2ps+q^2+2qr+2qs+r^2+2rs+s^2=1$

Now it becomes a plug 'n chug assignment; simply assign the frequencies and calculate.

Assuming $p$ is Adh-1, $q$ is Adh-2, etc., $p^2=0.0121$, $2pq=0.1848$, and so on.

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