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Consider a bi-allelic locus with alleles A and a. We denote the frequency of the A allele by $p$. Assuming absence of selection and panmixia, the expected number of heterozygotes is $2p(1-p)N$, where $N$ is the population size. We can also calculate the expected number of AA and aa individuals.

What is the joint probability distribution of the number of AA, Aa and aa individuals?

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  • $\begingroup$ The distribution would be binomial. $\endgroup$
    – WYSIWYG
    Aug 31, 2015 at 5:30
  • $\begingroup$ Thank you. I developed my question a little bit to improve it. I am looking for the joint distribution for each possible genotype. Probably some joint binomial distribution. If you know the answer, please make it an answer. $\endgroup$
    – Remi.b
    Aug 31, 2015 at 20:17

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It is a multinomial (trinomial) distribution with $N$ trials, $p_{AA}=p^2$, $p_{Aa}=2p(1-p)$, and $p_{aa}=(1-p)^2$. The probability to get exactly $n_{AA}$ AA genotypes, $n_{Aa}$ Aa genotypes and $n_{aa}=N-n_{AA}-n_{Aa}$ aa genotypes is

$$P\left(N_{AA} = n_{AA}, N_{Aa} = n_{Aa}\right) = \frac{N!}{n_{AA}!n_{Aa}!n_{aa}!}p^{2n_{AA}}\left(2p(1-p)\right)^{n_{Aa}}(1-p)^{2n_{aa}}.$$

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  • $\begingroup$ Thank you! I edited your answer to write out the actual distribution (let me know if I made a mistake). +1 $\endgroup$
    – Remi.b
    Dec 26, 2015 at 4:05

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