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So let's say I have a tetraploid species and I have 4 possible alleles for a particular locus. I have found that I can have 35 total possible genotypes and I know the allele frequencies for each of the 4 alleles, that I will randomly set as $$F_a = 0.2, F_b = 0.3, F_c = 0.4, F_d=0.1$$ for the sake of the example.

However, I am having a hard time wrapping my mind around how to calculate the number of occurrences of a particular genotype, $abc$, in a population of 1000 people.

I understand the allele frequencies, for instance, in the population of 1000 people, given all 1000 observed genotypes, 40% of them had allele $c$. But these are individual values and I have no information on how likely they are to be linked. (I am supposed to assume HW principle holds, but not sure how this helps).

How do I calculate the number of occurrences of the genotype $abc$ in the population?

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Under Hardy-Weinberg conditions, the frequency of the genotype abcd (in this particular order) is simply $F_a \cdot F_b \cdot F_c \cdot F_d = 0.24\%$.

If we don't consider the order, then we must ultiply the previous probability by 24. There are indeed 24 (=4*3*2*1) ways you can get this particular combination. The probability of abcd in any particular order is therefore $24 \cdot F_a \cdot F_b \cdot F_c \cdot F_d = 5.76\%$.

You might want to have a look at the post Solving Hardy Weinberg problems

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  • $\begingroup$ But that's only for that particular permutation of that genotype, correct? Hence, we'd have to take into account how many different ways that genotype can occur $\endgroup$ – Jonathan Jan 20 at 2:26
  • $\begingroup$ Or is that, since the allele frequency already account for those things, we can ignore the other permutations? $\endgroup$ – Jonathan Jan 20 at 2:31
  • $\begingroup$ @Jonathan Good point, I should have made that clear. If you are interest in the fraction of abcd genotype (in any particular order), then you have to multiply by the number of ways you can get this particular genotype. $\endgroup$ – Remi.b Jan 20 at 10:57
  • $\begingroup$ But that doesn't fully answer the question either, because then we would double counting if you just did $4!$. Hence, it would have to be the number of ways to get that particular genotype plus some constraint. Maybe for abcd it's straight forward. But if it was $a^2b^2$, then it wouldn't be. $\endgroup$ – Jonathan Jan 21 at 4:55
  • $\begingroup$ Could you please update your answer to include this? It would be very useful key point for everyone else with the same question. $\endgroup$ – Jonathan Jan 21 at 4:56

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