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Sewall Wright defined the $F_{ST}$ in a metapopulation as being:

$$F_{ST} = \frac{\text{Var}(p)}{\bar p (1-\bar p)}$$

, where $p$ is a vector of frequencies of a given allele and $\bar p$ and $\text{Var}(p)$ are the mean and variance of this vector.

For example, consider a metapopulation made of 4 subpopulations. The allele frequencies in these 4 subpopulations are p=[0.2, 0.5, 0.8, 0.3]. $\bar p$ is the mean of $p$ ($\bar p = 0.45$) and $\text{Var}(p)$ is the variance of $p$ ($\text{Var}(p )=0.07 \space$).

Question

Can we define $F_{ST}$ for a multiple allele locus? Or should $F_{ST}$ be defined for each allele independently?

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Does $\bar{p}$ refer to mean of allele frequecies from different populations? –  WYSIWYG May 30 at 11:29
    
@WYSIWYG I updated my post hoping to make the definition of $\bar p$ easier to understand. Let me know if it answered your comment. Thanks –  Remi.b May 30 at 11:38
    
For a bialleleic loci it is obviously not necessary. For multiallelic loci it is mathematically reasonable to assign frequency vector for n-1 alleles ( $\bar{p_{i}}$ ). Similarly $F_{ST_{i}}$ –  WYSIWYG May 30 at 13:29
    
And then, we repeat the same operation of the $n$ possible way of grouping $n-1$ alleles to $n$ different $F_{ST}$ for on locus? –  Remi.b May 30 at 13:39
1  
@daniel $F_{ST}$ is also called fixation index. It is an index used in population genetics in order to describe the population structure (non-random mating throughout the population) of a metapopulation. Here is the wiki page. I think there is only one definition of $F_{ST}$ but several ways to equivalently understand it. There are a bunch of ways to measure and approximate the $F_{ST}$ but this is the matter of another discussion. –  Remi.b May 30 at 15:46

1 Answer 1

Consider this as extended comment because I am not sure about the exact nature of the function.

If I am guessing correctly for multiallelic locus $F_{ST}$ (which is a function of an allele) should be:

$F_{ST_i} = \frac{Var(p_i)}{\bar{p_i}.\sum\limits_{j,\ \ j\neq i}^{n} \bar{p_j}}$

where $i$ is an allele at a given locus.

You can calculate this for any $n-1$ alleles because the $n^{th}$ allele can be known if you know the rest. You can choose to keep any allele out (No need for combinations) .

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This answer surprises me a bit given your comments! I would have expected that to compute the $F_{ST}$ for the allele $A_i$ we just consider $p$ to be the frequency of $A_i$ and therefore $\bar p$ to be $1/D \sum_{d=1}^{D}p_{id}$, where $p_{id}$ is the frequency of the allele $A_i$ in the deme $d$ and where $D$ is the total number of demes (or total number of subpopulations if you prefer). Therefore, I would have expected that the $F_{ST}$ for the allele $A_i$ is dependent on the frequencies of $A_i$ in all demes but independent of the number and frequencies of other alleles. –  Remi.b May 30 at 14:10
    
As I guess, the $(1-\bar{p})$ part refers to absence of the other allele(s). So I thought this one should be correct. Unless we know what $F_{ST}$ really is, it would be difficult to provide an exact answer. Sorry there was a typo ($i$ is allele not locus). This formula would reduce to the original one when you set $n=2$ –  WYSIWYG May 30 at 14:13
    
if $\bar p_i$ is the average frequency of the allele $A_i$ then, it is also equal to on minus the sum of the frequencies of all the other alleles ($1-\sum_{j=1}^{n-1}\bar p_j = 1-\bar p$) and therefore your denominator becomes $\bar p (1-(1-\bar p)) = \bar {p} ^2$. Or what do you mean by $\sum_{i=1}^{n-1}\bar p_i$? –  Remi.b May 30 at 14:22
    
edited.. but if you can please let me know what is the purpose and real definition of $F_{ST}$ –  WYSIWYG May 30 at 14:35
    
I personally think I have not really answered your question. Are you doing it to lower the answer:question ratio ?? –  WYSIWYG Jul 30 at 16:12

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