Additive genetic variance with $n$ loci

The genetic variance of a quantitative trait (the quantitative trait in question is fitness) can be express as the sum of two components, the dominance and additive variance:

$$\sigma_D^2 + \sigma_A^2 = \sigma^2$$

, where $\sigma$ is the genetic variance, $\sigma_D^2$ is the dominance variance and $\sigma_A^2$ is the additive variance. $\sigma_D^2$ and $\sigma_A^2$ are given by

$$\sigma_D^2 = x^2(1-x)^2(2 \cdot W_{12} - W_{11} - W_{22})^2$$

$$\sigma_A^2 = 2x(1-x)(xW_{11}+(1-2x)W_{12} - (1-x)W_{22})^2$$

, where $W_{11}$, $W_{12}$ and $W_{22}$ are the fitness of the three possible genotypes and $x$ and $1-x$ give the allele frequencies.

Question

The above definition makes sense for one bi-allelic locus only.

• How are $\sigma_D^2$, $\sigma_A^2$ and $\sigma^2$ defined for $n$ bi-allelic loci? Is it:

$$\sigma^2 = \sum_{i=1}^n \sigma_i^2$$ $$\sigma_A^2= \sum_{i=1}^n \sigma_{Ai}^2$$ $$\sigma_D^2 = \sum_{i=1}^n \sigma_{Di}^2$$

Here is a related question

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Not in general -- there can be linkage disequilibrium among the loci. For instance, say that there are two di-allelic loci, $A/a$ and $B/b$, and that the frequencies of the $A$ and $B$ alleles are both $1/2$ and that they have the same effect on the trait, with no dominance. If all haplotypes in the population are either $Ab$ or $aB$ (with no $AB$ or $ab$ haplotypes), then the genetic variance at each locus is high, but the total genetic variance for the trait is 0!
Thanks for your answer. Yes that totally makes sense. So the formula would be $$\sum_{i=1}^{n} \left( \sigma_i \frac{\sum_{j=1}^{n} 1-\text{cor}(i,j)}{n} \right)$$ , where $\text{cor}(i,j)$ is the correlation between the two loci... or something like that…? – Remi.b Apr 12 '14 at 20:06