Additive genetic variance with $n$ alleles

Question

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.

• How are $\sigma_D^2$, $\sigma_A^2$ and $\sigma^2$ defined for a locus that have $n$ alleles?

Here is a related question

Answer 1

Well, the total genetic variance is just, by the definition of the variance, $$ \sigma^2 =\sum_{i,j} f_i f_j (w_{ij}-\bar{w})^2 $$ (using $f_i$ and $w_{ij}$ for frequency and fitness, respectively), and $$\bar{w} = \sum_{i,j} f_i f_j w_{ij}$$ is just the average fitness.

You can calculate the additive genetic variance for different loci by simply assuming that there is no dominance effect, i.e. the alleles are independent. If it helps, think of it as a quantitative trait in a haploid organism. Thus,

$$ \sigma^2_A =\sum_{i,j} f_i f_j (w_iw_j-\bar{w}')^2 = \sum_i f_i(w_i-\bar{w}')^2. $$ with $$ f_i=\sum_jf_{ij}; w_i=\sum_j f_{ij}w_{ij}; \bar{w}' = \sum_i f_i w_i; $$