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.


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


1 Answer 1


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!

  • $\begingroup$ 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…? $\endgroup$
    – Remi.b
    Apr 12, 2014 at 20:06

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