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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?

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this might be a useful resource for you @Remi.b - I'll read the Falconer and Mackay book tomorrow at work too nitro.biosci.arizona.edu/zbook/NewVolume_2/newvol2.html –  GriffinEvo Mar 31 at 19:07
    
really like this question, spent some time at the whiteboard today trying to work it out! made good progress! –  GriffinEvo Apr 1 at 20:06
    
Chapter 4 of lynch and Walsh seems to deal with this - from around table 4.2 –  GriffinEvo Apr 3 at 18:13
    
@GriffinEvo At which page is it? I could find table 4.1 at page 11 and table 4.3 at page 22! table 4.2 is probably in between :D –  Remi.b Apr 3 at 18:26
    
In my copy it is page ~77, I think you should see equation 4.23a and some of the preceeding section on "average excess" - this is Lynch and Walsh Genetics and analysis of quantitative traits. I'm away for the next couple of days but will be trying to write this in an R code with data to test it out next week - perhaps we should move to the chat room to discuss more! –  GriffinEvo Apr 4 at 9:39

1 Answer 1

up vote 1 down vote accepted

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 $$ f_i=\sum_jf_{ij}; w_i=\sum_j f_{ij}w_{ij}; \bar{w}' = \sum_i f_i w_i; $$

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Thanks a lot for your answer. Could you please help me to develop the formula for $\sigma_A^2$ for a diploid population? –  Remi.b Apr 2 at 15:07
    
this should be true for a diploid population. The additive genetic variance is defined as the variance in a population where there are no interaction effects between genotypes, so $w_{ij}$ can be written as $w_iw_j$ –  bpeter Apr 2 at 18:33
    
I'm sorry I might be a bit slow to understand :)… Can you please show the development to find back $\sigma_A^2 = 2x(1-x)(xW_{11}+(1-2x)W_{12} - (1-x)W_{22})^2$ from your equations? Thank you. –  Remi.b Apr 3 at 8:17

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