Inbreeding depression and dominance

Question

From this article, second paragraph of the second page

A classic theoretical result is that the mean of a character controlled by a single locus i with two alleles A_i1 and A_i2 is only affected by the value of f if there is some degree of dominance.

where f is the Wright's inbreeding coefficient.

The authors then add:

Assume that the trait is scaled such that the values for the genotypes A_i1A_i1, A_i1A_i2, and A_i2A_i2 are -a_i, d_i, and a_i respectively. The quantities a_i and d_i measure the effect of the locus on the character and the degree of dominance of the locus, respectively. With two alleles, there is a linear decline in the mean of a trait with increasing f either with overdominance (d_i > a_i) or if the allele associated with an increased value of the trait is dominant or partially dominant (a_i ≥ d_i > 0). With additive joint effects of different loci on the trait, this conclusion can be extended to a polygenic character; inbreeding decline occurs if the average value of d_i over all i is positive.

I don't understand why "[..] there is a linear decline in the mean of a trait with increasing f either with overdominance (d_i > a_i) or if the allele associated with an increased value of the trait is dominant or partially dominant (a_i ≥ d_i > 0)"

Can you please explain why the mean of a character controlled by a single locus is only affected by the value of "f" if there is some degree of dominance.

Answer 1

Recap of the question:

Looking at a single locus trait ($A$) controlled by two alleles, $A_1$ and $A_2$, the phenotypic mean is only affected by inbreeding depression, $f$ (Wright's inbreeding coefficient), if there is some degree of dominance, $d$. Why?

Answer:

If we take inbreeding as a higher than expected frequency of homozygotes, such that if the frequencies of alleles $A_1$ and $A_2$ are $p = q = 0.5$ the expected frequencies of the genotypes in an outbred population, $P_{outbred}$, would be $p^2 = 0.25$, $2pq = 0.50$, and $q^2 = 0.25$.

Each genotype has the following phenotypic result on a trait $A_1A_1 = 5$, $A_1A_2 = 7.5$, $A_2A_2 = 10$... this is an additive trait with allele $A_1$ having an additive effect of 2.5, and $A_2$ an additive effect of 5.

We can then compare an inbred population of 100 individuals, $P_{inbred}$ where the frequency of $A_1A_1 = 0.5$, $A_1A_2 = 0$, $A2_A2 = 0.5$, to $P_{outbred}$ , where I use $\bar{X}$ to symbolise trait means.

$\bar{X}_{inbred} = \frac{(50 \cdot 5) + (0 \cdot 7.5) + (50 \cdot 10)}{100} = 7.5$

$\bar{X}_{outbred} = \frac{(25 \cdot 5) + (50 \cdot 7.5) + (25 \cdot 10)}{100} = 7.5$

So when the trait is additive the population means are equal. It is onlt when we add dominance to the situation that the population means will differ as a result of inbreeding. Continuing from the above example, the populations are the same size and structure, but $A_2$ is now completely dominant to $A_1$ such that $A_1A_2$ individuals now have a trait value of 10. Now

$\bar{X}_{inbred} = \frac{(50 \cdot 5) + (0 \cdot 10) + (50 \cdot 10)}{100} = 7.5$

$\bar{X}_{outbred} = \frac{(25 \cdot 5) + (50 \cdot 10) + (25 \cdot 10)}{100} = 8.75$

Thus inbreeding has affected the population mean in a trait which has dominance, it is basically because the heterozygotes are less common in an inbreeding population and this is where dominance effects are seen.

"there is a linear decline in the mean of a trait with increasing f... with overdominance (di > ai)"

This is because in the case of overdominance, the outbred population mean becomes even more elevated - if we say now the $A_1A2$ trait is a phenotype of 15 the means become

$\bar{X}_{inbred} = \frac{(50 \cdot 5) + (0 \cdot 15) + (50 \cdot 10)}{100} = 7.5$

$\bar{X}_{outbred} = \frac{(25 \cdot 5) + (50 \cdot 15) + (25 \cdot 10)}{100} = 11.25$

" there is a linear decline in the mean of a trait with increasing f... if the allele associated with an increased value of the trait is dominant or partially dominant (ai ≥ di > 0)"

The mean will increase if the dominant allele produces a smaller phenotype. Here we say $A_1$ is now completely dominant,

$\bar{X}_{inbred} = \frac{(50 \cdot 5) + (0 \cdot 5) + (50 \cdot 10)}{100} = 7.5$

$\bar{X}_{outbred} = \frac{(25 \cdot 5) + (50 \cdot 5) + (25 \cdot 10)}{100} = 6.25$

I hope that answers the question, I tried to answer last night but was too tired to think - this is just what I came up with on the walk to work today so I haven't had the time to check it/overthink it