# Inbreeding depression and dominance

A classic theoretical result is that the mean of a character controlled by a single locus i with two alleles Ai1 and Ai2 is only affected by the value of f if there is some degree of dominance.

where f is the Wright's inbreeding coefficient.

Assume that the trait is scaled such that the values for the genotypes Ai1Ai1, Ai1Ai2, and Ai2Ai2 are -ai, di, and ai respectively. The quantities ai and di 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 (di > ai) or if the allele associated with an increased value of the trait is dominant or partially dominant (aidi > 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 di 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 (di > ai) or if the allele associated with an increased value of the trait is dominant or partially dominant (aidi > 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.

• man i should be able to answer this. i didnt take that class though. I could write a doctor who knows the answer but i doubt he would respond. – user1357 May 14 '14 at 6:09

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?

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

• Yes, that answers my question. It is actually a very good "for dummies" explanation of a fairly complicated text. You answered so many of my questions this week. Thanks a lot! – Remi.b Dec 12 '14 at 16:45
• haha - there's been a bit of activity to try and get the unanswered Q's reduced, there was a comment in the chat room that probably the only person to answer some of your unanswered Q's was me so I set myself the challenge of answering 5 this week... two to go! @remi.b – rg255 Dec 12 '14 at 17:16