The dominance variance on a single locus

Question

I was reading the book "Genetics and Analysis of Quantitative Traits", by Lynch and Walsh. I how the covariance between two individuals with IBD $\Theta$ gets divided into just the additive variance and dominance variance component, even in the simple $1$ locus case.

Here my understanding of the modelling (for the simple one allele case):

Given a genotypic value $G_{i,j}$ of mean $0$, $i,j \in \{0,1\}$ we find numbers $\alpha_0$ and $\alpha_1$ minimising the least squares of the following form $\mathbb{E}(G_{i,j}-\alpha_i-\alpha_j)^2$, where the expectation is over the population.

We next define the error terms in each case as $\delta_{i,j}=G_{i,j}-\alpha_i-\alpha_j$. From the properties, viewed as functions of the population $\alpha_i$ is independent of $\delta_{i,j}$, and both have mean $0$.

The claim made in the book is that given two individuals, with IBD $\Theta$ and probability that the genotype is equal $\Delta$, the covariance of the genotypes $G_{i,j}$ and $G_{k,l}$ is given by,

$$\text{cov}(G_{i,j},G_{k,l}) = 2\Theta \sigma_A^2 +\Delta \sigma_D^2,$$

where $\sigma_A^2= \text{Var}(\alpha_i)$, and $\sigma_D^2 = \text{Var}(\delta_{i,j})$.

Expanding the LHS of the expression, showing that $\mathbb{E}[(\alpha_i +\alpha_j)(\alpha_k+\alpha_l)] =\Theta \sigma_A^2$ is quite easy. It also seems to follow from the that the the terms $\mathbb{E}[(\alpha_i +\alpha_j)\delta_{k,l}]=0$ from independence of errors from the $\alpha$.

On analysing $\mathbb{E}[\delta_{i,j}\delta_{k,l}]$, we see that if both genotypes are equal, which occurs with probability $\Delta$, then this reduces to $\sigma_D^2$. This gives us a term $\Theta \sigma_D^2$. Further, if both $i,j$ and $k,l$ are not IBD then the covariance is $0$. However when one of the two alleles are IBD, then it is not clear to me that this the covariance will still be $0$.

The book seems to claim that unless both alleles are IBD, $\delta_{i,j}$ and $\delta_{k,l}$ are independent. I do not see why this is the case. Am I missing anything here? I'd appreciate any help wrt this.

Answer 1

It looks like I could solve the question. My understanding is as follows.

Let $(X_1,Y_1)$, $(X_2,Y_2)$ (corresponds to $i,j$ and $k,l$ in the question) be the genotypes of two individuals at some location. We assume that neither individual is in-bred. In this case define, \begin{align*} \Delta_7&= \text{Pr}(\text{Both alleles $\{X_1,Y_1\}$ and $\{X_2, Y_2\}$ are IBD}),\\ \Delta_8&= \text{Pr}(\text{Exactly one allele pair of the four pairs is IBD}),\\ \Delta_9&= \text{Pr}(\text{ No IBD between the two individuals}). \end{align*} $\Delta_7$, $\Delta_8$ and $\Delta_9$ are calculated from the pedigrees. For example, for non-twin siblings, $\Delta_7 = \frac{1}{4}$, $\Delta_8=\frac{1}{2}$, $\Delta_9 =\frac{1}{4}$.

We note that the IBD coefficient (also called kinship coefficient) $\Theta$ is \begin{align*} \Theta = \frac{1}{2} \Delta_7 + \frac{1}{4} \Delta_8. \end{align*} Also the fraternity coefficient, $\Delta =\Delta_7$. Thus we have that the covariance between the genotypic value of two individuals is given by \begin{align*} \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)] =& \Delta_7 \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)| \text{both IBD}] + \\ &\Delta_8 \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)| \text{one IBD}] + \Delta_9 \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)| \text{no IBD}]. \end{align*}

Further note that, \begin{align*} \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)| \text{both IBD}] &= \mathbb{E}[G^2(X_1,Y_1)],\\ &= \sigma_A^2 +\sigma_D^2. \end{align*}

\begin{align*} \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)| \text{no IBD}] &= \mathbb{E}[G(X_1,Y_1)] \mathbb{E}[G(X_2,Y_2)],\\ &=0. \end{align*}

\begin{align*} \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)| \text{one IBD}] &= \mathbb{E}[(\alpha(X)+\alpha(Y_1)+\delta(X,Y_1))(\alpha(X)+\alpha(Y_2)+\delta(X,Y_2))],\\ &=\mathbb{E}[\alpha^2(X)] + 2 \mathbb{E}(\alpha(X) \delta(X,Y_1)) + \mathbb{E} [\delta(X,Y_1)\delta(X,Y_2)],\\ &= \frac{1}{2} \sigma_A^2 + 2 \mathbb{E} [ \alpha(X) \mathbb{E}[\delta(X,Y_1) | X] ] + \mathbb{E} [\mathbb{E}[\delta(X,Y_1)|X]\mathbb{E}[\delta(X,Y_2)|X]],\\ &= \frac{1}{2} \sigma_A^2. \end{align*}

Thus we have that, \begin{align*} \mathbb{E}[G(X_1,Y_1)G(X_2,Y_2)] =& \left(\frac{1}{2}\Delta_7 + \frac{1}{4}\Delta_8\right) 2 \sigma_A^2 + \Delta_7 \sigma_D^2,\\ &= 2\Theta \sigma_A^2 + \Delta \sigma_D^2, \end{align*} which is as claimed.