A numerical example

Now we'll return to an example we saw earlier (Table 2). This set of genotypes and phenotypes may look familiar. It is the same one we encountered earlier when we calculated additive and dominance components of variance. Let's assume that $p = 0.4$. Then we know that

$$\begin{eqnarray*} \bar x &=& 1.104 \\ V_a &=& 0.5192 \\ V_d &=& 0.0092 \quad . \end{eqnarray*}$$

We can also calculate the numerical version of Table 1, which you'll find in Table 3.

Table 2: An example of a non-additive relationship between genotypes and phenotypes.

Genotype	$A_1A_1$	$A_1A_2$	$A_2A_2$
Phenotype	0	0.8	2

Table 3: Mother-offspring combinations (half-sib) for the numerical example in Table 2.

Maternal		Offspring genotype
genotype	Frequency	$A_1A_1$	$A_1A_2$	$A_2A_2$
$A_1A_1$	0.16	0.4	0.6	0.0
$A_1A_2$	0.48	0.2	0.5	0.3
$A_2A_2$	0.36	0.0	0.4	0.6

So now we can follow the same approach we did before and calculate the numerical value of the covariance between half-sibs in this example:

$$\begin{eqnarray*} \mbox{Cov}(S_1,S_2) &=& [(0.4)^2(0.16) + (0.2)^2(0.48)](0 - 1... ...\ &=& 0.1298 \\ &=& \left({1 \over 4}\right)0.5192 \quad . \end{eqnarray*}$$