Sib analysis

Mate a number of males (sires) with a number of females (dams). Each sire is mated to more than one dam, but each dam mates only with one sire. Do an analysis of variance on the phenotype in the progeny, treating sire and dam as main effects. The result is shown in Table 5.

Table 5: Analysis of variance table for a full-sib analysis of quantitative genetic variation.

			Composition of
Source	d.f.	Mean square	mean square
Among sires	$s-1$	$MS_S$	$\sigma^2_W + k\sigma^2_D + dk\sigma^2_s$
Among dams	$s(d-1)$	$MS_D$	$\sigma^2_W + k\sigma^2_D$
(within sires)
Within progenies	$sd(k-1)$	$MS_W$	$\sigma^2_W$
$s = \hbox{number of sires}$
$d = \hbox{number of dams per sire}$
$k = \hbox{number of offspring per dam}$

Now we need some way to relate the variance components ($\sigma^2_W$, $\sigma^2_D$, and $\sigma^2_S$) to $V_a$, $V_d$, and $V_e$.⁵ How do we do that? Well,

$$V_p = \sigma^2_T = \sigma^2_S + \sigma^2_D + \sigma^2_W \quad .$$

$\sigma^2_S$ estimates the variance among the means of the half-sib familes fathered by each of the different sires or, equivalently, the covariance among half-sibs.⁶

$$\begin{eqnarray*} \sigma^2_S &=& \mbox{Cov}_{HS} \\ &=& \left(\frac{1}{4}\right)V_a \quad . \end{eqnarray*}$$

Now consider the within progeny component of the variance, $\sigma^2_W$. In general, it can be shown that any among group variance component is equal to the covariance among the members within the groups.⁷ Thus, a within group component of the variance is equal to the total variance minus the covariance within groups. In this case,

$$\begin{eqnarray*} \sigma^2_W &=& V_p - \mbox{Cov}_{FS} \\ &=& V_a + V_d + V_e ... ...c{1}{2}\right)V_a + \left({3 \over 4}\right)V_d + V_e \quad . \end{eqnarray*}$$

There remains only $\sigma^2_D$. Now $\sigma^2_W = V_p - Cov_{FS}$, $\sigma^2_S = Cov_{HS}$, and $\sigma^2_T = V_p$. Thus,

$$\begin{eqnarray*} \sigma^2_D &=& \sigma^2_T - \sigma^2_S - \sigma^2_W \\ &=& V... ...eft(\frac{1}{4}\right)V_a + \left(\frac{1}{4}\right)V_d \quad . \end{eqnarray*}$$

So if we rearrange these equations, we can express the genetic components of the phenotypic variance, the causal components of variance, as simple functions of the observational components of variance:

$$\begin{eqnarray*} V_a &=& 4\sigma^2_S \\ V_d &=& 4(\sigma^2_D - \sigma^2_S) \\ V_e &=& \sigma^2_W - 3\sigma^2_D + \sigma^2_S \quad . \end{eqnarray*}$$