An example: body weight in female mice

The analysis involves 719 offspring from 74 sires and 192 dams, each with one litter. The offspring were spread over 4 generations, and the analysis is performed as a nested ANOVA with the genetic analysis nested within generations. An additional complication is that the design was unbalanced, i.e., unequal numbers of progeny were measured in each sibship. As a result the degrees of freedom don't work out to be quite as simple as what I showed you.⁸ The results are summarized in Table 6.

Table 6: Quantitative genetic analysis of the inheritance of body weight in female mice (from Falconer and Mackay, pp. 169-170.)

			Composition of
Source	d.f.	Mean square	mean square
Among sires	70	17.10	$\sigma^2_W + k'\sigma^2_D + dk'\sigma^2_s$
Among dams	118	10.79	$\sigma^2_W + k\sigma^2_D$
(within sires)
Within progenies	527	2.19	$\sigma^2_W$
$d = 2.33$
$k = 3.48$
$k' = 4.16$

Using the expressions for the composition of the mean square we obtain

$$\begin{eqnarray*} \sigma^2_W &=& MS_W \\ &=& 2.19 \\ \sigma^2_D &=& \left({1 ... ...right)(MS_S - \sigma^2_W - k'\sigma^2_D) \\ &=& 0.48 \quad . \end{eqnarray*}$$

Thus,

$$\begin{eqnarray*} V_p &=& 5.14 \\ V_a &=& 1.92 \\ V_d + V_e &=& 3.22 \\ V_d &=& (0.00\hbox{-}1.64) \\ V_e &=& (1.58\hbox{-}3.22) \\ \end{eqnarray*}$$

Why didn't I give a definite number for $V_d$ after my big spiel above about how we can estimate it from a full-sib crossing design? Two reasons. First, if you plug the estimates for $\sigma^2_D$ and $\sigma^2_S$ into the formula above for $V_d$ you get $V_d = 7.96, V_e = -4.74$, which is clearly impossible since $V_d$ has to be less than $V_p$ and $V_e$ has to be greater than zero. It's a variance. Second, the experimental design confounds two sources of resemblance among full siblings: (1) genetic covariance and (2) environmental covariance. The full-sib families were all raised by the same mother in the same pen. Hence, we don't know to what extent their resemblance is due to a common natal environment.⁹If we assume $V_d = 0$, we can estimate the amount of variance accounted for by exposure to a common natal environment, $V_{Ec} = 1.99$, and by environmental variation within sibships, $V_{Ew} = 1.23$.¹⁰ Similarly, if we assume $V_{Ew} = 0$, then $V_d = 1.64$ and $V_{Ec} = 1.58$. In any case, we can estimate the narrow sense heritability as

$$\begin{eqnarray*} h^2_N &=& \left({1.92 \over 5.14}\right) \\ &=& 0.37 \quad . \end{eqnarray*}$$

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