1 Equal Gene Frequencies

Twin correlations may be derived in a number of different ways, but the most direct method is to list all possible twin-pair genotypes (taken as deviations from the population mean) and the frequency with which they arise in a random-mating population. Then, the expected covariance may be obtained by multiplying the genotypic effects for each pair, weighting them by the frequency of occurrence, and summing across all possible pairs. By this method the covariance among pairs is calculated directly. The overall mean for such pairs is, of course, simply the population mean, $\frac{1}{2}h$, in the case of equal gene frequencies, as shown in the previous section. There are shorter methods for obtaining the same result, but these are less direct and less intuitively obvious. The covariance calculations are laid out in Table 3.2 for MZ, DZ, and Unrelated pairs of siblings, the latter being included in order to demonstrate the expected zero covariance for genetically unrelated individuals. The nine possible combinations of genotypes are shown in column 1, with their genotypic effects, $x_{1i}$ and $x_{2i}$, in columns 2 and 3. From these values the mean of all pairs, $\frac{1}{2}h$, is subtracted in columns 4 and 5. Column 6 shows the products of these mean deviations. The final three columns show the frequency with which each of the genotype pairs occurs for the three kinds of relationship. For MZ twins, the genotypes must be identical, so there are only three possibilities and these occur with the population frequency of each of the possible genotypes. For unrelated pairs, the population frequencies of the three genotypes are simply multiplied within each pair of siblings since genotypes are paired at random. The frequencies for DZ twins, which are the same as for ordinary siblings, are more difficult to obtain. All possible parental types and the proportion of paired genotypes they can produce must be enumerated, and these categories collected up across all possible parental types. These frequencies and the method by which they are obtained may be found in standard texts (e.g., Crow and Kimura, 1970, pp. 136-137; Falconer, 1960, pp. 152-157; Mather and Jinks, 1971, pp. 214-215).

Table 3.2: Genetic covariance components for MZ, DZ, and Unrelated siblings with equal gene frequencies at a single locus ( $u=v=\frac{1}{2}$).

Genotype	Effect					Frequency
Pair	$x_{1i}$	$x_{2i}$	$x_{1i} - \mu_{1}$	$x_{2i} - \mu_{2}$	$(x_{1i} - \mu_{1})(x_{2i} - \mu_{2})$	MZ	DZ	U
$AA,AA$	$d$	$d$	$d-\frac{1}{2}h$	$d-\frac{1}{2}h$	$d^2 - dh + \frac{1}{4}h^2$	$\frac{1}{4}$	$\frac{9}{64}$	$\frac{1}{16}$
$AA,Aa$	$d$	$h$	$d-\frac{1}{2}h$	$\frac{1}{2}h$	$\frac{1}{2}dh - \frac{1}{4}h^2$	-	$\frac{3}{32}$	$\frac{1}{8}$
$AA,aa$	$d$	$-d$	$d-\frac{1}{2}h$	$-d-\frac{1}{2}h$	$-d^2 + \frac{1}{4}h^2$	-	$\frac{1}{64}$	$\frac{1}{16}$
$Aa,AA$	$h$	$d$	$\frac{1}{2}h$	$d-\frac{1}{2}h$	$\frac{1}{2}dh - \frac{1}{4}h^2$	-	$\frac{3}{32}$	$\frac{1}{8}$
$Aa,Aa$	$h$	$h$	$\frac{1}{2}h$	$\frac{1}{2}h$	$\frac{1}{4}h^2$	$\frac{1}{2}$	$\frac{5}{16}$	$\frac{1}{4}$
$Aa,aa$	$h$	$-d$	$\frac{1}{2}h$	$-d-\frac{1}{2}h$	$-\frac{1}{2}dh - \frac{1}{4}h^2$	-	$\frac{3}{32}$	$\frac{1}{8}$
$aa,AA$	$-d$	$d$	$-d-\frac{1}{2}h$	$d-\frac{1}{2}h$	$-d^2 + \frac{1}{4}h^2$	-	$\frac{1}{64}$	$\frac{1}{16}$
$aa,Aa$	$-d$	$h$	$-d-\frac{1}{2}h$	$\frac{1}{2}h$	$-\frac{1}{2}dh - \frac{1}{4}h^2$	-	$\frac{3}{32}$	$\frac{1}{8}$
$aa,aa$	$-d$	$-d$	$-d-\frac{1}{2}h$	$-d-\frac{1}{2}h$	$d^2 + dh + \frac{1}{4}h^2$	$\frac{1}{4}$	$\frac{9}{64}$	$\frac{1}{16}$
$\mu_{x_1} = \mu_{x_2} = \frac{1}{2}h$ in all cases; genetic covariance = $\sum_i f_i(x_{1i}-\mu_1)(x_{2i}-\mu_2)$

The products in column 6, weighted by the frequencies for the three sibling types, yield the degree of genetic resemblance between siblings. In the case of MZ twins, the covariance equals

$$\mbox{Cov(MZ)} = d^2(\frac{1}{4}+\frac{1}{4})+ dh(-\frac{1}{4} + \frac{1}{4}) + \frac{1}{4}h^2(\frac{1}{4} + \frac{2}{4} + \frac{1}{4})$$

$$= \frac{1}{2}d^2 + \frac{1}{4}h^2 \; ,$$ (11)

which is simply expression 3.2, the total genetic variance in the population. If we sum over loci, as we did in expression 3.4, we obtain $V_A + V_D$, the additive and dominance variance, as we would intuitively expect since identical twins share all genetic variance. The calculation for DZ twins, with terms in $d^2$, $dh$, and $h^2$ initially separated for convenience, and collected together at the end, is

$$\mbox{Cov(DZ)} = d^2(\frac{9}{64} - \frac{1}{64} - \frac{1}{64} + \frac{9}{64})$$

$$+ dh (-\frac{9}{64} + \frac{3}{64} + \frac{3}{64} - \frac{3}{64} - \frac{3}{64} + \frac{9}{64})$$

$$+ \frac{1}{4}h^2(\frac{9}{64} - \frac{6}{64} + \frac{1}{64} - \frac... ...4} + \frac{20}{64} - \frac{6}{64} + \frac{1}{64} - \frac{6}{64} + \frac{9}{64})$$

$$= \frac{1}{4}d^2 + \frac{1}{16}h^2$$ (12)

When summed over all loci, this expression gives $\frac{1}{2}V_A + \frac{1}{4}V_D$. The calculation for unrelated pairs of individuals yields a zero value as expected, since, on average, unrelated siblings have no genetic variation in common at all:

$$\mbox{Cov(U)} = d^2(\frac{1}{16} - \frac{1}{16} - \frac{1}{16} + \frac{1}{16})$$

$$= dh(-\frac{1}{16} + \frac{1}{16} + \frac{1}{16} - \frac{1}{16} - \frac{1}{16} + \frac{1}{16})$$

$$= \frac{1}{4}h^2(\frac{1}{16} - \frac{2}{16} + \frac{1}{16} - \frac... ...16} + \frac{4}{16} - \frac{2}{16} + \frac{1}{16} - \frac{2}{16} + \frac{1}{16})$$

$$= 0$$ (13)

It is the fixed coefficients in front of $V_A$ and $V_D$, 1.0 and 1.0 in the case of MZ twins and $\frac{1}{2}$ and $\frac{1}{4}$, respectively, for DZ twins that allow us to specify the Mx model and estimate $V_A$ and $V_D$, as will be explained in subsequent chapters. These coefficients are the correlations between additive and dominance deviations for the specified twin types. This may be seen easily in the case where we assume that dominance is absent. Then, MZ and DZ genetic covariances are simply $V_A$ and $\frac{1}{2}V_A$, respectively. The variance of twin 1 and twin 2 in each case, however, is the population variance, $V_A$. For example, the DZ genetic correlation is derived as

$$\begin{eqnarray*} r_{\mbox{DZ}} & = & \frac{\mbox{Cov(DZ)}}{\sqrt{V_{T1} V_{T2}}} = \frac{\frac{1}{2}V_A}{\sqrt{V_A V_A}} = \frac{1}{2} \end{eqnarray*}$$