An example: the genetic variance with known genotypes

We've been through a lot of algebra by now. Let's run through a couple of numerical examples to see how it all works. For the first one, we'll use the set of genotypic values in Table 2

Table 2: A set of perfectly additive genotypic values. Note that the genotypic value of the heterozygote is exactly halfway between the genotypic values of the two homozygotes.

Genotype	$A_1A_1$	$A_1A_2$	$A_2A_2$
Genotypic value	0	1	2

For $p = 0.4$

$$\begin{eqnarray*} {\bar x} &=& (0.4)^2(0) + 2(0.4)(0.6)(1) + (0.6)^2(2) \\ &... ... (0.0+1.0)]^2 + (0.6)^2[2 - 2(1.0)]^2 \\ &=& 0.00 \quad . \end{eqnarray*}$$

For $p = 0.2$, ${\bar x} = 1.60$, $V_g = V_a = 0.32$, $V_d = 0.00$. You should verify for yourself that $\alpha_1=0$ and $\alpha_2=1$ for $p = 0.2$. If you are ambitious, you could try to prove that $\alpha_1=0$ and $\alpha_2=1$ for any allele frequency.

For the second example we'll use the set of genotypic values in Table 3.

Table 3: A set of non-additive genotypic values. Note that the genotypic value of the heterozygote is closer to the genotypic value of $A_1A_1$ than it is to the genotypic value of $A_2A_2$.

Genotype	$A_1A_1$	$A_1A_2$	$A_2A_2$
Genotypic value	0	0.8	2

For $p = 0.4$

$$\begin{eqnarray*} {\bar x} &=& (0.4)^2(0) + 2(0.4)(0.6)(0.8) + (0.6)^2(2) \\ ... ...]^2 \\ &&+ (0.6)^2[2 - 2(0.968)]^2 \\ &=& 0.0092 \quad . \end{eqnarray*}$$

To test your understanding, it would probably be useful to calculate ${\bar x}$, $\alpha_1$, $\alpha_2$, $V_g$, $V_a$, and $V_d$ for one or two other allele frequencies, say $p = 0.2$ and $p=0.8$. Is it still true that $\alpha_1$ and $\alpha_2$ are independent of allele frequencies? If you are really ambitious you could try to prove that $\alpha_1$ and $\alpha_2$ are independent of allele frequencies if and only if $x_{12} = (x_{11}+x_{12})/2$, i.e., when heterozygotes are exactly intermediate.