The algebraic development

You should know by now that I'm not going to be satisfied with a numerical example. I now feel the need to do some algebra to describe this situation a little more generally.

Suppose we know allele frequencies in $k$ subpopulations. Let $p_i$ be the frequency of $A_1$ in the $i$th subpopulation. Then if we assume that all subpopulations contribute equally to combined population,² we can calculate expected and observed genotype frequencies the way we did above:

	$A_1A_1$	$A_1A_2$	$A_2A_2$
Expected	$\bar p^2$	$2\bar p\bar q$	$\bar q^2$
Observed	$\frac{1}{k}\sum p_i^2$	$\frac{1}{k}\sum 2p_iq_i$	$\frac{1}{k}\sum q_i^2$

where $\bar p = \sum p_i/k$ and $\bar q = 1 - \bar p$. Now

$$\frac{1}{k}\sum p_i^2 = \frac{1}{k}\sum (p_i - \bar p + \bar p)^2$$ (1)

$$= \frac{1}{k}\sum \left((p_i - \bar p)^2 + 2\bar p(p_i - \bar p) + \bar p^2\right)$$ (2)

$$= \frac{1}{k}\sum (p_i - \bar p)^2 + \bar p^2$$ (3)

$$= \hbox{Var}(p) + \bar p^2$$ (4)

Similarly,

$$\frac{1}{k}\sum 2p_iq_i = 2\bar p\bar q - 2\hbox{Var}(p)$$ (5)

$$\frac{1}{k}\sum q_i^2 = \bar q^2 + \hbox{Var}(p)$$ (6)

Since $\hbox{Var}(p) \ge 0$ by definition, with equality holding only when all subpopulations have the same allele frequency, we can conclude that

• Homozygotes will be more frequent and heterozygotes will be less frequent than expected based on the allele frequency in the combined population.

• The magnitude of the departure from expectations is directly related to the magnitude of the variance in allele frequencies across populations, $\hbox{Var}(p)$.

• The effect will apply to any mixing of samples in which the subpopulations combined have different allele frequencies.³

• The same general phenomenon will occur if there are multiple alleles at a locus, although it is possible for one or a few heterozygotes to be more frequent than expected if there is positive covariance in the constituent allele frequencies across populations.⁴

• The effect is analogous to inbreeding. Homozygotes are more frequent and heterozygotes are less frequent than expected.⁵

To return to our earlier numerical example:

$$\hbox{Var}(p) = \left((0.4 - 0.55)^2 + (0.7 - 0.55)^2\right)$$ (7)

$$= 0.0225$$ (8)

	Expected				Observed
$A_1A_1$	0.3025	+	0.0225	=	0.3250
$A_1A_2$	0.4950	-	2(0.0225)	=	0.4500
$A_2A_2$	0.2025	+	0.0225	=	0.2250