Lecture 17: Multi-locus population genetics

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Lecture overview

Multi-locus population genetics

1. Multi-locus population genetics

In Lecture 15 we learned how to find equilibria and determine their stability in nonlinear multivariate models. In Lecture 16 we looked at a continuous-time example from epidemiology. Here, we'll extend the discrete-time dipliod selection model of Lecture 4 to multiple loci (see Section 8.3 in the text).

Genomes contain many loci -- what new dynamics arise when we model more than one locus? Here we look at the simplest multi-locus model, with two loci each with two alleles. Let's denote the loci with letters, $A$ and $B$ , and the alleles at each with numbers, $1$ and $2$ . This gives a total of $2^{2} = 4$ haploid genotypes, which we'll give frequencies $x_{1}$ to $x_{4}$ like so

genotype	frequency
$A_{1} B_{1}$	$x_{1}$
$A_{1} B_{2}$	$x_{2}$
$A_{2} B_{1}$	$x_{3}$
$A_{2} B_{2}$	$x_{4}$

with the constraint that the total frequency sums to one, $x_{1} + x_{2} + x_{3} + x_{4} = 1$ .

To determine how these frequencies change from one generation to the next, let's first determine the order of events in a life-cycle diagram

In words, we'll census the population in the gamete (haploid) phase while selection happens in the diploid phase.

Next, to consider how the frequencies change through this life cyle, let's construct a life-cycle (mating) table. To do this, we need to consider what happens during meiosis when there are multiple loci.

In the 1-locus models we analyzed earlier, meiosis meant Mendelian segregation: each allele is present in 1/2 of the gametes. Here, with 2 loci, things are slightly more complicated and we need to consider recombination. Every meiosis there is a chance, $r$ , of an odd number of crossover events between the two loci ( $r$ will increase with the distance between the loci). When this happens the pairing of the alleles at loci A and B get swapped. This only has an effect in diploid individuals that are heterozygous at both loci ("double heterozygotes"), here $A_{1} B_{1}$ x $A_{2} B_{2}$ and $A_{1} B_{2}$ x $A_{2} B_{1}$ . Every time these individuals go through meiosis the original pairings are kept with probability $1 - r$ and the alternative pairings are created with probability $r$ .

We can now fill in the following table

union	frequency	frequency after selection	gamete frequency after meiosis ( $A_{1} B_{1}$ , $A_{1} B_{2}$ , $A_{2} B_{1}$ , $A_{2} B_{2}$ )
$A_{1} B_{1}$ x $A_{1} B_{1}$	$x_{1}^{2}$	$x_{1}^{2} w_{11} / \bar{w}$	1, 0, 0, 0
$A_{1} B_{1}$ x $A_{1} B_{2}$	$2 x_{1} x_{2}$	$2 x_{1} x_{2} w_{12} / \bar{w}$	1/2, 1/2, 0, 0
$A_{1} B_{1}$ x $A_{2} B_{1}$	$2 x_{1} x_{3}$	$2 x_{1} x_{3} w_{13} / \bar{w}$	1/2, 0, 1/2, 0
$A_{1} B_{1}$ x $A_{2} B_{2}$	$2 x_{1} x_{4}$	$2 x_{1} x_{4} w_{14} / \bar{w}$	$(1 - r) / 2$ , $r / 2$ , $r / 2$ , $(1 - r) / 2$
$A_{1} B_{2}$ x $A_{1} B_{2}$	$x_{2}^{2}$	$x_{2}^{2} w_{22} / \bar{w}$	0, 1, 0, 0
$A_{1} B_{2}$ x $A_{2} B_{1}$	$2 x_{2} x_{3}$	$2 x_{2} x_{3} w_{23} / \bar{w}$	$r / 2$ , $(1 - r) / 2$ , $(1 - r) / 2$ , $r / 2$
$A_{1} B_{2}$ x $A_{2} B_{2}$	$2 x_{2} x_{4}$	$2 x_{2} x_{4} w_{24} / \bar{w}$	0, 1/2, 0, 1/2
$A_{2} B_{1}$ x $A_{2} B_{1}$	$x_{3}^{2}$	$x_{3}^{2} w_{33} / \bar{w}$	0, 0, 1, 0
$A_{2} B_{1}$ x $A_{2} B_{2}$	$2 x_{3} x_{4}$	$2 x_{3} x_{4} w_{34} / \bar{w}$	0, 0, 1/2, 1/2
$A_{2} B_{2}$ x $A_{2} B_{2}$	$x_{4}^{2}$	$x_{4}^{2} w_{44} / \bar{w}$	0, 0, 0, 1

where $w_{i j} = w_{j i}$ is the fitness of the diploid that is composed of haploid genotypes $i$ and $j$ and $\bar{w}$ is the population mean fitness, which is the sum of the frequencies after selection.

We can build the recursion equations from this table by multiplying the frequency after selection by the gamete frequency after meiosis. For example, the frequency of $A_{1} B_{1}$ in the next generation is found by multiplying the first entry in the final column by the frequency after selection and summing this up over rows, giving

\begin{aligned} x_{1} (t + 1) & = x_{1} (t)^{2} w_{11} / \bar{w} + x_{1} (t) x_{2} (t) w_{12} / \bar{w} + x_{1} (t) x_{3} (t) w_{13} / \bar{w} + (1 - r) x_{1} (t) x_{4} (t) w_{14} / \bar{w} + r x_{2} (t) x_{3} (t) w_{23} / \bar{w} \\ = x_{1} (t) (x_{1} (t) w_{11} / \bar{w} + x_{2} (t) w_{12} / \bar{w} + x_{3} (t) w_{13} / \bar{w} + x_{4} (t) w_{14} / \bar{w}) - r (x_{1} (t) x_{4} (t) w_{14} / \bar{w} - x_{2} (t) x_{3} (t) w_{23} / \bar{w}) \\ = x_{1} (t) \sum_{i = 1}^{4} x_{i} (t) w_{1 i} / \bar{w} - r D^{*} \end{aligned}

where $D^{*} = x_{1} (t) x_{4} (t) w_{14} / \bar{w} - x_{2} (t) x_{3} (t) w_{23} / \bar{w}$ is called linkage disequilibrium (the asterisk differentiates it from the same quantity measured prior to selection, $D = x_{1} (t) x_{4} (t) - x_{2} (t) x_{3} (t)$ ). Linkage disequilibrium is an important term in population genetics that measures the deviation of the association of alleles at two loci from that expected by chance under random assortment. For example, when $A_{1}$ pairs with $B_{1}$ more often than expected by chance then $x_{1} (t) x_{4} (t) > x_{2} (t) x_{3} (t)$ and $D > 0$ .

The remaining equations are created in the same way, giving

\begin{aligned} x_{2} (t + 1) & = x_{2} (t) \sum_{i = 1}^{4} x_{i} (t) w_{2 i} / \bar{w} + r D^{*} \\ x_{3} (t + 1) & = x_{3} (t) \sum_{i = 1}^{4} x_{i} (t) w_{3 i} / \bar{w} + r D^{*} \\ x_{4} (t + 1) & = x_{4} (t) \sum_{i = 1}^{4} x_{i} (t) w_{4 i} / \bar{w} - r D^{*} \end{aligned}

Now we have a system of recursion equations to work with. This system is nonlinear and four dimensional, which makes things relatively complex. For instance, it is impossible to find all the equilibria analytically. Here we'll not worry about that and just deal with a particularly simple equilibrium where $A_{1} B_{1}$ is fixed, ${\hat{x}}_{1} = 1$ and ${\hat{x}}_{2} = {\hat{x}}_{3} = {\hat{x}}_{4} = 0$ . This implies $\bar{w} = w_{11}$ .

To determine the stability of this equilbrium we need to calculate the Jacobian

J = (\begin{matrix} \frac{\partial x_{1} (t + 1)}{\partial x_{1} (t)} & \frac{\partial x_{1} (t + 1)}{\partial x_{2} (t)} & \frac{\partial x_{1} (t + 1)}{\partial x_{3} (t)} & \frac{\partial x_{1} (t + 1)}{\partial x_{4} (t)} \\ \frac{\partial x_{2} (t + 1)}{\partial x_{1} (t)} & \frac{\partial x_{2} (t + 1)}{\partial x_{2} (t)} & \frac{\partial x_{2} (t + 1)}{\partial x_{3} (t)} & \frac{\partial x_{2} (t + 1)}{\partial x_{4} (t)} \\ \frac{\partial x_{3} (t + 1)}{\partial x_{1} (t)} & \frac{\partial x_{3} (t + 1)}{\partial x_{2} (t)} & \frac{\partial x_{3} (t + 1)}{\partial x_{3} (t)} & \frac{\partial x_{3} (t + 1)}{\partial x_{4} (t)} \\ \frac{\partial x_{4} (t + 1)}{\partial x_{1} (t)} & \frac{\partial x_{4} (t + 1)}{\partial x_{2} (t)} & \frac{\partial x_{4} (t + 1)}{\partial x_{3} (t)} & \frac{\partial x_{4} (t + 1)}{\partial x_{4} (t)} \end{matrix})

(we omit writing out the derivatives for brevity) and then evaluate it at the focal equilibrium, $J_{x_{1} = 1, x_{2} = x_{3} = x_{4} = 0}$ . Local stability of the focal equilibrium is determined by the eigenvalues of this matrix, which are

\begin{aligned} λ_{1} & = 0 \\ λ_{2} & = w_{12} / w_{11} \\ λ_{3} & = w_{13} / w_{11} \\ λ_{4} & = (1 - r) w_{14} / w_{11} \end{aligned}

The first eigenvalue, $λ_{1} = 0$ , indicates that there is an axis along which or system does not change. This is due to the fact that the frequencies always sum to one, $x_{1} + x_{2} + x_{3} + x_{4} = 1$ . We therefore effectively have a three dimensional model, e.g., we could just track $x_{1}$ , $x_{2}$ , and $x_{3}$ because we know that $x_{4} = 1 - x_{1} - x_{2} - x_{3}$ .

The second and third eigenvalues, $λ_{2} = w_{12} / w_{11}$ and $λ_{3} = w_{13} / w_{11}$ , are analogous to what we found in the 1 locus (univariate) case. Remembering that stability in discrete time requires that the eigenvalues are less than 1 in absolute value, we can interpret these eigenvalues as saying that the $B_{2}$ allele can invade (instability) when it has higher fitness than the $B_{1}$ allele ( $w_{12} > w_{11}$ ) and the $A_{2}$ can invade when it has higher fitness than the $A_{1}$ allele ( $w_{13} > w_{11}$ ).

The fourth eigenvalue, $λ_{4} = (1 - r) w_{14} / w_{11}$ , is the new part, which depends on recombination. Interestingly, here, even if $A_{2} B_{2}$ has higher fitness than $A_{1} B_{1}$ , meaning $w_{14} > w_{11}$ , it is possible that the $A_{2} B_{2}$ genotype cannot invade. This is because, for a rare $A_{2} B_{2}$ genotype, every generation it pairs with the common $A_{1} B_{1}$ genotype and therefore gets broken apart into $A_{1} B_{2}$ and $A_{2} B_{1}$ by recombination with probability $r$ . In other words, recombination can hinder the spread of an adaptive combination of alleles. This is epitiomized by the scenario where having a single "2" allele is deleterious $w_{12} < w_{11}$ and $w_{13} < w_{11}$ (making $λ_{2} < 1$ and $λ_{3} < 1$ ) but having two "2" alleles is beneficial, $w_{14} > w_{11}$ . Such a scenario is called a fitness valley because of the plot below

import matplotlib.pyplot as plt

w11=1
w12=0.9
w13=0.8
w14=1.1

fig,ax=plt.subplots()

ax.plot([0,1,2],[w11,w12,w14],marker='o')
ax.plot([0,1,2],[w11,w13,w14],marker='o')
ax.text(0,w11,r'$A_1B_1$',va='bottom')
ax.text(1,w12,r'$A_1B_2$',va='top')
ax.text(1,w13,r'$A_2B_1$',va='top')
ax.text(2,w14,r'$A_2B_2$',va='bottom',ha='right')

ax.set_ylabel('fitness')
ax.set_xlabel('number of "2" alleles')
ax.set_xticks([0,1,2])
plt.show()

png

Here we've seen how recombination can slow the spread of the optimal genotype, $A_{2} B_{2}$ , potentially preventing fitness-valley crossing. There is also, however, a constructive aspect of recombination, not explored in this simple model: when the deleterious genotypes $A_{1} B_{2}$ and $A_{2} B_{1}$ are both present, there is a chance that they pair and recombine, giving rise to the optimal genotype $A_{2} B_{2}$ . The role of recombination in fitness-valley crossing is therefore a relatively interesting and complex problem.