Drummond:PopGen

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Introduction

Here I will treat some basic questions in population genetics. For personal reasons, I tend to include all the algebra.

Per-generation and instantaneous growth rates

What is the relationship between per-generation growth rates and the Malthusian parameter, the instantaneous rate of growth?

Let $n i (t)$ be the number of organisms of type $i$ at time $t$ , and let $R$ be the per-capita reproductive rate per generation. If $t$ counts generations, then

$n_i(t+1) = n_i(t)R\!$

and

$n_i(t) = n_i(0)R^t.\!$

Now we wish to move to the case where $t$ is continuous and real-valued. As before,

$n_i(t+1) = n_i(t)R\!$

but now

$n_i(t+\Delta t)\!$	$=n_i(t)R^{\Delta t}\!$
$n_i(t+\Delta t) - n_i(t)\!$	$= n_i(t)R^{\Delta t} - n_i(t)\!$
$\frac{n_i(t+\Delta t) - n_i(t)}{\Delta t}$	$=\frac{n_i(t)R^{\Delta t} - n_i(t)}{\Delta t}$
$\frac{n_i(t+\Delta t) - n_i(t)}{\Delta t}$	$=n_i(t) \frac{R^{\Delta t} - 1}{\Delta t}$
$\lim_{\Delta t \to 0} \left[{n_i(t+\Delta t) - n_i(t) \over \Delta t}\right]$	$=\lim_{\Delta t \to 0} \left[ n_i(t) \frac{R^{\Delta t} - 1}{\Delta t}\right]$
$\frac{d n_i(t)}{dt}$	$=n_i(t) \lim_{\Delta t \to 0} \left[\frac{R^{\Delta t} - 1}{\Delta t}\right]$
$\frac{d n_i(t)}{dt}$	$=n_i(t) \ln R\!$

where the last simplification follows from L'Hôpital's rule. Explicitly, let $ε = Δ t$ . Then

$\lim_{\Delta t \to 0} \left[{R^{\Delta t} - 1 \over \Delta t}\right]$	$= \lim_{\epsilon \to 0} \left[\frac{R^{\epsilon} - 1}{\epsilon}\right]$
	$=\lim_{\epsilon \to 0} \left[\frac{\frac{d}{d\epsilon}\left(R^{\epsilon} - 1\right)}{\frac{d}{d\epsilon}\epsilon}\right]$
	$=\lim_{\epsilon \to 0} \left[\frac{R^{\epsilon}\ln R}{1}\right]$
	$=\ln R \lim_{\epsilon \to 0} \left[R^{\epsilon}\right]$
	$=\ln R\!$

The solution to the equation

$\frac{d n_i(t)}{dt} = n_i(t) \ln R$

is

$n_i(t) = n_i(0) e^{t\ln R} = n_i(0) R^{t}.\!$

Note that the continuous case and the original discrete-generation case agree for all integer values of

t

. We can define the instantaneous growth rate

r = ln R

for convenience.

Continuous rate of change

If two organisms grow at different rates, how do their proportions in the population change over time?

Let $r 1$ and $r 2$ be the instantaneous rates of increase of type 1 and type 2, respectively. Then

${dn_i(t) \over dt} = r_i n_i(t).$

With the total population size

$n(t) = n_1(t) + n_2(t)\!$

we have the proportion of type 1

$p(t) = {n_1(t) \over n(t)}$

Define the fitness advantage

$s \equiv s_{12} = r_1 - r_2\!$

Given our interest in understanding the change in gene frequencies, our goal is to compute the rate of change of

p (t)

.

${\partial p(t) \over \partial t}$	$= {\partial \over \partial t}\left({n_1(t) \over n(t)}\right)$
	$= {\partial n_1(t) \over \partial t}\left({1 \over n(t)}\right) + n_1(t){-1 \over n(t)^2}{\partial n(t) \over \partial t}$
	$= {\partial n_1(t) \over \partial t}\left({1 \over n(t)}\right) + n_1(t){-1 \over n(t)^2}\left({\partial n_1(t) \over \partial t} + {\partial n_2(t) \over \partial t}\right)$
	$= {r_1 n_1(t) \over n(t)} - {n_1(t) \over n(t)^2}\left(r_1 n_1(t) + r_2 n_2(t)\right)$
	$= {r_1 n_1(t) \over n(t)} - {n_1(t) \over n(t)^2}\left(r_1 n_1(t) + (r_1-s)(n(t)-n_1(t))\right)$
	$= {r_1 n_1(t) \over n(t)} - {n_1(t) \over n(t)^2}\left(r_1 n(t) -s n(t) + s n_1(t))\right)$
	$= {n_1(t) \over n(t)^2}\left(s n(t) - s n_1(t))\right)$
	$= s{n_1(t) \over n(t)}\left(1 - {n_1(t) \over n(t)}\right)$
	$= s p(t)(1-p(t))\!$

This result says that the proportion of type 1 $p$ changes most rapidly when $p = 0.5$ and most slowly when $p$ is very close to 0 or 1.

Evolution is linear on a log-odds scale

The logit function $\mathrm{logit} (p) = \ln {p \over 1-p}$ , which takes $p \in [0,1] \to \mathbb{R}$ , induces a more natural space for considering changes in frequencies. Rather than tracking the proportion of type 1 or 2, we instead track their log odds. In logit terms, with $L_p(t) \equiv \mathrm{logit} (p(t))\!$ ,

${\partial L_p(t) \over \partial t}$	$= {\partial \over \partial t}\left(\ln {p(t) \over 1-p(t)}\right)$
	$= {\partial \over \partial t}\left(\ln {n_1(t) \over n_2(t)}\right)$
	$= {\partial \over \partial t}\left(\ln {n_1(0) \over n_2(0)} e^{st}\right)$
	$= s. \!$

This differential equation $L p'(t) = s$ has the solution

$L_p(t) = L_p(0) + st\!$

showing that the log-odds of finding type 1 changes linearly in time, increasing if $s > 0$ and decreasing if $s < 0$ .

Diffusion approximation

Insert math here.

Statistical analysis of relative growth rates

We have three strains, $i$ , $j$ and $r$ , where $r$ is a reference strain. Strains $i$ and $j$ have fitness $w_i = e^{r_i}$ and $w_j=e^{r_j}$ . Define the selection coefficient $s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j$ as usual. We have data consisting of triples ( $g =$ number of generations, $n i =$ number of cells of type $i$ , $n r =$ number of cells of type $r$ ). We have data consisting of pairs ( $g =$ number of generations, $p i r = n i / n r$ ) where $n i$ =number of cells of type $i$ and $n r =$ number of cells of type $r$ .