Drummond:PopGen

From OpenWetWare

Revision as of 19:53, 14 July 2008 by Dadrummond (Talk | contribs)

(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)

the drummond lab

home people research publications news jobs protocols contact

Per-generation and instantaneous growth rates

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 values of

t

. We can define the instantaneous rate of increase

r = ln R

for convenience.

Continuous rate of change

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 $p (t)$ .

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