Drummond:PopGen

(Difference between revisions)
 Revision as of 22:49, 23 January 2009 (view source) (→Per-generation and instantaneous growth rates)← Previous diff Revision as of 08:29, 2 April 2009 (view source) (→Continuous rate of change)Next diff → Line 113: Line 113: |} |} - 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.  In logit terms, + 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 $L_p(t) \equiv \mathrm{logit} (p(t))$, + + ${\partial L_p(t) \over \partial t} = {\partial \over \partial t}\left(\ln {p(t) \over 1-p(t)}\right)$ + + ${\partial L_p(t) \over \partial t} = s t$. + + That is, $L_p(t)$, the log-odds of finding type 1 in a random draw from the population, changes linearly in time with slope $s$. ==Diffusion approximation== ==Diffusion approximation== Insert math here. Insert math here.

the drummond lab

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 ni(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 = Δt. 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 = lnR for convenience.

Continuous rate of change

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

Let r1 and r2 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
$p(t) = {n_1(t) \over n(t)}$
$s \equiv s_{12} = r_1 - r_2\!$
Given our interest in understanding the change in gene frequencies, our goal is to compute the rate of change of p(t).
 ${\partial p(t) \over \partial t}$ $= {\partial \over \partial t}\left({n_1(t) \over n(t)}\right)$ $= {\partial n_1(t) \over \partial t}\left({1 \over n(t)}\right) + n_1(t){-1 \over n(t)^2}{\partial n(t) \over \partial t}$ $= {\partial n_1(t) \over \partial t}\left({1 \over n(t)}\right) + n_1(t){-1 \over n(t)^2}\left({\partial n_1(t) \over \partial t} + {\partial n_2(t) \over \partial t}\right)$ $= {r_1 n_1(t) \over n(t)} - {n_1(t) \over n(t)^2}\left(r_1 n_1(t) + r_2 n_2(t)\right)$ $= {r_1 n_1(t) \over n(t)} - {n_1(t) \over n(t)^2}\left(r_1 n_1(t) + (r_1-s)(n(t)-n_1(t))\right)$ $= {r_1 n_1(t) \over n(t)} - {n_1(t) \over n(t)^2}\left(r_1 n(t) -s n(t) + s n_1(t))\right)$ $= {n_1(t) \over n(t)^2}\left(s n(t) - s n_1(t))\right)$ $= s{n_1(t) \over n(t)}\left(1 - {n_1(t) \over n(t)}\right)$ $= s p(t)(1-p(t))\!$

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 $L_p(t) \equiv \mathrm{logit} (p(t))$,

${\partial L_p(t) \over \partial t} = {\partial \over \partial t}\left(\ln {p(t) \over 1-p(t)}\right)$

${\partial L_p(t) \over \partial t} = s t$.

That is, Lp(t), the log-odds of finding type 1 in a random draw from the population, changes linearly in time with slope s.

Diffusion approximation

Insert math here.