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(Statistical analysis of relative growth rates)
Current revision (22:40, 28 March 2011) (view source)
(Continuous rate of change)
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|<math>= s p(t)(1-p(t))\!</math>
|<math>= s p(t)(1-p(t))\!</math>
This result says that the proportion of type 1 <math>p</math> changes most rapidly when <math>p=0.5</math> and most slowly when <math>p</math> is very close to 0 or 1.  
This result says that the proportion of type 1, <math>p</math>, changes most rapidly when <math>p=0.5</math> and most slowly when <math>p</math> is very close to 0 or 1.
==Evolution is linear on a log-odds scale==
==Evolution is linear on a log-odds scale==

Current revision


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\!
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
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)}
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 Lp'(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, ni = number of cells of type i, nr = number of cells of type r). We have data consisting of pairs (g = number of generations, pir = ni / nr) where ni=number of cells of type i and nr = number of cells of type r.

What is the best estimate, and error, on sij?


Assuming exponential growth, lnpir =

Let \Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij}).

Maximum-likelihood approach

Add text.

Bayesian approach

Add text.

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