# Difference between revisions of "Drummond:PopGen"

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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== | ||

− | The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, 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 <math>L_p(t) \equiv \mathrm{logit} (p(t))</math>, | + | The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, 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 <math>L_p(t) \equiv \mathrm{logit} (p(t))\!</math>, |

:{| | :{| | ||

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|- | |- | ||

| | | | ||

− | |<math>= {\partial \over \partial t}\left(\ln e^{st}\right)</math> | + | |<math>= {\partial \over \partial t}\left(\ln {n_1(0) \over n_2(0)} e^{st}\right)</math> |

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|} | |} | ||

− | + | This differential equation <math>L_p'(t) = s</math> has the solution | |

− | <math>L_p(t) = L_p(0) | + | :<math>L_p(t) = L_p(0) + st\!</math> |

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

==Diffusion approximation== | ==Diffusion approximation== | ||

Insert math here. | Insert math here. | ||

+ | |||

+ | ==Statistical analysis of relative growth rates== | ||

+ | We have three strains, <math>i</math>, <math>j</math> and <math>r</math>, where <math>r</math> is a reference strain. | ||

+ | Strains <math>i</math> and <math>j</math> have fitness <math>w_i = e^{r_i}</math> and <math>w_j=e^{r_j}</math>. Define the selection coefficient <math>s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j</math> as usual. | ||

+ | We have data consisting of triples (<math>g=</math>number of generations, <math>n_i=</math>number of cells of type <math>i</math>, <math>n_r=</math>number of cells of type <math>r</math>). | ||

+ | We have data consisting of pairs (<math>g=</math>number of generations, <math>p_{ir}= n_i/n_r</math>) where <math>n_i</math>=number of cells of type <math>i</math> and <math>n_r=</math>number of cells of type <math>r</math>. | ||

+ | |||

+ | What is the best estimate, and error, on <math>s_{ij}</math>? | ||

+ | |||

+ | ===Model=== | ||

+ | Assuming exponential growth, <math>\ln p_{ir} = </math> | ||

+ | |||

+ | Let <math>\Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij})</math>. | ||

+ | |||

+ | ===Maximum-likelihood approach=== | ||

+ | Add text. | ||

+ | |||

+ | ===Bayesian approach=== | ||

+ | Add text. |

## Latest revision as of 19:40, 28 March 2011

## 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 be the number of organisms of type at time , and let be the *per-capita reproductive rate per generation*. If counts generations, then

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

where the last simplification follows from L'Hôpital's rule. Explicitly, let . Then

The solution to the equation

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle n_i(t) = n_i(0) e^{t\ln R} = n_i(0) R^{t}.\!}**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t}**. We can define the

*instantaneous growth rate*

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 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 **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle r_1}**
and **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle r_2}**
be the instantaneous rates of increase of type 1 and type 2, respectively. Then

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle {dn_i(t) \over dt} = r_i n_i(t).}**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle n(t) = n_1(t) + n_2(t)\!}**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p(t) = {n_1(t) \over n(t)}}**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle s \equiv s_{12} = r_1 - r_2\!}**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p(t)}**.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle {\partial p(t) \over \partial t}}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {\partial \over \partial t}\left({n_1(t) \over n(t)}\right)}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {\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}}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {\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)}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {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)}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {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)}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {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)}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = {n_1(t) \over n(t)^2}\left(s n(t) - s n_1(t))\right)}****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle = s{n_1(t) \over n(t)}\left(1 - {n_1(t) \over n(t)}\right)}**

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

## Evolution is linear on a log-odds scale

The logit function , which takes , 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 ,

This differential equation has the solution

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

## Diffusion approximation

Insert math here.

## Statistical analysis of relative growth rates

We have three strains, , and , where is a reference strain. Strains and have fitness and . Define the selection coefficient as usual. We have data consisting of triples (number of generations, number of cells of type , number of cells of type ). We have data consisting of pairs (number of generations, ) where =number of cells of type and number of cells of type .

What is the best estimate, and error, on ?

### Model

Assuming exponential growth,

Let .

### Maximum-likelihood approach

Add text.