Difference between revisions of "Fluctuating Population Size and Genetic Drift"
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<math>\sigma^2 = \left(\frac{\sqrt{2N p (1-p)}}{2N}\right)^2 = \frac{2N p (1-p)}{4N^2} = \frac{p (1-p)}{2N}</math> | <math>\sigma^2 = \left(\frac{\sqrt{2N p (1-p)}}{2N}\right)^2 = \frac{2N p (1-p)}{4N^2} = \frac{p (1-p)}{2N}</math> | ||
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| + | Imagine the population fluctuating between two sizes, ''N''<sub>1</sub> and ''N''<sub>2</sub>, so that it spends half of its time at ''N''<sub>1</sub> and half at ''N''<sub>2</sub>. There is an effective population size, ''N''<sub>e</sub>, of constant size that has the same average variance in allele frequency change each generation as the fluctuating population. | ||
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| + | <math>\frac{p(1-p)}{2N_e}=\frac{1}{2}\times\frac{p(1-p)}{2N_1}+\frac{1}{2}\times\frac{p(1-p)}{2N_2}</math> | ||
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| + | We can cancel out a lot of components to get | ||
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| + | <math>\frac{1}{N_e}=\frac{1}{2}\times\frac{1}{N_1}+\frac{1}{2}\times\frac{1}{N_2}</math> | ||
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| + | and solve for ''N''<sub>e</sub>. | ||
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| + | <math>N_e=\frac{1}{\frac{1}{2}\times\frac{1}{N_1}+\frac{1}{2}\times\frac{1}{N_2}}</math> | ||
to be continued ... | to be continued ... | ||
Revision as of 07:16, 20 September 2018
In general the variance from a binomial process is [math]\sigma^2 = n p (1-p)[/math], where n is the number of trials and p is the probability of one of the two outcomes.
The variance for the sampling of two alleles each generation is [math]2N p (1-p)[/math], where N is the diploid population size and p is the allele frequency.
We like to work with allele frequencies rather than allele counts in a population. The variation is scaled by 2N so the underlying standard deviation is a fraction between zero and one.
[math]\frac{\sigma}{2N} = \frac{\sqrt{2N p (1-p)}}{2N}[/math]
[math]\sigma^2 = \left(\frac{\sqrt{2N p (1-p)}}{2N}\right)^2 = \frac{2N p (1-p)}{4N^2} = \frac{p (1-p)}{2N}[/math]
Imagine the population fluctuating between two sizes, N1 and N2, so that it spends half of its time at N1 and half at N2. There is an effective population size, Ne, of constant size that has the same average variance in allele frequency change each generation as the fluctuating population.
[math]\frac{p(1-p)}{2N_e}=\frac{1}{2}\times\frac{p(1-p)}{2N_1}+\frac{1}{2}\times\frac{p(1-p)}{2N_2}[/math]
We can cancel out a lot of components to get
[math]\frac{1}{N_e}=\frac{1}{2}\times\frac{1}{N_1}+\frac{1}{2}\times\frac{1}{N_2}[/math]
and solve for Ne.
[math]N_e=\frac{1}{\frac{1}{2}\times\frac{1}{N_1}+\frac{1}{2}\times\frac{1}{N_2}}[/math]
to be continued ...
