Mutation-Drift Equilibrium
Genetic variation is input by mutation and removed by genetic drift. A natural way to quantify genetic variation is by population heterozygosity, H. This is the frequency that two randomly selection gene copies from the population differ from each other.
The rate of mutation and drift for the same number of gene copies over the same unit of time, most convenient is two gene copies and per generation time, can be used to generate an equilibrium prediction. Two gene copies are sampled from gene copies in the previous generation. There is a 1/2N chance that they were both sampled from the same gene copy in the generation before. [add a diagram of sampling between generations] In this case they would be identical to each other and not contribute to heterozygosity. If the population size is kept constant, then if one gene copy gives rise to two copies in the next generation then another gene copy is lost from the population. The gene copy that was lost was potentially heterozygous with the gene copy that was amplified; genetic drift results in a reduction of heterozygosity at a rate of 1/2N per generation.
If a gene copy mutates then, assuming each mutation results in a new SNP in a DNA sequence, it results in a difference between gene sequences and increases heterozygosity. There are two chances, one for each of two alleles sampled from the previous generation, for mutation to occur with a total per generation rate of 2μ. [add a diagram of mutation of two lineages between generations] Population heterozygosity is increased at a rate of 2μ per generation.
At equilibrium the rate of increase of heterozygosity by mutation and the rate of decrease by drift are equal.
[math]2 \mu = H \frac{1}{2N}[/math]
If there is no variation in the population then genetic drift has no effect. Genetic drift can only act to reduce the heterozygosity that exists in the popualtion, so 1/(2N) is multiplied by H.
By rearranging this predicts
[math]H = 4 N \mu[/math].
[to be added ... show the alternative derivation from the coalescent, and visualizing this as a product of [math]H = 2 \mu \times 2 N[/math]]
[give examples of how this can be used to estimate N and the meaning of Ne]
Questions
- 1). You sequence 1,000 basepairs of a pseudogene from a single diploid individual of Montipora flabellata (blue rice coral) and find eight heterozygous SNPs. Based on a per basepair mutation rate of 10-8 per generation what is your best estimate of the effective population size of this coral.
Nucleotide heterozygosity in this case is
[math]H_n = 4N_e \mu = 8/1000 = 0.008[/math].
[math]N_e = \frac{H_n}{4 \mu} [/math]
[math]N_e = \frac{0.008}{4 \times 10^{-8}} = 200{,}000[/math]
The effective population size estimate is two hundred thousand diploid individuals.
- 2). ...
