Probability of fixation
This was derived in Kimura 1962.
[math]u(p)=\frac{1-e^{-4N_esp}}{1-e^{-4N_es}}[/math]
If we are considering the initial frequency of a single new mutation in the population p=1/(2Ne),
[math]u(p)_1=\frac{1-e^{-4N_es\frac{1}{2N_e}}}{1-e^{-4N_es}}=\frac{1-e^{-2s}}{1-e^{-4N_es}}[/math].
And if 4Nes is large
[math]u(p)_2\approx\frac{1-e^{-2s}}{1}=1-e^{-2s}[/math].
[math]e^{2s}\approx 1+2s[/math]
[math]u(p)_2 \approx 1-e^{-2s} \approx 1-1+2s = 2s[/math].
This agrees with the results of Fisher 1930 (pp. 215--218) and Wright 1931 (pp. 129--133).
It may be surprising at first the the probability of fixation of a new allele that confers a fitness advantage is only approximately 2s. So if it gives a 3% fitness advantage the probability of fixation is only about 6%. In other words there is a 94% chance the new adaptive allele will be lost due to genetic drift. This implies that adaptive evolution of a species is very inefficient and that adaptive alleles have to occur repeatedly by mutation, to be lost by drift, before they eventually fix.
Why is this process so inefficient? When an allele is rare, such as a single copy as a new mutation, the forces of drift are typically much larger than the forces of selection. As an example work out the probability of sampling zero copies of an allele starting at a count of one from one generation to the next with a Poisson distribution and a mean of λ = 1.
[math]P(k)=\frac{\lambda^k e^{-\lambda}}{k!}[/math]
There is a 0.3679 probability of loss in the next generation. The probability in the next generation of one copy is also 0.3679, two copies is 0.1839, three copies 0.0613, four copies 0.0153, five copies 0.00307, etc. With a greater than a third chance of loss in one generation due to neutral drift this can more than outweigh the small advantage (say on the order of single percents) added by selection.
Even if the allele survives into the next generation it is most likely at a count of one and still has a greater than 1/3 loss in the following generation. If it manages to make it to two copies there is a 13.5% chance of loss in the next generation, a 27.1% chance of reducing back down to a count of one again, etc. Multiplying all of the paths from one in the first generation to zero in the third generation up to a count of five in the second, 1->0->0, 1->1->0, 1->2->0, 1->3->0, 1->4->0, 1->5->0, using a Poisson distribution gives, 0.3679, 0.1353, 0.02489, 0.003053, 0.0002807, 0.00002066 respectively for a total probability of 0.5315. So there is a greater than 1/3 loss in the second generation and a greater than 1/2 loss in the third generation.
Notes
Kimura's derivation
This is derived from
[math]u(p) = \frac{\int_0^p G(x)\, \mbox{d} x}{\int_0^1 G(x)\, \mbox{d} x}[/math],
equation 3 of Kimura 1962.
[math]u(p,t)[/math] is the probability of fixation of an allele at frequency p within t generations.
The change in allele frequency ([math]\delta p[/math]) over short periods of time ([math]\delta t[/math]) is
[math]u(p, t+\delta t) = \int f(p, p+\delta p; \delta t) u(p+ \delta p, t) \, \mbox{d} (\delta p)[/math],
integrating over all values of changes in allele frequency ([math]\delta p[/math]).
A mean and variance of the change in allele frequency (p) per generation are defined as
[math]M_{\delta p}=\lim_{\delta t \to 0} \frac{1}{\delta t} \int (\delta p) f(p, p+\delta p; \delta t) \, \mbox{d} (\delta p)[/math]
[math]V_{\delta p}=\lim_{\delta t \to 0} \frac{1}{\delta t} \int (\delta p)^2 f(p, p+\delta p; \delta t) \, \mbox{d} (\delta p)[/math]
The probability of fixation given sufficient time for fixation to occur is
[math]u(p)=\lim_{t \to \infty} u(p,t)[/math]
[math]G(x) = e^{-\int \frac{2M_{\delta x}}{V_{\delta x}} \, \mbox{d} x}[/math]
(to be continued ... I need to work through this and my calculus is rusty.)
A different approach
[math]u(p)=\frac{1-e^{-4N_esp}}{1-e^{-4N_es}}[/math]
The numerator is the probability of not zero (loss) in a Poisson distribution with a mean of 4Nesp. 2Np is the number of copies of the allele in the population. This is multiplied by 2s. [math]2Np \times 2s = 4Nsp[/math]
Why is s multiplied by two here?
The denominator is the same thing but with p=1 (the largest value possible). This rescales the numerator to be a fraction out of one(?).
I suspect there is a more intuitive approach to understanding this by exploring this line of reasoning but I am not quite seeing it yet.
[math]1-e^{-4N_esp}\approx 1- 1 - 4Nsp = -4Nsp[/math] ?
[math]u(p)=\frac{1-e^{-4N_esp}}{1-e^{-4N_es}}\approx\frac{-4Nsp}{-4Ns}=p[/math] ?
[math]1-e^{-4N_esp}\approx 1- (1 - 2s)^{2Np}[/math] ?
There is more to go through in Fisher 1930 (pp. 215--218) and Wright 1931 (pp. 129--133). I need to start there and work forward.
