Two-way mutation and its equilibrium
Opposing mutation rates fix an equilibrium frequency, and real rates take tens of thousands of generations to reach it.
Read first: Genotype frequencies and Hardy-Weinberg
Two flows that balance
Allele A mutates to a at rate μ per copy per generation. Allele a mutates back to A at rate ν. The module follows the frequency p of A with nothing else acting: no selection, no drift, no migration. The population is infinite, so the result is exact and the same every time.
The recursion is p(t + 1) = p(1 − μ) + (1 − p)ν. The first term is the A copies that did not mutate. The second is the a copies that mutated to A. The loss of A is μp per generation and the gain is ν(1 − p).
The defaults are μ = 0.001, ν = 0.0005 and p = 0.5. In the first generation A loses 0.5 × 0.001 = 0.0005 and gains 0.5 × 0.0005 = 0.00025, so p goes to 0.49975. The flows are equal when μp = ν(1 − p). Solving gives the equilibrium p̂ = ν / (μ + ν) = 0.0005 / 0.0015 = 1/3. It depends only on the two rates.
Mutation acts on allele copies, so ploidy and mating system do not enter. At equilibrium mutation has not stopped. A still mutates to a and a still mutates back. The two flows cancel, so the frequency stays put.
The approach to 1/3
How fast it gets there
The gap between p and p̂ shrinks by the same factor each generation: p(t) − p̂ = (p(0) − p̂)(1 − μ − ν)^t. Here 1 − μ − ν = 0.9985. The gap is 0.1667 at the start. It is 0.0833 at generation 462, 0.0371 at generation 1000 and 0.00009 at generation 5000. After 1000 generations 78% of the gap is gone.
The gap halves every ln 2 / (μ + ν) generations, which is 462 for the defaults. Halving is the shape of every approach in this module. The curve is steep at first, when the gap is large, and flat near p̂.
The equilibrium does not depend on where p starts. The next figure begins above it.
The factor 1 − μ − ν also covers two special cases. If μ + ν = 1 the factor is 0 and p reaches p̂ in one generation: μ = 0.6, ν = 0.4 and p = 0.5 give 0.4 at once. If μ + ν is above 1 the factor is negative, so p alternates between the two sides of p̂ while the gap shrinks. With μ = 0.8 and ν = 0.6 it goes 0.5, 0.4, 0.44 and onward. The controls allow these values but they are not biological rates.
Any start, same end
Realistic rates
Exercise
Swap the rates
Set μ to 0.0005 and ν to 0.001, keep P(A) at 0.5 and run for 5000 generations.
What to expect. The equilibrium line moves to 2/3 and p ends at 0.6666. Mutation favours the allele that it creates faster. Here a mutates to A at 0.001 and A mutates to a at only 0.0005, so A accumulates. The sum μ + ν is still 0.0015, so the gap again halves in 462 generations: p is 0.5834 at generation 462, halfway from 0.5 to 2/3.
Open the walkthrough step →Exercise
Use equal rates
Set μ and ν both to 0.001, P(A) to 0.1, and run for 5000 generations.
What to expect. The equilibrium is 0.5. The frequency is 0.4460 at generation 1000 and 0.49998 at generation 5000. With equal rates the two flows are equal at p = 0.5, whatever the rate. A larger common rate only makes the approach faster.
Open the walkthrough step →Summary
- Each generation A loses μp copies and gains ν(1 − p) copies.
- The equilibrium is p̂ = ν / (μ + ν). With μ = 0.001 and ν = 0.0005 it is 1/3.
- Mutation acts on copies, so ploidy and mating do not enter the recursion.
- The starting frequency does not change p̂. From 0.5 or from 0.9 the frequency ends at 1/3.
- The gap to p̂ shrinks by (1 − μ − ν) per generation and halves every ln 2 / (μ + ν) generations: 462 at the defaults.
- At μ = 0.00001 and ν = 0.000001 the halving time is about 63,000 generations, and p reaches only 0.1362 after 200,000.