Molecular population genetics
Two estimates of the same mutation rate, one from segregating sites and one from pairwise differences, disagree when population size has changed.
Read first: Genetic drift
Two ways to count variation
Sequence n copies of a stretch of DNA that is L sites long. Two numbers summarise the variation. S is the number of segregating sites, the positions where the copies are not all the same. k̂ is the mean number of differences between two copies, averaged over all n(n − 1)/2 pairs. Dividing by length gives the nucleotide diversity π = k̂/L.
Both numbers estimate the same quantity when the population is neutral and has constant size. That quantity is θ = 4Nμ, where N is the number of diploid individuals and μ is the mutation rate per generation for the whole stretch. Watterson's estimator is θ_W = S/aₙ, with aₙ = 1 + 1/2 + 1/3 + ... + 1/(n − 1). The pairwise estimator is k̂ itself.
The Sim mode builds a sample with the coalescent. Going back in time, lineages merge in random pairs until one common ancestor is left. Mutations fall on the branches, and every copy below a mutated branch carries it. All sites share one tree, because the module has no recombination, so sites are not independent draws. For n = 12 the expected time back to the common ancestor is 2(1 − 1/n) = 1.83, in units of 2N generations.
A worked case from the Calc mode of the module: n = 10 gives aₙ = 2.829. A sample with S = 12 segregating sites has θ_W = 12/2.829 = 4.24. If the same sample has k̂ = 3.5, then π = 3.5/500 = 0.007 per site for L = 500. The two estimates, 4.24 and 3.5, differ. Tajima's D asks whether they differ by more than chance.
What Tajima's D compares
S counts every mutation once, whether one copy carries it or half of them do. k̂ weights each mutation by the fraction of pairs it separates. A mutation on a single copy (a singleton) separates n − 1 pairs and adds only 2/n to k̂. A mutation carried by 5 of 10 copies separates 25 of 45 pairs and adds 0.56. Rare variants therefore raise S much more than k̂, and common variants raise k̂ as much as S.
Tajima's D = (k̂ − θ_W) divided by its standard deviation under the neutral model. For n = 10 and S = 12 the possible k̂ run from 2.4, when every site is a singleton, to 6.67, when every site splits the sample 5 and 5. At k̂ = 3.5, D = −0.79. At k̂ = 6.6, D = 2.51. At k̂ = 2.4, D = −1.96. Negative D means an excess of rare variants, positive D an excess of intermediate ones.
One neutral sample
The same mutation rate after growth
A recent bottleneck raises D
Exercise
Push D to both ends in Calc mode
Open the module and switch to Calc. Keep n = 10, S = 12 and L = 500. Set k to 6.6, then 2.4, then 12.
What to expect. k = 6.6 gives D = 2.51 and the tag "D > 2". k = 2.4 gives D = −1.96, the lowest value this n and S allow, so the tag still reads neutral. k = 12 is rejected with a message, because the largest mean this n and S allow is 6.67.
Open the module →Exercise
Growth and an old bottleneck in Sim mode
In Sim mode set n = 12, θ = 10 and L = 300. Run 20 times with Model = Growth and r = 5, counting the runs with negative D. Then choose Bottleneck with t start 0.2, duration 0.1 and Nb 0.02 and repeat.
What to expect. Growth gives negative D in most runs, 125 of 150 in a seeded set with mean −0.64. The old, deep bottleneck also gives negative D, mean −1.08, because lineages merge inside the bottleneck and leave a star-shaped tree. A bottleneck can move D either way.
Open the walkthrough step →What to keep
- S counts mutations once each, k̂ weights them by the pairs they separate, and both estimate θ = 4Nμ when the population is neutral and constant.
- Tajima's D is negative when rare variants are in excess (growth, recent selection) and positive when intermediate variants are in excess (a recent bottleneck, structure).
- One sample scatters widely around the expected spectrum θ/i, because a single tree is shared by every site.
- Bottlenecks can push D either way, and mutation saturation pushes it up, so D alone cannot separate demography from selection.