L2 · Chapter 3

The coalescent traces a sample back to its common ancestor

Sandbox Physics · 2026-10-30

Pick six bacteria from today’s flask and follow each one to its parent, then to the parent’s parent. Sooner or later two of the lines land on the same cell, and from there they travel together. Keep going and all six meet in one ancestor. In a flask of fifty this takes about eighty generations on average. Running the model backwards like this is often easier than running it forwards, and it gives the same answers.

Two lineages meet when they pick the same parent

In the Wright–Fisher model every child picks its parent uniformly among $N$ candidates. Two different children therefore share a parent with probability $1/N$ . If they do not, their two parents are again two different individuals of the previous generation, and the same rule applies one step further back.

The number of generations $T_2$ until the two lines merge is geometric, with mean $N$ . The chance that they have not merged after $t$ generations is the same factor that multiplied the heterozygosity in chapter 1. That is no accident. Two individuals can differ only if their lines have not yet met, because a shared ancestor passes the same allele to both.

P(T_2>t)=\Bigl(1-\frac1N\Bigr)^{t}\approx e^{-t/N},\qquad \mathrm{E}[T_2]=N

The waiting time for two sampled lineages to coalesce, in generations.

Source: K. S. Korolev · N. Berestycki

Many lineages: Kingman’s coalescent

Now follow $k$ lineages. There are $\binom{k}{2}$ pairs, and each pair merges with probability $1/N$ per generation. When $N$ is large compared with the number of pairs, two mergers in the same generation are rare, so the next merger arrives after a waiting time with mean $N/\binom{k}{2}$ and the count drops from $k$ to $k-1$ .

Measured in units of $N$ generations, the genealogy of a Wright–Fisher sample converges to Kingman’s coalescent: every pair of lineages merges at rate one, independently of the others, and the merging pair is chosen uniformly. Berestycki states this limit as a theorem, for both the Moran and the Wright–Fisher model.

\mathrm{E}[T_k]\approx\frac{N}{\binom{k}{2}}=\frac{2N}{k(k-1)}

Mean time spent with

k

lineages, in generations.

Source: N. Berestycki

The last two lines take half the time

Adding the stages from $n$ lineages down to one gives a sum whose terms cancel in pairs. The mean age of the common ancestor of a sample is $2N(1-1/n)$ generations.

The stage with two lineages left lasts $N$ generations on average, more than half of the total for any sample. Going from two individuals to a sample of a thousand only raises the age from $N$ to almost $2N$ . Recent mergers happen fast because many pairs are available, while the final meeting waits on a single pair.

Chapter 1 met almost the same number from the other direction. A lone neutral mutant that fixes takes about $2N$ generations, because the conditional fixation time there tends to $2N$ as $p$ goes to zero. Once a mutant carried by one individual has fixed, that individual is an ancestor of everyone, so the two times agree.

\mathrm{E}[T_{\mathrm{MRCA}}]=\sum_{k=2}^{n}\frac{2N}{k(k-1)}=2N\sum_{k=2}^{n}\Bigl(\frac{1}{k-1}-\frac1k\Bigr)=2N\Bigl(1-\frac1n\Bigr)

The mean time back to the most recent common ancestor of a sample of

n

individuals.

Source: N. Berestycki

Worked example: six individuals among fifty

Problem. Sample $n=6$ individuals from a neutral Wright–Fisher population of $N=50$ . How many generations back is their most recent common ancestor, and how much of that time is spent with each number of lineages?

Stages. With six lineages there are $15$ pairs, so the first merger takes $100/30=3.3$ generations on average. Then $5$ , $8.3$ and $16.7$ generations for five, four and three lineages, and $50$ generations for the last pair.

Total. $3.3+5+8.3+16.7+50=83.3$ , which is $2N(1-1/6)$ . The last pair takes $60$ per cent of it.

Finite-population check. The coalescent assumes at most one merger per generation. Solving the exact chain on the number of lineages, which allows several mergers at once, gives $83.0$ generations. The two agree to within half a per cent here.

Simulation. Over $400$ seeded pedigrees the six sampled lines met after $82.1\pm2.6$ generations, half a standard error from both references. Pairs met after $48.7\pm2.3$ generations against $50$ .

Answer. About $83$ generations, most of them spent waiting for the last two lines.

Source: N. Berestycki · K. S. Korolev

Where the approximation strains, and what it leaves out

A sample of twelve in a population of fifty has $66$ pairs, more than $N$ . In its first generation back, at least one merger happens with probability $0.76$ , so the one-merger-at-a-time picture is poor at the start. Those early stages are short, though. The exact chain gives $91.1$ generations against Kingman’s $91.7$ , and the seeded pedigrees give $96.1\pm3.0$ , which is $1.7$ standard errors above the exact value.

Kingman’s coalescent holds for neutral alleles in a well-mixed population of constant size. Selection makes ancestry depend on which allele a lineage carries, and space makes it depend on where the lineage lives. At an expanding colony edge, the next chapter shows, ancestors are concentrated at the edge and the genealogy looks very different.

Source: N. Berestycki · O. Hallatschek

Try it in the Lab

Trace a sample back to one ancestor

Choose One neutral newcomer and watch the white lines climb the pedigree. Predict how many generations back the six traced individuals meet, then compare with the readout over several rounds.
Change the number of individuals traced back from six to two and then to twelve. Predict both averages from the formula before reading the Kingman reference.
Double the population size. Predict how the age of the common ancestor changes, and explain why one round can land far from the average.

Exercises with explained answers

A neutral population has $N=100$ . What is the mean age of the common ancestor of $n=10$ individuals, and what fraction of it is spent with two lineages?
Show the answer
$2\times100\times(1-1/10)=180$ generations. The two-lineage stage lasts $100$ of them, $56$ per cent.
The time for two lineages to meet is geometric with success probability $1/N$ , so its variance is $N(N-1)$ . For $N=50$ and $400$ independent pedigrees, what standard error should the simulated mean have?
Show the answer
The standard deviation is $\sqrt{2450}=49.5$ generations, so the standard error is $49.5/\sqrt{400}=2.5$ . The benchmark reports $2.3$ . A single pair can easily meet after $10$ or $150$ generations, which is why many pedigrees are needed.
With $k$ lineages in a population of $N$ , the probability that no two share a parent in one generation is $\prod_{j=0}^{k-1}(1-j/N)$ . Evaluate it for $k=6$ and $N=50$ , and compare the chance of a merger with $\binom{k}{2}/N$ .
Show the answer
The product is $0.732$ , so a merger happens with probability $0.268$ . The coalescent rate gives $15/50=0.30$ . The two are close because $0.3$ is still small. For twelve lineages the rate would be $1.32$ , larger than one, so only the exact chain works there.

Sources and what we recomputed

K. S. Korolev, M. Avlund, O. Hallatschek and D. R. Nelson, Genetic demixing and evolutionary forces in the one-dimensional stepping stone model, arXiv:0904.4625v2 (2011)
Version 2, full text. Section II, Population genetics in well-mixed populations: genetic drift as random sampling of a finite population, the heterozygosity as the chance that two sampled individuals differ, and its loss with time.
N. Berestycki, Recent progress in coalescent theory, arXiv:0909.3985v1 (2009), Ensaios Matemáticos, Sociedade Brasileira de Matemática
Version 1, full text. Section 2.1.1: the Kingman coalescent, in which each pair of lineages merges at rate one. Theorem 2.3: the Moran genealogy becomes this coalescent after rescaling time. Theorem 2.4: the Wright-Fisher genealogy does too when time is measured in units of $N$ generations. Theorem 2.8, Eqs. (36)-(38): a neutral allele fixes with probability equal to its frequency, with the diffusion formula for the mean absorption time; the formula after it gives the time conditioned on fixation.
O. Hallatschek, P. Hersen, S. Ramanathan and D. R. Nelson, Genetic drift at expanding frontiers promotes gene segregation, arXiv:0812.2345v1; PNAS 104, 19926 (2007)
Preprint, version 1, full text. Fig. 2B and the text around it: the mean-square wandering of 92 domain boundaries grows as $L^{2\zeta}$ with $\zeta=0.65\pm0.05$ . The scaling argument after it gives $\zeta=1+(\chi-1)/z$ , which is $2/3$ for the Eden model.