Lecture notes. Written from the material of MIT 8.591J 2014, licensed CC BY-NC-SA 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.
15. The Moran Process and Fixation¶
What this covers¶
This chapter builds the Moran process, the standard model for how a population of fixed size evolves when new types arise one individual at a time, and uses it to answer two questions: how fast do neutral (fitness-irrelevant) mutations spread and fix, and what determines whether a mutant with a real fitness advantage takes over the population at all. It assumes basic probability (conditional probability, expectation, a random walk with absorbing boundaries) and the population-genetics notion of relative fitness; it does not assume familiarity with the Wright-Fisher process, which is mentioned only for contrast, or with diffusion approximations.
The Moran process: a population that never changes size¶
Populations of interest for evolution are often enormous — \(10^9\) bacteria in a flask, say — and yet stochastic effects are never negligible, because every new mutant, however large the population, starts out as a single individual. Tracking what happens to that one individual is irreducibly a question about randomness, no matter how large \(N\) is.
The Moran process is the simplest model built to study this. It fixes population size at a constant \(N\) and considers an asexually reproducing population of two types, \(A\) and \(B\), differing (for now) only in some trait already present — no new mutation happens during the process itself. Rather than having all \(N\) individuals reproduce at once in synchronized generations, as in the Wright-Fisher process, the Moran process moves one birth at a time:
- One individual is chosen to reproduce, with probability proportional to fitness.
- Its offspring replaces one individual chosen uniformly at random from the whole population of \(N\) — including, possibly, the parent itself.
This keeps \(N\) fixed by construction: one birth, one death (or "replacement" — the two words mean the same thing here). It is a reasonable theoretical stand-in for a turbidostat, a device that holds population size constant by pulling out a random individual every time a cell divides, in contrast to a chemostat, which fixes the dilution rate instead.
Write \(i\) for the number of \(A\) individuals, so there are \(N-i\) individuals of type \(B\). (The letter \(i\), rather than something more mnemonic, follows Nowak's Evolutionary Dynamics, chapter 6 — the assigned reading behind this lecture.)
Neutral dynamics: a random walk that looks biased and isn't¶
Suppose \(A\) and \(B\) have equal fitness — the neutral case. Then the probability that an individual is chosen to reproduce is just proportional to how many of that type there are: an \(A\) is chosen to reproduce with probability \(i/N\), a \(B\) with probability \((N-i)/N\).
Going from \(i\) to \(i+1\) requires two things at once: an \(A\) chosen to reproduce and a \(B\) chosen to be replaced:
Going from \(i\) to \(i-1\) requires the mirror event, a \(B\) reproducing and an \(A\) being replaced:
These are the same product written in the other order, so they are exactly equal for every value of \(i\) — even though the population need not be split evenly between \(A\) and \(B\). This is the kind of fact that is easy to prove and hard to feel: if \(i/N=1/3\), it still doesn't matter, because the imbalance in who is likely to reproduce is exactly cancelled by the imbalance in who is likely to be replaced. So \(i\) performs an unbiased random walk, with two absorbing boundaries at \(i=0\) (only \(B\) left) and \(i=N\) (only \(A\) left): once the population hits either wall, it stays there.
Fixation under neutrality: an argument that needs no algebra¶
Given that \(i\) random-walks between two absorbing boundaries, what is the probability that \(A\) eventually fixes (takes over the whole population), starting from \(i\) copies?
The algebra is not needed for this one. Imagine tagging every individual with its own distinct colour, rather than lumping them into just two types. Because \(N\) is fixed forever, the lineage of some individual will eventually fill the entire population — someone's descendants must eventually occupy all \(N\) slots, purely as a consequence of the random walk hitting a boundary. By symmetry, since every individual is equally likely (under neutrality) to be that eventual ancestor, each one has probability exactly \(1/N\) of being it. Summing over the \(i\) individuals that happen to carry type \(A\):
That the answer depends on \(i/N\) so simply, when the underlying dynamics look symmetric in a different way (recall the two transition probabilities above were equal, independent of \(i\)), is exactly the kind of result that feels obvious and confusing at the same time — both things can be true together.
This is the same style of argument behind why a single, non-recombining lineage — the Y-chromosome, say, or mitochondrial DNA, neither of which recombines the way the rest of the genome does — is expected to trace back to a single common ancestor (colloquially, "Adam" or "Eve"): with no recombination shuffling lineages together, the population of gene copies along that line is exactly this random walk, and one lineage has to win. (With recombination, many ancestors contribute to any one descendant, and the argument no longer applies in the same form — sexually reproducing populations need a different treatment, not covered here.)
How much real time is one iteration?¶
One iteration of the Moran process is one birth and one replacement — and only one individual out of \(N\) actually gets to reproduce in that step. So if a whole population is to have had roughly one chance each to reproduce, that takes about \(N\) iterations. Equivalently: with \(N\) individuals each independently "waiting" to be the one to divide, a \(10\times\) larger population reaches its first division \(10\times\) faster. So the real time elapsed per iteration scales as \(1/N\), and one generation time (the typical interval between a cell's birth and its own division) corresponds to about \(N\) iterations of the model. This conversion is exactly what is needed to turn a target real-time duration — say, 100 hours of bacteria growing in a turbidostat — into a number of iterations to run in a simulation.
The molecular clock: the rate of neutral substitution¶
Now allow mutation. Let \(\mu\) be the probability that a birth event produces a mutant of a new, selectively neutral type — a mutation probability per birth, equivalently per generation. Two things must happen for a new neutral mutation to leave a permanent trace along a lineage: it must appear, and it must fix.
- Rate of appearance. In one generation (about \(N\) iterations, each with one birth), the expected number of new neutral mutants arising is \(\mu N\) — linear in population size, since more individuals means more birth events.
- Probability of fixation. A new mutant starts out as a single copy, \(i=1\), so by the neutral fixation result above its probability of eventually fixing is \(1/N\).
Multiplying, the rate at which neutral mutations fix — the rate of neutral molecular evolution — is
independent of population size, measured in substitutions per generation. This cancellation is the content of the (neutral) molecular clock: the two population-size dependences, one in how many mutants appear and one in how likely each is to fix, exactly cancel, leaving a rate governed only by the per-generation mutation rate. (The calculation implicitly assumes new mutants are rare enough that they don't have to compete against each other while fixing — a "separation of time scales" that breaks down once multiple lineages coexist, the subject of clonal interference, taken up later in the course.)
The model is too simple in one respect worth flagging: it predicts a substitution rate constant per generation, so organisms with the same \(\mu\) but very different generation times — humans and mice, say — should accumulate neutral substitutions at very different rates per year. That isn't what is observed; real substitution rates look closer to constant per year across quite different generation times. Something is missing from this simplest account, and the lecture flags it as a question to return to rather than resolving it here.
Reading the clock: how far back is that?¶
Treating the neutral substitution rate as roughly constant lets an observed count of neutral differences between two lineages be converted into an estimate of how long ago they diverged — this is the basis for molecular-clock dating. It helps to have a few memorized landmarks for the resulting timescales, so a new number has something to be compared against: the universe is about 13 billion years old, the earth about 4.5 billion years, life appears roughly a billion years after that, the dinosaurs disappear some 60-odd million years ago, agriculture begins about 12,000 years ago — and the human/chimpanzee divergence, argued over but usually quoted within a factor of two of \(7\times10^6\) years, sits among these mid-scale numbers.
A sharper application: dating when humans started wearing clothes. Direct archaeological evidence (needles, clothed figurines) only reaches back about 30,000 years, and body hair was apparently lost around a million years ago — leaving a wide, otherwise unrecoverable gap. The trick that closes it: a species of louse specializes in living in human clothing, distinct from the one that lives in human hair, and it presumably could not have speciated before there was any clothing to colonize. Sequencing the two louse lineages and counting accumulated neutral differences gives a molecular-clock estimate for when they split — bounded, sensibly, between the archaeological floor of 30,000 years and the human/chimpanzee divergence of about 7 million years. One early estimate (a researcher at the Max Planck Institute) put it at about 70,000 years; a later study (a University of Florida group) put it at about 170,000 years. The exact number is still debated, but the method — turning a question with no physical record at all into a testable estimate, using nothing but DNA — is the point.
Beyond neutrality: the general fixation probability¶
Now suppose \(A\) has a genuine fitness advantage, captured by a relative fitness \(r=\) (fitness of \(A\))/(fitness of \(B\)): \(r>1\) means \(A\) is advantageous, \(r<1\) deleterious. Only the reproduction step is affected by \(r\); replacement stays uniformly random. Nowak's book gives — without deriving it here, since the derivation is the assigned reading behind this lecture and is not repeated in it — the probability that \(A\) eventually fixes, starting from \(i\) copies out of \(N\):
The formula is compact but nearly opaque on sight. The standard remedy — and a habit worth adopting for any formula derived by hand, not just this one — is to check it against every limit whose answer is already known:
- \(i=0\): no \(A\) individuals at all should never fix, whatever \(r\) is. Indeed \(r^{-0}=1\), so the numerator is \(0\).
- \(i=N\): already fixed stays fixed. Indeed \(r^{-N}\) cancels top and bottom, giving \(x_N=1\).
- \(r\to\infty\): an overwhelmingly fit mutant should fix for sure, given at least one copy. Indeed every \(r^{-k}\to 0\) for \(k>0\), so \(x_i \to 1/1 = 1\).
- \(r\to 1\) (recovering the neutral case): substituting \(r=1\) directly gives \(0/0\). Applying L'Hopital's rule — differentiating numerator and denominator with respect to \(r\), and only then taking the limit — gives \(i\,r^{-i-1}\) over \(N\,r^{-N-1}\), which at \(r=1\) is \(i/N\): exactly the neutral result derived above by the symmetry argument, recovered here as a special case of the general formula. The useful thing to notice is that this limit is not automatically small, even though naively plugging \(r=1\) into the un-simplified expression looks like it gives \(0\).
That every one of these checks passes is not a proof that the formula is right, but it is exactly the kind of sanity check that catches a wrong formula quickly, and is worth running on any result before trusting it.
Weak selection: when is a mutation "nearly neutral"?¶
Real fitness advantages are usually small — a paper assigned for the next lecture reports selection coefficients of order 1–3% for mutations that let bacteria do better in a new environment. Write the relative fitness as \(r=1+s\), where \(s\) is the selection coefficient: \(s>0\) beneficial, \(s<0\) deleterious, and \(|s|\ll 1\) is the case of interest. Take the fixation probability of a single new mutant, \(x_1\), and consider two regimes.
Strong enough selection (\(sN\gg1\)). If the population is large enough that \(r^N\gg1\), the denominator \(1-r^{-N}\) is essentially \(1\), so
A beneficial mutation with a 1–3% advantage therefore has only a 1–3% chance of ever fixing — most copies of even a genuinely beneficial mutation are lost to chance while still rare. (The constant in front of \(s\) is model-dependent: other stochastic models of the same idea, such as branching-process approximations, can give \(2s\) instead. What's robust across models is that the fixation probability is of order \(s\), not of order \(1\).)
Weak relative to drift (\(sN\ll1\), "nearly neutral"). Expanding the exact formula for small \(s\) gives
A mutation whose fitness effect is small enough that \(|s|N\ll1\) behaves, for practical purposes, just like a truly neutral one — the population is too small, or the effect too weak, for selection to move the fixation probability far from its neutral baseline \(1/N\). This is the precise sense in which "nearly neutral" is defined: not that \(s\) is literally zero, but that \(|s|N\ll1\).
Plotting \(x_1\) against \(s\) makes the comparison visible at once: near \(s=0\) the curve sits at height \(1/N\) with a shallow local slope of \(s/2\) — half as steep as the line \(x_1=s\) it eventually approaches — and on the deleterious side it bends down and decays toward \(0\) once \(|s|N\gg1\), but does not reach it instantly. That last point — a deleterious mutation's fixation probability shrinks but never vanishes — is the content of the next section.
Deleterious mutations can fix too: Muller's ratchet¶
The formula above says a deleterious mutation (\(s<0\)) has small fixation probability when \(|s|N\gg1\), but not zero — and it is not even small once \(|s|N\) is only of order \(1\) or below, which happens whenever the population is small enough. Small populations are worse "filters" for selection: a mildly harmful mutation that a huge population would reliably purge can drift to fixation in a small one.
This is exploited deliberately in a mutation accumulation assay: grow a population of bacteria (so ordinary selection is acting — faster dividers are, in general, outcompeting slower ones), then plate it out into colonies, each founded by a single cell. Pick one colony at random, grow it up, and repeat. Because the colony propagated at each step is chosen at random rather than by how well it grew, the effect of selection is essentially removed across steps — a colony carrying a fitness-reducing mutation is just as likely to be picked as any other. Passing a population through this kind of single-cell bottleneck at every step drives its effective population size, \(N_{\text{effective}}\), down toward \(1\), regardless of how large the population grows in between bottlenecks — a fluctuating population's effective size, for questions about drift, is generally dominated by its smallest point, not its average. With \(N_{\text{effective}}\) this small, \(|s|N_{\text{effective}}\) is small for essentially any deleterious mutation, and deleterious mutations accumulate steadily rather than being filtered out.
The general phenomenon — that deleterious mutations can accumulate and fix in an asexual lineage, degrading its fitness over time, especially in small or bottlenecked populations — is called Muller's ratchet. It matters beyond the laboratory assay: it is one of the proposed explanations for why sexual reproduction persists despite its steep cost (an asexual population can, in principle, grow about twice as fast as a sexual one, since in the sexual case only half the population — the females — gives birth). The idea is that recombination in a sexual population can separate a beneficial mutation from a deleterious one that happened to arise on the same lineage, letting selection act on each independently — something a purely asexual population, whose lineages cannot exchange material, cannot do. The lecture flags this as an open question the field argues about, to be developed more quantitatively later in the course.
Sources¶
- Recording:
recordings/klrpm-beeoi-captions.srt, converted todocs/computational-biology/mit-ocw/8591j-2014/recordings/recordings/klrpm-beeoi.md— MIT 8.591J Systems Biology, Fall 2014, transcript timestamps [00:00]–[1:20:13]. This lecture is transcript-only: nothing written on the board survives in the source, so the fixation-probability formula, both figures, and the human-history timeline are reconstructed here from what the lecturer said about them, not copied from a slide or board. - Assigned reading referred to throughout but not supplied to this chapter: Martin Nowak, Evolutionary Dynamics, chapter 6 — the source of the Moran-process fixation formula \(x_i=(1-r^{-i})/(1-r^{-N})\) and of its derivation, which the lecture explicitly declines to repeat ([58:32]).
- A paper assigned for the following lecture, on selection coefficients of 1–3% in bacterial adaptation to a new environment, referred to at [1:06:32] but not identified further or supplied.
- Two lice-genomics studies used as a molecular-clock worked example ([54:00]–[57:28]): one from a researcher at the Max Planck Institute (~70,000 years), one from a University of Florida group (~170,000 years). Neither is identified by author or full citation in the transcript.
- Topics named as coming later in the course and not developed here: clonal interference ([41:48]), and the quantitative treatment of Muller's ratchet and the evolution of sex ([1:19:06]–[1:20:13]).
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