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.

4. When Does Clonal Interference Matter¶

What this covers¶

This chapter picks up right where the Moran-process fixation formula left off and asks what changes when two mutant lineages are alive in the population at once. A set of in-class quantitative questions builds the idea of when a beneficial mutation has become established — safely past the risk of early chance extinction — and that idea is then used to derive the two competing time scales that decide whether one lineage's fate is affected by another's at all: this is clonal interference. The second half applies the same machinery to a real measurement problem — why the distribution of effect sizes of beneficial mutations is much harder to measure than it sounds, for a reason that turns out to have nothing to do with competition between lineages. It assumes the Moran model and the fixation-probability formula from the previous lecture, $x_i = \dfrac{1-r^{-i}}{1-r^{-N}}$, where $r$ is a mutant lineage's fitness relative to the rest of the population, $N$ is the (fixed) population size, and $i$ is how many copies of the mutant are currently present.

Recap: fixation probability in the Moran model¶

$x_i$ is the probability that a lineage currently at $i$ copies eventually takes over the whole population of $N$. Writing $s = r - 1$ for the mutant's selective advantage, this formula has two limits worth having on hand for everything below:

Neutral ($s = 0$): $x_i \to i/N$, and in particular a single neutral mutant ($i=1$) fixes with probability $1/N$.
Moderately beneficial ($s > 0$ and $Ns \gg 1$): $x_1 \to s$.

Both limits come from the same formula; the second is only valid once $Ns \gg 1$, and forgetting that condition is the first trap below.

A single mutant against a single mutant, small population¶

Take a population of $N=10$, with the other eight individuals at fitness $1$, and introduce one copy each of two mutants: $A$ with $r_A = 1.01$ (slightly beneficial, $s_A = 0.01$) and $B$ with $r_B = 0.99$ (slightly deleterious, $s_B = -0.01$). Neither new mutation is allowed to appear once the clock starts. What is the probability that $A$ fixes?

The tempting shortcut is to plug $s_A = 0.01$ straight into the "moderately beneficial" limit and answer $1\%$. That answer is nonsensical, and the reason is instructive: it would say that $A$ fixes less often than a neutral mutation would ($1/N = 10\%$), even though $A$ is (weakly) beneficial. The $x_1 \to s$ limit is only the right tool when $Ns \gg 1$; here $s_A N = 0.01 \times 10 = 0.1$, which is small, not large. $A$ is not moderately beneficial at this population size — it is nearly neutral, the regime $|s|N \ll 1$, in which the fixation probability is (to good approximation) the neutral value regardless of the sign of $s$: $$x_1(A) \approx x_1(B) \approx \frac1N = 0.1.$$

The same population size that makes $s_A N = 0.1$ small would not do so for a much larger population: with a million individuals, an $s=0.01$ mutation would be solidly in the moderately beneficial regime instead. Whether a given selective advantage counts as "detectable" by drift is a statement about $sN$, not about $s$ alone.

This resolves a natural piece of intuition, that $A$ and $B$ should fix with roughly the same probability by symmetry (both nearly neutral, one just slightly on each side of neutral) — that part is right. What that intuition misses is that $A$ and $B$ fixing are not the only two outcomes. In fact, in the fully neutral limit, every one of the $N$ individuals present — the two mutants and the eight ordinary ones — is equally likely to be the eventual common ancestor of the whole population, each with probability $1/N$. $A$ and $B$ being nearly neutral just means they inherit this same $1/N$ share; the remaining $8/N = 80\%$ belongs, in total, to the eight individuals that are not mutants at all. The "50/50 between $A$ and $B$" answer is wrong not because the two are unequal, but because it silently assumes one of the two mutants has to win.

A single mutant against a single mutant, large population¶

Now change the numbers: $N = 10^6$, and two single mutants, $A$ with $r_A = 2$ and $B$ with $r_B = 1.01$ ($s_B = 0.01$). What is the probability that $B$ fixes?

First, a contrast worth keeping in mind throughout: if this population were instead modeled by a deterministic differential equation — no stochasticity at all — whichever mutant has the larger fitness would win with probability $1$, full stop. $A$ would fix, always. The entire content of this section is where that certainty breaks down once the dynamics are stochastic at low copy number.

For $B$ to fix, two things both have to happen: $B$ has to survive its own early stochastic extinction, and $A$ has to fail to survive its own. (If $A$ survives, it establishes and then spreads essentially deterministically, since it is far more fit — it wins.) So $$P(B \text{ fixes}) \approx P(B \text{ survives}) \times P(A \text{ does not survive}).$$

$B$ is moderately beneficial here ($s_B N = 0.01 \times 10^6 = 10^4 \gg 1$), so $P(B \text{ survives}) \approx x_1(B) \approx s_B = 0.01$.

$A$ is not a small perturbation ($s_A = 1$ is not small), so the $x_1 \to s$ shortcut does not apply to it at all — the exact formula must be used instead: $$x_1(A) = \frac{1 - r_A^{-1}}{1 - r_A^{-N}} \approx 1 - \frac12 = \frac12,$$ since $r_A^{-N} = 2^{-10^6}$ is utterly negligible. So $P(A \text{ does not survive}) = \tfrac12$.

Multiplying, $$P(B \text{ fixes}) \approx 0.01 \times 0.5 = 0.005 = 0.5\%.$$

Multiplying the two probabilities together treats $A$'s survival and $B$'s survival as independent events. That is justified here specifically because the population is enormous: while both lineages are still rare, each one is overwhelmingly interacting with the sea of $10^6$ ordinary individuals, not with each other, so to a very good approximation they do not "see" one another yet. This has nothing to do with how close $r_A$ and $r_B$ are — it would hold even if both mutants had exactly the same fitness. What it does depend on is that this is only true of the early, still-rare phase: if both $A$ and $B$ happen to survive stochastic extinction and both start spreading, the independence assumption breaks down and the two lineages genuinely compete — that joint event is itself rare (each survives with probability of order $s$, so the joint probability is of order $s_A s_B$), but it is exactly the event that clonal interference, discussed below, is about.

As a check: the three exhaustive outcomes — $A$ fixes, $B$ fixes, or both go extinct and the already-dominant background remains — should sum to $1$: $0.5 + 0.005 + (0.5)(0.99) \approx 1$. ✓.

Starting from ten copies instead of one: what "established" means¶

Same population, same two fitnesses, but now $10$ copies of $A$ and $10$ copies of $B$ appear at once instead of one copy each. Does the answer change much?

Redo the same calculation using $x_{10}$ instead of $x_1$. For $B$: $$x_{10}(B) = 1 - r_B^{-10} = 1 - 1.01^{-10} \approx 1 - \frac{1}{1.1} \approx 0.1,$$ using $(1+x)^n \approx 1+nx$ for small $nx$. Starting with ten copies instead of one raises $B$'s own survival chance tenfold, from $1\%$ to about $10\%$ — consistent with $x_i \approx is$ while $is \ll 1$.

For $A$: $$x_{10}(A) = 1 - r_A^{-10} = 1 - 2^{-10} \approx 1 - \frac{1}{1024} \approx 0.999,$$ so $P(A \text{ does not survive}) \approx 2^{-10} \approx 10^{-3}$ — starting $A$ at ten copies makes its extinction a thousand-to-one long shot rather than a coin flip.

\[P(B \text{ fixes}) \approx 0.1 \times 10^{-3} = 10^{-4}.\]

This is the opposite of what a naive "bigger head start helps" intuition suggests: giving both lineages the same tenfold head start actually makes $B$'s overall fixation probability roughly $50$ times smaller (from $0.5\%$ down to $0.01\%$), because it barely moves $B$'s survival chance but almost eliminates $A$'s risk of extinction. The reason the two respond so differently to the same head start is the real content of this problem: how much extra survival probability an additional copy buys you depends on how far $i$ already is from a characteristic threshold, $1/s$. For $B$ ($s_B = 0.01$), that threshold is around $100$ copies, and $i=10$ is still well below it — $B$ is still deep in the regime where each extra copy helps roughly linearly. For $A$ ($s_A = 1$), the threshold is around $1$ copy, and $i=10$ is already far past it — $A$ is essentially already safe.

That threshold is worth deriving properly, because it is the general notion of an established mutation: one that has grown large enough that further stochastic extinction is unlikely, so from that point on its spread is well described by a deterministic exponential, not by chance. Dropping the $r^{-N}$ term as before, $$1 - x_i \approx r^{-i} = (1+s)^{-i} \approx e^{-is}$$ for small $s$, using $\ln(1+s) \approx s$. This extinction probability is small once $is \gg 1$, so the number of copies needed to be established is $$n_{\text{established}} \sim \frac1s.$$ At exactly $i = 1/s$, the extinction probability is $e^{-1} \approx 0.37$ — the crossover point where a lineage becomes more likely than not to survive. For the beneficial mutations discussed later in this chapter, with $s$ of a few percent, $1/s$ is a few tens, so $n_{\text{established}}$ is of order a hundred individuals.

Two competing time scales, and when clonal interference can be ignored¶

Once a lineage is established, it grows roughly deterministically, from $n_{\text{established}}$ up to the full population $N$, at exponential rate $s$. That takes a time $$T_{\text{establish} \to \text{fix}} \sim \frac1s \ln\!\left(\frac{N}{n_{\text{established}}}\right) \sim \frac1s\ln(Ns).$$ The time to become established in the first place is comparatively short and does not much affect this, because it is a highly biased sample over trajectories — most attempts at establishment fail quickly, and the rare ones that do not, establish quickly too.

The other relevant clock is how often new established mutations appear at all. Let $\mu$ be the per-individual, per-generation probability of acquiring a beneficial mutation of the (single, for simplicity) magnitude $s$ under discussion. New mutations of this kind arise in the whole population at rate $\mu N$, so the time between successive appearances is exponentially distributed with mean $1/(\mu N)$. But only a fraction of order $s$ of those appearances go on to establish — the rest go extinct — so the time between successive established mutations is exponentially distributed with the longer mean $$T_{\text{mutation, established}} \sim \frac{1}{\mu N s}.$$ (Both $s$ and $\mu$ are already rates per generation, so both time scales above come out in units of generations — there is no unit mismatch, even though it can look that way at first.) As a concrete illustration, for $s = 0.02$ one has to wait for around $1/s = 50$ appearances of this mutation, on average, before one of them happens to be the lucky one that establishes.

Clonal interference is what happens once two lineages are both established and spreading at the same time: growing exponentially, a fitter lineage can catch up to and overtake a lineage that established earlier but is spreading more slowly, so the first mutation never gets to fix before being out-competed by the second.

Wild type A gives way as mutant B establishes and spreads; before B can fix, a fitter mutant C establishes on top of the same background and overtakes B in turn. Neither B nor A ever fixes — this competition between simultaneously-spreading lineages is clonal interference.

No clonal interference means the two time scales are far apart — mutations are established, spread, and fix one at a time, well-separated: $$T_{\text{mutation, established}} \gg T_{\text{establish}\to\text{fix}} \quad\Longleftrightarrow\quad \mu N \ll \frac{1}{\ln(Ns)},$$ which, up to the slowly-varying logarithm, is the familiar criterion $\mu N \ll 1$: the population-wide rate of new mutations, per generation, must be much less than one.

Measuring the distribution of fitness effects¶

This machinery was built to make sense of an experimental question: given a population dropped into a new environment, what is the probability distribution of the sizes of the beneficial mutations available to it?

Picture a thought experiment: take an E. coli genome (a few million base pairs, depending on strain) and make every possible point mutation — on the order of ten million distinct single-mutant strains — then measure each one's growth rate $\gamma$ relative to wild type $\gamma_{\text{wt}}$ in some fixed environment. What should the histogram of $\gamma/\gamma_{\text{wt}}$ look like? Two naive extremes are both wrong: it is not the case that almost nothing matters (a spike exactly at $1$), nor that almost everything is lethal (a spike at $0$). What is known independently — from whole-gene knockout screens, a much harsher perturbation than a single point mutation — is that only some $10$–$20\%$ of genes are individually essential, and only a small fraction of point mutations within a gene actually knock its function out, and not all of the genome even codes for protein; so the lethal fraction of point mutations is small (very roughly $10^{-4}$ to $10^{-3}$, an order of magnitude, not a measurement). The bulk of the distribution is a sharp peak at $\gamma/\gamma_{\text {wt}} = 1$ (most point mutations have no measurable effect at all), with a small tail down toward $0$ (rare lethal mutations) and — the part of interest here — a much smaller, even more tightly clustered, tail above $1$: the beneficial mutations. Writing $s = \gamma/\gamma_{\text{wt}} - 1$ and zooming into just that tail gives a density $p(s)$ for $s>0$: many mutations nearly neutral, falling off in some way as $s$ grows. This is the quantity a paper referred to in the lecture only as "Roy's paper" set out to measure directly, by evolving replicate populations in a new environment and tracking which beneficial mutations actually took hold.

The paper's puzzle was that several qualitatively different candidate shapes for $p(s)$ — for example a single fixed effect size, a uniform spread of effects, or an exponential tail — could each be tuned (by choosing a mutation rate and a mean effect size) to reproduce the same observed data. Bulk measurements of this kind cannot, by themselves, tell the shapes apart; the lecture calls this the equivalence principle, and returns to it in more detail in the next lecture.

A sharper question, though, is this: suppose clonal interference is eliminated entirely — take $\mu \to 0$, or work in a small enough population that only ever one beneficial mutation is present at a time. Would the distribution of mutations you then see spreading finally equal the true $p(s)$? The answer is no, and the reason has nothing to do with competition between lineages at all: even a mutation with no competitors still has to survive its own stochastic extinction before you ever see it spread, and the probability of surviving is proportional to $s$. So what you observe is not $p(s)$ but something shaped like $s\,p(s)$: small effects are thrown away disproportionately often, purely by early-generation bad luck, regardless of whether any other mutation is around to compete with. If the true $p(s)$ is, say, monotonically decreasing and maximal at $s=0$ (an exponential, for instance — there are arguments from extreme value theory for something like this shape), the observed, survival-weighted distribution $s\,p(s)$ starts at $0$, rises, and only then falls: it is peaked at some nonzero value of $s$, even with zero clonal interference. Add clonal interference back in and the effect compounds: what spreads is now (roughly) the larger of several independent draws from that already survival-biased distribution, which pushes the observed peak further to the right and makes it narrower and taller still.

The true effect-size distribution p(s) is maximal at s=0 (shaded). What is actually observed spreading is size-biased toward larger s even with no competing lineages, because only mutations that survive early stochastic extinction are ever seen, and survival probability scales with s; clonal interference (taking the winner among several established competitors) shifts the observed peak further right and sharpens it.

So there are two entirely separate reasons the measured distribution of fitness effects is not the true one: the survival-bias effect above, which is present even for a single mutation with no competition, and clonal interference proper, which only matters once more than one lineage establishes at a time. Untangling how much of the "equivalence principle" puzzle is due to each is exactly where the discussion picks up in the next lecture.

Sources¶

Recording: recordings/6pxncdxixne-captions.srt, converted to docs/computational-biology/mit-ocw/8591j-2014/recordings/recordings/6pxncdxixne.md — MIT 8.591J Systems Biology, Fall 2014, transcript timestamps [00:00]–[1:20:02]. This lecture is transcript-only: nothing written on the board or shown on a slide survives in the source. In particular, the exact wording and numeric options of the in-class multiple-choice questions, any board work for the arithmetic in the worked examples, and the two schematic drawings described at [45:48]–[46:56] (lineages spreading and being out-competed) and [1:05:44]–[1:20:02] (shapes of the fitness-effect histogram) are not in the transcript; both figures above are reconstructed here from what the lecturer said about them, not copied from a board or slide.
The Moran-process fixation-probability formula and its two limits are recapped from "last time" ([02:17]–[04:35]) but were derived in the previous lecture, which is not part of the material supplied for this chapter.
A paper referred to throughout only as "Roy's paper" ([01:10] onward), on measuring the distribution of effects of beneficial mutations in E. coli and introducing what the lecture calls the "equivalence principle" — not identified further (no title, author spelling, or venue is audible in the transcript).
Martin Nowak, Evolutionary Dynamics, chapter 6, referred to again at [35:26] as the source of the fixation formula being used ("that was the equation that was derived in chapter six") — not supplied to this chapter.
Michael Desai's work on evolving yeast populations with high-resolution lineage tracking, mentioned at [46:56] as "maybe Nature 2012, '13" — not identified further.
Charlie Boone's (Toronto) genome-wide pairwise gene-knockout screen in yeast, mentioned at [1:05:44] — not identified further.
Topics flagged as continuing "on Tuesday" and not developed further here: more discussion of "Roy's paper," and fitness landscapes and the rate of evolution ([1:20:02]).

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