Eventual fixation probability: t → ∞ solution to backward diffusion equation (w/ no selection or mutation) In the last slide, it looked like P(x=1|x o,,t)≡P fix (x o,t) was tending toward a linear function as t→∞. We prove this below… The key is to note that ∂P/∂t → 0 as t→∞. This is exactly the same argument we made when analyzing the continuous-time branching process (back then, we said P extinct (t=∞) = P extinct (t=∞+dt) ). If x o ≠ 0 and x o ≠ 1, we can divide both sides by N/(x o (1-x o )), resulting in This is a super-easy differential equation to solve. It’s best done “by inspection.” The eq. says that the second derivative = 0. And second derivative is like curvature. So P fix (x o,t→∞) must be a line: “P fix = mx o + b”. (I got tired of writing t→∞) But we also know the “boundary conditions”: P fix (x o =0)=b=0 and P fix (x o =1)=m+b=m=1. So, P fix (x o, t→∞) = x o

Interpretation and consequence of P fix =x o 30 second clicker question: what is the most interesting value of x o ? answer: x o =1/N (b/c new alleles originate as a single copy). In this case, P fix =1/N. Interpretation: After infinite time, one of the N lineages will have taken over the population, and the other N-1 lineages gone extinct. In the neutral case, all N lineages are equivalent (in terms of fitness). So, by symmetry, the probability that a new allele is the lucky one (that takes over the population)= 1/N. Consequence: few minute clicker question: Suppose that every time a genome is duplicated, U neut neutral mutations are introduced, on average. Each generation, how many mutations are introduced in the whole population? (assume we’re dealing with unicellulars, like bacteria) What fraction of the new, neutral mutations will eventually achieve fixation? Multiply your answer to obtain the substitution rate of neutral mutations. The exercise you just did has played a huge role in population genetics. It says that the tempo of evolutionary change is independent of everything except U neutral. So, U neutral sets the pace of a “molecular clock” that can date evolutionary divergence.

-- Warning: do not use forward diffusion equation to calculate P fix -- In the last few slides, we calculated P fix using the backward equation. Another approach that might have occurred to you is the following: set up the forward equation for t→ ∞ : solve this equation plug in x=1 to the solution This approach should work, but it is in fact unproductive The reason has to do with the fact that probability piles up at x=0 and x=1 (see previous slide: ** Solutions to forward equation: a mathematical portrait of genetic drift ** FYI, these “spikes” that pile up at x=0 and x=1 are called “Dirac delta functions” even though, technically, they’re not functions. They are infinitely high and infinitessimally thin.

Backward diffusion equation for Moran model with selection (but no mutation) Part 1 Recall, there is a recipe for setting up a diffusion equation: calculate 〈 (Δx) 2 〉 and 〈 Δx 〉. Then plug them into either: backward equation: or, forward equation: In both cases, I’ve abbreviated P(x|x o, t) as P. In forward equation, write 〈 (Δx) 2 〉 and 〈 Δx 〉 as functions of x (e.g. x(1-x)) In backward equation, write 〈 (Δx) 2 〉 and 〈 Δx 〉 as functions of x 0 (e.g. x 0 (1-x 0 ))

Back to the Moran transition matrix T(x|x 0 ), now with selection Birth Death A: (1+s)  x 0 a: (1-x 0 ) A: (x 0 ) (1+s)  x 0 2 T - (x) = x 0 (1-x 0 ) a: (1-x 0 ) T + (x) = (1+s)  x 0 (1-x 0 ) (1-x 0 ) 2 T + (x) = T(x|x 0 = x-1) = (1+s)  x 0 (1-x 0 ) T - (x) = T(x|x 0 = x+1) = x 0 (1-x 0 ) = (1+s)  x 0 (1-x 0 )/N – x 0 (1-x 0 )/N = sx 0 (1-x 0 )/N = (1+s)  x 0 (1-x 0 )/N 2 + x 0 (1-x 0 )/N 2 = (r A + 1)x 0 (1-x 0 )/N 2 = (2+s)x 0 (1-x 0 )/N 2 ≈ 2x 0 (1-x 0 )/N 2 Food for thought: do the entries in the above table sum to 1? Should we care? (This isn’t T, just the path to it.) Same as for drift alone

new term: selection drift: same as before This leads to the following backward diffusion equation: Backward diffusion equation for Moran model with selection (but no mutation) Part 2 As before, this can be solved analytically but it’s very complicated. Most intuition can be gained by looking at the t → ∞ limit (setting ∂P/∂t = 0) = (1+s)  x 0 (1-x 0 )/N – x 0 (1-x 0 )/N = (r A – 1)x 0 (1-x 0 )/N = sx 0 (1-x 0 )/N = (2+s)x 0 (1-x 0 )/N 2 ≈ 2x 0 (1-x 0 )/N 2 Also as before, we’ve rescaled time in units of generations by multiplying by 1/N

** Eventual fate of a new mutation under selection and drift (part 1)** If x o ≠ 0 and x o ≠ 1, this simplifies to: As far as differential equations go, this one is easy to solve this, especially if you (perhaps just mentally) make the substitution Q=dP/dx o. Then, This is the easiest differential equation of them all: the solution is an exponential:, where C is a “constant of integration” that we’ll deal with later. Now let’s get back to P, which was defined by dP/dx o = Q: This equation can be solved by integrating both sides: k, like C, is just a constant of integration that we’ll deal with in a minute…

** Eventual fate of a new mutation under selection and drift (part 2)** So, we’re most of the way there: Much like we determined the constants “m” and “b” when we solved the neutral case, we need to determine the constants “C” and “k” by considering the boundary conditions: 30 second clicker question: 1. P(x o =0) = ?. 2. What is P(x o =1) = ?. In your HW, you’ll be asked to work through the algebra of determining C and k. For now, we’ll just quote the final answer: 1 minute clicker question: Show that the equation above satisfies the boundary conditions. “Kimura’s formula”

N = 10 N = 25 N = beneficial s>0. denominator also > 0 a.Ns >>1: Pfix ≈ s. s<<1. N doesn’t matter. Selection dominates. But it’s subtle b/c Pfix<<1, still. b.Ns <<1: Pfix ≈ s/(Ns) = 1/N: neutral case is recovered from this more general formula. Drift dominates 2.deleterious s<0. denominator also < 0 a.|Ns|>>1: Pfix ≈ exp(-Ns): super negative. Purifying selection dominates. b.|Ns|<<1: Pfix ≈ s/(Ns-1/2N 2 s 2 ) = 1/N * 1/(1+|Ns|) ≈ 1/N: mildly deleterious allele almost as likely to fix as neutral one. Drift dominates. ** Interpreting Kimura’s formula** P fix ≈ s when s > 1)

Punchlines from Beneficial mutations don’t always fix – Even in the limit of N -> ∞, P fix ≈ s so long as Ns << 1. That’s because here, the stochasticity is being driven not by the total population size but by the population size of beneficial mutants. – There’s a “sticky barrier” in frequency space, x sticky x sticky dynamics are deterministic. – Reminiscent of the establishment probability result from Unit 1. Deleterious mutations aren’t always purged – Such “mildly deleterious” mutations can play an important role in evolution, as we’ll see later on.