Some mathematical models from population genetics 5: Muller s ratchet and the rate of adaptation Alison Etheridge University of Oxford joint work with Peter Pfaffelhuber (Vienna), Anton Wakolbinger (Frankfurt) Charles Cuthbertson (Oxford), Feng Yu (Bristol) New York, Sept. 07 p.1
Muller s ratchet Asexually reproducing population. All mutations deleterious. Chromosomes passed down as indivisible blocks. New York, Sept. 07 p.2
Muller s ratchet Asexually reproducing population. All mutations deleterious. Chromosomes passed down as indivisible blocks. Number of deleterious mutations accumulated along any ancestral line can only increase. New York, Sept. 07 p.2
Muller s ratchet Asexually reproducing population. All mutations deleterious. Chromosomes passed down as indivisible blocks. Number of deleterious mutations accumulated along any ancestral line can only increase. When everyone in the current best class has accumulated at least one additional bad mutation the ratchet clicks. New York, Sept. 07 p.2
Muller s ratchet Asexually reproducing population. All mutations deleterious. Chromosomes passed down as indivisible blocks. Number of deleterious mutations accumulated along any ancestral line can only increase. When everyone in the current best class has accumulated at least one additional bad mutation the ratchet clicks. How many generations will it take for an asexually reproducing population to lose its best class? New York, Sept. 07 p.2
Haigh s model Wright-Fisher model: Individuals in (t + 1)st generation select parent at random from generation t. Probability individual which has accumulated k mutations is selected as parent proportional to relative fitness (1 s) k. Number of mutations carried by offspring then k + J, where J Poiss(λ) (independent). New York, Sept. 07 p.3
Type frequencies: x(t) = (x k (t)) k=0,1,... Nx k (t) = No. of individuals carrying exactly k mutations at time t. New York, Sept. 07 p.4
Type frequencies: x(t) = (x k (t)) k=0,1,... Nx k (t) = No. of individuals carrying exactly k mutations at time t. H an N 0 -valued random variable. P[H = k] (1 s) k x k (t), New York, Sept. 07 p.4
Type frequencies: x(t) = (x k (t)) k=0,1,... Nx k (t) = No. of individuals carrying exactly k mutations at time t. H an N 0 -valued random variable. J Poiss(λ) independent of H. P[H = k] (1 s) k x k (t), New York, Sept. 07 p.4
Type frequencies: x(t) = (x k (t)) k=0,1,... Nx k (t) = No. of individuals carrying exactly k mutations at time t. H an N 0 -valued random variable. P[H = k] (1 s) k x k (t), J Poiss(λ) independent of H. K 1, K 2,..., K N independent copies of H + J. Random type frequencies in next generation are X k (t + 1) = 1 N #{i : K i = k}. New York, Sept. 07 p.4
Infinite populations As N, LLN x(t + 1) = E x(t) [X(t + 1)]. New York, Sept. 07 p.5
Infinite populations As N, LLN x(t + 1) = E x(t) [X(t + 1)]. Suppose x(t) Poiss(α). P[H = k] (1 s) k x k = (1 s) k αk e α. k! Then H Poiss(α(1 s)), J Poiss(λ), so H + J Poiss(α(1 s) + λ). New York, Sept. 07 p.5
Infinite populations As N, LLN x(t + 1) = E x(t) [X(t + 1)]. Suppose x(t) Poiss(α). P[H = k] (1 s) k x k = (1 s) k αk e α. k! Then H Poiss(α(1 s)), J Poiss(λ), so H + J Poiss(α(1 s) + λ). Poisson weights Poisson weights. New York, Sept. 07 p.5
For every initial condition with x 0 > 0, the solution to the deterministic dynamics converges as t to the stationary point π := Poiss(λ/s). New York, Sept. 07 p.6
Back to finite populations Write k for the number of mutations carried by individuals in fittest class. New York, Sept. 07 p.7
Back to finite populations Write k for the number of mutations carried by individuals in fittest class. forms a recurrent Markov chain. Y := (Y k ) k=0,1,... := (X k +k ) k=0,1,2... New York, Sept. 07 p.7
Back to finite populations Write k for the number of mutations carried by individuals in fittest class. Y := (Y k ) k=0,1,... := (X k +k ) k=0,1,2... forms a recurrent Markov chain. Condition on Y(t) = y(t). Size of new best class, y 0 (t + 1) Binom(N, p 0 (t)), with p 0 (t) probability of sampling parent from best class and not acquiring any additional mutations: p 0 (t) = y 0(t) W (t) e λ, W (t) = y i (t)(1 s) i. i=0 New York, Sept. 07 p.7
Back to finite populations Write k for the number of mutations carried by individuals in fittest class. Y := (Y k ) k=0,1,... := (X k +k ) k=0,1,2... forms a recurrent Markov chain. Condition on Y(t) = y(t). Size of new best class, y 0 (t + 1) Binom(N, p 0 (t)), with p 0 (t) probability of sampling parent from best class and not acquiring any additional mutations: p 0 (t) = y 0(t) W (t) e λ, W (t) = y i (t)(1 s) i. i=0 Evolution of best class determined by W (t), the mean fitness in the population. New York, Sept. 07 p.7
Elements of Haigh s analysis Immediately after a click, the type frequencies are π := 1 1 π 0 (π 1, π 2,...). New York, Sept. 07 p.8
Elements of Haigh s analysis Immediately after a click, the type frequencies are π := 1 1 π 0 (π 1, π 2,...). Phase one: deterministic dynamical system dominates, decaying exponentially fast towards its equilibrium New York, Sept. 07 p.8
Elements of Haigh s analysis Immediately after a click, the type frequencies are π := 1 1 π 0 (π 1, π 2,...). Phase one: deterministic dynamical system dominates, decaying exponentially fast towards its equilibrium Phase two: the bulk of the population changes only slowly. Mean fitness assumed constant and then No. of individuals in best class approximated by Galton-Watson branching process with Poisson offspring distribution. New York, Sept. 07 p.8
The Fleming Viot diffusion Relevant for applications are: large N, small s, small λ. New York, Sept. 07 p.9
The Fleming Viot diffusion Relevant for applications are: large N, small s, small λ. Dynamics then captured by dx k = j s(j k)x j X k + λ(x k 1 X k ) dt + j k 1 N X jx k dw jk, k = 0, 1, 2,... where X 1 = 0 and (W jk ) j>k array of independent Brownian motions, W kj := W jk. New York, Sept. 07 p.9
The Fleming Viot diffusion Relevant for applications are: large N, small s, small λ. Dynamics then captured by dx k = j s(j k)x j X k + λ(x k 1 X k ) dt + j k 1 N X jx k dw jk, k = 0, 1, 2,... where X 1 = 0 and (W jk ) j>k array of independent Brownian motions, W kj := W jk. As before Y k = X k +k, dy 0 = s(m 1 (Y) λ)y 0 (t)dt + 1 N Y 0(1 Y 0 )dw 0, M 1 (Y) = j jy j. New York, Sept. 07 p.9
Infinite population limit with x 1 = 0. dx k = (s(m 1 (x) k) λ)x k + λx k 1 ) dt, k = 0, 1, 2,... New York, Sept. 07 p.10
Infinite population limit dx k = (s(m 1 (x) k) λ)x k + λx k 1 ) dt, k = 0, 1, 2,... with x 1 = 0. Transform into system of equations for cumulants: log x k e ξk = k=0 k=1 κ k ( ξ) k k!. Assume x 0 > 0 and set κ 0 = log x 0. Then κ k = sκ k+1 + λ, k = 0, 1, 2,... New York, Sept. 07 p.10
System can be solved. In particular, κ 1 (t) = kx k (t) = ξ log k=0 k=0 x k (0)e ξk ξ=st + λ s (1 e st ). New York, Sept. 07 p.11
System can be solved. In particular, κ 1 (t) = kx k (t) = ξ log k=0 k=0 x k (0)e ξk ξ=st + λ s (1 e st ). Notice the exponential decay towards the equilibrium vaue of λ/s. The time τ = log(λ/s) s corresponds exactly to the end of Haigh s phase one. New York, Sept. 07 p.11
Approximations dy 0 = s(m 1 (Y) λ)y 0 (t)dt + 1 N Y 0(1 Y 0 )dw 0. Cannot solve for M 1 (Y). Instead seek a good approximation of M 1 given Y 0. New York, Sept. 07 p.12
Approximations dy 0 = s(m 1 (Y) λ)y 0 (t)dt + 1 N Y 0(1 Y 0 )dw 0. Cannot solve for M 1 (Y). Instead seek a good approximation of M 1 given Y 0. Simulations suggest a good fit to a linear relationship between Y 0 and M 1. New York, Sept. 07 p.12
Extending Haigh s approach Haigh assumes at click times π 0 distributed evenly over other classes. Suppose now that this holds in between click times too: given Y 0 approximate state of system by the PPA (Poisson Profile Approximation) Π(Y 0 ) = ( Y 0, 1 Y ) 0 (π 1, π 2,...). 1 π 0 New York, Sept. 07 p.13
Extending Haigh s approach Haigh assumes at click times π 0 distributed evenly over other classes. Suppose now that this holds in between click times too: given Y 0 approximate state of system by the PPA (Poisson Profile Approximation) Π(Y 0 ) = ( Y 0, 1 Y ) 0 (π 1, π 2,...). 1 π 0 Estimate M 1 not from PPA but from relaxed PPA obtained by evolving PPA according to the deterministic dynamical system for time Aτ := A log(λ/s)/s. This gives M 1 = θ + η e η 1 ( 1 Y ) 0, η = (λ/s) 1 A. π 0 New York, Sept. 07 p.13
Three one dimensional diffusions Substituting in the one-dimensional diffusion approximation for Y 0 gives: A small, dy 0 = λ(π 0 Y 0 )Y 0 dt + ( A = 1, dy 0 = 0.58s 1 Y ) 0 π 0 A large, ( dy 0 = s 1 Y ) 0 Y 0 dt + π 0 1 N Y 0dW, Y 0 dt + 1 N Y 0 dw, 1 N Y 0 dw, New York, Sept. 07 p.14
A rescaling Z(t) = 1 π 0 Y 0 (Nπ 0 t). Set γ = Nλ Ns log(nλ). New York, Sept. 07 p.15
A rescaling Z(t) = 1 π 0 Y 0 (Nπ 0 t). Set γ = Nλ Ns log(nλ). A small, dz = (Nλ) 1 2γ (1 Z)Zdt + ZdW, A = 1, 1 dz = 0.58 γ log(nλ) (Nλ)1 γ (1 Z)Zdt + Z dw, A large, dz = 1 γ log(nλ) (Nλ)1 γ (1 Z)Zdt + Z dw. New York, Sept. 07 p.15
Implications A small fast clicking: dz = (Nλ) 1 2γ (1 Z)Zdt + ZdW. The ratchet never clicks for γ < 1/2. New York, Sept. 07 p.16
Implications A small fast clicking: dz = (Nλ) 1 2γ (1 Z)Zdt + ZdW. The ratchet never clicks for γ < 1/2. A = 1 (A large) moderate clicking (slow clicking): dz = 0.58(1) 1 γ log(nλ) (Nλ)1 γ (1 Z)Zdt + Z dw. New York, Sept. 07 p.16
Implications A small fast clicking: dz = (Nλ) 1 2γ (1 Z)Zdt + ZdW. The ratchet never clicks for γ < 1/2. A = 1 (A large) moderate clicking (slow clicking): dz = 0.58(1) 1 γ log(nλ) (Nλ)1 γ (1 Z)Zdt + Z dw. In order for 0.58 1 γ log(nλ) (Nλ)1 γ to be > 5, γ 0.3 0.4 0.5 0.55 0.6 0.7 0.8 0.9 Nλ 20 10 2 9 10 2 4 10 3 2 10 4 4 10 6 2 10 11 8 10 26 New York, Sept. 07 p.16
Rule of thumb For biologically realistic parameters, transition from no clicks to moderate clicks (on evolutionary timescale) around γ = 0.5. New York, Sept. 07 p.17
Rule of thumb For biologically realistic parameters, transition from no clicks to moderate clicks (on evolutionary timescale) around γ = 0.5. The rate of the ratchet is of the order N γ 1 λ γ for γ (1/2, 1), whereas it is exponentially slow in (Nλ) 1 γ for γ < 1/2. New York, Sept. 07 p.17
Simulations New York, Sept. 07 p.18
New York, Sept. 07 p.19
Purely beneficial mutations We saw before that the probability of an isolated selected sweep 2σ. New York, Sept. 07 p.20
Purely beneficial mutations We saw before that the probability of an isolated selected sweep 2σ. What about overlapping sweeps? New York, Sept. 07 p.20
Purely beneficial mutations We saw before that the probability of an isolated selected sweep 2σ. What about overlapping sweeps? Interference reduces the chance of fixation as beneficial mutations compete with one another. New York, Sept. 07 p.20
Purely beneficial mutations We saw before that the probability of an isolated selected sweep 2σ. What about overlapping sweeps? Interference reduces the chance of fixation as beneficial mutations compete with one another. Is there a limit to the rate of adaptation? New York, Sept. 07 p.20
A mixture of mutations All mutations equal strength so ith individual s fitness characterized by net number, X i, of beneficial mutations. New York, Sept. 07 p.21
A mixture of mutations All mutations equal strength so ith individual s fitness characterized by net number, X i, of beneficial mutations. Mutation: For each individual i a mutation event occurs at rate µ. With probability 1 q, X i changes to X i 1 and with probability q, X i changes to X i + 1. Selection: For each pair of individuals (i, j), at rate σ N (X i X j ) +, individual i replaces individual j. Resampling: For each pair of individuals (i, j), at rate 1 N, individual i replaces individual j. New York, Sept. 07 p.21
A mixture of mutations All mutations equal strength so ith individual s fitness characterized by net number, X i, of beneficial mutations. Mutation: For each individual i a mutation event occurs at rate µ. With probability 1 q, X i changes to X i 1 and with probability q, X i changes to X i + 1. Selection: For each pair of individuals (i, j), at rate σ N (X i X j ) +, individual i replaces individual j. Resampling: For each pair of individuals (i, j), at rate 1 N, individual i replaces individual j. Technical modification: suppress mutations which would make the width of the population > L N 1/4. New York, Sept. 07 p.21
..., P 0 (t),..., P k (t),... - Proportion of individuals with k mutations at time t k max (k min ) - type of the fittest (least fit) individual New York, Sept. 07 p.22
..., P 0 (t),..., P k (t),... - Proportion of individuals with k mutations at time t k max (k min ) - type of the fittest (least fit) individual dp k = µ k (P ) dt + σ l Z(k l)p k P l dt + dm P k = [ µ k (P ) + σ(k m(p ))P k ] dt + dm P k µ k (P ) µ (qp k 1 P k + (1 q)p k+1 ) New York, Sept. 07 p.22
..., P 0 (t),..., P k (t),... - Proportion of individuals with k mutations at time t k max (k min ) - type of the fittest (least fit) individual dp k = µ k (P ) dt + σ l Z(k l)p k P l dt + dm P k = [ µ k (P ) + σ(k m(p ))P k ] dt + dm P k µ k (P ) µ (qp k 1 P k + (1 q)p k+1 ) M P is a martingale with [ ] M P 2µ k (t) N t + 1 N t 0 (2 + σ(k l) + + σ(l k) + )P k (s)p l (s) ds l Z New York, Sept. 07 p.22
Moments Mean fitness m(p ) = k kp k satisfies dm(p ) = ( µ(p ) + σc 2 (P )) dt + dm P,m µ(p ) µ(2q 1) New York, Sept. 07 p.23
Moments Mean fitness m(p ) = k kp k satisfies dm(p ) = ( µ(p ) + σc 2 (P )) dt + dm P,m µ(p ) µ(2q 1) Ignoring mutation terms, centred moments c k = k (k m(p))n P k satisfy dc 2 σc 3 dt + small noise terms dc 3 σ(c 4 3c 2 c 2 ) dt + small noise terms dc 4 σ(c 5 4c 3 c 2 ) dt + small noise terms dc 5 σ(c 6 5c 4 c 2 ) dt + small noise terms New York, Sept. 07 p.23
Heuristics Centred process has a stationary distribution. New York, Sept. 07 p.24
Heuristics Centred process has a stationary distribution. Approximate by a deterministic distribution. New York, Sept. 07 p.24
Heuristics Centred process has a stationary distribution. Approximate by a deterministic distribution. Equations for centred moments give: 0 σc 3 dt 0 σ(c 4 3c 2 c 2 ) 0 σ(c 5 4c 3 c 2 ) 0 σ(c 6 5c 4 c 2 )... New York, Sept. 07 p.24
Heuristics Centred process has a stationary distribution. Approximate by a deterministic distribution. Equations for centred moments give: 0 σc 3 dt 0 σ(c 4 3c 2 c 2 ) 0 σ(c 5 4c 3 c 2 ) 0 σ(c 6 5c 4 c 2 )... Stationary distribution approximately Gaussian. New York, Sept. 07 p.24
Suppose stationary distribution Gaussian with mean m(p ), variance b 2. New York, Sept. 07 p.25
Suppose stationary distribution Gaussian with mean m(p ), variance b 2. Front is at K where 1 2πb e K2 /2b 2 = 1 N K b 2 log N New York, Sept. 07 p.25
Suppose stationary distribution Gaussian with mean m(p ), variance b 2. Front is at K where 1 2πb e K2 /2b 2 = 1 N K b 2 log N If there is a single individual at m(p ) + K at time t = 0, how long until there is an individual at m(p ) + K + 1? New York, Sept. 07 p.25
Suppose stationary distribution Gaussian with mean m(p ), variance b 2. Front is at K where 1 2πb e K2 /2b 2 = 1 N K b 2 log N If there is a single individual at m(p ) + K at time t = 0, how long until there is an individual at m(p ) + K + 1? Ignoring beneficial mutations occurring to individuals at m(p ) + K 1, until the front advances Z(t) e (σk (1 q)µ)t. New York, Sept. 07 p.25
More heuristics Front advances as soon as individual of Z(t) accumulates a beneficial mutation. New York, Sept. 07 p.26
More heuristics Front advances as soon as individual of Z(t) accumulates a beneficial mutation. The probability that the front doesn t advance by time t is exp { t qµ 0 } Z(s) ds exp { } qµ σk (1 q)µ (e(σk (1 q)µ)t 1) New York, Sept. 07 p.26
More heuristics Front advances as soon as individual of Z(t) accumulates a beneficial mutation. The probability that the front doesn t advance by time t is exp { t qµ 0 } Z(s) ds exp Average time for front to advance by one is { } qµ σk (1 q)µ (e(σk (1 q)µ)t 1) 1 σk (1 q)µ log(σk (1 q)µ). New York, Sept. 07 p.26
More heuristics Front advances as soon as individual of Z(t) accumulates a beneficial mutation. The probability that the front doesn t advance by time t is exp { t qµ 0 } Z(s) ds exp Average time for front to advance by one is { } qµ σk (1 q)µ (e(σk (1 q)µ)t 1) 1 σk (1 q)µ log(σk (1 q)µ). But wave speed is µ(2q 1) + σc 2 (P ) µ(2q 1) + σb 2 µ(2q 1) + σk2 2 log N. New York, Sept. 07 p.26
Conclusion Consistency condition: σk (1 q)µ log(σk (1 q)µ) = µ(2q 1) + σk2 2 log N. New York, Sept. 07 p.27
Conclusion Consistency condition: σk (1 q)µ log(σk (1 q)µ) = µ(2q 1) + σk2 2 log N. For large K, this approximately reduces to K log(σk) = 2 log N. New York, Sept. 07 p.27
Conclusion Consistency condition: σk (1 q)µ log(σk (1 q)µ) = µ(2q 1) + σk2 2 log N. For large K, this approximately reduces to K log(σk) = 2 log N. If K = log N, then LHS > RHS; K = log 1 δ N, δ > 0, then LHS < RHS. New York, Sept. 07 p.27
Conclusion Consistency condition: σk (1 q)µ log(σk (1 q)µ) = µ(2q 1) + σk2 2 log N. For large K, this approximately reduces to K log(σk) = 2 log N. If K = log N, then LHS > RHS; K = log 1 δ N, δ > 0, then LHS < RHS. So K between log N and any fractional power of log N rate of adaptation, of order between log N and any fractional power of log N. New York, Sept. 07 p.27
Rigorous result Theorem. If q > 0, then for any β > 0, there exists a positive constant c µ,σ,q such that if N is sufficiently large. E π [m(1)] E π [c 2 ] c µ,σ,q log 1 β N New York, Sept. 07 p.28
Simulations With µ = 0.01, q = 0.01, σ = 0.01, N = 1000, 2000, 5000, 10000, 30000. First row: mean; second row: variance. 0 0.5 1 1.5 2.5 0.5 0 0.5 1 2 1 1.5 0 1.5 0.5 0.5 1 mean 2 mean mean 0.5 mean mean 2.5 1 0 0.5 1 3 0 3.5 1.5 1.5 0.5 0.5 4 0 500 1000 1500 time 2 0 500 1000 1500 time 2 0 500 1000 1500 time 1 0 500 1000 1500 time 1 0 500 1000 1500 time 2 2.5 2.5 1.8 1.6 1.8 1.6 1.4 1.6 2 2 1.4 1.2 1.4 1.2 1.2 1.5 1.5 1 1 var 1 var var var var 0.8 0.8 0.8 1 1 0.6 0.6 0.6 0.4 0.5 0.5 0.4 0.4 0.2 0.2 0.2 0 0 500 1000 1500 time 0 0 500 1000 1500 time 0 0 500 1000 1500 time 0 0 500 1000 1500 time 0 0 500 1000 1500 time New York, Sept. 07 p.29
6 x 10 3 5 4 Adaptation Rate 3 2 1 0 1 10 3 10 4 10 5 N Adaptation rate against population size. From top to bottom, q = 4%, 2%, 1%, 0.2%, µ = 0.01, σ = 0.01. New York, Sept. 07 p.30