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Long-term evolution on complex fitness landscapes when mutation is weak


Understanding evolution on complex fitness landscapes is difficult both because of the large dimensionality of sequence space and the stochasticity inherent to population-genetic processes. Here, I present an integrated suite of mathematical tools for understanding evolution on time-invariant fitness landscapes when mutations occur sufficiently rarely that the population is typically monomorphic and evolution can be modeled as a sequence of well-separated fixation events. The basic intuition behind this suite of tools is that surrounding any particular genotype lies a region of the fitness landscape that is easy to evolve to, while other pieces of the fitness landscape are difficult to evolve to (due to distance, being across a fitness valley, etc.). I propose a rigorous definition for this “dynamical neighborhood” of a genotype which captures several aspects of the distribution of waiting times to evolve from one genotype to another. The neighborhood structure of the landscape as a whole can be summarized as a matrix, and I show how this matrix can be used to approximate the expected waiting time for certain evolutionary events to occur and to provide an intuitive interpretation to existing formal results on the index of dispersion of the molecular clock.

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Many thanks to Arjan de Visser, Santiago Elena, Inês Fragata, and Sebastian Matuszewski for organizing this theme issue on fitness landscapes, and to two anonymous reviewers for many helpful comments on the manuscript.

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Conflict of interest

The authors declare that they have no conflict of interest.

Correspondence to David M. McCandlish.



(A) The dynamical neighborhood of i is connected and includes i

We want to show that for a fixed i, the set of genotypes j such that Z(i,j) > 0 is connected, in the sense that there is a path l1,…,lk such that Z(i,l1),…,Z(i,lk) > 0 and \(P\left( {j,l_1} \right)P\left( {l_1,l_2} \right) \cdots P\left( {l_k,i} \right) > 0\). In fact, we will show that for a fixed choice of genotype i and constant a, the set of genotypes j with EjTi < a is connected. Because EiTj = (Z(j,j) − Z(i,j))/π(j) and EπTj = Z(j,j)/π(j), this reduces to the previous condition for a = EπTi

Suppose to the contrary that the set A of genotypes j such that EjTi < a is not connected. Then let A′ be the component of A that contains i (i.e., let A′ be the maximal connected subset of A containing i), and let A′′ be some other component of A. Now, let k be a genotype in A′′ and consider EkTi. We can write EkTi = ET(1) + ET(2) where ET(1) is the expected time for a population that begins at k to first arrive at a genotype that is not in A and then ET(2) is the expected remaining waiting time for the population to first arrive at i. Of course, ET(1) > 0 since the population begins at genotype \(k\,\in\,{A}\). Furthermore, ET(2) ≥ a since at T(1) the population is no longer in A. Thus we have:

$$\begin{array}{*{20}{l}}E_kT_i & = & ET^{\left( 1 \right)} + ET^{\left( 2 \right)}\hfill{}\\ \hfill{} & \hfill{} & > ET^{\left( 2 \right)}\cr \hfill{} & \hfill{} & > a \end{array}$$

which contradicts our original assumption that \(k\,\in\,{A}\). Thus, the set of genotypes j such that EjTi < a is connected, as is the set of genotype j such that Z(i,j) > 0.

We also need to show that i is in its own dynamical neighborhood, i.e., Z(i,i) > 0. Let \(Z_t\left( {i,j} \right) \equiv \mathop {\sum}\nolimits_{m = 0}^{t - 1} \left( {P^m\left( {i,j} \right) - \pi \left( j \right)} \right)\). From Eq. (13), we can write Zt(i,t) as:

$$Z_t\left( {i,i} \right) = \mathop {\sum}\limits_{k \ge 2} \frac{{1 - \lambda _k^t}}{{1 - \lambda _k}}\;r_k\left( i \right){\kern 1pt} l_k\left( i \right) = \mathop {\sum}\limits_{k \ge 2} \frac{{1 - \lambda _k^t}}{{1 - \lambda _k}}\;q_k\left( i \right)^2$$

which is clearly nonnegative and increasing with t. Now, Z1(i, i) = 1 π(i), which is positive except in the trivial case, which we choose to exclude, of a fitness landscape with only one genotype. Because Zt is increasing with t, it follows that Z(i, i) = Z(i, i) ≥ 1 − π > 0, so that i is in its own dynamical neighborhood.

(B) Analytical formula for Z in the neutral case

In the neutral case, we have P = I − μL, where L is the Laplacian matrix of the l-dimensional Hamming graph on \(\left| {\cal A} \right|\) letters. Because the eigendecomposition of L is known and can be expressed in terms of the Krawtchouk polynomials (Stadler and Happel 1999), we can plug this eigendecomposition into Eq. (14) to get:

$$Z\left( {i,j} \right) = \mathop {\sum}\limits_{k = 1}^l \frac{1}{{\mu \,k\left| {\cal A} \right|}}\mathop {\sum}\limits_{q = 0}^{min\left\{ {k,d} \right\}} \frac{{\left( { - 1} \right)^q\left( {\left| {\cal A} \right| - 1} \right)^{k - q}}}{{\left| {\cal A} \right|^l}}\left( {\begin{array}{*{20}{c}} d \cr q \end{array}} \right)\left( {\begin{array}{*{20}{c}} {l - d} \cr {k - q} \end{array}} \right),$$

where d is the Hamming distance between sequence i and sequence j.

(C) Proof that the approximation in Eq. ( 11 ) is exact for any event that either always ends or always begins in a single genotype

We want to show that \(E_\pi T_V = \left( {1 + c^{{\mathrm{T}}}Zg} \right){\mathrm{/}}E_\pi g\) for any V that is all zeros except that either (A) a single row of V contains nonzero entries (the event always begins in a particular genotype i) or (B) a single column of V contains nonzero entries (an event always ends in a single genotype i).

First we will consider the expected waiting time until an event that always begins in genotype i. The expected time until such an event can be viewed as the sum of two waiting times: the waiting time to first reach i, and then the waiting time for the event to occur when starting at i. Thus, we have

$$E_\pi T_V = E_\pi T_i + E_iT_V.$$

Now EπTi is simply Z(i,i)/π(i), and so what we want to know is EiTV.

Given that the population is at i, the total number of times the population will be at i before TV is geometrically distributed, where the success probability is g(i). Thus, on average there are (1 − g(i))/g(i) failures before the first success. After a failure, the population is distributed as:

$$\frac{{\left( {P\left( {i,1} \right)\; \ldots \;P\left( {i,n} \right)} \right) - \left( {V\left( {i,1} \right)\; \ldots \;V\left( {i,n} \right)} \right)}}{{1 - g\left( i \right)}}$$

and one step has occurred. So after a failure, the expected time to return to i is:

$$\begin{array}{ccccc}\cr 1 + \mathop {\sum}\limits_j \frac{{P(i,j) - V(i,j)}}{{1 - g(i)}}{E_j}{T_i} &= & 1 + \frac{{Z(i,i)}}{{\pi (i)}} - \frac{{\mathop {\sum}\limits_j P(i,j)Z(j,i)}}{{\left( {1 - g(i)} \right)\pi (i)}} \\\cr && \mathbin{+}\frac{{\mathop {\sum}\limits_j V(i,j)Z(j,i)}}{{\left( {1 - g(i)} \right)\pi (i)}}\\\cr &= & 1 + \frac{{Z(i,i)}}{{\pi (i)}} - \frac{{Z(i,i) + \pi (i) - 1}}{{\left( {1 - g(i)} \right)\pi (i)}} \\\cr && \mathbin{+} \frac{{c^{{\mathrm{T}}}Zg}}{{\left( {1 - g(i)} \right)\pi (i)}}\\\cr &= & \frac{{1 - \pi (i)g(i) - Z(i,i)g(i) + c^{{\mathrm{T}}}Zg}}{{\left( {1 - g(i)} \right)\pi (i)}}.\cr \end{array}$$

In addition, there is exactly one success, which takes exactly one time step. Thus, multiplying the duration for each failure by the expected number of failures and adding one for the final success, we have:

$$\begin{array}{ccccc}\cr E_iT_V &= & \frac{{1 - \pi \left( i \right)g\left( i \right) - Z\left( {i,i} \right)g\left( i \right) + c^{{\mathrm{T}}}Zg}}{{\pi (i)g(i)}} + 1\\\cr &= & \frac{{1 - Z\left( {i,i} \right)g\left( i \right) + c^{{\mathrm{T}}}Zg}}{{\pi \left( i \right)g\left( i \right)}}.\cr \end{array}$$

Plugging Eq. (A7) into Eq. (A4) gives us:

$$\begin{array}{*{20}{l}} E_\pi T_V & = & \frac{{Z\left( {i,i} \right)}}{{\pi \left( i \right)}} + \frac{{1 - Z\left( {i,i} \right)g\left( i \right) + c^{{\mathrm{T}}}Zg}}{{\pi \left( i \right)g\left( i \right)}}\hfill{}\\ & = & \frac{{1 + c^{{\mathrm{T}}}Zg}}{{E_\pi g}}\hfill{} \end{array}$$

as required.

Next, let us consider an event V that always ends in genotype i. We first treat the special case that an event occurs each time the population transitions into some specific state i, (that is, we let V(k,j) = P(k,j) for j = i and 0 otherwise). Since the population is always at genotype i when an event occurs, c is the vector with a 1 in position i and 0 elsewhere. Furthermore, since an event happens with each transition into i, g is simply the ith column of P.

Now, we turn to calculating EπTV. Because an event can only occur on a transition into state i, no event can occur at t = 0, and so the first possible time for an event to occur is t = 1. Moreover, because the initial genotype is drawn from the stationary distribution π at t = 0 the location of the population is still described by the stationary distribution at t = 1 and so we have EπTV = 1 + EπTi, i.e.:

$$E_\pi T_V = \frac{{Z\left( {i,\,i} \right)}}{{\pi \left( i \right)}} + 1.$$

To show that this expression is equal to (1 + cTZg)/Eπg, we will need the following identity for Z(i,i):

$$Z\left( {i,i} \right) = \mathop {\sum}\limits_{t = 0}^\infty \left( {P^t\left( {i,i} \right) - \pi \left( i \right)} \right)$$
$$= 1 - \pi \left( i \right) + \mathop {\sum}\limits_{t = 1}^\infty \left( {P^t\left( {i,i} \right) - \pi \left( i \right)} \right)$$
$$= 1 - \pi \left( i \right) + \mathop {\sum}\limits_{t = 1}^\infty \left( {\mathop {\sum}\limits_k P^{t - 1}\left( {i,k} \right)P\left( {k,i} \right) - \mathop {\sum}\limits_k \pi \left( k \right)P\left( {k,i} \right)} \right)$$
$$= 1 - \pi \left( i \right) + \mathop {\sum}\limits_k \left( {\left( {\mathop {\sum}\limits_{t = 1}^\infty P^{t - 1}\left( {i,k} \right) - \pi \left( k \right)} \right)P\left( {k,i} \right)} \right)$$
$$= 1 - \pi \left( i \right) + \mathop {\sum}\limits_k Z\left( {i,k} \right)P\left( {k,i} \right),$$

where we have used the fact \(\pi \left( i \right) = \mathop {\sum}\nolimits_k \pi \left( k \right)P\left( {k,i} \right)\) and the definition of Z. Noting that for our choice of V we have \(c^{{\mathrm{T}}}Zg = \mathop {\sum}\nolimits_k Z\left( {i,k} \right)P\left( {k,i} \right)\) and Eπg = π(i), we can substitute the above expression for Z(i,i) into Eq. (A9) to obtain

$$E_\pi T_V = \frac{{1 - \pi \left( i \right) + \mathop {\sum}\limits_k Z\left( {i,k} \right)P\left( {k,i} \right)}}{{\pi \left( i \right)}} + 1$$
$$= \frac{{1 - E_{\mathrm{\pi }}g + c^{{\mathrm{T}}}Zg}}{{E_{\mathrm{\pi }}g}} + 1$$
$$= \frac{{1 + c^{{\mathrm{T}}}Zg}}{{E_{\mathrm{\pi }}g}}$$

as required.

To extend this result to events V where events only occur upon transition to state i but do not occur on every transition to i, we consider an augmented Markov chain with transition matrix P* and event V* chosen so that the distribution of waiting times for V* is identical to the distribution for V, but we add an auxiliary state n + 1 so that events only occur on transitions to state n + 1 and occur on each such transition.

In particular, we define P* as

$$\begin{array}{l}P^ \ast \left( {j,k} \right) = \cr \left\{ {\begin{array}{*{20}{c}} {P\left( {j,k} \right)} & {{\mathrm{for}}\,j \ne i,\,j \ne n + 1,\,k \ne i,\,k \ne n + 1} \cr {P\left( {i,k} \right)} & {{\mathrm{for}}\,j = i\,{\mathrm{or}}\,j = n + 1,\,k \ne i,\,k \ne n + 1} \cr {P\left( {j,i} \right) - V\left( {j,i} \right)} & {{\mathrm{for}}\,k = i} \cr {V\left( {j,i} \right)} & {{\mathrm{for}}\,k\, = \,n\, + \,1,} \end{array}} \right.\end{array}$$

and V* as

$$V^ \ast \left( {j,\,k} \right) = \left\{ {\begin{array}{*{20}{l}} {V\left( {j,i} \right)} \hfill & {{\mathrm{for}}\,k = n + 1,\,j \ne n + 1} \hfill \cr {V\left( {i,i} \right)} \hfill & {{\mathrm{for}}\,k{\mathrm{ = }}n + 1,\,j = n + 1} \hfill \cr 0 \hfill & {{\mathrm{otherwise}}.} \hfill \end{array}} \right.$$

Then P* with event V* has dynamics identical to P with event V except that we have split the state i into two states so that the population goes to n + 1 instead of i when an event occurs. Defining π*, c*, Z* and g* based on P* and V*, we then have (c*)TZ*g* = cTZg, \(E_{\pi ^ \ast }g^ \ast = E_\pi g\), and \(E_{\pi ^ \ast }T_{V^ \ast } = E_\pi T_V\). Thus,

$$E_\pi T_V = E_{\pi ^ \ast }T_{V^ \ast }$$
$$= \frac{{1 + \left( {c^ \ast } \right)^{{\mathrm{T}}}Z^ \ast g^ \ast }}{{E_{\pi ^ \ast }g^ \ast }}$$
$$= \frac{{1 + c^{{\mathrm{T}}}Zg}}{{E_\pi g}},$$

as required.

(D) The index of dispersion for generalized events

We will proceed in somewhat more generality than the main text and consider a general ergodic transition matrix P with events defined by a matrix V as before. To write down a formula for R, it will be helpful to write down formulas for EπIt, which is the probability that an event occurs at time t given that the initial genotype is a draw from the equilibrium distribution, and \(E_\pi \left( {I_{t_1}I_{t_2}} \right)\), which is the probability that an event occurs at both time t1 and time t2 given that the initial genotype is a draw from the equilibrium distribution. Of course,

$$E_\pi I_t = \pi ^{{\mathrm{T}}}P^{t - 1}g = \pi ^{{\mathrm{T}}}g = E_\pi g$$

and for t1 < t2

$$\begin{array}{ccccc}\cr E_\pi \left( {I_{t_1}I_{t_2}} \right) &= & \pi ^{{\mathrm{T}}}P^{t_1 - 1}VP^{t_2 - t_1 - 1}g\\\cr &= & \pi ^{{\mathrm{T}}}VP^{t_2 - t_1 - 1}g\\\cr &= & \left( {E_\pi g} \right)\left( {cP^{t_2 - t_1 - 1}g} \right)\cr \end{array}$$

since the probability that an event occurs at time t1 is Eπg, and given that an event occurs at t1 the distribution of the location of the population is given by c.

Writing St for the number of events that have occurred through time t, using Eq. (A23) we have:

$$\begin{array}{ccccc}\cr E_\pi S_t &= & E_\pi \left( {\mathop {\sum}\limits_{m = 1}^t I_m} \right)\\\cr &= & \mathop {\sum}\limits_{m = 1}^t E_\pi I_m\\\cr &= & tE_\pi g.\cr \end{array}$$

Furthermore using the general formula for the variance of a sum of random variables and noting that \(E_\pi \left( {I_{m_1}I_{m_2}} \right) = E_\pi \left( {I_{m_2}I_{m_1}} \right)\) depends only on |m2 − m1| and Eπ(Im)2 = EπIm, we have:

$$\begin{array}{ccccc}\cr Var_\pi S_t &= & Var_\pi \left( {\mathop {\sum}\limits_{m = 1}^t I_m } \right)\\\cr &= & \mathop {\sum}\limits_{m = 1}^t Var_\pi I_m + 2\mathop {\sum}\limits_{m_1 = 1}^{t - 1} \mathop {\sum}\limits_{m_2 = m_1 + 1}^t Cov_\pi \left( {I_{m_1},I_{m_2}} \right)\\\cr &= & \mathop {\sum}\limits_{m = 1}^t \left( {E_\pi I_m - \left( {E_\pi I_m} \right)^2} \right)\\\cr && \mathbin{+} 2\mathop {\sum}\limits_{m_1 = 1}^{t - 1} \mathop {\sum}\limits_{m_2 = m_1 + 1}^t \left( {E_\pi \left( {I_{m_1}I_{m_2}} \right) - E_\pi I_{m_1}E_\pi I_{m_2}} \right)\\\cr &= & tE_\pi g(1 - E_\pi g) + 2\left( {E_\pi g} \right)\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {t - m} \right)\left( {c^{{\mathrm{T}}}P^{m - 1}g - E_\pi g} \right)\\\cr &= & E_\pi S_t\left( {1 - E_\pi g + 2\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {1 - \frac{m}{t}} \right)\left( {c^{{\mathrm{T}}}P^{m - 1}g - E_\pi g} \right)} \right){\kern 1pt} .\cr \end{array}$$

Concentrating on the last term, we have Eπg = πTg = cTg, where ∏ ≡ 1πT is a projection matrix corresponding to the stationary distribution of P. Thus:

$$\begin{array}{l}\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {1 - \frac{m}{t}} \right)\left( {c^{{\mathrm{T}}}P^{m - 1}g - E_\pi g} \right)\cr = c^{{\mathrm{T}}}\left( {\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {1 - \frac{m}{t}} \right)\left( {P^{m - 1} - {\Pi}} \right)} \right)g.\end{array}$$

Pulling apart the sum in parentheses, we have

$$\begin{array}{*{20}{l}}\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {1 - \frac{m}{t}} \right)\left( {P^{m - 1} - {\Pi}} \right)\cr = \left( {\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {P^{m - 1} - {\Pi}} \right)} \right) - \left( {\mathop {\sum}\limits_{m = 1}^{t - 1} \frac{m}{t}.\left( {P^{m - 1} - {\Pi}} \right)} \right).\end{array}$$


$$\mathop {{lim}}\limits_{t \to \infty } \mathop {\sum}\limits_{m = 1}^{t - 1} \left( {P^{m - 1} - {\Pi}} \right) = Z$$

by the definition of Z and

$$\mathop {{lim}}\limits_{t \to \infty } \left( {\mathop {\sum}\limits_{m = 1}^{t - 1} \frac{m}{t}.\left( {P^{m - 1} - {\Pi}} \right)} \right) = 0$$

since Pt converges at least geometrically quickly to ∏ entry-wise for sufficiently large t under Perron–Frobenius theory (see, e.g., Theorem 1.2 of Seneta 2006) using the facts that P is primitive and that the Perron eigenvalue of P is 1.

Putting this all together, we have

$$\begin{array}{*{20}{l}} R_\infty \equiv \mathop {{lim}}\limits_{t \to \infty } \frac{{Var_\pi S_t}}{{E_\pi S_t}}& = & \mathop {{lim}}\limits_{t \to \infty } \left( {1 - E_\pi g + 2\mathop {\sum}\limits_{m = 1}^{t - 1} \left( {1 - \frac{m}{t}} \right)\left( {c^{{\mathrm{T}}}P^{m - 1}g - E_\pi g} \right)} \right)\hfill{}\\ & = & 1 - E_\pi g + 2c^{{\mathrm{T}}}Zg\hfill{} \end{array}$$

as required.

(E) Sufficient condition for overdispersion

If DV1 = (DV)T1 then c = Dg/Eπg. This means that we can write:

$$\begin{array}{*{20}{l}}c^{{\mathrm{T}}}\left( {\tau _kr_kl_k^{{\mathrm{T}}}} \right)g & = & \tau _kg^{\mathrm{T}}D\left( {r_kl_k^{{\mathrm{T}}}} \right)g/\left( {E_\pi g} \right)\hfill{}\\ & = & \tau _kg^{\mathrm{T}}\left( {l_kl_k^{\mathrm{T}}} \right)g{\mathrm{/}}\left( {E_\pi g} \right)\hfill{}\\ & = & \tau _k\left( {l_k^{\mathrm{T}}g} \right)^2{\mathrm{/}}\left( {E_\pi g} \right)\hfill{} \end{array}$$

which is nonnegative because the numerator is nonnegative and the denominator is positive. Thus, if DV1 = (DV)T1, cTZg is also nonnegative, by Eq. (15).

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