Crossing fitness canyons by a finite population

We consider the Wright-Fisher model of the finite population evolution on a fitness landscape defined in the sequence space by a path of nearly neutral mutations. We study a specific structure of the fitness landscape: One of the intermediate mutations on the mutation path results in either a large fitness value (climbing up a fitness hill) or a low fitness value (crossing a fitness canyon), the rest of the mutations besides the last one are neutral, and the last sequence has much higher fitness than any intermediate sequence. We derive analytical formulas for the first arrival time of the mutant with two point mutations. For the first arrival problem for the further mutants in the case of canyon crossing, we analytically deduce how the mean first arrival time scales with the population size and fitness difference. The location of the canyon on the path of sequences has a crucial role. If the canyon is at the beginning of the path, then it significantly prolongs the first arrival time; otherwise it just slightly changes it. Furthermore, the fitness hill at the beginning of the path strongly prolongs the arrival time period; however, the hill located near the end of the path shortens it. We optimize the first arrival time by applying a nonzero selection to the intermediate sequences. We extend our results and provide a scaling for the valley crossing time via the depth of the canyon and population size in the case of a fitness canyon at the first position. Our approach is useful for understanding some complex evolution systems, e.g., the evolution of cancer.

Figure 1

Fitness along the path of mutations for three intermediate sequences with $m=3$. For the $l\mathrm{th}$ mutant we take $1+S_l$ at the point $l+0.5$ on our graphics. The fitness of the end-point mutant is always much higher than that of the intermediate mutants. (a) Population climbing the fitness hill. (b) Population crosses the fitness canyon.

Figure 2

We generate the random number with a binomial distribution with a probability parameter ${\eta }_0$ and a number of trials $N$ to get ${\widehat{i}}_0$. Then we repeat this procedure for the next mutant type with a probability parameter ${\eta }_1$ with the number of trials $N-{\widehat{i}}_0$ to get $i_1$. We continue this procedure until the mutant type $m-1$ to get $i_{m-1}$ and then we get the number of replicators with the last type ${\widehat{i}}_m=N-{\sum }_{l=0}^{m-1}{\widehat{i}}_l$. (a) The mean first arrival time $T\equiv T_1$ for a double mutant is $m=2, N=10000, S_0=0$, and $S_1=-c/\sqrt{N}$. The solid line is given from our analytical formula (5) and the dots are obtained from the numerics. (b) The mean first arrival time $T\equiv T_1$ for half of the population is $m=2, N=10000, S_0=0, S_1=-c/\sqrt{N}$, and $S_2=1/\sqrt{N}$. The solid line shows a linear function that fits the numerical data and the dots correspond to the numerical data.

Figure 3

(a) Mean first arrival time $T\equiv T_1$ for $m=3, N=10000, S_0=0, S_1=-c/N^{3/4}$, and $S_2=0$. The solid line is the fit of numerics for the linear function and the dots correspond to the numerics. (b) Mean first arrival time $T\equiv T_1$ for $m=3, N=10000, S_0=0, S_1=0$, and $S_2=-c/N^{1/4}$. The solid line is the fit of numerics for the linear function and the dots correspond to the numerics.

Figure 4

Mean first arrival time for $m=3, N=10000, S_0=0, S_1=c/N^{3/4}$, and $S_2=0$.

Figure 5

Here $A$ is the maximal acceleration of the mean first arrival due to the intermediate high fitness at the position for $m=3, S_0=0, S_2=0$, and $S_1>0$ and $N$ is the population size. We are choosing the value of $S_1$ giving the maximal acceleration ($A$ times) compared with the neutral case $N_1=1$. The solid line is the fit of the data via logarithmic function and dots correspond to data. In addition, $m=3, N=10000, S_0=0, S_1=c/N^{3/4}$, and $S_2=0$.

Figure 6

Mean time of fixation for the third mutant for $m=3, N=1000, S_0=0, S_2=10, S_1=-c/N^{0.75}$, and $S_3=10/N$.

Figure 7

Mean first arrival time for the third mutant (solid line) and the fixation time (dashed line) versus the canyon depth $k$ for $m=3, S_0=0, S_1=-k/N, S_2=0, S_3=0$, and $u=1/N^2$.