Scaling Solution in the Large Population Limit of the General Asymmetric Stochastic Luria–Delbrück Evolution Process

Abstract

One of the most popular models for quantitatively understanding the emergence of drug resistance both in bacterial colonies and in malignant tumors was introduced long ago by Luria and Delbrück. Here, individual resistant mutants emerge randomly during the birth events of an exponentially growing sensitive population. A most interesting limit of this process occurs when the population size $N$ is large and mutation rates are low, but not necessarily small compared to $1/N$. Here we provide a scaling solution valid in this limit, making contact with the theory of Levy $\alpha$-stable distributions, in particular one discussed long ago by Landau. One consequence of this association is that moments of the distribution are highly misleading as far as characterizing typical behavior. A key insight that enables our solution is that working in the fixed population size ensemble is not the same as working in a fixed time ensemble. Some of our results have been presented previously in abbreviated form [12].

Appendix: Recovering \(P_N(m)\) from the Generating Function

The methodology we have adopted yields directly the generating function $P_N(x)$ and to retrieve $P_N(m)$ we need to perform a contour integral in the complex plane, as in Eq. (15) for the symmetric pure-birth case, which generalizes (see Eq. (28)) to

$$\begin{aligned} P_N(m) = \frac{1}{2\pi i} \oint \frac{dz}{z^{m+1}} \exp \left( -\mu N(1-z) \,{}_2F_1\!\left( 1,1;1+r;z\right) \right) . \end{aligned}$$

(110)

in the general asymmetric pure-birth case. The integrand has a branch cut, which we can place on the ray $z>1$. The contour circles the origin. It is convenient to deform the contour, expanding it outward except where it is “pinched” by the branch cut. The resulting contour is given by $z(t)=\lim _{\epsilon \rightarrow 0} 1 + t -\epsilon + i\epsilon \tanh t$, $-\infty <t<\infty$ plus vanishing contributions from infinity. Thus in the limit the integral reduces to the discontinuity of the integrand along the branch cut:

$$\begin{aligned} P_N(m) = \frac{1}{2\pi i} \int _0^\infty \frac{dt}{(1+t)^{m+1}}\,\text {Disc}\left[ \exp \left( \mu Nt \,{}_2F_1\!\left( 1,1;1+r;1+t\right) \right) \right] . \end{aligned}$$

(111)

Using standard identities involving hypergeometric functions, we have

$$\begin{aligned} - t\,{}_2F_1(1,1;1+r;1+t)&= \frac{r}{r-1}\left( \frac{t}{1+t}\right) {}_2F_1\!\left( 1,1-r;2-r;\frac{t}{1+t}\right) \nonumber \\&\quad {} + \frac{\pi r}{\sin \pi r}\left( \frac{-t}{1+t}\right) ^r{}_2F_1\!\left( r,0;r;\frac{t}{1+t}\right) \nonumber \\&= -\frac{r}{r-1}\left( \frac{t}{1+t}\right) {}_2F_1\!\left( 1,1-r;2-r;\frac{t}{1+t}\right) \nonumber \\&\quad {} + \frac{\pi r}{\sin \pi r}\left( \frac{-t}{1+t}\right) ^r. \end{aligned}$$

(112)

As the argument of the hypergeometric function on the right hand side lies between 0 and 1, it is continuous across the real axis. The only discontinuity then arises from the factor $(-t/(1+t))^r$ in the second term. This factor takes the values

$$\begin{aligned} \lim _{\epsilon \rightarrow 0^+}\left. \left( \frac{-t}{1+t}\right) ^{r}\right| _{t\pm i\epsilon } = \left( \frac{t}{1+t}\right) ^{r} e^{\mp ir\pi } \end{aligned}$$

(113)

Thus, we obtain

$$\begin{aligned} P_N(m)&= \frac{1}{\pi } \int _0^\infty \frac{dt}{(1+t)^{m+1}}\,\exp \left( \mu N\frac{rt}{(r-1)(1+t)} \,{}_2F_1\!\left( 1,1-r;2-r;\frac{t}{1+t}\right) \right) \nonumber \\&\times \exp \left( -\mu N\frac{\pi r}{\tan \pi r}\left( \frac{t}{1+t}\right) ^r\right) \sin \left( \mu N\pi r\left( \frac{t}{1+t}\right) ^r\right) . \end{aligned}$$

(114)

The calculation of Eq. (48) follows along the same lines.

For $r\rightarrow 1$, since

$$\begin{aligned} \lim _{r\rightarrow 1} \frac{r}{r-1}\left( \frac{t}{1+t}\right) {}_2F_1\!\left( 1,1-r;2-r;\frac{t}{1+t}\right) = \frac{t}{1+t}\left[ \frac{1}{r-1} + 1 + \ln \frac{1}{1+t}\right] \nonumber \\ \end{aligned}$$

(115)

and

$$\begin{aligned} \lim _{r\rightarrow 1} \frac{\pi r}{\tan \pi r} \left( \frac{t}{1+t}\right) ^r = \frac{t}{1+t}\left[ \frac{1}{r-1}+1+\ln \frac{t}{1+t}\right] \end{aligned}$$

(116)

we have in this case

$$\begin{aligned} P_N(m)&= \frac{1}{\pi } \int _0^\infty \frac{dt}{(1+t)^{m+1}}\,\exp \left( -\mu N \frac{t}{1+t}\ln t\right) \sin \left( \mu N \pi \frac{t}{1+t}\right) \nonumber \\&= \frac{1}{\pi } \int _0^\infty \frac{dt}{(1+t)^{m+1}}\,t^{-\mu Nt/(1+t)}\sin \left( \mu N \pi \frac{t}{1+t}\right) \end{aligned}$$

(117)

This of course also be obtained directly from Eq. (15) via similar methods, where there we label the distribution $P(m|N)$ since in the death case $N$ is itself a random variable being conditioned upon.