Feb23-2019

从前,我从广泛的阅读里获得了很多好处,但读的东西基本上不能帮助我进行工作,而稍微有用一些的东西,在放松的状态下,又读不进去。

我决定每周抽出几天来广泛地看一些文章。最好是一些有趣的研究,在生物或其他的背景下应用了物理的一些工具,得到了重要的结果。这样,训练了读写的能力,也可以提升对于整个领域的了解。大概一次写一篇或两篇。


Luria-Delbrueck Experiment

原文地址:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1209226/pdf/491.pdf

The Luria-Delbrueck experiment aims to test two contradictory hypothesis in the occurrence of a phage-proof gene:

h1: the mutation that lead to the gene happened after infections, i.e. the immunity was acquired.

h2: the mutation was independent of the infections.

The authors noticed that the variance of the number of mutants may help testing, which was a great step leaned in history of quantitative biology. Without much experimental exquisiteness, the second hypothesis was proved.

First, under h1 the N(mutants) at one specific time (after a contact with phages) must obey a Poisson distribution, meaning that averagevariance=1\frac{\mathrm{average}}{\mathrm{variance}} = 1.

Things under h2 however, are a bit more complicated. Assume that rr is the number of mutants and NN is the overall population. To make things simple, we just say that N=N0etN=N_0e^t and drdt=aN+r\frac{\mathrm{d}r}{\mathrm{d}t}=aN+r. Integrating the second equation from t0t_0 to tt, we have r=(tt0)aNr = (t-t_0)aN. The idea of integrating from t0t_0 is to make sure that integration begins at a time when there is at least one mutant, eliminating potential errors. Denoting number of cultures as CC, we can write a(Nt0N0)C=1a(N_{t_0}-N_0)C=1, in which N0N_0 could be ignored and tt0=ln(aNC)t-t_0=\ln(aNC) be acquired. With all these above, r=aNln(aNC)r = aN\ln (aNC).

Then we consider the average number of new mutations occurred in [tτ,tτ+dτ][t-\tau, t-\tau+\mathrm{d}\tau], which should be NteτadτN_te^{-\tau}a\mathrm{d}\tau. Since the mutation is a Poisson process, that should also be its variance, hence during the time [tτ,t][t-\tau, t], the variance caused by these mutations grows to NteτadτN_te^{\tau}a\mathrm{d}\tau. Integrating τ\tau in [0,tt0][0, t-t_0], we have that varr=aNt(ett01)=Ca2Nt2\mathrm{var}_r=aN_t(e^{t-t_0}-1)=Ca^2N_t^2.

Therefore, ln(aNC)aNC\frac{\ln(aNC)}{aNC} and 1 can be compared with the data to see which was closer. It turned out to be the former, suggesting a much larger variance. That is natural, since compared to acquired immunity in which all the mutants originate at the same time, independent mutation is more like

a slot machine, where the average return from a limited number of plays is probably considerably less than the input, and improbably, when the jackpot is hit, the return is much bigger than the input.