This classic Lenski paper computes the effective population size of an evolving E. coli population subjected to daily population bottlenecks as $N_e = N_0 * g$, where $N_e$ is the effective population size, $N_0$ is the population size directly after the bottleneck and $g$ is the number of generations between bottlenecks.

Unfortunately, the formula was not derived in the referenced paper and the referenced articles appear to not describe the formula directly, but only provide the fundamentals for deriving it.

Can someone explain how this formula comes about?


1 Answer 1


The effective population size is the harmonic mean (time average of $1/N$) of the time dependent population size:

$\frac{1}{N_e} = \frac{1}{t_f - t_0} \sum_{t = t_0}^{t_{f}} \frac{1}{N(t)}$

where $t_0$ is the foundation of the E. coli line, $t_f$ is some time much later when Lenski's poor grad students are sequencing them (with the understanding that $t_f - t_0 \ll N_e$, $t_0$ may be before the experiment began), and $N(t)$ is a deterministic function produced by removing individuals from the population with a pipette (throwing the rest in the freezer for the living fossil record) and letting them expand to cover a newly inhabited plate.

Since the population size is cyclical in Lenski's experiments (this procedure is done at regular time intervals, so at approximately consistent generation time), the average over one cycle should be the same effective population size as integrating over all time. In other words:

$\frac{1}{N_e} = \frac{1}{t_{nextBN} - t_{BN}}\sum_{t=t_{BN}}^{t_{nextBN}} \frac{1}{N(t)} $

where $t_{BN}$ is the start of the bottleneck and $t_{nextBN}$ is the generation before the following bottleneck (once a day I think? so, $\sim$6 generations per cycle if I remember correctly). In the Lenski lines, the population grows rapidly from inhabiting a new environment on the fresh plate, so we can assume the growth rate is exponential with some rate $\gamma$ (e-folds per generation). Now we have:

$\frac{1}{N_e} = \frac{1}{t_{nextBN}-t_{BN}} \sum_{t=t_{BN}}^{t_{nextBN}} \frac{1}{N_0 e^{\gamma t}} = \frac{1}{\Delta t N_0} \sum_{0}^{\Delta t} \frac{1}{e^{\gamma t}}$

In the second expression I have defined $\Delta t = (t_{nextBN} - t_{BN})$ as the time between bottlenecks, and the sums are equal because we are free to start counting generations at $t_{BN} = 0$. I also pulled the constant $N_0$ out of the sum.

Computing the sum of the series over a finite number of generations $\Delta t$, we find the following:

$\frac{1}{N_e} = \frac{1}{\Delta t N_0} \frac{ \left(e^{\gamma} - e^{-\Delta t \gamma}\right)}{\left(e^{\gamma }-1\right) }$

Now, I looked up Lenski's 1991 paper (here: https://www.jstor.org/stable/2462549?casa_token=dNMGpr9sBFcAAAAA:BrRjPHxwmxAuegCuN8eHFXr2w4slkPLJwTOrpD6nQRIHsx6ukS4kWaIYHtPuwgxWQ9j7G_6ICPpCTKZA7XW0izImSN_-7wwXpS_lnGWhJSN3PbWRiSU) and he states (in the same section) that $N_0 \sim 5\times 10^6$, $N_f \sim 5 \times 10^{8}$; if $ N_f = N_0 e^{\gamma \Delta t}$ and $\Delta t \sim 6$ (assuming the cycle is once per day), we find that the growth rate for each bottleneck is roughly $\gamma \sim 0.75$ (in that ballpark). This is extremely quick growth, so the $e^{\gamma}$ terms in the numerator of the above expression dominate over $e^{-\gamma \Delta t}$. This means we can approximate the inverse effective population size as follows.

$\frac{1}{N_e} \approx \frac{1}{\Delta t N_0} \frac{e^{\gamma} }{e^{\gamma}-1} \sim \frac{2}{\Delta t N_0}$

Note, there is a factor of roughly 2 here that I suppose Lenski is ignoring? Using Lenski's variable $g = \Delta t$, this reproduces the relationship he quoted, up to a factor of $\frac{e^{\gamma}-1}{e^{\gamma}} \approx \frac{1}{2}$:

$N_e \approx \frac{1}{2} g N_0$

As a little tip, if you ever have a population genetics question you're not sure about, just scour Kimura's papers. Everything you can ever think of is sitting in one of them, including this question: https://www.jstor.org/stable/2406157?casa_token=cWTvTL8k29QAAAAA:GVDLvS2SL7FJioYREhU0qjAIU5ynBpcncl7Z8r7xZh6S_PQWIVh2Di0bKt9BbVgfJIo-O-CMiFudnSnDMjmwzth1YDTA_EoPoQ7jATWxuzzR2gzcke4

EDIT: I initially introduced a sign error in the penultimate expression, which was originally $\frac{1}{N_e} \approx \frac{1}{\Delta t N_0} \frac{e^{\gamma} }{e^{\gamma}+1} \sim \frac{1}{\Delta t N_0}$, where there was a negligible factor of roughly 2/3 from $\gamma=0.75$ that can be treated as approximately 1. The above text is now corrected, and has an apparent factor of 2 that would need to be dropped to match Lenski's expression.

  • $\begingroup$ Thank you for this comprehensive answer. I guess we can drop the factor of ~2/3 because in this context the error it introduces is irrelevant due to the large uncertainty in all other parameters? $\endgroup$
    – Dahlai
    Jun 15, 2022 at 11:43
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    $\begingroup$ Additionally, I found this paper (onlinelibrary.wiley.com/doi/10.1111/j.0014-3820.2001.tb00772.x, also from Lenski) that derives $N_e = N_0\Delta t\gamma$. Again, I guess $\gamma$ can be neglected? $\endgroup$
    – Dahlai
    Jun 15, 2022 at 11:52
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    $\begingroup$ First, no problem! Second, thank you for citing that paper. I had not seen it and it's obviously relevant. There is also this paper, by the same authors: academic.oup.com/genetics/article/162/2/961/6049989?login=true. Regarding the difference between the two solutions, there is also a strange difference that comes up if you integrate rather than sum to compute the harmonic mean of the population size. I will post the result (only) in another comment in a few mins. Also, I made a dumb mistake and flipped the sign in the denominator of the second to last equation (oops!). $\endgroup$ Jun 16, 2022 at 18:19
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    $\begingroup$ The solution if you assume continuous generation time is the following: $\frac{1}{N_e} = \frac{1}{\Delta t} \int_0^{\Delta t} \frac{1}{N_0 e^{\gamma t}} dt = \frac{1-e^{-\gamma \Delta t}}{N_0 \gamma \Delta t}$. Since $\gamma \Delta t \approx 6 \times 0.75 = 4.5$, the exponential decay is negligible. This gives $N_e \approx N_0 \gamma \Delta t$, matching Wahl and Gerrish. I am not quite sure where the difference comes from, but it's entirely possible I made a mistake above. Wahl and Gerrish also claim that their results match Lenski (1991) (the original paper we are discussing). ctd below $\endgroup$ Jun 16, 2022 at 18:33
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    $\begingroup$ I'll try to look at the discrete case in my answer again and see if I can find some issue that alters the result (other than the sign error). I think difference has something to do with changing the discrete sum with the element having no units to the differential dt. I am going to edit the sign error in my original answer, but I will make it clear where the mistake was so you can track it. $\endgroup$ Jun 16, 2022 at 18:34

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