How long does it take for a bacterial culture in a Petri dish to experience all possible single base pair mutations? Can 12 hours be enough?

I want to get an intuition for whether a given mutation is a rare event we can only hope for or something that is all but guaranteed to happen.

  • $\begingroup$ Remember that not every mutation is beneficial or conveys some sort of survival advantage, especially in a rich environment like a Petri dish where you are theoretically growing the bacteria on some sort of nutrient agar or in a nutrient-containing solution. Also remember that populations evolve, not individuals. $\endgroup$
    – MattDMo
    Sep 7, 2021 at 14:52
  • $\begingroup$ Sure! Maybe I should delete the last paragraph. The surprising part for me just that mutation is not a matter of "if" but "when". Like in ncbi.nlm.nih.gov/pmc/articles/PMC6242871 I can see antibiotics resistance evolve in a matter of days. Super cool. $\endgroup$ Sep 7, 2021 at 15:03
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    $\begingroup$ I don't understand your question, as you give no value for P. Are you saying that the twit quoted some value and you wish to know what it is? In which case your question would appear to be "What is the frequency of mutations per base-pair of E.coli during cell division?" and by extension "How many mutations occur in a petri dish in a period of (say) 24h, and how does this compare with the number of base pairs of DNA in the E.coli genome?". If this is the case please say so explicitly. $\endgroup$
    – David
    Sep 7, 2021 at 22:28
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    $\begingroup$ It occurs to me that you might be interested also in our sister site, Skeptics. They specialise in challenging notable unreferenced claims in the various media. $\endgroup$ Sep 8, 2021 at 8:01
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    $\begingroup$ Note that your question, while it takes the usual framing of single base pair mutations, ignores a very large number of possible mutations in the form of structural rearrangements, transpositions, etc., which do most of the real work of changing genomes. Contrary to single-base mutations, there is no way to saturate the space of all such mutations in culture. $\endgroup$ Sep 8, 2021 at 10:47

1 Answer 1


We can calculate the number of base pair copy errors in a Petri dish per hour by multiplying the number of base pairs (B), the copy error rate (P), the number of divisions in an hour (D), and the number of cells in a Petri dish (N). (We assume a stable population in a saturated medium.)

$$ B \times P \times D \times N $$

I've looked up some articles and tried to pull out numbers from them:

  • B: E. coli has 4.6 million base pairs. [1]
  • P: Mutation rate per base pair per generation is around $10^{-10}$. [1][2]
  • D: The doubling time can be as little as 20 minutes. So 3 divisions per hour. [3]
  • N: Around a billion E. coli per ml. [4] A few billion in a "colony". [5] Let's call it 3 billion for the Petri dish.

$$ B \times P \times D \times N = 4.6 \times 10^6 \times10^{-10} \times 3 \times 3 \times 10^{9} $$

This comes out to 0.9 mutations on average per base pair per hour. Each base pair can change into 3 different base pairs. After 3 hours we will have had almost as many mutations as possible! We can get the same mutation multiple times. But even accounting for that, it's almost certain that a given mutation will happen within a day:

probability of a given mutation over time

(Click the chart for an interactive view where you can try different numbers.)

Indeed we can see research [6] where selection can give rise to new mutant strains every day:

Doxycycline resistance evolving on SAGE plates over 6 days

In general if the inverse of the base pair copy error rate is close to the population size, we will see a large subset of all possible single base pair mutations in each generation.

How likely it is that truly 100% of mutations have happened is known as the "coupon collector's problem". Its solution tells us that the expected number of mutations needed to cover all 3 × 4.6 million point mutations is $3 B \times log(3 B)$. With our numbers above we are doing around 4.1 million mutations per hour, so we would on average need around 55 hours.

  1. Spontaneous mutation rates come into focus in Escherichia coli, 2014 by A B Williams

  2. DNA replication fidelity in Escherichia coli: a multi-DNA polymerase affair, 2012 by I J Fijalkowska, R M Schaaper, and P Jonczyk

  3. Organization of sister origins and replisomes during multifork DNA replication in Escherichia coli, 2007 by S Fossum, E Crooke, and K Skarstad

  4. General calibration of microbial growth in microplate readers, 2016 by K Stevenson, A F McVey, I B N Clark, P S Swain, and T Pilizota

  5. Role of the RuvAB protein in avoiding spontaneous formation of deletion mutations in the Escherichia coli K-12 endogenous tonB gene, 2004 by K Mashimo, Y Nagata, M Kawata, H Iwasaki, and K Yamamoto

  6. Access to high-impact mutations constrains the evolution of antibiotic resistance in soft agar, 2018 by N Ghaddar, M Hashemidahaj, and B L Findlay

  • $\begingroup$ The tweet was from an expert and I don't know the first thing about any of this. So I'm pretty sure this answer is wrong. But which part? $\endgroup$ Sep 8, 2021 at 8:04
  • $\begingroup$ I got the whole formula wrong! That's error rate PER BASE PAIR. Then the claim is 100% good. I'll fix this! $\endgroup$ Sep 8, 2021 at 8:51
  • $\begingroup$ All fixed. I'm quite unsure about the number of cells. But doi.org/10.1016/j.bbrc.2004.08.078 also gives cell counts in the billions so I hope it's roughly correct. $\endgroup$ Sep 8, 2021 at 9:16
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    $\begingroup$ One needs to compare the total number of mutations, $P\times D\times N \times T$ with the total number of base pairs $B$. If $P\times D\times N \times T > B$, i.e., $P\times D\times N \times T/B >1$, then the mutations explore all of the genome. In fact the condition is even more stringent, if we take into account that it is a stochastic process, i.e., some bases will mutate multiple number of times. $\endgroup$ Sep 8, 2021 at 10:04
  • $\begingroup$ 10 mL is a very large volume of cells to be growing on a plate (solid media). Cells do not grow in the medium but on top of it. I would suggest instead using a number of cells in a colony 3.3e9 (bionumbers.hms.harvard.edu/…). Alternatively, one could instead assume that the culture is in liquid medium of 10 mL, in which case your 1 billion cell number is close (though other estimates are higher, ncbi.nlm.nih.gov/pmc/articles/PMC3146540). $\endgroup$ Sep 8, 2021 at 10:43

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