When a bacteria A divides it produces two cells A', A''. Each of them receives a copy of the chromosome/plasmids. Now, DNA replication occurs way before division in a semiconservative manner. That is, each new chromosome has an 'old' strand and a 'new' strand. Since the polymerase is error prone, my belief is that both genomes can potentially have mutations. Now when people refer to the mutation rate/genome/replication, e.g. 3x10-4 , Does this mean that:

  1. Only one of the resulting genomes have this mutation rate? (according to me not likely).

  2. The mutation rate takes into account the total number of mutations in both new genomes.

  3. Each of the genomes can potentially have independently as many mutations as the mutation rate specifies.

  • $\begingroup$ i'm referring to replication erros, and not the two strands in a single genome, but to the two new strands in each of the two genomes. $\endgroup$
    – raygozag
    Commented Jun 25, 2013 at 15:33
  • $\begingroup$ edit of previous comment: I believe it would be option 2 or 3 depending on what sort of error has occurred and when - i.e. if there is damage to the old strand it could affect a strand of dna in both cells, if it is a transcription error in the new strand it may only affect a strand dna in one cell (A') and the A'' cell may be identical to the parent... (just a quick estimate at an answer) $\endgroup$
    – rg255
    Commented Jun 25, 2013 at 15:36

1 Answer 1


Your question comes down to the definition of the units you state in your question - mutations genome-1 generation-1. In my opinion the answer is your choice (2).

How are mutation frequencies measured? Usually it is a case of scoring the appearance of a mutant phenotype as an indicator of underlying mutational events (see Drake,JW (1991)A constant rate of spontaneous mutation in DNA-based microbes. Proc. Natl. Acad. Sci. USA 88:7160-7164 for a detailed explanation.) Since, as you say, DNA replication is semi-conservative, the two daughter cells of a bacterial cell division each inherit one old DNA strand and one new DNA strand and so have the same probability of sustaining a given mutation. The overall mutational frequency has to be an aggregate of what has happened in both of them (and in all other cells in the population) since there is no way of distinguishing between them. In terms of your question the answer is (2) and this is why the mutation rate is normalised per genome.

In practice this figure is arrived at as follows. Suppose that you study a 1 kb gene, gene X, and measure the rate of spontaneous mutation in that gene (by some convenient assay) as 1 mutation per 107 cell divisions (again see the Drake reference for details). Assume also that the bacteria have a genome size of 2 x 106 bp. Since each cell division equates to the creation of one new genome we get:

mutations per gene X per generation = 1 x 10-7

mutations per bp per generation = 1 x 10-7/1000 (because it is a 1000 bp gene)

mutations per genome per generation = 1 x 10-7 * 2 x 106/1000

mutations per genome per generation = 0.0002

Now this is an underestimate because not all mutations will create a detectable change in phenotype. Drake uses a correction factor of 3.12 for this.

true mutations per genome per generation = 0.0002 * 3.12 = 0.0006 or 6 x 10-4

These were made-up numbers. See review in Drake JW et al.(1998) Rates of spontaneous mutation. Genetics 148: 1667-1686 for real examples.

  • $\begingroup$ Sorry for being so dense i think i understand why the rate is normalized by genome, but then can i say that daughter A' has N mutations and daughter A'' M mutations, then can i say that both N and M are distributed as a poisson variable with the mutation rate as the lambda? $\endgroup$
    – raygozag
    Commented Jun 26, 2013 at 18:48

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