When a bacteria A divides it produces two cells A\', A\'\'. Each of them receive

Question

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:

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

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

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

Explanation / Answer

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

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