Calculating bacterial mutation rate

I'm reading Molecular Biology of the Cell by Alberts et. al and at one point, the authors mention:

One finds that a single gene that encodes an average-sized protein (~$$10^3$$ coding nulceotide pairs) accumulates a mutation (not necessarily one that would inactivate the protein) approximately once in about $$10^6$$ bacterial cell generations. Stated differently, bacteria display a mutation rate of about three nucleotide changes per $$10^{10}$$ nucleotides per cell generation [emphasis mine].

I'm trying to figure out how they came up with the numbers in the text stated in bold. Here's what I have so far: According to the first sentence, we have $$\frac{1}{10^6}$$ mutations per $$10^3$$ nucleotides per cell generation. So per $$10^{10}$$ nucleotides per generation, we have $$\frac{10^{10}}{10^3}\times\frac{1}{10^6} = 10$$ mutations, which is not the 3 that they mention. What am I missing here?

One mutation per 10,000 bases per 1,000,000 generations.

1 mutation / 103 bases / 106 generations

= 1 mutation / 109 bases/generation

= 3 mutations / (3*109) bases/generation

= 3 mutations / 109.5 bases/generation

Since all these are rounded approximations, if you started with more precise numbers and round at the end (always round as your final step!), it would be very easy to get 3 mutations / 1010.

• Thanks for the answer. Why can't you just jump from the 1 mutation / $10^9$ bases/generation step to deduce 10 mutations / $10^{10}$ bases/generation i.e do what you did but multiply by 10 instead of 3? May 6, 2022 at 16:10
• @AnIgnorantWanderer I made it 3 to match their estimate. I assume they say 3/10^10 rather than 10/10^10 because that's closer to their actual estimate if they don't round too early. May 6, 2022 at 16:13