*** NOTE: This answer has been corrected ***************************************************************
On 31.01.2017 (11:23) Isato wrote: "I am not sure whether the answer s in the ex16_ecoli are correct..."

There was a mistake because it assumed that only one single mutation occurs in the entire population
every time it doubles. In reality, mutations are probabilistic-the larger the population, the more "new"
mutations you expect to see in a single generation.
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Mutations are introduced into the E. coli genome at the rate of 1 mutation per 10^9 base pairs per
generation. Imagine that you start with a population of 10^6 E. coli, none of which carry any mutations
in your gene of interest, which is 1000 nucleotides in length and not essential for bacterial growth and
survival. In the next generaion, after the population doubles in number, what fraction of the cells,
on average, would you expect to carry a mutation in your gene? After the population doubles again,
what would you expect the frequency of mutants in the population to be? What would the frequency
be after a third doubling?

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LOGIC:
Gene mutation rate (u) = (1000 bp / 10^9 bp) = 10^-6 per replication.
New mutants per doubling = (Number of new DNA strands synthesized) * (u).
Existing mutants also double in number along with the rest of the population.
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Gen 0: 10^6 cells
       0 mutants

Gen 1: 2*10^6 cells (after 1st doubling)
       New mutants: 10^6 replications * 10^-6 = 1
       Total mutants: 1
       Frequency: 1 of 2*10^6 = 0.5 * 10^-6

Gen 2: 4*10^6 cells (after 2nd doubling)
       Old mutants doubled: 1 * 2 = 2
       New mutants: 2*10^6 replications * 10^-6 = 2
       Total mutants: 4
       Frequency: 4 of 4*10^6 = 1.0 * 10^-6

Gen 3: 8*10^6 cells (after 3rd doubling)
       Old mutants doubled: 4 * 2 = 8
       New mutants: 4*10^6 replications * 10^-6 = 4
       Total mutants: 12
       Frequency: 12 of 8*10^6 = 1.5 * 10^-6

Answer:
The frequency of mutants increases linearly by 0.5 * 10^-6 each generation.