The mutation rate in bacteria is about 3 nucleotide changes per 10 billion nucleotides per cell generation. Under laboratory conditions, bacteria such as Escherichia coli can divide and double in number about every 40 minutes. If a single Escherichia coli cell is allowed to exponentially divide for 10 hours in this manner, how many mutations would you expect to observe on average in the genome (4.5 million nucleotide pairs) of each of the resulting bacteria compared to the original cell? Assume all mutations are neutral; that is, they do not affect the cell-division time.
A) Less than 0.001
B) About 0.02
C) One or two
D) About 10
E) About 100

Question

Quizplus · Accepted Answer

About 15 cell generations pass in 10 hours of growth. Thus, the expected average number of mutations is equal to:
(15 generations) × (3 × 10-¹⁰ mutations/generation/nucleotide) × (4.5 × 10⁶ nucleotides/genome)
= ~2 × 10-² mutations per genome.
This is equivalent to about 98% of the bacterial cells having zero mutations and the remaining 2% having only one mutation each. Please note that the number of mutations in different cells in a given culture will vary widely around the average calculated above.

The Mutation Rate in Bacteria Is About 3 Nucleotide Changes