During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine that this error was not corrected and that it has no effect on the ability of the progeny to grow and reproduce
A. After this original bacterium has divided once, what proportion of its progeny would you expect to contain the mutation?
B. What proportion of its progeny would you expect to contain the mutation after three more rounds of DNA replication and cell division?
A. One-half, or 50%. DNA replication in the original bacterium will create two new DNA molecules, one of which will now carry a mismatched C-T base pair. So one daughter cell of that cell division will carry a completely normal DNA molecule; the other cell will have the molecule with the mutation mispaired to a correct nucleotide.
B. One-quarter, or 25%. At the next round of DNA replication and cell division, the bacterium carrying the mismatched C-T will produce and pass on one normal DNA molecule from the undamaged strand containing the T and one mutant DNA molecule with a fully mutant C-G base pair. So at this stage, one out of the four progeny of the original bacterium is mutant. Subsequent cell divisions of these mutant bacteria will give rise only to mutant bacteria, whereas the other bacteria will give rise to normal bacteria. The proportion of progeny containing the mutation will therefore remain at 25%.
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