Chapter 12

Question 12.1

The cross-linked pyrimidine dimer causes a distortion in the DNA that prevents base pairing in the active site of the DNA polymerase.

Question 12.2

  1. where represents O6-meG.

Question 12.3

Any three of the following: defects in repair; UV light; TLS error-prone bypass; oxidative damage; spontaneous hydrolysis. For these lesions to become mutations, replication must occur (before repair), producing a naturally occurring base pair that is different from the original base pair.

Question 12.4

Serine auxotrophs can grow only on serine-containing medium; when plated on a serine-free medium, the cells will die—unless they have undergone a reversion mutation to restore the serine-synthesizing pathway. Treatment with a genotoxic agent, even though it kills most cells by producing mutations in vital genes, produces some of these particular reversion mutations.

Question 12.5

The survivors arise from spontaneous reversion mutations caused by spontaneous hydrolysis, oxidative damage, or natural sources of irradiation.

Question 12.6

(a) The colonies arose because background (spontaneous) mutations included reversion mutations that reversed the cells’ histidine dependency. (b) 2-Aminoanthracene causes DNA damage; some of the lesions escape repair before DNA replication, thus forming a mutation. Some of these are reversion mutations in the histidine-pathway gene, allowing cells to survive in a histidine-free medium. (c) 2-Aminoanthracene is a potential carcinogen, because it causes mutations in the Ames test.

Question 12.7

(a) G≡C to A=T. (b) C≡C to T=A. (c) A=T to G≡C

Question 12.8

(a) XP mutant cells usually contain a mutation in a gene required for NER, the main repair pathway for UV lesions in humans. Single-strand DNA breaks are produced during NER, accounting for the short ssDNA fragments observed in normal cells after irradiation. However, a defective NER system in the XPG mutant cells prevents formation of single-strand breaks. (b) XPG cells are defective at a step preceding the first strand incision of NER, which is needed to produce the fragmented ssDNA.

Question 12.9

The initiation step. In global NER, the XPC protein recognizes the lesion. In TCR, RNA polymerase recognizes the lesion, by stalling at the lesion.

Question 12.10

Both processes cleave the phosphodiester backbone, remove the pentose phosphate, then insert a correct dNTP and ligate the nick. In BER, a specific DNA glycosylase cleaves a damaged base from the pentose to form the abasic (AP) site.

Question 12.11

The most common DNA lesions leading to G–T mismatches are insertion of T opposite an O6-methylguanine lesion during replication and deamination of a 5-methylcytosine. In both cases, it is the T that is incorrect and potentially mutagenic if not repaired.

Question 12.12

Frameshift mutations would lead to alteration of many amino acid residues in the protein product and could be caused by template slippage of DNA polymerase in the region with the repeated A residues.

Question 12.13

Reactive oxygen species generated during normal aerobic metabolism are a major source of DNA damage, because they form free radicals that react with the DNA.

Question 12.14

With only one good copy of a key DNA repair gene, there is a good chance that the gene will be inactivated in one or more cells by normal mutagenic processes such as unrepaired oxidative damage. Whenever this happens, a DNA repair system is no longer present in that cell, and the mutation rate increases substantially. Eventually, mutations occur in—and inactivate—one or more genes that normally control cell division, and a cancer cell is born.

Question 12.15

About 1,700 phosphodiester bonds derived from dNTPs are expended in the repair: 850 in the DNA degraded between the mismatch and the GATC sequence, and another 850 in the DNA synthesis needed to fill the resulting gap. ATP is hydrolyzed by the MutL-MutS complex and by the UvrD helicase.

Question 12.16

NER and BER occur only in double-stranded DNA. Both processes excise the damaged base or bases from the damaged strand, leaving a gap that can be filled in—but only if an undamaged complementary strand is present.

S-13

Question 12.17

Each molecule of O6-methylguanine methyltransferase is used only once and is degraded after the repair reaction. Thus, the reaction expends all the energy needed to synthesize the protein, along with all the energy used to mark the protein for degradation and to carry out that degradation.

Question 12.18

(a) There may have been a trace contaminant of Pol III or Pol II in the preparation of UmuD′ and UmuC. (b) Pol III is the main replicative DNA polymerase in the cell. A strain with a completely inactivated Pol III would be dead. The temperature sensitivity allows the cells to be grown at a permissive temperature, but Pol III can be inactivated, when needed, by increasing the temperature. (c) Fraction 56 contains both Pol III and Pol V. The Pol V can replicate over the lesion, and Pol III can then extend the DNA. (d) The mutant Pol III is inactivated at 47°C. (e) Fraction 64 contains UmuC (and UmuD′, not shown) almost exclusively. The lack of temperature sensitivity provides evidence that UmuC and/or UmuD′ have a DNA polymerization activity.