Answer to DNA Repair Mid-term Exam Question
April, 1997

1. Draw a diagram of the structure of a T-C (6-4) dimer. The structure of a T-C (6-4) dimer is provided in the lecture notes. Click here to see the structure as drawn in the lecture notes. Of course, the diagram in the lecture notes is in color and has more information in it than is relevant to this question. Consequently, I didn't expect you to produce such a detailed diagram in response to the exam question. The greatest level of detail that any of you provided--which was more than I expected--is shown in this diagram:

Simple Diagram

I gave full credit for the following simpler diagram that many of you provided, because, despite its simplicity, this diagram nevertheless provides complete information about what distinguishes a TC (6-4) dimer from a normal TC dinucleotide:

Even more simple diagram

2. By which pathway would such a dimer be repaired: direct reversal, mismatch repair, base excision repair, or nucleotide excision repair? The answer, of course, is nucleotide excision repair. Although direct reversal might be possible, I'm not aware of any evidence for the existence of a direct reversal pathway for (6-4) dimers. So far as I'm aware, neither the mismatch repair pathway nor the base excision repair pathway is capable of repairing a lesion such as an intra-strand dimer that is larger than a single nucleotide.

3. Describe the steps in the pathway responsible for repairing (6-4) dimers. List all the proteins required for this pathway. If the role of the protein in repair is known, state what that role is. I apologize for the fact that I did not explain in the question that I would be satisfied with a summary of the nucleotide excision repair pathway in either E. coli or in mammalian cells. Here is the level of detail I was looking for:

In E. coli:

  1. a. Two molecules of UvrA bind 1 molecule of UvrB in a step requiring ATP hydrolysis.
  2. b. The A2B complex binds to the damaged site on the DNA molecule.
  3. c. With the aid of ATP hydrolysis, the UvrA molecules are released and UvrB remains tightly bound, possibly distorting the DNA.
  4. d. The next 3 steps (e-g) all require ATP binding but not ATP hydrolysis.
  5. e. The UvrC protein binds near UvrB, activating the nuclease activity of UvrB.
  6. f. UvrB nicks the DNA about 4 nucleotides downstream of the damage. This activates the nuclease activity of UvrC, possibly as a result of a DNA conformational change.
  7. g. UvrC nicks the DNA about 7 nucleotides upstream of the damage.
  8. h. Next, UvrD (also known as DNA helicase II) uses energy from ATP hydrolysis to unwind the oligonucleotide from between the two nicks, leaving a gap of about a dozen nucleotides. At the same time, the UvrC protein is displaced.
  9. i. The gap is filled in by DNA polymerase I or II, using the non-damaged strand as template. At the same time, UvrB is displaced.
  10. j. The resulting nick is sealed by DNA ligase.

In mammalian cells:

  1. a. XPA and RPA bind to the damaged DNA. Both can bind damaged DNA independently, but the XPA/RPA complex binds more strongly than either alone.
  2. b. XPA interacts with TFIIH (a complex of 9 polypeptides also important for transcription) and promotes its binding at the damaged site.
  3. c. XPA and TFIIH both interact with XPC plus HHR23B, recruiting them to the damaged site. XPC and HHR23B help to stabilize interactions among the proteins binding to the damaged site.
  4. d. Two of the TFIIH subunits (XPB and XPD) are DNA helicases of opposite polarities. These are thought to unwind the DNA around the damaged site, thus permitting the binding of three additional proteins.
  5. e. The three additional proteins are the structure-specific nucleases, XPF+ERCC1 and XPG. Their binding is aided by the facts that XPA interacts with ERCC1 and RPA interacts wtih XPG.
  6. f. The complex of XPF+ERCC1 nicks the damaged strand 22-24 nucleotides upstream of the damage, and XPG nicks about 5 nucleotides downstream of the damage.
  7. g. The oligonucleotide between the two nicks is displaced, possibly by the XPB and/or XPD helicases.
  8. h. The resulting gap is filled in by DNA polymerase delta or epsilon, using the undamaged strand as template.
  9. i. The resulting nick is sealed by DNA ligase I.

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