PS3 Key - MCB110 Problem 3 1. True or False_T_ Depurination of a base gives rise to an abasic site in the DNA strand_ F_

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Unformatted text preview: MCB110 Problem 3 1. True or False __T__ Depurination of a base gives rise to an abasic site in the DNA strand __ F__ Uracil DNA glycosylase makes a nick 3’ to an abasic site during base excision repair It cuts at 5’ to an abasic site __ F__ Mismatch repair pathway fixes chemically modified bases __ F__ Base (BER) and nucleotide excision repair (NER) pathways both require helicases __ F__ MutH/S/L are involved in the same repair pathway as the XP proteins __ T__ Deamination can result in the changing of one base to another, e.g. C ­ ­>U __ T__ A pyramidine ­pyramidine dimer is an example of intrastrand covalent linkage 2. Which DNA polymerase adds new nucleotides at the end of repairing a deamination event? Is it processive? PolI, it’s not processive since there’s only one nt to repair. 3. During mismatch repair, MutH needs to recognize the palindromic sequence 5’ ­ GATC ­3’ to make a nick at this site, then a helicase/exonuclease complex can remove the displaced strand for PolIII to fill in the ssDNA gap left behind. a) How does MutH know which strand to cut? In E. coli, what other enzyme(s) facilitates this recognition? MutH recognizes hemi ­methylated 5’ ­GA(me)TC ­3’ sites and cuts at these sites. DamI in E. coli methylates this site and facilitate strand recognition. b) What kind of polarity, 5’ ­>3’ or 3’ ­>5’, does the helicase/exonuclease complex associate with mismatch have? Both, depends on where the hemi ­methylated site is relative to the mismatch site to be repaired. c) Why can the displaced strand be up to 1kB in length? Because 1kB is the farthest possible distance in E. coli between a potential mismatch site and a GATC sequence that could become hemi ­methylated. 4. What mechanism/pathway have cells evolved to have for fixing depurination events? What enzyme would cut the resulting abasic site and where would it cut? Does this same enzyme participate in deamination events? Base excision repair. AP endonuclease cuts at the 5’ abasic site. Yes it also participates in deamination events, because deamination is recognized by glycosylases which would get rid of the base, creating an abasic site for AP endonuclease to recognize. 5. E. coli and humans both use MutS/L complex to recognize and bind mismatches, recruiting an endonuclease (MutH in bacteria) to cut the incorrect strand. The complex in bacteria is a homodimer (MutS2/MutL2) and in eukaryotes is a heterodimer composed of peptides coded by different genes. a) What kind of advantages do humans have in having heterodimeric conformation of MutS/L complexes? Heterodimeric conformation allows humans to recognize different types of mismatches b) Why don’t humans have a protein that resembles MutH? Because our GATC sites cannot be hemi ­methylated since we don’t have DamI methylase. We use a yet ­to ­be ­discovered mechanism to distinguish the correct strand as a template in repair. c) If you want to ascertain that MutS/L can thread DNA on both sides of the mismatch, where would you choose to start mutagenesis? Monomer ­monomer interface, residues responsible for ATP hydrolysis, or DNA binding site? Would you prefer to do this in E. coli or humans, and why? You would mutagenize residues in the ATP hydrolysis site. This way, the dimer can still form and bind DNA and mismatch, but one of the monomers can no longer thread DNA through its DNA interacting surface. This can only be achieved in humans since mutating one monomer in a heterodimer won’t affect the other monomer. 6. During SOS response for DNA repair in E. coli, Pol V also binds beta ­clamp, how can Pol III compete away Pol V once the mutation is patched up? Side note: would you expect Pol V active site to have a tight or loosened grip, and why? What potential advantage(s) can Pol V contribute to E. coli growth? Because Pol III favor correct base pairing more than Pol V, thus after the ds break is passed, Pol III can come in and continue replication. Pol V would have a loosened grip because it needs to accommodate mistakes. Because it’s error ­prone, E. coli can evolve characteristics such as antibiotic resistance from the new information that Pol V generates. 7. We learned in class that normally replicative ligase can only seal nicks in 1 strand of a DNA duplex, how then are dsDNA break re ­ligated in nonhomologous end ­ joining (NHEJ)? In NHEJ repair during SOS response, dsDNA breaks are bound by Ku proteins regardless of blunt or overhang ends. Then a specialized ligase, Ligase 4, recognizes these Ku ­bound ends and seals them—error prone. 8. Match the types of DNA damage and the proteins involved to the proper repair pathway, there might be more than one correct answer for each pathway a. deamination NER AP endonuclease b. incorrect base pairing BER MutS/L c. intrastrand crosslinking Mismatch repair Glycosylase d. pyramidine ­pyramidine dimers PolV UvrA/B e. depurination Direct repair XP/RAD proteins f. interstrand crosslinking UvrD g. base modification NER: c, d—UvrA/B, UvrD, XP/RAD proteins BER: a, e—Glycosylase, AP endonuclease Mismatch repair: b—MutS/L, UvrD PolV—f (that results in ds breaks) Direct repair: g 9. Supercoiled DNA can have the status of being overwound or underwound. Which of these states is associated with negatively supercoiled DNA and which is associated with positively supercoiled DNA? Also, describe for both negatively and positively supercoiled DNA whether unwinding or overwinding of the duplex is favored. Negatively supercoild DNA is underwound and positively supercoiled DNA is overwound. Negatively supercoiled DNA favors unwinding of the double helix, and positively supercoiled DNA favors overwinding. 10. What characteristics of a duplex DNA determine the Writhe introduced into a plasmid? The twist and linking number determine the value of the writhe that DNA is forced to assume. Wr = Lk – Tw. 11. The plasmid pcDNA is a closed circular double ­stranded DNA molecule with 5,500 base pairs. Assume that in B ­DNA duplex there are 10 bp/turn. A. How many helical turns are there in the relaxed molecule? # of turns (Tw) = #bp/pitch (bp per turn) = 5500/10 = 550 B. What is the linking number of the molecule when it is relaxed B ­DNA? Lk = Tw + Wr, since there’s no writhe because it’s a relaxed DNA, then Lk = Tw = 550. 12. The plasmid from problem 11 is transferred from aqueous solution to 70% ethanol. Under these conditions, the structure changes from B to A ­DNA due to the relatively lower water concentration. (A ­DNA has 11 bp/turn). A. What is the linking number now? Lk = 550, unchanged because no DNA strands was broken B. How many helical turns are there now? What is the Writhe for this molecule? # of turns (Tw) = 5500/11 = 500 Wr = Lk – Tw = +50 C. Which molecule (from problem 11 or 12) would have the more compact structure? Molecule in problem 12 would be more compact because it must become more writhed, whereas the molecule from problem 11 is relaxed and has no writhe under those conditions. 13. Linking number can only be changed by breaking one or both strands of the DNA, winding them tighter or looser, then rejoining the ends. Enzymes called Topoisomerases can change linking number. Explain the key difference between the 2 classes of Topoisomerases (Topo I and Topo II). What are the changes in Lk these two classes make (include the subtypes and explain in terms of mechanism)? Topo I cuts 1 strand Topo II cuts 2 strands TopoIA changes Lk by 1 through a pass and rejoin mechanism; TopoIB can change Lk by 1 or more (n) because after cutting, one end is allowed to free rotate (swivelase mechanism); TopoII changes Lk by 2 because cuts both strands. 14. Topoisomerase II was selected as a target for discovery of anticancer drugs and led to the development of active anticancer drugs such as etoposide and doxorubicin. Why is Topoisomerase II a good target for anticancer drugs? TopoII has served as a great anticancer target because inhibition of TopoII leads to cell death. Cancer cells are rapidly dividing and can thus be targeted by TopoII inactivation more readily than normal cells. The double helical structure of DNA creates topological problems during DNA replication. For example, during DNA unwinding positive supercoils are generated ahead of the replication fork, and they must be resolved for replication fork progression (TopoI and TopoII). In addition, the daughter replicated chromosomes need to be decatenated for chromosome segregation (TopoII) ...
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This document was uploaded on 09/12/2011.

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