Spring 2006 Biochemistry 302 Exam 1

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1 Name Spring 2006 Biochemistry 302 Exam 1 Directions: This exam has 36 questions/problems totaling 90 points. Check to make sure you have all six pages. Some questions have multiple parts so read each one carefully. Succinct answers are preferred. Continue answers on the back of the page if the space provided is not sufficient. Include drawings in your answers as you see fit. Partial credit will be given. You have up to two hours to complete the exam. Good luck.

2 I. Short answer. A few choice words or phrases should suffice. 1. The molecule shown below inhibits the synthesis of a certain type of polynucleotide. Based on your knowledge of the substrate requirements for DNA polymerase and RNA polymerase, predict which enzyme would be most susceptible to inhibition by this drug. Provide a biochemical rationale for your answer (3 points). RNA polymerase because the ribose has a hydroxyl group attached to the 2 carbon 2. Of the two ribose configurations shown below, which one would you expect to find in double-stranded RNA and why? (3 points) The N (anti) configuration because the hydroxyl group at the ribose 2 position is sterically unfavorable in the S (syn) configuration. 3. What is the biochemical basis for DNA strand separation under basic conditions (3 points)? Disruption of hydrogen-binding between base-pairs 4. How does the activity of an endonuclease differ from that of an exonuclease (3 points)? An endonuclease cleaves phosphodiester bonds at sites within a polynucleotide chain while exonucleases remove nucleotides from either the 5 or 3 end. 5. Write the chemical equations describing the synthesis of DNA and RNA (4 points). DNA: (dnmp) n + dntp (dnmp) n+1 + PP i d=deoxyribo, r=ribo RNA: (rnmp) n + rntp (rnmp) n+1 + PP i or (NMP) n + NTP (NMP) n+1 + PP i

3 II. Multiple choice. Write the letter of the best answer on the line provided. (2 points each) 6._B_ Self-complementary double-stranded RNA adopts this type of helical configuration. A. B-form B. A-form C. Z-form D. H-form 7._A_ The Meselson-Stahl experiment using density gradient centrifugation of differentially radioisotope-labeled genomic DNA from E. coli provided conclusive evidence of this feature of DNA replication. A. Semi-conservative B. Bidirectional C. Initiation from a specific origin D. Termination at a fixed site E. Discontinuous 8._D_ The three-dimensional thumb, finger, palm architecture is a characteristic feature of this enzyme. A. DNA N-glycosylase B. Cre recombinase C. RecBCD nuclease D. Klenow fragment of DNA polymerase 1 9._D_ One of the following is NOT a component of the nucleosome core particle. A. Histones H3 and H4 B. Histones H2A and H2B C. 146 bp DNA D. Histone H1 10._B_ This class of DNA modifying enzymes possess both nuclease and ligase activity. A. Helicase B. Topoisomerase C. Primase D. Histone acteyltransferase 11._A_ This heterocyclic base is normally found only in RNA. A. Uracil B. Thymine C. Cytosine D. Adenine

4 12._C_ DNA molecules with the theoretical propensity to adopt a cruciform configuration generally exhibit this type of sequence signature. A. Polypurine/polypyrimidine strand asymmetry B. Stretch of 5-methyl cytosine residues C. Inverted repeats 13._E_ A 100 base pair DNA molecule composed of 75% AT content will: A. Melt at a higher temperature than a 100 bp DNA molecule of 75% GC content B. Melt at a lower temperature than a 100 bp DNA molecule of 75% GC content C. Have a lower UV absorbance than its denatured counterpart D. Have a higher UV absorbance than its denatured counterpart E. Both B and C F. Both A and D 14._B_ The eukaryotic general transcription factor, TATA-binding protein (TBP), and the E. coli site-specific recombination protein, integration host factor (IHF), are functionally related in that their association with DNA induces this alteration in DNA structure. A. Melting B. Bending C. Supercoiling 15._D_ This protein does NOT possess ATPase activity. A. RecA recombination factor B. DnaA replication initiation factor C. Rho helicase D. SSB (Single-strand DNA-binding protein) 16._D_ This enzyme is involved in direct repair of damaged DNA. A. DNA photolyase B. O 6 -alkylguanine alkyltransferase C. α-ketoglutarate-fe 2+ -dependent dioxygenase D. All of the above E. None of the above 17._A_ The first step in base excision repair requires the action of this type of enzyme. A. DNA N-glycosylase B. AP endonuclease C. DNA polymerase D. DNA ligase

5 III. Matching. Write the letter of the term that best matches the description in the left column on the line provided (2 points each). 18. Synthesis of pre-ribosomal RNA _G_ A) DNA polymerase I 19. Resolution of Holliday junction _D_ B) Protein-DNA interaction 20. Recognition element in mismatch repair _J_ C) Telomerase 21. Guanine crosslinking agent _K_ D) RuvABC complex 22. 5-3 exonuclease activity _A_ E) RNA 23. Core promoter element _H_ F) DNA ligase 24. Nuclease protection assay _ B_ G) RNA polymerase I 25. Base-labile phosphodiester bond _E_ H) TATA box 26. Reverse transcriptase _C_ I) DNA 27. Supercoiled plasmid _I_ J) GA m TC 28. Adenylated enzyme intermediate _F_ K) Cisplatin IV. True/False. Write T or F on the line provided. If a statement is deemed false, provide a BRIEF statement explaining why you think it is false (2 points each). 29. Base-stacking and base-pairing interactions contribute to the thermal stability of the DNA double helix. _T_ 30. 2,3 dideoxy NTPs inhibit DNA polymerase-catalyzed DNA chain elongation. _T_ 31. The inability of E. coli RNA polymerase and eukaryotic RNA polymerase II to proofread RNA transcripts is due to an absence of intrinsic phosphodiesterase activity. _F_ GreB and TFIIS elongation cofactors can change active site to hydrolysis mode in stalled RNAP. 32. E. coli sigma factors (e.g. σ 70 ) enhance the DNA-binding affinity and specificity of RNA polymerase for core promoter elements of certain genes. _T_ 33. In human cells, thymine dimers are dealt with by a combination of mechanisms including translesional bypass catalyzed by DNA polymerase η (eta) and nucleotide excision repair initiated by members of the XP protein family. _T_

6 V. Questions requiring a few sentences to answer (6 points each). 34. Describe the structural biology underlying 8-oxoguanine recognition and removal by hogg1 (oxog N-glycosylase) in the context of double-helical DNA. The x-ray crystal structure of a mutant hogg1-dna complex suggests that this base excision repair enzyme flips 8-oxoG out of the DNA helix to presumably facility catalytic attack of the β-glycosyl bond. This unusual configuration appears to be stabilized by atomic interactions (H-bonding and van der Waals contacts) between certain amino acid side chains of the enzyme and the flipped out 8-oxoG and estranged cytosine residues. 35. Why are some mechanisms of DNA repair costly from a chemical energy perspective? Cite an example which supports your argument. Repair of a single damaged base may require the destruction of an entire protein (in the case of alkylguanine alkyltransferase) or the removal and replacement of an extended polynucleotide sequence in the case of nucleotide excision repair or mismatch repair. The former is costly in terms of chemical energy expended in synthesizing the protein (and the transcript encoding it for that matter) while the latter is costly in terms of consumption of high energy phosphates (dntps). 36. DNA and RNA polymerases are metalloenzymes. Describe how certain metal ions contribute to catalysis from both a structural and chemical perspective. In the active site of each enzyme, two magnesium (Mg 2+ ) ions are coordinated to the side chains of certain conserved aspartate residues and the phosphate groups of incoming dntp (DNA Pol) or rntp (RNA Pol) substrates. This structural arrangement facilitates the formation of a pentacovalent transition state initiated by nucleophilic attack of the terminal 3 OH group of the primer strand on the α phosphorus of the incoming NTP. One Mg 2+ probably serves to polarize the 3 OH promoting nucleophilic attack and phosphodiester bond formation while the other facilitates displacement of the PPi product.