Biochemistry Exam IV

Page 1 Exam IV Biochemistry I 1. I. MULTIPLE CHOICE. Choose the BEST answer to the question by CIRCLING the appropriate letter. No other answer will be accepted. You have been sitting in the sun a lot lately, though you know full well that the UV rays in sunlight can cause the formation of pyrimidine dimers in your DNA of your skin cells. Luckily, skin cells have repair mechanisms that will most likely fix the lesions formed by UV light. Which of the following describes enzymes or mechanisms used in such repair processes? A. B. C. D. E. 2. At pH 7.0, the rare protonation of adenine at N1 (pK a=3.8) can lead to: A. B. C. D. E. 3. deprotonation of N7 base pairing with thymidine A:T Æ G:C mutations A:T Æ A:G mutations A:T Æ T:A mutations The sigma factor of E. coli RNA polymerase: A. B. C. D. E. 4. ABC exinuclease of nucleotide excision repair, recognizes the lesion, and cuts out the damaged DNA. RNA polymerase III fill in the damaged area. DNA glycosylases remove the UV-caused lesions. The MutH protein marks the strand for repair. Methyltransferases transfer methyl groups from DNA to Cys residues on the enzyme. combines with the core enzyme to confer specific binding to a promoter. will catalyze synthesis of RNA from both strands of the DNA template in the absence of the core enzyme. is inseparable from the core enzyme. is required for termination of an RNA chain. has none of the above characteristics. Plasmids are: A. B. C. D. E. part of the bacterial chromosome found only in mitochondria. composed of RNA and protein. closed circular double-stranded DNA molecules. found only in viruses. 5. Page 2 Which of the following is NOT involved in the processing of mRNA precursors in eukaryotic cells? A. B. C. D. E. 6. Which of these statements about alternate forms of DNA is false ? A. B. C. D. E. 7. are very specific proteases that cleave peptides at only certain sequences. are highly specialized ribonucleases that degrade mRNA soon after its synthesis. are sequence-specific DNA endonucleases. catalyze the addition of a certain amino acid to a specific tRNA. act at the membrane to restrict the passage of certain molecules into the cell. The enzyme reverse transcriptase is useful in the generation of cDNA clones for which of the following reasons? A. B. C. D. E. 10. a purine nucleotide. a pyrimidine nucleotide. adenosine. AMP. a deoxyribonucleoside. Restriction enzymes: A. B. C. D. E. 9. B-form DNA predominates in aqueous solution; dehydration favors the A-form. Z-form DNA has a helical sense (left-handed) that is opposite that for A-form DNA. B-form DNA has 3.6 base pairs per turn, where the A-form has 2.6. A major difference between the A-form and the B-form is in the "pucker" of the ribose ring. Nucleotide sequence affects whether DNA takes the Z-form. The compound that consists of ribose linked by an N -glycosidic bond to N-9 of adenine is: A. B. C. D. E. 8. Capping of the 5'-end Addition of poly-A Excision of introns Splicing of exons Transport of the pre-mRNA to the cytoplasm It is sensitive to high temperatures and so can be readily "killed" by heat treatment when the reaction is completed. It does not require a primer to initiate polymerization as do most DNA polymerases. It is insensitive to high temperatures and so can survive the many cycles of heating required for the polymerase chain reaction. It is an RNA-dependent DNA polymerase. It lacks the proofreading function of most DNA polymerases and so is able to utilize mRNA from mutated genes as a template. The phosphodiester bond that links adjacent nucleotides in RNA and DNA: A. B. C. D. joins the 3'-OH of one nucleotide to the 5'-phosphate of the next. is susceptible to alkaline hydrolysis. is uncharged at neutral pH. always links A with T and G with C. 11. 12. 13. Page 3 II. STRUCTURE. Draw the structure of either an adenine-thymine OR a guanine-cytosine base pair as found in the Watson-Crick double-helical structure of DNA. Include all hydrogen bonds and identify atoms numbered 1 of each base. The fidelity of proper Watson-Crick base pairing uses interactions with the base pairs in one groove. Identify that groove and use arrows to denote the atoms that have hydrogen-bond interactions to the polymerase (10 pts). III. SHORT ANSWER. Give a brief answer or fill in the blanks as directed to each problem or question below. Compare and/or contrast DNA polymerase I and RNA polymerase from E. coli in regard to each of the following features. (4 pts) (a) Activated precursors (b) Direction of chain elongation (c) fidelity (d) Need for a primer Describe which enzymatic activities (e.g., activity should end in "ase") are responsible for the fidelity of each of the following and what the degree of fidelity is (e.g., 1 mistake in 100 events = 10-2). (7 points) 1. DNA replication a) b) c) 2. Transcription 3. tRNA activation for Ile-tRNAIle when Val is activated 4. tRNA activation for Phe-tRNAPhe 5. Translation Page 4 14. How does the ribose sugar pucker affect DNA structure? (2 pts) 15. Explain why DNA rich in G/C base pairs has a higher Tm than DNA that is rich in A/T base paris. (2 pts) 16. Explain why methylation at O6 of guanidine (methylation of the oxygen at the C6 position) is mutagenic whereas C5 methylation of cytosine is not. If might be helpful to draw a picture. (2 pts) 17. You wish to set up a system for in vitro protein translation from prokaryotes for an mRNA you have isolated. What components would such a system require? (4 pts) 18. The isoleucyl-tRNA synthetase has a proofreading function that ensures the fidelity of the aminoacylation reaction, but the histidyl-tRNA synthetase lacks such a proofreading function. Explain. (3 pts) 19. 20. Page 5 Prokaryotic topoisomerase II, also called DNA gyrase, is specifically inhibited by an antibiotic called novobiocin, which binds to the b subunit of the enzyme. This antibiotic inhibits bacterial DNA replication. What does DNA gyrase do and suggest why a compound that inhibits gyrase would be an effective antibiotic. (2 pts) Early evidence that helped researchers define nucleosome structure is illustrated by the agarose gel above, in which the thick bands represent DNA. It was generated by briefly treating chromatin with an enzyme that degrades DNA, then removing all protein and subjecting the purified DNA to electrophoresis. Numbers at the side of the gel denote the position to which a linear DNA of the indicated size would migrate. What does this gel tell you about chromatin structure? Why are the DNA bands thick and spread out rather than sharply defined? (3 pts) Page 6 IV. 21. LABORATORY You have performed an LDH purification and measured the volumes (corrected as though you used all of the sample from each of the preceding steps), activity concentration by your LDH assay, and protein concentration by your dyebinding assay. You generate the following table: From this data, calculate the total activity, total protein, specific activity, and yield for each step. Then decide which of your steps was the best or worst by putting a "B" and "W" in the last colum for each of these steps. (5 pts) 22. What would be the approximate activity concentration (Units/mL) for an enzyme stock solution when an LDH assay of that enzyme had an initial DA/min of -0.620, where the assay volume was 5 mL in a 1 cm cell? The assay contained 50 µL of a 1:500 diluted enzyme solution. [Use 6200 M-1cm-1 as e] (4 pt) A. B. C. D. E. 23. 5 U/mL 302 U/mL 5000 U/mL 3,125 U/mL more than 5,000 U/mL You decide to examine the effects of a new inhibitor on the properties of your LDH. After doing a number of kinetic assays using different concentrations of substrate and inhibitor, you obtain results that you plot to obtain the following double reciprocal (Lineweaver-Burk) plots shown below. a) Determine the K m and Vmax for the enzyme in the absence and in the presence of inhibitor (be sure to make clear which is which). b) What kind of inhibition is reflected in this experiment? Explain briefly. c) What is the K I for the inhibitor? d) Circle the data point corresponding to [S] = 0.05 mM in the uninhibited reaction. (7 pts) 4 1 v0 (µM/min) -1 [I] 3 1.0 mM 2 0 mM 1 -10 -5 5 10 1 (mM)-1 [S] 20 Page 7 24. For affinity chromatography of LDH, Cibracron-Blue F3GA immobilized on agarose beads were used because ______. A. B. C. D. E. IV. it only binds to LDH and absolutely no other proteins its structure closely resembles the AMP part of NADH it has a higher affinity than pyruvate both A and B both B and C MATCHING 25. Match the named experiment/discovery with the finding. Those listed on the left may have more than one item on the right to be correct. Only one response is required for full credit. (4 pts) _________1. Crick A. structure of double helical DNA _________2. Watson B. laws of inheritance _________3. Avery C. rule for the ratios of bases in DNA _________4. Mullis D. DNA sequence determination _________5. Hershey/Chase E. DNA replication is semi-conservative _________6. Chargaff F. Central Dogma of Molecular Biology _________7. Messelson/Stahl G. DNA carries the genetic information using 32P-DNA _________8. Sanger H. Made PCR possible by incorporating Taq polymerase I. DNA carries the genetic information using traits 26. Circle the letter (right) or number (left) for the condition wherein a DNA duplex would have the higher Tm. Unless stated, all other conditions are identical. (5 pts) _________1. GC content of 50% OR A. AT content of 60% _________2. 1X SSC OR B. 0.1X SSC _________3. 50% formamide OR C. no formamide _________4. 5X SSC, pH 12 OR D. 5X SSC, pH 7 _________5. 1X SSC, pH 1 OR E. 1X SSC, pH 7 27. For the items on the left, identify the BEST description of their use in recombinant DNA technology from the list on the right. Do not use items on the right more than once. Only one response is needed for full credit although some items on the left may have more than one correct response from the list on the right. (6 pts) _________1. cloning vector A. has a selection marker _________2. bacterial expression vector B. has a bacterial promoter adjacent to the cloning site _________3. S1-nuclease C. used to express eukaryotic proteins in bacteria _________4. Polynucleotide kinase D. used to study the structure & expression of a gene _________5. genomic clone E. used to label DNA for use as a radioactive probe _________6. cDNA clone F. used in generation of cDNA G. used to make DNA from mRNA H. used in generation of genomic DNA Answer Key for Exam No. on Test Correct Answer 1 2 3 4 5 6 7 8 9 10 A C A D E C C C D A 11 NH CH3 2 O H N N 1 H O N N N 1 A N HN 1 O N2N N N T N H N1 O C N N G The bottom end of the bp shown here should be labeled as the minor groove with arrows pointing to the N3 of the purine and the O6 of the pyrimidine. 12 13 (a) DNA polymerase uses deoxynucleotide triphosphate precursors (dNTPs) and RNA polymerase uses ribonucleotide triphosphate precursors (NTPs) (b) Both DNA polymerase and RNA polymerase elongated from 5' to 3' (c) DNA polymerase I has a 3' to 5' (backward) exo-nuclease to hydrolyze any mis-added nucleotides (error rate of 10- 8), whereas RNA polymerase does not (error rate of 10- 4). (d) DNA polymerase needs a primer, whereas RNA polymerase does not need a primer. ---1. 2. DNA replication a) b) c) Transcription the 5'->3' polymerase activity of DNA Pol III; 10-4 to 10-5 the 3'->5' exonuclease activity of DNA Pol III; 10-4 to 10-5 miswmatch repair OK (MutH endonuclease) the 5'->3' polymerase activity of RNA polymerase; 10-3 to 10-4 3. tRNA activation for Ile-tRNAIle -the editing or hydrolase activity; 10-2 , but overall with synthetase its 10-4 to 10-6 4. tRNA activation for Phe-tRNAPhe -the aminoacyl-tRNA synthetase; 10-4 to 10-6 5. Translation -the GTPase activity of EF-Tu; 10-4 this activity is slow to allow correct condon-anticodon interactions Page 1 No. on Test Correct Answer 14 15 The ribose pucker governs the relative orientations and distances between the phosphates attached to the 5' and 3' hydroxyl groups. The consequence of changing the pucker is a change from B-DNA to A-DNA (C2'-endo conformation to C3'-endo conformation; with Z-DNA having both, one for pyrimidines and the other for purines, respectively) same as 12-74 Base stacking interactions are higher for G/C base pairs. ---- 16 O6-methylguanine causes G:C to A:T mutations as shown by the better base-pair made with T in the figure (right) than either the C shifted down to make 2 Hbonds (middle), but would be deformed OR the proper form with only one normal H-bond. 5-Me cytosine will not interfere with base pairing as noted in the G:C base pair shown on the left. This is a very common epigenetic modification (a heritable factor affecting gene transcription that is not based on the nucleotide sequence). 17 ribosomes tRNAs 20 amino acids Aminoacyl tRNA synthetases ATP & GTP initiation factors (IF-1, IF-2, IF-3) elongation factors (EF-Tu, EF-Ts, and EF-G) termination factors (RF-1, RF-2, RF-3) Page 2 No. on Test Correct Answer 18 19 20 Isoleucine is similar in structure to several other amino acids, particularly valine. Distinguishing between valine and isoleucine in the aminoacylation process requires the second filter of a proofreading function. Histidine has a structure unlike that of any other amino acid, and this structure provides opportunities for binding specificity adequate to ensure accurate aminoacylation of the cognate tRNA. -Topoisomerase II or DNA gyrase decreases the supercoiling by two negative supercoils using ATP as an energy source. It breaks the DNA strands and under-winds one strand with the other while holding onto the ends using a covalent attachment to active site Tyr residues. -Inhibition of this enzyme makes it more difficult to unwind DNA during DNA replication, thus inhibiting the process and thus inhibiting bacterial growth. -The bands have a periodicity of about 200 bp (200, 400, 600 bp, etc.), showing that the chromatin is protected from nuclease digestion at regular intervals of 200 bp. This suggests that the nucleosomal cores (146 bp) provide the protection; this was verified in numerous subsequent investigations. Thus, the nucleosomes themselves are in a regular array, occurring about once every 200 bp. The nuclease cuts the regions of doublestranded DNA that link but do not bind to the nucleosome cores specifically, the spacer regions of about 60 bp. These regions are not always digested to completion; consequently, some bands correspond to the DNA from single nucleosomes (200 bp), others to two nucleosomes (400 bp), and so forth. If the nucleosomes were randomly distributed in the chromatin, a large number of differently sized DNA fragments would be generated by nuclease cleavage; in this case, a heterogeneous population of DNA fragments would have smeared through the gel. -The bands are thick because the spacer is long (60 bp in this example) relative to the nucleosomal core (146 bp). The nuclease can cut essentially anywhere in the spacer, so the band corresponding to, for example, mononucleosomes has DNAs ranging in length from 146 to 206 bp. 21 The best step was the last one as the SA went up over 10-fold with a 50% loss from the previous step. The worst step was #3 as the SA went DOWN 2-fold and you lost 75% of your protein. Your TF obviously lost it for you. 22 29 C -The number of units (µmole/min) in assay is, using Beer's Law: -0.62/6200 = 100 µM/min x 0.005 L = 0.5 µmole/min or U. -The activity concentration (Units/mL) of the enzyme added to the assay is: 0.5 U/0.05 mL = 10 U/mL -This enzyme was diluted 1:500 from the enzyme stock solution. So, the approximate activity concentration (Units/mL) for the enzyme stock solution is: 10 U/mL x 500 = 5000 U/mL (Answer C) 87- 23 Page 3 No. on Test Correct Answer a) Absence of inhibitor: 23 -1/K M = -10 mM-1 K M = 0.1 mM 1/Vmax = about 0.5 (µM/min)-1 Vmax = ~2 µM/min or ~2 x 103 Units/mL Presence of inhibitor: -1/K Mapp = -5 mM-1 K Mapp = 0.2 mM 1/Vmaxapp = same as above in the absence. b) Competitive, the Vmax does not change, only K m. c) K Mapp = K M (1+[I]/K I): 0.2 = 0.1 (1+[I]/K I) so, (1+[I]/K I) = 2 If (1+[I]/K I) = 2, then [I]/K I = 1, and since [I]=1mM, K I = 1 mM d) A circle should be on the no-inibitor line (I = 0 mM) at the 1/[S] = 20 (1/0.05 = 20).. 24 B ---25 26 27 1. 2. 3. 4. 5. 6. 7. 8. A or F A I H G C E D 1. 2. 3. 4. 5. 1 2 C D E 1. 2. 3. 4. 5. 6. A or C A or B or C F or G E D C Page 4