Exam II 2005 - VVVVVV General Biochemistry Examination 11...

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Unformatted text preview: VVVVVV General Biochemistry Examination 11 Thursday, November 17‘“, 2005 4:00 — 6:00 PM Put your name and identification number on the answer sheet. Mark only one answer for each question on the answer sheet. Use only the pencil provided for the examination Do not fold or bend the answer sheet. Make sure you have 13 question pages. Read each question carefully. Which protein is required for recognizing the TATA—box during transcription initiation in eukaryotes? A. Factor sigma B. TF—HD C. TF—IIH D. TF—IIF E. RNA polymerase II Which one ofthe following statements regarding the termination oftranscription in prokaryotes is correct? A. Pausing of RNA polymerase during termination oftranscription depends on a molecular feature in the newly synthesized RNA containing a self complementary GC rich stem—loop structure followed immediately by repeated U’s. B. Each termination site contains a unique base where transcription stops. C. All prokaryotic termination sites are identified by a stem—loop—forming sequence which must be recognized by protein rho. D. The rho protein participates in transcriptional termination by binding to RNA polymerase and inactivating its 5’—>3’ polymerase activity. A mutation in the DNA changing at codon to UAA is known as WCOW> Missense Silent Nonsense Frameshift None ofthe above Owing to “wobble,” which of the following occurs? Strict Watson—Crick base—pairing of codons and anticodons occur. Certain anticodons pair with codons which differ at the 3’-end. Certain codons pair with anticodons which differ at the 3’-end. Error in protein synthesis is kept to a minimum. Mutations in the DNA can occur without any phenotypic consequence. mc0w> How many high-energy phosphate—bond equivalents are utilized in the process of activation of amino acids for protein synthesis? mcow> 0 l 2 3 4 Which ofthe following combination of statements concerning inhibition of eukaryotic protein synthesis is true? 1. Loss of5’—CAP in mRNA 2. An increase in phosphorylation ofthe or—subunit ofinitiation factor eIF-2. 3. An increase in state ofphosphorylation ofinitiation factor elF—4E A. 1 only B. 2 only C. 3 only D. 2 and 3 only E. l and 2 only F. l and 3 only genetic code is said to be degenerate when A triplet nucleotide can code for more than one amino acid. A tRNA can be aminoacylated by more than one aminoacyl-tRNA synthetase. An amino acid can be coded for more than one triplet nucleotide. An aminoacyl-tRNA synthetase can charge more than one tRNA species. com» > You discovered a new chemical that acts by producing single base insertions or deletions. Which of the following would result from such a mutation? A. A single amino acid residue change would occur, giving rise generally to a functional protein. B. A protein would be produced that would differ in more than one amino acid residue change from the normal protein. C A single amino acid residue change would occur; giving rise to a nonfunctional protein. D. No detectable change would result in the protein produced. E No product would be generated from the mutant gene. 10. ll. 12. 3 You have been asked to determine the mechanism of action of a new antibiotic. Addition of this antibiotic to a prokaryotic translation system in which polyribonucleotide AUGUUUUUUUUU. . is used as the mRNA template shows only synthesis of fl\/Iet~Phe. You conclude that this antibiotic acts by inhibiting A. Peptidyl transferase B. GTP—dependent translocation C. Binding of elongation ternary complex to the A site D. Formation of the 708 initiation complex E. Binding of release factors Proteins that are secreted from cells COW? .111 Contains methionine as the N—terminal amino acid. Are produced from translation products that have a signal sequence at the C—terminus. Are synthesized on ribosomes that eventually must associate with the endoplasmic reticulum. Contain a hydrophobic sequence at the C—terminus that is embedded in the membrane of secretory vesicles. Contain carbohydrate residues that bind to receptors on the interior of endoplasmic reticulum membranes. The functions of many enzymes and structural proteins can be quickly activated or deactivated by covalent modification of specific amino acid residues. All ofthe following examples illustrate such a mechanism of control of eukaryotic protein synthesis except .5097? Phosphorylation ofthe a—subunit ofinitiation factor elF—2. ADP—ribosylation of elongation factor EF—Z. Phosphorylation of initiation factor elF—4E. Inactivation ofinitiation factor elF—4G by proteolysis. Puromycin inhibits protein synthesis by cow> F“ Inhibiting the mono—ADP—ribose transferase activity of diphtheria toxin. Inhibiting peptidyl transferase. Preventing the binding of mRNA to the ribosomal subunits. Acting as a structural analog of aminoacyl-tRNA, thereby providing a mechanism for inducing premature termination of protein synthesis. Inhibiting binding ofthe large ribosome to the 48S initiation complex. 13. 14. 15. 16. 17. Which of the following statements best illustrates the importance of 5’—UTR (5’—NTR) of a prokaryotic mRNA? A. B. C. D. It contains the initiation codon. The 5’-UTR contains a purine—rich sequence capable of base-pairing with the 3’—end of 16S rRNA. Interaction between 5 ’-UTR and the anticodon arm of fMet~tRNA defines the reading frame of translation. Recognition of 5’—UTR by one of the initiation factors is necessary for recruitment of prokaryotic mRNA. Each ofthe following molecular features or reactions is paired with their role(s) in protein synthesis EXCEPT A. 3’—adenosine at the CCA end oftRNA: site of amino acid attachment . B. Shine—Dalgarno sequence: Complementary binding with 3’—end 0f18S rRNA. C. Trinucleotides UAA, UAG, UGA: termination of protein synthesis. D. Formation of aminoacyl adenylate: an obligate intermediate in charging of tRNA. E. Recognition of leader synthesis by SRP: temporary arrest of synthesis of secretory proteins. All ofthe following are examples of translational control EXCEPT U0w> Inhibition of protein synthesis resulting from the phosphorylation of eTF—ZOL by elF—20t kinase. Mutation in the initiation codon. Enhanced degradation of mRNA by 2’, 5’—oligoadenylate—activated RNase L. Balanced synthesis ofthe 0c— and the B—globin chains. Recognition ofthe 5’—cap structure of eukaryotic mRNA is mediated by mmbow> eIF—Z elF—3 elF—4E e1]? -4A eIF—4G elF—S Translation in eukaryotes differs from that in prokaryotes with respect to WOW? F“ The use ofUGG as a termination codon instead of UAG. The use of AUA as the initial codon for methionine. The ability to utilize multiple AUG’s to mark the reading frame of mRNA. Use of an ATP—dependent RNA helicase to unwind secondary structures associated with the 5’—NTR of mRNA. The fact that eukaryotic translation utilizes the same proteins for initiation and elongation. 18. 19. 20. 21. 05:» WU . It is a substrate for the charging enzyme in protein synthesis. Each of the following enzymes is matched with its appropriate functions or activities except EF—Z: translocation eIF—2: GTP—dependent binding of initiator tRNA eIF—4A: ATP—dependent RNA helicase PKR (active form): phosphorylation of oc—subunit of elF~2 RNase L: degradation of mRNA eIF-3: binding of mRNA WWUOW?” Each of the following statements about the structure shown is 1 I A correct except . . Ammo acrd :- attachment site 5, It is involved in decoding of information contained in mRNA. Treatment ofthis structure with alkali will result in its complete degradation to nucleoside monophosphates and nucleoside. It serves as site of attachment by amino acids. Its synthesis in mammalian cells shows exquisite sensitivity to mushroom toxin, oc—amanitin. Variable arm K2... Genetic alteration of the B—globin mRNA that introduces a signal peptide sequence to the amino terminus of the protein will result in A. Translocation and processing of the signal peptide in the endoplasmic reticulum. B. Synthesis ofa defective B—globin. C. Increased binding of B—globin chain to the a—globin chain D. Greater stability of the globin chain. E. Increased susceptibility of B—globin to proteases. Decrease in protein synthesis of an individual consuming a diet deficient in tryptophan is best explained by Failure in charging of a_ll tRNAs Deficiency in tryptophan codons. Inability of mRNAs containing tryptophan codons to bind to ribosomes Reduction in the rate of elongation due to lack of tryptophanyl—tRNA COW? 22. You prepared a cell—free protein synthesizing system from E. coli capable of translating synthetic polyribonucleotides. This system does not require a Shine—Dalgamo sequence or an initiation codon, and it translates polyribonucleotides in all three reading frames. Translation of the repeating sequence CAA in this cell free system produces three homopolypeptides — polyglutamine, polyasparagine, and polythreonine. If the codons for glutamine and asparagines are CAA and AAC, respectively, which one ofthe following triplets is a codon for threonine? A. ACC B. CAC C. AAC D. ACA E. CCA 23. All of the following statements about the properties of RNA are correct EXCEPT for A. RNA is chemically very stable even in alkali. B. RNA is less stable than DNA. C. RNA contains ribose instead of deoxyribose. D. RNA is usually single stranded but can form double stranded regions. E. RNA contains A,G,C,U as bases. 24. Which of the following statements about topoisomerases is incorrect? A. They alter the linking numbers of topoisomers. B. They break and reseal phosphodiester bonds . C. They form covalent intermediates with their DNA substrates. D. DNA topoisomerase 1 requires ATP as an energy donor. E. These enzymes form a transient covalent bond with DNA. 25. Which of the following statements concerning DNA polymerase III is incorrect? The structure of the B s ubunit is shaped like a torus. The [3 subunit is a trimer. The [3 ring acts as a sliding clamp that holds the replication machinery on the DNA. The [5 subunit is critical for processivity. COW?” 26. Which of the statements about the replication of eukaryotic DNA is correct? Replication proceeds discontinuously on both strands. Replication takes place bi—directionally. Replication proceeds by chain growth in the 3’ to 5’ direction. DNA replicates without a template. COP“? 27. 28. 29. 30. 31. Methyl—directed mismatch repair WUOW? corrects replication errors before the daughter DNA strand has become methylated. requires nonhomologous end joining. triggers apoptosis. catalyzes the monomerization of pyrimidine dimers. triggers double stranded breaks. When the retinoblastoma protein pr is phosphorylated it COW? cannot bind and inactivate E2F. activates the cyclins. leads to cell cycle arrest. activates p53. Which of the following statements is not true? saw C. D. E The level of p53 increases after exposure to agents that damage DNA. Increase in p53 level causes G1 specific cell cycle arrest and allows time for DNA repair to occur. p53 is a guardian of the genome. p53 acts by reducing the expression of cyclin/cdk inhibitor proteins. The level of p21 in cells is regulated by p53. Which of the following statements is false? A. B. C. D. The looping of the template for the lagging strand enables a dimeric DNA polymerase to synthesize both daughter strands. The RNA primer is removed by the 5’ to 3’ exonuclease activity of E. coli DNA polymerase I. On agarose gel electrophoresis of circular DNA with different degrees of supercoiling, the more relaxed the DNA is, the faster it moves. During SV4O DNA replication, the DNA helicase at the fork is SV40 T antigen. The two strands in the DNA double helix are held together by WOO“? hydrogen bonds between the bases. n—electron interactions between the bases (stacking forces). Van de Waals forces. Both hydrogen bonds and stacking forces between the deoxyribose subunits. Both hydrogen bonds and stacking forces between the bases. 32. 33. 34. The ends of linear DNA molecules are successively shortened during each round of DNA replication because: A. B. C. D. a terminal cap protein prevents the DNA polymerase from reaching the end of the molecule. the primer RNA is removed by DNA polymerase leaving a DNA strand that cannot be extended at its 5’ end. the DNA polymerase cannot add nucleotides to the 3’end of DNA strands. the ends are shortened by exonucleases. Which of the following statements regarding Telomerase is correct? A. com contains an RNA molecule that serves as template for the extension of telomeres on chromosomes. functions during genetic recombination. is usually found to be absent in tumor cells. trims the ends of mRNA molecules. Which of the following is incorrect? A. A transition is the replacement of a purine by a pyrimidine or that of a pyrimidine by a purine. The rare tautomer of G can base pair with a T in the template to allow its incorporation into a growing DNA strand during replication. In addition to spontaneous mutagenesis, mutations are generated by many chemical mutagens. Runs of identical bases in the DNA may cause the DNA polymerase to slip and make either the template or the growing strand loop out, and generating frame shift mutations. 35. 36. 37. Which of the following statements about DNA polymerase III holoenzyme from E. coli are correct? 1. It elongates a growing DNA chain hundreds of times faster than does DNA polymerase I. 2. It associates with the parental template, adds a few nucleotides to the growing chain, and then dissociates before initiating another synthesis cycle. 3. It maintains a high fidelity of replication, in part by acting in conjunction with a subunit containing a 3’ to 5’ exonuclease activity. 4. When replicating DNA, it is a molecular assembly composed of at least 10 different kinds of subunits. A. 1 B. l, 2, 3 C. l, 3, 4 D. 1, 3, 3, E. non of the above. A patient is treated with ciprofloxacin for a Pseudomonas aeruginosa infection in his lungs. Which of the following emymatic activities is most directly affected by this drug? @305”? The synthesis of RNA primers. The adding of telomeres. The breaking and subsequent rejoining of the DNA backbone by DNA topoisomerase. The removal of RNA primers. The joining together of Okazaki fragments. Dideoxyinosine is a nucleoside analog sometimes used to treat HIV infections. This drug blocks DNA chain elongation when it is incorporated into viral DNA synthesized by reverse transcriptase. Why does DNA synthesis stop? ?> ECO?“ The analog becomes covalently bound to reverse transcriptase, thus inactivating the enzyme. There is no 3’-hydroxyl group to form the next phosphodiester bond. Proofreading is inhibited. The analog cannot hydrogen bond to the RNA template. Incorporation of the analog initiates rapid degradation of the newly synthesized strand. 38. 39. 40. 41. 42. 10 While studying the structure of a small gene that was recently sequenced during the Human Genome Project, an investigator notices that one strand of the DNA molecule contains 20 A’s, 25 G’s, 30 C’s, and 22 T’s. How many of each base is found in the complete double—stranded molecule? A240, G=50, C260, T=44 A=44, G=60, C250, T240 A—45, G—45, C—52, T—52 A=50, 0:47, C—50, T=47 A=42, G255, C255, T=42 muow> Which of the following statement/ statements is incorrect? A. Eukaryotic genes and chromosomes are very complex. B. About 30% of the genome consists of sequences included in genes coding for proteins. C. Much of the nongene DNA is in the form of repeated sequences of several kinds. D. The most abundant intermediate repetitive DNA sequences in mammals are LINES and SINES. E. Alu sequences cannot be used to study human population genetics. Which of the following sequences are Lot coded for in the DNA? A. lntrons B. Exons C. tRNA D. Poly A tail E. rRNA Which of the following does not undergo modification after transcription? Bacterial mRNA Bacterial rRNA Bacterial tRNA Yeast mRNA Yeast rRNA Yeast tRNA 71.111.50.50? All of the following are true regarding splicing of eukaryotic mRNA, except It brings together different exons. It involves a lariat intermediate. It involves transesterification reactions. It occurs in the nucleus. It is most effective with removal of long compared to short introns. mcow> 44. 47. Which of the following is true regarding RNA editing? It can change the sequence of a gene product. It occurs after translation It is another name for splicing. It involves a lariat intermediate It changes the size of the mRNA. mcow> Which of the following statements is correct for both prokaryotic and eukaryotic gene expression? A. Transcription and translation are coupled. B. An mRNA is often polycistronic, coding for several polypeptides. C. The mRNA is transcribed 5’ to 3’. D. After transcription, a 3’ poly A tail and a 5’ cap are added. Alternative splicing can lead to which of the following Multiple mRNAs from the same gene. Multiple proteins from the same gene. Two mRNAs which are identical at the 5’ and 3 ’ ends and differ in the middle. Different mRNAs formed from the same gene in different tissues. All the above. mcow> All of the following are true regarding prokaryotic mRNA except It is a minor percentage of the total RNA in bacteria. It has a very short half life. It is used for translation prior to completion of transcription. It contains U instead of T in the coding strand of the DNA. Only a minor portion (less than 5%) of transcription is devoted to formation of mRNA. mbhw> Recruitment of co—activators can affect transcription in eukaryotes by Competing for binding to the enhancer sequence. Leading to acetylation of the histones in nucleosomes. Leading to binding of RNA polymerase 11. Leading to phosphorylation of RNA polymerase 11. Leading to fewer DNAase sensitive sites on the DNA. WUOW? 48. 49. 50. 51. 52. 12 All the following are true regarding the lac operon of E. coli except A. It contains three structural genes that are co—regulated. B The lac repressor can combine with allolactose to form a complex that no longer binds to the operator region. C The lac repressor is a protein that binds to the operator region, D. Its transcription is up—regulated by decreased glucose levels. E The repressor and CRP protein compete for the same sites on the regulator gene. When bacteria are grown in the presence of lactose A Transcription of the lac repressor is reduced. B Translation of the lac repressor is reduced. C. Transcription of the lac operon is increased. D A new sigma factor is used for transcription. E Coactivators are recruited to the lac operon. Which of the following is true regarding zinc finger transcription factors? A They include nuclear receptors such as the estrogen receptor B. They dimerize via a leucine zipper structure. C. They must be phosphorylated to be activated D They regulate transcriptional initiation and elongation E They are required for initiation of transcription by RNA polymerase 11 Factors which affect chromatin remodeling and expression of eukaryotic genes include all the following except Methylation of DNA Modification of histone tails by acetylation Sensitivity to DNase Interaction of transcription factors with coactivators The transcription of histone genes £11,505”? All of the following are correct regarding the regulation of transcription by cAMP except A. Levels of CAMP can regulate initiation of transcription in both prokaryotic and eukaryotic cells The response to CAMP is restricted to a subset of genes Phosphorylation of CREB is involved in the response in eukaryotic cells CAMP directly binds the appropriate transcription factor in both prokaryotic and eukaryotic cells E. The transcription of the gene encoding beta—galactosidase is elevated by high CAMP. .5093” 53. All of the following statements are correct regarding the promoter of genes recognized by RNA polymerase 11 except: They contain a TATA sequence at about 25 nucleotides upstream of the start site. Mutation of any of the nucleotides in the promoter reduces transcription. They often contain a GC rich region. They affect the efficiency of transcriptional initiation. They can bind several trans acting factors. mcow> l3 _ {1333 TVIDOS 'ON ALIH 17100 WOICIEIWEIHOA AMEN 5). 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Exam II 2005 - VVVVVV General Biochemistry Examination 11...

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