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MCMP 208 Exam II Key - 1 Examination II Key MCMP 208 – Biochemistry for Pharmaceutical Sciences I March 7, 2017 Correct answers in multiple choice questions are indicated in RED and underlined. Correct answers to essay questions are indicated in RED in comic book font. In some cases and explanation is provided in BLUE/BLUE The following accurate genetic code table may be useful to you for some parts of this exam. U C A G UUU Phe UCU Ser UAU Tyr UGU Cys UUC Phe UCC Ser UAC Tyr UGC Cys U UUA Leu UCA Ser UAA Stop UGA Stop UUG Leu UCG Ser UAG Stop UGG Trp CUU Leu CCU Pro CAU His CGU Arg CUC Leu CCC Pro CAC His CGC Arg C CUA Leu CCA Pro CAA Gln CGA Arg CUG Leu CCG Pro CAG Gln CGG Arg AUU Ile ACU Thr AAU Asn AGU Ser AUC Ile ACC Thr AAC Asn AGC Ser A AUA Ile ACA Thr AAA Lys AGA Arg AUG Met ACG Thr AAG Lys AGG Arg GUU Val GCU Ala GAU Asp GGU Gly GUC Val GCC Ala GAC Asp GGC Gly G GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly MULTIPLE CHOICE. For problems 1 to 17, select from the list immediately following each question the single most correct choice to complete the statement, solve the problem, or answer the question. Mark that answer on your answer sheet. [3 points each] 1. Identify which set of proteins constitute the core protein components of nucleosomes. H2A, H2B, H3, H4 These are the 4 core histone proteins. leucine zipper proteins DNA polymerases ( and ), PCNA (sliding clamp), MCM DNA helicase, Primase Cdc6, Cdt1, MCM DNA helicase, ORC (origin replication complex) Uvr A, B, and C Mut S/L and MutH TBP (TATA box binding protein), RNA polymerase, transcription factors (TFII A, B, D, E, G, H) DNA polymerase I and DNA ligase MCMP 208 Exam II Key - 2 2. The posttranslational modified residue shown to the right is best described as hydroxylation. lipidation. farnesylation. acetylation. N-linked glycosylation. O-linked glycosylation. This residue displays a sugar modification through the oxygen (O) of a serine side chain. 3. A common function of DNA helicases is to introduce single or double strand breaks to relieve DNA supercoiling. to bind to and stabilize the DNA double helix. to bind to and stabilize single stranded DNA. to form phosphodiester linkages between 5’ phosphomonoesters and 3’ hydroxyls on two DNA strands. to separate the annealed strands of duplex DNA. Helicase activity is critical to both DNA replication and transcription as ssDNA templates are used in both processes. to synthesize RNA primers for DNA polymerases in the process of replication. to increase processivity of DNA polymerases. to form Okazaki fragments. 4. Chargaff’s rules were critical information that helped Watson and Crick construct their structural model of DNA. Which is the best description of Chargaff’s rules? adenine pairs with thymine via 2 hydrogen bonds; cytosine pairs with guanine via 3 hydrogen bonds phosphodiester bonds connect deoxyribose sugars DNA does not contain sulfur atoms bacteriophage always transfer DNA, not protein, to their host DNA contains deoxyribose sugars, not ribose sugars both proteins and DNA fold into helical conformations the ratio of adenine to thymine and the ratio guanine to cytosine are both 1:1 in all organisms While the relative ratio of A:G or A:C etc was variable among species, the A:T and G:T ratios were always one. Van der Waals interactions stabilize the double helix 5. A difference between bacterial genomes and eukaryotic genomes is bacterial genomes can be made of either DNA or RNA, while eukaryotes only use DNA. the DNA of bacterial genomes is not supercoiled, while eukaryote genomic DNA is supercoiled. the majority of the DNA in bacterial genomes does not encode for gene products, while in eukaryotes the major of the DNA does encode for gene products. prokaryotes can have a circular DNA molecule for a genome, while eukaryotes always have linear molecules. bacterial genomes are packaged in nucleosomes, while eukaryotes do not package their genomes. bacterial genomes are typically much larger, containing more total base pairs, than eukaryotic genomes. MCMP 208 Exam II Key - 3 6. In the mechanism of mRNA splicing by the spliceosome, what is the initial nucleophile in the reaction at the 5' splice site that forms the lariat intermediate. an OH in the gamma phosphate of ATP. the OH on the active site serine of the spliceosome ribonuclease. a water molecule activated in the active site of the spliceosome ribonuclease. the 2'-OH of an adenosine within the intron to be cleaved. the 3'-OH of a guanosine in the 5' exon. an OH of the 5' phosphate of a guanosine in the 3' exon the SH on the active site cysteine of the spliceosome ribonuclease. Mg2+ ions in the active site of the spliceosome ribonuclease. 7. Place the following steps of eukaryotic transcription in chronological order. i. TBP (TATA box binding protein) within TFIID bind to the core promoter elements (CPEs) on DNA. ii. A poly(A) signal is transcribed. iii. RNA polymerase II dissociates from the mediator complex. iv. 23 nucleotides of RNA are synthesized allowing the preinitiation complex to move away from the core promoter. v. General transcription factors (TFII A, B, E, G, H), RNA polymerase II, chromatin remodeler proteins, and the mediator complex bind at core promoter elements (CPEs) on DNA forming the preinitiation complex. iii, v, i, iv, ii v, iv, i, ii, iii v, i, iv, iii, ii i, v, iv, iii, ii i, v, iii, iv, ii i, v, iv, ii, iii 8. Which is not a mechanism involved in the regulation of transcription? chromatin remodeling. repressor proteins binding to gene operators. gene expression level of transcription factors. compartmentalization of transcription factors (sequestration in cytosol, e.g.). regulation of transcription factor activity by phosphorylation. DnaA proteins binding to oriC sites. This is part of the mechanism for DNA replication in prokaryotes. 9. Which of step would result in a stabilization of the following duplexed (double stranded) oligonucleotides? 5'-ATGGTATCAA-3' 3'-TACCATAGTT-5' exchanging a G:C base pair with an A:T base pair. exchanging an A:T base pair with a G:C base pair. G:C base pairs are more stable than A:T base pairs, as they have an additional hydrogen bond and form more stable base stacking interactions. increasing the temperature. decreasing the ionic strength of the solution. adding a denaturing, water miscible solvent, such as formamide. shortening each oligonucleotide by two bases. MCMP 208 Exam II Key - 4 10. Which of the following catalyzes the attachment of amino acids to tRNAs? elongation factor EF-Tu elongation factor EF Ts the ribosome aminoacyl tRNA synthetase reverse transcriptase RNA polymerase ribonucleases restriction endonucleases 11. This molecule directly provides the chemical energy used to translocate the ribosome down the mRNA as translation proceeds. ATP TTP GTP The GTPase eEF2 hydrolysis GTP causing a translocation of the ribosome along the mRNA. deoxyATP deoxyTTP deoxyGTP creatine phosphate dolichol phosphate 12. In the general procedures of molecular cloning of DNA molecules, what is the best description of the role of a vector? The vector contains a DNA polymerase making several copies of a DNA sequence. The vector amplifies DNA using short DNA sequences called primers. The vector cuts DNA at specific sites to generate sticky ends. The vector accepts the DNA to be cloned and is capable of being replicated in a host cell. The vector is also a DNA molecule which is typically ligated with a DNA to be replicated. The vector will contain the necessary sequences for the host cell to replicate. The vector is a host organism, such as E. coli, which is tasked with replicating a DNA sequence. The vector is a solid support, typically a nitrocellulose filter, that binds DNA prior to probing with radiolabeled complementary DNA sequences. 13. The mechanism of peptidyl transferase catalyzed by the ribosome has been described as a concerted proton shuttle mechanism. Where is the atom that acts as both a proton donor and acceptor in this reaction? The 2' oxygen of a rRNA uridine residue located at the P-site in the large ribosomal subunit. The 2' oxygen of a rRNA uridine residue located at the A-site in the large ribosomal subunit. The nitrogen of the incoming amino acid located on the aminoacyl tRNA within the A-site. The 3' oxygen of the peptidyl-tRNA within the ribosomal P-site. The 2' oxygen of the peptidyl-tRNA within the ribosomal P-site. The oxygen of a water molecule activated by hydrogen bonding to a rRNA uridine residue located at the P-site in the large ribosomal subunit. MCMP 208 Exam II Key - 5 14. Why is the level of A1c useful diagnostic information? It reflects the general health of an individual. It reflects the current amount of liver glycolysis It reflects the current amount of turnover of total body protein It reflects the amount of turnover of total body protein during the past several weeks It results from a chemical reaction and so indicates the average level of blood sugar for the past several weeks. It reflects whether a patient is anemic or not It reflects whether or not the patient is unable to degrade glycosaminoglycans and therefore has a form of mucopolysacchardosis It is the result of a reaction of carbohydrate with protein that reflects the average body temperature in the past several weeks. 15. Glycosaminoglycans are proteins with a lot of glycosylation are large glycolipids are linear polysaccharides containing acidic and amino sugars are short branched oligosaccharides are any oligosaccharide containing at least one sialic acid residue are branched polysaccharides used for energy storage are not able to be digested very well after being eaten by humans are homopolysaccharides that are common in invertebrate exoskeletons 16. Short heteroglycans attached to lipids and proteins function primarily as transporters of nutrients across the cell membrane structures used in molecular recognition and sensing storage of carbohydrates for use in energy generation by cells an important part of the mechanism that targets those proteins and lipids to subcellular organelles elements that stabilize proteins and lipids by protecting them from degradation an important mechanism for stabilizing the high order structure of all proteins 17. A point mutation that converts the code for an amino acid into a premature stop signal is best described as a transversion mutation. a silent mutation. a missense mutation. a misdirectional mutation. a missing mutation. a nonsense mutation. MCMP 208 Exam II Key - 6 ESSAY PROBLEMS. Write your answers to problems 18 to 22 in the space immediately below each problem. 18. A. [2 points] Electron micrographs of DNA in the process of general homologous recombination have shown that chi structures generally have two pairs of equal length arms (as shown below). Explain why chi structures exhibit this particular symmetry. Homologous recombination is an exchange of DNA between similar or identical molecules of DNA, such as homologous chromosomes. This symmetry arises because the similar sequences occur at the same locations within the two molecules of DNA. B. [2 points] The process of chi structure formation in general recombination involves assembly of two DNA molecules, which we could call arbitrarily strand A and strand B. In the picture above, label each arm of the chi structure as being from either the A or B strand. One arm is labeled for you. A B B B B A Proc. Natl. Acad. Sci. USA. 73 (1976) 3001. Each of the above two figures is a correct answer. 19. [6 points] Below are structures of the deoxyribonucleosides commonly present in DNA. A. Indicate if each deoxyribonucleoside contains a purine or pyrimidine base. B. The circle provided represents a view of the DNA double helix looking down the length of the helix from above. Choose two bases that normally form base pairs and draw them within the circle taking care to place the bases appropriately within the helix barrel. Draw just the bases with the glycosidic bond linking to the circle perimeter and indicate the proper hydrogen bonding pattern. C. Indicate the sides of the circle that represent the major groove and the minor groove in the double helix. MCMP 208 Exam II Key - 7 Either drawing in the red box is a correct answer for parts B and C. Note the also correct are if the base pairs are drawn inverted from right to left (i.e., C:G vs G:C). Either A:T or G:C base pairs are acceptable. The narrower angle (on the minor groove side) between glycosidic bonds should be clearly less than 180 and greater than 90 degrees. 20. [6 points total] This problem concerns the dsDNA gene sequence from E. coli shown below. 1 2 3 5’- GCCATTTCCCGTTAAGGAGGTGCCATCTATGGAAGCCAGCTACTAA -3’ (+) coding strand 3’- CGGTAAAGGGCAATTCCTCCACGGTAGATACCTTCGGTCGATGATT-5’ (-) noncoding strand A. [4 points] Site 1 is the transcriptional start site. Site 2 is a consensus sequence that when present in the corresponding RNA transcript is called the Shine-Dalgarno sequence. Explain the role of this sequence in transcripts in prokaryotes and explain where one would find the anti-Shine-Dalgarno sequence. The Shine-Dalgarno sequence is a ribosomal binding site found in bacteria. This site functions to ensure the ribosome is able to find the proper AUG start for translation initiation. The ribosome binds this site through Watson-Crick base pairing (RNA:RNA) to the anti-Shine-Dalgarno sequence within the in the ribosomal RNA (rRNA) of the small subunit. B. [2 points] Using the provided genetic code table, determine the peptide/protein product of this gene. Met-Glu-Ala-Ser-Tyr MCMP 208 Exam II Key - 8 21. [6 points total] The Meselson-Stahl experiment established the mechanism of DNA replication as being a semi-conservative. A. [2 points] Briefly explain how DNA replication is semiconservative. It is called semiconservative because upon replication two DNA copies are produced and each contains one of the original strands and one new strand. B. [4 points] The Meselson-Stahl experiment involved use of a heavy form of nitrogen (15N). Growing bacteria on 15N containing media adds significant mass to the base pairs. Heavy (15N) and light (14N) DNA can be distinguished on a density gradient with ultracentrifugation (shown below in the control sample test tubes). In the experiment, bacteria grown overnight in heavy media were transferred to light media and the DNA was analyzed by centrifugation after both one and two doublings (1 and 2 generations). Use the test tube pictures below to indicate the location of the DNA band(s) that Meselson and Stahl observed after 1 and 2 generations. Also, indicate the location of the DNA that would have been observed had the mechanism of replication been conservative. The relative intensity of the DNA bands in the answer (in the test tubes in the red boxes) won’t be required/graded, just the appropriate positions in the tube. In the semiconservative mechanism (actual results), the DNA after just one replication has one new strand (14N) and one old strand (15N). So, all observed DNA is at an intermediate density. After an additional replication, now two of the four strands contain all new DNA (14N) and two of the four contain one new (14N) and one old (15N) strand. If the mechanism were conservative, this would mean that an entirely new dsDNA is generated from a dsDNA template. Thus, the expected result with this mechanism yields the original DNA template at the lower tube position due to the 15N and the new strand being composed completely of 14N. With additional replications, all newly synthesized strands are composed of 14N and the original 15N containing template will remain. MCMP 208 Exam II Key - 9 22. [7 points total] This problem tests your knowledge of carbohydrate structure and nomenclature. A. [4 points] D-isomaltose is the disaccharide that can be produced by digestion of amylopectin by an enzyme that hydrolyzes all the linkages except the linkage involved uniquely at the branch points. Draw the structure of α-D-isomaltose employing a standard Haworth projection. B. [3 points] D-ribulose is the ketose with the same stereochemistry as D-ribose. Draw the structure of the open chain form of D-ribulose employing a standard Fischer projection.