MCB 102 Third Exam Spring 2015

MCB 102 Third Exam Spring 2015 Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 (14 points) (9 points) (10 points) (9 points) (5 points) (6 points) (7 points) (8 points) (9 points) (8 points) TOTAL (85 points) Page 1 of 14

The following figure is reproduced from your textbook and provided for your reference: The exam requests molecular structures in certain questions you may use organic chemistry conventions (unlabelled carbons and omitted hydrogens) when drawing these structures. When molecular structures are not requested, you can use schematics and abbreviations that convey the detail needed to answer the question. When an underlined blank is present in the exam, we are looking for a very brief answer that should fit into this blank. Calculators are allowed. Show your equations clearly and we will score your response based primarily on your demonstration of biological principles and facts, not your arithmetic. Page 2 of 14

Problem 1 (14 points) The chemical 5-bromouracil (5-BrU) depicted below is used as a mutagen. It is tautomeric and usually exists in the major (keto) form but occasionally adopts a minor (enol) form. 5-bromouracil major (keto) form 5-bromouracil minor (enol) form A. (0.5 points) Is 5-BrU a base analog, a nucleoside analog, or a nucleotide analog? base analog B. (0.5 points) How would 5-BrU need to be transformed by the cell before it could be incorporated into DNA? converted into 5-bromouridine triphosphate C. (1 point) Uracil is not incorporated into DNA, and is removed when it does occur. Propose why 5-BrU might be incorporated into DNA even though normal uracil is excluded? Bromine on 5 position is large like the 5-methyl group on thymine, not small like the hydrogen on uracil. D. (2 points) In the major (keto) form, 5-BrU can form the same base pair that uracil forms. Draw the molecular structure of this base pair, with 5-BrU. E. (2 points) The minor (enol) form of 5-BrU mis-pairs to a nucleotide other than the usual partner of uracil. Draw the molecular structure of the enol tautomer mis-pairing of 5-BrU. Page 3 of 14

The questions below ask you to draw base pairs in DNA sequences. Draw a base pair of N and M like this: N M Use standard abbreviations, and 5 for 5-BrU. The normal pairing of 5-BrU is like uracil, as in (D), and the mis-pairing of 5-BrU is the pairing of the enol form from (E). F. (4 points) 5-BrU could cause mutations at the time of its incorporation into DNA. Draw i. A base pair where mis-pairing during 5-BrU incorporation could be mutagenic ii. The two duplexes resulting after mis-paired 5-BrU incorporation during replication iii. The four duplexes resulting after a second round of normal replication. iv. What base pair change results here? G C G G 5 C G A G G C 5 C C GC to AT transition G. (4 points) 5-BrU could also cause mutations at the time it serves as a template for DNA replication. As above, draw i. Normally base-paired 5-BrU prior to DNA replication ii. The two duplexes resulting after mis-pairing with a 5-BrU template iii. The four duplexes resulting after a second round of normal replication iv. What base pair change results here? A 5 G A 5 T G A A A C 5 T T AT to GC transition Page 4 of 14

Problem 2 (9 points) You find that the DNA encoding a protein is 1,050 base pairs long, including the stop codon. A. (1 point) What is the rough physical length of the DNA encoding this gene (1 significant figure is fine but specify the units) and how many double helix turns are present in this DNA? 3.4 Å 1,050 = 3,600 Å = 360 nm 100 helical turns B. (1 point) How many amino acids are in this protein? (1,050-3) 3 = 1,047 3 = 349 amino acids You clone this gene into a plasmid in order to express the protein in bacteria. C. (1 point) What sequence element must be present in the plasmid in order to ensure that it is copied during bacterial cell growth? origin of replication D. (2 points) Transcription in bacteria i. What sequence element must be present in the plasmid in order to ensure that a gene is transcribed by bacteria? ii. promoter Name the protein in the cell that recognizes this sequence specifically? sigma 70 E. (3 points) Translation in bacteria i. What sequence element must be present in the plasmid in order to ensure that a gene is translated by bacteria? ii. Shine-Dalgarno What RNA-protein complex in the bacterial cell recognizes this sequence specifically? iii. small ribosomal subunit How does this recognition happen? base pairing F. (1 point) Draw how the sequence elements needed for transcription (in part D) and translation (in part E) should be arranged relative to the DNA encoding a protein, in order for the protein to be expressed in bacteria. promoter DNA encoding protein Shine-Dalgarno Page 5 of 14

Problem 3 (10 points) A. (3 points) What are the topological parameters (linking number, twist, and writhe) of a relaxed, closed circular DNA that is 2,100 bp long? Lk = +200 Tw = +200 Wr = 0 The CRISPR/Cas9 protein-rna complex unwinds ~21 bp of DNA in order to recognize its target sites in the genome. B. (1 point) Why is DNA unwinding required for CRISPR/Cas9 to recognize target sites? guide RNA (sgrna) must base pair with target sequence C. (3 points) How would the topological parameters (linking number, twist, and writhe) of the 2,100 bp relaxed circular DNA from (A) change when CRISPR/Cas9 binds the DNA and unwinds 21 bp? Qualitatively, how would this affect the shape of the DNA? Unwinding 21 bp Tw decreases by 2 Lk cannot change Wr increases by 2 DNA adopts plectonemic supercoiling (or left-handed supercoiling, etc.) D. (2 points) How would the topological parameters (linking number, twist, and writhe) of circular DNA from above, bound by CRISPR/Cas9, change when the enzyme cuts the DNA? DNA unlinked when it s cut no more linking number / writhe E. (1 point) Name one DNA repair pathway that can repair the double-stranded break created by CRISPR/Cas9. Any one of: Homology-directed repair Homologous recombination Nonhomologous end joining NHEJ Microhomology-mediated end joining MMEJ Page 6 of 14

Problem 4 (9 points) Mitochondria have their own distinct translational machinery (separate ribosomes, separate trnas, etc.). Often, mitochondrial translation uses a slightly different genetic code than the universal code found in all bacterial, archaeal, and eukaryotic nuclear genomes. In the universal genetic code, only AUG encodes methionine, and AUA encodes isoleucine. In some mitochondria, both AUG and AUA encode methionine. A. (1 point) Draw the anticodon decoding an AUG methionine codon in the universal genetic code, indicating the anticodon sequence and its direction in the trna clearly. 3 -UAC-5 anticodon 5 -AUG-3 B. (2 point) One single trna can decode both AUG and AUA methionine codons in the mitochondrion. Draw the single anticodon decoding an AUG and an AUA methionine codon, indicating the anticodon sequence and its direction in the trna clearly. 3 -UAU-5 anticodon 5 -AUG-3 3 -UAU-5 anticodon 5 -AUA-3 C. (2 points) Draw the anticodon decoding an AUA isoleucine codon in the universal genetic code. What problem does this anticodon create? 3 -UAU-5 anticodon 5 -AUA-3 Can also recognize AUG methionine codon! OR 3 -UAI-5 anticodon, pairing with all three AU(U/C/A) Ile codons D. (3 points) In addition to ribosomes and trnas, other factors are needed in order to synthesize proteins. Name the factors that carry out the listed functions in bacteria, and indicate any energy carrying molecules they use to perform their function. i. Charging amino acids onto trnas ii. iii. Factor is trna synthetase and uses energy carrying molecule ATP Delivering charged trnas to the ribosome Factor is EF-Tu and uses energy carrying molecule GTP Translocating the ribosome along the mrna by one codon Factor is EF-G and uses energy carrying molecule GTP D. (1 point) The ribosome itself catalyzes one step in translation elongation. What does the ribosome catalyze? peptide bond formation (or peptidyl transfer) Page 7 of 14

Problem 5 (5 Points) Both adenine and guanine are synthesized from the metabolic intermediate hypoxanthine. Bacteria control the transcription of genes encoding enzymes that synthesize hypoxanthine, such as purc, with a repressor protein, purr, that can bind hypoxanthine. High levels of hypoxanthine shut off expression of purc and many other genes. A. (2 points) Do you expect that purr binds DNA in the presence or in the absence of hypoxanthine? Why? Presence of hypoxanthine transcription shut off purr a repressor shuts off transcription when bound purr binds DNA in the presence of hypoxanthine B. (1 point) How will expression of purc change when purr is mutated to have lower affinity for DNA? purc expressed more even in high hypoxanthine (or purc constitutive) C. (1 point) How will expression of purc change when purr is mutated to have lower affinity for hypoxanthine? purc expressed more even in high hypoxanthine (or purc constitutive) D. (1 point) What is the term for the DNA sequence that purr binds? operator Page 8 of 14

Problem 6 (6 points) The expression of the enzymes that convert hypoxanthine into guanine, such as xanthine phosphoribosyltransferase (xpt), is also regulated by guanine levels. This regulation does not require any proteins, but instead involves a riboswitch in the xpt transcript. The xpt mrna itself can bind to guanine, and changes its conformation when it does so, as shown below: Gua Shine-Dalgarno Gua Shine-Dalgarno xpt gene xpt gene A. (1 point) What is the role of the Shine-Dalgarno sequence in the expression of xpt? Shine-Dalgarno recruits the ribosome and ensures translation B. (1 point) Do you expect that xpt expression is higher when guanine levels are low or high? xpt expression is high when [guanine] is low C. (1 point) At what stage of gene expression is xpt expression controlled? Translation D. (2 points) Biosynthesis of several amino acids, including tryptophan and histidine, is controlled by attenuation. i. What is the direct effect of low amino acid levels in regulation by attenuation? Low amino acid levels lead to ribosomal stalling while the ribosome waits for a trna charged with that amino acid ii. Given how low amino acids are sensed in attenuation, why is it impossible for guanine levels to regulate xpt expression by attenuation? Guanine is not an amino acid so low guanine cannot stall elongation E. (1 point) Name one reason why the xpt mrna would not be translated at all if it were extracted from bacteria and introduced into mammalian cells. Any one of: No 5 -methylguanosine cap No 3 poly-(a) tail May have upstream AUG in the 5 leader Page 9 of 14

Problem 7 (7 points) APOBEC enzymes are a family of enzymes that edit the sequence of an mrna after it is synthesized. The are named APOBEC because the first APOBEC enzyme, APOBEC1, was found because it edits the apolipoprotein B (ApoB), and named based on that function. Expression of ApoB is medically interesting because this protein is required for VLDL, which is important in cholesterol transport. A. (3 points) APOBEC1 is a cytidine deaminase that acts on a specific cytidine in the middle of the ApoB mrna. The target site is shown below, including the nucleotide sequence and the protein translation. 5 - AUA CAA UUU -3 Ile Gln Phe i. What is the mrna sequence after APOBEC catalyzes cytidine deamination? 5 - AUA UAA UUU -3 ii. How does the change in the mrna sequence affect the protein? Introduces a stop codon in the middle of the protein. The protein is shorter. While APOBEC1 edits the normal cellular ApoB mrna, other APOBEC enzymes are involved in anti-viral defense. APOBEC3G restricts the replication of HIV by modifying cytidines in viral genomes. However, HIV expresses a protein called Vif that can overcome APOBEC3G restriction. B. (2 points) The HIV protein Vif binds to APOBEC3G and brings it to an E3 ubiquitin ligase (cullin5-elonginbc in particular). What is the effect of recruiting a protein such as APOBEC3G to an E3 ubiquitin ligase, and how would this inactivate APOBEC3G? E3 ubiquitin ligase adds a K48-linked poly-ubiquitin chain to APOBEC3G. This polyubiquitylation leads to proteasomal degradation, and so APOBEC3G is inactivated by being degraded. Page 10 of 14

The drug zalcitabine was an early HIV treatment, though it has since been supplanted by other, chemically related drugs. The structure of zalcitabine is shown below. C. (2 points) How does zalcitabine inhibit the reverse transcriptase enzyme of HIV? Zalcitabine is a nucleoside analog drug 2, 3 -dideoxycytosine. Zalcitabine will be converted into ddctp in the cell and then incorporated into the first-strand cdna by reverse transcriptase. After zalcitabine incorporation, there is no 3 -O to continue extending the DNA being synthesized by the reverse transcriptase, and so DNA synthesis is terminated early. Page 11 of 14

Problem 8 (8 points) A. (4 points) Using molecular structures, draw the reaction mechanism for the reaction of adding one nucleotide to the end of a growing RNA chain. You do not need to draw the nucleobases or the template strand. What additional products, if any, are produced in this reaction? Inorganic pyrophosphate also produced Here is a section of a eukaryotic pre-mrna before spliceosomal splicing: exon #1 D intron exon #2 B. (2 points) Certain defects in splicing block the second chemical step of the reaction. Sketch the splicing intermediate that will accumulate when this second chemical step of spliceosomal splicing is blocked, noting any unusual features of the RNA(s) in that intermediate. OH exon #1 intron exon #2 2-5 linkage C. (2 points) What binds the splicing site labeled D in the pre-mrna and how does this recognize the labeled site? U1 snrna component of the U1 snrnp base pairs with this 5 splice site Page 12 of 14

Problem 9 (9 points) A. (4 points) Draw the newly-synthesized strands onto this replication fork and indicate the direction of DNA synthesis. There are two different names given to the two newlysynthesized strands label them with these names and briefly explain the distinction between them. 3 Leading Strand 5 Lagging Strand 5 3 template strands The leading strand is synthesized continuously. The lagging strand is synthesized discontinuously the direction of fork movement is the opposite of the direction of replication on this strand. B. (3 points) Three repair pathways, listed below, work by degrading and resynthesizing one strand of DNA. For each, indicate the molecular feature of DNA damage that they recognize in order to initiate DNA repair. 1. Base excision repair Chemically damaged (e.g., methylated) bases Uracil in DNA 2. Nucleotide excision repair Distorted DNA backbone 3. Mismatch repair Unmodified but mis-paired (non-watson-crick paired) nucleotides C. (2 points) What unique challenge arises for mismatch repair in identifying the DNA strand to degrade and resynthesize? What would be the effects of degrading and resynthesizing the wrong strand? In mismatch repair, neither strand is obviously damaged, and so other factors must be used to determine which strand is the older correct strand and which strand is newer and contains DNA synthesis errors. Degrading and resynthesizing the wrong strand would convert replication errors into permanent mutations instead of repairing them. Page 13 of 14

Problem 10 (8 points) A. (3 points) Below is a sketch of chromatin. nucleosome 1. How much DNA is in one nucleosome? 146 bp 2. What else is in a nucleosome? Histone octamer 2 each of H2A, H2B, H3, and H4 DNA B. (3 points) Draw the chromatin around an active promoter in the same style as the sketch above, labeling the promoter itself and any nucleosome modifications or other chromatin features important for active transcription. H3K9-Ac H3K14-Ac promoter Ac nucleosomefree region C. (2 points) Eukaryotic repressors generally function by recruiting co-repressors. Suggest two enzymatic activities that could be found in a eukaryotic co-repressor. Any two of: Histone deacetylase Histone methyltransferase DNA methyltransferase Chromatin remodeling factors Page 14 of 14