Exam 4

Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: Section 1. Multiple Choice (2 points each, 44 points total). Clearly circle the letter of the correct answer. Please read the questions CAREFULLY and understand what is asked of you. Answer all questions on these pages. If you come to a question and you do not know how to attack it immediately, do not panic. Skip the question and go on. When you are done with the remainder of the test, return to the question. 1) Tautomerization can result in all but which of the following mutations? a. Transversion mutation – due to normal base pairing, since ring structures are not altered, cannot switch a purine for a pyrimidine and vice versa b. Point mutation c. Transition mutation d. Frameshift mutation e. Nonsense mutation 2) Prokaryotic gene regulation can occur at which stage below: a. Regulation of mRNA processing b. Regulation of translation c. Regulation of mRNA transport d. Regulation of protein activity e. Regulation of gene transcription 3) You are studying protein translation and come across a cell with rough ER but curiously there are no proteins present at the cell surface. Of the components in the pathway for secreted and/or membrane bound proteins, which is most likely mutated that allows proper recruitment of the ribosome-mRNA complex to the ER but prevents entrance of the growing peptide chain into the ER? a. Signal sequence – arrived at ER ok b. Signal recognition particle – arrived at ER ok c. SRP receptor – arrived at and binding to ER ok d. Sec61 – ribosome binding site could be intact but have a mutation within the channel preventing protein passage into ER e. Signal peptidase – do not know, cannot enter ER to be cleaved 4) Which of the following is a false statement regarding aminoacyl tRNA synthetases? a. Facilitates the addition of an amino acid to the proper tRNA b. Requires ATP as an energy source to charge a tRNA c. Creates a covalent link between the amino acid and the 5’ end of the tRNA (AA added to 3’ end of tRNA) d. Creates a covalent link between the 5’ phosphate of ATP and the amino acid e. None of the above Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 5) Which of the following is not a component of the prokaryotic translation initiation complex? a. Small ribosomal subunit b. Protein factors c. Shine-Dalgarno sequence (Ribosome initially binds to mRNA through and then slides to start signal, initiation complex will not assemble on this DNA sequence) d. tRNA for f-met e. mRNA start codon 6) Which of the following steps during translation does not require GTP as an energy source? a. Addition of large ribosomal subunit to initiation complex b. Shifting of growing chain to P-site c. Release of ribosome from the mRNA d. Peptide bond formation between two amino acids e. None of the above 7) Which of the following would cause transcription from the lac Operon in the absence of lactose? a. Mutation that removes the DNA site bound by CAP-cAMP b. Mutation of the lac operator c. Mutation of a structural gene in the lac operon d. Mutation of the CAP gene e. A and D 8) The correct translation from nucleic acid language to amino acid language is primarily driven by: a. The base-pairing of the Shine-Dalgarno sequence to the anti-Shine-Dalgarno sequence. b. The base-pairing of the start codon sequence to the anti-codon sequence of the proper tRNA. – if do not match f-Met with start codon, protein will not be translated correctly. c. The selective introduction of the specific tRNA-f-Met by EF-Tu. – can introduce but may not recognize anticodon d. The correct interaction between a tRNA and its specific aminoacyl tRNA synthetase. e. The identification of the AUG start codon by the translation initiation complex Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 9) BAD QUESTION – MULTIPLE ANSWERS ACCEPTED Modified bases, present in the anticodon loop of a tRNA, are recognized by which pocket of the ribosome where tRNAs are matched with the mRNA codon? a. A site – only tRNAs in the A site are able to recruit appropriate tRNAs (modified bases or not) b. P site – first tRNA incorporated here c. E site – only uncharged tRNAs here (not an accepted answer) d. Both A and B e. Both B and C 10) Protein functional domains are determined by which level of protein structure? a. Primary b. Secondary c. Tertiary d. Quarternary e. Exon Sequence 11) When a part of the growing peptide chain in the P pocket, which portion of the amino acid is involved in peptide bond formation with the amino acid attached to the new tRNA in the A pocket? a. Carboxyl group – if growing peptide chain has an N-terminus (at end of Met) then the chain must be bound to the tRNA in the P pocket through the carboxyl end which then reacts with the N-term of the new AA in the A pocket to form a new peptide bond) b. Amino group c. Radical group d. Central carbon e. Hydrophilic residue 12) Of the components required for a proper Ames test, which is critical for testing substances that by themselves are not harmful but could become harmful in the body? a. Bacteria with known mutation b. Liver enzymes – need to metabolize chemical c. Disc coated with possible mutagen d. Bacterial culture plates e. Control plate 13) Which one of the following mutagenic events does not occur spontaneously? a. Deamination b. Tautomerizations c. Depurination d. Alkylation – need an agent e. Replication slippage Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 14) All of the following statements about mismatch repair are true except: a. Mismatch repair depends on the fact that parental strand DNA is more heavily methylated b. Mismatch repair requires the action of a methyl transferase c. Mismatch repair determines which base in a mismatch is the incorrect one d. DNA gap is filled in by DNA polymerase e. Mismatch repair recognizes methylated cytosine (adenines are methylated) 15) Attenuation regulates expression from the trp operon by which of the following mechanisms? a. Blocking initiation of transcription when tryptophan levels are high – transcription initiated before ribosome hits terminator b. Blocking translation of mRNA when tryptophan levels are high – translation proceed but creates non-functional product c. Terminating transcription when tryptophan levels are low - FALSE d. Synthesis of a repressor protein when tryptophan levels are high – repressor is made constitutively e. Terminating transcription when tryptophan levels are high 16) Which of the following are examples of positive control in operons? a. Regulation of the lac operon by the lac repressor. b. Action of the catabolite activator protein in the lac operon – all others are negative control c. Attenuation of the trp operon d. Regulation of the trp operon by the trp repressor e. Regulation of the trp operon by TRAP 17) Which of the following would result due to a mutation of the trp codons of the attenuator region in the trp operon? a. The anti-terminator loop could no longer form – still forms, mutation in loop 1 b. The terminator loop would no longer form – still forms, mutation in loop 1 c. The normal function of the attenuator is maintained – not possible…need trp codons d. Lack of tryptophan would not promote transcription – ribosome would not stall, terminator loop formed e. Lack of tryptophan would promote translation - cannot translate if not transcribed 18) In B. subtilis, the trp operon is regulated by TRAP rather than the attenuator. Which of the following is a true statement about the anti-TRAP protein? a. Provides a metabolic signal that there is enough tryptophan tRNA b. The AT gene is encoded by the trp operon c. AT interacts with TRAP when TRAP is not bound to tryptophan d. AT prevents TRAP from associating with leader sequence of mRNA e. AT prevents the trp operon from being transcribed Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 19) Which of the following is a gene regulatory mechanism in eukaryotes promotes a reduction in gene activity? a. Chromosome location in the center of the nucleus b. Methylation of histone tails within nucleosome c. Methylation of CpG islands on DNA d. Acetylation of histone tails within nucleosome e. Removal of nucleosomes to allow access to promoter 20) An silencer sequence – just like an enhancer a. must be located immediately upstream from the gene it affects. b. must be located immediately downstream from the gene it affects. c. must be located within the gene it affects. d. can be inverted without altering its functional ability. e. affects the same gene even if its position is altered. 21) Which of the following is not a mechanism by which mRNA can be regulated to affect gene expression? a. Alternative splicing of exons b. Exon shuffling in mRNA – only happens in DNA c. Shortening of the PolyA Tail d. Autoregulation by its protein product e. Response elements in UTRs 22) Which of the following is not true about the regulation of p53 protein stability? a. MDM2 binds to the transactivation domain of p53 b. MDM2 post-transcriptionally modifies p53 – post-translationally modifies! c. MDM2 targets p53 for degradation d. Phosphorylation of p53 prevents MDM2 binding e. p53 stimulated MDM2 gene expression Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: _ Short Answer/Problem Solving (Points noted for each problem). Organize your thoughts and your work, so that the graders can CLEARLY see what you are doing. Show ALL of your work, if you get a wrong answer but have the right process you may still be awarded partial credit. If there are multiple possibilities, list all possibilities! Answer all questions on these pages. If you need more room, write on the back of the same page, clearly labeling to which question the work belongs. If you come to a question and you do not know how to attack it immediately, do not panic. Skip the question and go on. When you are done with the remainder of the test, return to the question. 1) Glucose is brought into bacteria as glucose-6-phosphate. Prior to use of glucose-6- phosphate as an energy source, it must be converted to fructose-6-phosphate in order to enter the glycolysis pathway. You are interested in studying how loss of enzymes within this pathway affects the health of your favorite bacteria. You have identified several bacterial mutants which are unable to produce fructose-6-phosphate when plated on glucose. Based on the data below, assign the bacterial mutants to the step in the pathway where an enzyme is likely deficient by writing the letter of the mutant in the appropriate box (5 points). Growth Supplement Test 1 Mutant Minimal medium Glucose-6- phosphate Ribulose-5- phosphate Glyceraldehyde- 3-phosphate 6- phospho- gluconate 6-phopho- gluconolactone A - - + + + + B - - - - - - C - - + + + - D - - + + - - E - - - + - - A C D E B Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 2) Dr. Wood studies a protein called Pod1, a bHLH protein. Pod1 is involved in the regulation of the protein SF1. The Sf1 gene has seven exons. SF1 is essential in the transcription of genes which encode steroidogenic enzymes. In proposed stem cells of the adrenal, Pod1 is thought to inhibit Sf1 expression. Sf1 is also regulated by the USF1 and USF2 bHLH proteins which bind to an E-box sequence on the DNA just prior to the transcription start site. SF1 is regulated in a tissue-specific manner. A DNA element within the gene sequence of SF1 between exon 6 and 7 allows expression in the brain. A DNA element 5000 kb upstream of transcription start site allows expression of SF1 in the fetal adrenal. In cells where SF1 is not expressed, the E-box region is methylated. SF1’s ability to activate expression of steroidogenic enzymes depends on whether it is phosphoylated (required) or SUMOylated (inhibitory). Based on the new knowledge you gained above, answer the following questions (10 points total): NOTE TO GRADERS – IF IDENTIFY THE ITEMS BUT DO NOT PROVIDE ANSWER THE REST OF THE QUESTION, GIVE HALF CREDIT – USE YOUR JUDGEMENT a. There are three cis-acting elements discussed in Sf1 regulation. What are they and what is their function? i. E-box – transcription factor binding site in the promoter, allow USF1/2 binding (1 point) ii. Sequence between exon 6 and 7 – tissue specific enhancer for brain (1 point) iii. Element 5000 kb upstream of TSS – tissue specific enhancer for the adrenal (1 point) b. Three trans-acting factors are mentioned. What are they and identify which are activators and which are repressors? i. Pod1 – repressor (1 point) ii. USF1 – activator (1 point) iii. USF2 – activator (1 point) iv. SF1 – activator (1 bonus point) c. From the trans-acting factors identified above, can you propose a mechanism by which these factors may interact to regulate SF1? i. Pod1 inhibits binding of USF proteins to E-box either through directly binding to the E-box itself (competitive inhibition) OR through binding directly to USF proteins to prevent them from binding to DNA – researchers still do not know. (1 point) Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: d. Two additional methods of gene regulation are also mentioned regarding the regulation of Sf1. What are they? What type of regulation are they (ex. chromatin remodeling)? Identify if they are repressive or activating. i. E-box methylation – epigenetic modification, repressive (1 point) ii. Phosphorylation of SF1 – post-translational modification, activating (1 point) iii. SUMOylation of SF1 – post-translational modification, repressive (1 point) Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 3) Luria and Delbruck determined that most mutations are due to spontaneous mutations. Explain how they arose at this conclusion. What were their initial hypotheses and how did they test these hypotheses? Why did their data support this conclusion? (7 points) Hypothesis I: Adaptive Mutation - Bacteria mutate in response to incubation with bacteriophage, Every bacterium has small but constant probability of acquiring T1 resistance, Number of resistant cells will depend only on the number of bacteria and phages added to each plate (2 points) Hypothesis II: Spontaneous Mutation - Nothing to do with any stimulus from being incubated with bacteriophage, Rate of mutation low during incubation with medium prior to plating, Mutations that occur early in incubation, will result in large number of mutated offspring, Mutations that occur late in incubation, will result in few number of mutated offspring, Number of resistant cells will fluctuate greatly from one experiment to the other! (2 points) Experimental set-up: must mention incubation with T1 phage in different number, incubation in large cultures results in all bacteria with same mutation rate (control), incubation in small cultures results in variable mutation rate, number of phages does not matter (2 points). Data supported conclusion: because had large variation in number of mutants from tube to tube even when cultured in controlled conditions (1 point) Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 4) Understanding the lac operon. a. How does the lac operon sense the presence of lactose? In your answer, describe the molecules that act as sensors and explain mechanistically how they exert an effect on expression (i.e. what is going on when lactose is absent and then how does that change when lactose is present). 5 points. LacI encodes a repressor protein that binds to the lac operator as a dimer (1 point). The repressor protein must contact the DNA at two sites around the promoter causing it to loop (1 point). Binding of the repressor blocks transcription through inhibiting the binding of RNA polymerase and the CAP-cAMP complex (1 point). When lactose is present, it binds the repressor causing an allosteric change so that the repressor is no longer able to bind DNA (1 point). Repression is lost and RNA polymerase and CAP-cAMP complex are able to bind, allowing transcription (1 point). b. How does the lac operon determine the need for lactose as an energy source? In your answer describe the molecules that act as sensors and explain mechanistically how they affect expression of the Lac Operon (i.e. explain what happens when lactose is needed and what happens when lactose is not needed even though it might be present). 6 points When the cell does not have enough glucose, cAMP levels rise (1 point). cAMP binds to CAP causing an allosteric change that allows CAP-cAMP to bind to the promoter (1 point). Binding of RNA polymerase to the promoter is weak. The binding of CAP-cAMP stabilizes RNA polymerase binding to the promoter allowing transcription to proceed (1 point). When the cell has enough glucose available as an energy source, adenyl cyclase is inhibited (1 point) and cAMP levels are low (1 point). The lack of cAMP prevents CAP from binding to the promoter and thus RNA polymerase is not stable enough to allow transcription of the lac operon (1 point). Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 5) RNA Polymerase II transcribes many RNA products that do not result in proteins. These miRNAs have been found to regulate gene transcription. Describe how these miRNAs are processed to be regulatory factors within the cell and the three possible ways in which gene regulation occurs. In other words, how are the RNAs processed to eventually complex with the RISC and RITS complexes, what do these complexes do, and where do they act? Drawings of this process will be accepted if CLEARLY labeled and functions indicated properly. Drawings are NOT necessary if you prefer to describe it in words (10 points). miRNA forms a hairpin loop (1 point) due to complementary sequence. Drosha (1 point), cleaves the stem loop to form a pre-miRNA. miRNA moves to the cytoplasm and is cleaved by Dicer into short 20-23 nt fragments (1 point). Double stranded miRNA complexes with either the RISC or RITS (1 point) complex. Passenger strand is degraded and guide strand remains associated with the complex (not on diagram shown, must mention that one strand degraded – 1 point). When associated with the RISC complex in cytoplasm (0.5 points), if guide strand is perfectly complementary to mRNA then mRNA is marked for degradation (1 point for both key terms) through action of Argonaute (1 bonus point). If guide strand is not perfectly complementary, then translation of that mRNA is inhibited due to RISC complex association (1 point). When associated with the RITS complex in nucleus (0.5 points), complex is tethered to chromatin (1 point), mRNA is recognized as being transcribed and marked for degradation (1 point). Sp/Su 2011 Name: Exam 4 July 28, 2011 Access ID: 6) There are two types of excision repair mechanisms that we discussed in class: Base excision repair and nucleotide excision repair. a. Describe the steps of repair that are shared between the two mechanisms (5 points). i. Distortion or error present on one of two strands of DNA is recognized (1 point) ii. Enzymatic removal of distortion by endonuclease (1 point) 1. Nicks in phosphodiester backbone usually include some nucleotides around error as well (0.5 points) 2. Leave a gap (0.5 points) iii. DNA polymerase fills in gap (1 point) iv. DNA ligase seals the fragments (1 point) b. Complete the chart below. For each enzyme listed, state in which type of repair the enzyme is involved and the function of that protein. Be sure to specify prokaryotic or eukaryotic repair mechanism. Nucleotide excision repair can be indicated by an “N” and Base excision repair can be indicated with a “B.” If an enzyme has multiple functions or is involved in multiple repair mechanisms, indicate all possibilities. (8 points). Note to graders: each enzyme worth one point, award partial credit as you see fit. Enzyme Type of Excision Repair Function XPD N Helicase, unwind DNA (eukaryotic) uvr gene products N Recognize damage and clip out DNA lesions (prokaryotic) XPA N Recognize DNA damage, recruit XPB and XPD (eukaryotic) DNA glycosylase B Recognize and excise incorrect base (prokaryotic) DNA polymerase N and B Fill in removed DNA (both prokaryotic and eukaryotic) XPF N Nuclease, nick DNA (eukaryotic) DNA ligase N and B Reattach new DNA to original DNA strand (both prokaryotic and eukaryotic) AP endonuclease B Recognize lesion in DNA due to missing nitrogenous base and nick DNA (prokaryotic)