Midterm 2 - Masterfile

Name__________________________________ Perm Number __________________ Discussion Section __________________ MCDB 101A Fall 2023 Section 200 November 16, 2023 Midterm 1 Version X 20 Multiple Choice - 3 points each – 60 points total 7 Short answer – 4 – 6 points – 40 points total 1. Select the best answer possible 2. In your scantron, mark the version of the exam 3. Write your name, perm number and discussion section in this page as well as in the short answers pages 4. Write as legible as possible. 5. FOR THE SHORT ANSWERS, USE ONLY THE ALLOTED SPACE, points will be taken off otherwise.

Name__________________________________ Perm Number __________________ Discussion Section __________________ A B C 0 An s Question 1 1 4 5 1 B Which of the following enzymes is not part of the nucleotide excision repair mechanism? A. DNA polymerase B. Glycosylase C. UvrA D. UvrC E. All listed enzymes are required in excision repair 1 9 4 1 4 2 A Which of the following mechanisms could repair a double strand break? A. Nonhomologous end-joining (NHEJ) B. Base Excision Repair (BER) C. Methyl directed mismatch repair D. Repair by Photolyase E. More than one of the listed mechanisms 1 7 6 1 8 3 C A mutant of the adenine methylase that is part of the methyl directed mismatch repair mechanism mutates so that its efficiency and speed of methylation is 100 times faster than wildtype. As a consequence, DNA becomes methylated as it becomes synthetized. What would be a consequence of this mutation? A. Replication of the lagging strand cannot be completed B. The expression of some metabolic genes will be higher C. The mutation rate will be higher D. The function of MutS and MutL will be impaired E. None of the listed would be a consequence of this mutation 4 5 1 2 4 D The following table corresponds to the results of a complementation table. How many different complementation groups are affected by this mutations? Which mutations affect the same genes? A. 5 complementation groups: a/b/c/d/e/f B. 2 complementation groups: bcde / af C. 3 complementation groups: af/bd/ce D. 3 complementation groups: af/bde/c E. 2 complementation groups: afc/bde 6 1 0 1 6 5 E Based on the following mutation map, which of the following pairs of mutations could recombine to produce completely wild type recombinants? A. 1 and B B. 3 and B C. 1 and C D. B and C E. 2 and A

Name__________________________________ Perm Number __________________ Discussion Section __________________ 1 1 2 0 1 9 1 0 C After repeating Beadle and Tatum's experiments, you expose Neurospora to X-rays and isolate a mutant that cannot grow on minimal media alone but can grow when the media is supplemented with leucine. What conclusion can you draw from this result? A. The X-rays have impaired the Neurospora's ability to digest/metabolize leucine. B. Unlike the wild type, Leucine is lethal to the mutant Neurospora C. The mutant Neurospora has a genetic defect in the pathway that synthesizes leucine. D. The Neurospora has developed a new metabolic pathway that depends on external leucine. E. None of the listed options are correct 9 1 6 6 1 1 A Red/green color blindness is often cause by A. Unequal crossing over between red and green photoreceptors genes B. Point mutations (missense) in the rhodopsin gene C. Mismatch repair after recombination of the rhodopsin gene and the green photoreceptor gene D. Mismatch repair after recombination of the rhodopsin gene and the red photoreceptor gene E. Premature decay of rod cells in the retina 2 1 8 2 0 1 2 B An alien organism was found to have a similar transcription/translation mechanism than life on earth, except that it uses 5 nitrogenous bases instead of 4 and has 25 aminoacids instead of 20. How long do you predict their codons must be? A. 2 nucleotides B. 3 nucleotides C. 4 nucleotides D. 5 nucleotides E. 6 nucleotides 1 2 8 7 1 3 C Yanofsky’s observations that each point mutation affects only one amino acid led him to conclude that A. The genetic code is unambiguous B. The genetic code is redundant C. Each nucleotide is part of only one codon D. That the bases within a codon cannot recombine E. None of these are appropriate conclusions for Yanofsky’s experiments 1 3 7 1 7 1 4 E You induce a mutation in E coli using proflavin, an intercalating agent. The resulting mutant is incapable of growing in minimal media without supplementing it with arginine. You repeat the proflavin treatment in the mutant and obtain a strain that can survive without arginine. What can be concluded from these experiments? A. The first treatment likely caused a frameshift mutation that was restored by the second mutation. B. Both mutations likely affected the same gene C. If the first mutation consisted of an insertion, the second mutation was likely a deletion and vice versa.

Name__________________________________ Perm Number __________________ Discussion Section __________________ D. The sequence of amino acids located after the first mutation will be altered, but it will be partially restored to wild type in the aminoacids located after the second mutation E. All of the answers are appropriate conclusions of these experiments 1 4 9 2 1 5 D You repeat the Nirenberg and Matthaei experiment and add the trinucleotide poly-UAC to an in vitro translation system. What do you expect to obtain? A. One peptide only, whose identity could be EITHER Tyr-Tyr-Tyr…, Thr-Thr-Thr… or Leu…Leu…Leu B. One peptide with the sequence: Tyr-Thr-Leu-Tyr-Thr-Leu-Tyr-Thr- Leu… C. One peptide with the sequence: Tyr-Ser-Tyr-Ser-Tyr-Ser… D. Three peptides: Tyr-Tyr-Tyr…, Thr-Thr-Thr… and Leu…Leu…Leu E. None of the above 1 5 1 9 8 1 6 A Brenner notice that mutations to the sequence of certain codons to specific sequences resulted in truncated proteins. The specific mutated sequences were always the same three. This led him to conclude… A. That the sequences corresponded to the three stop codons B. That the induced mutations were always deletion mutations C. That the mutations were always frameshift mutations D. That the genetic code is redundant E. That the reading frame is established from a fix start point 1 8 1 2 1 3 1 7 B Which of the following elements are present in Eukaryotes but not in prokaryotes? A. Promoters B. Spliceosomes C. Shine-Dalgarno Box D. Rho-dependent termination E. Sigma factors 8 1 7 1 0 1 8 C Transcription terminators… A. refer to enzymes that signal the end of transcription B. refer to proteins that occupy the ribosome A site when a stop codon is read by the ribosome C. in bacteria can be either extrinsic (rho dependent) or intrinsic (rho independent) D. refer to stop codons that cannot be read by any tRNA E. More than one option is true 7 1 3 1 1 1 9 D Enhancers… A. Are part of the transcripts but not of the final protein B. Are the binding site of the RNA polymerase in the promoter C. Are sequences that allow for the proper alignment of the mRNA in the ribosome D. Facilitate the binding of the RNA polymerase to the promoter E. None of the above 1 1 1 2 B The peptidyl transferase…

Name__________________________________ Perm Number __________________ Discussion Section __________________ Short Answer 21. Explain the SOS DNA repair mechanisms used by bacteria. What is the enzyme that repairs the DNA in this process and how it is different from other enzymes of the same type? (2pts) Why is only used in cases of extreme damage to the DNA(2pts)? Given the downsides of this mechanism, why is it even used (2pts)? The SOS DNA repair system is a last-resort mechanism in bacteria, primarily done by a sloppy, error prone DNA polymerase. This system is reserved for critical situations because it can induce mutations. It is used because is a survival mechanisms in the face of lethal DNA damage, offering a chance for recovery at the cost of increased mutations. 22. To make a recombinant test for two rII mutants, you coinfected E.coli B with both mutants, do three 10X serial dilutions of the solution and plate 250 µL of this solution on a plate with E. coli B and observe 120 plaques. Separately, you do two 10X serial dilutions of the original lysate and plate 100 µL of this solution in E coli strain K and obtain 24 plaques. What is the map distance between mutants (3pts)? Show your work to get full credit (3 pts) RF = ( 10 2 ) ( 1 0.1 )( 24 colonies )( 2 ) ( 10 ¿¿ 3 )( 1 0.25 )( 120 colonies )= 0.1 ¿ Map distance = (RF)*100 = 10 mu 23. You repeat the experiments of Nirenberg and Leder to disambiguate the identities of the aminoacids produced by the in vitro translation of the repeating dinucleotide AGAGAGAG…. You obtain a product that has alternating Ser-Leu- Ser-Leu… Using a diagram, represent the setup of the experiment using AGA as the mini mRNA. Make sure to include all the components (2 pts) and clearly indicate what is marked with radioactivity in the two experiments and results (2 pts). Include the results of both experiments and the conclusions (2pts). Diagram for correct answer substitutes CUC for AGA 24. Describe 3 differences in translation between prokaryotes and eukaryotes (2 pts each) Any combination of three of the following :

Name__________________________________ Perm Number __________________ Discussion Section __________________  Initiation of Translation: In Prokaryotes translation begins at the Shine-Dalgarno sequence, a purine-rich sequence upstream of the start codon; in Eukaryotes translation begins at the 5' cap structure (methylguanosine cap) of the mRNA.  Ribosome Size and Composition: Prokaryotes ribosomes are smaller  Separation of transcription/translation in eukaryotes but is concurrent in prokaryotes  mRNA Structure: prokaryotes are often polycistronic, carrying information for several proteins while eukaryotes are generally monocistronic, carrying information for a single protein  mRNA Processing: For prokaryotes mRNA is often translated directly after transcription whereas in eukaryotes mRNA undergoes extensive processing, including 5' capping, 3' polyadenylation, and splicing. 25. The expression of the short DNA sequence below produced the peptide: Met-Ser-Gly-Pro-Arg. Which of the two strands was used as the template for this peptide? (2pts). Write the mRNA sequence (1pt) as well as the tRNA sequence of the anticodons (1pt) that correspond to this peptide including the 3’ and 5’ ends. Using the wobble rules, indicate which of the codons can only be bound by one tRNA (2 points). TCTCGGTACAGACCCGGTGCTACTCCGGCT AGAGCCATGTCTGGGCCACGATGAGGCCGA The template strand is the top. mRNA sequence : 5’ AUG UCU GGG CCA CGA UGA 3’ tRNA sequence: 3’ UAC AGA CCC GGU GCU 5’ Codons only bound by one tRNA: AUG and GGG 26. Crick and Brenner induced mutations using proflavine, an intercalating agent. What was the effect of this mutagenic agent on genes and their products? What did they observe when the mutant alleles were exposed a second time to proflavine? What did they concluded based on these results? Proflavine can insert or delete a single base pair, leading to frameshift mutations that change the identity of all the amino acids after the insertion or deletion. They also observed that if they deleted or inserted 3 bases at a time, there was no frameshift (2) When the mutants are exposed a second time to proflavine, the new insertion or deletion arises. In some cases, the second exposure to proflavine caused reversion back to the normal function of the affected genes(2) because they correct the reading frame. This suggested that the genetic code is read in groups of three nucleotides or codons (2). (alternatively they can also mentioned that they concluded that the genetic code is non-overlaping and linear) 26. What are the two mechanisms of transcription termination in prokaryotes? 2 points ea The Rho-dependent mechanism involves the Rho protein, a helicase that binds to the RNA at a specific recognition site (RUT) and turns. As it turns, Rho pulls the mRNA away from the RNA polymerase and causing the dissociation of the RNA polymerase and the DNA. (2) In the Rho-independent mechanism, also known as intrinsic termination, relies on the formation of a hairpin structure in the RNA transcript followed by a series of uracil residues. This structure forms because of complementary base-pairing within the RNA sequence. The hairpin causes the RNA polymerase to pause, because it generates friction against the cytosol which eventually lead to the release of the RNA transcript, thus terminating transcription. (2)