BIOL1001 Exam 4 FA'24 PDF

BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) Full Name: _____________________________________________________________ Discussion section: _________...

BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) Full Name: _____________________________________________________________ Discussion section: _________ TA name: ______________________ (add your name and discussion section if you want to have your exam paper returned to you) **Please note that there is a codon table on the last page of the exam. You should refer to this whenever you feel appropriate. Please turn in the genetic code with your exam. ** 1. The picture below shows nucleosomes along a DNA strand. In what ways will region A and B differ? a. Region A as an open conformation b. Genes in region A will be transcribed at a higher rate. c. Genes in region B will be transcribed at a higher rate. d. A and B e. A and C 2. What is the nucleotide sequence of the anticodon of the tRNA that normally carries tryptophan (trp). a. 5’ – ACC – 3’ b. 5’ – UGG – 3’ c. 3’ – ACC – 5’ d. 3’ – UGG – 5’ e. There is more than one anticodon for trp. 1 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) 3. RNA polymerase reads the ___________ strand during transcription. a. Polypeptide b. RNA coding strand c. RNA template strand d. DNA coding strand e. DNA template strand Use the image below for the next 2 questions. 4. Based on the lac operon, shown above, what are the current environmental conditions of the cell? a. High glucose, high lactose b. Low glucose, high lactose c. High glucose, low lactose d. Low glucose, low lactose 5. The cell becomes exposed to high levels of glucose, would the expression of the operon change and why? a. No, the picture is already showing the operon in a high glucose environment due to the repressor being inactive. b. No, the picture is already showing the operon in a high glucose environment due to the activator being active. c. Yes, the repressor would become inactive. 2 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) d. Yes, the activator would become active. e. Yes, the activator would become inactive. 6. The gene XLM1 is expressed in skin cells and muscle cells. This gene undergoes alternative splicing. One exon that is INCLUDED in skin cells is EXCLUDED in muscle cells. Researchers isolate DNA and mRNA from both types of cells. How do you expect the XLM1 gene to differ in those samples? a. The gene and the mRNA will be the same length in both types of cells. b. The gene and the mRNA will be shorter in skin cells. c. The gene will be the same length in both cells. The mRNA would be longer in muscle cells. d. The gene will be the same length in both cells. The mRNA will be longer in skin cells. e. The gene will be present in both cells, but the mRNA will only be present in muscle cells. Use the image below to answer the next 2 questions. 7. The picture above shows a ribosome in the process of translating mRNA. What is the sequence of the anticodon for the exposed codon in the A-site? a. 5’ – UUA – 3’ b. 3’ – CCU – 5’ c. 5’ – UCA – 3’ 3  BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) d. 3’ – AGU – 5’ e. 5’ – GGA – 3’ 8. After initiation, elongation, and termination of translation, what will be the nal polypeptide sequence be? a. Met – Gly – Pro – Leu b. Met – Gly – Pro – Leu – Ser – Gly c. Met – Gly – Pro – Leu – Pro – Gly – Met – Ser d. Met – Gly – Pro – Leu – Ser – Arg – Val – Ser e. Met – Gly – Pro – Leu – Leu – Ser – Gly 9. A cluster of genes under the control of one promoter is __________. a. a metabolic pathway b. a polycistronic mRNA c. a polysome d. a genome e. an operon 10. A segment of a chromosome becoming attached to a different chromosome is an example of _______________. a. Missense mutation b. Gene ampli cation c. Chromosomal translocation d. Retroviral insertion 11. The image to the right shows RNA polymerase (the gray oval) transcribing a region of DNA. The mRNA is identi ed in the picture by the arrow. For one strand of DNA some of the sequence is shown. What sequence would be found in the RNA for this region? 4 fi fi fi BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) a. 5’-TTACCCG-3' b. 5’-UUACCCG-3' c. 5’-AATGGGC-3' d. 5’-UUTGGGC-3' e. 5’-CGGGTUU-3' 12. A mutation deletes most of the operator sequence in the lac operon. Which option most accurately describes how this would affect the expression of the lac operon? Assume that no glucose is present. a. The operon will be expressed in the presence of lactose, and lactose can be metabolized by the cell. b. The operon will be expressed in the presence of lactose, but lactose cannot be metabolized by the cell. c. Lactose will not be able to enter the cell, and the operon will not be expressed. d. The operon will be expressed in the presence or absence of lactose. e. The operon will never be expressed. 13. A mutation in which location is least likely to impact gene expression? a. Promoter b. Transcriptional regulatory element/operator site c. Splice sites d. Translational regulatory element e. Intron Use the image of the operon below to answer the next 2 questions. 5 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) 14.In scenario 1, a(n) ___________, is causing _____________, in a ___________ cell. a. Repressor, negative control, eukaryotic cell b. Repressor, negative control, prokaryotic cell c. Activator, positive control, eukaryotic cell. d. Activator, positive control, prokaryotic cell. e. More information is required to answer. 15. Scenario 2 shows the same operon, but in the presence of A. How would you best describe what A is? a. Repressor b. Regulatory transcription factor c. Small effector molecule d. Catalyst e. Enhancer 16. In bacteria, an operon encodes several enzymes needed for the synthesis of the amino acid proline. Researchers notice that when proline levels are high in cells, the operon is not expressed. What would be an appropriate hypothesis based on this observation? 6 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) a. Proline binds to DNA and inhibits transcription of the operon. b. Proline is an effector that causes an activator to bind to the operon. c. Proline is a small effector molecule that causes a repressor protein to bind to the operon. d. The operon is under positive control by proline. e. Proline binds to the enzymes encoded in the operon and inactivates them. Use the image below to answer the next 2 questions. A wild-type RNA sequence is shown. The location of two possible mutations (mutation #1 and #2) are shown. 17. With mutation #1 (ignore mutation #2), how will the encoded protein be different from the wild-type? a. It will have the same conformation as the wild type. b. It will be shorter than the wild-type protein due to a premature termination of translation. c. It will not be transcribed. d. It will be transcribed, but not translated. e. It will have a different primary structure than the wild-type’s protein. 18. Mutation #2 is a __________ and the resulting protein will be ________ the wild-type protein (Ignore mutation #1). a. Nonsense mutation; the same length as b. Nonsense mutation; longer than c. Nonsense mutation; shorter than d. Frame shift; longer than 7 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) e. Frame shift; shorter than 19. Events in eukaryotic transcription are listed below. Put them in the order in which they occur. (Note that one of the events DOES NOT occur). I. Addition of the 5’-cap. II. RNA polymerase recognizes the start codon. III. Transcription factors bind to the promoter. IV. Addition of the poly-A tail. a. II, I, IV b. III, IV, I c. III, I, IV d. I, IV, II e. II, IV, I 20. Where do transcription, translation, and splicing occur in a PROKARYOTIC cell. a. Transcription and translation in the nucleus, splicing in the cytosol. b. Transcription and translation in the cytosol, splicing does not happen. c. Transcription in the cytosol, translation in the environment, splicing does not happen. d. Transcription and splicing in the nucleus, translation in the cytosol. e. Transcription in the nucleus, translation in the cytosol, splicing does not happen. 21. In gene expression, the input of a gene is __________, and the nal output is _________. a. DNA, mRNA b. mRNA, protein c. Protein, DNA d. DNA, protein e. mRNA, DNA 8 fi BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) 22. In eukaryotic cells, the small ribosomal subunit recognizes the __________, once attached it will scan the mRNA and _____________. a. Ribosomal binding sequence, tRNAfMet will recognize the start codon. b. 16s rRNA, tRNAMet will recognize the start codon. c. 5’ cap, tRNAMet will recognize the start codon. d. 16s rRNA, the large ribosomal subunit will be added. e. TATA box, the large ribosomal subunit will be added. 23. Researchers identify a mutation (a deletion of 10 base pairs) about 250 bases upstream of the transcription start site of the VRN4 gene of mice. The result of the mutation is that transcription and mRNA levels of the VRN4 gene are much lower in the mutant than in wildtype mice. What can you conclude about the mutation? a. It alters the amino acid sequence of the VRN4 protein. b. It deletes an enhancer sequence. c. It deletes the promoter sequence. d. It deletes a silencer sequence. e. It deletes an effector sequence. 24. It has been noted that individuals with colorectal cancer have a mutation in the KRAS gene increasing its activity and leading to excess cell proliferation. The KRAS gene is an example of what? a. Oncogene b. Tumor-suppressor gene c. Operon d. Silent mutation e. Small effector molecule 25. How would the expression of a eukaryotic gene change if an enzyme adds methyl groups to the core promoter region of the gene? a. Transcription will increase, resulting in more protein. b. Transcription will decrease, resulting in less protein. 9 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) c. Transcription will start in a different location, resulting in a different protein being produced. d. Transcription will not change, but translation will increase, resulting in more protein. e. Transcription will not change, but translation will decrease, resulting in less protein. Use the DNA sequence below to answer the following 2 questions. 26. What is the mRNA transcript of the above DNA strand? a. 5’ – AAUCGUGGGUGAUACUUUGUAAUUUAU – 3’ b. 5’ – UUAGCACCCACUAUGAAACAUUAA AUA – 3’ c. 5’ – AUGAAACAUUAAAUA– 3’ d. 5’ – AUGUUUUACAGUGGGUGCUAA – 3’ e. 5’ – UAUUUAAUGUUUCAUAGUGGGUGCUAA– 3’ 27. What is the polypeptide sequence that would result if this DNA strand was expressed? a. Met- Phe – His – Ser – Gly – Cys b. Tyr – Leu – Met- Phe – His – Ser – Gly – Cys c. Asn – Arg – Gly d. Met – Lys – Leu e. Leu – Ala – Pro – Thr – Met – Lys – Leu 28. A mutation results in the loss of the aminoacyl-tRNA synthetase responsible for attaching methionine (met) to tRNAmet. What would be the most likely effect of this mutation? (Note: Met = Methionine) a. Translation would not be initiated, and no protein would be made. b. Protein would be made, but the rst amino acid on every protein would be something other than methionine. c. A different codon, other than AUG, would be used as the start codon. d. Stop codons would not be recognized resulting in proteins being longer than normal. 10 fi BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) e. A different aminoacyl-tRNA synthetase would attach methionine to tRNAmet, allowing translation to proceed as normal. 29. Researchers develop a technique to isolate ribosomes from the cytoplasm of eukaryotic cells. Which of the following molecules would NEVER be found in one of these samples? a. DNA b. mRNA c. rRNA d. tRNA e. Protein 30. Which answer shows the mRNA transcript that the DNA coding strand below would make if it was transcribed? a.5’ – AUG UAC UCA UGU GGU CGU UAG – 3’ b. 3’ – UAC AUG AGU ACA CCA GCA AUC – 5’ c. 5’ – UAC AUG AGU ACA CCA GCA AUC – 3’ d. 3’ – AUG UAC UCA UGU GGU CGU UAG – 5’ e. 3’ – TAC ATG AGT ACA CCA GCA ATC – 5’ 31. Which of the following mutations would be expected to produce the largest change to the structure and function of a protein? (Assume all mutations occur in the coding region of the gene.) a. A missense mutation in the 5’ end of the gene (close to the promoter). b. A missense mutation in the 3’ end of the gene (close to the terminator). c. A nonsense mutation in the 5’ end of the gene. 11 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) d. A nonsense mutation in the 3’ gene of the gene. e. A silent mutation. 32. Which of the following options describes conditions when the trp operon will be expressed? a. Presence of tryptophan with an active repressor. b. Presence of tryptophan with an inactive repressor. c. Absence of tryptophan with an active repressor. d. Absence of tryptophan with an inactive repressor. 33. It is common to get a DNA sample from cheek cells collected by swabbing the inside of the mouth. Would it be possible to determine a person’s hair color from the cheek cell’s DNA? If not, why? a. Yes b. No, hair color genes would never be found in cheek cells c. No, cheek cells contain genomes that only contain genes that the cheek cell needs to function. d. No, as an embryo develops, hair and cheek cells alter their proteomes, and therefore alter their genomes so that they can ful ll their function. e. No, mutations would have resulted in many changes between the genetic information of hair and cheek cells. 34. Wonka is a type of cancer that involves the loss-of-function mutation of the gene, chocolate, a tumor suppressor gene. What is the function of this gene when it is working normally? a. To induce cells to replicate DNA. b. To include cells to enter mitosis. 12 fi  BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) c. To stop the cell cycle when DNA damage is detected. d. To repair DNA damage. e. To cause cancer. 35. Which of the following statements is NOT TRUE of the proteins synthesized by a polysome. a. They will all have the same primary structure. b. They will differ in length and amino acid sequence. c. They will have methionine on the N-terminus d. They will have the same amino acid at the C-terminus. 36. Prokaryotic genes have a promoter, coding region, and terminator. ___________ plays an important part in RNA polymerase binding to DNA, and allowing RNA polymerase to transcribe which of these sequences (promoter, coding region, terminator)? a. Sigma factor, all three are transcribed. b. TATA box, all three are transcribed. c. Sigma factor, only the coding region and terminator are transcribed. d. TATA box, only the coding region and terminator are transcribed. e. Sigma factor, only coding region transcribed. 37. The formation of a stem-loop is important to which of the following processes? a. Initiation of transcription in prokaryotes. b. Termination of transcription in prokaryotes. c. Structure and function of tRNA. d. A and C e. A, B, and C Use the information below for the following 2 questions. In the metabolic pathway each arrow is an enzymatic reaction (a reaction catalyzed by an enzyme), the starting reactant is A, and the nal product is E. 13 fi BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) You are characterizing the metabolic pathway and have isolated a series of mutations and are inspired by Beadle and Tatum’s work to decipher the steps of this metabolic pathway. You grow your mutants on media supplemented with A, B, C, D, or E of this metabolic pathway. Your results are below. Yes = the cells were rescued and grew. No = the cells were not rescued and could not grow. Media supplemented with… Gene encoding Knockout A B C D E enzyme Gene S Enzyme 1 No No Yes Yes Yes Genes P Enzyme 2 No No No No Yes Gene Q Enzyme 3 No No No Yes Yes Genes L Enzyme 4 No Yes Yes Yes Yes 38. Select the answer the shows the correct order of the enzymes in the metabolic pathway. a. b. c. d. e. 39. If gene P has a mutation, the cell could be rescued when supplemented with… a. E only b. D or E 14 BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) c. C, D, or E d. B, C, D, or E e. A, B, C, or E 40. All but one of the following statements are true of gene regulation in both prokaryotes and eukaryotes. Identify the one statement that is NOT true of both. a. Transcriptional regulation is the most common form of gene regulation. b. Activator proteins increase transcription when bound to DNA. c. Repressor proteins decrease transcription when bound to DNA. d. Gene regulation is in uenced by how tightly packed the chromatin is. e. Mutations in DNA sequences can alter transcription levels. ~END OF EXAM~ Ensure you circle answer key 1 on your scantron. 15 fl BIOL1001 (ABRAMOVICH) FA’24 EXAM 4 ANSWER KEY 1 (Bubble in on scantron) *Please turn in your genetic code with your exam. 16