Lodish 8e Ch09 TestBank PDF

9-1 9 Section 9.1 Transcriptional Control of Gene Expression 1. In bacteria, an operon: a. is a region of DNA that is transcribed as a single mRNA encoding several proteins. b. encod...

9-1 9 Section 9.1 Transcriptional Control of Gene Expression 1. In bacteria, an operon: a. is a region of DNA that is transcribed as a single mRNA encoding several proteins. b. encodes for miRNAs. c. contains a promoter unique for each individual gene in the operator. d. none of the above Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 2. Sigma factors present in bacteria are proteins required to: a. allow translation to proceed. b. terminate DNA replication. c. initiate transcription. d. terminate DNA replication and initiate transcription. Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 3. Operator constitutive mutants of the lac operon would a. express the lac repressor constitutively. b. block the binding of RNA polymerase to the promoter. c. express -galactosidase constitutively. d. prevent the inducer from binding to the repressor. Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 4. How does binding of the lac repressor to the lac operator block transcription initiation? a. lac repressor binding blocks RNA polymerase from interacting with DNA at the start site. b. lac repressor binding induces a DNase that cleaves the DNA at the transcription start site. c. lac repressor binding causes a conformational change in RNA polymerase. d. lac repressor binding induces a protease that degrades the sigma subunit of RNA polymerase. Ans: a Question Type: Multiple Choice  9-2 Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 5. Which of the following proteins does not “footprint” the lac operon control region? a. lac repressor b. -galactosidase c. RNA polymerase d. cAMP-CAP Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: : Remembering, Understanding Difficulty: Moderate Section 9.2 6. You are studying the expression of the gene that appears to be under the control of three different transcription- control regions during mouse embryonic development. Which one of the following is the BEST method to use to determine when each of these regions are active in the developing mouse embryo? a. DNA affinity chromatography b. polymerase chain reaction c. reporter gene assay d. DNAse 1 footprinting Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Applying, Analyzing Difficulty: Moderate 7. Reporter genes employ fragments of DNA encoding proteins that when translated do not have any obvious effects in the cells and tissues. Which one of the following is NOT a reporter protein? a. luciferase b. green fluorescent protein c. β-galactosidase d. RNA polymerase 1 Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 8. You are studying the effects of α-amanitin, a poisonous cyclic octapeptide, on eukaryotic cells and have noticed that following treatment, there is no miRNA transcription. Based on this evidence you conclude that α-amanitin must be inhibiting: a.RNA polymerase I. b. RNA polymerase II. c. RNA polymerase III. d. RNA polymerase I and RNA polymerase III. Ans: b Question Type: Multiple Choice  9-3 Chapter: 9 Blooms: Applying, Analyzing Difficulty: Difficult 9. Which of the following is a fundamental difference between gene regulation in bacteria compared with eukaryotes? a. In bacteria, but not eukaryotes, there is a specific sequence that specifies where RNA polymerase binds and initiates transcription. b. In eukaryotes, but not bacteria, transcription can be influenced by how effectively the DNA sequence of a promoter region interacts with histone octamers. c. Transcription regulation is the most widespread form of control of gene expression in bacteria but not in eukaryotes. d. Gene regulation is readily reversible in eukaryotes but not bacteria. Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 10. All of the following statements about the essential carboxy terminal domain (CTD) of RNA polymerase are true except: a. The CTD is present in RNA polymerase I, II, and III. b. The CTD can become phosphorylated. c. The CTD is critical for viability. d. The CTD of mammals contains more than 50 repeats of a heptapeptide. Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 11. Define the terms cis-acting DNA sequences and trans-acting proteins. Ans: Cis-acting DNA elements affect only the expression of genes on the same DNA molecule that are linked to the DNA element. In contrast, trans-acting proteins can diffuse through the cell to bind to their target DNA sequence. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 12. Describe the similarities and differences between prokaryotic and eukaryotic RNA polymerases. Ans: In prokaryotes, there is only one RNA polymerase, which consists of five subunits. In eukaryotes, there are three RNA polymerases, which are more complex than the bacterial RNA polymerase. RNA polymerase I synthesizes ribosomal RNA; RNA polymerase II synthesizes messenger RNA; and RNA polymerase III synthesizes tRNA and other small RNAs. All three contain two large subunits and 12–15 smaller subunits, which contain some sequence homology to the E. coli RNA polymerase subunits (, , and ´). Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 13. Describe the structure and function of the carboxy terminal domain (CTD) of RNA polymerase II.  9-4 Ans: The carboxy terminal domain (CTD) of RNA polymerase II consists of a heptapeptide repeat, with a consensus sequence of Tyr-Ser-Pro-Thr-Ser-Pro-Ser. Yeast RNA polymerase II contains 26 or more repeats of this sequence, while the mammalian RNA polymerase II contains 52 repeats. The CTD is critical for viability, and at least 10 copies of the repeat must be present for survival. During formation of the transcription initiation complex, the CTD is unphosphorylated. When the RNA polymerase transcribes downstream of the promoter, the CTD is phosphorylated at serine and threonine residues. One hypothesis is that phosphorylation of the CTD causes the release of RNA polymerase from the transcription initiation complex. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.3 14. You have identified a transcription factor and hypothesize that it binds to the promoter region of a gene that encodes a protein that causes cells to stop dividing. In order to test the interaction between the transcription factor and the DNA you will need to do a specific assay. Which one of the following would you use to test your hypothesis? a. fluorescent in situ hybridization b. chromatin immunoprecipitation c. immunocytochemistry d. high-throughput DNA sequencing Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Applying, Analyzing Difficulty: Moderate 15. The human genome encodes transcription factors that contain an acidic activation domain that is phosphorylated in response to increased levels of the second messenger cAMP. Which one of the following contains one of these activation domains? a. CBP b. CDK9 c. CREB d. CTD Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 16. Enhancers are considered transcription-control elements that regulate the expression of eukaryotic genes. Which one of the following is true regarding these elements? a. They are only found upstream of the transcription start site. b. They are never found more than one kilobase away of the transcription start site. c. They are only found in introns. d. They generally range in length from about 50–200 base pairs. Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate  9-5 17. Which one of the following terms is used to describe the protein:DNA complex containing several transcription factors bound to a single enhancer? a. nucleosome b. chromosome c. enhanceosome d. proteasome Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 18. The TATA box a. serves as a promoter sequence for genes transcribed by RNA polymerase III. b. is located approximately 100 base pairs upstream of the start site for mRNAs. c. is present in all eukaryotic genes. d. acts to position RNA polymerase II for transcription initiation. Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 19. All the following elements can function as eukaryotic promoters except a. a TATA box. b. an initiator element. c. CpG islands. d. an enhancer. Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 20. Which of the following is the correct order of binding of general transcription factors to initiate transcription at RNA polymerase II promoters? a. TFIID, TFIIB, Pol II, TFIIH b. PolII, TFIID, TFIIB, TFIIH c. TFIIB, PolII, TFIIH, TFIID d. TFIID, TFIIH, TFIIB, PolII Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 21. What is the function of TFIIH in the transcription initiation complex? a. binding to the TATA box b. unwinding the DNA duplex c. catalyzing the synthesis of RNA d. all of the above  9-6 Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 22. This serves as the promoter for 70% of eukaryotic genes and typically serves as a control region for genes that are transcribed at relatively low rates. a. TATA box b. enhancers c. CpG islands d. UAS (upstream activating sequences) Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 23. Describe the structure of the RNA polymerase II transcription initiation complex. Ans: The RNA polymerase II transcription initiation complex is a multiprotein complex. This complex consists of a DNA promoter element to which general transcription factors (i.e., TFIIA, TFIIB, TFIID, TFIIE, TFIIH) bind along with RNA polymerase II. This multisubunit nucleoprotein complex consists of 60−70 polypeptides with a mass of approximately 3 MDa and is nearly as large as a eukaryotic ribosome. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 24. Describe the functional properties of TFIID and TFIIH. Ans: TFIID is a large, multisubunit complex of approximately 750 kDa. TFIID consists of a 38 kDa TATA box- binding protein (TBP) and 11 TBP-associated factors (TAFs). TBP is the first protein to bind to a TATA box- containing promoter. TFIIH is the last protein to bind to the initiation complex. TFIIH contains helicase activity, which unwinds the DNA duplex at the start site. As the polymerase transcribes from the promoter, a subunit of TFIIH phosphorylates the carboxy terminal domain (CTD) of RNA polymerase II. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.4 25. An enhancer a. is a DNA element that stimulates transcription of eukaryotic promoters. b. binds to RNA polymerase and stimulates transcription. c. acts as a binding site for RNA polymerase. d. interacts with repressor proteins to enhance transcriptional repression. Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy  9-7 26. Which of the following is not used in the electrophoretic mobility shift assay (EMSA)? a. a radiolabeled DNA fragment b. a polyacrylamide gel c. a DNA binding protein d. DNase I Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 27. A leucine-zipper motif contains a. a stretch of five leucine residues in a row. b. a leucine residue at every seventh position. c. a leucine residue complexed with a zinc ion. d. an alternating leucine-alanine-proline structure. Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 28. Which of the following is not a structural motif found in a DNA-binding domain? a. homeodomain b. zinc-finger c. helix-loop-helix d. random-coil acidic domain Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 29. What is the functional difference between enhancers and promoter proximal elements? Ans: Enhancers can stimulate transcription from a promoter tens of thousands of base pairs away. In contrast, promoter-proximal elements are located 100 to 200 base pairs upstream of the start site and usually lose their ability to stimulate transcription from a promoter when moved only several tens of base pairs away. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 30. Describe how the electrophoretic mobility shift assay (EMSA) and the DNase I footprinting techniques are used to identify DNA-protein interactions. Ans: In the electrophoretic mobility shift assay (EMSA), DNA-protein interactions are detected by changes in the mobility of a DNA fragment bound to a protein. A DNA fragment containing a putative protein binding site is first radiolabeled and then incubated in the presence of sequence-specific DNA binding proteins. The DNA fragment containing a bound protein migrates slower in a gel, causing a shift in the location of the radiolabeled DNA detected by autoradiography. In the DNase I footprinting technique, a DNA fragment is first labeled at only one end with 32P.  9-8 The radiolabeled DNA fragment is incubated with a DNA binding protein and then digested with a limiting concentration of DNase I. The DNase I concentration is set such that on average each DNA molecule is cut only once. The resulting DNA fragments are separated by denaturing gel electrophoresis and visualized by autoradiography. In the absence of a DNA binding protein, a ladder of DNA bands is detected on the autoradiogram. Binding of a protein to the DNA prevents DNase I from digesting the radiolabeled DNA at the site of the DNA-protein interaction, resulting in a blank area (or “footprint”) in the DNA ladder. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 31. What is an enhanceosome? Ans: An enhanceosome is a large nucleoprotein complex bound to an enhancer element. This complex is formed by the cooperative assembly of transcription factors to their multiple binding sites in an enhancer. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 32. How can transcription factors be purified using sequence-specific DNA-affinity chromatography? Ans: Sequence-specific DNA-affinity chromatography is a technique that takes advantage of the binding specificity of a protein to a specific DNA sequence. Once the DNA sequence to which a transcription factor binds is identified, this DNA sequence can be coupled to a bead in a column. A protein mix containing the transcription factor is applied to this column. Proteins that do not bind to the DNA fragment are washed off the column. The bound transcription factor can then be eluted from the column in the presence of a high concentration of a salt. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 33. Describe the structure and function of a zinc-finger motif. Ans: A zinc finger is a structural motif found in DNA binding domains; it consists of a short length of the polypeptide chain folded around a Zn2+ ion. The two basic classes of zinc finger domains are the C2H2 and C4 structures. The C2H2 zinc finger domain consists of two cysteine (C) and two histidine (H) residues bound to one Zn2+ ion. The C4 zinc finger contains four cysteines bound to one Zn2+ ion. The three-dimensional structure of the zinc finger forms a compact domain, which can insert its  helix into the major groove of DNA. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.5 34. Bromodomains are found in chromosome-associated proteins that contribute to transcriptional activation. To facilitate this activation, the bromodomains bind to histones, specifically their lysine residues that have been post- translationally modified by: a. methylation. b. acetylation. c. phosphorylation. d. ubiquitination. Ans: b  9-9 Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 35. Which of the following is NOT true regarding the chromatin-remodeling SWI/SNF complex? a. It acts as tumor suppressor. b. It serves as a co-activator of transcription. c. It has homology to DNA helicases. d. It can stabilize DNA-histone interactions. Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 36. All the following statements about heterochromatin are true except: a. DNA dyes stain heterochromatin more darkly than euchromatin. b. The DNA of heterochromatin is more highly condensed than that of euchromatin. c. Heterochromatin is associated with inactive genes. d. Heterochromatin is more susceptible to DNaseI than is euchromatin. Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 37. All of the following events play a role in yeast-mating type switching except a. methylation of the silent-mating-type locus. b. transcription of the gene at the MAT locus. c. chromatin condensation at the silent mating type locus. d. a recombination event known as gene conversion. Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 38. The mediator complex a. can form a molecular bridge between activators of transcription and DNA replication machinery. b. can function to maintain a promoter in a hypoacetylated state. c. has histone acetylase activity. d. none of the above Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 39. Transcriptionally inactive genes a. are always located within euchromatin. b. are not located within nucleosomes.  9 - 10 c. often are methylated. d. are not resistant to DNase I. Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 40. Describe the role of histone deacetylation and hyperacetylation in yeast transcriptional control. Ans: Histone deacetylation/hyperacetylation is one mechanism for regulating transcriptional control in yeast. Repressor proteins can cause deacetylation of lysine residues in histone N-termini in nucleosomes. Unacetylated histones contain positive charges due to the N-terminal lysines and interact strongly with DNA phosphates. These strong interactions may restrict access of general transcription factors, thus leading to transcriptional repression. In contrast, histones with hyperacetylated lysines in their N-termini are neutral in charge, eliminating the strong electrostatic interactions with the DNA phosphates. This more open chromatin configuration facilitates access of general transcription factors and induces transcriptional activation. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.6 41. The nuclear-receptor superfamily consists of several proteins that bind to consensus sequences of DNA response elements. Which of the following is NOT considered a member of this superfamily? a. retinoic acid receptor b. acetylcholine receptor c. glucocorticoid receptor d. progesterone receptor Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 42. Which one of the following techniques would be best suited to follow how a thyroxine-bound receptor translocates from the cytoplasm to its DNA response element? a. in situ hybridization and a radioactive fragment of the DNA response element b. fluorescence microscopy and a GFP-tagged receptor fusion protein c. pulse-chase radiolabeling d. none of the above Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Applying, Analyzing Difficulty: Difficult 43. Lipid soluble hormones activate transcription by a. binding to specific cell-surface receptors. b. phosphorylating a protein kinase. c. binding to a nuclear receptor.  9 - 11 d. inhibiting a histone deacetylase. Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 44. Which protein domains are found in nuclear-receptor family members? a. variable region, DNA-binding domain, ligand-binding domain b. acetylase domain, DNA-binding domain, ligand-binding domain c. variable region, acetylase domain, ligand-binding domain d. variable region, DNA-binding domain, acetylase domain Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 45. Regulation of transcription by steroid hormones a. involves hormone receptors only found in the nucleus. b. involves cytoplasmic hormone receptors that can move to the nucleus. c. involves two ligase domains. d. always activates transcription. Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 46. Describe how lipid soluble hormones, glucocorticoid for example, regulate gene transcription acting through nuclear hormone receptors. Ans: Glucocorticoid is a lipid soluble hormone that binds to a member of the nuclear hormone receptor family, the glucocorticoid receptor (GR), which regulates gene transcription. In the absence of glucocorticoid, the GR in the cytoplasm is bound to the protein HSP90. When glucocorticoid diffuses through the cell membrane, it binds to the GR ligand-binding domain and causes a conformational change in the GR, releasing HSP90. The GR bound to glucocorticoid is then translocated into the nucleus, where it interacts with glucocorticoid response elements (GRE) and regulates transcription of responsive genes. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 47. Describe the mechanism of transcriptional control for the heat shock genes. What advantage does this type of control impart to the cell? Ans. During transcription of the heat-shock genes, RNA Pol II pauses after transcribing ≈25 nucleotides. Under stress conditions, where intracellular proteins are denatured or may become denatured, heat shock transcription factor (HSTF) is activated. In this state, HSTF binds to specific regions in the promoter of the heat shock genes, stimulating RNA Pol II to continue chain elongation. Binding also facilitates the rapid re-initiation by other RNA Pol II molecules, leading to a significant up-regulation in heat-shock-gene-expression. Thus, the mechanism of stalling the RNA Pol II and having partially completed transcripts ready to finish elongation and undergo translation when the need arises is a safeguard, protecting cells against unexpected, stressful conditions.  9 - 12 Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.7 48. Epigenetics marks refer to modifications to DNA and proteins that in turn regulate gene expression. Which statement is true regarding these specific types of modifications? a. They have the ability to both silence and activate genes. b. They can involve the methylation of cytosine bases. c. They are linked to the acetylation of histones. d. All of the above Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 49. What family of proteins plays an essential role in the repression of genes that help to direct the formation of specific tissues and organs in a developing embryo? a. Retinoblastoma b. Trithorax c. Polycomb d. Pax Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 50. X chromosome inactivation in mammals is mediated by a. micro RNAs (miRNA). b. long non-coding RNAs (ncRNA). c. messenger RNA (mRNA). d. short RNA-directed methylation of histones and DNA. Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate Section 9.8 51. Which of the following statement(s) regarding transcription initiation and RNA Pol III is (are) true? a. ATP hydrolysis is not required for initiation. b. Pol III is responsible for synthesizing tRNAs and 5S-rRNA. c. The promoter elements of tRNA genes lie entirely within the transcribed sequence. d. all of the above Ans: d Question Type: Multiple Choice  9 - 13 Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult