CELL2050 EXAM III

50. On the DNA strands shown below, two RNA polymerase enzymes are using the top strand as a template. In the boxes, label the 5′ and 3′ ends of the DNA molecules and the RNA molecules being made. With arrows, indicate the directions, left to right or right to left, that the polymerases are moving.

Answer: Section 13.2 Application Question

51. Draw a diagram of the transcription and translation of the trp operon under high tryptophan conditions. In addition to the trp operon, show RNA polymerase; the 5 UTR RNA with regions 1, 2, 3, 4, and the string of uracils; and a ribosome.

Answer: Section: 16.3 Comprehension

62. Histone acetyltransferases add acetyl groups to lysines, which are positively charged amino acids. How might this affect the association of a nucleosome with DNA?

Answer: (1) DNA is negatively charged. (2) Acetylating lysines would neutralize the positive charge, which would allow them to form ionic bonds with DNA. (3) Nucleosomes that contain histones with a reduced ability to bond with DNA would be less tightly associated with DNA. Section 13.4 Application Question

46. The discovery of ribozymes led to the theory that the evolution of life on Earth began with an "RNA world." What properties of proteins and DNA make them more suitable than RNA as enzymes and the cell's genetic material?

Answer: (1) Proteins are made of 20 different amino acids; RNA is made of relatively uniform nucleotides with only four different bases. Because they can be more chemically diverse, proteins are better suited to catalyzing a variety of cellular reactions. (2) DNA is double-stranded, with a structure that protects the molecule from degradation. The lack of 2′ OH makes the molecule less susceptible to self-hydrolysis. Because it is more stable, DNA is a better molecule for the cell's genetic material. Section 13.1 Comprehension Question

59. How are enhancers different from promoters?

Answer: (1) Unlike for promoters, the distance between enhancers and the gene they influence is not critical. (2) Enhancers may affect several genes in their vicinity. (3) Enhancers may be downstream or upstream of the genes whose transcription they affect. Section 13.4 Comprehension Question

44. The discovery of ribozymes led to the theory that the evolution of life on earth began with an "RNA world." Describe the chemical properties and functions of RNA that would allow it to be the basis of the first self replicating systems.

Answer: (1) As a nucleic acid, RNA can serve as a template for self-replication: An RNA can be copied into a complementary strand that is a template for generating more of the original RNA. (2) As a nucleic acid, RNA can carry genetic information in its base sequence. (3) RNA could have performed the reactions required of a self-replicating system because RNA ribozymes have catalytic activity. Section 13.1 Comprehension Question

60. Describe how the activities and requirements of prokaryotic RNA polymerase and eukaryotic RNA polymerase II are both similar and different.

Answer: Both transcribe mRNA in a 5′ to 3′ direction, use rNTPs as a substrate, use a single DNA strand as a template, and do not require a primer to begin RNA synthesis. They associate with different proteins and bind to different promoter sequences to begin transcription: the prokaryotic enzyme requires the sigma subunit and uses the -10 and -35 sequences; the eukaryotic enzyme requires TBP and a whole host of additional general transcription factors. Additionally, the eukaryotic enzyme uses several different binding sequences like the TATA box, TFIIB recognition element, etc. The eukaryotic enzyme may also interact with enhancer sequences. Sections 13.3 & 13.4 Application Question

50. Explain how RNA editing violates a general principle of molecular genetics.

Answer: Except for RNA viruses, genetic information resides exclusively within the nucleotide sequence of DNA. This fundamental principle is violated by RNA editing, a process whereby the coding sequence of an mRNA molecule is altered post-transcriptionally. This results in the translation of a polypeptide having an amino acid sequence that differs from the coding sequence contained in the gene. Section: 14.2 Challenge Question

45. Is the below sequence RNA or DNA? How can you tell? 5 ...GGAGCUCGUUGUAUU... 3

Answer: It is RNA because it contains Us (uracil) and no Ts (thymine). Section: 15.3 Application Question

47. What would be the effect on the below amino acid sequence if the sequence were changed to 5 GGAGACUCGUUGUAUU 3? Explain why the amino acid sequence changes. 5 ...GGAGCUCGUUGUAUU... 3

Answer: The amino acid sequence would be gly-asp-ser-leu-tyr. The reading frame, or grouping of bases used as codons, changes after the mutation (the addition of an A in the fifth position). Section: 15.3 Application Question

42. An E. coli strain of chromosomal genotype lacI lacP+ lacO+ lacZ+ lacY+ constitutively expresses the genes of the lac operon. The strain is converted to wild-type lac operon regulation by the addition of an extra piece of DNA. What gene or genes are contained on this extra DNA that explain this conversion to wild type? Include an explanation of cis- or trans-acting factors and how they work.

Answer: The extra piece of DNA contains a wild-type lacI+ gene. It encodes a wild-type trans-acting lac repressor that can bind to the chromosomal operator of the lac operon and restore wild-type regulation. Section: 16.2 Application

49. What is the RNA sequence transcribed from the DNA shown below? promoter +1 5′ GTAACTATAATTAACGTAAGACTAT 3′ 3′ CATTGATATTAATTGCATTCTGATA 5′

Answer: 5′ GUAAGACUAU 3′ Section 13.2 Application Question

34. An in vitro transcription system that contains a bacterial gene initiates transcription, but from random points on the DNA. Which of the following proteins most likely is missing from the reaction? a. Sigma factor b. Rho factor c. RNA polymerase II d. TATA binding protein (TBP) e. TFIID

- Answer: a Section 13.3 Application Question

53. The following diagram represents DNA that is part of the RNA coding sequence of a transcription unit. The sequence of the RNA product of transcription is 5′UGGCAUCGCCUAU3′. Label the diagram of the DNA molecule to indicate which strand serves as the template strand. Also, label the 5′ and 3′ ends of both strands.

Answer: Section 13.2 Application Question

50. Four different uracil auxotrophs of Neurospora, a eukaryotic mold, are tested for growth on uracil and uracil precursors. The data are shown in the following table. A plus sign (+) means growth. Compound A B C D Uracil Mutant 1 + + + - + Mutant 2 - - + - + Mutant 3 + - + - + Mutant 4 - - - - + Diagram the uracil pathway, showing the step at which each mutant is blocked.

Answer: 1 3 2 4 D B A C uracil Section: 15.1 Application Question

39. Write the anticodon, with correct polarity, of all tRNAs that will bind to the mRNA codon 5 UCG 3, considering wobblebase pairing rules.

Answer: 5 CGA 3 Section: 15.3 Application Question

Short-Answer Questions 36. What is the difference between a transcription regulation system that uses induction and a system that uses repression?

Answer: (1) Induction: Stimulates expression of a gene in response to a specific substrate. Repression: Prevents expression of a gene until a specific substrate is present. (2) Induction: Transcription is normally off and is stimulated in response to a specific substrate. Repression: Transcription is normally on and is shut off in response to a specific product. Section: 16.1 Application

59. Describe the events in prokaryotic translation elongation.

Answer: (1) A ribosome with a growing peptide attached to a tRNA in the P site accepts a charged tRNA with the next amino acid into the A site. The charged tRNA enters as a complex with EF-Tu and GTP. (2) If the anticodon of the charged tRNA matches the codon, GTP is cleaved, EF-Tu exits and is regenerated to EF-Tu-GTP by EF-Ts. (3) A peptide bond is formed by the peptidyl transferase activity of the large subunit rRNA. The polypeptide chain on the tRNA in the P site is transferred to the amino acid on the tRNA in the A site. (4) The ribosome translocates toward the 3 end of the mRNA with the aid of EF-G and GTP hydrolysis. The empty tRNA that was in the P site moves to the E site and exits. The tRNA with the polypeptide that was in the A site moves to the P site. Section: 15.3 Comprehension Question

54. Compare and contrast the mechanisms of splicing for nuclear pre-mRNA introns, group I introns, group II introns, and tRNA introns.

Answer: (1) Pre-mRNA introns: Spliceosome-mediated intron excision and exon splicing. Introns form lariat structures prior to their release. (2) Group I introns: All group I introns fold into a common secondary structure containing nine looped stems, which are necessary for transesterification and splicing. Group I introns self-splice exclusively and thus do not require spliceosome-related components. (3) Group II introns: The splicing mechanism of group II introns is very similar to the mechanism used by nuclear genes, involving snRNPs and two transesterifications, which generate a lariat structure. However, all group II introns form a specific secondary structure and self-splice. (4) tRNA introns: For both prokaryotes and eukaryotes, different tRNAs are processed differently, so there is no standard processing pathway to speak of. The splicing process for all tRNAs is not spliceosome-mediated. Instead, generally speaking, tRNA introns are excised by endonucleases and released for degradation, and then the two ends of the tRNA molecule—which are held in place by intramolecular bonding within the secondary tRNA structure—are ligated together to form the complete, processed tRNA molecule. Section: 14.2 and 14.3 Application Question

45. The discovery of ribozymes led to the theory that the evolution of life on earth began with an "RNA world." Describe how the current cellular roles of RNA support this theory.

Answer: (1) RNA is an intermediate between DNA, the permanent genetic information, and proteins. (2) RNA has a role as a primer for DNA replication. (3) RNA is required for protein synthesis as mRNA, rRNA, and tRNA. Section 13.1 Comprehension Question

57. What would you add to an in vitro transcription system that contains an E. coli gene for glyceraldehyde 3phosphate dehydrogenase, an enzyme in glycolysis, in order to get transcription that begins from the normal transcription start site?

Answer: (1) RNA polymerase (2) Sigma factor (3) Ribonucleoside triphosphates with A, C, G, and U bases (rNTPs) Section 13.3 Application Question

66. What would you add to an in vitro transcription system that contains a Drosophila gene for glyceraldehyde 3phosphate dehydrogenase, an enzyme in glycolysis, in order to get basal transcription?

Answer: (1) RNA polymerase II (2) Basal transcription factors (TFIIA, TFIIB, TFIID, TFIIE, TFIIF, and TFIIH) (3) Ribonucleoside triphosphates with A, C, G, and U bases (rNTPs) Section 13.4 Application Question

55. Compare rho dependent and rho independent transcription termination in terms of their use of RNA secondary structures.

Answer: (1) Rho dependent transcription termination relies on a lack of RNA secondary structure so that rho can bind the RNA, move up to the RNA polymerase, and remove the RNA from the DNA template. (2) Rho independent transcription termination relies on a hairpin in the transcribed RNA that causes RNA polymerase to pause, facilitating termination. Section 13.3 Comprehension Question

58. Explain why eukaryotic transcripts from the same RNA polymeraseIItranscribed gene may vary in the sequences at the 3′ end.

Answer: (1) The 3′ end is the termination end. Termination on polymerase-II-transcribed genes can occur at multiple sites. (2) RNA polymerase transcribes until the 3′ end is processed by a cleavage complex. The complex may prevent termination until the processing sequence is transcribed and the complex can cleave the 3′ end of the new RNA. Section 13.4 Comprehension Question

64. Describe three ways that eukaryotic transcription initiation is different from prokaryotic transcription initiation.

Answer: (1) There are many different eukaryotic RNA polymerases. Each recognizes the promoter of a different type of gene. (2) Eukaryotic RNA polymerases require general transcription factors to associate with a promoter. (3) Eukaryotic promoters do not have the same -10 and -35 consensus sequences that prokaryotic promoters have. (4) Eukaryotic RNA polymerases do not have the polymerase subunits that are required for promoter recognition. (5) Eukaryotic transcription initiation requires altering or removing nucleosomes at the gene to be transcribed. (6) The different types of eukaryotic polymerases recognize different promoter sequences. Section 13.4 Application Question

48. This DNA sequence represents an open reading frame (ORF) of a transcriptional unit. Transcribe and then translate this gene in the spaces provided below. 5 ATGGGAGCTCGTTGTATTTGA 3 3 TACCCTCGAGCAACATAAACT 5

Answer: (a) Transcribed mRNA sequence: 5 AUGGGAGCUCGUUGUAUUUGA 3 (b) Translated protein sequence: met-gly-ala-arg-cys-ile (c) Which amino acid is at the C-terminus of this peptide? Ile Section: 15.3 Application Question

45. Fill in the blanks in this table with "yes" or "no" for each condition of lac operon regulation. The strain is wild type, with no partial diploidy. The first line is filled in for reference. Condition Allolactose levels high? lac repressor bound to operator? cAMP levels high? CAP bound to CAP binding site? Transcription at highest level? high glucose no lactose no yes no no no no glucose high lactose high glucose high lactose no glucose no lactose

Answer: Condition Allolactose levels high? lac repressor bound to operator? cAMP levels high? CAP bound to CAP binding site? Transcription at highest level? high glucose no lactose no yes no no no no glucose high lactose yes no yes yes yes high glucose high lactose yes no no no no no glucose no lactose no yes yes yes no Section: 16.2 Application

39. Fill in the blanks in the following table with "yes" or "no" for each condition of trp operon regulation. The strain is wild type, with no partial diploidy. Medium trp repressor bound to trp? trp repressor bound to operator? Antiterminator hairpin formed in mRNA? Transcription attenuated? high tryptophan low tryptophan

Answer: Medium trp repressor bound to trp? trp repressor bound to operator? Antiterminator hairpin formed in mRNA? Transcription attenuated? high tryptophan yes yes no yes low tryptophan no no yes no Section: 16.2 Application

37. Describe one similarity and one difference in how the trp and lac repressor proteins function.

Answer: Similarity Difference • Both bind operator sequences in DNA. • When bound, transcription is repressed. • Both are encoded by genes NOT included in the operons they regulate. • When the lac repressor binds the inducer (lactose), it CANNOT bind to DNA. • When the trp repressor binds the co-repressor (tryptophan), it DOES bind to DNA. Section: 16.2 Comprehension

41. Fill in the blanks in the "strain genotype" column of the following table. Write chromosomal genotypes with no partial diploidy. (+) means transcription of the lac operon. (-) means no transcription of the operon. The first line is filled in for reference. Strain genotype Lactose absent Lactose present lacI+ lacP+ lacO+ lacZ+ lacY+ - + lacI+ lacP+ lacOc lacZ+ lacY+ OR lacI- lacP+ lacO+ lacZ+ lacY+ lacI+ lacP- lacO+ lacZ+ lacY+ OR lacIs lacP- lacO+ lacZ+ lacY+

Answer: Strain genotype Lactose absent Lactose present lacI+ lacP+ lacO+ lacZ+ lacY+ - + lacI+ lacP+ lacOc lacZ+ lacY+ OR lacI- lacP+ lacO+ lacZ+ lacY+ + + lacI+ lacP- lacO+ lacZ+ lacY+ OR lacIs lacP- lacO+ lacZ+ lacY+ - - Section: 16.2 Application

38. The following table shows several bacterial strain lac operon genotypes (some are partial diploids). a. Fill in the blanks in the "lactose absent" and "lactose present" columns in this table. (+) means significant levels of active ß-galactosidase enzyme can be detected. (-) means no significant levels of active ß-galactosidase enzyme. The first line is filled in for reference. Strain genotype Lactose absent Lactose present 1. lacI+ lacP+ lacO+ lacZ+ lacY+ - + 2. lacI+ lacP+ lacOc lacZ+ lacY+ 3. lacI+ lacP+ lacOc lacZ- lacY+ 4. lacIs lacP+ lacOc lacZ+ lacY+ 5. lacI- lacP+ lacO+ lacZ+ lacY+ / lacI+ 6. lacIs lacP+ lacO+ lacZ+ lacY+/ lac I+ b. If lactose is absent and glucose is present in the culture medium of strain 4, will significant levels of active ß-galactosidase enzyme be detected? Explain. c. Using your answers in the previous table, explain the dominance relationships among the three lacI alleles (lacI+, lacI−, lacIs).

Answer: a. Strain genotype Lactose absent Lactose present 1. lacI+ lacP+ lacO+ lacZ+ lacY+ - + 2. lacI+ lacP+ lacOc lacZ+ lacY+ + + 3. lacI+ lacP+ lacOc lacZ- lacY+ - - 4. lacIs lacP+ lacOc lacZ+ lacY+ + + 5. lacI- lacP+ lacO+ lacZ+ lacY+ / lacI+ - + 6. lacIs lacP+ lacO+ lacZ+ lacY+/ lac I+ - - b. Yes, despite the fact that the super-repressor allele is present (lacIs), it cannot bind to the operator due to the lacOc mutation at the operator—this allows for constitutive ß-galactosidase enzyme expression (expression whether lactose is present of absent). If glucose is present, the CAP protein will NOT be active (cAMP levels will be low). The combination of lack of repression and lack of activation by CAP will most likely result in transcription of the operon and production of active ß-galactosidase enzyme. c. lacIs is (trans) dominant to both lacI+ and lacI−; lacI+ is (trans) dominant to lacI−. When both lacI+ and lacI− are present in the cell, we see the lacI+ phenotype (see strain 5). When lacIs is present in the cell, we see the lacIs phenotype (see strain 6). Section: 16.2 Application

47. Imagine the following scenario: You take the regulatory region of the trp operon (including the promoter, operator, and 5′ UTR) and attach it upstream of the structural genes of the lac operon. You then introduce this artificial construct into a mutant in which the lac operon is completely nonfunctional. Indicate the level of ß-galactosidase activity in each of the following cases and explain why you expect that level of activity: a. No tryptophan, no lactose b. High tryptophan, high lactose

Answer: a. In the artificial construct, the normal regulatory apparatus of the lac operon is completely absent. Instead, the lacZ and lacY genes are now associated with the regulatory proteins and sequences of the trp operon. Therefore, the level of ß-galactosidase activity will be entirely dependent on tryptophan levels and entirely independent of lactose levels. In the absence of tryptophan, the trp repressor protein will be unbound, and the anti-terminator loop will form in the 5′ UTR due to ribosome stalling. Therefore, expression will be high, and ß-galactosidase activity will be high. b. When tryptophan levels are high, tryptophan will bind to the trp repressor and act as a co-repressor. Therefore, expression will be low. In addition, high tryptophan levels lead to read-through in the 5′ UTR, formation of the attenuation loop, and termination of transcription. As a result, lacZ will not be transcribed, ß-galactosidase will not be generated, and activity levels will be low. Section: 16.2 Application

49. Although the genetic code is nearly universal, some variations do exist. In vertebrate mitochondria, UGA codes for Trp (instead of termination), AUA codes for Met (instead of Ile), and AGA and AGG are stop codons (instead of coding for Arg) Translate the following coding strand DNA sequences using both the standard code and the vertebrate mitochondrial code. a. 5' ATGGCCATAAGATGA 3' b. 5' ATGGGGGATCGCTAA 3' c. 5' ATGTGATGGCATCTTATAAATTGATAA 3'

Answer: a. Standard: Met-Ala-Ile-Arg; Mitochondrial: Met-Ala-Met b. Standard: Met-Gly-Asp-Arg; Mitochondrial: Met-Gly-Asp-Arg c. Standard: Met; Mitochondrial: Met-Trp-Trp-His-Leu-Met-Asn-Trp Section: 15.3 Application Question

49. Use the mRNA sequence shown below to answer the following questions. mRNA: 5′ ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG 3′ a. Underline the 5′ and 3′ splice sites, then write the sequence of the spliced mRNA in the space provided. b. Predict what would happen if the G in the 5′ splice site were mutated to a C. c. We learned in this chapter that the 5′ cap in an mRNA plays a role in translation initiation. What do you think is one plausible mechanism by which a 5′ cap can enhance initiation? How can you experimentally demonstrate that a 5′ cap is important for this process?

Answer: a. mRNA: 5′ ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG 3′ spliced mRNA: 5′ ACUGGACAG/UCGGCACCACG 3′ ( / indicates an exon-intron junction) b. The mRNA would not be spliced properly, if at all, because the U1 snRNA would not be able to bind complementarily to the 5′ splice site, due to the mutation. A different site with a close consensus sequence might be found, resulting in a completely different mRNA sequence. Ultimately, the protein encoded by this gene would be different from the normal protein and probably would not have the normal function. c. One possibility is that the 5′ cap could serve as a recruiting site for proteins to bind and help assemble the ribosomes to facilitate translation initiation. In an in vitro translation system, translation initiation efficiency of an mRNA with and without cap can be measured by monitoring the levels of proteins synthesized. If the hypothesis that 5′ cap enhances translation initiation efficiency is correct, then in this experiment more protein will be produced from the mRNA with a cap.

58. Describe the events in prokaryotic translation initiation.

Answer: (1) IF-3 separates ribosome subunits so that a small subunit can bind mRNA through base pairing of the 16S rRNA and the Shine-Delgarno sequence on the mRNA. (2) An initiator tRNAformylmet binds the initiation codon, with the help of IF-1 and IF-2 complexed with GTP. The tRNAformylmet is positioned in the P site. (3) IF-3 dissociates, allowing a large subunit to bind the 30S initiation complex. Section: 15.3 Comprehension Question

57. List three differences between prokaryotic and eukaryotic translation.

Answer: (1) Initiation in prokaryotes begins with a formyl-methionine. In eukaryotes, it begins with methionine. (2) Prokaryotic and eukaryotic ribosomes are different sizes. (3) Prokaryotic ribosomes are sensitive to antibiotics that do not affect eukaryotic ribosomes. (4) Prokaryotic large subunits contain two rRNAs while eukaryotic large subunits contain three. (5) Prokaryotic mRNAs are short-lived. Eukaryotic mRNAs vary in their half-life. (6) Recognition of the start codon is different. Prokaryotes have an rRNA that binds to the mRNA. Eukaryotic recognition of the start codon involves the 5 cap, the 3 poly(A) tail, and the Kozak sequence near the start codon. (7) Prokaryotic initiation, elongation, and termination factors are different. (8) Transcription and translation routinely occur simultaneously in prokaryotes, but are not routine in eukaryotes. Section: 15.3 Comprehension Question

48. What are the three different RNA polymerases found in all eukaryotes and what types of genes do they transcribe?

Answer: (1) RNA polymerase I transcribes large rRNA. (2) RNA polymerase II transcribes pre-mRNA,some snRNA, snoRNAs, some miRNAs (3) RNA polymerase III transcribes tRNA, small rRNA, and some snRNAs, some miRNA Section 13.2 Comprehension Question

44. Fill in the blanks in the "level of transcription" column of this table with: (+) for high levels of transcription, (+/-) for medium levels of transcription, and (-) for minimal levels of transcription of the lac operon. Consider regulation by both the lac repressor and CAP (catabolite activator protein). The strain is wild type, with no partial diploidy. The first line is filled in for reference. Medium conditions Level of transcription high glucose, no lactose - no glucose, high lactose high glucose, high lactose no glucose, no lactose

Answer: Medium conditions Level of transcription high glucose, no lactose - no glucose, high lactose + high glucose, high lactose +/- no glucose, no lactose - Section: 16.2 Application

53. The wild-type sequence of a protein is ... asp-ile-cys-trp.... The sequence of a mutant form of the protein is ... asp-ile-tyr-trp.... Write all possible DNA sequences for the wild-type and mutant alleles of the gene for this protein; underline the changed nucleotide in the mutant allele.

Answer: Wild-type 5 GAT/C ATC/A/T TGT/C TGG 3 3 CTA/G TAG/T/A ACA/G ACC 5 Mutant 5 GAT/C ATC/A/T TAT/C TGG 3 3 CTA/G TAG/T/A ATA/G ACC 5 Section: 15.3 Challenge Question

61. The antibiotic streptomycin binds to the 30S subunit in bacteria and interferes with translation. Streptomycin resistance (SmR ) can occur through mutation in the rpsL gene, which encodes the ribosomal protein S12. The antibiotic can no longer bind to the altered protein, so translation can occur without interference from the antibiotic. However, in the absence of streptomycin, many SmR strains of bacteria grow less well than the non-mutant streptomycin-sensitive (SmS) bacteria. a. Propose a model for why SmR bacteria grow better than SmS bacteria in the presence of streptomycin but not as well in the absence of streptomycin. b. During translation, an incorrect amino acid is ocassionally incorporated into the polypeptide chain—that is, the amino acid added does not correspond to the codon at that location in the mRNA. You develop an assay that can assess this rate of incorrect amino acid incorporation and thereby estimate the accuracy of translation in bacteria. You find that SmR bacteria are more accurate than SmS bacteria in this experiment. How might this observation affect your model above?

Answer: a. In the SmR bacteria, streptomycin cannot bind the ribosome because of the mutation. This observation suggests a change in the shape of the ribosome caused by the mutation. However, the change in the ribosome must not be so severe as to prevent the ribosome from functioning entirely, since the bacteria can still grow. Nevertheless, in the absence of streptomycin, the altered ribosome likely works less well than the normal ribosome, resulting in slower growth than the SmS strain of bacteria. b. The model above proposed that the mutant ribosome works less well, but the observation of greater accuracy by the mutant seems to contradict that model. However, another parameter in the function of a ribosome (aside from accuracy) would be speed. If the translation rate of the mutant ribosome was slower than the normal ribosome, that could account for the slower growth of the mutant. Apparently, then, the bacteria must have a tradeoff between accuracy and speed—the mutant ribosome is more accurate, but at the cost of slower speed. Section: 15.1 Application Question

54. You identify a novel deep-sea vent (DSV) bacteria and wish to explore its molecular biology. You find that you can introduce DNA sequences into the bacteria and it will transcribe and translate those sequences. However, you notice some curious differences between the expected proteins and the actual proteins produced. a. A 9-bp DNA sequence that in E. coli encodes a short peptide of Met-Trp-Phe instead produces a peptide of Met-Cys-Val-Trp-Gly-Val-Phe in the DSV bacteria. What is the sequence of the coding strand of the short DNA fragment? Propose an explanation for the longer peptide produced by the DSV bacteria. b. Curiously, a 9-bp sequence that in E. coli produces a peptide Met-Thr-Asn does not produce any peptide in the DSV bacteria. What is the sequence of the coding strand of this DNA fragment? Propose an explanation for the absence of any peptide.

Answer: a. The DNA sequence must be 5'-ATGTGGTTN-3' to encode Met-Trp-Phe. These amino acids are present in the longer DSV peptide, too. However, two intervening amino acids separate them. The simplest explanation is that, somehow, the DSV ribosome only shifts one base after reading a codon. Thus, the DSV ribosome translates the codons AUG-UGU-GUG-UGG-GGU-GUU-UUN to produce the longer peptide. b. The DNA sequence is 5'-ATGACNAAY-3' (where N is any base and Y is either C or T). Consistent with the model above, the second codon the DSV ribosome would encounter is UGA, which is a stop codon. Therefore, no peptide would form. Section: 15.2 Challenge Question

52. An RNA molecule has the following percentage of bases: A = 15%, U = 30%, C = 20%, and G = 35%. a. Is this RNA single stranded or double stranded? How can you tell based on the base composition? b. What would be the percentage of bases in the template strand of the DNA that encodes this RNA molecule?

Answer: a. This RNA molecule must be single-stranded, like most RNA molecules. If the molecule were double-stranded, it would consist of complementary base pairs so that A = U and C = G, which is not the case. b. The template strand of DNA would be complementary to this RNA molecule, so the composition would be T = 15%, A = 30%, G = 20%, and C = 35%. Section 13.2 Application Question

40. Write the codon, with correct polarity, of all mRNA codons that will bind to the tRNA anticodon 5 GCU 3, considering wobblebase pairing rules.

Answer: 5 AGU 3 5 AGC 3 Section: 15.3 Application Question

42. In your own words, list a comprehensive definition for "gene" at the molecular level.

Answer: A gene is a sequence of DNA, including all structural and regulatory sequences, that encodes information dictating synthesis of one or more discrete polypeptides or RNA molecules. Section: 14.2 Comprehension/Application Question

52. Critique the idea that the information needed to encode a particular polypeptide resides within a single gene.

Answer: A major exception to this idea is RNA editing. For those transcripts that are edited, some of the information needed to encode the polypeptide is carried in the corresponding structural gene, and some of the information is carried in the gene that produces the guide RNA molecule. In some mRNAs that have been studied, more than 60% of the sequence is determined by RNA editing. Another exception is alternative RNA processing. Although all of the information needed to encode the various polypeptide products of a particular gene might be contained within the gene (unless RNA editing is at work), which particular product is made within a particular cell is determined by the processing mechanisms, the components of which are encoded by a variety of other genes. Section: 14.2 Challenge Question

40. Aside from the case of lacOc, where the operator is unable to bind to the repressor protein, briefly describe another mutant that could have caused constitutive expression of lac operon.

Answer: A mutation (cis acting) in the promoter sequence of lacI could abolish lacI RNA and protein synthesis. In the absence of lacI protein, lac operon will be constitutively expressed. Section: 16.2 Application

49. In the experiments described in the text, Jacob and Monad deciphered that the lacI+ gene product can function in trans and regulate the lac operon either on the plasmid or on the chromosome. What genotype of the partial diploid bacterial strain was key to their experiments, leading them to such a conclusion? Explain why this strain helped them to reach that conclusion.

Answer: A partial diploid strain with the genotype lacI+ lacZ−/lacI− lacZ+ was central to their efforts. This strain functioned normally, synthesizing β-galactosidase only when lactose was present. In this strain, the functional lacZ+ gene was not physically linked to the functional lacI+ gene. Despite that, since a lacI+ gene could regulate a lacZ+ gene located on a different DNA molecule, they concluded that a functional lacZ+ can act in trans to regulate the lac operon that is present either in the same chromosome. Section: 16.2 Challenge

44. What is a pulse-chase experiment? Explain how this technique can be used to follow the products of a short-term event.

Answer: A pulse-chase experiment involves transiently introducing labeled precursors into an active biological system ("pulse"), then removing the labeled precursor and adding unlabeled (or "cold") precursors to "chase" the label so that it can be monitored. By monitoring the movement of the label in the system, questions regarding the kinetics, stability, and location of labeled cellular components can be answered. Section: 14.2 Application Question

56. What would be the consequences of a two nucleotide-deletion mutation in the middle of the first exon of a protein-coding gene?

Answer: A reading frameshift would occur. All codons (and thus amino acids) downstream of the codon in which the deletion occurred would be different, including the stop signal. Section: 15.2 Application Question

43. What is an snRNP and what role does it play in the cell?

Answer: A small nuclear ribonucleoprotein is a complex of small nuclear RNAs (snRNAs)—U1, U2, etc.—and proteins that catalyze the transesterification reactions and splicing of exons during eukaryotic mRNA processing. Section: 14.2 Comprehension/Application Question

51. Neurospora can be grown as haploids or diploids. Haploid mutants 1 and 2 are fused to make a diploid. If both 1 and 2 carry recessive mutations, on which compounds will the diploid be able to grow?

Answer: All compounds: D, B, A, C, and uracil Section: 15.1 Application Question

54. Diagram an active transcription unit as it would appear at high magnification. On the diagram indicate each of the following: (a) The DNA molecule (b) The RNA molecule that was initiated earliest (c) The 5′ end of an RNA molecule (d) The 5′ end of the template strand of DNA (e) The approximate location of the promoter (f) A molecule of RNA polymerase (g) An arrow indicating the direction of transcription

Answer: An acceptable answer would be: Section 13.2 Application Question

51. Transcription of a gene using what is normally the nontemplate strand can be accomplished using standard genetic engineering techniques to move the promoter. The resulting RNA is called "antisense RNA." Antisense RNA is often used to prevent production of a protein. Explain how antisense RNA would interact with RNA normally made by the gene, and how this would interfere with protein production.

Answer: Antisense RNA would bind to the normal RNA from a gene because it would be antiparallel and complementary. The double-stranded RNA would not be able to be translated, so the protein encoded by the gene would not be produced. Section 13.2 Application Question

48. You have isolated two mutations linked to the lac operon, which you designate Lac1- and Lac2- that cause constitutive expression of the operon. You construct strains carrying a lac operon with a mutant lacY gene on an F′. You test both ß-galactosidase activity and Lac permease activity in the strains you constructed using the artificial inducer IPTG. ß-galactosidase activity Lac permease activity - IPTG + IPTG - IPTG + IPTG Lac1- lacZ- lacY+/ F′ lacZ+ lacY - - + + + Lac2- lacZ- lacY+/ F′ lacZ+ lacY - + + + + Are Lac1- and Lac2- dominant or recessive? Do they act in cis or in trans? Indicate how you can determine both of these properties for each mutation. What type of lac mutations best fit the properties of Lac1- and of Lac2-?

Answer: Because only the bacterial chromosome has a functional lacY gene and only the F′ has a functional lacZ gene, we can think of permease activity (from lacY+) and ß-galactosidase activity (from lacZ+) as reporters for lac operon expression from the bacterial chromosome and the F′, respectively. That is, permease activity indicates whether the operon on the bacterial chromosome is regulated, constitutive, or uninducible; and ß-galactosidase activity indicates whether the operon on the F′ is regulated, constitutive, or uninducible. Lac1- is a dominant, cis-acting mutation. It is dominant because permease activity is still constitutive in the partial diploid with one wild-type allele (the F′ should contain a Lac1+ allele) and the chromosomal Lac1- allele. Lac1- is cis-acting because it can only confer constitutive expression on the reporter in cis (on the same piece of DNA, lacY+), while the reporter in trans (on a physically separate piece of DNA, lacZ+) is regulated appropriately. Lac2- is a dominant, trans-acting mutation. It is dominant because permease activity is still constitutive in the partial diploid with the wild-type allele on the F′ and the chromosomal Lac2- mutation. Lac2- is trans-acting because it can confer constitutive expression on both the reporter in cis (lacY+) and the reporter in trans (lacZ+). Lac1- best fits the properties of lacOC mutations (cis-acting dominant). A mutation in the operator sequence prevents binding of the LacI repressor, causing constitutive expression, but only for genes on the same chromosome. A trans-acting factor is usually a protein, whereas a cis-acting factor is usually a DNA sequence. Therefore, the Lac2- mutation appears to be a mutation affecting a protein. Besides the structural genes, the only protein component of the lac operon is the LacI repressor protein. However, most of the time, mutations in lacI are recessive, because a functional copy of lacI on an F′ can substitute for a lacI mutation in the bacterial chromosome. For this allele to be dominant, we have to propose that the mutant form of LacI protein generated by the Lac2- mutation has to actually interfere with the function of the normal LacI protein produced from the F′. A mutant allele that can interfere with the function of a normal allele is typically referred to as a dominant negative, and mutations of this kind that have been identified in the lac operon are known as lacId mutations. Section: 16.2 Challenge

70. Compare and contrast the operation of promoters and enhancers in affecting transcription of genes in eukaryotes. Include a discussion of how they operate and from what positions with respect to the transcription initiation site and the roles of each in bringing about and regulating transcription.

Answer: Both promoters and enhancers are DNA sequences that regulate transcription of adjacent DNA sequences. Both are needed for a high level of transcription of a gene. Both serve as binding sites for proteins that are involved in bringing about and regulating transcription. Promoters are found just upstream of the transcription start site and need to be in a specific orientation with respect to the transcription start site. Eukaryotic promoters typically include a core promoter located just upstream of the transcription start site, as well as a regulatory promoter located immediately upstream of the core promoter. The core promoter allows for binding of the basal transcription apparatus (including RNA polymerase). The core promoter by itself usually can only bring about a low level of transcription. The regulatory promoter allows for the binding of a variety of additional transcription factors that can interact with basal transcription factors at the core promoter to bring about regulated gene expression at a higher level. The activity of promoters is constrained in that they can only operate from a specific position and in a specific orientation with respect to the transcription start site. Enhancers are DNA sequences that can be found near or far from and upstream or downstream of the transcription initiation site. Regulatory proteins bind to enhancers and interact with the transcription apparatus at a promoter to bring about a higher level of transcription. Because of the ability of enhancers to act at various distances and positions with respect to the promoter, a large number of enhancers can be found on a single gene, allowing the gene to be responsive to a variety of different signals and conditions in the cell. Section 13.4 Application Question

56. Discuss the features of C. elegans that makes it an important model system for studies of animal genetics and development. Include in your answer some unique features of C. elegans compared to other model systems.

Answer: C. elegans is a small nematode worm that has served well as a model system for studying animal genetics and development. It has all of the common benefits of other model organisms, including small size (1 mm), short generation time (3 days), fecundity (each female lays 200-1000 eggs), and ease of lab culture. Like Mendel's peas and some other genetic model systems, C. elegans is capable of self-fertilization or crossing. The genome is relatively small and has been completely sequenced. Approximately 25% of the genes in C. elegans are also found in the human genome. C. elegans is especially useful in the study of development due to its transparent body, which makes it easier to observe all stages of development, and due to its regular pattern of cell divisions that produce the adult body. RNA interference allows for particular genes to be turned off in order to assess gene functions. Also, transgenic technology is available for allow for further assessment of gene functions. C. elegans has been especially useful in the study of development, apoptosis, genetic control of behavior, and aging. Section: 14. 5 Challenge Question

51. Devise a strategy to prove that splicing occurs in the nucleus.

Answer: Collect cells and fractionate them into nuclear and cytoplasmic fractions. Isolate the proteins from each of the fractions. Add a pre-mRNA (made in vitro) with introns to each of the protein fractions. Monitor the progressive decrease in the size of pre-mRNA in the nuclear fraction by separating the RNA products on a gel, which can resolve size differences between pre-mRNA (longer) and spliced mRNA (shorter). The cytoplasmic fraction will be incapable of splicing, and therefore the pre-mRNA remains intact over a period of time without undergoing splicing. The difference in the ability to splice suggests that the nuclear fraction contains the machinery essential for splicing. Section: 14.2 Challenge Question

43. Explain why glucose-dependent catabolite repression in E.coli is important and how it is possible to achieve this repression without influencing glucose metabolism.

Answer: First, energy conservation is the key aspect that drives catabolite repression. Given the availability of carbon sources, whichever carbon sources will result in the least amount of energy expended to obtain their stored energy as well as any carbon sources that will result in net positive energy will be used by the E. coli. Second, the genes that are essential to utilize glucose are not dependent on c-AMP-CAP complex to promote their transcription. This nondependency allows glucose to shut transcription of other operons that are required to utilize alternate carbon sources. Section: 16.2 Application

47. What are guide RNAs (gRNAs) and what do they do?

Answer: Guide RNAs play a crucial role in some RNA editing strategies by containing sequences that are partly complementary to segments of specific pre-edited mRNA molecules. After the gRNA associates with the corresponding mRNA, the mRNA is cleaved and specific nucleotides are added, deleted, or altered depending on the template provided by the gRNA. After the mRNA is modified, its ends are ligated back together, and the intact, now altered, mRNA can be translated. Section: 14.2 Application Question

55. E. coli only has a single tryptophan tRNA gene. Why, given this arrangement, might it be difficult to identify mutants that alter the anticodon of this tRNA gene?

Answer: In addition to there only being the single tryptophan tRNA gene, there is only a single tryptophan codon in the genetic code. Any changes to the tryptophan tRNA anticodon would make the bacteria unable to recognize the tryptophan codon in mRNA. Therefore, any proteins that ought to contain the tryptophan amino acid will be unable to be translated successfully, which would be lethal to the organism. Other amino acids have multiple codons that code for them, and, in many cases, multiple copies of the tRNA gene. In this event, a mutation in the anticodon of a single tRNA gene would not prevent the incorporation of these amino acids, and so mutants in these tRNA genes can be identified (see Chapter 18), unlike the case for the tryptophan tRNA gene. Section: 15.3 Challenge Question

71. Organisms belonging to the archaea group possess a TATAbinding protein (TBP), but they are structurally very similar to members of the eubacteria (i.e., contain no nucleus, are single celled). What does this conundrum mean regarding the evolutionary relationships among the archaea, eubacteria, and eukaryotes?

Answer: It suggests that the archaea and eukaryotes are more closely related to each other than the archaea are to the eubacteria. Section 13.5 Application Question

60. Prokaryotic ribosomes are inhibited by antibiotics like tetracyclines. As eukaryotes, humans are unaffected by these antibiotics. However, ribosomes within human mitochondria more closely resemble prokaryotic ribosomes than eukaryotic ribosomes. Why aren't tetracyclines more toxic to humans if they can, in theory, inhibit mitochondrial ribosomes?

Answer: Mitochondrial ribosomes are protected within several membranes. Antibiotics need to be transported across the plasma membrane, the mitochondrial outer membrane, and the mitochondrial inner membrane to reach them. Section: 15.3 Challenge Question

61. If you wanted to express a eukaryotic gene in E. coli, would the normal promoter be sufficient for transcription? Why or why not?

Answer: No. Prokaryotic and eukaryotic promoters have different sequences and different proteins that associate with them. Sections 13.3 & 13.4 Application Question

67. If you remove the TATA box and place it immediately upstream of a transcription start site of a eukaryotic gene, and subsequently transcription of the mRNA is assayed, will you still achieve transcription from the same start site?

Answer: No. The TATA box needs to be present a certain number of nucleotides upstream of the transcription start site to allow enough space for the assembly of TATA-binding protein and other transcription factors on the core promoter. The RNA polymerase can then be placed appropriately over the transcription start site. Section 13.4 Application Question

47. If you were asked to isolate total RNA from two unknown samples and then were required to identify if the RNA was from prokaryotes or eukaryotes, what aspects regarding the classes of RNA present would help you distinguish one from the other?

Answer: RNA from prokaryotes will contain mRNA, tRNA, and rRNA. In addition to these three types of RNA, eukaryotic samples will contain pre-mRNA, snRNA, snoRNA, scRNA, miRNA, and siRNA. If the eukaryotic sample happens to be from mammalian testes, it will also contain piRNA. Section 13.1 Application Question

53. Draw the tRNA cloverleaf secondary structure, labeling the various loops and stems. Indicate the functions of the acceptor stem and the anticodon. How are modified bases produced in tRNAs?

Answer: See Figure 14.20 for tRNA structure. • Acceptor stem • Acceptor arm • Anticodon arm • DHU arm • T C arm The 3′ end of all tRNAs (prokaryotic and eukaryotic) contains the trinucleotide sequence 5′-CCA-3′ (the acceptor stem). Amino acids are covalently attached to the 3′ adenines in the acceptor stems of their cognate tRNAs by highly specialized enzymes called aminoacyl tRNA synthetases. Anticodons are trinucleotide sequences located at the bottom of the tRNA cloverleaf structure that interact, through specific complementary base pairing, with given anticodons contained in the genetic code carried by mRNA. Codon-anticodon specificity alone dictates which amino acids are incorporated into a given polypeptide during ribosome-mediated translation. Transfer RNAs contain rare bases, in addition to the standard four bases found in RNA. The rare bases in tRNA are initially transcribed as standard bases, but they are post-transcriptionally modified through chemical changes mediated by specific tRNA-modifying enzymes. Section: 14.3 Application Question

44. A yeast strain was exposed to chemical mutagen. As expected, exposure to mutagen resulted in DNA sequence change in an essential gene you examined. Yet, this mutation did not result in any lethal phenotype. How could you explain this apparent discrepancy?

Answer: The DNA sequence change occurred in a synonymous nucleotide position of an amino acid and as a result the protein sequence remained unaltered. Since the protein sequence was not affected, no lethal phenotype manifested. Another possibility is that the DNA sequence change resulted in an amino acid change, but that amino acid change had no negative effect on protein function (i.e., was a neutral change). Section: 15.3 Application Question

50. The bacterium, Bacillus subtilis, can grow on minimal media with a variety of sugars as carbon source. One such sugar is mannose, metabolized by the products of the man operon. Expression of the operon is controlled by a regulatory protein encoded in a separate gene, manR. Depending on conditions the regulatory protein may bind at one of two sites in the operon, as follows: (i) When mannose is absent from the cell, the regulatory protein is in a conformation called R1. R1 can bind specifically at an operator site manO. Binding of R1 at manO reduces transcription of the operon four-fold from a basal level of 20 units. (ii) When mannose is present in the cell, it binds to the regulatory protein, causing it to undergo an allosteric transition from conformation R1 to a new conformation, called R2. R2 cannot bind at manO. However, R2 can bind specifically at a different site called the initiator, manI. Binding of R2 at manI increases transcription of the operon two-fold from the basal level. Mutations, m1-m3, which affect expression of this operon, were identified. Each mutation affects only a single component of the operon. Levels of operon activity were measured in haploids. They were also measured in partial diploids with an F′ carrying the wild-type alleles of all genes and regulatory elements described above. Operon activity Haploids Partial Diploids () mannose (+) mannose () mannose (+) mannose wild type 5 40 10 80 m1 20 20 10 80 m2 5 20 10 60 m3 20 40 25 80 For each mutation, describe which component is affected. In addition, explain the observed activity in the haploid and partial diploid in each case.

Answer: The basal level of activity is 20 units of activity, but in the absence of the mannose substrate, activity is reduced four-fold, to 5 units. In the presence of mannose, the basal level of activity is doubled to 40 units. In the partial diploid, the repressed level of expression is 10 units (5 units × 2) because there are two copies of the operon, and the induced level is 80 units (40 units × 2). m1: In the haploid, there is neither repression nor activation of the basal level of activity, as indicated by the 20 units of activity both in the presence and absence of substrate. This phenotype is corrected in trans by a trans-acting factor from the F′. The manR regulatory protein is the only trans-acting factor described in the system. Because the partial diploid is phenotypically wild type, the added manR protein must be sufficient for normal activity. Therefore, m1 is a mutant of the manR gene, such that the gene product can neither activate nor repress expression, and introduction of a wild-type version restores both activation and repression. m2: In the haploid, repression is normal in the absence of mannose, but there is no activation beyond the basal level with the addition of mannose. In the partial diploid, the defect is not corrected in trans, such that the addition of mannose only allows for a basal 20 units of activity from the chromosomal operon, and the full 40 units of activity from the F′ operon. Because the defect is cis-acting, affecting the ability to activate transcription, the mutation is likely in the manI initiator element such that the R2 form of the regulatory protein cannot bind and increase transcription. m3: In the haploid, there is no repression in the absence of mannose, but activation remains normal. In the partial diploid, there is no correction of the defect as levels of uninduced activity appear to be the sum of the regulated F′ (5 units) and the unregulated chromosomal copy (20 units). This mutation is thus in a cis-acting element affecting the ability to repress transcription. m3 is therefore a mutation in the manO operator site. Section: 16.2 Challenge

68. An in vitro system to detect transcription initiation functions if protein extracts from either eukaryotic or archaebacterial cells are added. Initiation does not occur if extracts from eubacterial cells are added. What does the protein extract contain? Explain the results using the different sources of protein extracts.

Answer: The extracts that work contain a TATA-binding protein (TBP). Eukaryotic and archaebacterial cells use a TBP (and its associated transcription factors, TFIIB); eubacterial cells do not. Section 13.4 Application Question

43. An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription. Explain why the mutant lysine tRNA allele from the previous question might affect overall cell growth.

Answer: The mutant tRNAlysine will occasionally cause translation to continue beyond the normal stop codon on mRNAs that normally use UAG as a stop codon. The presence of some longer versions of normal proteins may affect cell growth. Section: 15.3 Challenge Question

56. Two sequences of a prokaryotic promoter are shown below. Sequence 1 is the wild type sequence, and sequence 2 is a mutant sequence (mutation shown in bold) that results in increased transcription of the gene. Explain why sequence 2 shows this phenotype. Sequence 1 Sequence 2 -10 -10 ...CGAATAATGAA... ...CGTATAATGAA...

Answer: The mutation changes the -10 Pribnow box so that it exactly fits the consensus sequence for RNA polymerase binding. Section 13.3 Application Question

42. An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription. E. coli that have both the original ade1- mutant allele shown above AND a mutant allele for a gene that encodes a lysine tRNA are able to make some normal ade1 enzyme. The mutant lysine tRNA allele makes a tRNA that has one base that is different from the wild-type tRNA. Explain how the lysine tRNA mutation allows synthesis of the normal ade1 enzyme.

Answer: The mutation in the lysine tRNA gene changes the sequence of the anticodon. The anticodon now pairs with UAG, a stop codon. When the mRNA from the ade1- allele is being translated, if a mutant tRNAlysine enters the ribosome at the UAG, the lysine carried by the tRNA will be incorporated into the protein, bypassing the mutant UAG. Section: 15.3 Challenge Question

30. Which of the following nitrogenous bases is frequently modified enzymatically to become rare type of nitrogenous base upon transcription of tRNA gene? a. Adenine b. Uracil c. Thymine d. Cytosine e. Guanine

Answer: b Section: 14.3 Comprehension Question

63. A new mutation in Saccharomyces cerevisiae, a eukaryotic yeast, causes the cells to be unable to produce the amino acid histidine. Specifically, they cannot catalyze the first reaction in the histidine biosynthesis pathway. When examined closely, they are producing a completely wildtype enzyme for this reaction, but at greatly reduced levels. Explain the mutation.

Answer: The mutation is either in the regulatory promoter for the gene or in an enhancer, reducing transcription. Section 13.4 Application Question

53. The lysC gene in Bacillus subtilis encodes the first enzyme in the metabolic pathway that generates lysine from its precursor molecule. In the presence of high concentrations of lysine, lysC expression is reduced. Researchers have generated mutant bacteria that produce the LysC enzyme constitutively, even in the presence of lysine. A mutant identified in this manner had two nucleotide changes in the 5′ UTR of the lysC gene itself. These nucleotide changes are thought to affect the function of a riboswitch regulating transcription of lysC. Describe two possible ways in which the mutations might affect the function of this riboswitch.

Answer: The nucleotide changes could change two aspects of the riboswitch: 1) they could disrupt the secondary structure of the mRNA, thereby preventing proper formation of a terminator structure, and causing constitutive transcription regardless of whether lysine is present or not; or 2) they could prevent binding of lysine (or another small regulatory molecule) to the mRNA, again preventing terminator formation and allowing constitutive transcription. Section: 16.4 Challenge

55. What attributes make miRNA a potent agent for silencing gene expression?

Answer: The pre-miRNA is cleaved into one or more smaller RNA molecules with a hairpin. Dicer binds to this hairpin structure and removes the terminal loop. One of the miRNA strands is incorporated into the RISC; the other strand is released and degraded. The RISC attaches to a complementary sequence on the mRNA, usually in the 3′ untranslated region of the mRNA. The region of close complementarity, called the seed region, is quite short, usually only about seven nucleotides long. Because the seed sequence is so short, each miRNA may potentially pair with sequences on hundreds of different mRNAs. Thus, miRNAs potentially can silence a large number of genes simultaneously. Section: 14.5 Application Question

41. An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription. How does this mutation change the mRNA, and how does it affect the enzyme?

Answer: The third codon in the mRNA is changed from AAG (lysine) to UAG (stop). So, the protein won't be synthesized and translation will stop at the third codon. Section: 15.3 Application Question

48. For a time, a gene was commonly considered to be a sequence of DNA that encodes a polypeptide. Critique this outdated notion of what a gene is and propose a more comprehensive definition.

Answer: There are many problems with this simple concept of a gene. First, many genes encode functional RNA molecules rather than polypeptides. These functional RNAs (tRNA, rRNA, snRNA, etc.) produce RNA molecules that have some function other than encoding polypeptides. Second, many genes (perhaps most in at least some species) can encode more than one polypeptide through mechanisms such as alternative processing and RNA editing. Third, all genes contain 5′ and 3′ untranslated regions and many genes (especially in eukaryotes) contain introns that are transcribed but do not encode polypeptides. Finally, transcription of a DNA sequence is regulated by upstream and downstream sequences that are not transcribed. Since these sequences are important to the normal expression and function of the gene product, they probably should be considered to be part of the gene. A more comprehensive definition of a gene would include a transcribed region that produces a functional RNA or an mRNA that encodes one or more polypeptides, along with adjacent DNA sequences that are involved in regulating expression of the gene. This definition is not perfect given that a single regulatory element can regulate more than one transcription unit, making it part of more than one gene by this definition. Also, this definition does not account for the fact that some genetic information carried on a particular RNA molecule might not be provided by the transcription unit but rather by other genes that control alternative RNA processing and RNA editing. Section: 14.1 and 14.2 Challenge Question

62. What was Beadle and Tatum's definition of a gene? Give a modern definition and explain how it differs from that of Beadle and Tatum.

Answer: They defined it as one gene-one enzyme. A modern version may be one gene-one polypetide because we now know that enzymes can be made up of multiple polypeptide chains, each encoded by a gene. However, our modern definition of the gene must also take into account that many genes encode functional RNA products, like rRNA. Section: 15.1 Application Question

69. You analyze a newly cloned eukaryotic gene by comparing the DNA sequence of the gene to the sequence of the transcribed RNA and find that it does not contain recognizable promoter sequences upstream of its coding sequence. How can you explain this result, and how might you confirm your hypothesis?

Answer: This gene is most likely a polymerase-III-transcribed gene that contains an internal promoter. In order to confirm this hypothesis, look for known polymerase III promoter sequences within the coding sequence. Section 13.4 Application Question

63. What was the key technical deficiency that prevented Nirenberg and colleagues from deciphering the genetic code? What were the two important breakthroughs that still allowed them to break the genetic code?

Answer: To solve the genetic code all they had to do was to generate an mRNA of a particular sequence and then in vitro translate it to make a corresponding protein. Then, if they could have sequenced both the mRNA and protein, they would have been able to break the code. However, in the early 1960s, RNA and protein sequencing technology did not exist. First, Nirenberg and colleagues exploited an important aspect of an enzyme called polynucleotide phosphorylase, which is that this enzyme can make an RNA polymer without the need of a template (an essential requirement for RNA polymerase). Using this enzyme, Nirenberg and colleagues made a variety of synthetic RNAs to solve some of the codons. However, because of the limitation in making a variety of random polymers in this approach, it was still not possible to solve many other codons. They overcome this problem by using ribosome bound tRNAs. They found that a very short sequence of mRNA—even one consisting of a single codon—would bind to a ribosome. By synthesizing a variety of very short mRNAs, they then isolated the tRNAs that were bound to the mRNA and ribosomes and determined which amino acids were present on the bound tRNAs. Section: 15.1 Application Question

52. The 5 UTR of the trp operon RNA contains several UGG tryptophan codons. What would be the effect on transcription-attenuation gene regulation if the trp codons were converted to UGC cysteine codons?

Answer: Translation through the 5 UTR would no longer be dependent on the availability of the amino acid tryptophan. (However, it would now be dependent on levels of cysteine.) Regardless of levels of tryptophan, translation would continue through the 5 UTR, the attenuator hairpin would form, and the antiterminator loop would form. The entire trp operon would not be transcribed (unless levels of cysteine were low). Section: 16.3 Challenge

65. Your goal is to achieve a high level of transcription—a level similar to its in vivo levels—of a eukaryotic gene in an in vitro transcription system. You add the DNA template for this gene, which contains the following, to the in vitro transcription system: a TATA box sequence located -25 to -30 bp upstream of transcription start site, the sequence for the gene, and terminator sequences. How would you interpret the result if you do not get high levels of RNA transcription, and what corrective measures would you take to achieve your goal?

Answer: Two causes contributed to the low level of transcription and need to be corrected: (a) The template did not include regulatory promoter regions including enhancers. (b) The regulatory proteins that bind to the enhancers and activate high levels of transcription are absent in the in vitro system. Section 13.4 Application Question

45. Devise and describe a strategy to conduct a pulse-chase experiment to determine if newly transcribed eukaryotic RNA actually moves from the nucleus to the cytoplasm.

Answer: Uracil is present only in RNA (and not DNA). Therefore, add labeled (e.g., using radioactivity) uracil to the growth medium of an actively growing eukaryotic cell culture. The uracil will be taken into the cells and incorporated into newly synthesized RNA. After a short time, gently spin down the cells, wash away the old (labeled) growth medium, and transfer the cells to new media containing "cold" (unlabeled) uracil. At various time intervals, take a small sample of cells from the culture and fractionate (separate) the cells into nuclear and cytosolic components. Isolate RNA from each component and fractionate the RNA using gel electrophoresis. Monitor all soluble samples and gels for radioactivity. At first, radioactivity will be present only in the nuclear fraction, but over time you will measure increasing radioactivity in the cytosolic component, which suggests that RNA is moving from the nucleus to the cytoplasm. To verify this assumption and confirm that the radioactivity measured is not simply unincorporated labeled uracil diffusing out of the nucleus, perform autoradiography on the RNA gels. The presence of high-molecular-weight radioactivity will confirm incorporation into RNA, and comparing the time-course results will confirm RNA movement. Note that RNA movement can alternatively be monitored using fluorescent labels, coupled with sophisticated confocal microscope-imaging techniques, which optically section intact cells. Section: 14.2 Application Question

46. Describe the evidence that suggested that RNA was the carrier of genetic information from the nucleus to the ribosome.

Answer: Using radioactively labeled uracil, Gros and coworkers performed pulse-chase experiments demonstrating that labeled phage RNA, which was synthesized in the nucleus, later associated with the ribosomes and was distinct from them. This RNA component was called messenger RNA. Section: 14.2 Application Question

46. The dotted line in the following graph shows levels of glucose in a culture of wild-type E. coli grown in medium that initially contains both glucose and lactose. The solid line shows levels of transcription of the lac operon. Describe what is happening to the culture and the lac operon, referring to the lac repressor, allolactose, cAMP, and CAP (catabolite activator protein).

Answer: When glucose and lactose are present, E. coli prefer to use glucose. When lactose is high, allolactose binds the lac repressor, removing it from the operator and allowing transcription of the lac operon. However, when glucose is high, cAMP is low, so CAP does not activate transcription of the lac operon to its highest levels. As glucose is used up, shown by the dotted line going down, cAMP levels increase, cAMP binds CAP, and transcription of the lac operon is activated to its highest levels. Section: 16.2 Application

Short-Answer Questions 38. The genetic code uses three bases to encode one amino acid. Why can't the code use only two bases to encode each amino acid?

Answer: With four bases in RNA, there are only 16 combinations of two bases. There are 20 different amino acids, so this would not be enough to encode all amino acids. Section: 15.2 Application Question

3. Over time, DNA replaced RNA as the primary carrier of genetic information, and the chemical stability of DNA is believed to be the key reason for this. Which attribute of DNA is the reason behind its chemical stability? a. DNA lacks a free hydroxyl group on the 2′ carbon atom of its sugar. b. Unlike RNA, DNA is usually double stranded. c. DNA does not usually form hairpin loops. d. One of the two pyrimidines found in DNA does not involve uracil. e. DNA contains thymines, which make it more chemically stable.

Answer: a Section 13.1 Comprehension Question

4. This molecule is synthesized using nucleotides containing the bases adenine, guanine, cytosine, and uracil. a. RNA only b. DNA only c. Both RNA and DNA d. Neither RNA nor DNA

Answer: a Section 13.1 Comprehension Question

27. If the sequence of an RNA molecule is 5'GGCAUCGACG3', what is the sequence of the nontemplate strand of DNA? a. 5'GGCATCGAC3' b. 3'GGCATCGACG5' c. 5'CCGTAGCTGC3' d. 3'CCGTAGCTGC5' e. 3'CGTCGATGCC5'

Answer: a Section 13.2 Application Question

10. Where are promoters usually located? a. Upstream of the start site b. Downstream of the start site c. Near nucleotide +25 d. Near the hairpin loop e. Downstream of the terminator

Answer: a Section 13.2 Comprehension Question

13. In transcription, to which end of the elongating strand are nucleotides always added? a. 3′ b. 5′ c. 3′ in prokaryotes and 5′ in eukaryotes d. It depends on which RNA polymerase is being used. e. It depends on which DNA strand is being used as the template.

Answer: a Section 13.2 Comprehension Question

22. What is the function of eukaryotic RNA polymerase I? a. Transcription of rRNA genes b. Transcription of mRNA genes c. Transcription of tRNA genes d. Transcription of snRNAs e. Initiation of transcription (but not elongation)

Answer: a Section 13.2 Comprehension Question

24. Which process is illustrated in the diagram below? a. Transcription b. Translation c. RNA processing d. Replication e. Nucleosome assembly

Answer: a Section 13.2 Comprehension Question

37. Which of the following statements is TRUE regarding the termination of transcription? a. In some organisms, transcription terminates thousands of nucleotides past the coding sequence. b. Transcription typically terminates precisely at the hairpin loop terminator sequence. c. In prokaryotes, transcription terminates as soon as rho has bound to the RNA. d. In yeast, transcription terminates as soon as Rat1 has bound to the RNA. e. Both C and D are correct answers.

Answer: a Section 13.4 Comprehension Question

40. In eukaryotic cells, where does the basal transcription apparatus bind? a. Core promoter b. Regulatory promoter c. Terminator d. Enhancer e. Ribozyme

Answer: a Section 13.4 Comprehension Question

1. Which of the following statements correctly describes the facts about introns and exons? a. The number of introns is always less than the number of exons in a gene. b. Introns are degraded in the cytoplasm. c. All eukaryotic genes contain an intron. d. Mitochondrial and chloroplast genes do not contain introns. e. Introns do not contain sequence-specific information.

Answer: a Section: 14.1 Comprehension Question

22. The list of events below describes eukaryotic pre-mRNA processing. Please select the choice that lists the events in correct sequential order. 1. Recognition and binding the 3′ AAUAAA sequence by specific protein factors 2. Cleavage at the poly(A) site 3. Addition of the 5′ cap 4. Export to the cytoplasm 5. Addition of the poly(A) tail a. 3, 1, 2, 5, 4 b. 2, 3, 4, 5, 1 c. 4, 2, 3, 1, 5 d. 1, 3, 5, 4, 1 e. 5, 4, 1, 3, 2

Answer: a Section: 14.2 Application Question

23. The 5′ cap on an mRNA is important for all the processes listed below except for the _________ of an mRNA molecule. a. transcription b. intron removal c. stability d. initiation of translation e. ribosomal interaction

Answer: a Section: 14.2 Application Question

26. During the post transcriptional processing of pre-mRNA, a 5′ cap is added to an mRNA in step by step manner. Which of the following reason prevents the possibility of 5′ capping process involving methylation occurring on a DNA strand? a. Lack of OH group on 2′ carbon of the deoxyribose b. Lack of OH group on 3′ carbon of the deoxyribose c. Lack of uracil nitrogenous base on the DNA strand d. Lack of GTP hydrolysis associated with DNA transcription e. Lack of H on 4′ carbon of the deoxyribose

Answer: a Section: 14.2 Application Question

10. Which of the following best explains why only pre-mRNA is recognized and receives a 5′ cap? a. The enzyme that initiates the capping step is known to associate with RNA polymerase II, which generate mRNAs. b. Only pre-mRNAs contain proper sequence for the cap to be added on. c. The tail of the pre-mRNA can recruit the right combination of enzymes for capping. d. Nuclear pore complex only recognize pre-mRNA to be allowed out to the cytoplasm for capping process to begin. e. rRNA and tRNAs do not exit the nucleus to receive the cap via enzyme sin the cytoplasm.

Answer: a Section: 14.2 Comprehension Question

8. The 5′ and 3′ untranslated regions (UTRs) of processed mRNA molecules are derived from a. exons. b. introns. c. promoter. d. terminator. e. protein-coding region.

Answer: a Section: 14.2 Comprehension Question

33. To which part on tRNAs would an amino acid attach to during tRNA charging? a. 3′ acceptor arm b. Anticodon arm c. TψC arm d. DHU arm e. Extra arm

Answer: a Section: 14.3 Application Question

1. Which of the following statements about proteins is incorrect? a. All proteins are made up of some combination of 20 essential amino acids. b. Like nucleic acids, polypeptides have polarity. c. A single polypeptide has primary, secondary and tertiary structures. d. α-helix and β-pleated sheets do not require specific sequence of amino acids to form. e. Some proteins contain more than one polypeptide chain.

Answer: a Section: 15.1 Comprehension Question

27. The genetic code is said to be "degenerate" because: a. there are more codons than amino acids. b. there are more amino acids than codons. c. different organisms use different codons to encode the same amino acid. d. some codons specify more than one amino acid. e. there are more tRNAs than amino acids.

Answer: a Section: 15.2 Application Question

18. The amino acid sequence of a polypeptide is referred to as the ______ sequence of the polypeptide. a. primary b. secondary c. tertiary d. quarternary

Answer: a Section: 15.2 Comprehension Question

12. If the bottom strand of the DNA from the diagram above serves as the template strand, the RNA sequence, left to right 5 to 3, is a. AUAGGCAGU b. UCCCAGGUG c. CACCUGGGA d. AGGGUCCAC e. GACAUUAGA

Answer: a Section: 15.3 Application Question

21. The next step in the translation of this mRNA will be the formation of a peptide bond between which two amino acids? a. amino acid 2 and amino acid 3 b. amino acid 2 and amino acid 4 c. amino acid 1 and amino acid 3 d. amino acid 1 and amino acid 2 e. amino acid 3 and amino acid 4

Answer: a Section: 15.3 Application Question

15. During initiation, the _____subunit is the first part of the ribosome to associate with the mRNA. a. small b. large c. intermediary d. secondary e. tertiary

Answer: a Section: 15.3 Comprehension Question

34. Mechanisms that exist to detect and deal with errors in mRNA in order to ensure the accurate transfer of genetic information are collectively referred to as a. mRNA surveillance b. Proofreading function c. RNA interference d. Alternative processing e. RNA transition

Answer: a Section: 15.4 Comprehension Question

3. Which of the following types of eukaryotic gene regulation is at the level of DNA? a. Alternation of chromatin structure b. mRNA processing c. RNA interference d. mRNA stability e. Post-translational modification

Answer: a Section: 16.1 Comprehension

5. Proteins with DNA binding motifs predominantly bind to the ____________ of DNA. a. major grooves b. minor grooves c. paired nitrogenous bases d. phosphate groups e. deoxyribose sugar

Answer: a Section: 16.1 Comprehension

31. Anticodons are found on _______ molecules. a. mRNA b. tRNA c. rRNA d. snRNA e. miRNA

Answer: b Section: 14.3 Comprehension Question

40. Which of the following small RNA types is unique to prokaryotes? a. siRNA b. crRNA c. miRNA d. piRNA e. lncRNA

Answer: b Section: 14.5 and 14.6 Application Question

18. When a structural gene is under negative inducible control, what would be the result of a mutation that eliminates the repressor protein? a. The structural gene will be constitutively expressed due to the lack of negative inducible control. b. The transcription of the structural gene will not be affected, as a repressor is not required. c. The mutation will lead to activation of an activator upon the lack of a repressor protein, which will allow the transcription to continue. d. As the transcription will require a repressor protein, the transcription will be turned off. e. More cAMP will be produced in a cell to compensate for the lack of a repressor protein.

Answer: a Section: 16.2 Application

20. A promoter that affects only genes that are on the same piece of DNA is referred to as a ____-acting promoter. a. cis b. trans c. enhancer d. positive e. negative

Answer: a Section: 16.2 Application

22. E. coli lac operon control by lacI is a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. attenuation.

Answer: a Section: 16.2 Application

25. If there are mutations that inactivate lacP and lacI, which of the following is TRUE? (I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator) a. These are mutations that are, respectively, cis- and trans-acting on lac operon expression. b. These are mutations that are, respectively, trans- and cis-acting on lac operon expression. c. These will affect the expression of I only. d. These will affect the expression of only Z, Y, and A. e. These mutations will have no effect.

Answer: a Section: 16.2 Application

13. An operon is controlled by a repressor. When the repressor binds to a small molecule, it is released from binding to DNA near the operon. The operon is never expressed if a mutation prevents the repressor from binding to the small molecule. The type of control illustrated is a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. attenuation.

Answer: a Section: 16.2 Comprehension

14. What is the function of allolactose in regulation of the lac operon? a. Inducer b. Repressor c. Activator d. Promoter e. Regulatory protein

Answer: a Section: 16.2 Comprehension

43. In which of the following organisms would transcription be the LEAST similar to archaea? a. E. coli b. Yeast c. Plants d. Mice e. Humans

Answer: a Section 13.5 Application Question

1. Which of the following is a type of RNA that gets translated? a. rRNA b. mRNA c. tRNA d. miRNA e. A, B, and C all get translated.

Answer: b Section 13.1 Comprehension Question

2. Which of the following RNA molecules are required for the process of translation? a. crRNA b. tRNA c. snRNA d. snoRNA e. siRNA

Answer: b Section 13.1 Comprehension Question

5. In eukaryotes, tRNAs are a. transcribed in the nucleus and function in the nucleus. b. transcribed in the nucleus but function in the cytoplasm. c. transcribed in the cytoplasm and function in the cytoplasm. d. transcribed in both the nucleus and the cytoplasm and function in the cytoplasm. e. transcribed in the cytoplasm and function in the nucleus.

Answer: b Section 13.1 Comprehension Question

16. In eukaryotes, which RNA polymerase transcribes the genes that encode proteins? a. RNA polymerase I b. RNA polymerase II c. RNA polymerase III d. RNA polymerase IV e. RNA polymerase V

Answer: b Section 13.2 Comprehension Question

18. When this molecule is synthesized, both strands of a DNA molecule are used as a template. a. RNA only b. DNA only c. Both RNA and DNA d. Neither RNA nor DNA

Answer: b Section 13.2 Comprehension Question

23. Which of the following is NOT required for transcription? a. Ribonucleotides b. RNA primers c. DNA template d. RNA polymerase e. Promoter

Answer: b Section 13.2 Comprehension Question

36. An in vitro transcription system transcribes a bacterial gene but terminates inefficiently. What is one possible problem? a. There is a mutation in the -10 consensus sequence, which is required for efficient termination. b. Rho factor has not been added. c. Sigma factor has not been added. d. A hairpin secondary structure has formed at the 3′ end of the mRNA, interfering with termination. e. Histones were added prematurely and interfered with termination.

Answer: b Section 13.3 Application Question

28. Which of the following is NOT necessary for RNA polymerase to recognize the promoter of a bacterial gene? a. Sigma factor b. Origin of replication c. -10 consensus sequence d. -35 consensus sequence

Answer: b Section 13.3 Comprehension Question

29. Which of the following features is primarily responsible for rho protein to cause termination of transcription? a. Recognizing unstructured RNA b. Helicase activity c. Migrating behind RNA polymerase d. RNAbinding activity e. Polymerase activity

Answer: b Section 13.3 Comprehension Question

32. When does sigma factor normally dissociate from RNA polymerase? a. After transcription has terminated b. After the process of initiation c. After the addition of nucleosomes d. After the binding of rho e. Following the addition of nucleosomes

Answer: b Section 13.3 Comprehension Question

3. How many introns are present on a gene that consists of 4 exons? a. 2 b. 3 c. 4 d. 5 e. The number cannot be determined from the information provided.

Answer: b Section: 14.1 Comprehension Question

12. Which of the following spliceosomal components specifically recognizes and binds to the branch point of the intron during pre-mRNA splicing? a. U1 b. U2 c. U5 d. U6 e. Spliceosomal proteins

Answer: b Section: 14.2 Comprehension Question

5. Which of the following statements about bacterial mRNA transcript is TRUE? a. Unlike eukaryotes, bacterial mRNA transcripts do not typically contain untranslated regions. b. The Shine−Dalgarno box associates with an RNA component in the small subunit of ribosomes. c. Transcription and translation take place sequentially in bacterial cells. d. Most of bacterial genes contain a large number of introns and small number of exons. e. The 5′ end and 3′ end of mRNA transcripts are modified in bacteria.

Answer: b Section: 14.2 Comprehension Question

26. To translate an mRNA requires two other types of RNA. These are: a. tRNA and mRNA. b. tRNA and miRNA. c. tRNA and rRNA. d. rRNA and siRNA. e. snRNA and snoRNA

Answer: c Section: 15.3 Comprehension Question

28. The genetic code is universal except for: a. prokaryotes, which use a different genetic code than eukaryotes. b. a few mitochondrial genes, which substitute one sense codon for another. c. viruses, which use an entirely different genetic code. d. archaebacteria, which have their own genetic code. e. animal species whose cells are more advanced and complex.

Answer: b Section: 15.2 Application Question

20. Which of the following statement about the formation of the peptide bond between amino acids is incorrect? a. The formation of peptide bond results in formation of a water molecule. b. The amino group of the first amino acid and the carboxyl group of the second amino group are involved in forming a peptide bond. c. Carboxyl group of the first amino acid reacts with the amino group of the second amino group to form a peptide bond. d. A polypeptide formed by a series of peptide bonds will result in two distinct free ends, one with a free amino group and the other with a free carboxyl group. e. The number of peptide bond formed in a polypeptide varies from protein to protein.

Answer: b Section: 15.2 Comprehension Question

13. If the bottom strand of the DNA serves as the template, the amino acid sequence of the protein produced from the RNA would be a. met-leu-ser. b. arg-val-his. c. thr-ile-phe d. pro-gly-trp e. lys-val-his

Answer: b Section: 15.2 and 15.3 Application Question

24. A tRNA anticodon is 5GCU3. What amino acid does it carry? a. ala b. arg c. ser d. pro e. thr

Answer: b Section: 15.3 Application Question

32. An mRNA has the codon 5' UAC 3'. What tRNA anticodon will bind to it? a. 5 AUG 3 b. 5 GUA 3 c. 5 ATC 3 d. 5 CTA 3 e. 5' CAU 3'

Answer: b Section: 15.3 Application Question

2. During initiation of translation, ________. a. the initiator tRNAmet binds to the A site of a ribosome. b. specific rRNA base pairs with a sequence in mRNA to position a ribosome at the start codon. c. IF-3 must be recruited to the 30S ribosome in order for the 70S initiation complex to assemble. d. there is no energy expenditure as the tRNA binding to mRNA is via complementary base pairing. e. Both 70S and 30S ribosome subunits must simultaneously recognize an mRNA to bind.

Answer: b Section: 15.3 Comprehension Question

16. During elongation, an incoming charged tRNA enters at the __ site of the ribosome. a. peptidyl (P) b. aminoacyl (A) c. Exit (E) d. Shine-Delgarno e. Kozak

Answer: b Section: 15.4 Comprehension Question

2. Which of the following generally get transcribed constitutively? a. Regulatory gene b. Structural gene c. Operator element d. Promoter element e. Operon

Answer: b Section: 16.1 Comprehension

16. In the absence of tryptophan, what happens to the genes within the trp operon? a. The regulator without tryptophan-binding prevents the genes from being transcribed. b. The regulator falls off the operator and structural genes get transcribed. c. Lack of tryptophan increases the level of cAMP high, which leads to activation of CAP protein and gene expression. d. The active repressor binds to the operator and genes do not get transcribed. e. The active activator binds to the operator and transcription of structural gene takes place.

Answer: b Section: 16.2 Application

26. A mutant E. coli strain, grown under conditions that normally induce the lac operon, produces high amounts of ß-galactosidase. What is a possible genotype of the cells? a. lacI+ lacP+ lacO+ lacZ- lacY+ lacA+ b. lacI+ lacP+ lacOc lacZ+ lacY+ lacA+ c. lacI- lacP+ lacO+ lacZ- lacY+ lacA+ d. lacI+ lacP- lacO+ lacZ+ lacY+ lacA+ e. lacI- lacP- lacO+ lacZ- lacY+ lacA-

Answer: b Section: 16.2 Application

12. An operon is controlled by a repressor. When the repressor binds to a small molecule, it binds to DNA near the operon. The operon is constitutively expressed if a mutation prevents the repressor from binding to the small molecule. The type of control illustrated is a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. catabolite repression.

Answer: b Section: 16.2 Comprehension

15. What is the function of cAMP in regulation of the lac operon? a. It activates a repressor protein. b. It activates an activator protein. c. It inactivates a repressor protein. d. It inactivates an activator protein. e. It causes attenuation.

Answer: b Section: 16.2 Comprehension

7. Which of the following statements about gene regulation concerning operon is INCORRECT? a. A repressible gene is controlled by a regulatory protein that inhibits transcription. b. For a gene under negative repressible control, a small molecule is required to prevent the gene's repressor from binding to DNA. c. For a gene under positive repressible control, the normal state is transcription of a gene, stimulated by a transcriptional activator. d. A regulator gene has its own promoter and is transcribed into an independent mRNA. e. Presence of operon where genes of related functions are clustered is common in bacteria, but not in eukaryotes.

Answer: b Section: 16.2 Comprehension

8. When binding of the inducer to the repressor causes a conformational change, which then prevents the repressor from binding to DNA, the repressor is called a(n) __________ protein. a. coactivator b. allosteric c. structural d. operating e. responsive

Answer: b Section: 16.2 Comprehension

9. The________ is a type of regulator protein that binds to a region of DNA in the promoter of a gene called the operator and prevents transcription from taking place. a. inducer b. repressor c. activator d. inactivator e. terminator

Answer: b Section: 16.2 Comprehension

32. The formation of 1+2 and 3+4 secondary structures of 5′ UTR region mRNA from the trp operon is triggered when a. the tryptophan level inside the bacterial cell is extremely low. b. the tryptophan level inside the bacterial cell is high. c. the repressor protein fails to bind to the operator. d. there is a spontaneous mutation introduced into the 5′ UTR. e. the structural gene transcription within the trp operon gets initiated.

Answer: b Section: 16.3 Application

15. What types of bonds are created between nucleotides during the process of transcription? a. Ionic b. Oxygen c. Phosphodiester d. Hydrogen e. Both C and D

Answer: c Section 13.2 Comprehension Question

17. The DNA replication enzyme that most closely resembles RNA polymerase is a. DNA polymerase I. b. DNA polymerase III. c. primase. d. telomerase. e. helicase.

Answer: c Section 13.2 Comprehension Question

19. This molecule is synthesized using triphosphate nucleotides as a substrate for a polymerase enzyme that forms phosphodiester bonds. a. RNA only b. DNA only c. Both RNA and DNA d. Neither RNA nor DNA

Answer: c Section 13.2 Comprehension Question

20. The polymerase that synthesizes this molecule uses DNA as a template and synthesizes new strands from 5′ to 3′. a. RNA only b. DNA only c. Both RNA and DNA d. Neither RNA nor DNA

Answer: c Section 13.2 Comprehension Question

8. Which of the following statements are TRUE regarding transcription in most organisms? a. All genes are transcribed from the same strand of DNA. b. Both DNA strands are used to transcribe a single gene. c. Different genes may be transcribed from different strands of DNA. d. The DNA template strand is used to encode double stranded RNA. e. The DNA nontemplate strand is used to encode single stranded RNA.

Answer: c Section 13.2 Comprehension Question

35. An in vitro transcription system that contains a bacterial gene does not initiate transcription. What is one possible problem? a. Histones that were on the DNA when it was isolated from E. coli are blocking access to the template. b. There is a mutation in the inverted repeat sequence that prevents a hairpin secondary structure from forming. c. There is a mutation at -10, where a promoter consensus sequence is located. d. Rho factor has not been added. e. TATA binding protein (TBP) has not been added.

Answer: c Section 13.3 Application Question

30. Prokaryotic promoters contain the sequence TATAAT at the position -10 from the transcription start. a. +1 b. -1 c. -10 d. -25 e. -35

Answer: c Section 13.3 Comprehension Question

33. In prokaryotes, rho independent transcription termination depends on a secondary structure formed in a. the RNA polymerase that is transcribing the gene. b. the DNA template. c. the RNA that is being transcribed. d. a protein factor that binds to RNA polymerase. e. a protein factor that binds to the RNA that is being transcribed.

Answer: c Section 13.3 Comprehension Question

38. Which one of the following statements regarding eukaryotic transcription is NOT true? a. Eukaryotic transcription involves a core promoter and a regulatory promoter. b. There is no one generic promoter. c. A group of genes is transcribed into a polycistronic RNA. d. Chromatin remodeling is necessary before certain genes are transcribed. e. There are several different types of RNA polymerase.

Answer: c Section 13.4 Comprehension Question

4. Which of the following statements regarding gene structure is NOT false? a. The amino acid sequence of a polypeptide can be precisely predicted by the nucleotide sequence of the gene that encodes it. b. The number of introns found in organisms is species specific. c. The number of exons and introns generally correlate to the complexity of the organisms. d. Intron cleavage and exon splicing are both mediated by protein enzymes. e. The number of exons is always less than the number of introns in a gene.

Answer: c Section: 14.1 Application Question

21. The list of events below describes intron removal and splicing during pre-mRNA processing. Please select the choice that lists the events in correct sequential order. 1. Attachment of snRNP U1 to the 5′ splice site 2. Transcription of the DNA template into the pre-mRNA molecule 3. Release of lariat structure 4. Splicing together of exons 5. Transesterification reaction at the branch point adenine a. 1, 2, 3, 4, 5 b. 4, 1, 3, 5, 2 c. 2, 1, 5, 3, 4 d. 3, 5, 1, 2, 4 e. 5, 3, 4, 1, 2

Answer: c Section: 14.2 Application Question

27. Which of the following consensus sequences are NOT found in nuclear introns? a. GU at the 5′ splice site at the beginning of the intron b. AG at the 3′ splice site at the end of the intron c. CCA at the 3′ site downstream of the branch point d. A at the lariat branch point site e. 3′ CAGG consensus sequence at the 3′ splice junction

Answer: c Section: 14.2 Application Question

28. What do group I and group II introns have in common? a. Both are found in mitochondrial genes. b. Both are found in bacteriophages. c. Both are known to be self-splicing introns. d. Both are found in protein-coding genes of chloroplasts. e. Both are frequently found in eukaryotic genes.

Answer: c Section: 14.2 Application Question

15. Which mechanism allows for more than one polypeptide to be encoded by a single gene? a. Regulated transcription b. RNA interference c. Alternative RNA processing d. Self-splicing of introns e. RNA methylation

Answer: c Section: 14.2 Comprehension Question

36. Which class of RNA is most abundant in cells? a. mRNA b. tRNA c. rRNA d. snRNA e. miRNA

Answer: c Section: 14.2, 14.3, 14.4, and 14.5 Comprehension Question

38. siRNAs and miRNAs function in which of the following processes? a. Transcription b. Translation c. RNA interference d. RNA editing e. RNA splicing

Answer: c Section: 14.5 Comprehension Question

31. When codons that code for the same amino acid differ in their ________, a single tRNA might bind both of them through wobble base pairing. a. 5 base b. middle base c. 3 base

Answer: c Section: 15.2 Comprehension Question

4. Which of the following is observed in prokaryotes but not in eukaryotes? a. UGG is an example of a stop codon only found in prokaryotes. b. An mRNA can be translated by only one ribosome at a time in prokaryotes. c. The 5' end of a prokaryotic mRNA can be translated while the 3 end is still being transcribed. d. Translation does not require any protein factors in prokaryotes. e. In prokaryotes, ribosomes move along an mRNA in the 3' to 5' direction.

Answer: c Section: 15.2 Comprehension Question

7. For a sequence of nucleotide, how many reading frame is possible? a. One b. Two c. Three d. Five e. Ten

Answer: c Section: 15.2 Comprehension Question

11. Process 2 represents ___________. a. replication b. transcription c. translation d. RNA processing e. RNA interference

Answer: c Section: 15.3 Application Question

22. After the peptide bond forms, what will happen? a. tRNA A will be carrying the polypeptide and it will shift to the P site. b. tRNA A will be carrying the polypeptide and it will shift to the A site. c. tRNA B will be carrying the polypeptide and it will shift to the P site. d. tRNA B will be carrying the polypeptide and it will shift to the A site. e. Ribosome disassembles to release the tRNAs and allow new tRNA to enter.

Answer: c Section: 15.3 Application Question

29. The function of aminoacyl-tRNA synthetases is to: a. transcribe tRNA genes. b. match tRNA anticodons and mRNA codons at the ribosome. c. attach appropriate amino acids to corresponding tRNAs. d. form the peptide bond between amino acids at the ribosome. e. synthesize and transport amino acids to the ribosomes.

Answer: c Section: 15.3 Comprehension Question

35. Which of the following mechanisms specifically allows detection and rapid degradation of mRNA containing premature termination codon? a. RNA interference b. no-go decay c. nonsense-mediated mRNA decay d. transfer-messenger RNA mediated ribosomal removal e. nonstop mRNA decay

Answer: c Section: 15.4 Comprehension Question

4. Which of the following DNA binding motifs are composed of three alpha helices? a. Zinc-finger b. Leucine-zipper c. Homeodomain d. Helix-turn-helix e. Helix-loop-helix

Answer: c Section: 16.1 Comprehension

11. E. coli lac operon control by CAP is a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. regulated by riboswitches.

Answer: c Section: 16.2 Application

19. What would happen to the lac operon in the absence of allolactose? a. The structural genes within the lac operon the will be constitutively transcribed. b. The activator protein will be bound to the operator, which will turn on the structural gene behind it. c. The repressor regulator protein binds to the operator and prevents the transcription of the structural gene. d. The catabolite activator protein becomes inactivated and no transcription occurs. e. The cAMP level rises in the absence of allolactose, which in turn inactivates the transcription.

Answer: c Section: 16.2 Application

21. It is possible for a repressor to negatively regulate the expression of an operon because a. the repressor induces the expression of the inducer by binding to the promoter that comes before the inducer gene. b. one of the structural genes expressed in the operon negatively regulates the repressor. c. the repressor-binding site overlaps the promoter site of the operon, allowing it to physically block the binding of RNA polymerase. d. the repressor-binding site on the DNA overlaps with the translation start site, hence preventing the transcription. e. the repressor physically blocks where the activator should be binding on the operator region.

Answer: c Section: 16.2 Application

10. An example of a gene product encoded by a regulatory gene is a. beta-galactosidase enzyme. b. allolactose. c. repressor protein. d. an operator. e. a terminator.

Answer: c Section: 16.2 Comprehension

30. If a mutation prevents the formation of the antiterminator 2+3 loop in the trp operon, what would be the effect? a. Transcription only when tryptophan is absent b. Transcription only when tryptophan is present c. Constitutive attenuation of transcription d. Constitutive transcription e. No effect, as 2+3 loop has no function

Answer: c Section: 16.3 Application

33. RNA molecules that are complementary to particular sequence on mRNA are called a. complementary RNA. b. sense RNA. c. antisense RNA. d. riboswitches. e. ribozymes.

Answer: c Section: 16.4 Comprehension

9. Which of the following is not required during the process of tRNA charging? a. amino acid b. tRNA c. GTP d. ATP e. aminoacyl-tRNA synthetase

Answer: c Section: 15.3 Comprehension Question

26. If the following DNA strand were used as a template, what would the sequence of an RNA be? 5′ GTACCGTC 3′ a. 5′ GUACCGUC 3′ b. 5′ GACGGTAC 3′ c. 5′ CAUGGCAG 3′ d. 5′ GACGGUAC 3′ e. 5′ GUCGGUAC 3′

Answer: d Section 13.2 Application Question

12. Whereas the nucleotide strand used for transcription is termed the _______, the nontranscribed strand is called the _________. a. promoter; terminator b. terminator; promoter c. transcription apparatus; TATA box d. template strand; nontemplate strand e. nontemplate strand; template strand

Answer: d Section 13.2 Comprehension Question

14. In a transcription reaction, two phosphate groups are cleaved from the incoming a. deoxyribonucleoside diphosphate. b. deoxyribonucleoside triphosphate. c. ribonucleoside diphosphate. d. ribonucleoside triphosphate. e. ribozyme.

Answer: d Section 13.2 Comprehension Question

21. This molecule is made of nucleotides joined by phosphodiester bonds that connect the 2′ OH to the 5′ phosphate. a. RNA only b. DNA only c. Both RNA and DNA d. Neither RNA nor DNA

Answer: d Section 13.2 Comprehension Question

25. Which statement about RNA polymerase is NOT true? a. RNA polymerase adds a ribonucleotide to the 3' end of a growing RNA molecule. b. RNA polymerase binds to a promoter to initiate transcription. c. During transcription of a gene, RNA polymerase reads only one strand of DNA. d. RNA polymerase reads a template strand of DNA 5' to 3'. e. RNA polymerase has many subunits.

Answer: d Section 13.2 Comprehension Question

7. During transcription, which parts of a DNA molecule are transcribed into RNA? a. All of the nucleotides in DNA on both strands b. All of the nucleotides on one strand of DNA c. Only parts of the DNA that encode mRNA d. Only regions of the DNA that contain genes e. Only regions of the DNA that encode rRNA

Answer: d Section 13.2 Comprehension Question

39. The TATA binding protein (TBP) binds to the TATA box sequence in eukaryotic promoters. What is its function in transcriptional initiation? a. It blocks access of RNA polymerase to the promoter, until removed by general transcription factors. b. It is the subunit of prokaryotic RNA polymerase that is required to recognize promoters. c. It modifies histones so that nucleosomes can be removed from DNA for transcription. d. It bends and partly unwinds DNA at a promoter. e. It creates a phosphodiester bond between the nucleotides.

Answer: d Section 13.4 Comprehension Question

25. A bacterial protein is encoded by the following mRNA sequence: 5'-AUGGUGCUCAUGCCCTAA-3'. The second methionine codon (AUG) in this mRNA sequence will: a. serve as the initiation codon. b. encode N-formylmethionine. c. encode methionine that will eventually be removed. d. encode unformylated methionine. e. be skipped as the translation progresses.

Answer: d Section: 15.3 Application Question

52. A random sequence of synthetic RNA is made using equal proportions of A and C. When it is translated, which amino acids will be incorporated into the protein?

Answer: pro, lys, thr, gln, his, thr, and asn Section: 15.2 Application Question

2. In 1958, Francis Crick proposed that genes and their corresponding polypeptides are "colinear." Which of the following statements concerning the concept of colinearity is incorrect? a. Colinearity means that the linear nucleotide sequence of a given gene corresponds directly to the linear amino acid sequence in the corresponding polypeptide. b. The number of nucleotides in a gene should be precisely proportional to the number of amino acids present in the corresponding polypeptide. c. Colinearity generally holds true for the coding regions of prokaryotic viral genes. d. The vast majority of eukaryotic genes also follow the concept of coliniarity although the size of genes may be larger. e. The exception to colinearity between genes and polypeptides is the presence of untranslated sequences (UTRs).

Answer: d Section: 14.1 Comprehension Question

20. Which mechanism allows for the production of polypeptides that are not entirely encoded by DNA? a. Regulated transcription b. RNA interference c. Alternative RNA processing d. RNA editing e. None of the above

Answer: d Section: 14.2 Application Question

24. A key modification in the 3′ end of eukaryotic mRNA is the addition of 50 to 250 adenine nucleotides at the 3′ end, forming a poly(A) tail. Which of the following statements is not a salient feature of this polyadenylation reaction? a. The stability of mRNA transcripts in the cytoplasm is affected by poly(A) tail. b. Poly(A) tail facilitates the attachment of the ribosomes to the mRNA. c. Proper poly(A) tail is important for proper nuclear export of mRNA. d. Poly(A) tail at the 3′ end translates to a long stretch of repeated amino acids. e. Multiple proteins will recognize and bind to poly(A) tail in the cytoplasm.

Answer: d Section: 14.2 Application Question

25. During the post transcriptional processing of pre-mRNA, a 5′ cap is added to an mRNA. Arrange the following events in correct order for the capping process. 1. Addition of guanine nucleotide via 5′-5′ bond 2. Removal of a phosphate from a ribonucleotide triphosphate at the 5′ head of pre-mRNA 3. Methylation to the 2′ position of the sugar in the second and the third nucleotide 4. Methylation at the 5′ end and addition of methyl group to the position 7 of the terminal guanine base a. 1, 2, 3, 4 b. 2, 4, 1, 3 c. 4, 1, 3, 2 d. 2, 1, 4, 3 e. 3, 2, 4, 1

Answer: d Section: 14.2 Application Question

14. Guide RNAs are needed in a. transcription. b. translation. c. RNA interference. d. RNA editing. e. RNA splicing.

Answer: d Section: 14.2 Comprehension Question

16. Which of the following elements would not be found on an mRNA molecule? a. Protein-coding region b. 3′ untranslated region c. 5′ untranslated region d. Promoter e. Start and stop codons

Answer: d Section: 14.2 Comprehension Question

17. Which of the following is found on the primary product of transcription but not on a mature mRNA molecule? a. Start codon b. Promoter c. Exons d. Introns e. Stop codon

Answer: d Section: 14.2 Comprehension Question

19. Scientists once believed that each gene can encode a single polypeptide. We now know that __________ and ___________ allow a single gene to encode more than one polypeptide. a. transcription; translation b. polyadenylation; RNA transport c. DNA methylation; chromatin condensation d. alternative processing; RNA editing e. gene silencing; RNA interference

Answer: d Section: 14.2 Comprehension Question

6. Which of the following statements about ribosomes and ribosomal RNA is NOT true? a. Ribosomes typically contain about 80% of the total cellular RNA. b. Ribosomal RNA is processed in both prokaryotes and eukaryotes. c. In eukaryotes, genes for rRNA are usually present within tandem repeats. d. Each ribosomal RNA component is encoded by separate gene. e. In eukaryotes, the rRNA transcripts are processed further by snoRNAs within the nucleus.

Answer: d Section: 14.2 Comprehension Question

7. The spliceosome is a large, ribonucleoprotein complex located in the a. cytoplasm. b. endoplasmic reticulum. c. Golgi. d. nucleus. e. nucleolus.

Answer: d Section: 14.2 Comprehension Question

9. Which of the following statements about group I and group II introns is NOT true? a. Both group I and II introns form elaborate and characteristic secondary structures with loops. b. The splicing mechanism of group II introns is similar to that of spliceosome mediated nuclear pre-mRNA splicing. c. The length of group I and group II introns is much longer than the exons within the structures. d. Group I and group II introns are exclusively found in mitochondrial and chloroplast encoded genes. e. Both Group I and group II introns are both found in bacterial genes.

Answer: d Section: 14.2 Comprehension Question

32. Which of the following choices does not support the notion that the gene is much more than DNA sequences that are transcribed into a single RNA molecule that encodes a single polypeptide? a. Alternative splicing—a single gene can yield multiple mRNA and protein products. b. A single ribosomal RNA transcripts can liberate several RNA molecules via further processing c. RNAs can be the functional product of a gene without being translated into protein product d. Protein coding region—each codon represents specific amino acid that will be linked to form a polypeptide e. Regulatory elements are part of a gene that regulate timing, degree, and specificity of gene expression but are not transcribed.

Answer: d Section: 14.2 and 14.3 Application Question

37. Which of the following classes of RNAs is unique to the eukaryotes? a. Messenger RNA (mRNA) b. Ribosomal RNA (rRNA) c. Transfer RNA (tRNA) d. Small nuclear RNAs (snRNAs) e. CRISPR RNAs (crRNAs)

Answer: d Section: 14.3, 14.4, and 14.5 Comprehension Question

39. What is the similarity among miRNAs, siRNAs, and piRNAs? a. All three types originate from transposons or viruses and are found in all organisms. b. They target and degrade the gene from which they were transcribed. c. All three are generated from a single-stranded RNA that gets cleaved. d. All of them can influence chromatin structure, which in turn, can influence gene expression. e. All three associate with Piwi proteins to be able to function in RNA degradation.

Answer: d Section: 14.5 Application Question

5. Which of the following codon codes different amino acid from the rest? a. CUU b. CUC c. UUA d. UUU e. CUA

Answer: d Section: 15.2 Application Question

17. Codons that specify the same amino acid are said to be a. wobbly b. isoaccepting c. hypothetical d. synonymous e. anonymous

Answer: d Section: 15.2 Comprehension Question

3. Which of the following statements about translation is correct? a. A special tRNA that does not have an attached amino acid binds to stop codons to terminate translation. b. The first three bases at the 5 end of an mRNA are the AUG at which translation begins. c. The codon for methionine appears only at the beginning of the mRNA for a protein, not in the middle or in the end. d. In eukaryotes, the 5 cap and the 3 poly(A) tail are involved in translation initiation. e. Ribosomes move along an mRNA in the 3 to 5 direction.

Answer: d Section: 15.3 and 15.4 Comprehension Question

37. Which of the following statements about protein folding and post translational modifications of proteins is correct? a. All nascent polypeptide chains have intrinsic ability to fold into the active conformation based on the primary structure. b. Only eukaryotic proteins undergo alterations following translation. c. Amino acids within a protein may be modified by molecular chaperones. d. Signal sequence of a protein helps direct a protein to a specific location within the cell. e. Attachment of a protein called ubiquitin directs proteins to enter into the nucleus.

Answer: d Section: 15.4 Application Question

36. Which molecule allows the release of mRNA from a stalled ribosome? a. miRNA b. snoRNA c. incRNA d. tmRNA e. siRNA

Answer: d Section: 15.4 Comprehension Question

17. When a structural gene is under positive inducible control, what would be the result of a mutation that eliminates the activator protein? a. The structural gene to be constitutively expressed due to the lack of inducible control. b. The transcription of structural gene will not be affected, as an activator is not required. c. The mutation will lead to activation of a repressor upon the lack of an activator protein, which will block transcription. d. d. As the transcription will require an activator protein, the transcription will be turned off. e. e. More cAMP will be produced in a cell to compensate for the lack of an activator protein.

Answer: d Section: 16.2 Application

24. Which parts of the DNA region shown in the diagram encode proteins? (I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator) a. P b. I, P, O c. P, O, Z, Y, A d. I, Z, Y, A e. I, P, O, Z, Y, A

Answer: d Section: 16.2 Application

27. A mutant E. coli strain, grown under conditions that normally induce the lac operon, does not produce ß-galactosidase. What is a possible genotype of the cells? a. lacI+ lacP+ lacO+ lacZ+ lacY- lacA+ b. lacI+ lacP+ lacOc lacZ+ lacY+ lacA+ c. lacl+ lacP+ lacO+ lacZ+ lacY+ lacA+ d. lacI+ lacP- lacO+ lacZ+ lacY+ lacA+ e. lacI_ lacP+ lacO—lacZ+ lacY+ lacA-

Answer: d Section: 16.2 Application

31. The trp operon is known to operate by both negative repressible regulation of operator and attenuation. Which of the following does NOT support the reason as to why dual control exists to regulate the operon? a. The repression alone is never complete, and some transcription can be initiated. b. Combined mechanism provides a much finer tuning of tryptophan synthesis regulation. c. Attenuation and repression allow the cell to more sensitively respond to the tryptophan level. d. It is most likely due to the fact that the attenuation is the evolutionary relic, which by accident has remained. e. Repression responds to the cellular levels of tryptophan, while attenuation responds to the number of tRNA charged with tryptophan.

Answer: d Section: 16.3 Application

34. Which of the following facts about riboswitches is INCORRECT? a. Binding of certain molecules to the riboswitches results in the formation of specific secondary structures of mRNA. b. Certain molecules that bind to riboswitches may act as repressors or inducers of transcription. c. Riboswitches are only found in bacterial cells but not in archaeal, fungal, or plant cells. d. Riboswitches are typically found in the 3′ UTR of the mRNA structure. e. The secondary structure that forms riboswitches typically contains a base stem and several branching hairpins.

Answer: d Section: 16.4 Comprehension

6. ____________________ probably began the evolution of life on Earth. a. DNA b. RNA promoters c. DNA polymerases d. RNA polymerases e. Ribozymes

Answer: e Section 13.1 Comprehension Question

9. Which of the following is a sequence of DNA where transcription is initiated? a. Hairpin loop b. TBP c. Initiator d. Sigma factor e. Promoter

Answer: e Section 13.2 Comprehension Question

31. In prokaryotic RNA polymerases, the holoenzyme consists of the core enzyme and the a. rho factor. b. TFIID. c. TBP. d. omega subunit. e. sigma factor.

Answer: e Section 13.3 Comprehension Question

41. Which of the following statements is FALSE regarding TFIID? a. It contains a TATA binding protein. b. It aids in initiation of transcription. c. It binds to the core promoter. d. It binds to the TATA box. e. It is a transcriptional activator protein.

Answer: e Section 13.4 Comprehension Question

42. What is the function of general transcription factors? a. They are DNA sequences to which RNA polymerase binds. b. They direct nucleosome assembly. c. They bind to regulatory promoters to increase the rate of transcription. d. They bind to enhancers to allow minimal levels of transcription. e. They are a part of the basal transcription apparatus.

Answer: e Section 13.4 Comprehension Question

29. Which of the following correctly describes the concept of alternative splicing? a. Eukaryotic gene and protein sequences are precisely colinear. b. With the rare exception of RNA editing, every nucleotide contained in a processed mRNA molecule is derived from exon sequences. c. Every other intron is removed in alternate manner to generate functional mRNA transcript. d. Only a subset the same mRNA transcripts are specifically selected for splicing in the nucleus. e. Multiple protein products are often produced from single eukaryotic genes.

Answer: e Section: 14.2 Application Question

11. Which of the following phenomena is not affected by the presence of alternative splicing? a. Speciation b. Development c. Organismal complexity d. Tissue specificity e. RNA interference

Answer: e Section: 14.2 Comprehension Question

11. Which of the following statements is NOT true? a. Both DNA and RNA are synthesized in a 5′ to 3′ direction. b. During RNA synthesis, the DNA template strand is read in a 3′ to 5′ direction. c. During RNA synthesis, new nucleotides are added to the 3′ end of the growing RNA molecule. d. RNA polymerase has 5′ to 3′ polymerase activity. e. RNA molecules have the same 5′ to 3′ orientation as the DNA template strands to which they are complementary.

Answer: e Section 13.2 Comprehension Question

13. The human gene encoding for calcitonin contains six exons and five introns and is located on chromosome 11. The pre-mRNA transcript from this gene can generate either calcitonin or calcitonin gene related peptide (CGRP) in a tissue-specific manner. Calcitonin produced from the thyroid gland is 32 amino acids long and functions to regulate the calcium while CGRP, which contains 37 amino acids, is produced by the brain cells and involved in transmission of pain. Which of the following processes makes production of two functionally and structurally different proteins from the same gene possible? a. Self-spicing introns b. Differential trnascription c. Alternative replication d. 5′ capping and polyadenylation e. Alternative RNA processing

Answer: e Section: 14.2 Comprehension Question

18. The information needed during RNA editing comes most directly from a. pre-mRNA. b. mRNA. c. rRNA. d. tRNA. e. guide RNA.

Answer: e Section: 14.2 Comprehension Question

35. The list of events below describes the processing of ribosomal RNAs. Please select the choice that lists the events in correct sequential order. 1. Methyl groups are added to specific bases and the 2′-carbon atom of some ribose sugars 2. Transcription of the rRNA precursors from DNA 3. Cleavage of precursor rRNA 4. Individual rRNA molecules ready for ribosome assembly 5. Trimming of precursor rRNA a. 3, 1, 2, 5, 4 b. 2, 3, 4, 5, 1 c. 4, 2, 3, 1, 5 d. 1, 3, 5, 4, 1 e. 2, 1, 3, 5, 4

Answer: e Section: 14.4 Application Question

34. Which of the following rRNA components originates from a separate gene transcript rather than as a cleaved product of a long single precursor rRNA transcript? a. Prokaryotic 16S rRNA b. Prokaryotic 23S rRNA c. Eukaryotic 18S rRNA d. Eukaryotic 5.8S rRNA e. Eukaryotic 5S rRNA

Answer: e Section: 14.4 Comprehension Question

41. Which of the following regulatory RNA types is different from the rest in terms of its length? a. siRNA b. crRNA c. miRNA d. piRNA e. lncRNA

Answer: e Section: 14.5 and 14.6 Application Question

14. If the bottom strand of the DNA is the template, the tRNA anticodon sequence, left to right 5 to 3, for the first RNA codon is a. GGA b. AUG c. CAC d. UCC e. CCU

Answer: e Section: 15.2 Application Question

8. Which amino acid is coded by the stop codons in most organisms? a. met b. pro c. trp d. cys e. none

Answer: e Section: 15.2 Application Question

19. There are __ different codons, which encode 20 amino acids and 3 stop codons. a. 16 b. 20 c. 23 d. 61 e. 64

Answer: e Section: 15.2 Comprehension Question

6. Which of the following statement describes the 'wobble' rules correctly? a. There is a flexible pairing between tRNA and amino acid as there are more tRNAs than the number of amino acids. b. The number of genetic code exceeds the number of amino acids available in the cell. c. There are multiple tRNA that may bind to the same amino acids. d. There are multiple codons that may code for the same amino acids. e. The third base pairing between the tRNA and mRNA is relaxed.

Answer: e Section: 15.2 Comprehension Question

23. What is the function of peptidyl transferase activity? a. it charges tRNAs b. it acetylates the end of a protein after translation c. it cleaves the polypeptide from the last tRNA during termination d. it moves ribosomes along mRNA during translation e. it forms peptide bonds

Answer: e Section: 15.3 Application Question

33. An mRNA has the stop codon 5' UAA 3'. What tRNA anticodon will bind to it? a. 5 ATT 3 b. 5 AUC 3' c. 5' ACU 3 d. 5' UUA 3' e. None

Answer: e Section: 15.3 Application Question

10. Which of the following does not enhance the biding of the ribosome to the 5' end of the mRNA? a. 5' cap b. 3' poly-A tail c. cap binding proteins d. poly(A) proteins e. enhancer

Answer: e Section: 15.3 Comprehension Question

30. What is the minimum number of different aminoacyl-tRNA synthetases required by a cell? a. 64, one for each codon b. 61, one for each sense codon c. 30, one for each different tRNA d. 50, one for each different tRNA e. 20, one for each amino acid

Answer: e Section: 15.3 Comprehension Question

1. Which of the following statement about regulation of gene expression is correct? a. An inducible gene is transcribed when a specific substance is absent. b. A gene is any DNA sequence that is transcribed into an mRNA molecule only. c. All genes are transcribed at all times as long as they have a functional promoter. d. The regulation of gene expression is the same in both eukaryotes and prokaryotes. e. The regulation of gene expression is critical for the control of life processes in all organisms.

Answer: e Section: 16.1 Comprehension

6. Which of the following statements about DNA binding protein is NOT true? a. Specific amino acids within the motif form hydrogen bonds with DNA. b. These proteins can affect the expression of a gene. c. Most DNA binding proteins bind dynamically. d. Some of these proteins incorporate metal ion such as zinc. e. Once bound, most of DNA binding proteins remain on DNA permanently.

Answer: e Section: 16.1 Comprehension Question

23. Where would the lac repressor be bound in a (nonmutant) E. coli cell that is growing in low glucose and high lactose? (I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator) a. P b. O c. P and O d. I, P, O e. The repressor would not be bound.

Answer: e Section: 16.2 Application

28. Transcriptional control that acts by regulating the continuation of transcription is called a. riboswitching. b. antitermination. c. negative control. d. operator mutation. e. attenuation.

Answer: e Section: 16.3 Comprehension

29. Which of the following secondary structures causes attenuation of structural genes of the trp operon? a. 1+2 b. 1+3 c. 2+4 d. 2+3 e. 3+4

Answer: e Section: 16.3 Comprehension

35. RNA-mediated repression is carried out by a. nonsense RNA. b. sense RNA. c. antisense RNA. d. riboswitches. e. ribozymes.

Answer: e Section: 16.4 Comprehension

46. Which amino acids are encoded, if the reading frame is as shown below, starting from the correct end? 5 ...GGAGCUCGUUGUAUU... 3

Answer: gly-ala-arg-cys-ile... Section: 15.3 Application Question