Question 1

Lactose utilization – lac operon enzymes: Which enzyme set is classically required for bacterial lactose fermentation and transport across the membrane in the lac operon paradigm?

Accepted Answer

Introduction / Context: The lac operon is a canonical model for gene regulation and carbohydrate utilization in bacteria. Understanding which enzymes are necessary for lactose uptake and catabolism underpins assays such as ONPG tests, MacConkey agar differentiation, and fermentation profiles in clinical microbiology. Given Data / Assumptions: LacZ encodes beta-galactosidase (hydrolyzes lactose to glucose and galactose; also performs transgalactosylation). LacY encodes galactoside permease (membrane transporter that imports lactose). LacA encodes transacetylase (thiogalactoside transacetylase; role in fermentation is auxiliary but part of the operon). Concept / Approach: Efficient lactose fermentation in standard lab contexts requires transport into the cell (LacY) and hydrolysis to utilizable monosaccharides (LacZ). LacA is co-expressed and, while not essential for lactose hydrolysis per se, is considered part of the enzymatic set associated with lactose metabolism in the classic operon description and is commonly included in exam questions on the topic. Therefore, the inclusive choice is appropriate. Step-by-Step Solution: Identify transport (LacY) and hydrolysis (LacZ) as core functions. Acknowledge LacA as an operon enzyme often taught with LacZ/LacY. Select the option that includes all three lac structural genes. Verification / Alternative check: Mutants lacking LacY fail to import lactose efficiently; LacZ mutants cannot cleave lactose; IPTG/X-gal reporter systems rely on LacZ activity directly. Why Other Options Are Wrong: Any single enzyme alone is insufficient for full lactose fermentation and transport; lactate dehydrogenase relates to pyruvate reduction in fermentation, not lactose import/hydrolysis. Common Pitfalls: Assuming LacA is dispensable and thus excluding it; many curricula still expect all three structural genes as the “set.” Final Answer: All of these.

Question 2

Transcription machinery – eukaryotic RNA polymerases: In eukaryotic nuclei, which RNA polymerase is primarily responsible for transcribing tRNA genes (and also 5S rRNA)?

Accepted Answer

Introduction / Context: Eukaryotes partition transcriptional duties among multiple nuclear RNA polymerases. Knowing which polymerase handles tRNA genes is critical for interpreting promoters, transcription factors, and drug/toxin sensitivities in cell biology and genetics. Given Data / Assumptions: Pol I → large rRNA (28S/18S/5.8S in nucleolus). Pol II → mRNA and most snRNA/snoRNA precursors. Pol III → tRNA and 5S rRNA (and some other small RNAs). Concept / Approach: tRNA genes typically use internal promoter elements (A and B boxes) recognized by TFIIIC/TFIIIB complexes to recruit RNA polymerase III. Pol III transcription produces short, structured RNAs required for translation and RNA processing. This functional division ensures efficient regulation and specialized transcription factor usage across RNA classes. Step-by-Step Solution: Match RNA classes to polymerases. Identify tRNA → Pol III. Select “RNA polymerase III.” Verification / Alternative check: Inhibitors like alpha-amanitin strongly affect Pol II at low doses, while Pol III is inhibited only at higher concentrations; Pol I is resistant—functional differences that support assignments. Why Other Options Are Wrong: Pol I does rRNA; Pol II does mRNA; “Any of these” is false; mitochondrial RNA polymerase transcribes mitochondrial genes, not the nuclear tRNA genes asked here. Common Pitfalls: Confusing 5S rRNA (Pol III) with the larger rRNAs (Pol I). Final Answer: RNA polymerase III.

Question 3

Transcription initiation – promoter recognition: In gene expression, what is the name of the specific DNA region to which RNA polymerase binds to initiate transcription of a downstream gene?

Accepted Answer

Introduction / Context: Transcription begins with the recognition of promoter DNA sequences by RNA polymerase and/or associated factors. Correct terminology is essential for discussing gene regulation, vector design, and regulatory mutations in both prokaryotes and eukaryotes. Given Data / Assumptions: Promoters contain sequence motifs (e.g., −10/−35 in bacteria; TATA box and others in eukaryotes) that position polymerase for accurate start-site selection. Terminators end transcription; operators modulate promoter output in prokaryotic operons but are not the polymerase binding site itself. Concept / Approach: The promoter is the dedicated DNA region required for transcription initiation. It provides orientation and defines the transcription start site. Additional regulatory elements (operators, enhancers, silencers) modulate promoter activity but are conceptually distinct. Therefore, the polymerase recognition/binding site is properly called the promoter. Step-by-Step Solution: Identify the function: polymerase binding and initiation. Map this to “promoter.” Exclude unrelated terms (receptor/facilitator) and “terminator,” which functions at transcription termination. Verification / Alternative check: Footprinting and ChIP-seq experiments show polymerase and transcription factor occupancy precisely over promoter regions during initiation. Why Other Options Are Wrong: “Receptor/facilitator” are not standard DNA elements; “terminator” ends transcription; “operator” is a repressor binding site adjacent to promoters in bacteria but not the RNAP docking site itself. Common Pitfalls: Conflating promoter with operator in operon diagrams; they are adjacent yet distinct functional elements. Final Answer: Promoter.

Question 4

Bacterial RNA polymerase – promoter specificity: Which subunit of the prokaryotic RNA polymerase holoenzyme primarily confers recognition of promoter sequences during transcription initiation?

Accepted Answer

Introduction / Context: Bacterial RNA polymerase exists as a core enzyme that elongates RNA and a holoenzyme that includes a specificity factor. Understanding the role of the sigma (σ) factor is central to grasping how bacteria rapidly reprogram transcription in response to stress and environmental cues. Given Data / Assumptions: Core enzyme (α2ββ′ω) lacks efficient promoter recognition on its own. Sigma factors bind core to form holoenzyme and direct it to specific promoter motifs. Different σ factors target different promoter consensus sequences. Concept / Approach: The σ subunit contacts the −10 and −35 regions (or alternative motifs) in promoter DNA, enabling closed-complex formation and promoter melting to the open complex. Swapping σ factors retargets RNAP to stress-responsive or housekeeping promoters, a key regulatory strategy in prokaryotes. Step-by-Step Solution: Define core vs holoenzyme. Assign promoter recognition role to σ subunit. Select the option naming sigma (σ). Verification / Alternative check: Genetic deletion or overexpression of specific σ factors changes promoter usage; in vitro reconstitution confirms σ-dependent promoter binding. Why Other Options Are Wrong: α assists assembly and activation; β/β′ form the catalytic center; ω aids assembly/stability; γ is not a standard bacterial RNAP subunit. Common Pitfalls: Assuming core RNAP alone targets promoters efficiently; without σ, specificity is poor. Final Answer: Sigma (σ) subunit.

Question 5

Transcription termination in prokaryotes – common features: Considering both rho-independent and rho-dependent mechanisms, which element is present in ALL prokaryotic transcription terminators?

Accepted Answer

Introduction / Context: Bacterial transcription terminates via two mechanistic classes: rho-independent (intrinsic) and rho-dependent. Exams often probe whether a single sequence or factor is universal to termination. The subtlety is that the two classes use different signals, so no single listed feature appears in both. Given Data / Assumptions: Intrinsic (rho-independent) terminators typically have a GC-rich dyad symmetry that forms a stem-loop followed by a poly-U tract in the RNA. Rho-dependent termination requires the ATP-dependent helicase Rho and a rut site on the RNA but lacks the obligatory intrinsic hairpin/poly-U signature. NusA can enhance intrinsic termination but is not universally required. Concept / Approach: Because rho-dependent terminators do not require a poly-U tract or a strong hairpin, and intrinsic terminators do not require Rho, there is no single structural or protein factor among the options that is shared by all terminators. Therefore, when asked to pick an element present in “all” terminators, the correct response is that none of the listed features is universal. Step-by-Step Solution: Evaluate option a: poly-U → intrinsic only. Evaluate option b: Rho factor → rho-dependent only. Evaluate option c: hairpin → intrinsic hallmark, not rho-dependent requirement. Conclude that no option fits both mechanisms; choose “None of these.” Verification / Alternative check: Genomic mapping shows distinct terminator classes: many lack poly-U or hairpins yet are rho-dependent; others terminate efficiently without Rho via hairpin/poly-U. Why Other Options Are Wrong: Each corresponds to a single mechanism, not universal termination signals. Common Pitfalls: Assuming every terminator has a hairpin; this is true only for intrinsic terminators. Final Answer: None of these.

Question 6

Eukaryotic transcription initiation – TATA box binding: During assembly of the pre-initiation complex at a TATA-containing promoter, which general transcription factor first binds the TATA box to nucleate RNA polymerase II recruitment?

Accepted Answer

Introduction / Context: Transcription initiation by RNA polymerase II requires an ordered assembly of general transcription factors (GTFs). The TATA box is a core promoter element in many, though not all, eukaryotic genes. Identifying the first factor that recognizes TATA is key to understanding pre-initiation complex architecture. Given Data / Assumptions: TFIID contains the TATA-binding protein (TBP) plus TBP-associated factors (TAFs). Binding of TBP to the minor groove of the TATA box bends DNA and seeds PIC assembly. Subsequent factors (TFIIA, TFIIB, Pol II with TFIIF, then TFIIE/TFIIH) are recruited. Concept / Approach: The canonical order places TFIID (TBP) at the start, directly engaging TATA. TFIIA stabilizes TBP–DNA interaction; TFIIB bridges TBP–DNA with Pol II/TFIIF. TFIIH later provides helicase and kinase activities for promoter opening and CTD phosphorylation. Therefore, the “first binder” to the TATA sequence is TFIID through its TBP subunit. Step-by-Step Solution: Identify the element: TATA box. Recall that TBP (within TFIID) recognizes and binds TATA. Place TFIIA/TFIIB and Pol II recruitment after TBP binding. Select TFIID (via TBP) as the correct answer. Verification / Alternative check: Structural studies show TBP inducing an ~80° DNA bend at TATA before other GTFs load; TBP mutants disrupt PIC assembly at TATA promoters. Why Other Options Are Wrong: TFIIA/TFIIB bind after TBP; Pol II does not directly recognize TATA; TFIIE acts later with TFIIH to open the promoter. Common Pitfalls: Assuming all Pol II promoters are TATA-dependent; many are TATA-less and use TAF-mediated recognition of other core elements. Final Answer: TFIID (via TBP).

Question 7

Molecular biology basics (gene expression): Which enzyme synthesizes RNA by using a DNA template and direction provided by the DNA sequence (promoter to terminator)?

Accepted Answer

Introduction / Context: Transcription is the first stage of gene expression, where the information encoded in DNA is copied into RNA. The central enzyme that performs this task is RNA polymerase, which reads the DNA template strand and builds a complementary RNA strand. Understanding which enzyme is responsible is foundational for microbiology, genetics, and biotechnology. Given Data / Assumptions: The process in question is synthesis of RNA directed by a DNA template. Promoters and terminators on DNA guide where transcription starts and stops. We are considering the core catalytic activity, not accessory factors. Concept / Approach: RNA polymerase binds to promoter sequences (often with help from sigma factors in bacteria or general transcription factors in eukaryotes), locally unwinds the DNA, and catalyzes the formation of phosphodiester bonds as ribonucleoside triphosphates are added to the growing RNA chain. The enzyme moves 3' to 5' along the DNA template while the RNA product grows 5' to 3'. Termination mechanisms then release the finished RNA. Step-by-Step Solution: Identify the biochemical transformation: DNA template → RNA product. Recall the enzyme class that catalyzes RNA synthesis: RNA polymerase. Exclude DNA polymerase (uses DNA template to make DNA, not RNA) and ligases (join breaks in nucleic acids). Select RNA polymerase as the only enzyme that fits the definition. Verification / Alternative check: In vitro transcription reactions require DNA template, ribonucleoside triphosphates, appropriate ions, and RNA polymerase to produce RNA. Removing RNA polymerase abolishes RNA synthesis, confirming the enzyme's essential role. Why Other Options Are Wrong: DNA ligase: Seals nicks in DNA; does not synthesize RNA. DNA polymerase: Extends DNA, not RNA. RNA ligase: Joins RNA fragments; not template-directed synthesis from DNA. Reverse transcriptase: Uses RNA to synthesize DNA, the opposite direction. Common Pitfalls: Confusing transcription (DNA → RNA) with replication (DNA → DNA) or reverse transcription (RNA → DNA). Also, accessory factors aid binding and regulation but do not replace RNA polymerase's catalytic function. Final Answer: RNA polymerase.

Question 8

Structure of a bacterial mRNA transcript: Which elements are typically present in a mature prokaryotic mRNA derived from a single gene (considering coding region and untranslated ends)?

Accepted Answer

Introduction / Context: Bacterial messenger RNA (mRNA) carries the information that ribosomes translate into protein. Recognizing the key landmarks on mRNA helps explain how translation begins, how it ends, and how the transcript itself is released from RNA polymerase during transcription. Given Data / Assumptions: Mature bacterial mRNA usually includes untranslated regions flanking the coding sequence. Translation starts at a start codon within the coding region and ends at a stop codon. Intrinsic (rho-independent) termination encodes a GC-rich palindromic segment followed by U residues that forms a hairpin in the RNA. Concept / Approach: The coding region of mRNA begins with a start codon (commonly AUG) and ends with a stop codon (UAA, UAG, or UGA). Upstream, a ribosome-binding site aids initiation. Downstream, sequences transcribed from a DNA terminator can form RNA structures (such as a hairpin followed by a uridine tract) that destabilize the transcription complex, resulting in release of the mRNA. Thus, features corresponding to start, stop, and termination are represented within or on the mRNA transcript. Step-by-Step Solution: Check for translation initiation: a start codon must be present in the coding sequence. Check for translation termination: a stop codon defines the end of the open reading frame. Check for transcription termination: sequences that give rise to a terminator hairpin are part of the RNA at its 3' end. Conclude that all listed elements can be found on the mature mRNA. Verification / Alternative check: Mapping experiments show that purified bacterial mRNAs contain 5' UTR, start codon, coding region, stop codon, 3' UTR, and, in intrinsic termination, a terminal hairpin-uridine tract region derived from the DNA terminator. Why Other Options Are Wrong: Any single element alone is incomplete; true mRNA typically contains all functional landmarks. None of these: Contradicted by standard prokaryotic gene architecture. Common Pitfalls: Confusing the DNA terminator with the resulting RNA structure; while the terminator is encoded in DNA, its sequence is transcribed and produces an RNA hairpin that functions in termination. Final Answer: All of these.

Question 9

Products of transcription: Which RNA species are direct primary products of transcription by RNA polymerase in cells?

Accepted Answer

Introduction / Context: Transcription is the synthesis of RNA from a DNA template. Cells produce several functional RNA types, each with a distinct role in gene expression and protein synthesis. Recognizing that multiple classes of RNA arise from transcription clarifies why different RNA polymerases or accessory factors exist in various domains of life. Given Data / Assumptions: RNA polymerase copies DNA into RNA without requiring a primer. Different genes encode different noncoding and coding RNAs. In bacteria, one RNA polymerase makes mRNA, rRNA, and tRNA; in eukaryotes, multiple nuclear polymerases share this work. Concept / Approach: Messenger RNA carries protein-coding information to ribosomes. Transfer RNA delivers amino acids during translation by base pairing with codons. Ribosomal RNA forms the catalytic and structural core of ribosomes. All three originate as transcripts from DNA. Additional RNAs such as small RNAs and snRNAs are also transcription products, but the essential triad mRNA, tRNA, and rRNA are the classical outputs. Step-by-Step Solution: Identify the process: transcription produces RNA from DNA. List canonical RNA types synthesized: mRNA, tRNA, and rRNA. Confirm each is transcribed by RNA polymerase. Select the inclusive option. Verification / Alternative check: Pulse-labeling and Northern blot experiments detect nascent transcripts corresponding to mRNA, tRNA, and rRNA genes, confirming transcriptional origin for all. Why Other Options Are Wrong: Each individual RNA type alone is incomplete. None of these: False because all listed RNAs are products of transcription. Common Pitfalls: Assuming only mRNA is transcribed because it encodes proteins; noncoding RNAs are equally essential and also arise from transcription. Final Answer: All of these.

Question 10

Gene regulation in bacteria: In the tryptophan (trp) operon of Escherichia coli, which regulatory mechanism prematurely terminates transcription when intracellular tryptophan is abundant?

Accepted Answer

Introduction / Context: Bacteria tune gene expression rapidly in response to metabolite levels. The trp operon provides a classic example where transcription can be curtailed early when tryptophan is plentiful, conserving resources by preventing unnecessary synthesis of tryptophan biosynthetic enzymes. Given Data / Assumptions: Tryptophan abundance increases charging of tRNATrp. Leader peptide coding region and attenuator sequences lie upstream of the structural genes. Coupled transcription and translation occur in bacteria. Concept / Approach: Attenuation relies on the interplay between ribosome speed on the leader peptide and formation of alternative RNA secondary structures in the leader transcript. When tryptophan is abundant, the ribosome rapidly translates the leader peptide, favoring an RNA terminator hairpin that causes RNA polymerase to release prematurely. When tryptophan is scarce, ribosome stalling promotes an anti-terminator structure, allowing transcription of the operon to continue. Step-by-Step Solution: Identify the condition: high tryptophan and highly charged tRNATrp. Link rapid leader translation to formation of a terminator hairpin. Recognize the resulting early termination of transcription. Name this mechanism: attenuation. Verification / Alternative check: Mutations that disrupt the leader peptide or attenuator structures abolish regulation, demonstrating that RNA structure and translation speed control transcriptional readthrough in the trp operon. Why Other Options Are Wrong: Activation: Increases transcription; opposite effect. Corepression: Refers to repressor-operator binding modulated by corepressor; in trp operon it acts at initiation, not the attenuation step. Desiccation: Environmental stress, not a regulatory mechanism. Riboswitch cleavage: Not the classical mechanism for trp control. Common Pitfalls: Confusing initiation control by the Trp repressor with attenuation; both operate, but attenuation specifically involves premature termination based on translation of the leader. Final Answer: Attenuation.

Question 11

Definition check: In microbial genetics, which statement best defines a replicon?

Accepted Answer

Introduction / Context: The term replicon is central to understanding plasmids, chromosomes, and certain viral genomes. It helps distinguish DNA elements that can autonomously replicate in a host from those that cannot persist without integration. Given Data / Assumptions: A replicon has an origin of replication and the necessary control sequences. Maintenance implies faithful propagation to progeny cells. Examples include bacterial chromosomes and many plasmids. Concept / Approach: A replicon is any DNA (or sometimes RNA in RNA viruses) molecule or region that behaves as a unit of replication, initiating replication at specific origins and being stably maintained across cell generations. Plasmids are classic replicons because they replicate independently of the chromosome and segregate to daughter cells via dedicated systems. Step-by-Step Solution: Identify the core requirement: autonomous replication capability. Recognize that maintenance across generations is part of the definition. Match the wording: a DNA molecule able to replicate and be maintained. Exclude options describing functions unrelated to replication competence. Verification / Alternative check: Engineering plasmid replicons requires compatible origins and host factors; incompatibility groups reflect control of replication and maintenance, underscoring the operational definition of a replicon. Why Other Options Are Wrong: Pili genes: Conjugation traits, not the definition of a replicon. Transcription template: All DNA can be a template; not synonymous with replicon. Transposition enzyme: A protein, not a DNA unit of replication. Protein-only element: Not applicable to DNA replicons. Common Pitfalls: Equating mobility (conjugation) with replication autonomy; a conjugative plasmid is a replicon, but conjugation is not required to be considered one. Final Answer: A DNA molecule that is able to replicate and be maintained.

Question 12

Prokaryotic promoter elements: What is the usual consensus sequence of the Pribnow box (−10 element) in bacterial promoters?

Accepted Answer

Introduction / Context: Promoters are DNA sequences that position RNA polymerase for accurate initiation of transcription. In bacteria, two key hexamer elements are commonly recognized: the −35 region and the −10 region, also known as the Pribnow box. Given Data / Assumptions: The Pribnow box lies approximately 10 base pairs upstream of the transcription start site. Consensus sequences reflect the most common nucleotides at each position among strong promoters. We are dealing with DNA sequences written in the coding strand sense. Concept / Approach: The canonical −10 element consensus is TATAAT. This AT-rich sequence facilitates DNA unwinding during open complex formation by RNA polymerase. The −35 element has a different consensus, typically TTGACA, and provides an additional binding site for sigma factor within the holoenzyme. Step-by-Step Solution: Recall that the −10 (Pribnow) consensus is AT-rich. Match options with known consensus: TATAAT fits the −10 box. Recognize TTGACA as the −35 element, not the −10 element. Select TATAAT as the correct answer. Verification / Alternative check: Mutational analyses show that deviations from TATAAT at key positions reduce promoter strength, confirming the functional importance of this consensus. Why Other Options Are Wrong: TTGACA: −35 consensus, not −10. AUAUA / UUUUU: RNA-like sequences, not DNA promoter elements. CGCGCG: GC-rich and does not serve as a standard −10 element. Common Pitfalls: Confusing RNA sequences with DNA promoter consensus or mixing up the −35 and −10 elements. Final Answer: TATAAT.

Question 13

Name the process: The synthesis of messenger RNA (mRNA) using a DNA template is known as what?

Accepted Answer

Introduction / Context: Information flow in the central dogma proceeds from DNA to RNA to protein. The first conversion, making RNA from DNA, is called transcription and is catalyzed by RNA polymerase. Given Data / Assumptions: The product asked about is mRNA. A DNA template strand is read by an enzyme to produce RNA. We are distinguishing transcription from replication and translation. Concept / Approach: Transcription copies the DNA sequence of a gene into an RNA transcript that can be translated or can function directly (for noncoding RNAs). The process involves initiation at a promoter, elongation through the gene body, and termination at specific sequences or via protein factors. Step-by-Step Solution: Identify the substrates and products: DNA template → RNA. Match this transformation to the named cellular process. Recognize the correct term is transcription, not replication or translation. Choose transcription. Verification / Alternative check: Inhibitors such as rifampicin specifically block bacterial transcription by targeting RNA polymerase, demonstrating that mRNA synthesis is a distinct process from replication and translation. Why Other Options Are Wrong: Replication: DNA copied to DNA. Translation: mRNA decoded to protein. DNA restriction: Endonuclease cleavage; unrelated to mRNA synthesis. Splicing: Eukaryotic mRNA processing step post-transcription. Common Pitfalls: Using terms replication and transcription interchangeably; they describe different enzymes, substrates, and outcomes. Final Answer: Transcription.

Question 14

Role of sigma factor in bacterial transcription: Which statement correctly describes the primary function of the sigma (σ) factor in RNA polymerase holoenzyme?

Accepted Answer

Introduction / Context: Bacterial RNA polymerase consists of a core enzyme and a sigma factor that together form the holoenzyme. Sigma factors are specificity subunits that determine which promoters are efficiently recognized and transcribed under given conditions. Given Data / Assumptions: Sigma factor is required for efficient promoter recognition and transcription initiation. Elongation is carried out primarily by the core enzyme after sigma dissociates. Termination involves separate sequence features or factors (e.g., rho protein). Concept / Approach: The sigma factor binds characteristic promoter elements (−35 and −10 regions). By lowering the affinity of RNA polymerase for non-promoter DNA and increasing its affinity for promoter DNA, sigma factors ensure accurate start-site selection. Different sigma factors redirect RNA polymerase to distinct promoter subsets in response to stress or developmental cues. Step-by-Step Solution: Define sigma's role as promoter-targeting specificity. Exclude catalytic elongation functions and termination recognition, which are separate. Choose the statement describing promoter recognition and initiation. Verification / Alternative check: In vitro, core RNA polymerase without sigma binds DNA nonspecifically and initiates poorly; addition of sigma restores correct promoter-dependent initiation. Why Other Options Are Wrong: Termination recognition: Not sigma's principal role. Pyrophosphate cleavage: Chemistry of phosphodiester formation is intrinsic to core enzyme, not sigma. DNA restriction / Helicase: Different enzymes entirely. Common Pitfalls: Assuming sigma stays bound throughout elongation; it most often dissociates after initiation, allowing promoter escape. Final Answer: Recognizes and binds promoter elements to direct initiation.

Question 15

Alternative sigma factors and timing: A specific ________ factor enables RNA polymerase to bind late promoters and transcribe late genes (for example, during certain phage infections or stress responses).

Accepted Answer

Introduction / Context: Bacteria and bacteriophages use alternative sigma factors to redirect RNA polymerase to distinct promoter sets. This allows temporal control of gene expression, such as early versus late genes in phage development or stress-responsive regulons in the bacterial host. Given Data / Assumptions: RNA polymerase core enzyme pairs with different sigma subunits. Late promoters have sequence motifs recognized by a specific sigma factor. Changing sigma changes promoter specificity without altering the core catalytic machinery. Concept / Approach: Sigma factors act as interchangeable specificity modules. During phage infection, a phage-encoded sigma or a host sigma modified by the phage can target late promoters that control capsid assembly genes, lysis genes, and other late functions. In bacteria, alternative sigmas (for heat shock, stationary phase, flagella, etc.) similarly reprogram transcriptional priorities. Step-by-Step Solution: Recognize that promoter specificity is governed by sigma, not beta, delta, or gamma subunits. Recall classic examples where late gene transcription requires a different sigma factor. Select “Sigma.” Verification / Alternative check: Biochemical reconstitution shows that swapping sigma subunits alters which promoter sequences are efficiently recognized in vitro and in vivo. Why Other Options Are Wrong: Beta: A core catalytic subunit; does not confer alternate promoter specificity by itself. Delta / Gamma / Theta: Not the standard promoter-recognition factors in bacterial RNA polymerase. Common Pitfalls: Confusing transcription factor proteins that bind DNA with sigma factors; sigma is a component of the holoenzyme, not a separate DNA-binding regulator acting alone. Final Answer: Sigma.

Question 16

Catalysis of transcription: Which enzyme catalyzes the synthesis of RNA from a DNA template during transcription?

Accepted Answer

Introduction / Context: Transcription converts genetic information stored in DNA into RNA. The catalytic engine of this process is RNA polymerase, an enzyme conserved across bacteria, archaea, and eukaryotes, albeit with differences in subunit composition and accessory factors. Given Data / Assumptions: The reaction is template-directed polymerization of ribonucleotides into RNA. DNA remains as the template; product is RNA, not DNA. Accessory enzymes like topoisomerases may assist but do not catalyze RNA synthesis. Concept / Approach: RNA polymerase binds promoters, melts a transcription bubble, and elongates the RNA chain in the 5' to 3' direction by forming phosphodiester bonds. DNA gyrase and topoisomerase I modulate DNA supercoiling ahead of and behind the transcription complex, and DNA ligase seals DNA nicks in replication or repair, but none of these synthesize RNA. Step-by-Step Solution: Define the biochemical task: DNA-templated RNA synthesis. Identify the enzyme class performing this reaction: RNA polymerase. Eliminate enzymes with different functions (gyrase, ligase, DNA polymerase). Choose RNA polymerase. Verification / Alternative check: Inhibitors like rifampicin block bacterial RNA polymerase and halt RNA synthesis, providing functional proof that RNA polymerase is the catalytic enzyme for transcription. Why Other Options Are Wrong: DNA polymerase: Catalyzes DNA synthesis, not RNA. DNA ligase: Joins DNA fragments; no polymerization from nucleoside triphosphates into RNA. DNA gyrase / Topoisomerase I: Change DNA topology; do not form RNA chains. Common Pitfalls: Assuming any nucleic acid enzyme can make RNA; catalytic specificities are distinct and highly regulated. Final Answer: RNA polymerase.

Question 17

Prokaryotic transcription – promoter architecture In bacterial (prokaryotic) promoters, the first conserved hexameric sequence (the “-35 element,” usually similar to TTGACA) is typically located at which position relative to the transcription start …

Accepted Answer

Introduction / Context: Prokaryotic promoters contain two core consensus elements that guide RNA polymerase holoenzyme (via sigma factor) to initiate transcription precisely. Understanding where these hexameric motifs sit relative to +1 is a foundational concept in bacterial gene regulation and biotechnology vector design. Given Data / Assumptions: The transcription start site is designated +1. Promoter elements are described by their approximate upstream positions (negative numbers). The two canonical hexamers are commonly called the -35 element and the -10 (Pribnow) box. Concept / Approach: The -35 element (consensus similar to TTGACA) is positioned about 35 nucleotides upstream of +1 and is crucial for initial recognition by the sigma subunit. The -10 element (consensus similar to TATAAT) lies about 10 nucleotides upstream and facilitates DNA unwinding. These distances are approximate but conserved enough to serve as landmarks in promoter mapping. Step-by-Step Solution: Identify the “first hexameric sequence” in standard descriptions: the -35 element.Recall its location: roughly 35 bp upstream of +1.Differentiate from the -10/Pribnow box: ~10 bp upstream.Select the option stating “approximately 35 bases upstream.” Verification / Alternative check: Sigma-70 dependent promoters in E. coli are routinely annotated with -35 and -10 elements at these positions; mutational studies confirm their functional placement for efficient initiation. Why Other Options Are Wrong: At +1: this is the start site, not a promoter hexamer. Approximately 10 bases upstream: that is the Pribnow (-10) element. Approximately 25 bases upstream: typical of the eukaryotic TATA box location, not bacterial. Approximately 50 bases upstream: outside the canonical -35 spacing. Common Pitfalls: Confusing eukaryotic TATA positioning with bacterial promoter architecture; mixing up the -10 and -35 elements due to both being hexamers. Final Answer: Approximately 35 bases upstream of the transcription start site (the -35 region).

Transcription Questions