Question 1

In prokaryotes, which mechanism correctly terminates transcription? Select the best answer, noting that bacteria use more than one termination pathway.

Accepted Answer

Introduction / Context: Bacterial transcription termination occurs by two main mechanisms: rho-dependent termination, which requires the hexameric ATPase rho, and rho-independent (intrinsic) termination, driven by an RNA hairpin followed by a U-rich tract. Understanding these distinguishes transcription from translation signals. Given Data / Assumptions: Stop and start codons regulate translation, not transcription. Poly(A) tails are hallmarks of eukaryotic mRNA processing, not bacterial termination. Rho protein can terminate certain transcripts by catching RNA polymerase. Concept / Approach: In rho-dependent termination, rho binds rut sites on nascent RNA, translocates along RNA using ATP, and disengages RNA polymerase at a paused site. This is a bona fide transcription termination pathway in bacteria. Step-by-Step Solution: Eliminate translation signals (stop/start codons, release factors) as unrelated to transcription termination.Eliminate eukaryotic-style polyadenylation as a termination trigger in bacteria.Identify rho as a valid bacterial termination factor. Verification / Alternative check: Biochemical assays show rho-dependent termination sensitivity to rho inhibitors and mutation of rut sites; intrinsic termination is observed with strong GC hairpins followed by U runs. Why Other Options Are Wrong: Stop/start codons are decoded by ribosomes. Poly(A) plays a different role in bacteria. Release factors act in translation termination, not transcription. Common Pitfalls: Conflating transcription termination with translation termination events occurring on the same polycistronic mRNA. Final Answer: Terminated by a protein called rho (ρ) in rho-dependent termination.

Question 2

In eukaryotic cells, which RNA polymerases are present in the nucleus and what do they transcribe? Choose the best overall statement.

Accepted Answer

Introduction / Context: Eukaryotic nuclei contain three specialized RNA polymerases, each dedicated to different RNA classes. Recognizing their co-existence in the nucleus underpins understanding of ribosome biogenesis, mRNA synthesis, and tRNA production. Given Data / Assumptions: RNA polymerase I, II, and III reside in the nucleus (with Pol I concentrated in nucleoli). Pol I → 45S rRNA precursor (28S, 18S, 5.8S). Pol II → mRNA and many sn/snoRNAs. Pol III → tRNA, 5S rRNA, and other small RNAs. Concept / Approach: Because all three polymerases function in the nuclear compartment, the comprehensive answer is that the nucleus contains I, II, and III. Each has distinct promoter recognition and sensitivity to inhibitors (for example, alpha-amanitin). Step-by-Step Solution: Identify location: nucleus includes nucleoplasm and nucleolus.Assign roles: Pol I (rRNA), Pol II (mRNA), Pol III (tRNA/5S rRNA).Select 'All of these' as correct. Verification / Alternative check: Cell fractionation and immunolocalization studies show distinct nuclear distributions of each polymerase. Why Other Options Are Wrong: Limiting to a single polymerase ignores the established division of labor. Mitochondrial RNA polymerase operates in mitochondria, not the nucleus. Common Pitfalls: Confusing nucleolus (Pol I-rich) with entire nucleus, leading to the mistaken exclusion of Pol II and Pol III. Final Answer: All of these (RNA polymerase I, II, and III) are present in the nucleus.

Question 3

Eukaryotic mRNA 3′ end formation: where are most protein-coding transcripts cleaved relative to the polyadenylation signal? Select the correct statement.

Accepted Answer

Introduction / Context: Most eukaryotic mRNAs undergo co-transcriptional 3′ end processing, including endonucleolytic cleavage and addition of a poly(A) tail. The location of cleavage relative to the AAUAAA polyadenylation signal is a core principle in RNA processing. Given Data / Assumptions: Canonical polyadenylation signal: AAUAAA (with variants). Cleavage typically occurs 10–30 nucleotides downstream of this signal. Poly(A) polymerase adds a poly(A) tail after cleavage. Concept / Approach: Cleavage and polyadenylation specificity factor (CPSF), cleavage stimulation factor (CstF), and other factors recognize the signal and downstream elements, specifying cleavage after the signal in the 3′ untranslated region. Step-by-Step Solution: Locate AAUAAA in pre-mRNA.Identify the cleavage site 10–30 nt downstream (toward the 3′ end) of the signal.Poly(A) is then added to the newly formed 3′ hydroxyl. Verification / Alternative check: Experimental mapping of 3′ ends by 3′ RACE and global studies confirm cleavage downstream of the signal, not before it and not at the 5′ end. Why Other Options Are Wrong: Cleavage occurs after the signal downstream, not before. The 5′ end language is irrelevant to polyadenylation. Sites are not random; they depend on specific sequence and protein factor recognition. Common Pitfalls: Confusing the AAUAAA position with the exact cleavage nucleotide; the cut is typically downstream by a defined distance with auxiliary elements (for example, U/GU-rich regions). Final Answer: At the 3′ end just after a polyadenylation signal.

Question 4

Transfer RNA (tRNA) architecture: how many stem–loop arms are present in the canonical cloverleaf secondary structure? Choose the best answer.

Accepted Answer

Introduction / Context: tRNAs adopt a conserved secondary structure often depicted as a cloverleaf. Understanding the standard arms and loops is essential for decoding, aminoacylation, and ribosome interactions. Given Data / Assumptions: Common arms: acceptor stem, D (dihydrouridine) arm, anticodon arm, TψC arm. Many tRNAs also have a variable arm of differing length. Concept / Approach: In the cloverleaf, there are three principal stem–loop structures: the D-loop, the anticodon loop, and the TψC loop. The acceptor arm is a stem without a loop at the top. A variable arm may be present but is not universal as a distinct stem–loop. Step-by-Step Solution: Identify loops: D-loop, anticodon loop, TψC loop → total three.Recognize acceptor stem is not a loop; it terminates in the 3′ CCA tail for amino acid attachment.Account for variable arm as optional and not always forming a canonical loop counted among the three core loops. Verification / Alternative check: Structural studies show L-shaped tertiary structure arises from coaxial stacking of these elements, consistent with three core loops. Why Other Options Are Wrong: Two loops undercounts the standard architecture; four or five are not consistently present across tRNAs. 'Either (a) or (b)' is imprecise. Common Pitfalls: Counting the acceptor stem as a loop or treating the variable arm as universally looped. Final Answer: Three stem loops.

Question 5

Bacterial RNA polymerase characteristics: which statement is correct about the enzymes that transcribe bacterial DNA? Select the best option.

Accepted Answer

Introduction / Context: In bacteria, a single type of RNA polymerase transcribes rRNA, tRNA, and mRNA (with different sigma factors for promoter specificity). Its subunit composition and biochemical properties are conserved across many species. Given Data / Assumptions: Core enzyme subunits: typically two alpha (α), one beta (β), one beta prime (β′), and an omega (ω) subunit; sigma (σ) associates for initiation. DNA polymerases are distinct enzymes responsible for DNA replication and repair. Concept / Approach: The bacterial RNA polymerase is a multisubunit complex (holoenzyme when associated with σ). It is not interchangeable with DNA polymerases, is not glycosylated as a rule, and remains active in vitro (cell-free systems), so not 'only inside the cell.' Step-by-Step Solution: Recall subunit architecture of the bacterial RNA polymerase.Eliminate claims about glycosylation and interchangeability with DNA polymerases.Select 'Multisubunit enzymes' as correct. Verification / Alternative check: Purified bacterial RNA polymerase performs transcription in vitro, confirming activity outside cells and its multisubunit nature. Why Other Options Are Wrong: DNA polymerases and RNA polymerases differ in function and structure. Glycosylation is uncommon for bacterial cytosolic enzymes. 'Only active inside the cell' is false; robust in vitro transcription systems exist. Common Pitfalls: Confusing the single bacterial polymerase type with a single polypeptide chain; it is one enzyme type but multimeric. Final Answer: Multisubunit enzymes.

Question 6

Phases of transcription by Escherichia coli RNA polymerase: how is the process classically divided? Choose the correct description.

Accepted Answer

Introduction / Context: Transcription is commonly described in phases to simplify a complex, dynamic pathway. In bacteria, the canonical phases are initiation, elongation, and termination. Given Data / Assumptions: Initiation includes promoter binding, isomerization to open complex, and initial transcript synthesis. Elongation involves processive RNA synthesis and RNA polymerase pausing mechanisms. Termination ends RNA synthesis via intrinsic or rho-dependent pathways. Concept / Approach: The standard tripartite division effectively captures the major mechanistic transitions without inventing a separate 'propagation' phase. Sub-steps exist, but the top-level framework remains three phases. Step-by-Step Solution: List recognized phases: initiation → elongation → termination.Exclude nonstandard labels such as 'propagation.'Select the option describing three phases. Verification / Alternative check: Biochemistry texts, lecture notes, and primary literature adopt this three-phase terminology for bacterial transcription. Why Other Options Are Wrong: Adding 'propagation' is nonstandard. Two-phase and 'none of the above' options do not accurately describe established pedagogy. Common Pitfalls: Confusing sub-steps within initiation (open complex formation) as separate phases. Final Answer: Three phases known as initiation, elongation, and termination.

Question 7

In eukaryotes, which polymerase transcribes the multiple 5S rRNA genes located separately from the main rRNA gene clusters? Pick the correct enzyme.

Accepted Answer

Introduction / Context: Eukaryotic ribosomes are assembled from rRNAs transcribed by different enzymes. The large ribosomal RNA precursor (45S) is made by RNA polymerase I, whereas the 5S rRNA is transcribed separately, typically from distinct chromosomal loci. Given Data / Assumptions: 5S rRNA genes are often organized as tandem arrays independent of the 45S rDNA repeats. RNA polymerase III specializes in small RNAs including tRNAs and 5S rRNA. Concept / Approach: Assign the appropriate transcription machinery based on RNA class: Pol I → 45S rRNA precursor; Pol III → 5S rRNA; Pol II → mRNA. Therefore, the enzyme for 5S rRNA is RNA polymerase III. Step-by-Step Solution: Identify target RNA: 5S rRNA.Map to polymerase: Pol III.Exclude Pol I (28S/18S/5.8S precursor) and Pol II (mRNA). Verification / Alternative check: Promoter analyses show internal control regions (A- and C-boxes) for Pol III transcription of 5S genes, distinct from Pol I promoters. Why Other Options Are Wrong: Pol II makes mRNA; Pol I does not transcribe 5S; DNA polymerase I replicates DNA, not RNA transcription. Common Pitfalls: Assuming all rRNAs come from Pol I; 5S is the major exception handled by Pol III. Final Answer: RNA polymerase III.

Question 8

What precursor is produced when RNA polymerase I transcribes a set of eukaryotic rRNA genes? Select the correct rRNA precursor molecule.

Accepted Answer

Introduction / Context: Ribosome biogenesis in eukaryotes begins with transcription of a large polycistronic rRNA precursor by RNA polymerase I. This precursor is processed to yield multiple mature rRNAs incorporated into ribosomal subunits. Given Data / Assumptions: RNA polymerase I operates primarily in the nucleolus. The initial Pol I transcript (45S pre-rRNA in higher eukaryotes) is processed to 28S, 18S, and 5.8S rRNAs. 5S rRNA is transcribed separately by Pol III. Concept / Approach: Match polymerase to product: Pol I → 45S rRNA precursor; Pol III → 5S rRNA; Pol II → mRNA. Therefore, the correct precursor from Pol I is the 45S pre-rRNA. Step-by-Step Solution: Identify polymerase: RNA polymerase I.Recall product: 45S pre-rRNA.Exclude single mature rRNAs as initial products; they arise via processing. Verification / Alternative check: Pulsed-labeling and processing maps show the 45S transcript is cleaved and modified to produce the mature rRNAs. Why Other Options Are Wrong: Options listing 5S, 18S, or 28S as single direct products misrepresent that these are processed derivatives, not the primary Pol I transcript. 5S is a Pol III product. Common Pitfalls: Equating bacterial 16S/23S nomenclature with eukaryotic processing; eukaryotic large precursor organization differs. Final Answer: A single 45S rRNA precursor molecule.

Question 9

Central dogma focus: RNA is synthesized using a DNA template in a process called what, and which enzyme catalyzes this reaction in cells?

Accepted Answer

Introduction / Context: This question tests a core piece of the central dogma of molecular biology: how information flows from DNA to RNA. It asks you to correctly name the process that uses DNA as a template to make RNA and to identify the enzyme that performs this task. Given Data / Assumptions: Cells contain DNA, RNA, and proteins. Template-directed synthesis follows base-pairing rules (A–U in RNA, A–T in DNA; G–C). Standard cellular processes are considered (no unusual viral enzymes). Concept / Approach: Information typically moves DNA → RNA → protein. The step DNA → RNA is called transcription. The enzyme that catalyzes transcription is RNA polymerase, which binds to promoters, unwinds a short region of DNA, and polymerizes ribonucleotides into an RNA chain using the DNA template strand for base pairing. Step-by-Step Solution: Identify the process that makes RNA from DNA: transcription.Identify the enzyme that synthesizes RNA: RNA polymerase.Match to the option that pairs both correctly.Select: transcription, RNA polymerase. Verification / Alternative check: Replication uses DNA polymerase and produces DNA from DNA. Translation uses ribosomes to build polypeptides from mRNA; there is no “translation polymerase.” Thus the only accurate pairing is transcription with RNA polymerase. Why Other Options Are Wrong: translation, RNA polymerase: translation does not synthesize RNA. transcription, DNA polymerase: DNA polymerase makes DNA, not RNA. replication, DNA polymerase: correct pairing, but wrong process for making RNA. Common Pitfalls: Confusing replication with transcription or assuming all polymerases make all nucleic acids. Each polymerase has specificity. Final Answer: transcription, RNA polymerase

Question 10

Transcription start site vs. start codon: In gene expression, where does transcription initiate relative to the translation start codon (AUG)?

Accepted Answer

Introduction / Context: Understanding the spatial relationship between transcription and translation landmarks is fundamental. The mRNA contains untranslated regions (UTRs) that are transcribed but not translated. This question probes whether you know that transcription begins before the coding region where translation starts. Given Data / Assumptions: Start codon (usually AUG) marks translation initiation. Promoters and transcription start sites (TSS) lie on DNA upstream of the coding sequence. mRNAs have 5′ UTRs that precede the start codon. Concept / Approach: RNA polymerase initiates at the promoter near the TSS, producing a primary transcript that includes the 5′ UTR, the coding sequence, and often a 3′ UTR. Because 5′ UTR sequences lie before the start codon, the TSS is upstream relative to the coding AUG. Therefore, transcription begins upstream from the translation start codon. Step-by-Step Solution: Recall: Promoter → TSS → 5′ UTR → AUG (start codon).Transcription initiates at TSS, not at AUG.Therefore, transcription begins upstream of the start codon. Verification / Alternative check: Mapping cDNA to genomic DNA reveals that the 5′ end of mRNA aligns upstream of the start codon, confirming transcription starts earlier. Why Other Options Are Wrong: At the start codon: that is where translation begins, not transcription. Downstream from the start codon: would omit 5′ UTRs, which contradicts abundant experimental evidence. None of these: incorrect because “upstream” is correct. Common Pitfalls: Conflating “start” of transcription with “start” of translation due to similar terminology. Always distinguish TSS from AUG. Final Answer: upstream from the start codon

Question 11

Bacterial transcription initiation: What is the principal function of the sigma (σ) factor subunit of RNA polymerase?

Accepted Answer

Introduction / Context: In bacteria, the RNA polymerase holoenzyme includes a σ factor that confers promoter specificity. Correctly assigning σ factor function is central to understanding promoter recognition and gene regulation in prokaryotes. Given Data / Assumptions: Bacterial core RNA polymerase (α2ββ′ω) requires σ for promoter binding. “Start sequence” refers to promoter elements such as −35 and −10 boxes. Translational signals (start/stop codons) are distinct from transcriptional signals. Concept / Approach: σ factors recognize and bind promoter DNA, positioning the core enzyme at the correct transcription start site (TSS). After initiation, σ often dissociates, allowing elongation. Different σ factors target distinct promoter classes, enabling adaptive gene expression. Step-by-Step Solution: Identify the relevant stage: initiation of transcription.Assign σ role: promoter recognition and TSS selection.Therefore, σ recognizes transcriptional start sequences. Verification / Alternative check: Mutations in −10/−35 elements reduce σ-dependent binding and initiation, confirming its role in start-site recognition. Why Other Options Are Wrong: Translational start/stop sequences concern ribosomes, not RNA polymerase σ. Transcriptional stop sequences are recognized by rho-dependent or intrinsic terminator mechanisms, not σ. Common Pitfalls: Mixing translation signals with transcription signals due to similar “start/stop” wording. Keep the processes distinct. Final Answer: recognition of the transcriptional start sequence

Transcription and Regulation Questions