Transcription and RNA Processing (1B)
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MCAT Biological and Biochemical Foundations of Living Systems › Transcription and RNA Processing (1B)
A team uses an in vitro transcription system with purified RNA polymerase II and general transcription factors to transcribe a DNA template encoding a stress-response enzyme. They add either (i) all four rNTPs, (ii) all four dNTPs, or (iii) all four rNTPs plus an RNase that degrades single-stranded RNA. Which condition is expected to yield the highest amount of full-length RNA transcript?
All four dNTPs plus RNase, because RNase removes RNA primers and enhances elongation by Pol II.
All four rNTPs plus RNase, because degrading nascent RNA prevents polymerase stalling and increases yield.
All four dNTPs, because RNA polymerase II can substitute dNTPs to synthesize RNA with deoxyribose.
All four rNTPs, because RNA polymerase II requires ribonucleotides and the transcript is not degraded during synthesis.
Explanation
This question tests fundamental knowledge of RNA polymerase II substrate requirements and the chemical differences between RNA and DNA synthesis. RNA polymerase II exclusively uses ribonucleoside triphosphates (rNTPs) as substrates to synthesize RNA, incorporating the 2'-OH group that distinguishes RNA from DNA, and cannot utilize deoxyribonucleoside triphosphates (dNTPs) which lack this hydroxyl group. In an in vitro transcription system with all necessary components, providing all four rNTPs allows continuous RNA synthesis without degradation, yielding the maximum amount of full-length transcript. The presence of general transcription factors ensures proper initiation and elongation. Choice A incorrectly suggests Pol II can use dNTPs, which would be like asking a chef to bake bread with metal instead of flour - the enzyme's active site specifically requires the 2'-OH for catalysis. Adding RNase (choice C) would degrade the RNA product as it's synthesized, reducing yield rather than increasing it. Remember that RNA polymerases are absolutely specific for rNTPs due to active site geometry and catalytic mechanism requirements.
A lab compares two alleles of a metabolic regulator gene differing by a single nucleotide at the 5' splice site of intron 1. Allele M weakens complementarity to U1 snRNA. Cells carrying Allele M show increased intron 1 retention and reduced protein levels, despite similar transcription rates.
Which intervention would most directly rescue mature mRNA production from Allele M?
Mutating the start codon so that translation does not compete with splicing in the nucleus.
Enhancing U1 snRNP recognition of the mutated 5' splice site (e.g., by compensatory U1 snRNA) to restore spliceosome assembly.
Inhibiting RNA polymerase II initiation to reduce the amount of unspliced RNA produced.
Blocking ribosomal translocation to prevent intron retention during translation.
Explanation
This question tests rescue strategies for splicing defects. The weakened 5' splice site impairs U1 recognition, causing retention; compensatory U1 restores assembly. Allele M shows retention and reduced protein despite normal transcription. Choice D is correct as enhancing U1 recognition directly fixes the defect. Choice B fails by suggesting initiation inhibition, misunderstanding that reducing unspliced RNA doesn't rescue maturation. For similar issues, identify splice site weaknesses and propose compensatory interactions. Assess mRNA and protein to confirm rescue.
In a hepatocyte-derived cell line, investigators test whether a nutrient-sensitive transcription factor (TF-X) coordinates expression of a glycolytic enzyme gene. Cells are transfected with a reporter plasmid in which luciferase is driven by the enzyme’s promoter containing a TF-X binding site. TF-X is activated by dephosphorylation during high glucose. Four conditions are compared for 6 hours: control, high glucose, high glucose + TF-X siRNA, and high glucose + an inhibitor of spliceosome assembly (blocks U1 snRNP recruitment). Luciferase activity and the fraction of spliced luciferase mRNA (measured by RT-qPCR across the exon–exon junction) are recorded.
Which conclusion is most consistent with the data pattern expected if TF-X primarily increases transcription initiation rather than RNA processing efficiency?
High glucose increases luciferase activity only when spliceosome assembly is inhibited, indicating TF-X acts post-transcriptionally on ribosome loading.
High glucose increases luciferase activity without changing the spliced mRNA fraction, and TF-X siRNA abolishes the activity increase while spliceosome inhibition lowers the spliced fraction but does not mimic TF-X siRNA.
High glucose increases the spliced mRNA fraction with minimal change in luciferase activity, and TF-X siRNA reverses only the splicing change.
High glucose decreases luciferase activity while increasing the spliced mRNA fraction, indicating TF-X represses RNA polymerase II elongation.
Explanation
This question tests understanding of transcription initiation versus RNA processing in gene expression regulation. Transcription initiation involves recruitment of RNA polymerase II to the promoter, while RNA processing includes splicing of pre-mRNA to form mature mRNA. In this scenario, TF-X is activated by high glucose and binds the promoter, with experiments comparing luciferase activity and spliced mRNA fractions under various conditions. Choice D is correct because if TF-X primarily enhances initiation, high glucose increases activity without altering splicing efficiency, siRNA blocks the initiation boost, and spliceosome inhibition affects splicing but not the initiation-driven activity. A common misunderstanding is that transcription factors directly affect splicing, as in choice B, but here siRNA does not reverse splicing changes since TF-X acts at initiation. To apply this reasoning elsewhere, check if perturbations affect total transcript levels versus processed fractions. Additionally, distinguish initiation effects by seeing if nascent RNA or activity changes independently of processing inhibitors.
A mechanistic description focuses on 3' end processing of an mRNA encoding an insulin-responsive transporter. A point mutation disrupts the AAUAAA polyadenylation signal but leaves the coding region intact. In mutant cells, nuclear RNA accumulates as a longer transcript that extends beyond the normal 3' end, and cytosolic mRNA levels drop.
Which outcome would be expected from this mutation?
Enhanced splicing at upstream introns compensates for the missing poly(A) signal, increasing cytosolic mRNA levels.
Increased cleavage at the 5' cap causes loss of the start codon and reduced translation, while nuclear RNA length remains unchanged.
No effect on cytosolic mRNA because polyadenylation occurs only in prokaryotes and is not required for export.
Reduced cleavage and polyadenylation lead to inefficient transcript termination and decreased mRNA export/stability, lowering cytosolic mRNA.
Explanation
This question addresses 3' end processing's role in mRNA maturation and stability. The polyadenylation signal enables cleavage and poly(D) addition, facilitating termination and export; mutation causes read-through and reduced cytosolic mRNA. The vignette describes longer nuclear RNA and dropped cytosolic levels due to the mutation. Choice D is correct as it links impaired processing to termination defects and instability. Choice C is wrong, assuming splicing compensation, misunderstanding 3' processing independence. To reason alike, compare nuclear versus cytosolic RNA upon 3' mutations. Check transcript lengths to detect read-through.
A lab examines why a glycolysis gene shows reduced protein despite normal levels of mature mRNA. They discover a mutation that creates a new 5' splice site within an exon, causing partial exon truncation in a subset of transcripts. The truncated mRNA lacks the normal start codon.
Which outcome is most consistent with this RNA processing change?
Protein levels decrease because aberrant splicing can remove the start codon, reducing translation even if total mature mRNA levels appear normal.
Protein levels are unchanged because start codons are added during polyadenylation at the 3' end.
Nascent transcription must decrease because splicing errors directly inhibit RNA polymerase binding at the promoter.
Protein levels increase because shorter mRNAs are always translated more efficiently due to faster RNA polymerase movement.
Explanation
This question probes consequences of aberrant splicing on translation. The new splice site truncates the exon, removing the start codon in some transcripts, reducing protein despite normal mRNA levels. The vignette highlights protein reduction with splicing alteration. Choice D is correct as it links splicing error to lost start codon and decreased translation. Choice B fails by assuming shorter mRNAs increase efficiency, a misunderstanding. In similar cases, check for coding disruptions in spliced variants. Compare mRNA and protein to detect translation impacts.
Researchers test whether a transcription factor (TF-M) that responds to amino acid availability activates transcription of an amino acid transporter gene by recruiting Mediator. In cells expressing TF-M, ChIP shows increased Mediator occupancy at the promoter and increased RNA polymerase II at the transcription start site. A TF-M mutant that binds DNA but cannot interact with Mediator shows normal DNA binding but low transcription.
Which statement best describes TF-M’s role?
TF-M activates transcription by increasing peptide bond formation on translating ribosomes, which feeds back to the promoter.
TF-M activates transcription by facilitating assembly of the preinitiation complex via Mediator recruitment, increasing initiation frequency.
TF-M activates transcription by binding the poly(A) signal and preventing transcript cleavage, thereby increasing mRNA length.
TF-M activates transcription by directly catalyzing intron removal from the transporter pre-mRNA.
Explanation
This question evaluates transcription factor mechanisms via Mediator. TF-M recruits Mediator to assemble the preinitiation complex, increasing initiation; mutant fails despite DNA binding. ChIP shows increased occupancy with wild-type TF-M. Choice A is correct as it describes Mediator-dependent activation. Choice B is wrong, attributing to splicing, confusing with post-transcriptional roles. For application, use ChIP and mutants to dissect recruitment. Monitor polymerase occupancy for initiation effects.
An experiment tests the effect of inhibiting RNA polymerase II CTD Ser5 phosphorylation (required early in transcription) on expression of a detoxification enzyme gene. After inhibitor treatment, total nascent RNA from the gene decreases, and the remaining transcripts show reduced 5' capping efficiency.
Which additional change would be expected as a direct consequence of reduced Ser5 phosphorylation?
Decreased recruitment of the capping machinery to early transcripts, increasing susceptibility to nuclear degradation and reducing productive mRNA formation.
Increased polyadenylation at the 5' end, which compensates for missing caps.
Enhanced U2 snRNP binding to branch points genome-wide because Ser5 phosphorylation blocks splicing.
Increased recruitment of ribosomes to the nucleus, enhancing co-transcriptional translation.
Explanation
This question examines CTD phosphorylation's role in co-transcriptional capping. Ser5 phosphorylation recruits capping enzymes; inhibition reduces capping, leading to degradation and lower mRNA. The vignette shows decreased nascent RNA and capping efficiency. Choice A is correct as it predicts reduced stability and export from poor capping. Choice C is incorrect, assuming splicing enhancement, confusing phosphorylation functions. To apply, link CTD marks to processing steps. Assay capping and stability upon perturbations.
Researchers compare transcription of a peroxisomal beta-oxidation enzyme gene in two conditions: high fatty acids vs low fatty acids. A ligand-activated nuclear receptor (NR) binds an enhancer upstream. When NR is activated, enhancer RNA (eRNA) production increases and promoter–enhancer looping (measured by 3C) increases. Mature mRNA increases.
Which inference is most consistent with these observations?
NR activation likely increases mature mRNA by converting introns into exons through RNA editing at the enhancer.
NR activation likely increases mature mRNA by directly improving ribosomal proofreading of codon–anticodon pairing.
NR activation likely increases transcription initiation by promoting enhancer–promoter communication and recruitment of transcriptional machinery.
NR activation likely decreases initiation but increases mRNA because eRNAs are translated into transcription factors.
Explanation
This question tests enhancer-mediated transcriptional activation. NR activation increases eRNA and looping, facilitating initiation and mature mRNA increase. High fatty acids induce the changes via NR. Choice A is correct as it infers enhanced initiation through enhancer-promoter interactions. Choice D fails by suggesting decreased initiation, contradicting mRNA increase. For parallels, measure eRNA and looping for activation evidence. Compare conditions to infer regulatory mechanisms.
A mechanistic description concerns nonsense-mediated decay (NMD) triggered by RNA processing. An alternatively spliced isoform of an ATP synthase assembly factor includes an exon that introduces a premature termination codon upstream of the final exon–exon junction. Cells expressing this isoform show reduced steady-state mRNA despite normal transcription rate.
Which outcome would be expected if exon junction complex (EJC)-dependent NMD is inhibited?
Steady-state levels of the premature-stop isoform increase because the transcript is less efficiently degraded after export.
Transcription initiation decreases because NMD inhibition blocks RNA polymerase II recruitment.
Protein levels increase because NMD inhibition converts the premature stop codon into a start codon.
Splicing of the premature-stop isoform is prevented because EJCs are required to remove introns.
Explanation
This question assesses NMD's role in mRNA quality control. The premature stop triggers EJC-dependent NMD, reducing mRNA; inhibition stabilizes the isoform. The vignette notes low steady-state despite normal transcription. Choice A is correct as NMD block increases the isoform's levels. Choice C is wrong, assuming EJCs required for splicing, a misconception. In similar scenarios, inhibit NMD and monitor isoform abundance. Distinguish transcription from degradation effects.
A lab studies how inhibition of the U1 snRNP affects expression of a mitochondrial import receptor gene with multiple introns. After treatment, total nuclear RNA from the gene increases slightly, but mature cytosolic mRNA decreases sharply. RT-qPCR shows widespread intron retention.
Which outcome would be expected if U1 snRNP inhibition is the primary perturbation?
Enhanced polyadenylation at internal sites increases mature mRNA because U1 snRNP normally prevents 3' end formation.
Reduced spliceosome assembly at 5' splice sites leads to intron retention and decreased mature mRNA export, even if transcription continues.
Increased RNA polymerase II initiation occurs because U1 snRNP normally blocks promoter binding.
Increased translation initiation increases intron retention because ribosomes stall on introns in the cytosol.
Explanation
This question tests understanding of RNA processing, specifically the role of snRNPs in splicing during eukaryotic mRNA maturation. U1 snRNP is essential for recognizing the 5' splice site and initiating spliceosome assembly, which removes introns from pre-mRNA to produce mature mRNA. In this scenario, inhibition of U1 snRNP disrupts splicing, leading to intron retention in nuclear transcripts and reduced export of mature mRNA to the cytosol, while total nuclear RNA accumulates slightly due to ongoing transcription. The correct answer, A, logically follows because impaired spliceosome assembly directly causes the observed intron retention and decreased cytosolic mRNA, aligning with the primary function of U1 snRNP in splicing. A common misunderstanding addressed in distractor D is that U1 snRNP's role in preventing premature polyadenylation would increase mature mRNA upon inhibition, but it actually leads to aberrant transcripts and reduced full-length mRNA. To apply this reasoning elsewhere, check if the perturbation affects splicing machinery by examining intron retention via RT-qPCR. Additionally, verify mRNA export by comparing nuclear and cytosolic levels, as unspliced transcripts are typically retained in the nucleus.