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“Master Transcription with Case-Based MCQs: Test Your Understanding!”
1. A previously healthy 8-year-old develops symptoms of fatigue and anemia. Bone marrow biopsy reveals decreased erythropoiesis, and further molecular analysis shows that the transcription of β-globin mRNA is markedly reduced due to a defect in the assembly of the pre-initiation complex. However, RNA polymerase II is intact and enzymatically functional. Which of the following is the most likely defective component?
A. Elongation factor SII
B. RNA polymerase I
C. TATA-binding protein
D. Transcription termination factor
E. U1 small nuclear ribonucleoprotein (snRNP)
Correct Answer: C. TATA-binding protein
Explanation:
The TATA-binding protein is a critical component of the transcription factor complex TFIID in eukaryotes. It binds specifically to the TATA box, a core promoter element found upstream of many genes, including the β-globin gene. TBP serves as a platform for the recruitment of other general transcription factors and RNA polymerase II, facilitating the assembly of the pre-initiation complex (PIC). If TBP is defective, this complex cannot assemble properly, preventing the initiation of transcription—even if RNA polymerase II is present and enzymatically active. This would lead to reduced expression of β-globin and contribute to anemia due to impaired hemoglobin synthesis.
Incorrect Options:
A. Elongation factor SII
This factor (also known as TFIIS) stimulates the RNA cleavage activity of RNA polymerase II and helps resolve transcriptional pausing during the elongation phase. It plays no role in the assembly of the pre-initiation complex. If this factor were defective, initiation would still occur normally, but transcript elongation might be inefficient or error-prone.
B. RNA polymerase I
RNA polymerase I is responsible for transcribing ribosomal RNA (rRNA) genes in the nucleolus, not protein-coding genes like β-globin, which are transcribed by RNA polymerase II. Therefore, a defect in RNA polymerase I would not affect β-globin mRNA transcription.
D. Transcription termination factor
Termination factors help properly end transcription at specific termination sequences. A defect here would result in readthrough transcription, not a failure to initiate. Transcripts may be improperly terminated or longer than normal, but initial transcription would still occur. This is inconsistent with the finding that the transcription of β-globin is markedly reduced.
E. U1 small nuclear ribonucleoprotein (snRNP)
U1 snRNP is involved in RNA splicing, specifically in recognizing the 5′ splice site of introns during the post-transcriptional processing of pre-mRNA. It has no role in transcription initiation or PIC assembly. A defect here would result in improperly spliced mRNA, possibly leading to dysfunctional proteins, but initial transcription levels would remain unaffected.
2. A researcher creates a eukaryotic cell line with a targeted mutation that prevents CTD (carboxy-terminal domain) phosphorylation of RNA polymerase II. Which of the following downstream events will most likely be impaired as a direct consequence?
A. DNA unwinding at the transcription start site
B. Loading of transcription factors at the enhancer region
C. Promoter clearance and transcriptional elongation
D. Polyadenylation of the 3’ mRNA end
E. Termination of RNA synthesis
Correct Answer: C. Promoter clearance and transcriptional elongation
The carboxy-terminal domain (CTD) of RNA polymerase II contains multiple repeats of a heptapeptide sequence (Tyr-Ser-Pro-Thr-Ser-Pro-Ser) that must be phosphorylated to transition the enzyme from transcription initiation to elongation. Specifically, TFIIH phosphorylates the CTD, triggering promoter clearance — the release of RNA polymerase II from the promoter — and enabling productive elongation of the RNA transcript. If CTD phosphorylation is blocked, RNA polymerase II remains stalled at the promoter, and the transcript either does not elongate or is aborted prematurely. This is a direct and essential step in eukaryotic gene transcription.
Incorrect Options:
A. DNA unwinding at the transcription start site
Unwinding of DNA at the transcription start site is part of transcription initiation and is mediated by helicase activity of TFIIH, independent of CTD phosphorylation. This step occurs before promoter clearance and is not directly regulated by CTD status. Therefore, this event can still proceed in the presence of the mutation.
B. Loading of transcription factors at the enhancer region
Enhancer regions are distant regulatory sequences where activator proteins (e.g., transcriptional activators like CREB or steroid hormone receptors) bind. These factors interact with mediator complexes and help recruit or stabilize the general transcription machinery at the core promoter, but their loading is not dependent on the CTD or its phosphorylation. Therefore, enhancer-related activity remains unaffected.
D. Polyadenylation of the 3’ mRNA end
Polyadenylation is a post-transcriptional modification that occurs after elongation and cleavage at the polyadenylation site. While the phosphorylated CTD does play a scaffolding role in recruiting processing factors, polyadenylation depends more on signals present in the mRNA sequence and cleavage and polyadenylation specificity factor (CPSF). It is indirectly affected if transcription fails, but not as an immediate downstream consequence of blocked CTD phosphorylation.
E. Termination of RNA synthesis
Termination is a complex process involving the recognition of polyadenylation signals or termination sequences and the recruitment of exonucleases. While CTD interactions help coordinate this, termination is a late event in transcription. A block in CTD phosphorylation prevents RNA polymerase from even reaching this stage, so termination per se is not directly impaired — it’s precluded by the earlier block in promoter escape.
3. A pharmaceutical company develops a novel compound that specifically inhibits the ability of sigma factors to bind to promoter regions. This inhibition disrupts transcription initiation in a targeted fashion. Which of the following outcomes would be expected in prokaryotic cells but not in eukaryotic cells?
A. Impairment of RNA elongation
B. Loss of transcriptional termination
C. Inability to initiate transcription at core promoters
D. Reduction in splicing efficiency
E. Accumulation of uncapped mRNA in the cytoplasm
Correct Answer: C. Inability to initiate transcription at core promoters
Explanation:
In prokaryotes, sigma (σ) factors are essential subunits of RNA polymerase holoenzyme that enable specific recognition of promoter sequences (e.g., –10 and –35 elements). The sigma factor helps RNA polymerase bind tightly to the promoter and initiate transcription. Without sigma factor binding, the core enzyme cannot locate or stably bind to promoter sequences, leading to a complete failure of transcription initiation.
In eukaryotes, promoter recognition is accomplished by a set of general transcription factors, including TFIID (which contains the TATA-binding protein), not sigma factors. Therefore, this drug would have no effect on eukaryotic transcription initiation, making the consequence prokaryote-specific.
Incorrect Options:
A. Impairment of RNA elongation
Elongation is primarily a function of the core RNA polymerase in both prokaryotes and eukaryotes. Once transcription has been initiated, and the sigma factor dissociates (as it normally does shortly after initiation), elongation proceeds independently. Blocking the sigma factor only affects initiation, not elongation. Therefore, this is not the earliest or most specific defect you would see from sigma factor inhibition.
B. Loss of transcriptional termination
Termination involves separate factors in both systems: Rho-dependent and Rho-independent mechanisms in prokaryotes and cleavage/polyadenylation signals in eukaryotes. Sigma factor function is limited to the initiation phase and does not directly impact how or when transcription stops.
D. Reduction in splicing efficiency
Splicing is a eukaryotic-specific process that removes introns from pre-mRNA via the spliceosome, involving snRNPs like U1 and U2. Prokaryotic genes lack introns and, therefore, do not require splicing at all. Thus, inhibition of sigma factors in prokaryotes would have no relationship to splicing — and this is not a relevant downstream consequence.
E. Accumulation of uncapped mRNA in the cytoplasm
The 5’ methylguanosine cap is a post-transcriptional modification unique to eukaryotic mRNAs. It helps with stability, nuclear export, and ribosomal recognition. Prokaryotes do not cap their mRNAs at all. Therefore, this phenomenon can occur in certain eukaryotic disorders or processing defects, but it is irrelevant to prokaryotes and to sigma factor inhibition.
4. A 27-year-old patient has an inherited defect in an enzyme responsible for methylating the 5’ cap of mRNA transcripts. Although transcription and splicing occur normally, protein synthesis is significantly impaired. Which of the following best explains this clinical observation?
A. Reduced recognition by ribosomes
B. Impaired transcription factor binding
C. Failure to splice out introns
D. Defective transcription termination
E. Unwinding of RNA duplexes
Correct Answer: A. Reduced recognition by ribosomes
Explanation:
In eukaryotic cells, the 5’ cap structure consists of a 7-methylguanosine added to the first nucleotide of the mRNA via a 5’-5’ triphosphate linkage. This cap is methylated by guanine-7-methyltransferase. The methylated cap is critical for ribosome binding during translation initiation. It facilitates interaction with eukaryotic initiation factors (especially eIF4E) that help recruit the small ribosomal subunit to the mRNA. If methylation fails, the mRNA cap is not recognized properly, and ribosomes cannot initiate translation efficiently, leading to the global suppression of protein synthesis despite intact upstream processes like transcription and splicing.
Incorrect Options:
B. Impaired transcription factor binding
Transcription factors bind upstream DNA regulatory regions (e.g., promoters, enhancers) to regulate the initiation of transcription. This process occurs before the addition of the 5’ cap and is unrelated to mRNA capping or its methylation. In this case, transcription is reported to be normal, making this explanation inconsistent with the findings.
C. Failure to splice out introns
Splicing is performed by the spliceosome, which recognizes conserved sequences at intron-exon boundaries. Although some spliceosomal components interact with the CTD of RNA polymerase II (which also helps recruit capping enzymes), splicing itself does not require cap methylation. In the clinical scenario, splicing is stated to be intact, ruling this out.
D. Defective transcription termination
Termination involves signals within the gene (e.g., polyadenylation sequences) and specialized factors like CPSF and XRN2. While some of these factors bind to the CTD, termination is not directly regulated by cap methylation. Moreover, defective termination would likely result in elongated or unstable transcripts, not impaired translation per se.
E. Unwinding of RNA duplexes
RNA duplex unwinding is performed by RNA helicases, which play roles in splicing, export, and ribosome scanning. However, these helicases act on RNA-RNA interactions or secondary structures, not on the 5’ cap. This process is not affected by the methylation status of the cap and does not directly explain impaired initiation of translation.
5. Which of the following components is required for the initiation of transcription in prokaryotes?5
A. Rho factor
B . Sigma factor
C. TATA-binding protein (TBP)
D. RNA Polymerase III
E. Pribnow box
Correct Answer: B) Sigma factor
Explanation:
In prokaryotes (such as E. coli), the sigma factor is a crucial subunit of the RNA polymerase holoenzyme. It enables specific recognition of promoter sequences, particularly the –10 (Pribnow box) and –35 regions upstream of the transcription start site. The sigma factor facilitates the binding of RNA polymerase to DNA and the melting (unwinding) of the DNA duplex to allow transcription to begin. After initiation, the sigma factor often dissociates, and the core enzyme continues elongation. Therefore, the sigma factor is essential for the initiation of transcription in prokaryotes.
Incorrect Options:
A. Rho factor
Rho is a transcription termination factor, not involved in initiation. It binds to the nascent RNA transcript and, in a Rho-dependent termination mechanism, helps dislodge the RNA polymerase from the DNA template. It has no role in promoter recognition or transcription initiation.
C. TATA-binding protein (TBP)
TBP is a component of the general transcription factor TFIID in eukaryotic transcription, not prokaryotic. TBP binds to the TATA box in eukaryotic promoters to initiate pre-initiation complex formation. Prokaryotes do not use TBP or TATA boxes; they rely on sigma factors and different promoter sequences.
D. RNA Polymerase III
This is one of the three eukaryotic RNA polymerases (I, II, and III). RNA Polymerase III transcribes small RNAs like tRNAs and 5S rRNA. It is not present in prokaryotes, which use a single type of RNA polymerase. Thus, this option is both irrelevant to prokaryotes and incorrect in the context of transcription initiation.
E. Pribnow box
The Pribnow box is a DNA sequence located around the –10 position in prokaryotic promoters. It is not a component or factor — it’s a sequence element, not a protein or enzymatic component that actively initiates transcription. The sigma factor is the required protein that binds to the Pribnow box, enabling transcription initiation.
6. In prokaryotic transcription, which region of the promoter does the sigma factor recognize and bind to?
A. –25 TATA box
B. Silencer region
C. –10 and –35 regions
D. Enhancer region
E. Initiator site (+1)
Correct Answer: C. –10 and –35 regions
Explanation:
In prokaryotes, sigma factors guide RNA polymerase to specific promoter sequences. These promoters have two highly conserved elements:
• The –35 region (consensus: TTGACA)
• The –10 region (also called the Pribnow box, consensus: TATAAT)
The sigma factor binds to both regions, properly positioning RNA polymerase for transcription initiation at the +1 site. This interaction is critical for specific and efficient promoter recognition in prokaryotic cells.
Incorrect Options:
A. –25 TATA box
This refers to a eukaryotic promoter element, found around 25 base pairs upstream of the transcription start site, recognized by the TATA-binding protein (TBP), a component of TFIID. It is not present in prokaryotic promoters and is not recognized by sigma factors.
B. Silencer region
Silencers are eukaryotic regulatory DNA elements that bind repressor proteins to inhibit transcription. Prokaryotic transcription does not involve silencers in the same way, and sigma factors do not bind to these types of sequences.
D. Enhancer region
Enhancers are distal regulatory elements in eukaryotes that increase transcription efficiency by interacting with promoter-bound complexes via looping. Sigma factors do not interact with enhancers, and prokaryotes generally do not use enhancers in their regulatory architecture.
E. Initiator site (+1)
The +1 site marks the start of transcription — the first nucleotide transcribed into RNA. While it is where RNA synthesis begins, sigma factors do not bind at +1. Instead, they help align RNA polymerase to the correct position by recognizing upstream –10 and –35 regions.
7. A 24-year-old medical student is diagnosed with a genetic disorder affecting RNA Polymerase II function. Which of the following cellular processes will be directly affected?
A. rRNA synthesis
B. mRNA synthesis
C. tRNA synthesis
D. DNA replication
E. Ribosome assembly
Correct Answer: B. mRNA synthesis
Explanation:
RNA Polymerase II is the enzyme in eukaryotic cells responsible for the synthesis of messenger RNA (mRNA). It transcribes protein-coding genes into precursor mRNA (pre-mRNA), which is then processed (capping, splicing, polyadenylation) to form mature mRNA. A defect in RNA Polymerase II function would directly impair mRNA synthesis, severely affecting gene expression and protein production across nearly all cellular processes.
Incorrect Options:
A. rRNA synthesis
This is carried out by RNA Polymerase I, which transcribes the genes for the large ribosomal RNA precursor (45S rRNA), later processed into 18S, 5.8S, and 28S rRNAs. A defect in RNA Polymerase II does not affect rRNA synthesis.
C. tRNA synthesis
RNA Polymerase III is responsible for transcribing tRNA genes, 5S rRNA, and other small RNAs. Therefore, a defect in RNA Polymerase II does not impair tRNA production.
D. DNA replication
DNA replication is carried out by DNA polymerases, not RNA polymerases. While mRNA synthesis indirectly influences the availability of replication proteins, replication itself is not directly dependent on RNA Polymerase II.
E. Ribosome assembly
Ribosome assembly involves the synthesis of rRNA (via RNA Polymerase I and III), ribosomal proteins (via mRNA translation), and nucleolar organization. While mRNA synthesis is indirectly related (since mRNA encodes ribosomal proteins), RNA Polymerase II is not directly responsible for assembling ribosomes.
8. A researcher studies a genetic mutation affecting nucleolar transcription. The patient presents with hematological abnormalities due to impaired ribosome production. Which RNA polymerase is most likely affected?
A. RNA Polymerase I
B. RNA Polymerase II
C. RNA Polymerase III
D. Telomerase
E. DNA Polymerase
Correct Answer: A. RNA Polymerase I
Explanation:
RNA Polymerase I is located in the nucleolus and is responsible for transcribing a single large precursor rRNA (45S), which is processed into 18S, 5.8S, and 28S rRNA—all essential components of the ribosome. A defect in RNA Polymerase I disrupts ribosomal RNA synthesis, leading to impaired ribosome biogenesis, which can manifest as hematologic abnormalities (e.g., due to defective protein synthesis in rapidly dividing cells like hematopoietic precursors).
Incorrect Options:
B. RNA Polymerase II
This enzyme transcribes mRNA and certain small nuclear RNAs (snRNAs). It is located in the nucleoplasm, not the nucleolus. Although vital for gene expression, it is not directly responsible for ribosome production.
C. RNA Polymerase III
This enzyme synthesizes tRNAs, 5S rRNA, and some small RNAs. While it contributes to ribosome function by producing 5S rRNA, it is not the primary enzyme for rRNA transcription and does not operate in the nucleolus.
D. Telomerase
Telomerase is a reverse transcriptase enzyme that extends telomeres in stem and germ cells. It plays a role in maintaining chromosome ends but has no role in rRNA synthesis or ribosome production.
E. DNA Polymerase
DNA polymerases synthesize DNA during replication, not RNA during transcription. They are not involved in nucleolar function or ribosome biogenesis.
9. A scientist is developing a drug that inhibits the transcription of 5S rRNA and tRNA. Which enzyme should the drug target?
A. RNA Polymerase I
B. RNA Polymerase II
C. RNA Polymerase III
D. RNA Helicase
E. RNA Ligase
Correct Answer: C. RNA Polymerase III
Explanation:
RNA Polymerase III is the eukaryotic RNA polymerase responsible for the transcription of tRNA, 5S rRNA, and other small RNAs (such as U6 snRNA). It is located in the nucleoplasm (not nucleolus) and is essential for synthesizing components needed for protein translation and ribosomal structure. A drug targeting RNA Polymerase III would specifically block the synthesis of 5S rRNA and tRNAs — exactly as described in the question.
Incorrect Options:
A. RNA Polymerase I
This enzyme is found in the nucleolus and transcribes the 45S rRNA precursor, which is cleaved into 18S, 5.8S, and 28S rRNAs — not 5S rRNA. Therefore, it would not affect 5S rRNA or tRNA synthesis.
B. RNA Polymerase II
This enzyme synthesizes mRNA, snRNA, and microRNAs. It does not produce 5S rRNA or tRNA. While essential for gene expression, it is not the correct target for inhibiting 5S rRNA and tRNA synthesis.
D. RNA Helicase
RNA helicases are involved in unwinding RNA secondary structures during RNA processing and translation, not in transcription. They are not RNA polymerases and do not initiate or elongate RNA transcripts.
E. RNA Ligase
RNA ligases are involved in RNA repair and splicing in some systems (especially in certain organisms), but they play no role in the transcription of rRNA or tRNA. They are not targets for inhibiting RNA synthesis.
10. A patient presents with severe liver damage after consuming Amanita phalloides (death cap mushroom). The toxin α-amanitin primarily inhibits which of the following enzymes?
A. RNA Polymerase I
B. RNA Polymerase II
C. RNA Polymerase III
D. DNA Polymerase
E. Reverse Transcriptase
Correct Answer: B. RNA Polymerase II
Explanation:
α-amanitin, a potent toxin found in Amanita phalloides, specifically binds to and inhibits RNA Polymerase II in eukaryotic cells. RNA Polymerase II is responsible for transcribing mRNA and several small nuclear RNAs. Inhibition of this enzyme leads to the cessation of mRNA synthesis, which halts protein production and results in cell death, especially in metabolically active tissues like the liver. This is why hepatotoxicity is a hallmark of death cap mushroom poisoning.
Incorrect Options:
A. RNA Polymerase I
This enzyme transcribes ribosomal RNA (rRNA) in the nucleolus, not mRNA. It is not the primary target of α-amanitin.
C. RNA Polymerase III
RNA Polymerase III transcribes tRNA and 5S rRNA, and it is only minimally affected by α-amanitin at very high concentrations.
D. DNA Polymerase
DNA polymerase is involved in DNA replication, not RNA transcription. It is not inhibited by α-amanitin.
E. Reverse Transcriptase
Reverse transcriptase is an enzyme used by retroviruses (like HIV) to convert RNA into DNA. It is not present in human cells under normal conditions and is not affected by α-amanitin.
11. A patient has a genetic mutation affecting TFIID, a transcription factor required for RNA polymerase function. This mutation most likely disrupts which of the following?
A. DNA replication at replication forks
B. Binding of RNA Polymerase to the promoter
C. Splicing of pre-mRNA
D. Ribosomal assembly
E. Protein degradation
Correct Answer: B) Binding of RNA Polymerase to the promoter
Explanation:
TFIID is a general transcription factor complex critical for transcription initiation in eukaryotic cells. It contains the TATA-binding protein (TBP), which recognizes and binds to the TATA box in the promoter region of genes transcribed by RNA Polymerase II. Once TFIID is bound, it recruits other general transcription factors and RNA Polymerase II to form the pre-initiation complex. A mutation affecting TFIID would, therefore, impair the initial binding of RNA Polymerase II to the promoter, halting transcription initiation.
Incorrect Options:
A. DNA replication at replication forks
This process involves DNA polymerases, helicases, primases, and ligases, not transcription factors like TFIID. TFIID is not involved in DNA replication.
C. Splicing of pre-mRNA
Splicing is carried out by the spliceosome, a complex made of snRNPs and proteins. While transcription and splicing are coordinated, TFIID’s role is limited to initiating transcription, not RNA processing.
D. Ribosomal assembly
Ribosome assembly involves rRNA (transcribed by RNA Polymerase I and III) and ribosomal proteins. TFIID functions in mRNA gene transcription via RNA Polymerase II and is unrelated to ribosome biogenesis.
E. Protein degradation
Protein degradation is mediated by the ubiquitin-proteasome system and lysosomal pathways. It has no connection with TFIID, which acts in transcription initiation, not protein turnover.
12. A mutation in a gene coding for small nuclear RNAs (snRNAs). Which of the following processes is most likely disrupted?
A. Addition of the 5’ cap
B. Polyadenylation at the 3’ end
C. Removal of introns from pre-mRNA
D. Ribosome assembly
E. DNA replication
Correct Answer: C. Removal of introns from pre-mRNA
Explanation:
Small nuclear RNAs (snRNAs) are essential components of the spliceosome, the molecular complex responsible for splicing — the removal of introns from precursor mRNA (pre-mRNA) transcripts in eukaryotic cells. snRNAs such as U1, U2, U4, U5, and U6 pair with specific sequences at the exon-intron boundaries and catalyze the precise removal of introns.
A mutation affecting snRNAs would compromise spliceosome function, leading to the retention of introns and nonfunctional or misfolded proteins, thus directly impairing pre-mRNA processing.
Incorrect Options:
A. Addition of the 5’ cap
This occurs co-transcriptionally and involves capping enzymes (e.g., guanylyltransferase and methyltransferase), not snRNAs. Cap addition occurs before splicing begins.
B. Polyadenylation at the 3’ end
This process is mediated by cleavage and polyadenylation specificity factor (CPSF) and poly(A) polymerase, not by snRNAs or the spliceosome. It adds a poly-A tail to stabilize the mRNA and aid in nuclear export.
D. Ribosome assembly
This involves rRNA transcription, processing, and association with ribosomal proteins, largely occurring in the nucleolus. snRNAs are not involved in ribosome biogenesis.
E. DNA replication
DNA replication is carried out by DNA polymerases, helicases, and other replication proteins. It is entirely unrelated to snRNAs, which act after transcription during RNA processing.
13. A patient has a mutation in the splice donor site (5’ GU sequence) of a gene involved in collagen production. What is the most likely consequence?
A. Increased mRNA stability
B) Exon skipping or retention of introns
C) Enhanced translation of the mRNA
D) Overproduction of functional protein
E) Increased DNA replication rate
Correct Answer: B) Exon skipping or retention of introns
Explanation:
The splice donor site (commonly 5’ GU sequence) is located at the 5’ end of introns and is essential for accurate splicing of pre-mRNA. The spliceosome (made of snRNPs like U1) recognizes this conserved sequence to initiate the removal of introns.
A mutation in the 5’ GU donor site can prevent recognition of the splice site, resulting in:
• Exon skipping (the exon before the intron is skipped)
• Intron retention (the intron is not removed)
Both outcomes lead to abnormal or truncated proteins that are often nonfunctional, which is particularly significant in collagen disorders (e.g., Ehlers-Danlos syndrome, osteogenesis imperfecta).
Incorrect Options:
A. Increased mRNA stability
Disruption of the splice donor site usually leads to defective or unstable mRNA, often targeted for nonsense-mediated decay, not increased stability.
C. Enhanced translation of the mRNA
Improperly spliced mRNA is less likely to be efficiently translated and may encode dysfunctional proteins. Translation is often reduced, not enhanced.
D. Overproduction of functional protein
The mutation leads to abnormal mRNA and defective protein, not the overproduction of functional protein.
E. Increased DNA replication rate
This mutation affects RNA splicing, a post-transcriptional process. DNA replication is unrelated and would not be affected by a splice site mutation.
14. Which of the following best explains how a single gene can code for multiple protein isoforms?
A. Post-translational modifications
B. Alternative splicing
C. DNA recombination
D. RNA editing
E. Frameshift mutations
Correct Answer: B. Alternative splicing
Explanation:
Alternative splicing is a regulated process during gene expression in which different combinations of exons are joined together from a single pre-mRNA transcript, allowing the production of multiple distinct mRNA variants (and thus protein isoforms) from one gene. This mechanism greatly increases the diversity of the proteome without requiring additional genes. It’s especially common in complex tissues like the brain and muscles.
Incorrect Options:
A. Post-translational modifications
These changes (like phosphorylation, glycosylation, or ubiquitination) alter the function or localization of a protein after translation but do not generate different isoforms from the same transcript.
C. DNA recombination
While this can lead to genetic variation (e.g., during meiosis or in immunoglobulin genes), it is not the primary mechanism by which one gene creates multiple protein products under normal circumstances.
D. RNA editing
RNA editing involves specific nucleotide changes in an mRNA molecule (e.g., deamination of adenosine to inosine) and can affect protein sequence, but it is less common and does not explain most isoform diversity.
E. Frameshift mutations
These are errors caused by insertions or deletions that disrupt the reading frame of mRNA. They typically result in nonfunctional or truncated proteins, not functional isoforms. This is a pathologic rather than physiological mechanism.
15. A 56-year-old man is being treated for active pulmonary tuberculosis. His treatment regimen includes isoniazid, pyrazinamide, ethambutol, and rifampicin. After two weeks, his symptoms improved, and follow-up sputum cultures showed a reduced Mycobacterium tuberculosis load. Rifampicin exerts its therapeutic effect by targeting which of the following processes?
A. Inhibition of RNA polymerase II–II-mediated mRNA synthesis in host cells
B. Inhibition of bacterial RNA polymerase–mediated transcription initiation
C. Blocking ribosomal translocation during bacterial protein synthesis
D. Inhibiting topoisomerase activity during transcription elongation
E. Disruption of 5’ capping of pre-mRNA in infected macrophages
Correct Answer: B) Inhibition of bacterial RNA polymerase–mediated transcription initiation
Explanation:
Rifampicin specifically binds to the β-subunit of bacterial DNA-dependent RNA polymerase, inhibiting the initiation of RNA synthesis. This prevents the transcription of bacterial genes and effectively halts mRNA production, leading to bactericidal activity. Its selectivity for prokaryotic RNA polymerase makes it effective in targeting Mycobacterium tuberculosis without inhibiting eukaryotic RNA polymerases.
Incorrect Options:
A. Inhibition of RNA polymerase II–mediated mRNA synthesis in host cells
This describes the mechanism of α-amanitin, not rifampicin. Eukaryotic RNA polymerase II is not affected by rifampicin.
C. Blocking ribosomal translocation during bacterial protein synthesis
This is the action of macrolides (like erythromycin), not rifampicin. Rifampicin acts at the transcriptional level, not translation.
D. Inhibiting topoisomerase activity during transcription elongation
This is the mechanism of quinolones (e.g., ciprofloxacin), which inhibit DNA gyrase or topoisomerase IV, affecting DNA replication and transcription indirectly. Rifampicin acts directly on RNA polymerase.
E. Disruption of 5’ capping of pre-mRNA in infected macrophages
This is a post-transcriptional modification that occurs in eukaryotic cells. Rifampicin does not affect mRNA processing in host cells.
