Q.1 What is the primary role of RNA polymerase during transcription?
Synthesizing DNA from an RNA template
Unwinding the DNA double helix
Catalyzing the synthesis of a complementary RNA strand from DNA
Ligating DNA fragments together
Explanation - RNA polymerase reads the DNA template strand and adds ribonucleotides to form a messenger RNA (mRNA) molecule.
Correct answer is: Catalyzing the synthesis of a complementary RNA strand from DNA
Q.2 In prokaryotes, the promoter region typically contains which conserved DNA sequence upstream of the transcription start site?
TATA box
Pribnow box (TATAAT)
CAAT box
GC-rich enhancer
Explanation - The -10 region, known as the Pribnow box, has the consensus sequence TATAAT and is recognized by the sigma factor of RNA polymerase.
Correct answer is: Pribnow box (TATAAT)
Q.3 Which of the following RNA molecules is involved in the removal of introns during RNA processing?
tRNA
rRNA
snRNA
miRNA
Explanation - Small nuclear RNAs (snRNAs) are core components of the spliceosome, the complex that excises introns from pre‑mRNA.
Correct answer is: snRNA
Q.4 During transcription, the DNA strand that is read by RNA polymerase is called the:
Coding strand
Template strand
Leading strand
Lagging strand
Explanation - RNA polymerase synthesizes RNA complementary to the template (or antisense) strand, while the coding strand has the same sequence as the RNA (except T→U).
Correct answer is: Template strand
Q.5 Which of the following statements about the TATA box is correct?
It is located downstream of the transcription start site.
It is recognized by the sigma factor in prokaryotes.
It is a DNA sequence found in many eukaryotic promoters.
It encodes the start codon for translation.
Explanation - The TATA box, usually positioned ~25-35 bp upstream of the transcription start site, helps position RNA polymerase II in eukaryotes.
Correct answer is: It is a DNA sequence found in many eukaryotic promoters.
Q.6 What is the function of the 5' cap added to eukaryotic mRNA?
Signal for transcription termination
Facilitates ribosome binding and protects mRNA from degradation
Marks the site of polyadenylation
Directs the mRNA to the nucleus
Explanation - The 7‑methylguanosine cap protects the mRNA from exonucleases and is recognized by the translation initiation machinery.
Correct answer is: Facilitates ribosome binding and protects mRNA from degradation
Q.7 Which enzyme adds a poly(A) tail to the 3' end of eukaryotic mRNA?
RNA polymerase II
Poly(A) polymerase
RNA helicase
DNA ligase
Explanation - Poly(A) polymerase catalyzes the addition of a stretch of adenine nucleotides after transcription termination.
Correct answer is: Poly(A) polymerase
Q.8 In the lac operon, the presence of allolactose leads to:
Activation of the repressor protein
Inhibition of RNA polymerase binding
Derepression of the operon
Enhancement of the promoter's affinity for sigma factor
Explanation - Allolactose binds the lac repressor, causing a conformational change that releases it from the operator, allowing transcription.
Correct answer is: Derepression of the operon
Q.9 Which of the following best describes an operon?
A single gene that encodes multiple proteins
A cluster of genes under the control of a single promoter and regulatory sequences
A DNA region that enhances transcription from a distant gene
A protein complex that synthesizes RNA
Explanation - An operon is a functional unit of DNA containing a promoter, operator, and one or more structural genes transcribed together.
Correct answer is: A cluster of genes under the control of a single promoter and regulatory sequences
Q.10 What distinguishes a sigma factor in bacterial transcription?
It catalyzes peptide bond formation.
It provides the catalytic activity of RNA polymerase.
It directs RNA polymerase to specific promoter sequences.
It terminates transcription by cleaving RNA.
Explanation - Sigma factors recognize promoter elements (e.g., -35 and -10 boxes) and help RNA polymerase bind DNA.
Correct answer is: It directs RNA polymerase to specific promoter sequences.
Q.11 Which of the following RNA polymerases is responsible for transcribing rRNA genes in eukaryotes?
RNA polymerase I
RNA polymerase II
RNA polymerase III
RNA polymerase IV
Explanation - RNA polymerase I synthesizes most ribosomal RNA (except 5S rRNA, which is transcribed by polymerase III).
Correct answer is: RNA polymerase I
Q.12 In eukaryotes, transcription termination for RNA polymerase II primarily depends on:
Rho factor
Hairpin loop formation
Polyadenylation signal (AAUAAA)
Sigma factor release
Explanation - The AAUAAA signal downstream of the coding region signals cleavage and polyadenylation, leading to transcription termination.
Correct answer is: Polyadenylation signal (AAUAAA)
Q.13 Which of the following is NOT a typical feature of prokaryotic mRNA?
Polycistronic transcripts
5' cap structure
Lack of introns
Rapid degradation after transcription
Explanation - Prokaryotic mRNAs are not capped; capping is a eukaryotic modification.
Correct answer is: 5' cap structure
Q.14 During transcription elongation, the RNA-DNA hybrid within the transcription bubble is typically:
10 base pairs long
12 base pairs long
8-9 base pairs long
5 base pairs long
Explanation - The transcription bubble maintains an RNA-DNA hybrid of about 8-9 nucleotides, allowing the enzyme to translocate.
Correct answer is: 8-9 base pairs long
Q.15 Which of the following best explains the concept of transcriptional attenuation in the tryptophan operon?
Repressor binding is enhanced by high tryptophan levels.
RNA polymerase terminates early due to the formation of a hairpin structure in the leader peptide mRNA.
The operon is activated by an inducer molecule.
A riboswitch directly binds tryptophan to inhibit transcription.
Explanation - When tryptophan is abundant, the ribosome quickly translates the leader peptide, allowing formation of a terminator hairpin that stops transcription.
Correct answer is: RNA polymerase terminates early due to the formation of a hairpin structure in the leader peptide mRNA.
Q.16 In eukaryotic transcription, which of the following factors is required for the formation of the pre‑initiation complex?
TFIIH, TFIIA, TFIIB, TFIID, TFIIE, TFIIF
Sigma factor only
Rho factor
DNA ligase
Explanation - These general transcription factors assemble with RNA polymerase II at the promoter to form the pre‑initiation complex.
Correct answer is: TFIIH, TFIIA, TFIIB, TFIID, TFIIE, TFIIF
Q.17 Which of the following is the correct order of events in eukaryotic transcription initiation?
Promoter melting → TFIIH recruitment → TBP binding → RNA synthesis
TBP binding → TFIIH recruitment → promoter melting → RNA synthesis
RNA synthesis → promoter melting → TBP binding → TFIIH recruitment
Promoter melting → TBP binding → TFIIH recruitment → RNA synthesis
Explanation - TATA‑binding protein (TBP) binds the TATA box, TFIIH is recruited to unwind DNA (promoter melting), then RNA polymerase initiates synthesis.
Correct answer is: TBP binding → TFIIH recruitment → promoter melting → RNA synthesis
Q.18 What is the primary purpose of a terminator sequence in bacterial transcription?
To recruit the ribosome for translation
To signal RNA polymerase to release the newly made RNA
To initiate DNA replication
To bind the sigma factor
Explanation - Terminator sequences cause RNA polymerase to dissociate from DNA, ending transcription.
Correct answer is: To signal RNA polymerase to release the newly made RNA
Q.19 Which of the following mechanisms allows eukaryotic cells to rapidly increase transcription of a specific gene in response to a signal?
DNA methylation of the promoter
Histone acetylation leading to chromatin relaxation
RNA interference
Rho-dependent termination
Explanation - Acetylation of histone tails reduces their positive charge, loosening DNA–histone interactions and making promoters more accessible.
Correct answer is: Histone acetylation leading to chromatin relaxation
Q.20 In the context of transcription, what does the term "processivity" refer to?
The speed at which RNA polymerase moves along DNA
The ability of RNA polymerase to synthesize a long RNA without dissociating
The accuracy of nucleotide incorporation
The number of promoters recognized by a polymerase
Explanation - Processivity measures how many nucleotides a polymerase adds before it falls off the template.
Correct answer is: The ability of RNA polymerase to synthesize a long RNA without dissociating
Q.21 Which of the following is a hallmark of RNA polymerase II promoters in metazoans?
Pribnow box
TATA box
CAAT box
Operator sequence
Explanation - Many RNA polymerase II promoters contain a TATA box ~25–35 bp upstream of the transcription start site.
Correct answer is: TATA box
Q.22 What is the role of the Mediator complex in eukaryotic transcription?
Directly catalyzes RNA synthesis
Bridges transcription factors and RNA polymerase II
Terminates transcription by releasing RNA polymerase
Splices introns from pre‑mRNA
Explanation - Mediator serves as a co‑activator that transmits regulatory signals from transcription factors to the polymerase machinery.
Correct answer is: Bridges transcription factors and RNA polymerase II
Q.23 Which of the following RNA polymerases synthesizes most small nuclear RNAs (snRNAs) in eukaryotes?
RNA polymerase I
RNA polymerase II
RNA polymerase III
RNA polymerase IV
Explanation - Most snRNAs are transcribed by RNA polymerase II, though some are produced by polymerase III.
Correct answer is: RNA polymerase II
Q.24 During transcription, the enzyme that adds uracil opposite adenine in the DNA template is:
DNA polymerase
RNA polymerase
Reverse transcriptase
DNA ligase
Explanation - RNA polymerase reads the DNA template and incorporates ribonucleotides; adenine pairs with uracil in RNA.
Correct answer is: RNA polymerase
Q.25 In eukaryotes, which modification occurs first on the nascent RNA transcript?
Polyadenylation
Splicing
5' capping
RNA editing
Explanation - The 7‑methylguanosine cap is added within seconds of initiation, before splicing and polyadenylation.
Correct answer is: 5' capping
Q.26 Which of the following best explains the concept of 'transcriptional pausing'?
RNA polymerase stops transcription due to lack of nucleotides.
RNA polymerase temporarily halts elongation near the promoter to allow regulatory factor binding.
RNA polymerase terminates prematurely at a hairpin loop.
RNA polymerase pauses to recruit ribosomes for translation.
Explanation - Promoter-proximal pausing gives transcription factors time to modulate gene expression before productive elongation.
Correct answer is: RNA polymerase temporarily halts elongation near the promoter to allow regulatory factor binding.
Q.27 In bacterial transcription termination, which factor is essential for rho-dependent termination?
Rho protein
Sigma factor
RNA polymerase sigma subunit
DNA helicase
Explanation - Rho is an ATP‑dependent helicase that binds nascent RNA and moves toward RNA polymerase to cause dissociation.
Correct answer is: Rho protein
Q.28 Which of the following RNA polymerases is responsible for the transcription of tRNA genes in eukaryotes?
RNA polymerase I
RNA polymerase II
RNA polymerase III
RNA polymerase IV
Explanation - RNA polymerase III transcribes tRNA, 5S rRNA, and other small RNAs.
Correct answer is: RNA polymerase III
Q.29 What is the primary function of the transcription factor TFIIB?
Bind the TATA box
Recruit RNA polymerase II to the promoter
Phosphorylate the RNA polymerase C‑terminal domain
Terminate transcription
Explanation - TFIIB interacts with TBP and RNA polymerase II, helping position the polymerase at the start site.
Correct answer is: Recruit RNA polymerase II to the promoter
Q.30 Which of the following is NOT a component of the bacterial transcription termination hairpin structure?
A GC‑rich stem
A uridine-rich (U) tract in the RNA
A consensus -35 box
A stable RNA secondary structure
Explanation - -35 boxes are promoter elements, not termination structures.
Correct answer is: A consensus -35 box
Q.31 During transcription elongation, the nascent RNA strand exits the RNA polymerase through:
The entry channel
The secondary channel
The RNA exit channel
The DNA-binding cleft
Explanation - RNA polymerase has a dedicated tunnel for the growing RNA chain to leave the enzyme.
Correct answer is: The RNA exit channel
Q.32 Which of the following best describes an enhancer?
A DNA sequence located immediately upstream of the start codon
A protein that binds RNA polymerase
A cis‑regulatory element that can increase transcription from a distant promoter
A region where transcription terminates
Explanation - Enhancers can act thousands of base pairs away, looping to contact promoters and boost transcription.
Correct answer is: A cis‑regulatory element that can increase transcription from a distant promoter
Q.33 In eukaryotic cells, which RNA processing step is coupled with transcription?
Splicing
Polyadenylation
5' capping
All of the above
Explanation - Capping, splicing, and polyadenylation often occur co‑transcriptionally as the RNA emerges from polymerase II.
Correct answer is: All of the above
Q.34 Which of the following is true about the bacterial sigma factor σ⁷⁰?
It remains bound to RNA polymerase throughout transcription.
It recognizes the -10 and -35 promoter elements.
It terminates transcription.
It adds the 5' cap to mRNA.
Explanation - σ⁷⁰ is the primary sigma factor in E. coli, directing RNA polymerase to promoters containing those consensus sequences.
Correct answer is: It recognizes the -10 and -35 promoter elements.
Q.35 What is the function of the CTD (C‑terminal domain) of RNA polymerase II?
Binding DNA during initiation
Catalyzing phosphodiester bond formation
Recruiting RNA processing factors via reversible phosphorylation
Terminating transcription
Explanation - The heptapeptide repeats of the CTD are phosphorylated to coordinate capping, splicing, and polyadenylation.
Correct answer is: Recruiting RNA processing factors via reversible phosphorylation
Q.36 Which of the following statements about the 'RNA polymerase clamp' is correct?
It opens to allow DNA entry and closes during elongation to increase processivity.
It synthesizes the first phosphodiester bond.
It binds the sigma factor.
It is only present in eukaryotic polymerases.
Explanation - The clamp domain stabilizes the transcription complex and its closing enhances processivity.
Correct answer is: It opens to allow DNA entry and closes during elongation to increase processivity.
Q.37 Which of the following best explains why bacterial operons are often polycistronic?
Ribosomes can translate multiple proteins from a single mRNA.
RNA polymerase can only bind one promoter per operon.
Eukaryotic cells lack operons.
DNA replication occurs simultaneously with transcription.
Explanation - A polycistronic mRNA contains several open reading frames, each of which can be independently translated.
Correct answer is: Ribosomes can translate multiple proteins from a single mRNA.
Q.38 In the context of transcriptional regulation, a 'silencer' is:
A DNA element that enhances transcription when bound by activators.
A DNA element that represses transcription when bound by repressors.
A protein that terminates transcription.
A mutation that eliminates promoter activity.
Explanation - Silencers function oppositely to enhancers, decreasing transcriptional output.
Correct answer is: A DNA element that represses transcription when bound by repressors.
Q.39 Which of the following is a characteristic of a 'strong' promoter?
High affinity for RNA polymerase and transcription factors, leading to high transcription rates.
Presence of a TATA box only.
Lack of upstream regulatory sequences.
Requires a co‑activator for any transcription.
Explanation - Strong promoters contain optimal consensus sequences that efficiently recruit the transcriptional machinery.
Correct answer is: High affinity for RNA polymerase and transcription factors, leading to high transcription rates.
Q.40 What is the main difference between 'constitutive' and 'inducible' genes?
Constitutive genes are expressed only under stress; inducible genes are always on.
Constitutive genes are constantly expressed; inducible genes require a specific signal.
Constitutive genes lack promoters; inducible genes have multiple promoters.
Constitutive genes are found only in eukaryotes.
Explanation - Constitutive genes provide basic cellular functions, while inducible genes are turned on in response to environmental cues.
Correct answer is: Constitutive genes are constantly expressed; inducible genes require a specific signal.
Q.41 Which of the following is the correct base-pairing rule during RNA synthesis?
A pairs with T, G pairs with C
A pairs with U, G pairs with C
A pairs with C, G pairs with U
A pairs with G, C pairs with U
Explanation - In RNA, adenine pairs with uracil, while guanine pairs with cytosine.
Correct answer is: A pairs with U, G pairs with C
Q.42 The process by which a newly synthesized RNA molecule is edited to change its nucleotide sequence is called:
Splicing
RNA editing
Polyadenylation
Capping
Explanation - RNA editing alters specific nucleotides after transcription, e.g., A-to-I editing in some transcripts.
Correct answer is: RNA editing
Q.43 In the lac operon, the catabolite activator protein (CAP) enhances transcription when:
Glucose levels are high
cAMP levels are high
Allolactose is absent
The operator is bound by the repressor
Explanation - High cAMP (low glucose) enables CAP to bind the promoter and stimulate RNA polymerase activity.
Correct answer is: cAMP levels are high
Q.44 Which of the following statements about the bacterial rho factor is FALSE?
It is an ATP‑dependent helicase.
It binds to a specific rut site on the nascent RNA.
It is required for all transcription termination in bacteria.
It moves 5'→3' along the RNA to catch up with RNA polymerase.
Explanation - Bacteria also use rho‑independent (intrinsic) termination; rho is only required for rho‑dependent termination.
Correct answer is: It is required for all transcription termination in bacteria.
Q.45 The presence of a poly‑U tract in bacterial terminators is important because:
U‑U base pairs are weaker than G‑C, facilitating strand separation.
It signals the ribosome to start translation.
It recruits sigma factor.
It stabilizes the RNA‑DNA hybrid.
Explanation - The weak A‑U interactions help the RNA transcript dissociate from DNA during termination.
Correct answer is: U‑U base pairs are weaker than G‑C, facilitating strand separation.
Q.46 Which of the following best describes the 'RNA polymerase II pause release' step?
RNA polymerase dissociates from DNA.
RNA polymerase resumes elongation after promoter-proximal pausing.
RNA polymerase initiates transcription.
RNA polymerase terminates transcription.
Explanation - Pause release involves factors like P‑TEFb phosphorylating the CTD, allowing productive elongation.
Correct answer is: RNA polymerase resumes elongation after promoter-proximal pausing.
Q.47 Which of the following sequences is most commonly found at the 3' end of bacterial mRNA to signal termination?
AAUAAA
GCCGCC
U-rich hairpin followed by a poly‑U tract
TATAAT
Explanation - Intrinsic terminators consist of a GC‑rich hairpin followed by a stretch of uracils.
Correct answer is: U-rich hairpin followed by a poly‑U tract
Q.48 During transcription, the term 'abortive initiation' refers to:
RNA polymerase releasing short RNA fragments before escaping the promoter.
Premature termination due to a hairpin structure.
Failure of ribosomes to bind mRNA.
Splicing errors in the nascent transcript.
Explanation - RNA polymerase often synthesizes 2‑10 nucleotide RNAs repeatedly before clearing the promoter.
Correct answer is: RNA polymerase releasing short RNA fragments before escaping the promoter.
Q.49 Which of the following is a key difference between eukaryotic and prokaryotic transcription termination?
Only eukaryotes use rho factor.
Prokaryotes use polyadenylation signals.
Eukaryotic termination often involves cleavage and polyadenylation, whereas bacteria may use rho‑dependent or intrinsic mechanisms.
Both rely on a TATA box for termination.
Explanation - Eukaryotic RNA polymerase II termination is linked to the poly(A) site; bacteria have distinct mechanisms.
Correct answer is: Eukaryotic termination often involves cleavage and polyadenylation, whereas bacteria may use rho‑dependent or intrinsic mechanisms.
Q.50 In a eukaryotic gene, the region downstream of the transcription start site that contains multiple exons and introns is called:
Promoter
5' UTR
Coding sequence (CDS)
Pre‑mRNA
Explanation - Pre‑mRNA is the primary transcript containing both exons and introns before processing.
Correct answer is: Pre‑mRNA
Q.51 Which of the following enzymes is responsible for removing the 5' cap from mRNA during decay?
Decapping enzyme Dcp2
RNase H
Poly(A) polymerase
Spliceosome
Explanation - Dcp2 hydrolyzes the 7‑methylguanosine cap, marking mRNA for degradation.
Correct answer is: Decapping enzyme Dcp2
Q.52 What is the primary function of the bacterial transcription factor NusG?
Promote transcription termination by interacting with Rho.
Recruit ribosomes to the mRNA.
Add the 5' cap to nascent RNA.
Facilitate DNA replication.
Explanation - NusG enhances Rho-dependent termination and also influences elongation rates.
Correct answer is: Promote transcription termination by interacting with Rho.
Q.53 Which of the following statements about 'chromatin remodeling' is TRUE?
It permanently changes DNA sequence.
It involves ATP‑dependent complexes that reposition nucleosomes to regulate access to DNA.
It only occurs during DNA replication.
It is exclusive to prokaryotes.
Explanation - Chromatin remodelers like SWI/SNF use ATP to slide or evict nucleosomes, affecting transcription.
Correct answer is: It involves ATP‑dependent complexes that reposition nucleosomes to regulate access to DNA.
Q.54 In the context of transcription, the term 'bursting' refers to:
A rapid series of transcription events followed by a silent period.
The release of RNA polymerase from a paused state.
The simultaneous transcription of multiple genes.
The cleavage of RNA at the poly(A) site.
Explanation - Transcriptional bursting describes stochastic pulses of gene expression in single cells.
Correct answer is: A rapid series of transcription events followed by a silent period.
Q.55 Which of the following best explains why eukaryotic genes often contain introns?
Introns are required for translation initiation.
Introns allow for alternative splicing, increasing proteomic diversity.
Introns stabilize mRNA.
Introns code for regulatory RNAs.
Explanation - Alternative splicing of introns enables a single gene to produce multiple protein isoforms.
Correct answer is: Introns allow for alternative splicing, increasing proteomic diversity.
Q.56 Which of the following is NOT a typical feature of eukaryotic promoters?
TATA box
CAAT box
Pribnow box
GC-rich region
Explanation - The Pribnow box is a bacterial -10 promoter element; eukaryotic promoters use TATA and CAAT boxes.
Correct answer is: Pribnow box
Q.57 The factor that phosphorylates the RNA polymerase II CTD to trigger promoter clearance is:
TFIIA
TFIIB
TFIIF
TFIIB-associated kinase (TFIIH)
Explanation - TFIIH possesses kinase activity that phosphorylates Ser5 of the CTD, allowing promoter escape.
Correct answer is: TFIIB-associated kinase (TFIIH)
Q.58 What is the main functional difference between a 'sigma factor' and a 'transcription factor'?
Sigma factors are found only in eukaryotes.
Sigma factors are part of the core RNA polymerase, whereas transcription factors are separate regulatory proteins.
Transcription factors synthesize RNA.
Sigma factors terminate transcription.
Explanation - Sigma factors bind directly to RNA polymerase to confer promoter specificity in bacteria.
Correct answer is: Sigma factors are part of the core RNA polymerase, whereas transcription factors are separate regulatory proteins.
Q.59 Which of the following best describes the function of a riboswitch?
A DNA element that enhances transcription.
A protein that binds RNA polymerase.
A segment of mRNA that binds a metabolite to regulate its own transcription or translation.
A sequence that signals polyadenylation.
Explanation - Riboswitches undergo structural changes upon ligand binding, affecting gene expression.
Correct answer is: A segment of mRNA that binds a metabolite to regulate its own transcription or translation.
Q.60 During transcription, the 'scrunching' model refers to:
RNA polymerase pulling DNA into its active site, generating a loop of unwound DNA.
RNA polymerase sliding along DNA without unwinding.
RNA polymerase rotating around DNA.
RNA polymerase dissociating from DNA.
Explanation - Scrunching explains how energy is stored during initiation before promoter escape.
Correct answer is: RNA polymerase pulling DNA into its active site, generating a loop of unwound DNA.
Q.61 Which of the following is a characteristic of an 'intrinsic' (rho‑independent) terminator in bacteria?
Requires the Rho protein to terminate transcription.
Consists of a GC‑rich hairpin followed by a stretch of U residues.
Contains a polyadenylation signal.
Is located upstream of the promoter.
Explanation - The hairpin destabilizes the RNA‑DNA hybrid, and the poly‑U tract promotes dissociation.
Correct answer is: Consists of a GC‑rich hairpin followed by a stretch of U residues.
Q.62 In eukaryotes, the process that removes the 5' cap from a specific subset of mRNAs during meiosis is called:
Decapping
Capping
Splicing
Polyadenylation
Explanation - Specific decapping enzymes target certain transcripts, leading to regulated degradation.
Correct answer is: Decapping
Q.63 Which of the following best explains why transcription and translation are coupled in prokaryotes?
Prokaryotic mRNA is immediately exported to the nucleus.
Ribosomes can bind to nascent mRNA as it emerges from RNA polymerase.
Prokaryotes lack ribosomes.
RNA polymerase synthesizes proteins directly.
Explanation - The absence of a nuclear membrane allows simultaneous transcription and translation.
Correct answer is: Ribosomes can bind to nascent mRNA as it emerges from RNA polymerase.
Q.64 What is the main purpose of the 'spacer' region between the -35 and -10 promoter elements in bacterial promoters?
To provide a binding site for the sigma factor.
To allow proper alignment of RNA polymerase with the transcription start site.
To encode the first few amino acids of the protein.
To serve as the terminator sequence.
Explanation - A spacer of ~17 bp positions the sigma factor correctly for initiation.
Correct answer is: To allow proper alignment of RNA polymerase with the transcription start site.
Q.65 Which of the following techniques is commonly used to map transcription start sites genome‑wide?
RNA‑seq
ChIP‑seq
CAGE (Cap Analysis of Gene Expression)
ATAC‑seq
Explanation - CAGE captures the 5' capped ends of transcripts, identifying transcription start sites.
Correct answer is: CAGE (Cap Analysis of Gene Expression)
Q.66 The presence of which histone modification is most strongly associated with active transcription start sites?
H3K27me3
H3K9me3
H3K4me3
H4K20me1
Explanation - Trimethylation of histone H3 at lysine 4 marks promoters of actively transcribed genes.
Correct answer is: H3K4me3
Q.67 Which of the following best describes the function of the bacterial transcription factor 'GreA'?
Promotes transcription initiation.
Stimulates RNA polymerase cleavage activity to rescue backtracked complexes.
Recruits ribosomes to mRNA.
Acts as a sigma factor.
Explanation - GreA induces endonucleolytic cleavage of the nascent RNA, allowing the polymerase to resume elongation.
Correct answer is: Stimulates RNA polymerase cleavage activity to rescue backtracked complexes.
Q.68 During transcription, the term 'backtracking' refers to:
RNA polymerase moving forward faster than normal.
RNA polymerase sliding backwards on DNA, extruding the 3' end of RNA.
Ribosome moving backwards on mRNA.
DNA polymerase reversing its direction.
Explanation - Backtracking can pause transcription; factors like GreA/B help reactivate the polymerase.
Correct answer is: RNA polymerase sliding backwards on DNA, extruding the 3' end of RNA.
Q.69 Which of the following statements about the 'RNA polymerase II C-terminal domain (CTD) code' is correct?
The CTD is composed of repeats that are never modified.
Specific patterns of serine phosphorylation coordinate transcription with RNA processing.
CTD phosphorylation inhibits transcription elongation.
CTD is only present in bacterial RNA polymerases.
Explanation - Ser2 and Ser5 phosphorylation states recruit capping, splicing, and polyadenylation factors at distinct stages.
Correct answer is: Specific patterns of serine phosphorylation coordinate transcription with RNA processing.
Q.70 Which of the following best explains why 'alternative promoters' contribute to gene regulation diversity?
They allow a gene to be transcribed from different start sites, producing transcripts with distinct 5' UTRs.
They cause the gene to be silenced completely.
They change the coding sequence of the gene.
They only function in prokaryotes.
Explanation - Alternative promoters can generate isoforms with different regulatory elements, affecting translation and stability.
Correct answer is: They allow a gene to be transcribed from different start sites, producing transcripts with distinct 5' UTRs.
Q.71 In the context of transcriptional regulation, the term 'co‑activator' refers to a protein that:
Directly binds DNA at the promoter.
Catalyzes peptide bond formation.
Interacts with transcription factors to enhance transcription without binding DNA itself.
Degrades mRNA after transcription.
Explanation - Co‑activators bridge transcription factors to the basal transcriptional machinery and often possess enzymatic activities (e.g., histone acetyltransferases).
Correct answer is: Interacts with transcription factors to enhance transcription without binding DNA itself.
Q.72 Which of the following mechanisms allows bacteria to rapidly shut down transcription of a specific operon in response to a metabolite?
RNA interference
Rho‑dependent termination triggered by the metabolite
Allosteric repression of a repressor protein binding the operator
DNA methylation of the promoter
Explanation - Metabolite binding can activate a repressor that blocks RNA polymerase access to the promoter.
Correct answer is: Allosteric repression of a repressor protein binding the operator
Q.73 Which of the following statements about 'enhancer RNAs (eRNAs)' is TRUE?
eRNAs are translated into proteins.
eRNAs are short, non‑coding RNAs transcribed from active enhancer regions and can aid transcriptional activation.
eRNAs are only found in bacteria.
eRNAs terminate transcription.
Explanation - eRNAs are a hallmark of active enhancers and may help stabilize enhancer‑promoter loops.
Correct answer is: eRNAs are short, non‑coding RNAs transcribed from active enhancer regions and can aid transcriptional activation.
Q.74 During transcription initiation in bacteria, the formation of the closed complex refers to:
RNA polymerase bound to DNA without DNA unwinding.
RNA polymerase synthesizing the first phosphodiester bond.
RNA polymerase releasing the newly formed RNA.
RNA polymerase bound to the terminator.
Explanation - The closed complex involves RNA polymerase–DNA binding; the open complex follows DNA melting.
Correct answer is: RNA polymerase bound to DNA without DNA unwinding.
Q.75 Which of the following best describes the function of a 'pause button' element in a gene promoter?
A DNA sequence that induces promoter-proximal pausing of RNA polymerase II.
A protein that terminates transcription.
A ribosomal binding site.
A polyadenylation signal.
Explanation - Pause button sequences cause RNA polymerase II to pause shortly after initiation, allowing regulatory input.
Correct answer is: A DNA sequence that induces promoter-proximal pausing of RNA polymerase II.
Q.76 Which of the following is the most common method for measuring transcriptional activity of a specific promoter in vivo?
Western blot
Luciferase reporter assay
Northern blot
Chromatin immunoprecipitation
Explanation - A reporter gene (e.g., luciferase) placed downstream of the promoter provides a quantifiable readout of transcriptional strength.
Correct answer is: Luciferase reporter assay
Q.77 Which of the following RNA polymerase subunits is directly involved in DNA binding during transcription initiation?
β' subunit
α subunit
ω subunit
δ subunit
Explanation - The β' subunit contains the active site and contacts the DNA template strand.
Correct answer is: β' subunit
Q.78 In eukaryotes, the process by which a nascent RNA transcript is cleaved at a specific site and then polyadenylated is called:
Capping
Splicing
3' end processing
RNA interference
Explanation - Cleavage at the polyadenylation signal followed by addition of a poly(A) tail constitutes 3' end processing.
Correct answer is: 3' end processing
Q.79 Which of the following statements about 'transcription factories' in the nucleus is correct?
They are sites where DNA replication occurs.
They are discrete nuclear foci enriched in RNA polymerase II and transcription factors where active transcription takes place.
They are structures that degrade RNA.
They are exclusive to prokaryotic cells.
Explanation - Transcription factories are thought to concentrate the transcriptional machinery for efficient gene expression.
Correct answer is: They are discrete nuclear foci enriched in RNA polymerase II and transcription factors where active transcription takes place.
Q.80 Which of the following best explains the term 'RNA polymerase proofreading'?
RNA polymerase removes incorrectly incorporated nucleotides via intrinsic hydrolytic activity.
RNA polymerase recruits DNA polymerase for error correction.
RNA polymerase does not have any proofreading ability.
RNA polymerase adds extra nucleotides at the 5' end.
Explanation - RNA polymerase can backtrack and cleave the nascent RNA to correct misincorporations.
Correct answer is: RNA polymerase removes incorrectly incorporated nucleotides via intrinsic hydrolytic activity.
Q.81 Which of the following RNA elements can act as a ribosome entry site (IRES) to initiate translation independently of the 5' cap?
tRNA
snRNA
5' UTR internal ribosome entry site (IRES)
Poly(A) tail
Explanation - IRES elements allow cap‑independent translation initiation, especially under stress conditions.
Correct answer is: 5' UTR internal ribosome entry site (IRES)
Q.82 Which of the following best describes the role of the bacterial transcription factor 'FIS' (Factor for Inversion Stimulation)?
It terminates transcription at rho‑dependent sites.
It binds to specific DNA sites to stimulate transcription of rRNA operons and facilitate DNA recombination.
It adds a 5' cap to bacterial mRNA.
It functions as a sigma factor.
Explanation - FIS is a nucleoid‑associated protein that enhances transcription of certain genes and participates in DNA topology changes.
Correct answer is: It binds to specific DNA sites to stimulate transcription of rRNA operons and facilitate DNA recombination.
Q.83 In eukaryotic transcription, which of the following processes is directly coupled with the phosphorylation state of the RNA polymerase II CTD?
DNA replication
RNA splicing
Protein folding
Membrane transport
Explanation - Phosphorylation of Ser2 on the CTD recruits spliceosomal components, linking elongation to splicing.
Correct answer is: RNA splicing
Q.84 Which of the following best explains why eukaryotic transcription initiation is generally slower than bacterial initiation?
RNA polymerase II lacks catalytic activity.
Eukaryotic promoters are less conserved.
Chromatin structure must be remodeled and multiple transcription factors must assemble before elongation.
Eukaryotic cells do not have sigma factors.
Explanation - Nucleosome positioning and the requirement for a large set of general transcription factors increase initiation time.
Correct answer is: Chromatin structure must be remodeled and multiple transcription factors must assemble before elongation.
Q.85 Which of the following is a hallmark of a 'bidirectional promoter'?
It drives transcription in opposite directions for two adjacent genes.
It only initiates transcription in the sense direction.
It contains a strong terminator downstream.
It is found exclusively in prokaryotes.
Explanation - Bidirectional promoters have a shared regulatory region that can initiate transcription of divergently oriented genes.
Correct answer is: It drives transcription in opposite directions for two adjacent genes.
Q.86 Which of the following best describes the role of 'Mediator subunit MED23' in transcription?
It directly binds DNA at the TATA box.
It bridges transcriptional activators with RNA polymerase II, influencing gene‑specific activation.
It terminates transcription by releasing RNA polymerase.
It adds the poly(A) tail to mRNA.
Explanation - MED23 is part of the Mediator complex that transduces regulatory signals to the polymerase.
Correct answer is: It bridges transcriptional activators with RNA polymerase II, influencing gene‑specific activation.
Q.87 Which of the following is the most common form of RNA polymerase regulation in response to cellular stress in eukaryotes?
Sigma factor switching
Phosphorylation of the CTD and recruitment of stress‑responsive transcription factors
DNA methylation of promoters
Rho factor activation
Explanation - Stress signals often modify the CTD and activate transcription factors like p53 to alter gene expression.
Correct answer is: Phosphorylation of the CTD and recruitment of stress‑responsive transcription factors
Q.88 In the context of transcription, the term 'cryptic promoter' refers to:
A promoter that is hidden by DNA methylation and becomes active only under specific conditions.
A promoter that drives expression of non‑coding RNAs.
A promoter located within a gene that can initiate transcription from an unexpected site.
A promoter that functions only in prokaryotes.
Explanation - Cryptic promoters can lead to aberrant transcripts when chromatin is altered.
Correct answer is: A promoter located within a gene that can initiate transcription from an unexpected site.
Q.89 Which of the following is true about the 'RNA polymerase clamp' movement during transcription elongation?
The clamp opens to allow the nascent RNA to exit the enzyme.
The clamp remains static throughout transcription.
The clamp closes around the DNA‑RNA hybrid, increasing processivity and preventing dissociation.
The clamp only functions during termination.
Explanation - Clamp closure stabilizes the transcription complex, allowing efficient elongation.
Correct answer is: The clamp closes around the DNA‑RNA hybrid, increasing processivity and preventing dissociation.
Q.90 Which of the following best describes the mechanism of action of the antibiotic rifampicin?
It binds to the bacterial ribosome and blocks translation.
It inhibits the sigma factor binding to promoters.
It binds to the β subunit of bacterial RNA polymerase, preventing initiation of RNA synthesis.
It interferes with DNA gyrase activity.
Explanation - Rifampicin blocks the RNA exit channel of bacterial RNA polymerase, halting transcription initiation.
Correct answer is: It binds to the β subunit of bacterial RNA polymerase, preventing initiation of RNA synthesis.
Q.91 Which of the following processes is most directly affected by the presence of a strong polyadenylation signal downstream of a gene?
RNA splicing
RNA capping
Transcription termination and mRNA stability
DNA replication
Explanation - The poly(A) signal directs cleavage of the pre‑mRNA and addition of the poly(A) tail, influencing termination and stability.
Correct answer is: Transcription termination and mRNA stability
Q.92 In eukaryotes, which of the following chromatin modifications is commonly associated with transcriptional repression?
H3K4me3
H3K9me3
H3K27ac
H3K36me3
Explanation - Trimethylation of histone H3 at lysine 9 is a mark of heterochromatin and transcriptionally silent regions.
Correct answer is: H3K9me3
Q.93 Which of the following is a characteristic feature of the bacterial transcription termination factor Rho?
It is a DNA helicase.
It requires ATP to translocate along RNA.
It binds to the -35 promoter element.
It adds a poly(A) tail to bacterial mRNA.
Explanation - Rho is an ATP‑dependent RNA helicase that moves 5'→3' on the nascent transcript to catch up with RNA polymerase.
Correct answer is: It requires ATP to translocate along RNA.
Q.94 Which of the following best describes the function of the 'transcriptional activator' protein GAL4 in yeast?
It binds to the GAL4 operator to repress transcription.
It binds upstream activating sequences (UAS) to recruit the transcriptional machinery for galactose‑responsive genes.
It terminates transcription of GAL genes.
It modifies DNA methylation patterns.
Explanation - GAL4 is a classic zinc‑finger transcription factor that activates genes involved in galactose metabolism.
Correct answer is: It binds upstream activating sequences (UAS) to recruit the transcriptional machinery for galactose‑responsive genes.
Q.95 Which of the following mechanisms can generate a single gene product with multiple functional domains without alternative splicing?
Use of alternative promoters
Use of a polycistronic operon
RNA editing
Transcriptional read‑through
Explanation - Different promoters can produce transcripts with distinct 5' exons, leading to proteins with varied N‑terminal domains.
Correct answer is: Use of alternative promoters
Q.96 Which of the following statements about 'RNA polymerase backtracking' is FALSE?
Backtracking can pause transcription.
The 3' end of the RNA is extruded into the secondary channel.
Gre factors stimulate cleavage to resume elongation.
Backtracking always leads to transcription termination.
Explanation - Backtracking is reversible; factors like GreA/B rescue the polymerase by cleaving the extruded RNA.
Correct answer is: Backtracking always leads to transcription termination.
Q.97 Which of the following best describes the 'ribosome‑shunting' mechanism?
A translation initiation strategy where the ribosome bypasses a structured region of 5' UTR to start translation downstream.
A process where ribosomes transcribe DNA.
A method of transcription termination.
A type of RNA splicing.
Explanation - Ribosome shunting allows translation initiation despite highly structured 5' leaders, common in some viral RNAs.
Correct answer is: A translation initiation strategy where the ribosome bypasses a structured region of 5' UTR to start translation downstream.
Q.98 Which of the following RNA polymerase II CTD phosphorylation patterns is primarily associated with transcriptional elongation and splicing factor recruitment?
Ser5 phosphorylation only
Ser2 phosphorylation
Tyr1 phosphorylation
Ser7 phosphorylation
Explanation - Ser2‑P marks the elongation phase and creates binding sites for splicing and 3' end processing factors.
Correct answer is: Ser2 phosphorylation
Q.99 What is the primary reason that bacterial transcription termination can occur without any protein factors?
DNA is single‑stranded.
Intrinsic terminators form a stable hairpin followed by a poly‑U tract, destabilizing the RNA‑DNA hybrid.
RNA polymerase lacks a catalytic domain.
Rho protein is always present.
Explanation - The hairpin and weak A‑U pairing cause the transcript to dissociate from DNA, ending transcription.
Correct answer is: Intrinsic terminators form a stable hairpin followed by a poly‑U tract, destabilizing the RNA‑DNA hybrid.
Q.100 Which of the following best explains why eukaryotic transcription requires a 5' cap before splicing?
The cap directly catalyzes intron removal.
Cap‑binding complexes recruit and coordinate splicing factors with the transcription machinery.
The cap prevents RNA polymerase from re‑initiating transcription.
The cap is needed for translation only.
Explanation - The cap-binding complex (CBC) interacts with spliceosomal components, linking capping to splicing.
Correct answer is: Cap‑binding complexes recruit and coordinate splicing factors with the transcription machinery.
Q.101 Which of the following statements about the bacterial transcription factor 'LexA' is correct?
LexA activates the SOS response genes.
LexA is a repressor that is cleaved during the SOS response, allowing DNA repair genes to be transcribed.
LexA functions as a sigma factor.
LexA terminates transcription at the lac operon.
Explanation - DNA damage triggers RecA‑mediated LexA autocleavage, derepressing SOS genes.
Correct answer is: LexA is a repressor that is cleaved during the SOS response, allowing DNA repair genes to be transcribed.
Q.102 Which of the following is a key difference between the bacterial sigma factor σ⁵⁴ and the primary sigma factor σ⁷⁰?
σ⁵⁴ does not require activator proteins for transcription initiation.
σ⁵⁴ binds to a distinct -12/-24 promoter consensus and requires an activator ATPase for open complex formation.
σ⁵⁴ is involved in transcription termination.
σ⁵⁴ is found in eukaryotes.
Explanation - σ⁵⁴-dependent promoters need an activator to remodel the closed complex before transcription can begin.
Correct answer is: σ⁵⁴ binds to a distinct -12/-24 promoter consensus and requires an activator ATPase for open complex formation.
Q.103 Which of the following best describes the function of the bacterial protein 'NusA' during transcription?
It terminates transcription at rho‑dependent sites.
It stabilizes the transcription elongation complex and can influence pausing and termination.
It adds the 5' cap to bacterial mRNA.
It serves as a sigma factor.
Explanation - NusA interacts with RNA polymerase and nascent RNA, modulating elongation dynamics.
Correct answer is: It stabilizes the transcription elongation complex and can influence pausing and termination.
Q.104 Which of the following methods allows direct detection of nascent RNA transcripts at single‑molecule resolution within cells?
RNA‑seq
Northern blot
smFISH (single‑molecule fluorescence in situ hybridization)
Western blot
Explanation - smFISH uses fluorescent probes to visualize individual RNA molecules in fixed cells.
Correct answer is: smFISH (single‑molecule fluorescence in situ hybridization)
Q.105 Which of the following best explains why the poly(A) tail enhances mRNA stability?
It prevents ribosome binding.
It binds to poly(A) binding proteins that protect the 3' end from exonucleases.
It recruits RNase H to degrade the mRNA.
It adds extra codons to the mRNA.
Explanation - Poly(A) binding proteins (PABPs) shield the tail and interact with translation factors, stabilizing the transcript.
Correct answer is: It binds to poly(A) binding proteins that protect the 3' end from exonucleases.
Q.106 Which of the following best describes the function of the bacterial protein 'H-NS' (histone‑like nucleoid structuring protein)?
It enhances transcription of all genes.
It silences transcription of AT‑rich DNA regions by oligomerizing and forming repressive nucleoprotein structures.
It adds a cap to bacterial mRNA.
It acts as a ribosomal protein.
Explanation - H‑NS binds preferentially to AT‑rich sequences and can repress transcription by altering DNA topology.
Correct answer is: It silences transcription of AT‑rich DNA regions by oligomerizing and forming repressive nucleoprotein structures.
Q.107 Which of the following statements about 'transcriptional interference' is correct?
It occurs when overlapping transcription units collide, affecting each other's expression.
It is a mechanism used only in eukaryotes.
It enhances transcription of both overlapping genes.
It refers to the binding of transcription factors to enhancers.
Explanation - Transcriptional interference can result from promoter occlusion, polymerase collisions, or antisense transcription.
Correct answer is: It occurs when overlapping transcription units collide, affecting each other's expression.
Q.108 Which of the following is an example of a 'cis‑acting' regulatory element?
A transcription factor protein that binds DNA.
A microRNA that regulates target mRNA.
An enhancer sequence located upstream of a gene.
A histone modification enzyme.
Explanation - Cis‑acting elements are DNA sequences that influence transcription of neighboring genes.
Correct answer is: An enhancer sequence located upstream of a gene.
Q.109 In bacteria, which of the following is true about the relationship between transcription and replication forks?
They never intersect due to spatial separation.
Head‑on collisions between transcription and replication can cause DNA damage.
Replication always occurs before transcription.
RNA polymerase displaces DNA polymerase.
Explanation - Concurrent transcription and replication can lead to conflicts, especially when moving in opposite directions.
Correct answer is: Head‑on collisions between transcription and replication can cause DNA damage.
Q.110 Which of the following best describes the role of the bacterial protein 'Rho' in transcription termination?
It binds to the -35 promoter region.
It recognizes a rut site on the nascent RNA and uses ATP to translocate toward RNA polymerase, causing termination.
It adds a poly(A) tail to the RNA.
It acts as a sigma factor.
Explanation - Rho binds to C‑rich, G‑poor sequences (rut sites) and, using ATP, moves along RNA to terminate transcription.
Correct answer is: It recognizes a rut site on the nascent RNA and uses ATP to translocate toward RNA polymerase, causing termination.
Q.111 Which of the following is a characteristic of a eukaryotic gene that is transcribed by RNA polymerase III?
It contains introns that are spliced out.
It encodes a ribosomal RNA (5S rRNA) or tRNA.
It requires a TATA box located 30 bp upstream of the start site.
It is regulated by sigma factors.
Explanation - RNA polymerase III transcribes small structural RNAs such as 5S rRNA and tRNAs.
Correct answer is: It encodes a ribosomal RNA (5S rRNA) or tRNA.
Q.112 Which of the following best explains why the poly(A) tail is added after transcription termination in eukaryotes?
The poly(A) tail is required for RNA polymerase to release DNA.
Polyadenylation occurs co‑transcriptionally after cleavage at the poly(A) site, which is downstream of the termination signal.
Polyadenylation is a form of transcription termination itself.
Poly(A) tails are synthesized by RNA polymerase during elongation.
Explanation - Cleavage at the AAUAAA signal releases the nascent transcript, which is then polyadenylated.
Correct answer is: Polyadenylation occurs co‑transcriptionally after cleavage at the poly(A) site, which is downstream of the termination signal.
Q.113 Which of the following best describes the role of the bacterial protein 'RNA polymerase II subunit RPB9'?
It functions as a sigma factor.
It contributes to the fidelity of transcription by influencing proofreading.
It terminates transcription at rho‑dependent sites.
It adds the 5' cap to bacterial mRNA.
Explanation - RPB9 (in eukaryotes) enhances the accuracy of RNA synthesis and is involved in transcriptional fidelity.
Correct answer is: It contributes to the fidelity of transcription by influencing proofreading.
Q.114 Which of the following statements about 'RNA polymerase pausing' in bacteria is correct?
Pausing only occurs at terminators.
Pausing can be regulated by transcription factors such as NusA and NusG, influencing downstream events like termination.
Pausing is independent of any regulatory proteins.
Pausing always leads to transcriptional activation.
Explanation - NusA and NusG modulate elongation speed and can promote termination or antitermination.
Correct answer is: Pausing can be regulated by transcription factors such as NusA and NusG, influencing downstream events like termination.
Q.115 Which of the following best characterizes a 'leader peptide' in the context of transcription attenuation?
A short peptide encoded upstream of the structural genes that influences RNA secondary structure formation during transcription.
A peptide that directly binds RNA polymerase to inhibit transcription.
A peptide that is secreted outside the cell.
A peptide that is added to the 5' cap.
Explanation - The leader peptide's translation rate affects the formation of terminator/antiterminator hairpins in attenuation mechanisms.
Correct answer is: A short peptide encoded upstream of the structural genes that influences RNA secondary structure formation during transcription.
Q.116 In eukaryotic transcription, which of the following factors is directly responsible for the recruitment of the splicing machinery to nascent transcripts?
TFIIH
Mediator
Cap‑binding complex (CBC)
DNA polymerase δ
Explanation - CBC binds the 5' cap and interacts with spliceosomal components, linking capping to splicing.
Correct answer is: Cap‑binding complex (CBC)
Q.117 Which of the following best describes the function of the bacterial transcription factor 'DksA'?
It stimulates transcription initiation at promoters with GC‑rich -10 regions.
It binds to RNA polymerase and modulates transcription in response to the alarmone ppGpp during the stringent response.
It adds a poly(A) tail to bacterial mRNA.
It functions as a sigma factor.
Explanation - DksA cooperates with ppGpp to reprogram transcription under nutrient limitation.
Correct answer is: It binds to RNA polymerase and modulates transcription in response to the alarmone ppGpp during the stringent response.
Q.118 Which of the following is true about the 'RNA polymerase II pausing' observed near the transcription start site in many eukaryotic genes?
Pausing is caused by the presence of a strong terminator sequence.
Pausing facilitates rapid and coordinated activation of gene expression in response to signals.
Pausing leads to immediate termination of transcription.
Pausing occurs only in prokaryotes.
Explanation - Promoter‑proximal pausing creates a poised polymerase ready for swift release upon activation.
Correct answer is: Pausing facilitates rapid and coordinated activation of gene expression in response to signals.
Q.119 Which of the following best explains why the polyadenylation signal (AAUAAA) is critical for proper mRNA 3' end formation?
It serves as a ribosome binding site.
It is recognized by the cleavage and polyadenylation specificity factor (CPSF) to direct endonucleolytic cleavage and poly(A) tail addition.
It encodes a stop codon.
It initiates transcription.
Explanation - CPSF binds AAUAAA, positioning the endonuclease and poly(A) polymerase for 3' processing.
Correct answer is: It is recognized by the cleavage and polyadenylation specificity factor (CPSF) to direct endonucleolytic cleavage and poly(A) tail addition.