Transcription Questions in English

(B) In prokaryotic transcription,the $RNA$ polymerase enzyme consists of a core enzyme (composed of subunits $\alpha_2, \beta, \beta', \omega$) and a sigma $(\sigma)$ factor.
The core enzyme is capable of catalyzing the polymerization of ribonucleotides into $RNA$ but lacks the specificity to initiate transcription at the correct site.
The sigma $(\sigma)$ factor is responsible for the recognition of the promoter sequence on the $DNA$ template.
Once the $RNA$ polymerase holoenzyme (core enzyme + $\sigma$ factor) binds to the promoter,the $\sigma$ factor facilitates the initiation of transcription and is subsequently released.
Therefore,the identification and binding of $RNA$ polymerase to the promoter is the function of the sigma factor.

(D) In prokaryotes,transcription occurs in the cytoplasm,and because there is no nuclear membrane,it is often coupled with translation. Prokaryotic mRNA does not require splicing,capping,or tailing (polyadenylation). Therefore,statement $(a)$ and $(b)$ are correct,while statement $(c)$ is incorrect because capping is not essential in prokaryotes. Thus,only $(c)$ is incorrect.

(A) The Pribnow box,also known as the $-10$ element,is a specific sequence of nucleotides found in the promoter region of prokaryotic genes,particularly in $E. coli$.
It consists of the consensus sequence $TATAAT$.
This sequence is essential for the binding of the $RNA$ polymerase holoenzyme to initiate transcription.
Therefore,the correct option is $A$.

(B) Tailing is a post-transcriptional modification process in eukaryotes. During this process,$200-300$ adenylate residues (poly-$A$ tail) are added to the $3'$ end of the $mRNA$ precursor. This addition occurs in a template-independent manner,catalyzed by the enzyme poly-$A$ polymerase. The poly-$A$ tail helps in the stabilization of $mRNA$ and its transport from the nucleus to the cytoplasm.

(C) In eukaryotes, there are three distinct types of $RNA$ polymerases: $RNA$ polymerase $I$ (transcribes $rRNAs$), $RNA$ polymerase $II$ (transcribes $mRNA$ precursor), and $RNA$ polymerase $III$ (transcribes $tRNA$, $5s$ $rRNA$, and $snRNAs$). Thus, the Assertion is correct.
Regarding the Reason, $RNA$ polymerase in prokaryotes (specifically the holoenzyme) consists of a core enzyme $(\alpha_2 \beta \beta' \omega)$ and a sigma $(\sigma)$ factor. The rho $(\rho)$ factor is a separate protein involved in the termination of transcription in prokaryotes, not a structural component of the $RNA$ polymerase enzyme itself. Therefore, the Reason is incorrect.

(D) In eukaryotes,there are three types of $RNA$ polymerases ($RNA$ polymerase $I$,$II$,and $III$) that transcribe different types of $RNA$. In bacteria (prokaryotes),a single $DNA$-dependent $RNA$ polymerase catalyses the transcription of all types of $RNA$ ($mRNA$,$tRNA$,and $rRNA$). Thus,the Assertion is incorrect because eukaryotes use multiple polymerases,not a single one.
Structural genes in bacteria are polycistronic (a single promoter regulates multiple genes),whereas in eukaryotes,they are monocistronic. Thus,the Reason is also incorrect.

(C) $1$. The Assertion is correct: In prokaryotes, the $RNA$ polymerase enzyme consists of a core enzyme and a sigma ($\sigma$) factor. The sigma factor specifically recognizes the promoter site and initiates transcription.
$2$. The Reason is incorrect: The promoter region is defined with respect to the coding strand. It is located at the $5'$ end of the coding strand (upstream), not the template strand. Therefore, the statement in the Reason is scientifically inaccurate.

(A) $RNA$ processing (also known as post-transcriptional modification) is a series of events that occur after the transcription of $RNA$ from $DNA$.
In eukaryotic cells,the primary transcript (pre-$mRNA$) undergoes several modifications before it is exported from the nucleus to the cytoplasm for translation.
These modifications include:
$1$. $5'$-capping: Addition of a $7$-methylguanosine cap to the $5'$ end.
$2$. Polyadenylation: Addition of a poly-$A$ tail to the $3'$ end.
$3$. Splicing: Removal of non-coding introns and joining of coding exons.
Therefore,$RNA$ processing is an event that occurs after $RNA$ is transcribed.

(B) Option $B$ is correct. In bacteria,the process of transcription is terminated when the $RNA$ polymerase enzyme encounters the $Rho$ factor ($ρ$ factor),which helps in the dissociation of the $RNA$ polymerase from the $DNA$ template.
Option $A$ is incorrect because in capping,methyl guanosine triphosphate is added to the $5^{\prime}$ end of $hnRNA$,not the $3^{\prime}$ end.
Option $C$ is incorrect because the template strand (not the coding strand) is copied to an $mRNA$ during transcription.
Option $D$ is incorrect because split gene arrangement (presence of introns and exons) is a characteristic feature of eukaryotes,not prokaryotes.

(B) In eukaryotes,there are three types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase $I$ transcribes $rRNAs$ ($28S, 18S,$ and $5.8S$).
$2$. $RNA$ polymerase $II$ transcribes the precursor of $mRNA$,which is called heterogeneous nuclear $RNA$ $(hnRNA)$.
$3$. $RNA$ polymerase $III$ is responsible for the transcription of $tRNA, 5S$ $rRNA,$ and small nuclear $RNA$ $(snRNA)$.
Therefore,the correct option is $B$.

(B) In prokaryotes,the process of transcription is carried out by a single type of enzyme known as $DNA$ dependent $RNA$ polymerase.
This enzyme is responsible for all three stages of transcription:
$1$. Initiation: The enzyme binds to the promoter site on the $DNA$ template with the help of a sigma $(\sigma)$ factor.
$2$. Elongation: The enzyme facilitates the polymerization of ribonucleotides to form the $RNA$ chain.
$3$. Termination: Upon reaching the terminator sequence,the enzyme dissociates from the $DNA$ template with the help of a rho $(\rho)$ factor.
Therefore,$DNA$ dependent $RNA$ polymerase is the only enzyme required for the entire process.

(D) Magnesium $(Mg^{2+})$ is an essential cofactor for the enzyme $RNA$ polymerase,which is the primary enzyme responsible for the process of transcription. It helps in the stabilization of the enzyme-substrate complex and is crucial for the catalytic activity of $RNA$ polymerase during the synthesis of $RNA$ from a $DNA$ template.

(B) The coding strand of $DNA$ has the same sequence as the $mRNA$ transcript,except that $Thymine$ $(T)$ in $DNA$ is replaced by $Uracil$ $(U)$ in $mRNA$.
Given coding strand sequence: $AAGCCTATCAG$.
Replacing $T$ with $U$ gives the $mRNA$ sequence: $AAGCCUAUCAG$.

(D) In a transcription unit,the promoter is located at the $5'$ end of the coding strand (upstream).
In the given diagram,$P$ represents the promoter region located at the $5'$ end of the upper strand.
The upper strand,which runs from $3'$ to $5'$ (relative to the promoter's direction),acts as the template strand.
The lower strand,which runs from $5'$ to $3'$,is the coding strand.
Thus,$S$ represents the coding strand.
Therefore,the coding strand is $S$ and the promoter is $P$.

(B) Capping is a post-transcriptional modification process in eukaryotes.
In this process,an unusual nucleotide,$7$-methyl guanosine triphosphate,is added to the $5'$ end of the heterogeneous nuclear $RNA$ $(hnRNA)$.
This cap protects the $5'$ end of the $mRNA$ from degradation by exonucleases and helps in the initiation of translation by facilitating ribosome binding.
Therefore,the correct option is $B$.

(D) Transcription is the process of copying genetic information from one strand of the $DNA$ into $RNA$.
This process is catalyzed by the enzyme $DNA$-dependent $RNA$ polymerase.
It uses $DNA$ as a template to synthesize a complementary $RNA$ strand.
Therefore,the correct option is $D$.

(B) The provided figure illustrates the process of transcription in eukaryotes,specifically highlighting post-transcriptional modifications.
Key features shown in the diagram include:
$1$. Transcription of $DNA$ into primary transcript (pre-mRNA).
$2$. Capping: Addition of a methyl guanosine triphosphate cap at the $5'$ end.
$3$. Splicing: Removal of introns and joining of exons.
$4$. Tailing (Polyadenylation): Addition of a poly-$A$ tail at the $3'$ end.
These modifications are characteristic of eukaryotic gene expression,as prokaryotes do not undergo such extensive processing of mRNA.

(B) In eukaryotes, there are three main types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase-$I$ is responsible for the transcription of $rRNA$ ($18S, 28S,$ and $5.8S$). Thus, $P-I$.
$2$. $RNA$ polymerase-$II$ transcribes the precursor of $mRNA$, which is called heterogeneous nuclear $RNA$ $(hnRNA)$. Thus, $Q-III$.
$3$. $RNA$ polymerase-$III$ is responsible for the transcription of $tRNA$, $5S rRNA$, and $snRNAs$ (small nuclear $RNAs$). Thus, $R-II$.
Therefore, the correct matching is $(P-I), (Q-III), (R-II)$.

(C) The process of copying genetic information from one strand of $DNA$ into $RNA$ is known as $Transcription$.
$Replication$ is the process of duplicating $DNA$.
$Translation$ is the process of polymerizing amino acids to form a polypeptide chain based on the sequence of nucleotides in $mRNA$.
Therefore,the correct answer is $Transcription$.

(A) In a transcription unit,the promoter is located at the $5'$ end of the structural gene.
It is referred to as being upstream with respect to the coding strand.
The promoter provides the binding site for $RNA$ polymerase and defines the direction of transcription.

(A) In a transcription unit,the structural gene is flanked by the promoter and the terminator.
$1$. The promoter is located towards the $5'$ end (upstream) of the structural gene with respect to the coding strand.
$2$. The terminator is located towards the $3'$ end (downstream) of the structural gene with respect to the coding strand.
Therefore,the terminator defines the end of the process of transcription.

(A) The provided figure illustrates the process of transcription in prokaryotes.
$1$. The process involves three main stages: Initiation, Elongation, and Termination.
$2$. In prokaryotes, the $RNA$ polymerase enzyme binds to the promoter region with the help of the sigma ($\sigma$) factor to initiate transcription.
$3$. During elongation, the $RNA$ polymerase synthesizes the $RNA$ strand.
$4$. Termination occurs when the $RNA$ polymerase encounters a terminator sequence and uses the rho ($\rho$) factor to release the $RNA$ transcript and the enzyme from the $DNA$ template.

(D) In prokaryotes,transcription is simpler than in eukaryotes.
$1$. Only a single type of $RNA$ polymerase is involved in the transcription of all types of $RNA$ ($mRNA$,$tRNA$,and $rRNA$) in bacteria.
$2$. Prokaryotic $mRNA$ does not require processing such as splicing (removal of introns),capping (addition of $7$-methylguanosine at the $5'$ end),or polyadenylation (addition of poly-$A$ tail at the $3'$ end).
$3$. These processes are characteristic of eukaryotic transcription.
Therefore,all the given options ($A$,$B$,and $C$) are incorrect regarding prokaryotic transcription,making $D$ the correct choice.

(D) In prokaryotes,transcription and translation are coupled because both processes occur in the same compartment (cytoplasm) and there is no nuclear membrane to separate them.
$Lactobacillus$ is a bacterium (prokaryote),whereas yeast,Amoeba,and Hibiscus are eukaryotes.
In eukaryotes,transcription occurs in the nucleus and translation occurs in the cytoplasm,separated by the nuclear envelope,preventing coupling.

(D) In eukaryotic cells,the primary transcript produced by $RNA$ polymerase $II$ is known as $hnRNA$ (heterogeneous nuclear $RNA$).
$hnRNA$ contains both coding sequences (exons) and non-coding sequences (introns).
To become functional,$hnRNA$ undergoes a process called post-transcriptional modification or processing.
This process includes splicing (removal of introns and joining of exons),capping (addition of $7$-methylguanosine at the $5'$ end),and tailing (addition of a poly-$A$ tail at the $3'$ end).
After these modifications,the processed $RNA$ is known as mature $mRNA$ (messenger $RNA$),which is then transported to the cytoplasm for translation.

(C) In eukaryotes,there are three distinct types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase-$I$ is responsible for the transcription of $rRNA$ ($28S, 18S,$ and $5.8S$).
$2$. $RNA$ polymerase-$II$ is responsible for the transcription of the precursor of $mRNA$,known as heterogeneous nuclear $hnRNA$ (heterogeneous nuclear $RNA$).
$3$. $RNA$ polymerase-$III$ is responsible for the transcription of $tRNA$,$5S$ $rRNA$,and $snRNA$ (small nuclear $RNA$).
Therefore,the correct enzyme for $tRNA$ synthesis is $RNA$ polymerase-$III$.

(A) In the process of transcription,the $RNA$ polymerase enzyme is responsible for synthesizing $RNA$ from a $DNA$ template.
This enzyme recognizes and binds to a specific sequence on the $DNA$ called the promoter region to initiate transcription.
The promoter acts as the starting signal for the $RNA$ polymerase to begin the synthesis of the $RNA$ strand.

(C) In eukaryotes,there are three major types of $RNA$ polymerases involved in transcription:
$RNA$ polymerase $I$ transcribes $rRNAs$ ($5.8S, 18S,$ and $28S$).
$RNA$ polymerase $II$ transcribes $hnRNA$ (precursor of $mRNA$).
$RNA$ polymerase $III$ transcribes $tRNAs$,$5S$ $rRNA$,and $snRNA$ (small nuclear $RNA$).

(D) In the process of transcription,the coding strand of $DNA$ has the same sequence as the mRNA,except that thymine $(T)$ is present in the $DNA$ instead of uracil $(U)$ in the $RNA$.
Given mRNA sequence: $5' AUCGAUCGAUCGAUCGAUCGAUCGAUCG 3'$
By replacing uracil $(U)$ with thymine $(T)$,we obtain the sequence of the coding strand:
Coding strand sequence: $5' ATCGATCGATCGATCGATCGATCGATCG 3'$
Therefore,the correct option is $D$.

(C) transcription unit of $DNA$ is defined primarily by three regions in the $DNA$:
$(i)$ $A$ promoter
$(ii)$ The structural gene
$(iii)$ $A$ terminator
The promoter is located towards the $5'$-end (upstream) of the structural gene (the reference is made with respect to the polarity of the coding strand).
The terminator is located towards the $3'$-end (downstream) of the coding strand.

(D) The process of transcription involves the synthesis of $RNA$ from a $DNA$ template strand.
The $DNA$ template strand is given as: $3'TACATGGCAAATATCCATTCA5'$.
According to the base-pairing rules for $RNA$ synthesis:
$A$ (Adenine) in $DNA$ pairs with $U$ (Uracil) in $RNA$.
$T$ (Thymine) in $DNA$ pairs with $A$ (Adenine) in $RNA$.
$C$ (Cytosine) in $DNA$ pairs with $G$ (Guanine) in $RNA$.
$G$ (Guanine) in $DNA$ pairs with $C$ (Cytosine) in $RNA$.
Applying these rules to the template $3'TACATGGCAAATATCCATTCA5'$:
$T \rightarrow A$
$A \rightarrow U$
$C \rightarrow G$
$A \rightarrow U$
$T \rightarrow A$
$G \rightarrow C$
$G \rightarrow C$
$C \rightarrow G$
$A \rightarrow U$
$A \rightarrow U$
$A \rightarrow U$
$T \rightarrow A$
$A \rightarrow U$
$T \rightarrow A$
$C \rightarrow G$
$C \rightarrow G$
$A \rightarrow U$
$T \rightarrow A$
$T \rightarrow A$
$C \rightarrow G$
$A \rightarrow U$
Thus,the resulting $mRNA$ sequence in the $5' \rightarrow 3'$ direction is: $5'AUGUACCGUUUAUAGGUAAGU3'$.

(C) The correct matching is as follows:
- $A$. $RNA$ polymerase $III$ is responsible for the transcription of $snRNAs$,$tRNA$,and $5s$ $rRNA$. Thus,$A-IV$.
- $B$. The termination of transcription in prokaryotes is facilitated by the Rho factor. Thus,$B-III$.
- $C$. Splicing of exons involves the removal of introns and joining of exons,a process performed by $snRNPs$ (small nuclear ribonucleoproteins). Thus,$C-I$.
- $D$. The $TATA$ box is a specific $DNA$ sequence found in the promoter region of the transcription unit. Thus,$D-II$.
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(B) In eukaryotic cells,the primary transcript (hnRNA) undergoes several post-transcriptional modifications before it is exported to the cytoplasm as mature mRNA.
$1$. Splicing: The non-coding introns are removed,and the coding exons are joined together $(B)$.
$2$. Capping: $A$ methyl guanosine triphosphate is added to the $5$' end of the hnRNA $(C)$.
$3$. Tailing (Polyadenylation): Adenylate residues are added at the $3$' end of the hnRNA $(D)$.
Transport of pre-mRNA to the cytoplasm occurs only after these processing steps are completed,not prior to splicing ($A$ is incorrect).
Base pairing of complementary RNAs is not a standard post-transcriptional processing event for mRNA maturation ($E$ is incorrect).
Therefore,the correct events are $B, C,$ and $D$.

(C) In prokaryotic transcription,the process is carried out by $RNA$ polymerase. The $RNA$ polymerase enzyme consists of a core enzyme and a sigma factor $(\sigma)$. The sigma factor is responsible for the initiation of transcription by recognizing the promoter site. Once the elongation phase is complete,the termination of transcription occurs. This termination is facilitated by the rho factor $(\rho)$. Therefore,the $\rho$ factor is essential for the termination of transcription.

(A) In bacteria,the genetic material is not enclosed within a membrane-bound nucleus; it is present in the cytoplasm.
Because there is no nuclear membrane,the processes of transcription (synthesis of $\text{mRNA}$ from $\text{DNA}$) and translation (synthesis of protein from $\text{mRNA}$) occur in the same cellular compartment.
Furthermore,bacterial $\text{mRNA}$ does not require complex post-transcriptional processing (like splicing,capping,or tailing) to become active.
As a result,translation can begin even before the transcription of the $\text{mRNA}$ molecule is fully completed.
This phenomenon is known as coupled transcription and translation.
Therefore,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(A) In eukaryotes, there are three types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase-$I$ transcribes $rRNA$ ($28S$, $18S$, and $5.8S$).
$2$. $RNA$ polymerase-$II$ transcribes the precursor of $mRNA$ (heterogeneous nuclear $RNA$ or $hnRNA$).
$3$. $RNA$ polymerase-$III$ is responsible for the transcription of $tRNA$, $5S$ $rRNA$, and $snRNA$.
Therefore, $5S$ $rRNA$ is synthesized by $RNA$ polymerase-$III$, not $RNA$ polymerase-$I$.

(C) Statement $(i)$ is correct: During transcription,multiple $\text{RNA}$ molecules can be synthesized from a single $\text{DNA}$ template.
Statement $(ii)$ is incorrect: The segment of $\text{DNA}$ that is transcribed into $\text{RNA}$ is called a transcription unit,not a translational unit.
Statement $(iii)$ is correct: The promoter is a sequence of $\text{DNA}$ located at the $5'$ end (upstream) of the structural gene,which provides the binding site for $\text{RNA}$ polymerase.
Statement $(iv)$ is correct: $A$ transcription unit consists of a promoter,a structural gene,and a terminator. The promoter and terminator flank the structural gene.
Therefore,statements $(i), (iii),$ and $(iv)$ are correct.

(A) $1$. In transcription,only one strand of the $dsDNA$ (double-stranded $DNA$) is copied into $RNA$ because if both strands were copied,they would produce two $RNA$ molecules with complementary sequences,which would form a double-stranded $RNA$ and prevent translation.
$2$. Option $B$ is incorrect because eukaryotes use $RNA$ polymerases $(I, II, III)$,not $DNA$ polymerase,for transcription.
$3$. Option $C$ is incorrect because in prokaryotes,there is no nucleus,so transcription and translation occur in the same compartment (cytoplasm).
$4$. Option $D$ is incorrect because the primary transcript in eukaryotes contains both exons and introns (it is a pre-$mRNA$ that undergoes splicing to remove introns).
Therefore,the correct statement is $A$.

(C) In a transcription unit,the promoter is a $DNA$ sequence that provides a binding site for $RNA$ polymerase.
It is located towards the $5'$ end (upstream) of the coding strand (also known as the sense strand).
Since the template strand runs in the $3' \rightarrow 5'$ direction,the promoter is situated at the $5'$ end of the coding strand,which corresponds to the upstream region of the transcription start site.

(C) Option $A$ is incorrect because tailing involves the addition of adenylate residues at the $3'$-end, not the $5'$-end.
Option $B$ is incorrect because the strand that does not code for anything is called the template strand (or antisense strand).
Option $C$ is correct. Split-gene arrangements (presence of introns) are considered an ancient feature of the genome, reflecting the '$RNA$ world' hypothesis.
Option $D$ is incorrect because the sigma factor ($\sigma$) is responsible for the initiation of transcription, while the rho factor ($\rho$) is responsible for termination in bacteria.

(D) Assertion $(A)$ is correct: In eukaryotes, the primary transcript, known as $hnRNA$ (heterogeneous nuclear $RNA$), is non-functional because it contains both coding sequences (exons) and non-coding sequences (introns). It must undergo splicing to remove introns and join exons to become functional $mRNA$.
Reason $(R)$ is correct: In bacteria (prokaryotes), there is no defined nucleus, and transcription and translation occur in the same compartment (cytoplasm). Furthermore, bacterial $mRNA$ does not require complex processing like splicing because it lacks introns.
Conclusion: While both statements are scientifically accurate, the Reason $(R)$ describes the nature of bacterial gene expression and does not explain why eukaryotic $hnRNA$ contains introns. Therefore, Reason is not the correct explanation of Assertion.

(D) In a $\text{DNA}$ molecule,the coding strand has the same sequence as the $\text{mRNA}$ transcript,except that thymine $(T)$ in $\text{DNA}$ is replaced by uracil $(U)$ in $\text{RNA}$.
Given coding strand: $\text{5'-ATCGATCGACTG-3'}$.
Since the $\text{mRNA}$ is complementary to the template strand and identical to the coding strand (with $U$ instead of $T$),the $\text{mRNA}$ sequence will be $\text{5'-AUCGAUCGACUG-3'}$.
Therefore,the correct option is $D$.

(A) During the process of transcription,$\text{RNA}$ polymerase catalyzes the synthesis of $\text{RNA}$ from a $DNA$ template.
This enzyme utilizes ribonucleoside triphosphates $(\text{NTPs})$ as substrates.
These $\text{NTPs}$ are polymerized in a template-dependent manner,meaning the sequence of the newly synthesized $\text{RNA}$ is complementary to the $DNA$ template strand.
Deoxyribonucleoside triphosphates $(\text{dNTPs})$ are used by $DNA$ polymerase during $DNA$ replication,not by $\text{RNA}$ polymerase during transcription.

(B) In the process of transcription,the $DNA$ strand with polarity $5' \rightarrow 3'$ is known as the coding strand (or sense strand) because its sequence is identical to the mRNA produced (except for Uracil replacing Thymine).
The $DNA$ strand with polarity $3' \rightarrow 5'$ acts as the template for $RNA$ polymerase and is known as the non-coding strand,template strand,or antisense strand.
Looking at the diagram:
- Strand $A$ has $5' \rightarrow 3'$ polarity,so it is the coding strand (statement $a$ is correct).
- Strand $B$ has $3' \rightarrow 5'$ polarity,so it is the template/antisense strand (statement $c$ is correct).
Therefore,statements $a$ and $c$ are correct.

(A) In eukaryotes,the primary transcript (pre-mRNA) contains both exons (coding sequences) and introns (non-coding sequences).
This arrangement is known as a split gene.
Because introns do not code for proteins,they must be removed through a process called splicing.
Additionally,the pre-mRNA undergoes capping and tailing to become functional mRNA.
Since the presence of split genes (introns) necessitates the removal of non-coding sequences,$RNA$ processing is essential in eukaryotes.
Therefore,both the assertion and the reason are true,and the reason correctly explains why $RNA$ processing occurs.

(A) In eukaryotes,there are three main types of $\text{RNA}$ polymerases involved in transcription:
$1$. $\text{RNA}$ polymerase-$I$ transcribes $\text{r-RNA}$ ($28\text{S}, 18\text{S},$ and $5.8\text{S}$).
$2$. $\text{RNA}$ polymerase-$II$ transcribes the precursor of $\text{m-RNA}$,called heterogeneous nuclear $\text{RNA}$ $(\text{hn-RNA})$.
$3$. $\text{RNA}$ polymerase-$III$ is responsible for the transcription of $\text{t-RNA}$,$5\text{S r-RNA}$,and $\text{sn-RNA}$ (small nuclear $\text{RNA}$).
Therefore,the correct option is $A$.

(C) In eukaryotes, the primary transcript known as $hnRNA$ (heterogeneous nuclear $RNA$) undergoes post-transcriptional modifications such as splicing, capping, and tailing.
These processes occur exclusively within the $nucleus$ before the mature $mRNA$ is transported into the cytoplasm for translation.

(C) In eukaryotic cells,the primary transcript $(hnRNA)$ undergoes three major post-transcriptional modifications to become functional $mRNA$:
$1$. $Capping$: $A$ methyl guanosine triphosphate $(m^7G)$ is added to the $5'$-end of the $hnRNA$.
$2$. $Splicing$: The non-coding introns are removed,and the coding exons are joined together in a defined order.
$3$. $Tailing$ $(Polyadenylation)$: Poly-$A$ tails (approx. $200-300$ adenine residues) are added to the $3'$-end of the transcript.
Therefore,the correct sequence is $Capping$,$Splicing$,$Tailing$.

(C) In a transcription unit,the $DNA$ strand with polarity $3' \rightarrow 5'$ acts as the template strand,and the strand with $5' \rightarrow 3'$ polarity acts as the coding strand.
Option $A$ is correct: The promoter is located at the $5'$ end of the structural gene (upstream).
Option $B$ is correct: $DNA$ dependent $RNA$ polymerase catalyzes polymerization only in the $5' \rightarrow 3'$ direction.
Option $C$ is incorrect: The transcribed $RNA$ is complementary to the template strand (antisense strand) and identical to the coding strand (sense strand),except for the presence of Uracil instead of Thymine. It is not similar to the antisense strand.
Option $D$ is correct: The terminator is located at the $3'$ end of the coding strand (downstream).

(D) The correct matches are as follows:
$i$. Cistron: It is a segment of $DNA$ coding for a polypeptide $(c)$.
$ii$. Exon: These are coding sequences that appear in mature or processed $mRNA$ $(d)$.
$iii$. Intron: These are non-coding sequences (intervening sequences) which are removed during the process of splicing $(b)$.
$iv$. Capping: It involves the addition of a methylated guanosine triphosphate $(GTP)$ to the $5'$-end of $hnRNA$ $(a)$.
Therefore,the correct matching is $i-c, ii-d, iii-b, iv-a$.