Transcription Questions in English

(B) During the process of transcription in prokaryotes,$RNA$ polymerase is the enzyme responsible for synthesizing $RNA$ from a $DNA$ template.
$RNA$ polymerase by itself is only capable of catalyzing the process of elongation.
It requires specific initiation factors (like the $\sigma$ factor) to initiate the process at the promoter site and termination factors (like the $\rho$ factor) to terminate the process.
These factors associate with the core enzyme of $RNA$ polymerase to alter its specificity,allowing it to recognize specific promoter or terminator sequences on the $DNA$ strand.

(D) In eukaryotic cells (such as animal cells,plant cells,and Amoeba),the primary transcript $(hnRNA)$ undergoes processing like splicing,capping,and tailing to form functional $m-RNA$.
However,in prokaryotic cells (such as Blue-green algae or Cyanobacteria),transcription and translation are coupled,and the $m-RNA$ does not require any post-transcriptional processing (splicing) because prokaryotic genes are not split (they lack introns).
Therefore,Blue-green algae is the correct answer.

(B) In prokaryotes,the $RNA$ polymerase enzyme consists of a core enzyme and a sigma factor $(\sigma)$.
The core enzyme is responsible for the polymerization of ribonucleotides.
The sigma factor $(\sigma)$ specifically recognizes the promoter site on the $DNA$ template.
When the sigma factor binds to the $RNA$ polymerase,it facilitates the binding of the enzyme to the promoter region,thereby initiating the process of transcription.
Therefore,the sigma factor is essential for the initiation of transcription.

(B) In eukaryotic cells,there are three main types of $RNA$ involved in protein synthesis: $mRNA$ (messenger $RNA$),$tRNA$ (transfer $RNA$),and $rRNA$ (ribosomal $RNA$).
Correspondingly,there are three distinct types of $RNA$ polymerases responsible for their transcription:
$1$. $RNA$ polymerase $I$ transcribes $rRNAs$ ($28S, 18S,$ and $5.8S$).
$2$. $RNA$ polymerase $II$ transcribes the precursor of $mRNA$,the heterogeneous nuclear $RNA$ $(hnRNA)$.
$3$. $RNA$ polymerase $III$ transcribes $tRNA$,$5S$ $rRNA$,and $snRNAs$ (small nuclear $RNAs$).
Therefore,there are $3$ types of $RNA$ and $3$ types of $RNA$ polymerases.

(A) $snRNAs$ stands for $Small\, Nuclear\, RNAs$.
These are small $RNA$ molecules found within the nucleus of eukaryotic cells.
They play a crucial role in the processing of pre-messenger $RNA$ $(pre-mRNA)$ into mature $mRNA$ by participating in the splicing process within the spliceosome complex.

(A) In eukaryotes,there are three 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 precursor of $mRNA$ (heterogeneous nuclear $RNA$ or $hnRNA$).
$3$. $RNA$ polymerase $III$ is responsible for the transcription of $tRNA$,$5S$ $rRNA$,and $snRNA$.
Therefore,$18S$ $rRNA$ is synthesized by $RNA$ polymerase $I$.

(C) 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 $mRNA$,which is $hnRNA$ (heterogeneous nuclear $RNA$).
$3$. $RNA$ polymerase $III$ is responsible for the transcription of $tRNA$,$5S$ $rRNA$,and $snRNA$ (small nuclear $RNA$).
Therefore,the correct option is $C$.

(D) In eukaryotes,there are three main types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase $I$ transcribes $r-RNA$ ($28S, 18S,$ and $5.8S$).
$2$. $RNA$ polymerase $II$ transcribes the precursor of $m-RNA$,which is known as heterogeneous nuclear $RNA$ $(hn-RNA)$.
$3$. $RNA$ polymerase $III$ transcribes $t-RNA$,$5S-rRNA$,and $Sn-RNA$ (small nuclear $RNA$).

(B) In eukaryotic cells,the primary transcript produced by $RNA$ polymerase $II$ is known as heterogeneous nuclear $RNA$ $(hnRNA)$.
This transcript undergoes processing,including splicing,capping,and tailing,to become mature $mRNA$ (messenger $RNA$).
Therefore,$hnRNA$ stands for heterogeneous nuclear $RNA$.

(D) In eukaryotic cells,the primary transcript $hnRNA$ (heterogeneous nuclear $RNA$) is non-functional and undergoes three major post-transcriptional modifications to become functional $mRNA$:
$1$. Capping: $A$ methyl guanosine triphosphate is added to the $5'$ end of $hnRNA$.
$2$. Splicing: The non-coding introns are removed,and the coding exons are joined in a defined order.
$3$. Tailing (Polyadenylation): Adenylate residues are added at the $3'$ end in a template-independent manner.
Therefore,all these processes are essential for the maturation of $hnRNA$.

(C) In eukaryotic cells,the primary transcript (pre-$mRNA$) contains both coding sequences (exons) and non-coding sequences (introns).
To produce a functional $mRNA$ molecule,the non-coding introns must be removed and the coding exons must be joined together in a specific order.
This process is known as $Splicing$.
$Capping$ involves the addition of a methylguanosine triphosphate to the $5'$ end,while $Tailing$ involves the addition of poly-$A$ residues to the $3'$ end.

(A) Capping is a post-transcriptional modification process in eukaryotic cells.
During this process,an unusual nucleotide,$7$-methylguanosine $(m^7G)$,is added to the $5'$-end of the primary transcript (pre-$mRNA$).
This addition is mediated by the enzyme guanylyltransferase.
The cap protects the $mRNA$ from degradation by exonucleases and assists in ribosome binding during translation.

(B) In eukaryotes,the primary transcript produced by $RNA$ polymerase $II$ is known as heterogeneous nuclear $RNA$ $(hnRNA)$.
$hnRNA$ undergoes two main processing steps: splicing (removal of introns and joining of exons) and capping/tailing (addition of $5'$ methyl guanosine cap and $3'$ poly-$A$ tail).
After these processing steps,the mature and functional $RNA$ is called messenger $RNA$ $(mRNA)$,which is then transported out of the nucleus for translation.

(B) In eukaryotic cells,the process of post-transcriptional modification involves 'tailing'.
During this process,$200-300$ adenylate residues (poly-$A$ tail) are added at the $3'$-end of the primary transcript (hnRNA) in a template-independent manner.
This polyadenylation helps in the stability of mRNA and its transport from the nucleus to the cytoplasm.

(A) The process of post-transcriptional modification in eukaryotes involves three main steps:
$1$. Splicing: The process of removing non-coding sequences called introns and joining the coding sequences called exons is known as splicing. Thus, $(a-3)$.
$2$. Capping: In this process, an unusual nucleotide, methyl guanosine triphosphate, is added to the $5'$ end of the $hnRNA$. Thus, $(b-4)$.
$3$. Tailing: In this process, adenylate residues (about $200-300$) are added at the $3'$ end in a template-independent manner. Thus, $(c-2)$.
$4$. $mRNA$: After these modifications, the fully processed $hnRNA$ is called $mRNA$, which contains only exons. Thus, $(d-1)$.
Therefore, the correct matching is $a-3, b-4, c-2, d-1$.

(D) The given figure represents a transcription unit in $DNA$.
In a transcription unit,the promoter is located towards the $5'$ end (upstream) of the structural gene.
It is a $DNA$ sequence that provides binding sites for $RNA$ polymerase and is the presence of a promoter in a transcription unit that also defines the template and coding strands.
In the diagram,$X$ represents the promoter region,which initiates the process of transcription.

(C) In the process of transcription,a transcription unit in $DNA$ is defined primarily by three regions: a promoter,a structural gene,and a terminator.
$1$. The promoter is located towards the $5'$ end (upstream) of the structural gene.
$2$. The structural gene is the segment of $DNA$ that is transcribed into $RNA$.
$3$. The terminator is located towards the $3'$ end (downstream) of the coding strand.
In the provided diagram,$X$ represents the promoter (at the $5'$ end) and $Y$ represents the terminator (at the $3'$ end).

(C) In the process of transcription,a transcription unit consists of three regions: a promoter,a structural gene,and a terminator.
In the provided diagram,$X$ represents the promoter,which is located at the $5'$ end of the coding strand.
$Y$ represents the terminator,which is located at the $3'$ end of the coding strand.
$Z$ refers to the structural gene,which is the segment of $DNA$ that is transcribed into $RNA$.

(B) The provided figure shows the transcription process in bacteria.
In the figure, the $RNA$ polymerase enzyme is associated with the $\sigma$ factor.
The $\sigma$ factor is responsible for the initiation of transcription by recognizing the promoter site on the $DNA$ template.
Once the initiation is complete, the $\sigma$ factor dissociates from the $RNA$ polymerase, and the elongation process begins.
Therefore, the presence of the $\sigma$ factor clearly indicates the initiation stage of transcription.

(D) The figure illustrates the process of transcription in eukaryotes.
Key features shown include:
$1$. The presence of introns and exons in the primary transcript (pre-mRNA).
$2$. Post-transcriptional modifications such as Capping (addition of methyl guanosine triphosphate at the $5'$ end),$RNA$ splicing (removal of introns and joining of exons),and Polyadenylation (addition of a poly-$A$ tail at the $3'$ end).
These processes are characteristic of eukaryotic gene expression,as prokaryotes do not undergo such complex post-transcriptional processing.

(B) The $TATA$ box (also known as the $Goldberg-Hogness$ box) is a $DNA$ sequence found in the promoter region of genes in eukaryotes and archaea. It serves as a binding site for transcription factors and $RNA$ polymerase,playing a crucial role in the initiation of transcription.

(C) Transcription is the process of copying a segment of $DNA$ into $RNA$.
It is most similar to $DNA$ replication because both processes involve the synthesis of a new nucleic acid strand based on a template strand of $DNA$.
In both cases,the enzyme ($DNA$ polymerase in replication and $RNA$ polymerase in transcription) reads the template strand in the $3' \rightarrow 5'$ direction and synthesizes the new strand in the $5' \rightarrow 3'$ direction,following the principle of complementary base pairing.

(A) The given sequence shows the flow of genetic information from $DNA$ to protein.
Step $(1)$: The $DNA$ sequence $(TAC \, AAG \, GCG \, AUA \, CGA)$ is converted into an mRNA sequence $(AUG \, UUC \, CGC \, UAU \, GCU)$. This process of synthesizing mRNA from a $DNA$ template is known as Transcription.
Step $(2)$: The mRNA sequence is then decoded into a polypeptide chain of amino acids $(Met - Phe - Arg - Tyr - Ala)$. This process of protein synthesis from mRNA is known as Translation.
Therefore,process $(1)$ is Transcription and process $(2)$ is Translation.

(C) The process of transcription in bacteria consists of three stages: Initiation, Elongation, and Termination.
$1$. In image $(a)$, the $RNA$ polymerase binds to the promoter region with the help of the sigma factor $(\sigma)$, which marks the initiation of transcription.
$2$. In image $(c)$, the $RNA$ polymerase moves along the $DNA$ template, facilitating the elongation of the $RNA$ chain.
$3$. In image $(b)$, the Rho factor $(\rho)$ binds to the $RNA$ polymerase at the terminator region, leading to the termination of transcription and the release of the nascent $RNA$.

(B) In the process of transcription,the coding strand of $DNA$ $(5' \to 3')$ has the same sequence as the $mRNA$ transcript,except that $Thymine$ $(T)$ is replaced by $Uracil$ $(U)$ in $RNA$.
Given coding strand: $5'AATTCAAATTAGG3'$.
Replacing $T$ with $U$,the $mRNA$ sequence is $5'AAUUCAAAUUAGG3'$.
Thus,the correct option is $(B)$.

(A) Prokaryotes are microscopic unicellular organisms that possess the simplest type of cellular organization.
In these cells,there is no nuclear membrane to separate the genetic material from the cytoplasm.
Therefore,the process of transcription (synthesis of $mRNA$ from $DNA$) and translation (synthesis of protein from $mRNA$) occurs in the same compartment,i.e.,the cytoplasm.

(B) The messenger $RNA$ or $mRNA$ is synthesized by the process of transcription. In this process,the genetic information from $DNA$ is copied into $mRNA$.

(C) The process of transcription in bacteria occurs in three stages: initiation, elongation, and termination.
$1$. In stage $(a)$, the $RNA$ polymerase enzyme binds to the promoter region with the help of the sigma factor $(\sigma)$, marking the initiation of transcription.
$2$. In stage $(b)$, the $RNA$ polymerase moves along the $DNA$ template, facilitating the elongation of the $RNA$ chain.
$3$. In stage $(c)$, the $RNA$ polymerase reaches the terminator region, where the rho factor $(\rho)$ helps in the termination of transcription and the release of the nascent $RNA$ strand.

(C) In the process of capping,an unusual nucleotide,$7$-methyl guanosine triphosphate,is added to the $5'$ end of $hn-RNA$.
This modification protects the $RNA$ from degradation by exonucleases and helps in the recognition of the ribosome during translation.
Therefore,the addition of methyl guanosine triphosphate at the $5'$ end is known as capping.

(A) During the process of $RNA$ splicing,the non-coding sequences known as introns are removed from the primary transcript $(hnRNA)$.
The remaining coding sequences,known as exons,are then joined together in a defined order to form mature $mRNA$.

(B) In eukaryotes,there are three main 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 $mRNA$,which is known as heterogeneous nuclear $RNA$ $(hnRNA)$.
$3$. $RNA$ polymerase $III$ transcribes $tRNA$,$5S$ $rRNA$,and $snRNA$ (small nuclear $RNA$s).
Since $mRNA$ is synthesized from $hnRNA$ (the product of $RNA$ polymerase $II$),the correct enzyme is $RNA$ polymerase $II$.

(D) In eukaryotes,the primary transcript contains both exons (coding sequences) and introns (intervening sequences). During the process of splicing,introns are removed and exons are joined together to form mature $RNA$. Therefore,intervening sequences (introns) do not appear in mature $RNA$.

(D) The process of forming $RNA$ from a $DNA$ template is known as transcription,not replication. Replication is the process of $DNA$ synthesis from a $DNA$ template.

(D) $mRNA$ (messenger $RNA$) provides the template for the synthesis of proteins,not $r-RNA$. Therefore,statement $i$ is incorrect.
$t-RNA$ (transfer $RNA$) acts as an adapter molecule that brings specific amino acids to the ribosome and reads the genetic code on the $mRNA$ template. Therefore,statement $ii$ is correct.
$RNA$ polymerase is the enzyme that binds to the promoter region of the $DNA$ to initiate the process of transcription. Therefore,statement $iii$ is correct.
$A$ segment of $DNA$ that codes for a polypeptide is called an exon,while non-coding sequences are called introns. Therefore,statement $iv$ is incorrect.
Thus,statements $ii$ and $iii$ are correct.

(D) In eukaryotes, there are three types of $RNA$ polymerases:
$1$. $RNA$ polymerase $I$ transcribes $rRNAs$ $(28S, 18S, \text{ and } 5.8S)$.
$2$. $RNA$ polymerase $II$ transcribes the precursor of $mRNA$, called heterogeneous nuclear $RNA$ $(hnRNA)$.
$3$. $RNA$ polymerase $III$ is responsible for the transcription of $tRNA$, $5sRNA$, and small nuclear $RNAs$ $(snRNAs)$.
Therefore, statements $(i)$ and $(iv)$ are correct.

(B) Transcription is the process of copying genetic information from one strand of the $DNA$ into $RNA$.
$RNA$ polymerase is the primary enzyme responsible for this process.
During transcription,$RNA$ polymerase uses a $DNA$ strand as a template to synthesize a complementary $RNA$ molecule.
Therefore,the template required for the transcription of genes is $DNA$.

(A) The sigma factor $\sigma$ is a subunit of the $RNA$ polymerase holoenzyme. Its primary role is to recognize the promoter site on the $DNA$ template to initiate transcription. Once the transcription process moves from the initiation phase to the elongation phase, the sigma factor dissociates from the $RNA$ polymerase core enzyme. Therefore, during the elongation of the polypeptide chain, the sigma factor is functionless in the context of the elongation complex.

(C) The process of synthesizing $RNA$ from a $DNA$ template is known as transcription.
In a transcription unit,the process begins at the $promoter$ region,which provides the binding site for $RNA$ polymerase.
The process concludes at the $terminator$ region,where the $RNA$ polymerase detaches from the $DNA$ template.
The enzyme primarily responsible for transcribing structural genes in eukaryotes is $RNA$ polymerase-$II$.

(C) The $mRNA$ formed after the transcription of a eukaryotic gene is shorter than the $DNA$ sequence because the gene contains non-coding sequences called introns.
During the process of post-transcriptional modification,these introns are removed through a process known as splicing,and the coding sequences called exons are joined together.
Therefore,the mature $mRNA$ is significantly shorter than the original $DNA$ template.

(B) During the process of transcription,the $DNA$ template strand is used to synthesize a complementary $mRNA$ strand.
In $RNA$,uracil $(U)$ replaces thymine $(T)$ found in $DNA$.
The base pairing rules are as follows:
$A$ (Adenine) in $DNA$ pairs with $U$ (Uracil) in $mRNA$.
$T$ (Thymine) in $DNA$ pairs with $A$ (Adenine) in $mRNA$.
$C$ (Cytosine) in $DNA$ pairs with $G$ (Guanine) in $mRNA$.
$G$ (Guanine) in $DNA$ pairs with $C$ (Cytosine) in $mRNA$.
Given the $DNA$ sequence $AACGTAACG$:
$A \rightarrow U$
$A \rightarrow U$
$C \rightarrow G$
$G \rightarrow C$
$T \rightarrow A$
$A \rightarrow U$
$A \rightarrow U$
$C \rightarrow G$
$G \rightarrow C$
Therefore,the resulting $mRNA$ sequence is $UUGCAUUGC$.

(A) The process of protein synthesis begins with the initiation codon $AUG$ on the $mRNA$ strand,which codes for the amino acid methionine.
Since the $mRNA$ is synthesized from the template strand of $DNA$ through complementary base pairing,the $DNA$ sequence must be complementary to $AUG$.
In $DNA$,$A$ pairs with $T$ and $U$ (which is $T$ in $DNA$) pairs with $A$,and $G$ pairs with $C$.
Therefore,the $DNA$ template sequence corresponding to $AUG$ is $TAC$.

(D) During the process of transcription,only one of the two strands of $DNA$ is copied into $RNA$. This strand,which acts as a template for the synthesis of $RNA$ by $RNA$ polymerase,is known as the template strand. It is also referred to as the antisense strand or the non-coding strand. The other strand,which has the same sequence as the $RNA$ (except for $U$ instead of $T$),is called the coding strand or sense strand.

(A) In eukaryotic genes,the coding sequences are interrupted by non-coding sequences called introns. During transcription,the entire gene is transcribed into $pre-mRNA$. The introns are removed during splicing to form mature $mRNA$. When mature $mRNA$ is hybridized with its template $DNA$ strand,the $DNA$ regions corresponding to the introns do not find complementary sequences in the $mRNA$ and thus form single-stranded loops. Therefore,these loops represent the introns present in the $DNA$.

(D) In eukaryotes,the primary transcript (pre-$mRNA$) undergoes post-transcriptional modifications.
One of these modifications is polyadenylation,where a poly-$A$ tail (consisting of $200-300$ adenine residues) is added to the $3'$ end of the $mRNA$ molecule.
This tail protects the $mRNA$ from degradation and assists in its export from the nucleus to the cytoplasm.
Therefore,the poly-$A$ tail is a characteristic feature of eukaryotic $mRNA$.

(A) In eukaryotic cells,there are three types of $RNA$ polymerases. $RNA$ polymerase $I$ transcribes $rRNAs$ ($28S, 18S,$ and $5.8S$). $RNA$ polymerase $II$ transcribes the precursor of $mRNA$,which is $hnRNA$. $RNA$ polymerase $III$ is responsible for the transcription of $tRNA$,$5S rRNA$,and $snRNA$ (small nuclear $RNA$).

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

(C) In bacteria,transcription and translation occur in the same compartment,which is the cytoplasm. Because there is no nuclear membrane separating the genetic material from the ribosomes,translation can begin even before the $mRNA$ is fully transcribed. This process is known as coupled transcription and translation.

(C) Post-transcriptional processing in eukaryotes involves capping,tailing,and splicing.
$1$. Capping requires Methyl guanosine triphosphate.
$2$. Splicing is mediated by the spliceosome complex,which consists of $SnRNA$ (Small nuclear $RNA$) and proteins.
$3$. Ligase is required for joining the exons after splicing.
$4$. $ScRNA$ (Small cytoplasmic $RNA$) is involved in protein trafficking and secretion,not in the processing of pre-$mRNA$ within the nucleus.

(B) The term $cDNA$ stands for complementary $DNA$,which is synthesized from mature $mRNA$ using the enzyme reverse transcriptase.
Mature $mRNA$ has already undergone the process of splicing,during which all non-coding sequences known as introns are removed.
Since $cDNA$ is synthesized from this processed $mRNA$ template,it contains only exons and lacks any introns.
Therefore,the number of introns present in $cDNA$ is $0$.

(B) Splicing is the process of removing introns and joining exons in a primary transcript (pre-mRNA) to form a mature mRNA.
This process is catalyzed by a large complex called the spliceosome.
The spliceosome is composed of small nuclear ribonucleoproteins $(snRNPs)$.
Each $snRNP$ consists of small nuclear $RNA$ $(snRNA)$ molecules and a specific set of proteins.
Therefore,$snRNA$ and proteins are essential for the splicing process.