Transcription Questions in English

(B) 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 a $DNA$ template to synthesize a complementary $RNA$ strand.
Therefore,the correct option is $B$.

(A) The process of copying genetic information from one strand of the $DNA$ into $RNA$ is termed as $Transcription$.
In this process,only a segment of $DNA$ and only one of the two strands is copied into $RNA$.
$Translation$ is the process of polymerization of amino acids to form a polypeptide based on the sequence of nucleotides in $mRNA$.

(A) Transcription is the process of copying genetic information from one strand of the $DNA$ into $RNA$. The enzyme responsible for this process is $RNA$ polymerase. It catalyzes the polymerization of ribonucleotides in a $5' \rightarrow 3'$ direction using a $DNA$ template. $DNA$ polymerases are primarily involved in $DNA$ replication,not transcription.

(B) In eukaryotic genes,the coding sequences are known as exons,which are expressed in the final protein product.
Introns are the intervening sequences that are transcribed into $pre-mRNA$ but are removed during the process of $RNA$ splicing.
Therefore,introns are transcribed but not translated into proteins.

(C) In eukaryotic cells,the primary transcript (pre-mRNA) contains both coding sequences called $exons$ and non-coding sequences called $introns$.
During the process of post-transcriptional modification,the $introns$ are removed and the $exons$ are joined in a defined order.
This process is known as $splicing$.
$Capping$ refers to the addition of a methylguanosine triphosphate at the $5'$ end,and $Tailing$ refers to the addition of poly-$A$ residues at the $3'$ end.

(C) In the process of transcription,only one strand of $DNA$ acts as a template for $mRNA$ synthesis.
This strand is known as the template strand or antisense strand.
The other strand,which has the same sequence as the $mRNA$ (except for $U$ instead of $T$),is called the coding strand or sense strand.
The coding strand does not directly code for the $RNA$ molecule during transcription; therefore,it is often referred to as the non-template strand.

(D) During the initiation of transcription in eukaryotes,the $RNA$ polymerase holoenzyme recognizes and binds to a specific promoter sequence on the $DNA$ template.
This specific sequence is rich in $Adenine$ and $Thymine$ bases and is known as the $TATA$ box (or $Goldberg-Hogness$ box).
It is located approximately $25$ to $30$ base pairs upstream of the transcription start site.
The binding of $RNA$ polymerase to this site creates a characteristic 'saddle' or 'bubble' structure,which facilitates the unwinding of the $DNA$ double helix to initiate transcription.

(D) During transcription,only one segment of $DNA$ is copied into $RNA$.
If both strands of $DNA$ were to act as templates,they would code for $RNA$ molecules with different sequences because the two strands of $DNA$ are complementary to each other,not identical.
Furthermore,if these two different $RNA$ molecules were produced simultaneously,they would be complementary to each other,forming a double-stranded $RNA$ $(dsRNA)$.
This $dsRNA$ would prevent the translation of $RNA$ into proteins,rendering the transcription process futile.
Therefore,both statements $(a)$ and $(b)$ are correct.

(A) In eukaryotic organisms,the coding sequences of a gene are interrupted by non-coding sequences.
These coding sequences are known as $exons$,and the non-coding sequences that intervene between them are known as $introns$.
$A$ gene that possesses both $exons$ and $introns$ is referred to as a $split$ $gene$.
This structure is a characteristic feature of eukaryotic genes,where the primary transcript (pre-mRNA) undergoes splicing to remove $introns$ and join $exons$ to form mature $mRNA$.

(C) During the process of transcription,only one segment of $DNA$ and only one of the two strands of $DNA$ acts as a template. This strand is known as the template strand or the antisense strand. The other strand,which does not code for anything,is called the coding strand or the sense strand. Therefore,only one of the two polynucleotide chains of a gene acts as a template.

(A) The central dogma of molecular biology states that genetic information flows from $DNA$ to $RNA$ (specifically $mRNA$) and then to protein.
During the process of transcription,the genetic information stored in $DNA$ is copied into $mRNA$.
Therefore,the flow of genetic information is from $DNA$ to $mRNA$.

(C) The process of synthesizing $m-RNA$ from a $DNA$ template is known as transcription.
In eukaryotic cells,this process occurs within the nucleus,where the enzyme $RNA$ polymerase uses one of the $DNA$ strands as a template to synthesize a complementary $m-RNA$ strand.
Therefore,$DNA$ acts as the template or the structural basis for $m-RNA$ synthesis.

(A) $m-RNA$ (messenger $RNA$) acts as a template for protein synthesis.
It carries the genetic information transcribed from $DNA$ in the form of a sequence of nucleotides.
This sequence is translated into a specific sequence of amino acids to form a protein.
$t-RNA$ (transfer $RNA$) brings amino acids to the ribosome.
$r-RNA$ (ribosomal $RNA$) forms the structural component of ribosomes.
Therefore,the correct option is $A$.

(A) The process of copying genetic information from one strand of the $DNA$ into $RNA$ is termed as transcription.
In this process,the enzyme $RNA$ polymerase catalyzes the polymerization of ribonucleotides based on the template provided by the $DNA$ segment (gene).
Therefore,$m-RNA$ is synthesized from the genetic information present in the $DNA$ segment.

(C) During the process of transcription,one of the two strands of $DNA$ acts as a template for the synthesis of $RNA$. This specific strand is known as the template strand. It is a polynucleotide chain that provides the sequence of nucleotides for the complementary $RNA$ strand to be synthesized by $RNA$ polymerase. Therefore,the most accurate description in the context of molecular biology is that it is a $DNA$ strand that serves as a template for transcription.

(B) Assertion $(A)$ is true because $DNA$ acts as a template for the transcription of all three major types of $RNA$ ($mRNA$,$tRNA$,and $rRNA$) in the cell.
Reason $(R)$ is also true because $mRNA$ provides the genetic code,$tRNA$ brings amino acids to the ribosome,and $rRNA$ forms the structural and catalytic components of the ribosome,all of which are essential for protein synthesis.
However,the reason $(R)$ explains the functional importance of $RNA$ types,not the mechanism of their synthesis by $DNA$. Therefore,$R$ is not the correct explanation of $A$.

(C) : Spliceosomes are complexes of small nuclear ribonucleoproteins $(snRNPs)$ that facilitate the removal of introns from pre-$mRNA$ during $RNA$ splicing.
Since bacteria are prokaryotes,their genes do not contain introns.
Consequently,prokaryotic $mRNA$ does not undergo splicing,and therefore,spliceosomes are absent in bacterial cells.

(A) : The strand of $DNA$ on which $RNA$ polymerase binds to catalyse transcription is called the template strand.
It is also known as the master or antisense strand.
It has the polarity of $3' \rightarrow 5'$.

(A) During the process of transcription, the enzyme $RNA$ polymerase catalyzes the synthesis of $RNA$ in the $5'$ to $3'$ direction.
The $DNA$ template strand, which is used for complementary base pairing, has a polarity of $3'$ to $5'$.
Therefore, the $RNA$ polymerase reads the template $DNA$ strand in the $3'$ to $5'$ direction to synthesize a new $RNA$ strand in the $5'$ to $3'$ direction.
Thus, option $A$ is the correct choice.

(D) The correct answer is $D$.
In eukaryotic cells,the primary transcript (pre-$mRNA$) contains both coding sequences called exons and non-coding sequences called introns.
The process of removing these non-coding introns and joining the coding exons together in a specific order to form a mature $mRNA$ molecule is known as splicing.
This process is essential for the production of functional proteins,as introns do not code for amino acids.

(B) In $DNA$,the base pairing rules are $A$ pairs with $T$,and $C$ pairs with $G$.
During the process of transcription,$DNA$ is transcribed into $RNA$.
In $RNA$,the base uracil $(U)$ replaces thymine $(T)$.
Therefore,the complementary $RNA$ sequence for the $DNA$ template strand $ATCTG$ is determined as follows:
$A$ pairs with $U$
$T$ pairs with $A$
$C$ pairs with $G$
$T$ pairs with $A$
$G$ pairs with $C$
Thus,the resulting $RNA$ sequence is $UAGAC$.

(A) The correct answer is $A$.
$A$ transcription unit in $DNA$ is defined primarily by three regions:
$1$. $A$ promoter: The site where $RNA$ polymerase binds to initiate the process of transcription.
$2$. The structural gene: The segment of $DNA$ that codes for an $RNA$ molecule.
$3$. $A$ terminator: The region that signals the termination of the transcription process and the release of the newly synthesized $RNA$ strand.
The inducer is a regulatory molecule (often a small metabolite) that interacts with a repressor protein to regulate gene expression,but it is not a structural component of the transcription unit itself.

(A) In eukaryotes,there are three main types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase $I$ is responsible for the synthesis of $rRNA$ ($28S, 18S,$ and $5.8S$).
$2$. $RNA$ polymerase $II$ is responsible for the synthesis of $hnRNA$ (precursor of $mRNA$).
$3$. $RNA$ polymerase $III$ is responsible for the synthesis of $tRNA$,$5S$ $rRNA$,and $snRNA$.
Therefore,the removal of $RNA$ polymerase $III$ will specifically affect the synthesis of $tRNA$.

(B) : Unlike in prokaryotes where transcription and translation take place in the same compartment,in eukaryotes,the primary transcript is first processed in the nucleus and then transported outside of the nucleus.
Since the primary transcripts of eukaryotes contain both expressing genes (exons) and non-expressing genes (introns),they undergo splicing of introns,followed by capping and tailing at the $5'$-end and $3'$-end,respectively.

(D) : $mRNA$ is not made directly in a eukaryotic cell. It is transcribed as heterogeneous nuclear $RNA$ $(hnRNA)$ in the nucleus. $hnRNA$ contains introns and exons. The introns are removed by $RNA$ splicing,leaving behind the exons,which contain the coding information. The exonic regions of $RNA$ are joined together to produce a single chain $RNA$ required for functioning as a translational template.

(C) In eukaryotic cells,the primary transcript (pre-$mRNA$) contains both coding sequences called exons and non-coding sequences called introns.
Splicing is a post-transcriptional modification process where the non-coding introns are removed from the pre-$mRNA$,and the coding exons are joined together in a specific order to form the mature $mRNA$.
Therefore,the correct definition of splicing is that introns are removed and exons are joined.

(C) In eukaryotic cells,there are three main types of $RNA$ polymerases responsible for the transcription of different types of $RNA$ molecules:
$1$. $RNA$ polymerase-$I$ is responsible for the transcription of $rRNAs$ ($28S$,$18S$,and $5.8S$).
$2$. $RNA$ polymerase-$II$ is responsible for the transcription of the precursor of $mRNA$,which is known as heterogeneous nuclear $RNA$ $(hnRNA)$.
$3$. $RNA$ polymerase-$III$ is responsible for the transcription of $tRNA$,$5S$ $rRNA$,and $snRNAs$ (small nuclear $RNAs$).
Therefore,$RNA$ polymerase-$III$ is the enzyme responsible for the transcription of $tRNA$ and $snRNAs$.

(A) In prokaryotic transcription, the $RNA$ polymerase enzyme consists of a core enzyme and a sigma factor $(\sigma)$.
The core enzyme is responsible for the elongation of the $RNA$ chain.
The sigma factor $(\sigma)$ specifically recognizes the promoter region on the $DNA$ template.
By binding to the promoter, the sigma factor facilitates the initiation of transcription.
Once initiation is complete, the sigma factor dissociates from the core enzyme, allowing elongation to proceed.

(D) transcription unit in $DNA$ is defined primarily by three regions in the $DNA$ molecule:
$1$. $A$ promoter
$2$. The structural gene
$3$. $A$ terminator
$RNA$ polymerase is an enzyme that catalyzes the process of transcription,but it is not a structural region or a part of the $DNA$ sequence of the transcription unit itself. Therefore,$RNA$ polymerase is the correct answer.

(A) In eukaryotes,there are three main types of $RNA$ polymerases involved in transcription:
$1$. $RNA$ polymerase $I$ is responsible for the transcription of $r-RNA$ ($28S$,$18S$,and $5.8S$).
$2$. $RNA$ polymerase $II$ is responsible for the transcription of the precursor of $m-RNA$,which is $hnRNA$ (heterogeneous nuclear $RNA$).
$3$. $RNA$ polymerase $III$ is responsible for the transcription of $t-RNA$,$5S$ $r-RNA$,and $snRNA$ (small nuclear $RNA$).
Therefore,the removal of $RNA$ polymerase $I$ will specifically affect the synthesis of $r-RNA$.

(A) Capping is a post-transcriptional modification process in eukaryotes.
During this process,an unusual nucleotide,$7$-methylguanosine triphosphate $(m^7G)$,is added to the $5'$ end of the $hnRNA$ (heterogeneous nuclear $RNA$).
This cap protects the $mRNA$ from degradation by exonucleases and helps in the initiation of translation by facilitating ribosome binding.

(A) In eukaryotes,the primary transcript (pre-$mRNA$) contains both coding sequences called exons and non-coding sequences called introns.
During the process of post-transcriptional modification,specifically splicing,the introns are removed,and the exons are joined together in a defined order.
Therefore,the mature $mRNA$ consists only of exons,which are translated into proteins.

(B) The figure shows the addition of a methyl guanosine triphosphate $(m^7Gppp)$ at the $5'$ end of the $mRNA$ molecule. This process is known as capping. In capping,an unusual nucleotide,methyl guanosine triphosphate,is added to the $5'$ end of the heterogeneous nuclear $RNA$ $(hnRNA)$. Polyadenylation,on the other hand,involves the addition of adenylate residues at the $3'$ end. Therefore,the correct option is $B$.

(B) The process of protein synthesis involves the transcription of $DNA$ into $m-RNA$ within the nucleus.
$m-RNA$ (messenger $RNA$) acts as a template that carries the genetic code from the $DNA$ in the nucleus to the ribosomes in the cytoplasm.
At the ribosomes,the genetic information is translated into a polypeptide chain.
Therefore,$m-RNA$ is the molecule responsible for transferring genetic information from the nucleus to the ribosomes.

(B) The process of copying genetic information from one strand of the $DNA$ into $RNA$ is termed as $Transcription$.
In this process, only one segment of $DNA$ and only one of the two strands is copied into $RNA$.
$Replication$ refers to the duplication of $DNA$.
$Protein$ $synthesis$ (Translation) is the process where $mRNA$ is used to synthesize a polypeptide chain.
$Synthesis$ of $primer$ is a specific step during $DNA$ replication.

(B) The process of copying genetic information from one strand of $DNA$ into $RNA$ is known as $Transcription$.
In this process,only a segment of $DNA$ and only one of the two strands is copied into $RNA$.
$Replication$ is the process of duplicating $DNA$.
$Translation$ is the process of polymerizing amino acids to form a polypeptide based on the sequence of $mRNA$.
$Translocation$ refers to the movement of the ribosome along the $mRNA$ during protein synthesis.

(A) During transcription,the $RNA$ polymerase enzyme synthesizes $RNA$ using the template strand of $DNA$ in the $3' \rightarrow 5'$ direction.
According to the principle of complementarity,$Adenine$ $(A)$ pairs with $Uracil$ $(U)$ in $RNA$,$Thymine$ $(T)$ pairs with $Adenine$ $(A)$,$Cytosine$ $(C)$ pairs with $Guanine$ $(G)$,and $Guanine$ $(G)$ pairs with $Cytosine$ $(C)$.
Given template strand: $3'-C-T-G-A-T-A-G-C-5'$.
Complementary $RNA$ sequence: $5'-G-A-C-U-A-U-C-G-3'$.
Therefore,the correct sequence is $5'-GACUAUCG-3'$.

(A) During the process of transcription,$DNA$ acts as a template to synthesize $mRNA$.
According to the principle of complementarity,$Adenine$ $(A)$ pairs with $Uracil$ $(U)$ in $RNA$,$Thymine$ $(T)$ pairs with $Adenine$ $(A)$,$Cytosine$ $(C)$ pairs with $Guanine$ $(G)$,and $Guanine$ $(G)$ pairs with $Cytosine$ $(C)$.
Given $DNA$ template sequence: $A-T-T-C-G-A-T-G$.
Complementary $mRNA$ sequence: $U-A-A-G-C-U-A-C$.
Therefore,the correct sequence is $UAAGCUAC$.

(A) During the process of transcription,the enzyme $RNA$ polymerase synthesizes $m-RNA$ by reading the $DNA$ template strand in the $3' \to 5'$ direction.
As a result,the newly synthesized $m-RNA$ strand is always formed in the $5' \to 3'$ direction due to the antiparallel nature of the nucleic acid strands.

(C) During the process of $RNA$ splicing, the non-coding sequences known as introns are removed, and the coding sequences known as exons are joined together.
This joining process is facilitated by the enzyme $RNA$ ligase, which catalyzes the formation of phosphodiester bonds between the $3'-OH$ and $5'-phosphate$ ends of the adjacent exon segments.
Therefore, the correct option is $C$.

(C) During the process of transcription in prokaryotes and eukaryotes,the $RNA$ polymerase enzyme recognizes and binds to a specific sequence on the $DNA$ template strand known as the promoter. The promoter is located towards the $5'$ end (upstream) of the structural gene and provides the binding site for $RNA$ polymerase to initiate transcription.

(C) During the process of transcription,$DNA$ acts as a template for the synthesis of $m-RNA$.
According to the base-pairing rules,$Adenine$ $(A)$ pairs with $Uracil$ $(U)$ in $RNA$,and $Thymine$ $(T)$ pairs with $Adenine$ $(A)$.
Given $DNA$ sequence: $ATACG$.
- $A$ pairs with $U$
- $T$ pairs with $A$
- $A$ pairs with $U$
- $C$ pairs with $G$
- $G$ pairs with $C$
Therefore,the resulting $m-RNA$ sequence is $UA UGC$.

(D) During the initiation of transcription in eukaryotes,the $RNA$ polymerase holoenzyme recognizes and binds to a specific promoter sequence located upstream of the gene.
This specific sequence is rich in $Adenine$ and $Thymine$ bases and is known as the $TATA$ box (or $Goldberg-Hogness$ box).
Binding of the $RNA$ polymerase to the $TATA$ box causes the $DNA$ double helix to unwind and bend,creating a characteristic saddle-like structure that facilitates the formation of the transcription initiation complex.

(C) Cell differentiation is the process by which a less specialized cell becomes a more specialized cell type. This process is primarily controlled by the differential expression of genes. The regulation of gene expression,which determines which genes are turned on or off in a specific cell type,is largely mediated by $Transcription$ $factors$. These proteins bind to specific $DNA$ sequences to regulate the transcription of genetic information from $DNA$ to messenger $RNA$ $(mRNA)$,thereby controlling the protein synthesis profile of the cell.

(D) In eukaryotic organisms,the primary transcript (pre-mRNA) contains both coding sequences called $Exons$ and non-coding sequences called $Introns$.
During the process of post-transcriptional modification,the $Introns$ are removed and the $Exons$ are joined together in a specific order to form mature $mRNA$.
This process is known as $Splicing$.

(A) In $DNA$,the nitrogenous bases are Adenine $(A)$,Thymine $(T)$,Cytosine $(C)$,and Guanine $(G)$.
According to the base-pairing rules,$A$ pairs with $T$ and $C$ pairs with $G$.
However,in $RNA$,Thymine $(T)$ is replaced by Uracil $(U)$.
Therefore,when transcribing $DNA$ to $RNA$,the base-pairing rules are:
$A$ pairs with $U$ (instead of $T$),
$T$ pairs with $A$,
$C$ pairs with $G$,
$G$ pairs with $C$.
Given the $DNA$ sequence $ATCTG$:
$A$ pairs with $U$
$T$ pairs with $A$
$C$ pairs with $G$
$T$ pairs with $A$
$G$ pairs with $C$
Thus,the complementary $RNA$ sequence is $UAGAC$.

(A) transcription unit in $DNA$ is defined primarily by three regions in the organism:
$1$. $A$ promoter
$2$. The structural gene
$3$. $A$ terminator
An inducer is a molecule that regulates gene expression (often associated with operons like the $lac$ operon) but is not a structural component of the transcription unit itself.

(A) 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 $m-RNA$,called $hn-RNA$ (heterogeneous nuclear $RNA$).
$3$. $RNA$ polymerase $III$ is responsible for the transcription of $t-RNA$,$5S$ $r-RNA$,and $sn-RNA$ (small nuclear $RNA$).
Therefore,removing $RNA$ polymerase $III$ will directly affect the synthesis of $t-RNA$.

(A) During the process of transcription,$DNA$-dependent $RNA$ polymerase catalyzes the polymerization of ribonucleotides in the $5' \rightarrow 3'$ direction.
This enzyme reads the $DNA$ template in the $3' \rightarrow 5'$ direction.
The specific $DNA$ strand that acts as a template for $RNA$ synthesis is known as the template strand.
The other strand,which has the same polarity as the $RNA$ (except $T$ is replaced by $U$),is called the coding strand.