Tools of recombinant DNA technology Questions in English

(B) The statement 'Sticky ends can be joined by using $DNA$ ligases' is technically misleading in the context of standard textbook questions because while $DNA$ ligase joins the phosphodiester backbone,the term 'sticky ends' refers to the single-stranded overhangs created by restriction enzymes. However,in many competitive exam contexts,the statement that is often considered 'wrong' or 'incorrect' regarding the fundamental definition of restriction enzymes is that they 'inspect the length' of the sequence; they actually inspect the specific 'nucleotide sequence' (recognition site),not the length. Re-evaluating the provided options: Option $B$ is the most scientifically inaccurate because restriction enzymes recognize specific sequences,not lengths. If the question implies that sticky ends are joined by ligase,that is actually a correct biological process. Therefore,the statement 'Each restriction enzyme functions by inspecting the length of a $DNA$ sequence' is the incorrect statement.

$(B)$ Ligases are enzymes that act as 'molecular glue' by catalyzing the formation of phosphodiester bonds between the $3'-OH$ end of one $DNA$ fragment and the $5'-phosphate$ end of another, thereby joining two $DNA$ molecules together. Exonucleases remove nucleotides from the ends of $DNA$, whereas endonucleases make cuts at specific positions within $DNA$. Polymerases are involved in the synthesis of $DNA$ strands, not in breaking them.

(A) The restriction enzyme $EcoRI$ is derived from $Escherichia \text{ } coli \text{ } RY13$.
It recognizes the specific palindromic base sequence $5'-GAATTC-3'$ and $3'-CTTAAG-5'$.
$A$ palindromic sequence in $DNA$ is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same (e.g., $5' \rightarrow 3'$ direction).
Therefore, the correct option is $A$.

(A) Restriction enzymes are endonucleases that recognize specific $DNA$ sequences known as palindromic sequences and cut the $DNA$ at these sites.
Option $A$ is incorrect because restriction enzymes do not inspect the length of the $DNA$ sequence; rather,they recognize specific nucleotide sequences (recognition sites) regardless of the total length of the $DNA$ molecule.
Option $B$ is correct as they cut at palindromic sites.
Option $C$ is correct as they are fundamental tools in genetic engineering.
Option $D$ is correct as $DNA$ ligase is used to join the sticky ends produced by restriction enzymes.

(A) $1$. Ligases are enzymes that catalyze the joining of two $DNA$ molecules by forming phosphodiester bonds. This is a correct statement.
$2$. Polymerases are enzymes that synthesize $DNA$ strands,not cut them.
$3$. Nucleases are enzymes that degrade $DNA$ by breaking phosphodiester bonds. Helicases are responsible for separating the $DNA$ double helix.
$4$. Exo-nucleases remove nucleotides from the ends of $DNA$ molecules,whereas Endo-nucleases cut $DNA$ at specific internal positions.

(N/A) Viruses act as efficient biological factories for the production of viral proteins. By utilizing recombinant $DNA$ technology,scientists can insert specific genes into viral vectors. When these modified viruses infect host cells,they hijack the host's machinery to express the inserted gene,leading to the mass production of desired proteins. This process is widely used in biotechnology for producing vaccines,therapeutic enzymes,and other pharmaceutical products.

(N/A) $(1)$ Sticky ends: These are single-stranded extensions of a $DNA$ fragment that have been cleaved at a specific sequence,often by a restriction endonuclease,in a staggered cut. These extensions are complementary to those of similarly cleaved $DNAs$,allowing them to base-pair with each other.
$(2)$ Transgene: $A$ gene that is isolated from one organism and inserted into the genome of another organism,such that it is replicated as part of the host genome and is present in all the recipient's cells. The resulting organism is described as transgenic.

(B) The construction of the first recombinant $DNA$ $(r-DNA)$ was achieved by Stanley Cohen and Herbert Boyer in $1972$.
They isolated the antibiotic resistance gene by cutting out a piece of $DNA$ from a plasmid which was responsible for conferring antibiotic resistance in the bacterium $Salmonella$ $typhimurium$.
This piece of $DNA$ was then linked to a plasmid of $E. coli$ to create the first artificial recombinant $DNA$ molecule.

(D) The first recombinant $DNA$ $(r-DNA)$ was constructed by Stanley Cohen and Herbert Boyer in $1972$. They isolated an antibiotic resistance gene from the plasmid of the bacterium $Salmonella typhimurium$ and linked it with a native plasmid of $Escherichia coli$ using the enzyme $DNA$ ligase. This allowed the antibiotic resistance gene to be replicated and expressed in the host $E. coli$ cells.

(B) The first recombinant $DNA$ $(r-DNA)$ molecule was constructed by $Stanley \ Cohen$ and $Herbert \ Boyer$ in $1972$.
They accomplished this by isolating the antibiotic resistance gene by cutting out a piece of $DNA$ from a plasmid which was responsible for conferring antibiotic resistance in the bacterium $Salmonella \ typhimurium$.

(D) Restriction enzymes are known as molecular scissors because they have the ability to cut $DNA$ molecules at specific recognition sequences.
These enzymes are essential tools in recombinant $DNA$ technology as they allow for the precise cleavage of $DNA$ strands to create recombinant molecules.

(B) The enzymes known as molecular scissors are $Restriction$ $Endonucleases$.
These enzymes recognize specific nucleotide sequences in $DNA$ molecules,known as recognition sites,and cut the $DNA$ at these specific locations.
This property is essential in recombinant $DNA$ technology for isolating genes and creating recombinant molecules.

(C) vector is a $DNA$ molecule used as a vehicle to artificially carry foreign genetic material into another cell,where it can be replicated and/or expressed.
Plasmids are small,circular,extrachromosomal $DNA$ molecules that are physically separated from chromosomal $DNA$ and can replicate independently.
Due to their small size and ability to replicate independently,plasmids are widely used in biotechnology as vectors to transfer genes of interest into host organisms.

(A) Recombinant $DNA$ $(r-DNA)$ technology involves the formation of new combinations of genetic material by inserting a foreign $DNA$ fragment into a vector,typically a plasmid.
Therefore,$r-DNA$ is formed by the combination of a plasmid (vector) and a foreign $DNA$ fragment.

(C) In $r-DNA$ technology,$Escherichia coli$ $(E. coli)$ is the most widely used host organism.
This is because $E. coli$ has a well-mapped genome,grows rapidly,is easy to handle,and can be easily transformed with foreign $DNA$.
Its simple genetic structure and the availability of various cloning vectors make it the preferred choice for molecular cloning experiments.

(D) $DNA$ ligase is the enzyme that acts as a molecular glue. It catalyzes the formation of phosphodiester bonds between the $3'$-hydroxyl end of one $DNA$ fragment and the $5'$-phosphate end of another,thereby joining the foreign $DNA$ fragment into the plasmid vector.

(D) The tools of recombinant $DNA$ $(r-DNA)$ technology include restriction enzymes,polymerases,ligases,vectors,and the host organism.
Restriction enzymes are used to cut $DNA$ at specific sites.
Cloning vectors are used to carry the foreign $DNA$ into the host.
The host organism is the cell where the recombinant $DNA$ is introduced and replicated.
Pepsin is a digestive enzyme found in the stomach of animals and is not used in $r-DNA$ technology.
Therefore,$Pepsin$ enzyme is the correct answer as it is not a tool of $r-DNA$ technology.

(C) The two enzymes responsible for restricting the growth of bacteriophage in $E. coli$ were isolated in the year $1963$.
These enzymes are known as restriction endonucleases.
One of these enzymes added methyl groups to $DNA$,while the other cut the $DNA$.
This discovery was a foundational step in the development of recombinant $DNA$ technology.

(B) The first restriction endonuclease to be discovered and characterized was $Hind \,II$. It was isolated from the bacterium $Haemophilus \,influenzae$. It always cuts $DNA$ molecules at a particular point by recognizing a specific sequence of $6$ base pairs. This specific recognition sequence is known as the recognition site.

(C) The restriction enzyme $Hind-II$ was the first restriction endonuclease to be isolated.
It always cuts $DNA$ molecules at a particular point by recognizing a specific sequence of $6$ base pairs.
This specific base pair sequence is known as the recognition sequence or palindromic sequence.
The palindromic sequence recognized by $Hind-II$ is $5'-GTPyCGAC-3'$ and $3'-CAPuGCTG-5'$,which consists of $6$ base pairs.

(A) According to the $NCERT$ textbook,today we know more than $900$ restriction enzymes that have been isolated from over $230$ strains of bacteria.
These enzymes are essential tools in recombinant $DNA$ technology,used to cut $DNA$ at specific recognition sequences.

(A) Restriction enzymes are a type of enzyme that recognizes specific nucleotide sequences in $DNA$ and cuts the $DNA$ at those sites.
These enzymes are classified under the category of nucleases because they act on nucleic acids ($DNA$/$RNA$).
Specifically,they are known as restriction endonucleases,which are a sub-category of nucleases that cleave the phosphodiester bonds within a $DNA$ molecule.

(C) Restriction enzymes are primarily classified into $3$ main types based on their structure,cofactor requirements,and the nature of their target sequences:
$1$. Type $I$ restriction enzymes: These are complex enzymes that require $ATP$,$Mg^{2+}$,and $S$-adenosylmethionine $(SAM)$ for their activity. They cleave $DNA$ at sites distant from their recognition sequence.
$2$. Type $II$ restriction enzymes: These are the most commonly used enzymes in biotechnology. They require only $Mg^{2+}$ as a cofactor and cleave $DNA$ at or near specific recognition sequences.
$3$. Type $III$ restriction enzymes: These enzymes require $ATP$ and $Mg^{2+}$ and cleave $DNA$ at a short distance from their recognition sequence.
Therefore,there are $3$ types of restriction enzymes.

(C) The naming convention for restriction enzymes follows a specific pattern:
$1$. The first letter is derived from the genus name of the organism from which the enzyme is isolated (e.g.,$E$ for $Escherichia$).
$2$. The next two letters are derived from the species name (e.g.,$co$ for $coli$).
$3$. The fourth letter represents the strain of the organism (e.g.,$R$ for $RY13$).
$4$. The Roman numeral indicates the order in which the enzyme was isolated from that strain (e.g.,$I$).
Therefore,$E$ stands for the genus and $R$ stands for the strain.

(C) Restriction enzymes are endonucleases that recognize specific $DNA$ sequences to cut the $DNA$ molecule. These specific sequences are known as recognition sites. Most of these recognition sites are palindromic sequences,which read the same in both directions $(5' \rightarrow 3')$ on the two strands of $DNA$. Therefore,the correct term for these specific sequences is a palindromic sequence.

(A) A palindromic $DNA$ sequence is a sequence of base pairs in double-stranded $DNA$ that reads the same when the orientation of reading is kept the same (e.g., $5' \rightarrow 3'$).
In the given options, the sequence $5'-GAATTC-3'$ paired with $3'-CTTAAG-5'$ is a palindrome because reading from $5' \rightarrow 3'$ on both strands yields the same sequence: $GAATTC$.
This specific sequence is the recognition site for the restriction enzyme $EcoRI$.

(A) Palindromic sequences are specific nucleotide sequences in $DNA$ that read the same in both the $5' \rightarrow 3'$ and $3' \rightarrow 5'$ directions on the two complementary strands. These sequences are recognized by restriction endonucleases,which cut the $DNA$ at specific sites. Therefore,palindromic sequences are a fundamental feature of $DNA$ molecules used in recombinant $DNA$ technology.

(B) The enzymes that cut $DNA$ are called restriction enzymes.
These are of two types:
$1$. Exonucleases: These enzymes remove nucleotides from the ends of the $DNA$ molecule.
$2$. Endonucleases: These enzymes make cuts at specific positions within the $DNA$ molecule.
Therefore,the enzyme that removes nucleotides from the ends of $DNA$ is an exonuclease.

(D) palindromic $DNA$ sequence is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same (e.g.,$5' \rightarrow 3'$).
For the restriction enzyme $EcoRI$,the specific recognition site is $5'-GAATTC-3'$.
The complementary strand reads $3'-CTTAAG-5'$.
When read in the $5' \rightarrow 3'$ direction,both strands show the sequence $GAATTC$,which satisfies the definition of a palindrome in molecular biology.

(B) The restriction enzyme $EcoR\, I$ recognizes the specific palindromic sequence $5'-GAATTC-3'$ in $DNA$.
It cuts the $DNA$ between the $G$ and $A$ bases on both strands.
This staggered cut results in single-stranded overhanging sequences at the ends of the $DNA$ fragments.
These overhanging sequences are known as 'sticky ends' or 'cohesive ends' because they can form hydrogen bonds with complementary sequences on other $DNA$ fragments.

(A) Restriction enzymes are classified based on their cutting pattern.
$1$. Some restriction enzymes cut the $DNA$ strands at the center of the palindromic sequence,resulting in blunt ends or flush ends.
$2$. Other restriction enzymes cut the $DNA$ strands a little away from the center of the palindromic sites,between the same two bases on the opposite strands,which leaves single-stranded portions at the ends. These overhanging stretches are called sticky ends or cohesive ends.
Therefore,cutting at the center of the palindromic site produces blunt ends.

(A) Sticky ends are short,single-stranded overhangs of $DNA$ produced by the action of restriction endonucleases.
These sticky ends form hydrogen bonds with complementary sequences on another piece of $DNA$ cut by the same restriction enzyme.
$DNA$ ligase then acts to seal the phosphodiester backbone between these fragments,facilitating the formation of recombinant $DNA$ molecules.
Therefore,sticky ends are specifically helpful in the action of $DNA$ ligase to join $DNA$ fragments.

(C) The restriction endonuclease $EcoR\,I$ is derived from the bacterium $Escherichia\,coli$ strain $RY13$.
It recognizes a specific palindromic $DNA$ sequence of $6$ base pairs.
The recognition sequence for $EcoR\,I$ is $5'-GAATTC-3'$ and $3'-CTTAAG-5'$.
Therefore,the total number of nucleotides in the recognition sequence is $6$.

(B) Restriction endonucleases are enzymes that recognize specific nucleotide sequences in $DNA$ and cut the $DNA$ at these specific sites,known as recognition sequences.
Exonucleases remove nucleotides from the ends of $DNA$ molecules.
Taq polymerase is used in $PCR$ for $DNA$ amplification.
Gyrase is involved in $DNA$ supercoiling and replication.
Therefore,the correct enzyme that cuts $DNA$ at specific internal sites is the restriction endonuclease.

(C) Gel electrophoresis is a laboratory technique used to separate $DNA$ fragments based on their size and charge.
In this process,$DNA$ samples are loaded into wells of an agarose gel and an electric current is applied.
Since $DNA$ molecules are negatively charged,they migrate towards the positive electrode (anode).
The smaller fragments move faster through the pores of the gel matrix compared to larger fragments,resulting in their separation.

(C) Gel electrophoresis is a technique used to separate $DNA$ fragments based on their size. In this process,the $DNA$ samples are loaded into wells made in a gel matrix. Agarose,a natural polymer extracted from seaweeds,is the most commonly used matrix for this purpose. It acts as a molecular sieve,allowing smaller $DNA$ fragments to move faster through the gel than larger ones when an electric current is applied.

(B) After the process of gel electrophoresis,the separated $DNA$ fragments cannot be seen directly under normal light.
To visualize the $DNA$ fragments,the gel is stained with a fluorescent dye called Ethidium Bromide $(EtBr)$.
When this stained gel is exposed to $UV$ radiation,the $DNA$ fragments appear as bright orange-colored bands.

(B) Agarose is a natural polymer extracted from sea weeds (marine algae),specifically from species like Gelidium and Gracilaria. It is widely used in biotechnology for gel electrophoresis to separate $DNA$ fragments based on their size.

(D) Agarose is a natural polysaccharide extracted from seaweeds. It is a linear polymer made up of repeating units of agarobiose,which consists of $D$-galactose and $3,6$-anhydro-$L$-galactopyranose. In biotechnology,it is widely used as a matrix in gel electrophoresis to separate $DNA$ fragments based on their size.

(D) plasmid is a small,circular,double-stranded $DNA$ molecule that is distinct from a cell's chromosomal $DNA$.
Plasmids naturally exist in bacterial cells and carry their own origin of replication $(ori)$,which allows them to replicate independently of the host's chromosomal $DNA$.
$r-DNA$ (recombinant $DNA$) is created by inserting a foreign $DNA$ fragment into a vector (like a plasmid).
Since the $r-DNA$ molecule is constructed using a plasmid vector,it also retains the ability to replicate independently within the host bacterial cell.
Therefore,both plasmids and $r-DNA$ molecules possess the ability to replicate independently within a bacterial cell.

(B) In a bacterial cell,the copy number of $DNA$ molecules varies significantly based on the type of vector or genetic element.
$1$. Genomic $DNA$: Usually,there is only $1$ copy of the bacterial chromosome per cell.
$2$. Plasmid: Plasmids are extrachromosomal $DNA$ molecules that can have a copy number ranging from $1$ to $100$ per cell.
$3$. Bacteriophage: Bacteriophages (like $\lambda$ phage or $M13$ phage) can replicate to produce a very high number of copies (often $10$ to $100$ or more) within the host cell during the lytic cycle.
Therefore,the order of copy number capacity is: Bacteriophage > Plasmid > Genomic $DNA$.

(B) The replication of a plasmid is controlled by a specific $DNA$ sequence known as the origin of replication,abbreviated as $ori$.
This sequence is responsible for initiating the replication process within the host cell.
Without the $ori$ sequence,the plasmid cannot replicate independently,and therefore,it cannot be maintained in the host cell.
$rop$ codes for proteins involved in the regulation of plasmid copy number,while $amp^R$ and $tet^R$ are antibiotic resistance genes used as selectable markers.

(C) The origin of replication $(ori)$ is a specific $DNA$ sequence where replication begins.
Genomic $DNA$ in prokaryotes and eukaryotes contains $ori$ sites to initiate replication.
Plasmids are extrachromosomal circular $DNA$ molecules that also contain an $ori$ site,allowing them to replicate independently within the host cell.
$RNA$ is a single-stranded molecule that does not contain an origin of replication sequence,as it is a product of transcription,not a template for autonomous replication in this context.
Therefore,$RNA$ does not possess an origin of replication.

(D) cloning vector must possess an origin of replication $(ori)$ to allow autonomous replication.
It must have a selectable marker to identify and eliminate non-transformants.
It should have unique cloning sites for restriction enzymes to allow the insertion of foreign $DNA$.
If a vector has more than one recognition site for a single restriction enzyme,it will lead to multiple fragments of the vector upon digestion,which complicates the gene cloning process. Therefore,it should not possess multiple recognition sites for the same restriction enzyme.

(A) selectable marker is a gene introduced into a cell,especially a bacterium or to cells in culture,that confers a trait suitable for artificial selection.
In recombinant $DNA$ technology,selectable markers (such as antibiotic resistance genes like $amp^R$,$tet^R$,etc.) are used to identify and eliminate non-transformants and selectively permit the growth of the transformants.
Therefore,the correct function is to allow the growth of transformants while inhibiting the growth of non-transformants.

(C) In the context of cloning vectors like $pBR322$,the $rop$ gene stands for 'repressor of primer'.
It codes for proteins that are involved in the regulation of the replication of the plasmid.
Specifically,the $rop$ protein helps in controlling the copy number of the plasmid within the host cell.

(D) selectable marker is a gene introduced into a cell,especially a bacterium or to cells in culture,that confers a trait suitable for artificial selection.
Common selectable markers in cloning vectors include genes encoding resistance to antibiotics such as ampicillin $(amp^R)$,tetracycline $(tet^R)$,and kanamycin $(kan^R)$.
$ori$ stands for the 'origin of replication',which is a sequence from where replication starts and is not a selectable marker.

(D) Normal $E. coli$ cells do not naturally possess antibiotic resistance genes such as $amp^R$ (ampicillin resistance),$tet^R$ (tetracycline resistance),or $kan^R$ (kanamycin resistance).
These genes are typically introduced into $E. coli$ via cloning vectors (like $pBR322$) during genetic engineering processes to act as selectable markers.
Therefore,none of these genes are present in the wild-type or normal $E. coli$ genome.

(A) $pBR322$ is a widely used cloning vector in biotechnology.
It is an artificially constructed plasmid derived from $E. coli$.
The name $pBR322$ stands for:
$p$ = Plasmid
$BR$ = Bolivar and Rodriguez (the scientists who constructed it)
$322$ = $A$ specific number assigned to distinguish it from other plasmids developed in the same laboratory.
Therefore,it is a plasmid.

(B) The plasmid $pBR322$ is one of the most commonly used cloning vectors in biotechnology.
It contains two distinct antibiotic resistance genes: the ampicillin resistance gene $(amp^R)$ and the tetracycline resistance gene $(tet^R)$.
These genes act as selectable markers,allowing researchers to identify and select transformants that have successfully taken up the recombinant plasmid.