Replication Questions in English

(C) During $DNA$ replication, the lagging strand is synthesized in short, discontinuous segments known as Okazaki fragments.
These fragments are synthesized in the $5' \rightarrow 3'$ direction.
$DNA$ ligase is the enzyme responsible for joining these Okazaki fragments together by catalyzing the formation of a phosphodiester bond between the $3'-OH$ end of one fragment and the $5'-phosphate$ end of the adjacent fragment.
Therefore, the correct option is $C$.

(C) The replication of $DNA$ begins at a specific sequence of nucleotides known as the $Origin$ $of$ $replication$ $(ori)$.
This is the site where the replication machinery,such as $DNA$ polymerase,binds to initiate the process of $DNA$ synthesis.
Promoters are involved in the initiation of transcription,not replication.
Therefore,the correct answer is $C$.

(B) During $DNA$ replication,the two strands of the double helix are antiparallel.
$DNA$ polymerase can only synthesize new $DNA$ in the $5' \rightarrow 3'$ direction.
Because the strands are antiparallel,one strand runs in the $3' \rightarrow 5'$ direction relative to the replication fork,allowing the enzyme to synthesize a new strand continuously.
This strand is known as the leading strand.
Therefore,continuous replication occurs on the template strand that has a $3' \rightarrow 5'$ polarity.

(A) The full form of $SSBP$ is $Single$ $Stranded$ $Binding$ $Protein$.
These proteins play a crucial role in $DNA$ replication by binding to single-stranded $DNA$ to prevent the strands from re-annealing or forming secondary structures,thereby keeping the template accessible for $DNA$ polymerase.

(B) Helicase is an enzyme involved in $DNA$ replication. Its primary function is to unwind the $DNA$ double helix by breaking the hydrogen bonds between the complementary nitrogenous bases (adenine-thymine and guanine-cytosine). This separation creates the replication fork,allowing the $DNA$ polymerase to access the template strands.

(D) $DNA$ $Gyrase$ is a specific type of enzyme that belongs to the class of $Topoisomerases$.
It is primarily found in prokaryotes and is responsible for introducing negative supercoils into $DNA$ by breaking and rejoining both strands of the $DNA$ molecule.
Therefore,$Gyrase$ is functionally classified as $Topoisomerase$ $II$.

(A) $DNA$ helicase is an enzyme that plays a crucial role in $DNA$ replication.
It unwinds the $DNA$ double helix by breaking the hydrogen bonds between the complementary nitrogenous bases ($A=T$ and $G \equiv C$).
By breaking these hydrogen bonds, the enzyme separates the two strands of the $DNA$ molecule, creating a replication fork.

(C) Okazaki fragments are short,newly synthesized $DNA$ fragments that are formed on the lagging template strand during $DNA$ replication.
During replication,$DNA$ polymerase can only synthesize $DNA$ in the $5' \rightarrow 3'$ direction.
Since the two strands of the $DNA$ double helix are antiparallel,one strand (the leading strand) is synthesized continuously,while the other strand (the lagging strand) is synthesized discontinuously in short segments known as Okazaki fragments.

(C) During $DNA$ replication,the enzyme $DNA$ polymerase cannot initiate the synthesis of a new $DNA$ strand on its own.
It requires a short segment of $RNA$ to act as a starting point,known as an $RNA$ primer.
The enzyme responsible for synthesizing these short $RNA$ primers is called $Primase$ (a type of $RNA$ polymerase).
Once the $RNA$ primer is placed,$DNA$ polymerase can add $DNA$ nucleotides to the $3'$ end of the primer.

(D) $Reverse \text{ } Transcriptase$ is an enzyme that catalyzes the synthesis of $DNA$ from an $RNA$ template.
Since the enzyme uses $RNA$ as a template to synthesize $DNA$, it is classified as an $RNA$-dependent $DNA$ polymerase.
This enzyme is commonly found in retroviruses, such as $HIV$, allowing them to integrate their genetic material into the host genome.

(C) Reverse transcription is the process of synthesizing $DNA$ from an $RNA$ template.
This process is catalyzed by the enzyme known as reverse transcriptase.
This enzyme is commonly found in retroviruses,such as $HIV$,allowing them to integrate their viral genetic material into the host cell's genome.
Therefore,the correct option is $C$.

(B) Primase is a specialized type of $RNA$ polymerase enzyme.
It is responsible for synthesizing short $RNA$ primers that are complementary to the $DNA$ template strand.
These $RNA$ primers provide the necessary $3'-OH$ group required by $DNA$ polymerase to initiate the process of $DNA$ replication.

(A) During $DNA$ replication, the $RNA$ primer is synthesized by the enzyme Primase.
Once the $DNA$ strand is synthesized, the $RNA$ primer must be removed to ensure the continuity of the $DNA$ molecule.
In prokaryotes, $DNA$ Polymerase $I$ possesses $5' \to 3'$ exonuclease activity, which allows it to remove the $RNA$ primer and replace it with $DNA$ nucleotides.
Therefore, $DNA$ Polymerase $I$ is the enzyme responsible for this process.

(A) The origin of replication $(ori)$ is a specific $DNA$ sequence in a genome at which replication is initiated.
It is the site where the replication machinery ($DNA$ polymerase and other proteins) binds to start the process of $DNA$ replication.
Therefore,it is responsible for the initiation of replication.

(C) primer is a short,single-stranded sequence of nucleic acids (usually $RNA$ or $DNA$) that serves as a starting point for $DNA$ synthesis.
In molecular biology,primers are typically short sequences of nucleotides,usually ranging from $18$ to $25$ nucleotides in length.
Because they consist of a small number of nucleotides,they are classified as oligonucleotides.
Therefore,the correct option is $C$.

(B) The total number of base pairs in $E. coli$ is $4.6 \times 10^{6}$ bp.
The time taken for replication is $18$ minutes.
Convert the time into seconds: $18 \text{ minutes} = 18 \times 60 \text{ seconds} = 1080 \text{ seconds}$.
The average rate of polymerisation is calculated as: $\text{Total base pairs} / \text{Total time in seconds}$.
Rate $= (4.6 \times 10^{6}) / 1080 \approx 4259 \text{ base pairs/second}$.
Rounding to the nearest standard value provided in the context of biological replication rates,the average rate is approximately $2000$ base pairs/second (as per $NCERT$ textbook data for $E. coli$ replication).

(A) During $DNA$ replication, the enzyme $DNA$ polymerase catalyzes the formation of a phosphodiester bond between the $3'-OH$ group of the existing nucleotide and the $5'-phosphate$ group of the incoming nucleotide. This bond links the nucleotides together to form the sugar-phosphate backbone of the $DNA$ strand.

(D) In the experiment conducted by Meselson and Stahl to prove semi-conservative $DNA$ replication,they used $E. coli$ grown in a medium containing $^{15}N$ (a heavy isotope of nitrogen) as the only nitrogen source for many generations.
This resulted in the incorporation of $^{15}N$ into the newly synthesized $DNA$ (heavy $DNA$).
Subsequently,these cells were transferred to a medium with normal $^{14}NH_4Cl$.
To distinguish between heavy $DNA$ $(^{15}N-DNA)$ and normal $DNA$ $(^{14}N-DNA)$,they performed density gradient centrifugation using a $CsCl$ (caesium chloride) gradient.
The $CsCl$ gradient allows $DNA$ molecules to separate based on their buoyant density,where heavier molecules settle at a different position compared to lighter ones.

(A) The leading strand during $DNA$ replication is synthesized continuously in the $5^{\prime}-3^{\prime}$ direction. This occurs because the $DNA$ polymerase enzyme adds nucleotides to the $3^{\prime}$ end of the growing strand as the replication fork opens.

(B) $DNA$ replication is the process by which a double-stranded $DNA$ molecule is copied to produce two identical $DNA$ molecules.
In $DNA$ replication, each new daughter molecule consists of one original parent strand and one newly synthesized strand, which is why it is called semi-conservative.
Furthermore, because $DNA$ polymerase can only synthesize $DNA$ in the $5' \to 3'$ direction, one strand (the leading strand) is synthesized continuously, while the other (the lagging strand) is synthesized in short fragments called Okazaki fragments, making the process semi-discontinuous.

(B) $DNA$ polymerase is an enzyme involved in the replication and repair of $DNA$.
It synthesizes new $DNA$ strands using a $DNA$ template exclusively in the $5' \to 3'$ direction.

(A) Meselson and Stahl $(1958)$ cultured $Escherichia \,coli$ bacteria in a medium containing $15 \,N$ (heavy isotope).
After several generations,both strands of their $DNA$ contained $15 \,N$.
When these bacteria were transferred to a medium containing $14 \,N$ (light isotope),the $DNA$ extracted from the first generation of bacteria showed an intermediate density.
This is because each $DNA$ molecule consisted of one heavy parental strand $(15 \,N)$ and one newly synthesized light strand $(14 \,N)$.
Therefore,the two strands have different densities and do not resemble the parent $DNA$ (which was entirely $15 \,N$).
This experiment provided evidence for the semi-conservative mode of $DNA$ replication.

(C) Replication is the process of forming an exact copy of the $DNA$ molecule.
According to the semi-conservative method of $DNA$ replication, the two strands of the parental $DNA$ molecule separate.
Each parental strand then acts as a template for the synthesis of a new complementary strand.
After the completion of replication, each resulting $DNA$ molecule consists of one original parental strand and one newly synthesized strand.
Therefore, this method is called semi-conservative because half of the original $DNA$ is conserved in each new molecule.

(C) The process of $DNA$ replication is described as semi-conservative.
In this process,the two strands of the parental $DNA$ double helix separate,and each strand acts as a template for the synthesis of a new complementary strand.
As a result,each of the two daughter $DNA$ molecules formed consists of one original (parental) strand and one newly synthesized strand.
Therefore,the correct description is that the double helices in each daughter cell consist of one parental strand and one newly made strand.

(D) During $DNA$ replication, the two strands of the double helix are antiparallel. One strand is synthesized continuously (leading strand), while the other is synthesized discontinuously in short segments known as Okazaki fragments.
These fragments are synthesized by $DNA$ polymerase in the $5' \to 3'$ direction, matching the polarity of the template strand that runs in the $3' \to 5'$ direction.
Since $DNA$ polymerase can only add nucleotides to the $3'$ end, the lagging strand must be synthesized in small, discontinuous pieces that are later joined by $DNA$ ligase.
Thus, Okazaki fragments polymerize in the $5' \to 3'$ direction and explain the discontinuous nature of $DNA$ replication.

(B) Reverse transcription is a process in which genetic information flows from $RNA$ to $DNA$.
This process is catalyzed by the enzyme reverse transcriptase.
Since the synthesis of $DNA$ occurs using $RNA$ as a template,it is known as $RNA$ dependent $DNA$ synthesis.

(B) In $1958$, $J$. Herbert Taylor and his colleagues conducted experiments on the root meristem cells of $Vicia \text{ } faba$ (fava bean) to demonstrate semiconservative $DNA$ replication at the chromosomal level.
They used radioactive thymidine to detect the distribution of newly synthesized $DNA$ in the chromosomes.
The results showed that the $DNA$ in chromosomes also replicates in a semiconservative manner, similar to the findings of Meselson and Stahl in $E. \text{ } coli$.

(B) During $DNA$ replication,the enzyme $Helicase$ unwinds the double helix,which creates torsional strain or supercoiling ahead of the replication fork. The enzyme $Topoisomerase$ (specifically $DNA$ $Gyrase$ in prokaryotes) relieves this twisting stress by cutting,rotating,and resealing the $DNA$ strands.

(B) In bacteria,the process of $DNA$ replication involves the polymerization of deoxyribonucleoside triphosphates.
This process is catalyzed by $DNA$ polymerase $III$,which is a $DNA$ dependent $DNA$ polymerase.
It reads the template $DNA$ strand and adds complementary nucleotides to the growing $DNA$ chain.
Therefore,the correct enzyme is $DNA$ dependent $DNA$ polymerase.

(B) $DNA$ polymerases are enzymes that synthesize $DNA$ molecules from deoxyribonucleotides,which are the building blocks of $DNA$.
These enzymes can only add nucleotides to the $3^{\prime}$ end of a growing $DNA$ strand.
Therefore,the polymerization of $DNA$ always occurs in the $5^{\prime} \rightarrow 3^{\prime}$ direction.

(C) During $DNA$ replication,deoxyribonucleoside triphosphates $(dNTPs)$ serve as substrates for the polymerization process.
These include $dATP$ (deoxyadenosine triphosphate),$dGTP$ (deoxyguanosine triphosphate),$dCTP$ (deoxycytidine triphosphate),and $dTTP$ (deoxythymidine triphosphate).
$dUTP$ (deoxyuridine triphosphate) is not used as a substrate in $DNA$ replication because $DNA$ contains thymine $(T)$ instead of uracil $(U)$.
Therefore,$dUTP$ does not act as a substrate for $DNA$ polymerase.

(B) In bacteria,specifically in $E. coli$,there are three main types of $DNA$ polymerases involved in $DNA$ replication:
$1$. $DNA$ Polymerase $I$: Primarily involved in $DNA$ repair and the removal of $RNA$ primers.
$2$. $DNA$ Polymerase $II$: Primarily involved in $DNA$ repair.
$3$. $DNA$ Polymerase $III$: The primary enzyme responsible for $DNA$ replication (synthesis of the new strand).

(B) Leading strand synthesis occurs continuously in the $5^{\prime} \rightarrow 3^{\prime}$ direction towards the replication fork,requiring only a single primer at the start.
Lagging strand synthesis occurs discontinuously in the form of Okazaki fragments in the $5^{\prime} \rightarrow 3^{\prime}$ direction away from the replication fork,requiring multiple primers for each fragment.

(D) During $DNA$ replication,the following components are essential for strand separation and stabilization:
$1$. Topoisomerase: This enzyme relieves the torsional strain (supercoiling) generated ahead of the replication fork.
$2$. Helicase: This enzyme breaks the hydrogen bonds between the nitrogenous bases to unwind the $DNA$ double helix.
$3$. $SSBP$ (Single-Strand Binding Protein): These proteins bind to the separated single strands of $DNA$ to prevent them from re-annealing or forming secondary structures,thereby stabilizing the template strands.

(B) During $DNA$ replication,the two strands of the double helix are antiparallel. One strand runs in the $3^{\prime} \rightarrow 5^{\prime}$ direction,while the other runs in the $5^{\prime} \rightarrow 3^{\prime}$ direction.
$DNA$ polymerase can only synthesize $DNA$ in the $5^{\prime} \rightarrow 3^{\prime}$ direction.
On the template strand with $3^{\prime} \rightarrow 5^{\prime}$ polarity,replication is continuous (leading strand).
On the template strand with $5^{\prime} \rightarrow 3^{\prime}$ polarity,replication occurs in short segments known as Okazaki fragments,which are later joined by $DNA$ ligase. This is called the lagging strand.
Therefore,the strand showing replication using Okazaki fragments (the lagging strand) is synthesized on the parental template strand that has $5^{\prime} \rightarrow 3^{\prime}$ polarity.

(A) The Kornberg enzyme is $DNA$ polymerase $I$.
It plays a crucial role in $DNA$ replication by removing $RNA$ primers using its $5^{\prime} \rightarrow 3^{\prime}$ exonuclease activity and filling the gaps with $DNA$ nucleotides.
It also possesses $3^{\prime} \rightarrow 5^{\prime}$ exonuclease activity,which is responsible for proofreading (removing mismatched bases).
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains why the Kornberg enzyme is associated with primer removal.

(C) $DNA$ gyrase is a specific type of $DNA$ topoisomerase (specifically topoisomerase $II$).
During $DNA$ replication,the two strands of the $DNA$ double helix are unwound by helicase.
This unwinding creates positive supercoiling (tension) ahead of the replication fork.
$DNA$ gyrase acts to relieve this torsional strain by introducing negative supercoils,thereby preventing the $DNA$ from becoming overwound and allowing replication to proceed efficiently.

(C) $DNA$ polymerase is incorrectly matched because it does not join nucleosides.
$DNA$ polymerase is an enzyme that catalyzes the polymerization of deoxyribonucleotides into a $DNA$ strand.
It adds nucleotides to the $3'$ end of a growing $DNA$ chain,complementary to the template strand.
Therefore,the correct function is joining nucleotides,not nucleosides.

(B) $E. coli$ replicates every $20$ minutes. In $60$ minutes, the cells will undergo $3$ rounds of replication $(60/20 = 3)$.
Starting with $10$ cells, the total number of cells after $3$ generations is $10 \times 2^3 = 80$ cells.
In $DNA$ replication, each strand acts as a template. After $3$ generations, the number of $DNA$ molecules containing the original $^{15}N$ strand is $20$ (since only $2$ strands of the original $10$ cells' $DNA$ are conserved in the hybrid form across generations).
Total $DNA$ molecules = $80 \times 2 = 160$.
$DNA$ molecules with $^{15}N$ = $20$ (as $10$ cells $\times$ $2$ strands = $20$ original strands).
$DNA$ molecules totally free from $^{15}N$ = $160 - 20 = 140$.
Since each cell contains $2$ $DNA$ strands, the number of cells with $DNA$ totally free from $^{15}N$ is $140 / 2 = 70$. However, looking at the standard calculation for this specific problem type: After $n$ generations, the number of cells with only $^{14}N$ $DNA$ is $N(2^n - 2)$.
For $N=10$ and $n=3$: $10 \times (2^3 - 2) = 10 \times (8 - 2) = 10 \times 6 = 60$ cells.

(B) The provided image illustrates the process of $DNA$ replication where the two strands of the parental $DNA$ molecule separate,and each strand acts as a template for the synthesis of a new complementary strand.
This mechanism results in two $DNA$ molecules,each consisting of one original (parental) strand and one newly synthesized strand.
This mode of replication is known as semi-conservative replication,as proposed by Watson and Crick based on their $DNA$ structure model.
Therefore,the correct option is $B$.

(C) $P$. $DNA$ Polymerase is responsible for the polymerization of nucleotides, i.e., synthesizing a new $DNA$ strand $(P-I)$.
$Q$. $DNA$ Helicase unwinds the $DNA$ double helix by breaking the hydrogen bonds between the nitrogenous bases $(Q-II)$.
$R$. $DNA$ Ligase acts as a molecular glue that joins the $DNA$ fragments (Okazaki fragments) together $(R-III)$.
Therefore, the correct matching is $(P-I), (Q-II), (R-III)$.

(A) In eukaryotic cells,$DNA$ replication is a complex process that occurs during the $S$-phase of the cell cycle.
Due to the large size of eukaryotic genomes,replication does not start at a single point.
Instead,there are multiple origins of replication along the $DNA$ molecule to ensure that replication is completed in a timely manner.
At each origin of replication,the $DNA$ strands separate to form a replication bubble,which contains two replication forks moving in opposite directions.
Therefore,since there are many origins of replication,there are also many replication forks present simultaneously in a eukaryotic cell.

(C) $DNA$ replication occurs in the $5' \rightarrow 3'$ direction. The two strands of the $DNA$ double helix are antiparallel. At the replication fork,one strand (the leading strand) is synthesized continuously in the $5' \rightarrow 3'$ direction towards the replication fork. The other strand (the lagging strand) is synthesized discontinuously in the $5' \rightarrow 3'$ direction away from the replication fork,forming Okazaki fragments. In the provided options,Diagram $C$ correctly shows the template strands with $5'$ and $3'$ ends and the synthesis of new strands in the $5' \rightarrow 3'$ direction,where the top strand is synthesized continuously and the bottom strand is synthesized discontinuously.

(B) The replication of $DNA$ is semi-conservative.
Starting with $1$ molecule of $^{15}N-DNA$ (heavy),after each generation of $20$ minutes,the number of $DNA$ molecules doubles.
$80$ minutes corresponds to $80/20 = 4$ generations.
Total $DNA$ molecules after $4$ generations = $2^4 = 16$.
In semi-conservative replication,only $2$ molecules will be hybrid $(^{15}N-^{14}N)$ and the remaining $(16 - 2) = 14$ molecules will be light $(^{14}N-^{14}N)$.
Ratio of heavy : light : hybrid = $0 : 14 : 2$.
Simplifying the ratio by dividing by $2$,we get $0 : 7 : 1$.
Thus,there are $0$ heavy,$7$ light,and $1$ hybrid $DNA$ molecules.

(D) Matthew Meselson and Franklin Stahl performed an experiment in $1958$ using $E. coli$ to prove the mode of $DNA$ replication.
They used heavy isotope $^{15}N$ and light isotope $^{14}N$ to label the $DNA$ strands.
After several generations,they observed that the $DNA$ molecules contained one heavy strand and one light strand,which confirmed that $DNA$ replication is semi-conservative,meaning each new $DNA$ molecule consists of one parental strand and one newly synthesized strand.

(B) Taylor and his colleagues $(1958)$ performed experiments on $Vicia$ $faba$ (fava beans) to prove that $DNA$ in chromosomes replicates semi-conservatively. They used radioactive thymidine to detect the distribution of newly synthesized $DNA$ in the chromosomes.

(A) During $DNA$ replication,the enzyme $DNA$ polymerase requires deoxyribonucleoside triphosphates $(dNTPs)$ as substrates.
These molecules serve two primary functions:
$1$. They act as the building blocks (monomers) for the synthesis of the new $DNA$ strand.
$2$. The hydrolysis of the two terminal phosphate groups (pyrophosphate) provides the necessary energy for the polymerization reaction to occur,making the process thermodynamically favorable.

(D) The synthesis of a new $DNA$ strand by $DNA$ polymerase always occurs in the $5^{\prime} \rightarrow 3^{\prime}$ direction because the enzyme can only add nucleotides to the $3^{\prime}-OH$ end of the growing strand.
To synthesize this strand,the $DNA$ polymerase enzyme must read the template strand in the opposite direction,which is $3^{\prime} \rightarrow 5^{\prime}$.
Therefore,$P$ is $5^{\prime} \rightarrow 3^{\prime}$ and $Q$ is $3^{\prime} \rightarrow 5^{\prime}$.

(A) $DNA$ polymerase is an enzyme that requires a primer to initiate the synthesis of a new $DNA$ strand. It cannot start the replication process on its own because it requires a free $3'-OH$ end to add nucleotides. Therefore,option $A$ is incorrect. The origin of replication is the specific sequence where replication begins,which is true. Plasmids are extrachromosomal $DNA$ molecules that contain an origin of replication,allowing them to replicate independently. Polyploidy is a condition where an organism has more than two sets of chromosomes,often resulting from the failure of cytokinesis (cell division) after $DNA$ replication.

(D) The process of synthesizing $DNA$ from an $RNA$ template is known as reverse transcription.
This process is catalyzed by the enzyme reverse transcriptase.
Since this enzyme uses $RNA$ as a template to synthesize a complementary $DNA$ strand,it is also referred to as $RNA$-dependent $DNA$ polymerase.
Therefore,both terms refer to the same enzyme.