Replication Questions in English

(A) Statement $I$ is correct: $DNA$ polymerases are enzymes that synthesize $DNA$ by adding nucleotides only to the $3^{\prime}$ end of a growing strand. Therefore,polymerization always occurs in the $5^{\prime} \rightarrow 3^{\prime}$ direction.
Statement $II$ is correct: Because $DNA$ strands are antiparallel and $DNA$ polymerase can only synthesize in the $5^{\prime} \rightarrow 3^{\prime}$ direction,one strand (the leading strand) is synthesized continuously toward the replication fork,while the other strand (the lagging strand) is synthesized discontinuously in short segments called Okazaki fragments away from the replication fork.
Conclusion: Both statements are correct.

(C) In prokaryotes like $E. coli$,the process of $DNA$ replication is mediated by $DNA$-dependent $DNA$ polymerase enzymes.
These enzymes are highly specific and can only catalyze the polymerization of nucleotides in the $5^{\prime} \rightarrow 3^{\prime}$ direction.
This unidirectional synthesis is a fundamental characteristic of $DNA$ replication,ensuring the accurate addition of nucleotides to the growing strand.

(C) $DNA$ polymerase enzyme can only catalyze the polymerization of nucleotides in the $5'$ to $3'$ direction.
Since the two strands of the $DNA$ double helix are antiparallel (one runs $3'$ to $5'$ and the other $5'$ to $3'$),the replication fork moves in one direction.
On the template strand with $3'$ to $5'$ polarity,the new strand is synthesized continuously in the $5'$ to $3'$ direction (leading strand).
On the template strand with $5'$ to $3'$ polarity,the new strand must be synthesized in short fragments (Okazaki fragments) in the $5'$ to $3'$ direction because the replication fork opens in the opposite direction relative to the synthesis requirement (lagging strand).
Therefore,the constraint of $5'$ to $3'$ synthesis is the fundamental reason for discontinuous replication.

(A) During $DNA$ replication,the two strands of the double helix are antiparallel and $DNA$ polymerase can only add nucleotides in the $5'$ to $3'$ direction.
$1$. The leading strand is synthesized continuously in the $5'$ to $3'$ direction towards the replication fork.
$2$. The lagging strand is synthesized discontinuously in short segments known as Okazaki segments,also in the $5'$ to $3'$ direction,moving away from the replication fork.
$3$. Because the lagging strand is synthesized in segments,it requires multiple $RNA$ primers to initiate each segment.
$4$. $DNA$ polymerase $III$ is the primary enzyme responsible for synthesizing both the leading and lagging strands.
Therefore,the statement 'Synthesis of Okazaki segments is continuous' is incorrect,as they are synthesized discontinuously.

(B) During $\text{DNA}$ replication,the enzyme $\text{DNA}$ polymerase can only synthesize a new strand in the $5^{\prime} \longrightarrow 3^{\prime}$ direction.
Because the two strands of the $\text{DNA}$ double helix are antiparallel,one strand (the leading strand) is synthesized continuously.
The other strand (the lagging strand) is synthesized discontinuously in short segments known as Okazaki fragments.
Even though these fragments are formed on the lagging strand,each individual Okazaki fragment is still synthesized by $\text{DNA}$ polymerase in the $5^{\prime} \longrightarrow 3^{\prime}$ direction.

(C) During $\text{DNA}$ replication,the lagging strand is synthesized in short,discontinuous segments known as Okazaki fragments.
These fragments are synthesized by $\text{DNA}$ polymerase $III$.
Once the $\text{RNA}$ primers are removed and replaced by $\text{DNA}$ nucleotides,the enzyme $\text{DNA}$ ligase catalyzes the formation of a phosphodiester bond between the adjacent fragments,effectively joining them into a continuous strand.
Therefore,the correct enzyme is $\text{DNA}$ ligase.

(B) $DNA$ replication is semi-conservative in nature,meaning each new $DNA$ molecule consists of one parental strand and one newly synthesized strand.
In the replication fork,$DNA$ polymerase can only synthesize $DNA$ in the $5' \rightarrow 3'$ direction.
Because the two strands of the $DNA$ double helix are antiparallel,one strand (the leading strand) is synthesized continuously.
The other strand (the lagging strand) is synthesized in short,discontinuous segments known as Okazaki fragments.
Therefore,$DNA$ replication in the lagging strand is semi-conservative and discontinuous.

(A) During $\text{DNA}$ replication,the enzyme $\text{DNA}$ polymerase is responsible for synthesizing a new $\text{DNA}$ strand by adding nucleotides that are complementary to the template strand.
$\text{DNA}$ polymerase catalyzes the polymerization of deoxyribonucleotides,which is essential for the accurate replication of the genetic material.
Therefore,the correct option is $A$.

(B) In the Meselson and Stahl experiment,the semi-conservative replication of $\text{DNA}$ is observed.
Starting with $N^{15}-N^{15}$ (heavy) $\text{DNA}$,after $n$ generations,the total number of $\text{DNA}$ molecules is $2^n$.
The number of heavy $\text{DNA}$ molecules $(N^{15}-N^{15})$ remains constant at $2$ after the first generation,but since we start with one molecule,the heavy strand is distributed into two hybrid molecules $(N^{15}-N^{14})$ in the first generation.
Specifically,in any generation $n$ (where $n \ge 1$),the number of hybrid $\text{DNA}$ molecules is always $2$,and the number of light $\text{DNA}$ molecules $(N^{14}-N^{14})$ is $2^n - 2$.
The number of heavy $\text{DNA}$ molecules $(N^{15}-N^{15})$ becomes $0$ after the first generation because the original heavy strands are separated and paired with light strands.
Therefore,in the $4^{\text{th}}$ generation,the percentage of heavy $\text{DNA}$ is $0 \%$.

(B) The generation time of $E. coli$ is $20$ minutes.
After $60$ minutes, the number of generations $(n)$ = $60 / 20 = 3$.
Initially, there is $1$ molecule of light $DNA$ $(^{14}N-^{14}N)$.
After $1$ generation: $2$ molecules ($1$ hybrid, $1$ light is not possible, both become hybrid $^{14}N-^{15}N$).
Actually, in semi-conservative replication, the original strands remain as templates.
After $n$ generations, the total number of $DNA$ molecules = $2^n = 2^3 = 8$.
The number of $DNA$ molecules containing original light strands = $2$ (since only the two original strands from the parent cell will remain as light templates).
Percentage of light $DNA$ = $(\text{Number of light } DNA / \text{Total } DNA) \times 100 = (2 / 8) \times 100 = 25 \%$.

(A) The replication of $\text{DNA}$ is semi-conservative.
Starting with one hybrid $\text{DNA}$ molecule $(N^{15}N^{14})$:
Generation $0$: $1$ hybrid $(N^{15}N^{14})$.
Generation $1$: The $N^{15}$ strand and $N^{14}$ strand separate. Each acts as a template for a new $N^{14}$ strand. Result: $2$ hybrid $\text{DNA}$ $(N^{15}N^{14})$.
Generation $2$: The $2$ hybrid molecules produce $2$ hybrid $(N^{15}N^{14})$ and $2$ light $(N^{14}N^{14})$ $\text{DNA}$ molecules.
Generation $3$: The $2$ hybrid molecules produce $2$ hybrid $(N^{15}N^{14})$ and $2$ light $(N^{14}N^{14})$ $\text{DNA}$ molecules,while the $2$ light molecules produce $4$ light $(N^{14}N^{14})$ $\text{DNA}$ molecules.
Total $\text{DNA}$ molecules after $3$ generations = $2^3 = 8$.
Number of $\text{DNA}$ molecules containing the original $N^{15}$ strand = $2$ (these are the $2$ hybrid molecules).

(A) In the process of $DNA$ replication,the $DNA$ polymerase enzyme can only synthesize $DNA$ in the $5' \rightarrow 3'$ direction.
Because the two strands of the $DNA$ double helix are antiparallel,one strand (the leading strand) is synthesized continuously in the direction of the replication fork movement.
In the given figure,$A$ represents the leading strand where continuous synthesis occurs.
The other strand (the lagging strand) is synthesized in short fragments known as Okazaki fragments,which represents discontinuous synthesis,indicated by $B$.
$C$ points to the original parental $DNA$ strands which act as the template for replication.
$D$ represents the newly synthesized $DNA$ strands formed during the process.
Therefore,the correct identification is $A \rightarrow$ Continuous synthesis,$B \rightarrow$ Discontinuous synthesis,$C \rightarrow$ Template strand,$D \rightarrow$ Newly synthesised strands.

(C) Assertion $(A)$ is correct because $\text{DNA}$ replication begins at a specific site called the origin of replication,where the $\text{DNA}$ helix unwinds to form a $Y$-shaped structure known as the replication fork.
Reason $(R)$ is incorrect because the joining of discontinuously synthesized $\text{DNA}$ fragments (Okazaki fragments) by the enzyme $\text{DNA}$ ligase occurs during $\text{DNA}$ replication,not during transcription.
Transcription involves the synthesis of $\text{RNA}$ from a $\text{DNA}$ template,not the synthesis of $\text{DNA}$ fragments.

(D) $1$. Ligase: This enzyme is responsible for joining short segments of $\text{DNA}$ (Okazaki fragments) by forming phosphodiester bonds. Thus,statement $a$ is correct.
$2$. $\text{DNA}$ polymerase: This enzyme is responsible for synthesizing a new $\text{DNA}$ strand by adding nucleotides complementary to the template strand. It does not cut $\text{RNA}$. Cutting $\text{RNA}$ at specific sequences is typically the function of ribonucleases or specific ribozymes. Thus,statement $b$ is false.
$3$. Helicase: This enzyme unwinds the $\text{DNA}$ double helix by breaking the hydrogen bonds between complementary nitrogenous base pairs during $\text{DNA}$ replication. Thus,statement $c$ is correct.
Conclusion: $a$ and $c$ are correct,while $b$ is false.

(D) Statement $I$ is incorrect because the template strand with a free $3'$ end is the leading template,and the strand with a free $5'$ end is the lagging template. The original statement provided in the question had these reversed.
Statement $II$ is correct. $DNA$ polymerase synthesizes the new strand in the $5' \rightarrow 3'$ direction. Since the new strand is antiparallel to the template,the replication process proceeds along the template strand from its $3'$ end towards its $5'$ end.

(B) During $DNA$ replication, the two strands of the $DNA$ double helix are antiparallel.
One strand has the polarity $3' \rightarrow 5'$, which acts as the template for the leading strand.
The synthesis of the leading strand occurs in the $5' \rightarrow 3'$ direction continuously towards the replication fork.
Therefore, the leading template strand is the one with $3' \rightarrow 5'$ polarity, and the replication process on this strand is continuous.

(C) During $DNA$ replication, the $RNA$ primers are removed by the $5' \to 3'$ exonuclease activity of $DNA$ polymerase-$I$ in prokaryotes.
After the removal of $RNA$ primers, the gaps are filled by the $DNA$ polymerase activity of $DNA$ polymerase-$I$ in prokaryotes.
In eukaryotes, the removal of $RNA$ primers and the subsequent filling of gaps is primarily carried out by $DNA$ polymerase-$\alpha$ (which has primase activity) and $DNA$ polymerase-$\delta$ or $\epsilon$.
Therefore, the correct sequence is: $i = \text{DNA polymerase-}I$, $ii = \text{DNA polymerase-}I$, and $iii = \text{DNA polymerase-}\alpha$.

(B) Statement $I$ is correct. Prokaryotic $DNA$ replication starts from a single origin of replication $(ori)$,meaning they have only one replicon. In contrast,eukaryotic $DNA$ is linear and very long,containing multiple origins of replication,resulting in several replicons in tandem.
Statement $II$ is incorrect. The enzyme helicase is responsible for unwinding the $DNA$ double helix by breaking the hydrogen bonds between the nitrogenous bases. The separated strands are prevented from rejoining (re-annealing) by Single-Strand Binding $(SSB)$ proteins,not by helicase.

(D) Initially,the $DNA$ is hybrid $(^{14}N-^{15}N)$.
After the first replication in a medium with $^{15}N$: Two $DNA$ molecules are formed,one hybrid $(^{14}N-^{15}N)$ and one heavy $(^{15}N-^{15}N)$.
After the second replication: $A$ total of four $DNA$ molecules are formed.
The hybrid $(^{14}N-^{15}N)$ molecule produces one hybrid $(^{14}N-^{15}N)$ and one heavy $(^{15}N-^{15}N)$ molecule.
The heavy $(^{15}N-^{15}N)$ molecule produces two heavy $(^{15}N-^{15}N)$ molecules.
Thus,the final composition is $0$ light,$3$ heavy,and $1$ hybrid $DNA$ molecules.
Therefore,the correct proportion is $0:3:1$.

(A) $1$. $DNA$ replication always occurs in the $5'$ to $3'$ direction.
$2$. In the given diagram,$A$ represents the $RNA$-primer,which initiates the synthesis of the new $DNA$ strand.
$3$. $B$ represents the Okazaki fragments,which are short,newly synthesized $DNA$ segments formed on the lagging strand.
$4$. $C$ represents the leading template strand,where synthesis occurs continuously.
$5$. $D$ represents the lagging template strand,where synthesis occurs discontinuously in the form of Okazaki fragments.
$6$. Therefore,the correct identification is: $A$ - $RNA$-primer,$B$ - Okazaki fragment,$C$ - Leading template,$D$ - Lagging template.

(C) $DNA$ polymerase can only synthesize $DNA$ in the $5' \rightarrow 3'$ direction.
On the lagging strand,the template strand runs in the $5' \rightarrow 3'$ direction,which means the new strand must be synthesized in the opposite direction relative to the movement of the replication fork.
As the replication fork opens,the lagging strand is synthesized in short segments known as Okazaki fragments.
These fragments are synthesized in the $5' \rightarrow 3'$ direction,moving away from the replication fork.

(D) $DNA$ replication is a semi-conservative process.
In this process,each of the two daughter $DNA$ molecules consists of one original (parental) strand and one newly synthesized strand.
Since one strand out of two in each daughter molecule is from the mother (parental) $DNA$,the contribution of the original nucleotides is $50\%$.

(D) Meselson and Stahl performed an experiment in $1958$ to prove that $DNA$ replication is semiconservative.
They cultured $E. coli$ in a medium containing $^{15}NH_4Cl$ ($^{15}N$ is the heavy isotope of nitrogen) as the only nitrogen source for many generations.
This resulted in the incorporation of $^{15}N$ into the newly synthesized $DNA$.
Subsequently,they transferred the cells into a medium with normal $^{14}NH_4Cl$ and took samples at various definite time intervals as the cells multiplied.
The extracted DNAs were centrifuged in a cesium chloride $(CsCl)$ density gradient to measure the densities of $DNA$.
Thus,they used the heavy isotope $^{15}N$ and $E. coli$ bacteria for their experiment.

(D) During $DNA$ replication,the two strands of the $DNA$ double helix are antiparallel ($5' \rightarrow 3'$ and $3' \rightarrow 5'$).
$DNA$ polymerase can only synthesize $DNA$ in the $5' \rightarrow 3'$ direction.
On the leading strand,synthesis occurs continuously towards the replication fork.
On the lagging strand,synthesis occurs in short segments known as $Okazaki$ fragments,moving away from the replication fork.
Because these fragments are synthesized in short,separate bursts,the process on the lagging strand is described as discontinuous.

(D) The generation time of $E. coli$ is $20$ minutes.
In $80$ minutes,the number of generations $(n)$ is $80 / 20 = 4$.
The total number of $DNA$ molecules after $4$ generations is $2^n = 2^4 = 16$.
In the Meselson and Stahl experiment,the starting $DNA$ is heavy $(^{15}N^{15}N)$.
After $1$ generation: $2$ intermediate $(^{15}N^{14}N)$.
After $2$ generations: $2$ intermediate $(^{15}N^{14}N)$ and $2$ light $(^{14}N^{14}N)$.
After $3$ generations: $2$ intermediate $(^{15}N^{14}N)$ and $6$ light $(^{14}N^{14}N)$.
After $4$ generations: $2$ intermediate $(^{15}N^{14}N)$ and $14$ light $(^{14}N^{14}N)$.
Thus,the ratio of light to intermediate $DNA$ molecules is $14: 2$.

(B) In the Meselson and Stahl experiment,$E. coli$ divides every $20$ minutes.
Starting with one heavy $(^{15}N^{15}N)$ $DNA$ molecule:
After $20$ minutes (first generation): $2$ hybrid $(^{15}N^{14}N)$ $DNA$ molecules are formed.
After $40$ minutes (second generation): $2$ hybrid $(^{15}N^{14}N)$ and $2$ light $(^{14}N^{14}N)$ $DNA$ molecules are formed.
After $60$ minutes (third generation): The $2$ hybrid molecules produce $2$ hybrid and $2$ light molecules,and the $2$ light molecules produce $4$ light molecules.
Total $DNA$ molecules = $8$.
Number of hybrid molecules = $2$.
Number of light molecules = $6$.
Ratio of light to hybrid = $6: 2$,which simplifies to $3: 1$.

(C) The genome of $E. coli$ contains $4.6 \times 10^6 \text{ bp}$.
$E. coli$ completes its replication process in approximately $38$ minutes ($2280$ seconds).
The rate of replication is calculated as:
$\text{Rate} = \frac{\text{Total base pairs}}{\text{Total time in seconds}}$
$\text{Rate} = \frac{4.6 \times 10^6}{2280} \approx 2017 \text{ bp/second}$.
Rounding this to the nearest standard value provided in the options, we get $2000 \text{ bp/second}$.

(B) The correct answer is $B$.
Matthew Meselson and Franklin Stahl performed an experiment in $1958$ using $Escherichia \text{ } coli$ to prove that $DNA$ replication is semiconservative.
They grew $E. \text{ } coli$ in a medium containing $^{15}N$ (a heavy isotope of nitrogen) for many generations, then transferred them to a medium with $^{14}N$ (a lighter isotope).
By analyzing the density of the $DNA$ using cesium chloride density gradient centrifugation, they observed that the $DNA$ molecules were hybrids of both isotopes, confirming the semiconservative model of replication.
This mechanism was subsequently shown to occur in higher organisms, including plants and human cells.

(C) The correct answer is $C$.
According to the semi-conservative model of $DNA$ replication demonstrated by Meselson and Stahl,the parental $DNA$ strands (labeled with $^{15}N$) separate,and each acts as a template for the synthesis of a new strand using $^{14}N$ from the medium.
In the first generation,each $DNA$ molecule consists of one heavy strand $(^{15}N)$ and one light strand $(^{14}N)$.
Since the two strands have different densities (one heavy and one light),the resulting $DNA$ molecule is a hybrid.
Because the hybrid $DNA$ contains one $^{14}N$ strand,it does not resemble the original parent $DNA$ (which was entirely $^{15}N$).

(B) The correct answer is $(B)$.
$DNA$ polymerase enzymes can only catalyze the addition of nucleotides in the $5^{\prime} \rightarrow 3^{\prime}$ direction.
Even during the synthesis of the lagging strand,where $DNA$ is synthesized in short fragments known as Okazaki fragments,each individual fragment is synthesized in the $5^{\prime} \rightarrow 3^{\prime}$ direction.

(D) The correct answer is $D$.
According to the Meselson and Stahl experiment,$E. coli$ cells grown in a medium containing $^{15}N$ (heavy isotope) incorporate it into their $DNA$.
When these cells are transferred to a medium containing $^{14}N$ (light isotope),the $DNA$ replication follows a semi-conservative model.
After the $1^{st}$ generation ($20$ minutes),all $DNA$ molecules are hybrid $(^{15}N^{14}N)$.
After the $2^{nd}$ generation ($40$ minutes),the hybrid $DNA$ molecules replicate to produce one hybrid $DNA$ molecule and one light $DNA$ molecule $(^{14}N^{14}N)$.
Therefore,the resulting population contains both hybrid and light $DNA$ in equal proportions.

(A) $DNA$ gyrase is a type of $DNA$ topoisomerase that relieves the torsional strain caused by the unwinding of $DNA$ during replication.
It introduces negative supercoils to counteract the positive supercoils that form ahead of the replication fork.

(D) The correct answer is $D$.
During $DNA$ replication,the process begins with the breaking of hydrogen bonds between the nitrogenous bases of the two $DNA$ strands.
This unwinds the double helix and allows each strand to serve as a template for the synthesis of a new complementary strand.

(A) The experiment to demonstrate the semi-conservative replication of $DNA$ in eukaryotes was performed by Taylor and his colleagues in $1958$.
They used radioactive thymidine to detect the distribution of newly synthesized $DNA$ in the chromosomes.
This experiment was performed on the root meristem cells of the plant $Vicia$ $faba$ (fava bean).
By using autoradiography, they observed that the radioactive label was distributed in both chromatids of the chromosome, confirming the semi-conservative mode of $DNA$ replication.

(A) The correct answer is $A$ (zero).
$E. coli$ replicates its $DNA$ every $20$ minutes.
Initially,the $E. coli$ cells contain $^{14}N / ^{14}N$ $DNA$.
When these cells are transferred to a medium containing $^{15}N$,the new $DNA$ strands synthesized will incorporate $^{15}N$.
After $20$ minutes (one generation),the $DNA$ molecules will be hybrid $(^{15}N / ^{14}N)$.
After $40$ minutes (two generations),the $DNA$ molecules will be a mix of hybrid $(^{15}N / ^{14}N)$ and heavy $(^{15}N / ^{15}N)$.
Since the medium only contains $^{15}N$,no new $^{14}N / ^{14}N$ $DNA$ can be formed,and the original $^{14}N$ strands will be paired with $^{15}N$ strands.
Therefore,the number of $^{14}N / ^{14}N$ $DNA$ molecules will be zero.

(C) Meselson and Stahl used $CsCl$ (Cesium chloride) density gradient centrifugation to distinguish between $^{15}N$ (heavy) and $^{14}N$ (light) $DNA$ isotopes based on their density differences. The $CsCl$ solution creates a density gradient that allows $DNA$ molecules to settle at positions corresponding to their buoyant density.