Q.Meselson and Stahl carried out centrifugation in CsCl2 density gradient to separate : (A) DNA from RNA (B) DNA from protein (C) The normal DNA from 15N-DNA (D) DNA from tRNA

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set 57/2/11 markMCQ
   Q.A bacterium having radioactive thymidine in its DNA is allowed to multiply in a medium having non-radioactive thymidine for two generations. What percentage of bacteria will have radioactive thymidine in its DNA ? (A) 100 (B) 50 (C) 75 (D) 25

   ›Reveal solutionSolution

   After two generations of growth in a non-radioactive medium, only 50% of the bacteria retain radioactive thymidine in their DNA — one strand per original double helix remains labelled.

   This question is a classic application of Chargaff’s rule and, more importantly, of semiconservative replication — the mechanism by which DNA copies itself. The key idea, first demonstrated by Meselson and Stahl, is that each new DNA molecule gets one old (parental) strand and one newly synthesised strand. Radioactive thymidine acts as a label on the parental strand; once the bacterium moves to a non-radioactive medium, any new strand built will use only non-radioactive thymidine.

   Let’s trace what happens. You start with one bacterium whose DNA is fully labelled — both strands contain radioactive thymidine. This is your generation zero. When this bacterium divides once (first generation), each daughter cell receives one original radioactive strand and one newly made non-radioactive strand. So after one generation, every bacterium has radioactive DNA — but only one strand per cell is labelled. That gives 100% of bacteria still radioactive.

   Now comes the second generation. Each of those two bacteria divides again. For each, the single radioactive strand acts as a template to build a new complementary strand (non-radioactive). The other, already non-radioactive strand also acts as a template, producing another non-radioactive strand. So from each first-generation cell, you get two daughter cells: one retains the original radioactive strand, the other gets only non-radioactive strands.

   Note

   This is exactly the pattern Meselson and Stahl saw: after two generations in a light medium, half the DNA molecules were hybrid (one heavy, one light) and half were fully light. Here, “heavy” is replaced by “radioactive” and “light” by “non-radioactive.”

   So after two generations, you have four bacteria in total. Two of them carry the original radioactive strand; the other two have no radioactivity at all. That means 50% of the bacteria are radioactive.

   Important

   The percentage refers to the proportion of bacteria that still contain any radioactive thymidine, not the proportion of radioactive DNA strands. Each radioactive bacterium has exactly one labelled strand; the other strand is unlabelled.

   The NCERT textbook explains this under the topic of DNA replication, specifically the semiconservative model. It does not give a numerical example with thymidine, but the logic follows directly from the Meselson-Stahl experiment described in Chapter 6 of Class 12 Biology. The rule is simple: after n generations in non-radioactive medium, the fraction of DNA molecules retaining the original label is 1/2^(n-1) — but here we count bacteria, not molecules. Since each original radioactive strand ends up in a separate bacterium, after two generations, two out of four bacteria carry it.

   ✓Final answer

   In short, after two generations, 50% of the bacteria will have radioactive thymidine in their DNA — option (B) is correct.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which enzyme helps in the separation of both strands of DNA in DNA replication?

   (a) Helicase

   (b) DNA polymerase

   (c) Topoisomerase

   (d) DNA ligase

   ›Reveal solutionSolution

   Helicase breaks the hydrogen bonds between complementary bases, unwinding and separating the two DNA strands so replication machinery can access them.

   During DNA replication, several enzymes act in coordination:

   • Helicase unwinds the double helix by breaking hydrogen bonds between base pairs, creating a replication fork with two separated single strands that serve as templates.
   • DNA polymerase then synthesises the new complementary strand by adding nucleotides in the 5'→3' direction, reading the template.
   • Topoisomerase (DNA gyrase) relieves the torsional/supercoiling strain generated ahead of the replication fork as the helix unwinds.
   • DNA ligase seals nicks/joins Okazaki fragments on the lagging strand, forming a continuous strand.

   Since the question asks specifically about separating the two strands, that is helicase's job.

   ✓Final answer

   (a) Helicase.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.What is the role of enzyme Helicase in DNA Replication?

   (a) Joining of discontinuously synthesised fragments of DNA.

   (b) Unwinding of the double helix of DNA.

   (c) Synthesises RNA primer.

   (d) Formation of H-bond between complementary base pairs.

   ›Reveal solutionSolution

   Helicase opens up (unwinds) the double helix ahead of the replication fork so the two strands can act as templates.

   During DNA replication, the double helix must first be separated into two single strands before DNA polymerase can copy them. Helicase is the enzyme that does this: it moves along the DNA at the origin of replication, breaking the hydrogen bonds between complementary base pairs, and unwinds the helix to create the replication fork.

   The other options belong to different enzymes: joining Okazaki fragments is done by DNA ligase; synthesising the RNA primer is done by primase; hydrogen bonding between base pairs is a physical property of the bases, not an enzymatic action.

   ✓Final answer

   (b) Unwinding of the double helix of DNA.

4. CBSE 2025Set 57/4/11 markMCQ
   Q.The substrate used during DNA replication by the enzyme DNA-dependent DNA polymerase is : (A) Deoxyribonucleotide triphosphate (B) Deoxyribonucleoside triphosphate (C) Ribonucleotide triphosphate (D) Ribonucleoside triphosphate

   ›Reveal solutionSolution

   DNA polymerase adds nucleotides to a growing DNA chain by using deoxyribonucleoside triphosphates (dNTPs) as substrates, which provide both the building block and the energy for phosphodiester bond formation. The answer is (B).

   Why deoxyribonucleoside triphosphates?

   DNA replication requires two things: the correct chemical building blocks that match the template strand, and energy to drive the formation of new bonds. DNA polymerase accomplishes both with a single substrate molecule.

   The key is understanding what "nucleoside" versus "nucleotide" means, and why the triphosphate form matters.

   A nucleoside = sugar + nitrogenous base

   A nucleotide = sugar + nitrogenous base + phosphate group(s)

   In biochemistry textbooks, you'll often see "nucleotide" used loosely to mean any of these forms, but the precise terminology matters here. When we say deoxyribonucleoside triphosphate, we mean:

   • Deoxyribose sugar (the "deoxyribo-" part)
   • A nitrogenous base (A, T, G, or C)
   • Three phosphate groups attached to the 5′ carbon of the sugar

   How DNA polymerase uses dNTPs

   1. The substrate arrives as a triphosphate

      DNA polymerase recognizes and binds deoxyribonucleoside triphosphates: dATP, dTTP, dGTP, and dCTP. The three phosphate groups are designated α, β, and γ (counting outward from the sugar).

   2. Base-pairing determines which dNTP is selected

      The enzyme checks which incoming dNTP correctly pairs with the template strand (A with T, G with C). Only the complementary dNTP is accepted.

   3. The polymerization reaction

      DNA polymerase catalyzes a nucleophilic attack by the 3′-OH group of the growing strand on the α-phosphate of the incoming dNTP. This forms a new phosphodiester bond and releases pyrophosphate (PPi, the β and γ phosphates together).

      DNAn + dNTP → DNAn+1 + PPi (pyrophosphate)

   4. Energy comes from breaking the high-energy bond

      The release of pyrophosphate (which is then hydrolyzed to two inorganic phosphates) makes the reaction thermodynamically favorable. The triphosphate form is essential because it provides the energy currency.

   Watch out

   Option (A) says "deoxyribonucleotide triphosphate." Technically, once the three phosphates are attached, you could call it a nucleotide — but the standard biochemical nomenclature for the substrate is nucleoside triphosphate, not nucleotide triphosphate. The IUPAC convention and all major biochemistry texts (Lehninger, Berg, Voet) use "deoxyribonucleoside triphosphate" or dNTP.

   Why not the other options?

   Option (C) and (D): Ribonucleotide/ribonucleoside triphosphates (rNTPs) contain ribose sugar, not deoxyribose. These are used by RNA polymerase during transcription, not by DNA polymerase during replication. DNA strands have deoxyribose; incorporating ribose would be a replication error.

   Option (A): While "deoxyribonucleotide triphosphate" might seem equivalent, the accepted term in enzymology is deoxyribonucleoside triphosphate. The substrate is named as a nucleoside derivative (base + sugar) with three phosphates attached, not as a nucleotide with additional phosphates.

   ✓Final answer

   The correct option is (B) Deoxyribonucleoside triphosphate.

5. CBSE 2025Set 57/4/11 markMCQ
   Q.The correct depiction of the experiment performed by Matthew Meselson and Franklin Stahl to prove that DNA replicates semi-conservatively on separation of DNA by centrifugation after 40 minutes is : (A) [centrifuge tube diagram] (B) [centrifuge tube diagram] (C) [centrifuge tube diagram] (D) [centrifuge tube diagram]

   ›Reveal solutionSolution

   E. coli replicates once about every 20 minutes, so 40 minutes = TWO generations. After two rounds of semi-conservative replication the tube shows two bands — a hybrid (intermediate-density) band and a light band — so the correct depiction is the diagram with two bands (option A), not a single band.

   Meselson and Stahl grew E. coli for many generations in medium containing heavy nitrogen (¹⁵N) so that all the DNA was 'heavy', then transferred the cells to medium with light nitrogen (¹⁴N). Newly made strands from that point on incorporate only ¹⁴N. The DNA is then separated by density-gradient (CsCl) centrifugation, where heavier molecules settle lower and lighter molecules band higher.

   Why 40 minutes means two generations

   The crucial fact is the generation time: E. coli divides roughly every 20 minutes. So:

   • After 20 minutes (one generation): every molecule is a hybrid (one ¹⁵N strand + one ¹⁴N strand) → a single band at intermediate density.
   • After 40 minutes (two generations): replication of those hybrids produces equal numbers of hybrid molecules and fully light molecules → two bands, one at intermediate density and one at the light position.

   Reading the tube after 40 minutes

   After 40 minutes the gradient shows two distinct bands: an intermediate (hybrid) band in the same position as the 20-minute band, and a light band higher up. This pattern is exactly what semi-conservative replication predicts after two generations, and it rules out conservative replication (which would keep an original heavy band) and dispersive replication (which would give a single, progressively lighter, diffuse band).

   Important

   The single intermediate band belongs to the 20-minute (one-generation) sample. At 40 minutes (two generations) the answer is two bands — intermediate + light. Treating 40 minutes as one generation is the error to avoid.

   ✓Final answer

   After 40 minutes (two generations) the correct centrifuge-tube depiction shows TWO bands — a hybrid/intermediate-density band and a light-density band — i.e. the option with two bands (option A). This confirms semi-conservative replication.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following is a substrate for DNA replication?

   (a) ATP

   (b) DNA polymerase

   (c) Helicase

   (d) Deoxyribonucleoside triphosphate

   ›Reveal solutionSolution

   DNA replication needs the four deoxyribonucleoside triphosphates as raw material - they supply both the nucleotide monomers and, through cleavage of their phosphate bonds, the energy for the polymerisation reaction.

   DNA polymerase catalyses replication by joining deoxyribonucleotides in a chain using the parental strand as template, but the enzyme itself is a catalyst, not a substrate. Helicase unwinds the double helix at the replication fork; it too is an enzyme, not consumed as a building block. ATP is the general energy currency of the cell but is a ribonucleotide, not directly incorporated into DNA. The actual substrates polymerised into the new DNA strand are the deoxyribonucleoside triphosphates (dATP, dGTP, dCTP, dTTP) - each contributes a nucleotide to the growing chain, and the energy released on cleaving off the pyrophosphate (two phosphates) drives the bond formation.

   ✓Final answer

   (d) Deoxyribonucleoside triphosphate.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.In DNA replication, the Okazaki fragments on the lagging strand are joined by

   (a) primase

   (b) DNA polymerase

   (c) helicase

   (d) DNA ligase

   ›Reveal solutionSolution

   DNA ligase acts as the 'molecular glue' that joins the short Okazaki fragments of the lagging strand into one continuous DNA strand.

   On the lagging strand of a replication fork, DNA is synthesised discontinuously in short segments called Okazaki fragments, since DNA polymerase can only extend DNA in the 5'→3' direction. After each fragment's RNA primer is removed and replaced with DNA, the enzyme DNA ligase forms the phosphodiester bond that joins the adjacent fragments into a single, continuous strand. Primase (a) synthesises the RNA primers, DNA polymerase (b) extends the DNA chain, and helicase (c) unwinds the DNA double helix at the replication fork — none of these performs the final joining/sealing step.

   ✓Final answer

   (d) DNA ligase

8. CBSE 2024Set 57/3/11 markMCQ
   Q.In an experiment, E. coli is grown in a medium containing 14NH4Cl. (14N is the light isotope of Nitrogen) followed by growing it for six generations in a medium having heavy isotope of nitrogen (15N). After six generations, their DNA was extracted and subjected to CsCl density gradient centrifugation. Identify the correct density (Light/Hybrid/Heavy) and ratio of the bands of DNA in CsCl density gradient centrifugation. (A) Hybrid : Heavy, 1 : 16 (B) Light : Heavy, 1 : 31 (C) Hybrid : Heavy, 1 : 31 (D) Light : Heavy, 1 : 05

   ›Reveal solutionSolution

   After six generations of semi-conservative DNA replication in a heavy nitrogen medium, the original two light strands will always form hybrid DNA, while all other DNA will be heavy. This results in a Hybrid : Heavy DNA ratio of 1 : 31.

   The core concept here is the semi-conservative nature of DNA replication, famously demonstrated by the Meselson-Stahl experiment. When DNA replicates, each new double helix consists of one original (parental) strand and one newly synthesized strand. This principle dictates how the nitrogen isotopes are distributed in the DNA molecules over successive generations, which in turn affects their density.

   Nitrogen is a key component of DNA bases.

   • 14N is the common, "light" isotope of nitrogen.
   • 15N is a heavier isotope. DNA containing 15N will be denser than DNA containing 14N. DNA with one 14N strand and one 15N strand (a hybrid molecule) will have an intermediate density.

   CsCl density gradient centrifugation separates molecules based on their density. When DNA is spun in a CsCl solution, it forms bands at positions where its density matches the density of the CsCl solution.

   • Light DNA (14N/14N) will form a band at the highest position (least dense).
   • Hybrid DNA (14N/15N) will form a band at an intermediate position.
   • Heavy DNA (15N/15N) will form a band at the lowest position (most dense).

   Let's trace the DNA composition over the generations:

   1. Initial State (Generation 0):

      The E. coli is initially grown in a medium containing 14NH4Cl. This means all the nitrogen incorporated into their DNA is the light isotope, 14N.

      So, all DNA molecules are 14N/14N (Light).

      Let's assume we start with 1 DNA molecule. It has two 14N strands.

   2. First Generation (after 1 replication in 15N medium):

      The E. coli is transferred to a medium containing 15NH4Cl and allowed to replicate once.

      According to semi-conservative replication, the original 14N/14N DNA molecule unwinds. Each 14N strand serves as a template for a new strand synthesized using 15N.

      This results in two DNA molecules, each consisting of one 14N strand and one 15N strand. These are Hybrid DNA molecules.

      • Total DNA molecules: 2^1 = 2
      • Hybrid DNA molecules: 2
      • Heavy DNA molecules: 0

   3. Second Generation (after 2 replications in 15N medium):

      The two hybrid DNA molecules from the first generation replicate again in the 15N medium.

      Each hybrid molecule (14N/15N) unwinds.

      • The 14N strand acts as a template, pairing with a new 15N strand to form a new Hybrid molecule (14N/15N).
      • The 15N strand acts as a template, pairing with a new 15N strand to form a new Heavy molecule (15N/15N). Since there were 2 hybrid molecules, they will produce 2 hybrid and 2 heavy molecules.
      • Total DNA molecules: 2^2 = 4
      • Hybrid DNA molecules: 2
      • Heavy DNA molecules: 2

   4. Generalizing for 'n' Generations:

      Notice a pattern:

      • The two original 14N strands (from the very first DNA molecule) will always serve as templates. Each time they replicate in the 15N medium, they will pair with a newly synthesized 15N strand. This means there will always be exactly 2 Hybrid DNA molecules, regardless of how many generations pass, as long as the replication continues in the 15N medium.
      • The total number of DNA molecules doubles with each generation. After 'n' generations, the total number of DNA molecules will be 2^n.
      • The remaining DNA molecules, which are not hybrid, must be heavy.
      • Number of Heavy DNA molecules = (Total DNA molecules) - (Hybrid DNA molecules) = 2^n - 2.

      After n generations of replication in a heavy isotope medium, starting from one light DNA molecule:

      Number of Hybrid DNA molecules = 2

      Number of Heavy DNA molecules = 2^n - 2

      Total DNA molecules = 2^n

   5. Applying for Six Generations (n = 6):

      We need to find the composition after six generations.

      • Total DNA molecules = 2^6 = 64.
      • Number of Hybrid DNA molecules = 2.
      • Number of Heavy DNA molecules = 2^6 - 2 = 64 - 2 = 62.

   6. Determining the Ratio and Bands:

      The DNA will be separated into two bands: Hybrid and Heavy. There will be no Light DNA band because all original 14N strands are now paired with 15N strands.

      The ratio of Hybrid : Heavy DNA molecules is 2 : 62.

      Simplifying this ratio by dividing both sides by 2, we get 1 : 31.

      Watch out

      A common mistake is to forget that the original two 14N strands persist and always form hybrid molecules. Some might incorrectly assume that after many generations, all DNA becomes heavy, which is not true due to semi-conservative replication.

   The bands observed will be Hybrid and Heavy, in a ratio of 1 : 31.

   ✓Final answer

   The correct density bands are Hybrid and Heavy, with a ratio of (C) Hybrid : Heavy, 1 : 31.

9. CBSE 2024Set E1 markMCQ
   Q.Which of the following statements is incorrect about DNA replication?

   (a) DNA replication is semi-conservative

   (b) Main enzyme for DNA replication is DNA polymerase

   (c) Mutation appears due to error in replication

   (d) Replication on both strands of DNA is continuous

   ›Reveal solutionSolution

   DNA replication is discontinuous on the lagging strand, so the statement that both strands replicate continuously is wrong.

   DNA replication is semi-conservative (each new molecule has one old and one new strand), the key enzyme is DNA polymerase, and errors during replication can lead to mutations — so statements (a), (b) and (c) are correct. However, because DNA polymerase synthesises new DNA only in the 5'→3' direction, only the leading strand is made continuously; the lagging strand is synthesised discontinuously in short Okazaki fragments that are later joined by DNA ligase. Therefore replication is NOT continuous on both strands.

   ✓Final answer

   (d) Replication on both strands of DNA is continuous — this statement is incorrect.

10. CBSE 2023Set 57/3/11 markMCQ
    Q.Given below is a list of steps Meselson and Stahl carried out in their experiment to prove that DNA replication is semi-conservative. Select the option that gives the correct sequence of steps followed by them.

    (i) Bacteria transferred to a N14 medium and sampled every 20 minutes.

    (ii) All bacteria contain hybrid DNA (N14 DNA and N15 DNA).

    (iii) Bacteria grown in N15 medium for many generations.

    (iv) All bacteria contain N15 DNA.

    (v) Bacteria contain either all N14 DNA or all hybrid DNA.

    (a)

    (ii) →

    (iv) →

    (iii) →

    (i) →

    (v)

    (b)

    (i) →

    (ii) →

    (v) →

    (iv) →

    (iii)

    (c)

    (iii) →

    (iv) →

    (i) →

    (ii) →

    (v)

    (d)

    (iv) →

    (iii) →

    (ii) →

    (v) → (i)

    ›Reveal solutionSolution

    Meselson and Stahl's experiment proved DNA replication is semi-conservative by tracking heavy nitrogen (15N) labeled DNA through successive generations in a light nitrogen (14N) medium.

    The discovery of DNA's double helix structure by Watson and Crick in 1953 immediately raised a crucial question: how does this molecule replicate itself to pass genetic information accurately from one generation to the next? The structure itself, with its complementary base pairing (adenine always pairing with thymine, and guanine with cytosine), offered a strong hint. If the two strands of the double helix could separate, each strand could then serve as a template for the synthesis of a new complementary strand. This idea formed the basis of what is known as semi-conservative replication.

    However, before Meselson and Stahl's groundbreaking work, there were three main hypotheses for how DNA might replicate:

    • Conservative Replication: In this model, the original double helix would remain intact after replication, and a completely new double helix would be synthesized from scratch. So, after one round, you'd have one old DNA molecule and one new DNA molecule.
    • Semi-conservative Replication: Here, each new DNA molecule would consist of one original (parental) strand and one newly synthesized strand. This is the model suggested by Watson and Crick.
    • Dispersive Replication: This model proposed that the parental DNA molecule would break into fragments, and new DNA would be synthesized in between these fragments. The resulting DNA molecules would be a mosaic of old and new DNA on both strands.

    To definitively distinguish between these possibilities, Matthew Meselson and Franklin Stahl designed an elegant experiment in 1958 using Escherichia coli bacteria and isotopes of nitrogen. Their method relied on the fact that DNA contains nitrogen, and different isotopes of nitrogen have different atomic masses, which would affect the density of the DNA molecule.

    Note

    Isotopes are atoms of the same element that have different numbers of neutrons, and thus different atomic masses. 15N (heavy nitrogen) has one more neutron than 14N (light nitrogen), making DNA containing 15N denser than DNA containing 14N.

    Here's how Meselson and Stahl carried out their experiment, following a precise sequence of steps:

    • Step 1: Labeling the parental DNA with heavy nitrogen.

      • (iii) Bacteria grown in N15 medium for many generations. They first grew E. coli in a culture medium where the only nitrogen source was 15NH4Cl (ammonium chloride containing heavy nitrogen). They allowed the bacteria to divide for many generations (multiple cell cycles).
      • (iv) All bacteria contain N15 DNA. After many generations in the 15N medium, virtually all the nitrogenous bases in the bacterial DNA incorporated 15N. This made the DNA molecules significantly denser than normal DNA. They confirmed this by centrifuging the DNA in a cesium chloride (CsCl) density gradient, where the heavy DNA settled at a lower position in the tube.

    • Step 2: Transferring to light nitrogen medium and observing the first generation.

      • (i) Bacteria transferred to a N14 medium and sampled every 20 minutes. They then transferred these 15N-labeled bacteria to a fresh culture medium containing only 14NH4Cl (light nitrogen). E. coli divides approximately every 20 minutes, so they took samples at this interval.
      • (ii) All bacteria contain hybrid DNA (N14 DNA and N15 DNA). After one generation (20 minutes), they extracted the DNA from the bacteria. When this DNA was centrifuged in a CsCl gradient, it formed a single band at an intermediate density, precisely between the positions of pure 15N DNA and pure 14N DNA. This "hybrid" band indicated that each DNA molecule contained both heavy (15N) and light (14N) nitrogen. This result immediately ruled out the conservative replication model, which would have predicted two distinct bands: one heavy and one light.

    • Step 3: Observing the second generation.

      • (v) Bacteria contain either all N14 DNA or all hybrid DNA. They allowed the bacteria to replicate for a second generation (another 20 minutes, totaling 40 minutes from the transfer to 14N medium). When DNA from this second generation was analyzed, it showed two distinct bands in the CsCl gradient: one band at the intermediate (hybrid) density and another band at the lighter (14N) density. The two bands were present in approximately equal proportions.

    Important

    The presence of two bands in the second generation (one hybrid and one light) was the definitive evidence. If replication were dispersive, all DNA molecules would still be a mosaic of 15N and 14N, resulting in a single band of intermediate density, but progressively lighter with each generation. The observation of distinct hybrid and light bands confirmed semi-conservative replication.

    The results perfectly aligned with the semi-conservative model:

    • In the first generation, each original 15N strand served as a template for a new 14N strand, resulting in all DNA molecules being hybrids (15N-14N).
    • In the second generation, these hybrid molecules separated. The 15N strands again templated new 14N strands (forming more hybrids), while the 14N strands templated new 14N strands (forming entirely light DNA molecules). This led to 50% hybrid DNA and 50% light DNA.

    Based on this detailed understanding, the correct sequence of steps followed by Meselson and Stahl is:

    (iii) → (iv) → (i) → (ii) → (v)

    This matches option (c).

    ✓Final answer

    In short, Meselson and Stahl first labeled bacterial DNA with heavy nitrogen, then transferred it to a light nitrogen medium, observing the formation of hybrid DNA in the first generation and both hybrid and light DNA in the second, thereby proving semi-conservative replication. The correct sequence of steps is (c) (iii) → (iv) → (i) → (ii) → (v).

11. CBSE 2023Set ANNUAL1 markMCQ
    Q.In which phase of the cell division does DNA replication take place?

    (a) G1 phase

    (b) S-phase

    (c) G2 phase

    (d) G0 phase

    ›Reveal solutionSolution

    The cell cycle's interphase has three parts (G1, S, G2); DNA is duplicated only in the S (Synthesis) phase, which is why it is named that.

    The cell cycle is divided into interphase and M-phase (division). Interphase itself has three sub-phases:

    • G1 (Gap 1): cell grows, prepares enzymes/materials needed for DNA synthesis.
    • S (Synthesis): the DNA content of the cell doubles — each chromosome, made of one DNA molecule (one chromatid), replicates to become two sister chromatids joined at the centromere.
    • G2 (Gap 2): cell continues to grow and prepares for mitosis.
    • G0: a resting/quiescent stage some cells exit into, with no active division.

    Since replication (doubling of DNA) is the defining event of the S-phase, this is when it takes place.

    ✓Final answer

    (b) S-phase.

12. CBSE 2023Set ANNUAL1 mark
    Q.What are Okazaki fragments?

    ›Reveal solutionSolution

    Because DNA polymerase can only synthesise DNA in the 5'→3' direction, the lagging strand is made in short, discontinuous stretches called Okazaki fragments, which are later joined together.

    During DNA replication, the two parental strands are antiparallel, but DNA polymerase can add new nucleotides only in the 5'→3' direction. On one template strand (the leading strand), synthesis of the new strand can proceed continuously in the same direction as the replication fork moves. On the other template strand (the lagging strand), synthesis must occur discontinuously, in short stretches, each begun with its own RNA primer, moving away from the replication fork; these short DNA segments are called Okazaki fragments. They are subsequently joined into a continuous strand by the enzyme DNA ligase, after the RNA primers are removed and replaced with DNA.

    ✓Final answer

    Okazaki fragments are the short, discontinuous DNA segments synthesised on the lagging strand during replication; they are later joined together by DNA ligase to form a continuous strand.