Rdna Tools Restriction Ligase

8 MCQs9-step worked example
Source: NCERT BiomoleculesPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The high-frequency trap in this topic: confusing what restriction enzymes, ligase, and polymerase actually do at the molecular level — and specifically, confusing sticky ends with blunt ends when naming specific enzymes.

Restriction enzymes (molecular scissors) are endonucleases that recognise specific palindromic sequences and cut both strands of DNA. Type II restriction enzymes are the workhorses of rDNA technology because they cut at defined sites within or near the recognition sequence. EcoRI recognises GAATTC and cuts between G and A on both strands, producing 5′ AATT overhangs — these are sticky ends (cohesive ends). Not all restriction enzymes produce sticky ends: HaeIII and AluI cut to leave blunt ends (no overhangs).

DNA ligase seals the sugar-phosphate backbone by forming phosphodiester bonds between adjacent nucleotides. In rDNA work, ligase joins the sticky ends of the insert DNA to the complementary sticky ends of the cut vector — this is the "paste" step. Without ligase, the hydrogen bonds between complementary overhangs are too weak to hold permanently.

DNA polymerase synthesises new DNA strands using a template — it "copies." In rDNA cloning, Klenow fragment (a modified DNA polymerase I) is sometimes used to fill in recessed 3′ ends or label probes.

The functional mnemonic: Cut = restriction enzyme. Paste = ligase. Copy = polymerase. The Cohen-Boyer experiment (1973) used EcoRI to cut both foreign DNA and the plasmid vector, then ligase to join them — the first recombinant DNA molecule.

Vectors (vehicles) carry foreign DNA into host cells. Essential features of a cloning vector: origin of replication (ori), selectable marker(s), and a restriction site (cloning site) within a marker gene for insertional inactivation. Common vectors: pBR322 (ampR + tetR), pUC (lacZ′ blue-white screening).

Watch out: EcoRI makes sticky ends, not blunt ends. If a question says "blunt ends," think HaeIII or AluI — never EcoRI.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which enzyme is called the "molecular scissors" in recombinant DNA technology?

MCQ 2Easy RecallPractice

EcoRI cuts the DNA sequence GAATTC to produce:

MCQ 3Easy RecallPractice

The function of DNA ligase in recombinant DNA technology is to:

MCQ 4Direct ApplicationPractice

Which of the following is NOT an essential feature of a cloning vector?

MCQ 5Direct ApplicationPractice

In the plasmid vector pBR322, insertional inactivation of the tetracycline resistance gene by cloning at the BamHI site means that recombinants can be identified because they:

MCQ 6Easy RecallPractice

Which restriction enzyme produces blunt ends?

MCQ 7Direct ApplicationPractice

A restriction enzyme recognises GAATTC on one strand. The sequence on the complementary strand, read 5′→3′, is:

MCQ 8Concept TrapPractice

In the Cohen-Boyer experiment (1973), foreign DNA was inserted into a plasmid vector using EcoRI and then sealed with DNA ligase. Which statement correctly describes why ligase was essential?

Worked Example

Pattern: NEET pattern: unit bundle (recall, similar-term-confusion distractor type; frequency 4 across 2023–2025)

  1. 1

    Given

    - Vector: pBR322 (has ampR and tetR genes; BamHI site within tetR; EcoRI site within ampR). - Both insert and vector cut with EcoRI. - Cells plated on ampicillin AND tetracycline. - All colonies grow on both antibiotics.

  2. 2

    Required

    Determine whether recombinants are present among the colonies.

  3. 3

    Concept

    Insertional inactivation: when foreign DNA inserts at a restriction site within a marker gene, that gene is disrupted. In pBR322, EcoRI cuts within the ampR gene. Therefore, recombinants lose ampicillin resistance.

  4. 4

    Reasoning framework

    If insert disrupts ampR → recombinants are ampR⁻ tetR⁺ (grow on tetracycline only). If colonies grow on BOTH → ampR is intact → no insert → non-recombinants.

  5. 5

    Application

    All colonies grow on both ampicillin and tetracycline → ampR gene intact in all → foreign DNA has NOT inserted at the EcoRI site → these are all non-recombinants (self-ligated vector).

  6. 6

    Conclusion

    No, she has not obtained recombinants. All colonies are non-recombinant (re-circularised vector without insert).

  7. 7

    Final answer

    **No recombinants obtained.** Growth on both antibiotics confirms intact ampR, meaning no insertional inactivation occurred at the EcoRI site within ampR.

  8. 8

    Common trap

    Confusing which marker gene is disrupted by which enzyme in pBR322: EcoRI disrupts ampR; BamHI disrupts tetR. If she had used BamHI, recombinants would grow on ampicillin but NOT tetracycline (trap: trap: rdna tool functions — tool-function confusion extends to which site is in which marker).

  9. 9

    Similar NEET-style question

    "A gene of interest is cloned at the BamHI site of pBR322. On which medium will recombinant colonies grow, and on which will they fail?" Answer: Grow on ampicillin (ampR intact), fail on tetracycline (tetR disrupted by insert at BamHI site). ---

Before solving, remember these

Key Fact

Tools

Restriction enzymes (EcoRI — palindromic site GAATTC, sticky ends). Ligase. Vectors: plasmid (pBR322), bacteriophage (λ), cosmid, BAC, YAC. Selection markers (antibiotic resistance, lacZ blue-white).

-- NCERT Class 12 Biology, Ch. 10, p. 220

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Bt vs RNAi vs herbicide-tolerant GM crop confusion

Category: Similar Terms

Bt: insect resistance (Cry toxin). RNAi: nematode resistance via gene silencing. Herbicide-tolerant: bar/EPSPS gene.

When it triggers

Question on which gene/strategy gives which trait.

How to avoid

Cry → insects. RNAi → nematodes. EPSPS → herbicide tolerance.

Trap

PCR vs blotting role confusion

Category: Similar Terms

PCR amplifies (makes copies); blotting transfers + visualises specific bands. Different stages of analysis.

When it triggers

Question asks which technique amplifies vs detects.

How to avoid

Amplify = PCR. Detect specific = blot (Southern DNA, Northern RNA, Western protein).

Trap

Restriction enzyme vs ligase vs polymerase

Category: Similar Terms

Restriction enzymes CUT at recognition sites; ligase JOINS sticky ends with phosphodiester bonds; polymerase synthesises strands.

When it triggers

Question asks which tool performs which step.

How to avoid

Cut = restriction. Paste = ligase. Copy = polymerase. Cohen-Boyer used EcoRI + ligase for first rDNA.

Past Year Questions

26 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2024

The following diagram shown restriction sites in E. coli cloning vector pBR322. Find the role of ‘X’ and ‘Y’ gens :

1The gene ‘X’ is responsible for resistance to antibiotics and ‘Y’ for protein involved in the replication of Plasmid.
2The gene ‘X’ is responsible for controlling the copy number of the linked DNA and ‘Y’ for protein involved in the replication of Plasmid.
3The gene ‘X’ is for protein involved in replication of Plasmid and ‘Y’ for resistance to antibiotics.
4Gene ’X’ is responsible for recognitions sites and ‘Y’ is responsible for antibiotic resistance
NTA Answer: Option 2(final)
NEET 2022

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Polymerase chain reaction is used in DNA amplification. Reason (R) : The ampicillin resistant gene is used as a selectable marker to check transformation In the light of the above statements, choose the correct answer from the options given below :

1(A) is not correct but (R) is correct
2Both (A) and (R) are correct and (R) is the correct explanation of (A)
3Both (A) and (R) are correct but (R) is not the correct explanation of (A)
4(A) is correct but (R) is not correct
NTA Answer: Option 3(final)
NEET 2022

Given below are two statements: Statement I: Restriction endonucleases recognise specific sequence to cut DNA known as palindromic nucleotide sequence. Statement II: Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site. In the light of the above statements, choose the most appropriate answer from the options given below:

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)
NEET 2020

¬˝ÁÃ’ œŸ ∞ ¡Êß◊Ê ∑ § Áfl ÿ ◊ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞– (4) After zygote formation

1Áø¬Áø¬ Á‚⁄ U «UË.∞Ÿ.∞. ‹Êߪ ¡ mÊ⁄UÊ ¡Ê «∏ ¡Ê ‚∑§Ã „Ò – 112. Identify the wrong statement with regard to Restriction Enzymes.
2¬˝àÿ ∑§ ¬˝ÁÃ’ œŸ ∞ ¡Êß◊ «UË.∞Ÿ.∞. ∑˝§◊ ∑§Ë ‹ê’Ê߸ ∑§Ê (1) Sticky ends can be joined by using DNA ÁŸ⁄UˡÊáÊ ∑§⁄U∑ § ∑§Êÿ¸ ∑§⁄Uà „Ò – ligases.
3ÿ «UË.∞Ÿ.∞. ∑§Ë ‹«∏Ë ∑§Ê ¬ÒÁ‹ã«˛UÊ Á◊∑§ SÕ‹Ê ¬⁄U (2) Each restriction enzyme functions by ∑§Ê≈Uà „Ò – inspecting the length of a DNA sequence. (3) They cut the strand of DNA at palindromic
4ÿ •ÊŸÈfl Á‡Ê∑§ ß ¡ËÁŸÿÁ⁄ Uª ◊ ©¬ÿÊ ªË „Ò – sites.
NTA Answer: Option 1(final)
NEET 2020

The sequence that controls the copy number of the 115. ∞∑§ fl Ä≈U⁄U ◊ ‚„‹ÇŸË «UË.∞Ÿ.∞. ∑§Ë ¬˝Áà ∑§Ë ‚ ÅÿÊ ∑§Ê linked DNA in the vector, is termed : ÁŸÿ ÁòÊà ∑§⁄UŸ flÊ‹ •ŸÈ∑˝§◊ ∑§Ê ÄÿÊ ∑§„Ê ¡ÊÃÊ „Ò?

1Recognition site (1) Á⁄U∑§ÊÚǟˇʟ (¬„øÊŸ) ‚Êß≈U
2Selectable marker
3Ori site (2) øÿŸÿÈÄà ◊Ê∑¸§⁄U
4Palindromic sequence (3) •Ê ⁄UË ‚Êß≈U (4) ¬Ò‹Ë «˛UÊ Á◊∑§ •ŸÈ∑˝§◊ 116. Name the enzyme that facilitates opening of DNA helix during transcription.
NTA Answer: Option 3(final)
NEET 2020

In gel electrophoresis, separated DNA fragments 124. ¡ ‹ ß‹ Ä≈˛UÊ »§Ê ⁄ UÁ‚‚ ◊ , ¬ÎÕ∑§ „È∞ «UË.∞Ÿ.∞. ∑ § πá«UÊ ∑§Ê can be visualized with the help of : Á∑§‚∑§Ë ‚„ÊÿÃÊ ‚ Œ πÊ ¡Ê ‚∑§ÃÊ „Ò?

1Ethidium bromide in infrared radiation (1) •fl⁄UÄà ÁflÁ∑§⁄UáÊ ◊ ∞ÁÕÁ«Uÿ◊ ’˝Ê ◊Êß«U ‚
2Acetocarmine in bright blue light (2) ø◊∑§Ë‹ ŸË‹ ¬˝∑§Ê‡Ê ◊ ∞ ‚Ë≈UÊ ∑§ÊÁ◊¸Ÿ ‚
3Ethidium bromide in UV radiation (3) UV ÁflÁ∑§⁄UáÊ ◊ ∞ÁÕÁ«Uÿ◊ ’˝Ê ◊Êß«U ‚
4Acetocarmine in UV radiation (4) UV ÁflÁ∑§⁄UáÊ ◊ ∞‚Ë≈UÊ ∑§ÊÁ◊¸Ÿ ‚
NTA Answer: Option 3(final)

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