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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which enzyme is called the "molecular scissors" in recombinant DNA technology?
Answer: A. Restriction endonucleases cut DNA at specific recognition sequences and are referred to as molecular scissors (NCERT Class 12 Biology Chapter 9, page 165).
Why B is wrong: B is wrong because DNA polymerase synthesises new DNA strands — it copies, not cuts.
Why C is wrong: C is wrong because DNA ligase joins DNA fragments (the 'paste' function), not cuts them.
Why D is wrong: D is wrong because reverse transcriptase synthesises DNA from RNA template — it is used in cDNA library construction, not for cutting DNA.
EcoRI cuts the DNA sequence GAATTC to produce:
Answer: D. EcoRI cuts between G and A on both strands of the palindrome GAATTC, generating 5′ AATT overhangs — these are sticky (cohesive) ends (NCERT Class 12 Biology Chapter 9, pages 166–167).
Why A is wrong: A is wrong because EcoRI leaves 4-nucleotide 5′ overhangs; blunt-end cutters include HaeIII and AluI (trap: mistake: restriction enzyme type — confusing EcoRI with blunt-end enzymes).
Why B is wrong: B is wrong because restriction enzymes produce linear fragments with defined ends, not circular single-stranded DNA.
Why C is wrong: C is wrong because EcoRI produces 5′ overhangs, not 3′ overhangs. Enzymes like PstI produce 3′ overhangs.
The function of DNA ligase in recombinant DNA technology is to:
Answer: B. DNA ligase seals nicks in the sugar-phosphate backbone by catalysing phosphodiester bond formation between the 3′-OH and 5′-phosphate of adjacent nucleotides, joining insert to vector (NCERT Class 12 Biology Chapter 9, page 166).
Why A is wrong: A is wrong because cutting at palindromic sequences is the function of restriction endonucleases, not ligase (trap: rdna tool functions — cut vs paste confusion).
Why C is wrong: C is wrong because strand synthesis from a template is the function of DNA polymerase.
Why D is wrong: D is wrong because unwinding the double helix is the function of helicase, not ligase.
Which of the following is NOT an essential feature of a cloning vector?
Answer: C. A basic cloning vector requires ori (for autonomous replication), a selectable marker (to identify transformants), and a restriction site (for inserting foreign DNA). A eukaryotic promoter is needed in expression vectors, not in all cloning vectors (NCERT Class 12 Biology Chapter 9, page 169).
Why A is wrong: A is wrong because ori is essential — without it the vector cannot replicate inside the host cell.
Why B is wrong: B is wrong because a selectable marker (e.g., antibiotic resistance) is essential to distinguish transformed from non-transformed cells.
Why D is wrong: D is wrong because at least one restriction site within or adjacent to the marker gene is essential for inserting foreign DNA.
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:
Answer: C. Inserting foreign DNA at BamHI disrupts tetR. Recombinants retain ampR (intact) but lose tetR. They grow on ampicillin plates but fail on tetracycline plates (NCERT Class 12 Biology Chapter 9, page 169).
Why A is wrong: A is wrong because growth on both antibiotics indicates an intact vector with no insert — a non-recombinant.
Why B is wrong: B is wrong because losing ampR while retaining tetR would require insertion into ampR, not into tetR at BamHI.
Why D is wrong: D is wrong because the ampR gene is intact and the transformed cell can grow on ampicillin; total failure on both plates indicates no uptake of vector at all.
A student cuts a plasmid vector with EcoRI (a sticky-end cutter) and cuts the foreign DNA to be inserted with HaeIII (a blunt-end cutter), then mixes the two and adds DNA ligase. What is the most likely outcome, and why?
Answer: B. Sticky ends form because their single-stranded overhangs hydrogen-bond with a complementary overhang, and this stickiness is what facilitates ligase's action (NCERT Class 12 Biology Chapter 9, page 167). NCERT further states that "unless one cuts the vector and the source DNA with the same restriction enzyme, the recombinant vector molecule cannot be created" (NCERT Class 12 Biology Chapter 9, page 168). A blunt HaeIII end has no overhang at all, so it cannot base-pair with the EcoRI sticky end on the vector — the two fragments do not anneal, and ligation of this intended recombinant fails.
Why A is wrong: A is wrong because ligase seals ends that are already held together by base-pairing (sticky-sticky) or held in blunt-blunt contact; it does not create pairing between a sticky end and an unrelated blunt end, so mismatched fragments are not reliably joined.
Why C is wrong: C is wrong because restriction enzymes act only on their own specific recognition sequence; nothing about ligating two different fragments recreates an EcoRI or HaeIII site for the enzymes to re-cut.
Why D is wrong: D is wrong because ligase only catalyses phosphodiester bond formation between existing 3′-OH and 5′-phosphate ends — it does not add nucleotides to build a new overhang on a blunt end.
A restriction enzyme recognises GAATTC on one strand. The sequence on the complementary strand, read 5′→3′, is:
Answer: A. Palindromic recognition sequences read the same on both strands in the 5′→3′ direction. The complement of 5′-GAATTC-3′ is 3′-CTTAAG-5′, which when read 5′→3′ is GAATTC — identical to the first strand. This palindromic nature is what allows the enzyme to recognise and cut both strands symmetrically.
Why B is wrong: B is wrong because CTTAAG is the complement read 3′→5′, not 5′→3′. The question asks for the 5′→3′ reading of the complementary strand.
Why C is wrong: C is wrong because AATTCG is neither the complement nor a palindromic reading — it results from misaligning the strands.
Why D is wrong: D is wrong because CTTAGG has no complementarity relationship to GAATTC and is not a valid reading of the opposite strand.
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?
Answer: D. Sticky ends can base-pair via hydrogen bonds alone, but these are temporary. Ligase catalyses formation of covalent phosphodiester bonds between the 3′-OH and 5′-phosphate groups, permanently sealing the backbone — without this step the recombinant molecule falls apart (NCERT Class 12 Biology Chapter 9, page 167).
Why A is wrong: A is wrong because replication inside the host depends on the vector's ori and the host's DNA polymerase machinery, not ligase.
Why B is wrong: B is wrong because hydrogen bonds between complementary overhangs form spontaneously by base-pairing — ligase is not needed for this. Ligase forms the stronger covalent bonds that make the join permanent (trap: rdna tool functions — confusing temporary H-bonding with permanent ligation).
Why C is wrong: C is wrong because restriction enzymes are not 'removed' by ligase; they simply dissociate after cutting. Ligase has no role in enzyme removal.
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Pattern: NEET pattern: unit bundle (recall, similar-term-confusion distractor type; frequency 4 across 2023–2025)
Given
• Vector: pBR322 (has ampR and tetR genes; BamHI and SalI sites within tetR; PstI and PvuI within ampR; the EcoRI site lies OUTSIDE both resistance genes).• Both insert and vector cut with EcoRI.• Cells plated on ampicillin AND tetracycline.• All colonies grow on both antibiotics.
Required
Determine whether recombinants are present among the colonies.
Concept
Insertional inactivation: when foreign DNA inserts at a restriction site within a marker gene, that gene is disrupted. The cloning site has to sit INSIDE a marker gene for this to work. In pBR322 the EcoRI site does not — NCERT’s own worked case is the BamHI site inside tetR, where recombinants lose tetracycline resistance.
Reasoning framework
Insertional inactivation can only report on a marker the insert actually interrupts.
Cloning at EcoRI interrupts neither ampR nor tetR, so recombinants and non-recombinants alike stay ampR⁺ tetR⁺ and both grow on both antibiotics — the screen has no way to tell them apart.
Application
All colonies grow on both ampicillin and tetracycline — which is exactly what BOTH recombinants and non-recombinants would do after cloning at EcoRI. The observation therefore separates nothing.
Conclusion
No, she has not obtained recombinants. All colonies are non-recombinant (re-circularised vector without insert).
Final answer
Nothing can be concluded from this screen. Growth on both antibiotics does not mean there are no recombinants; it means insertional inactivation was never available, because the EcoRI site is not inside either resistance gene. To screen this way the gene must be cloned at BamHI (disrupting tetR) or at PstI (disrupting ampR) — or a different method used, such as blue-white selection.
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: rdna tool functions — tool-function confusion extends to which site is in which marker).
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).
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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, Chapter 9, p. 165These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Restriction enzymes CUT at recognition sites; ligase JOINS sticky ends with phosphodiester bonds; polymerase synthesises strands.
Question asks which tool performs which step.
Cut = restriction. Paste = ligase. Copy = polymerase. Cohen-Boyer used EcoRI + ligase for first rDNA.
Root cause: concept gap
EcoRI: GAATTC palindrome, cuts to leave 5' AATT overhangs (sticky ends). Blunt-ending enzymes: HaeIII, AluI.
More in Biotechnology and its Applications: 5 exam traps and mistakes · 1 question pattern from its other lessons.
14 questions from NEET 2020, 2022, 2023, 2024, 2025, 2026. Answers verified against NTA official keys.
The “Ti plasmid” of Agrobacterium tumefaciens stands for
Which of the following is not a cloning vector?
Which of the following is not a desirable feature of a cloning vector?
Identify the wrong statement with regard to Restriction Enzymes.
The sequence that controls the copy number of the linked DNA in the vector, is termed :
Choose the correct pair from the following :
All 40 past-paper questions from Biotechnology and its Applications →
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