Recombinant DNA technology rests on one foundational principle: a DNA fragment from any organism can be inserted into a self-replicating DNA molecule (a vector) and propagated inside a host cell. This is possible because DNA is chemically universal — the same four bases, the same phosphodiester backbone, the same base-pairing rules operate across all life forms. A restriction enzyme that recognises GAATTC in human DNA will recognise GAATTC in bacterial DNA just as faithfully (NCERT Class 12 Biology Chapter 9, page 165).
The principle has three conceptual pillars that NEET questions target:
1. Cutting and joining are separate, specific events. Restriction endonucleases cut DNA at defined palindromic sequences. DNA ligase seals the sugar-phosphate backbone. These are not interchangeable — confusing their roles is a common NEET distractor strategy.
2. A vector is not just any DNA — it must replicate autonomously. Plasmids, bacteriophages, and cosmids qualify because they carry an origin of replication (ori). A random DNA fragment ligated to another random fragment will not propagate — it needs the ori-containing vector backbone.
3. The recombinant molecule must be introduced into a competent host. Transformation (chemical/heat shock), transfection, and microinjection are delivery methods. The host cell's own replication machinery then copies the recombinant DNA along with the vector.
A frequent NEET approach is to present a list of molecular tools and ask which one performs a specific step. The principle-level understanding tested here is whether you can distinguish the role of each tool within the overall rDNA workflow — not the detailed mechanism of any single tool (those belong to neighbouring topics on restriction enzymes, vectors, and PCR).
Watch out: questions on "principles of rDNA technology" often test whether you understand why the technique works (chemical universality of DNA, palindromic recognition, autonomous replication of vectors) rather than how to perform a specific protocol step.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The basis of recombinant DNA technology is that DNA from different organisms can be combined because:
Answer: D. D is correct. The chemical universality of DNA — the same four nucleotide bases (A, T, G, C) and the same phosphodiester linkage — allows DNA from any source to be cut, joined, and replicated together (NCERT Class 12 Biology Chapter 9, page 167).
Why A is wrong: A is wrong. While organisms do share a common amino acid set, the basis of rDNA technology is DNA-level compatibility (same bases, same backbone), not protein-level similarity.
Why B is wrong: B is wrong. Plasmids are NOT found in all living cells. They occur naturally in bacteria and some yeasts. Eukaryotic animal and plant cells do not natively carry plasmids.
Why C is wrong: C is wrong. Organisms do NOT all share identical restriction enzyme systems. Restriction enzymes are produced by specific bacteria as a defence mechanism; they are tools we borrow, not a universal feature.
Which of the following is a necessary feature of a cloning vector used in recombinant DNA technology?
Answer: B. B is correct. A cloning vector must possess an origin of replication (ori) so it can replicate autonomously inside the host cell. Without ori, the recombinant DNA cannot be propagated (NCERT Class 12 Biology Chapter 9, page 164).
Why A is wrong: A is wrong. An antibiotic resistance gene is useful as a selectable marker but is not the defining requirement — a vector without ori cannot replicate regardless of how many marker genes it carries.
Why C is wrong: C is wrong. A promoter for RNA polymerase is needed for expression vectors (to transcribe the insert), but the fundamental requirement for ANY cloning vector is the ability to replicate, which requires ori.
Why D is wrong: D is wrong. Telomeric sequences are features of linear eukaryotic chromosomes, not a standard requirement of cloning vectors (most common vectors are circular plasmids).
Palindromic nucleotide sequences in DNA are significant in recombinant DNA technology because they are:
Answer: A. A is correct. Restriction endonucleases recognise specific palindromic sequences (e.g., GAATTC for EcoRI) and cut the DNA at or near these sites, generating fragments that can be recombined (NCERT Class 12 Biology Chapter 9, pages 165–166).
Why B is wrong: B is wrong. DNA ligase joins DNA fragments by forming phosphodiester bonds between adjacent nucleotides — it does not specifically require palindromic sequences. It acts on any nick in the backbone where a 3'-OH and 5'-phosphate are adjacent.
Why C is wrong: C is wrong. Origins of replication (ori) are specific sequences that allow autonomous replication — they are not defined by being palindromic.
Why D is wrong: D is wrong. Promoter regions are sequences where RNA polymerase binds to initiate transcription. They have their own consensus sequences (e.g., TATA box) unrelated to restriction-site palindromes.
A researcher wants to insert a human gene into a plasmid vector. Both the human DNA and the plasmid are cut with the same restriction enzyme that produces sticky ends. What enzyme is then required to seal the human gene into the plasmid?
Answer: A. A is correct. After compatible sticky ends from the insert and vector anneal by base pairing, DNA ligase catalyses the formation of phosphodiester bonds to permanently seal the recombinant molecule (NCERT Class 12 Biology Chapter 9, page 167).
Why B is wrong: B is wrong. Reverse transcriptase synthesises cDNA from an RNA template. It is used in cDNA library construction, not in sealing a gene into a vector.
Why C is wrong: C is wrong. DNA polymerase I synthesises new DNA strands by adding nucleotides — it does not seal two pre-existing DNA fragments together.
Why D is wrong: D is wrong. Topoisomerase relieves supercoiling in DNA by cutting and re-joining strands, but it is not the enzyme used to ligate an insert into a vector in standard cloning.
If a plasmid vector lacks an origin of replication (ori) but contains a selectable marker and a cloning site, what will happen when recombinant DNA is introduced into a host bacterium?
Answer: B. B is correct. Without ori, the plasmid cannot initiate autonomous replication inside the host. During successive cell divisions, daughter cells will not inherit copies of the plasmid, and it will be diluted out and lost.
Why A is wrong: A is wrong. The host chromosomal ori directs replication of the host's own chromosome — it does not extend replication capability to a separate, ori-less plasmid molecule.
Why C is wrong: C is wrong. Random integration into the host chromosome is not a reliable or default outcome of transformation. Standard cloning relies on the vector's own ori for propagation, not on chromosomal integration.
Why D is wrong: D is wrong. A selectable marker (e.g., antibiotic resistance) allows identification of cells that have taken up the plasmid — it has no role in DNA replication.
In recombinant DNA technology, the term "competent host cell" refers to a cell that:
Answer: D. D is correct. A competent cell is one that has been chemically treated (e.g., with CaCl₂) or physically treated (e.g., heat shock, electroporation) to make its membrane permeable to exogenous DNA, enabling transformation (NCERT Class 12 Biology Chapter 9, page 169).
Why A is wrong: A is wrong. A competent cell does not already contain the gene of interest — the entire point of transformation is to introduce that gene via the recombinant vector.
Why B is wrong: B is wrong. Competence refers to the ability to take up foreign DNA, not to growth rate. Fast-growing cells are not necessarily competent for transformation.
Why C is wrong: C is wrong. Antibiotic resistance genes are features of the vector (as selectable markers), not a property that defines cell competence.
A student claims: "Any two DNA fragments can be joined using DNA ligase, even if they were cut by different restriction enzymes producing incompatible ends." This claim is:
Answer: C. C is correct. DNA ligase seals nicks by forming phosphodiester bonds, but it requires that the ends are either compatible sticky ends (complementary overhangs that can base-pair) or blunt ends. Two incompatible sticky ends (non-complementary overhangs) cannot be directly ligated because they do not anneal.
Why A is wrong: A is wrong. Ligase cannot join incompatible sticky ends because the overhangs will not base-pair — there is no stable alignment for the enzyme to act on.
Why B is wrong: B is wrong. This is too restrictive. Ligase can join ends from DIFFERENT enzymes if those enzymes produce compatible (complementary) overhangs (e.g., BamHI and BglII both produce 5'-GATC overhangs). It can also join blunt ends. The requirement is end compatibility, not enzyme identity.
Why D is wrong: D is wrong. Ligase does NOT have polymerase activity. It cannot fill gaps. It only seals nicks where a 3'-OH and a 5'-phosphate are already adjacent.
A circular plasmid of 4.0 kb has a single EcoRI site at position 1.0 kb. A linear insert of 2.0 kb with EcoRI-compatible sticky ends is ligated into this site. The resulting recombinant plasmid is then cut with a second enzyme, HindIII, which has a single site at position 3.0 kb on the original plasmid (no HindIII site exists within the insert). How many linear fragment(s) will result from the HindIII digestion?
Answer: C. C is correct. The recombinant plasmid is circular (4.0 kb original + 2.0 kb insert = 6.0 kb total). Cutting a circular molecule at a single restriction site (HindIII at the single site on the plasmid; the insert has no HindIII site) linearises it, producing exactly one linear fragment of 6.0 kb.
Why A is wrong: A is wrong. Three fragments would require three cut sites. With only one HindIII site in the entire recombinant plasmid, only one linear molecule is produced.
Why B is wrong: B is wrong. Two fragments would result if there were two HindIII sites. A circular molecule cut once gives one linear fragment, not two.
Why D is wrong: D is wrong. Four fragments would require four cut sites. This answer likely confuses the number of enzymes or sites mentioned in the problem with the number of fragments produced.
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Given
• Original circular plasmid: 4.2 kb, one EcoRI site, one BamHI site• BamHI site is 1.5 kb clockwise from EcoRI site on original plasmid• Insert: 3.5 kb, no internal EcoRI or BamHI sites• Recombinant plasmid formed by EcoRI ligation
Required
Number and sizes of fragments after double digestion (EcoRI + BamHI).
Concept
A circular DNA molecule cut at n sites produces n linear fragments. The insert disrupts the original EcoRI site region: upon ligation, the insert sits between the two original EcoRI half-sites. The recombinant plasmid is circular with total size = 4.2 + 3.5 = 7.7 kb. It has one EcoRI site on each side of the insert (reconstituted during ligation) and one BamHI site — wait: actually, when a fragment is inserted into a single EcoRI site, the single site is consumed and two EcoRI sites are reconstituted (one at each insert-vector junction). So the recombinant has 2 EcoRI sites + 1 BamHI site = 3 cut sites total.
Formula/Principle
Number of fragments from cutting a circular molecule = number of distinct cut sites.
Substitution
Total cut sites = 2 (EcoRI, at each end of the insert) + 1 (BamHI) = 3.
Recombinant plasmid total = 7.7 kb (circular).
Map positions clockwise on the recombinant circle:• EcoRI site 1 (left junction of insert): position 0• Insert spans 0 to 3.5 kb• EcoRI site 2 (right junction): position 3.5 kb• BamHI site was 1.5 kb clockwise from the original EcoRI on the vector backbone. Since the insert (3.5 kb) was added at the EcoRI site, the BamHI site is now at position 3.5 + 1.5 = 5.0 kb from EcoRI site 1 (clockwise).
Calculation
Three cuts on a 7.7 kb circle produce 3 fragments:• Fragment 1: EcoRI site 1 → EcoRI site 2 = 3.5 kb (this is the insert)• Fragment 2: EcoRI site 2 → BamHI site = 1.5 kb (vector segment)• Fragment 3: BamHI site → EcoRI site 1 = 7.7 − 3.5 − 1.5 = 2.7 kb (remaining vector segment)
Check: 3.5 + 1.5 + 2.7 = 7.7 kb ✓
All values here (4.2, 3.5, 1.5) are problem-defined exact quantities — they do not limit significant figures in this context since this is a mapping problem, not a measurement calculation.
Final answer
3 fragments: 3.5 kb, 2.7 kb, and 1.5 kb.
Common trap
A frequent error is forgetting that inserting a fragment into a single restriction site reconstitutes TWO sites (one at each junction), not one. If you count only one EcoRI site, you get 2 fragments instead of 3 — and the sizes are wrong.
Similar NEET-style question
A 5.0 kb plasmid has single sites for EcoRI and BamHI, separated by 2.0 kb. A 1.8 kb gene with EcoRI-compatible ends is cloned into the EcoRI site. How many fragments result from double digestion with EcoRI + BamHI? (Answer: 3 fragments — 1.8 kb, 2.0 kb, 3.0 kb.)
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Recombinant DNA: isolation of DNA → fragmentation by restriction enzymes → ligation into vector → host transformation → screening → expression. First rDNA: Cohen + Boyer 1972 (E. coli antibiotic resistance gene).
-- NCERT Class 12 Biology, Chapter 9, p. 164More in Biotechnology and its Applications: 7 exam traps and mistakes · 1 question pattern from its other lessons.
3 questions from NEET 2023, 2024, 2026. Answers verified against NTA official keys.
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