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 10, page 216).
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:
Which of the following is a necessary feature of a cloning vector used in recombinant DNA technology?
Palindromic nucleotide sequences in DNA are significant in recombinant DNA technology because they are:
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?
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?
In recombinant DNA technology, the term "competent host cell" refers to a cell that:
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:
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?
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.) ---
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, Ch. 10, p. 216These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Bt: insect resistance (Cry toxin). RNAi: nematode resistance via gene silencing. Herbicide-tolerant: bar/EPSPS gene.
Question on which gene/strategy gives which trait.
Cry → insects. RNAi → nematodes. EPSPS → herbicide tolerance.
Category: Similar Terms
PCR amplifies (makes copies); blotting transfers + visualises specific bands. Different stages of analysis.
Question asks which technique amplifies vs detects.
Amplify = PCR. Detect specific = blot (Southern DNA, Northern RNA, Western protein).
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
Eli Lilly's Humulin: A and B chains expressed SEPARATELY in two E. coli strains; chains then combined in vitro and joined by disulfide bonds.
Root cause: concept gap
PCR requires TWO primers — one for each end of target DNA, anneal to opposite strands, point inward. Together they bracket the amplicon.
Root cause: concept gap
EcoRI: GAATTC palindrome, cuts to leave 5' AATT overhangs (sticky ends). Blunt-ending enzymes: HaeIII, AluI.
Root cause: term confusion
Bt cotton expresses Cry protein INSIDE its tissues — endogenous; insects ingesting plant tissue are killed. Not externally sprayed.
26 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
Why can’t insulin be given orally to diabetic patients?
Silencing of specific mRNA is possible via RNAi because of
Identify the part of a bio-reactor which is used as a foam braker from the given figure.
Polymerase chain reaction (PCR) amplifies DNA following the equation.
The “Ti plasmid” of Agrobacterium tumefaciens stands for
Upon exposure to UV radiation, DNA stained with ethidium bromide will show
Which of the following is not a cloning vector?
Which one of the following statement is not true regarding gel electrophoresis technique?
Transposons can be used during which one of the following ?
¬˝ÁÃ’ œŸ ∞ ¡Êß◊Ê ∑ § Áfl ÿ ◊ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞– (4) After zygote formation
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
similar term confusion
Biology relies on precise terminology; close terms tempt selection.
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