Biotechnology Class 12 NCERT Exemplar Solutions Chapter 9

Correct option: (b) Production of CO2.

Concept used. Bread dough is fermented by the budding yeast Saccharomyces cerevisiae, also known as baker's yeast. Under anaerobic conditions inside the kneaded dough, yeast metabolises the sugars released from wheat starch through ethanolic fermentation: C6H12O6 ->[yeast] 2 C2H5OH + 2 CO2. The CO2 gas produced gets trapped inside the gluten network of the dough, inflating it like a balloon. This is what makes the dough rise.

1. Yeast cells take up glucose (and maltose released by amylases acting on starch). Under low oxygen, glucose is broken down to pyruvate by glycolysis, then to ethanol + CO2.
2. Each glucose molecule yields 2 molecules of CO2. The gas cannot escape the elastic gluten matrix, so it forms millions of tiny bubbles.
3. These bubbles expand on warming, almost doubling the volume of the dough. The ethanol mostly evaporates during baking.

Why the other options are wrong. (a) Yeast does multiply in the dough, but cell division alone adds negligible volume — the inflation is due to gas release. (c) Emulsification is the dispersion of one liquid in another and is irrelevant here. (d) Starch hydrolysis only releases sugars; it doesn't inflate the dough by itself.

Option (b): Production of CO2.

Correct option: (b) exonuclease.

Concept used. Nucleases are enzymes that hydrolyse the phosphodiester bond of a nucleic-acid strand. They fall in two camps:

• 0pt
• Exonucleases: act only at the ends of a DNA strand and snip nucleotides off one at a time (5' or 3' end, depending on the enzyme).
• Endonucleases: cut within the DNA, often at a specific sequence (these include restriction enzymes like Hind II).

1. Match the verb in the question — ``removal from the ends'' — to the definition of an exonuclease. By definition only an exonuclease can act from an end.
2. Eliminate (a): an endonuclease cuts internally, not at the ends.
3. Eliminate (c): DNA ligase joins two DNA pieces by forming a phosphodiester bond. It is the opposite reaction.
4. Eliminate (d): Hind II is a restriction endonuclease; it makes an internal cut at a specific 6 bp palindrome.

Option (b): exonuclease.

Correct option: (a) Transduction.

Concept used. Bacteria exchange DNA by three natural mechanisms:

• 0pt
• Transformation — uptake of free, naked DNA from the surroundings (discovered by Griffith with Streptococcus pneumoniae).
• Conjugation — direct cell-to-cell transfer of DNA through a sex pilus (Lederberg & Tatum's F⁺ × F^- cross).
• Transduction — transfer of DNA from one bacterium to another packaged inside a bacteriophage (a viral vector). Discovered by Zinder and Lederberg in Salmonella.

1. Read the discriminator: ``mediation of a viral vector''. Only one of the three natural transfer mechanisms uses a virus, namely transduction.
2. Eliminate (b) conjugation: needs physical pilus contact, not a virus.
3. Eliminate (c) transformation: needs free DNA, not a virus.
4. Eliminate (d) translation: that is the ribosome-mediated synthesis of protein from mRNA, not a DNA-transfer process at all.

Option (a): Transduction.

Correct option: (d) Ethidium bromide stained DNA can be seen under exposure to UV light.

Concept used. DNA is colourless and its absorption peak is at 260 nm (deep UV) so it is invisible to the naked eye in white light. To see DNA bands on an agarose gel we stain with ethidium bromide (EtBr): a planar aromatic molecule that intercalates between adjacent base pairs of the DNA double helix. Once intercalated, EtBr absorbs UV (∼ 300 nm) and re-emits as bright orange visible light at ∼ 605 nm (fluorescence). On a UV transilluminator the DNA bands therefore glow orange against a dark background.

1. State the two facts the answer depends on:
   (i) DNA + visible light = invisible.
   (ii) DNA + EtBr + UV light = orange fluorescence.
2. Test each option against these facts. Options (a), (b) and (c) all claim that DNA can be seen in visible light, which contradicts fact (i).
3. Only option (d) combines the stain (needed for any visualisation) with the right light source (UV).

Option (d): Ethidium bromide stained DNA can be seen under exposure to UV light.

Correct option: (c) Prevention of the multiplication of bacteriophage by the host bacteria.

Concept used. The word restriction is historical, not mechanistic. In the 1950s, Salvador Luria observed that phages grown on one E. coli strain could not infect another strain — their growth was restricted. The molecular cause, discovered by Werner Arber, Hamilton Smith and Daniel Nathans (Nobel 1978), is a bacterial defence system: enzymes that recognise and chop foreign (phage) DNA at specific palindromic sequences while sparing the host's own DNA, which is protected by methylation. So the cut-action is the mechanism, but restriction is what the bacterium achieves by it — restricting phage multiplication.

1. Distinguish action from purpose. Phosphodiester cleavage (a) and sequence-specific cutting (b) are the action of the enzyme. ``Restriction'' refers to the purpose it serves the bacterium.
2. That purpose is to chop incoming bacteriophage DNA into harmless pieces before it can hijack the host. The host therefore restricts (prevents) phage multiplication.
3. Option (d) is wrong because it lumps together the mechanism and the purpose under one label. Only (c) captures the historical meaning.

Option (c): Prevention of phage multiplication by the host bacteria.

Correct option: (d) E. coli.

Concept used. A recombinant DNA molecule is built in vitro by combining a fragment of foreign DNA with a vector (usually a plasmid). The minimum kit on the bench is:

• 0pt
• A restriction endonuclease to cut both the vector and the foreign DNA at the same recognition sequence, producing compatible sticky ends.
• A DNA ligase to seal the two phosphodiester gaps that hold the fragments together.
• The DNA fragments themselves — the gene of interest and the linearised vector.

E. coli is the host cell used to propagate (multiply) the recombinant molecule after it has been assembled. It is not needed during the in-vitro ligation step.

1. Separate the two stages of rDNA work — molecule construction (in vitro) versus molecule propagation (inside a host).
2. The question asks what is not required for construction. Map each option: (a), (b), (c) all sit on the bench, (d) sits in the incubator.
3. So E. coli is the odd one out.

Option (d): E. coli.

Correct option: (b) Size only.

Concept used. In agarose gel electrophoresis, all DNA molecules carry essentially the same charge per unit length — every nucleotide contributes one negatively charged phosphate group. So the electric force on every fragment per unit length is the same. What varies between fragments is how easily they can squeeze through the porous agarose matrix: smaller fragments wriggle through quickly, larger ones get held back. The result is that fragments migrate at a rate that depends almost entirely on their size.

1. Confirm uniform charge density: DNA backbone = alternating sugar and phosphate; one phosphate per nucleotide ⇒ uniform charge-to-mass ratio. So charge alone cannot discriminate.
2. Confirm size dependence: the agarose matrix acts as a sieve. Migration distance ∝ ₁₀(1/size in bp) over the linear range.
3. Compare options: (a) wrong (same charge density), (c) wrong (the ratio is roughly constant for DNA), (d) wrong (it includes a, c), (b) right.

Option (b): Size only.

Correct option: (a) Origin of replication (ori).

Concept used. A cloning vector must possess four features:

• 0pt
• Origin of replication (ori) — the DNA sequence where host DNA polymerase initiates replication. Any piece of DNA linked to an ori will be replicated by the host machinery.
• Selectable marker — usually an antibiotic-resistance gene, to identify host cells that have taken up the vector.
• Cloning sites — unique restriction sites where the foreign DNA can be inserted.
• Small size — easier to manipulate in vitro.

All four matter, but without an ori, the vector cannot replicate, so the foreign gene cannot be multiplied — defeating the whole purpose of cloning. Hence ori is the most important.

1. Define the purpose of cloning: make many identical copies of a gene inside a host cell. ``Many copies'' requires replication.
2. Without an ori, the vector is a dead piece of DNA inside the cell — it will be diluted out as the cell divides.
3. Selectable marker, cloning sites and small size are practical conveniences; they don't determine replicability.

Option (a): Origin of replication (ori).

Correct option: (c) Deoxyribonuclease.

Concept used. The goal of DNA isolation is to liberate intact DNA from the cell while removing every other macromolecule. The roles of the four enzymes:

• 0pt
• Lysozyme — digests peptidoglycan in the bacterial cell wall, cracking the cell open.
• Ribonuclease (RNase) — chops up RNA, which would otherwise co-purify with DNA.
• Protease — digests cellular proteins, including DNA-binding histone-like proteins and contaminating enzymes.
• Deoxyribonuclease (DNase) — chops up DNA. Adding it would destroy the very molecule we are trying to isolate!

1. Re-state the aim: keep DNA whole, remove everything else.
2. Match each enzyme to a class of molecules to remove (cell wall, proteins, RNA) — three are useful.
3. DNase removes DNA, which is the exact opposite of what we want. Hence it is the one enzyme that must never be added.

Option (c): Deoxyribonuclease.

Correct option: (d) Availability of `Thermostable' DNA polymerase.

Concept used. PCR cycles repeatedly between three temperatures: denaturation at ∼ 95 ^∘C (to separate the strands), annealing at ∼ 55 ^∘C (primers bind), extension at ∼ 72 ^∘C (polymerase synthesises). The denaturation step destroys ordinary DNA polymerases (e.g. E. coli polymerase I). Mullis's original PCR required adding fresh polymerase after every cycle — slow, expensive, labour-intensive. The breakthrough was Taq polymerase, isolated from the hot-spring bacterium Thermus aquaticus, which survives the ∼ 95 ^∘C denaturation step. One dose of Taq lasts the whole 30-cycle reaction.

1. Identify the bottleneck before Taq: the polymerase had to be refreshed every cycle, making PCR impractical.
2. Identify the fix: a heat-stable polymerase that survives 95 ^∘C. Taq, with optimum ∼ 75 ^∘C, fits.
3. Eliminate (a)–(c): template DNA, primers and dNTPs were already available before PCR became routine. They were enabling, but not popularising.

Option (d): Availability of `Thermostable' DNA polymerase.

Correct option: (c) Recombinant bacterial cells.

Concept used. In modern rDNA technology vectors carry two antibiotic-resistance genes flanking the cloning site (e.g. pBR322 carries amp^R and tet^R). When the foreign DNA is inserted into one of these genes (say tet^R) it is inactivated by insertion — the gene no longer works. So:

• 0pt
• Cells that picked up the recombinant plasmid are amp^R but tet-sensitive.
• Cells that picked up the empty self-ligated plasmid are still amp^R tet^R.

The differential antibiotic response is what lets us select the recombinants from the non-recombinants. (Mere transformation is selected on the first antibiotic alone.)

1. Recall the standard pBR322 selection: plate on ampicillin first to get all transformants, then replica-plate onto tetracycline. Colonies that grow on amp but die on tet are the recombinants.
2. Note that the question says ``usually helps in the selection of'' — this matches the recombinant-selection use case.
3. Eliminate (a) competent cells: those are made chemically, not selected by antibiotic. Eliminate (b) transformed cells: that takes one antibiotic, but the differential value of a second marker is recombinant selection.

Option (c): Recombinant bacterial cells.

Correct option: (c) Uptake of DNA through transient pores in the bacterial cell wall.

Concept used. Bacterial cells do not normally take up DNA. To make them competent, we incubate them on ice with divalent Ca^2+ ions, which neutralise the negative charges on both the cell wall and the DNA so they no longer repel each other and the DNA sticks to the cell surface. A brief heat shock (42 ^∘C for ∼ 90 s, then back to ice) creates transient pores in the wall through which the surface-bound DNA slips inside. Without the heat shock the DNA sits on the outside; without Ca^2+ it never gets close enough to slip in.

1. Recall the three-step competent-cell protocol: (i) ice-cold CaCl2 treatment, (ii) DNA addition + further ice incubation, (iii) 42 ^∘C heat shock for 90 s, (iv) recovery in growth medium.
2. The heat shock physically perturbs the cell envelope, opening transient pores. The temperature gradient (ice → warm → ice) is what drives the perturbation.
3. Eliminate (a): binding to wall is the role of Ca^2+, not heat. Eliminate (b): bacteria have no DNA-import membrane proteins. Eliminate (d): heat shock doesn't switch on antibiotic genes.

Option (c): Uptake of DNA through transient pores.

Correct option: (a) Formation of phosphodiester bond between two DNA fragments.

Concept used. A phosphodiester bond is the covalent linkage between the 3'-OH of one nucleotide and the 5'-phosphate of the next. The sugar-phosphate backbone of a DNA strand is held together by these bonds. When a restriction enzyme cuts DNA, it breaks exactly these bonds — leaving free 3'-OH and 5'-phosphate ends. DNA ligase re-seals these breaks in an ATP-dependent reaction. After two compatible sticky ends base-pair (via hydrogen bonds, which happen spontaneously), the ligase covalently joins them with two new phosphodiester bonds (one per strand).

1. Recall the chemistry of the cut: restriction enzyme breaks the phosphodiester bond between the sugar and the next phosphate.
2. Recall the chemistry of the repair: ligase consumes one ATP per bond, and forms a new phosphodiester bond between 3'-OH and 5'-PO₄.
3. Distinguish from hydrogen-bonding: the H-bonds between sticky ends form spontaneously without ligase. Ligase's unique job is the covalent step.
4. Eliminate (b): H-bonding is base-pairing, not ligation. Eliminate (c): ligase does not act on bases. Eliminate (d): (a) is correct.

Option (a): Formation of phosphodiester bond between two DNA fragments.

Correct option: (c) Entamoeba coli.

Concept used. Restriction enzymes are named after the bacterium they come from: the first letter of the genus, the next two letters of the species, plus a strain letter and a Roman numeral (e.g. EcoR I from Escherichia coli RY13). Crucially, restriction enzymes occur in bacteria as a defence against phage DNA. Entamoeba coli is not a bacterium; it is a protozoan (a single-celled eukaryote) that lives in the human gut. So it cannot be a source of restriction enzymes.

1. Recall the source organisms of common restriction enzymes: Hind III ← Haemophilus influenzae strain Rd; EcoR I ← Escherichia coli RY13; BamH I ← Bacillus amyloliquefaciens H.
2. Test Entamoeba coli: this is a protozoan (not to be confused with the pathogen Entamoeba histolytica). It lacks the restriction-modification defence system that bacteria evolved.
3. So three of four options are bacterial; Entamoeba coli is the misfit.

Option (c): Entamoeba coli.

Correct option: (c) Extension of primer end on the template DNA.

Concept used. PCR cycles between three temperatures, and each temperature does one job; not all three jobs need an enzyme.

• 0pt
• Denaturation (95 ^∘C) — heat alone breaks the H-bonds between the two DNA strands. No enzyme needed.
• Annealing (50–60 ^∘C) — primers find their complementary sequences and base-pair spontaneously. No enzyme needed.
• Extension (72 ^∘C) — Taq polymerase adds dNTPs to the 3'-OH of each annealed primer, copying the template. This is the only enzyme-catalysed step.

1. Pair each PCR step with its driver: denaturation → heat; annealing → complementary base-pairing; extension → Taq polymerase.
2. Eliminate (a) and (b): heat and H-bonding don't need a polymerase.
3. Confirm (c): Taq adds nucleotides at 72 ^∘C (its optimum), so extension is its job.

Option (c): Extension of primer end on the template DNA.

Correct option: (d) All of the above.

Concept used. Expressing a human gene in a bacterial host runs into several mismatch issues:

• 0pt
• Introns — human genes have introns that must be spliced out by the eukaryotic spliceosome. Bacteria lack splicing machinery, so they cannot make a functional mRNA from a raw human genomic gene.
• Codon usage bias — although the genetic code is universal, humans and bacteria prefer different synonymous codons. A human gene loaded with rare-in-E. coli codons translates poorly.
• Proteolysis — even if a human protein is produced, the bacterial proteases may recognise it as foreign and chop it up.

1. Walk a human gene through bacterial expression: transcription → no splicing → mRNA still has introns → non-functional protein. Reason (a) confirmed.
2. If you supply a cDNA (intron-free): transcription → translation, but rare codons stall ribosomes → poor yield. Reason (b) confirmed.
3. If translation succeeds: bacterial proteases may degrade the foreign protein. Reason (c) confirmed.
4. All three reasons are valid contributors. Hence (d) is the right answer.

Option (d): All of the above.

Correct option: (c) A continuous culture system.

Concept used. For industrial-scale protein production, three reactor geometries are possible:

• 0pt
• Batch culture (a closed flask, even a giant one) — nutrients run out and waste builds up; growth stops.
• Stirred-tank batch bioreactor — same problem at industrial scale; one batch, then clean and restart.
• Continuous culture system — fresh medium flows in, used medium and product flow out, so the cells stay at log phase (exponentially growing) for days or weeks.

A continuous system keeps the bioreactor permanently in the optimal growth phase, giving the highest steady-state yield of recombinant protein.

1. Compare yield over time. A batch flask peaks once, then dies — total protein per litre is limited by initial nutrient pool.
2. A continuous system keeps replenishing nutrients and removing waste, so it produces protein continuously at the cells' maximum rate.
3. Eliminate (a): flask size cannot fix the batch-culture problem. Eliminate (b): a closed stirred tank is still a batch reactor. Eliminate (d): not ``any of the above'' — (c) is strictly better.

Option (c): A continuous culture system.

Correct option: (c) Kary Mullis.

Concept used. Kary B. Mullis invented PCR in 1983 while working at Cetus Corporation, and was awarded the 1993 Nobel Prize in Chemistry (shared with Michael Smith). The other three scientists are famous, but for different work:

• 0pt
• Herbert Boyer — co-developed the first recombinant DNA molecule (with Stanley Cohen, 1973); founded Genentech.
• Har Gobind Khorana — Nobel 1968 for cracking the genetic code (translation, not PCR).
• Arthur Kornberg — Nobel 1959 for discovering DNA polymerase I (a different polymerase, not the PCR technique).

1. Match each name to its primary contribution. Only Mullis matches the PCR keyword.
2. Eliminate the distractors by their actual Nobel work.

Option (c): Kary Mullis.

Correct option: (c) It is isolated from viruses.

Concept used. Restriction enzymes are produced by bacteria as a defence against viruses (bacteriophages) — they are not isolated from viruses. The classic source organisms are Haemophilus influenzae (Hind III), Escherichia coli (EcoR I) and Bacillus amyloliquefaciens (BamH I).

The other three statements are true:

• 0pt
• (a) Type-II restriction enzymes recognise short palindromic sequences (e.g. EcoRI recognises 5'-GAATTC-3' which reads the same on the complementary strand).
• (b) They cut within DNA (between bases of a recognition site), so they are endonucleases.
• (d) Cutting two different DNAs with the same enzyme yields identical sticky ends, which is what makes them ligatable to each other — the very basis of rDNA technology.

1. Spot the falsehood. Restriction enzymes come from bacteria, not viruses. They evolved to restrict (prevent) viral infection.
2. Verify each true statement quickly: palindrome (yes), endonuclease (yes), uniform sticky ends across substrates (yes — the whole point of cloning).
3. So (c) is the statement that does not hold.

Option (c): It is isolated from viruses.

Concept used. The copy number of a plasmid is the average number of plasmid molecules per host cell. Recombinant protein yield is roughly proportional to the number of mRNA transcripts the cell can make from the gene, which in turn depends on how many copies of the gene template are present. So more plasmid copies → more mRNA → more protein.

1. State the chain: copy number ∝ gene-template count ∝ mRNA produced ∝ protein yield (assuming translation is not the bottleneck).
2. For pUC-type high-copy plasmids (∼ 500 copies per cell) yield is far higher than for low-copy BAC vectors (∼ 1 copy).

Yield is directly (approximately linearly) proportional to plasmid copy number.

Concept used. An exonuclease chews nucleotides off the ends of a DNA strand. During rDNA construction we deliberately create ends — the sticky ends produced by restriction enzymes — and these are exactly what we need to ligate the foreign DNA into the vector. An exonuclease would chew those ends back, destroying the sticky overhangs and making ligation impossible.

1. Identify the ends in play: sticky 5' or 3' overhangs left by the restriction enzyme on both vector and insert.
2. If exonuclease acts: overhangs trimmed → blunt or no end → no complementary base-pairing → no ligation.

No. An exonuclease would destroy the sticky ends needed for ligation.

Concept used. Restriction-enzyme naming is a four-part code based on the source organism and order of discovery (Smith & Nathans convention, 1973):

• 0pt
• H — first letter of the genus name (Haemophilus).
• in — first two letters of the species name (influenzae).
• d — the strain from which it was isolated (H. influenzae strain Rd).
• III — Roman numeral for the order of discovery (the third restriction enzyme purified from that strain).

H = Haemophilus; in = influenzae; d = strain Rd; III = third enzyme of that strain (Smith & Nathans convention).

Concept used. A cloning vector is designed so that exactly one cut by the chosen restriction enzyme linearises the vector at the cloning site, leaving room to ligate the insert. If the enzyme had more than one site within the vector, cutting would chop the vector into multiple pieces:

1. One useful piece (the backbone with ori and marker) and one or more unwanted fragments would be released.
2. Re-ligation in the presence of the insert could produce a chaotic soup of multiple recombinant forms, most of which are non-functional.
3. The ori or marker gene might itself be excised, destroying the vector's ability to replicate or to be selected.

Multiple cut sites would shatter the vector instead of linearising it, destroying the ori or marker and producing useless ligation products.

Concept used. A competent cell is one whose envelope has been pre-treated so it can take up exogenous DNA from the surroundings. Bacterial cells are normally impermeable to DNA. After treatment with ice-cold CaCl2 (which neutralises the negative charges on the cell wall and on the DNA), the cell becomes competent — i.e. able to be transformed.

`Competent' means the cell has been chemically prepared (e.g. with CaCl2) so its envelope is permeable enough to take in foreign DNA during transformation.

Concept used. Cellular DNA is heavily coated with proteins — most prominently histones (in eukaryotes) and histone-like proteins (in bacteria), plus countless transcription factors and packaging proteins. Proteases chop these proteins into amino acids, freeing the DNA and also destroying nucleases that would otherwise degrade it.

1. Free DNA from protein scaffold → better recovery.
2. Destroy contaminating nucleases (which are themselves proteins) → better DNA integrity.

Proteases strip away DNA-binding proteins and destroy contaminating nucleases, so the isolated DNA is both pure and intact.

Concept used. Denaturation at 95 ^∘C melts the double-stranded DNA into single strands. Only single-stranded template can be bound by the primers in the next (annealing) step. If denaturation is skipped, the DNA stays double-stranded, primers cannot anneal, and Taq polymerase has no primer-template junction to extend.

1. No denaturation ⇒ no single-stranded template.
2. No template ⇒ no annealing ⇒ no extension.
3. Net result: zero amplification — no PCR product.

No amplification will occur because the template stays double-stranded, blocking primer binding and extension.

Concept used. A recombinant vaccine is one in which the antigen is produced by inserting the corresponding gene into a host cell (yeast or bacterium) and then purifying the protein. The most widely used recombinant vaccine is the Hepatitis B vaccine: the gene encoding the hepatitis B surface antigen (HBsAg) is expressed in Saccharomyces cerevisiae, and the purified antigen is the vaccine.

Hepatitis B vaccine — produced by recombinant HBsAg expression in yeast.

Concept used. The three-dimensional shape of every biomolecule — the DNA double helix, the folded protein, the hydrated enzyme active site — depends on a continuous hydration shell of water molecules that forms hydrogen bonds with polar groups. Remove the water and the molecule collapses or unfolds, losing biological activity. This is why dehydrated seeds, dried blood spots and lyophilised enzymes are stable but inactive until rehydrated.

No. Biomolecules need water for their active 3D structure and enzyme catalysis; in anhydrous conditions they are biologically inactive (though they may be chemically stable).

Concept used. The Ti (tumour-inducing) plasmid of Agrobacterium tumefaciens naturally transfers a segment called T-DNA into the plant genome at the wound site, causing crown-gall tumours. To use this plasmid as a cloning vector, the tumour-inducing genes within the T-DNA region are removed (disarmed) and replaced with the gene of interest plus a selectable marker. The vir (virulence) genes that drive T-DNA transfer are kept intact, so the plasmid still delivers the cargo to the plant.

The tumour-inducing genes of the T-DNA region are deleted and replaced with the gene of interest; the disarmed Ti plasmid then carries the gene into the plant genome via its retained vir system.

Concept used. Gene cloning is the production of many genetically identical copies of a specific DNA segment (the gene of interest) by:

1. 0pt
2. Cutting the gene out of the source DNA with restriction enzymes;
3. Ligating it into a cloning vector that has its own origin of replication;
4. Introducing the recombinant vector into a competent host cell by transformation;
5. Letting the host divide so each daughter cell carries (and replicates) a copy of the gene.

After overnight growth a single host cell yields millions of daughter cells, each with many plasmid copies — a clone of the original gene.

1. State the goal: amplify a single gene to millions of copies in vivo.
2. Outline the workflow: cut → ligate → transform → grow → harvest.
3. Contrast with PCR (in vitro amplification): cloning produces a living, expressible copy, while PCR produces a pure molecular copy.

Gene cloning = producing many identical copies of a gene by inserting it into a self-replicating vector and propagating the recombinant vector in a host cell.

Concept used. Biotechnology is defined by the European Federation of Biotechnology (EFB) as ``the integration of natural sciences and organisms, cells, parts thereof, and molecular analogues for products and services''. By this broad definition, any use of living cells to make a product is biotechnology. Wine fermentation by Saccharomyces cerevisiae (a centuries-old traditional process) and recombinant Hepatitis B vaccine production (a modern molecular process) both qualify.

1. Apply the EFB definition: wine making uses live yeast to convert sugars to ethanol + CO2 — biotechnology.
2. Vaccine making uses recombinant yeast expressing HBsAg — biotechnology.
3. Therefore both claims are correct. Traditional biotechnology and modern (rDNA-based) biotechnology are both subsets of the discipline.

Both are correct. Traditional fermentation (wine) and modern recombinant production (vaccine) are equally biotechnology.

Concept used. An exonuclease chews nucleotides off the ends of any linear DNA exposed in the tube. A correctly ligated recombinant plasmid is circular (no free ends), so an exonuclease shouldn't touch it. But the tube usually contains a mixture of:

• 0pt
• Successfully ligated circular recombinant molecules (resistant).
• Un-ligated linear vector and insert fragments (vulnerable).
• Nicked or partially ligated species (vulnerable).

1. Exonuclease degrades all linear DNA in the tube; only intact circles survive.
2. If ligation was efficient, intact recombinant circles still transform, but at lower yield (because their preparation has been net-degraded).
3. If ligation was incomplete, the recombinant fraction is mostly eaten and transformation efficiency crashes; few or no recombinant colonies appear.

Linear DNA is degraded; circular recombinant molecules survive but yield is much reduced, so transformation efficiency falls sharply (and may fail outright if ligation was incomplete).

Concept used. The whole cut-and-paste logic of rDNA relies on predictability: an enzyme that recognises a specific palindrome (e.g. 5'-GAATTC-3' for EcoRI) cuts every occurrence of that palindrome and only those. This generates uniform, predictable sticky ends.

If restriction enzymes cut at random:

1. Fragments would have random end sequences and random lengths → no way to predict where the cut falls in either vector or insert.
2. Sticky ends from the two halves of the same cut would no longer be guaranteed to match the ends from a different cut on a different molecule → no ligation.
3. The vector might be cut inside the ori or selectable marker, ruining the cloning vehicle.
4. Even the cut sites cannot be reproduced from one experiment to the next, so cloning becomes a one-off lottery rather than a routine procedure.

Without sequence specificity, cuts would be random — vector and insert would have unpredictable, mismatched ends; ligation would fail; the ori/marker could be destroyed; the cut would be irreproducible. Cloning would be impossible.

Concept used. A circular plasmid cut at a single site is converted into one linear molecule of the same total length — geometry-wise, one cut on a closed loop produces one line. A linear molecule cut at a single site, however, is split into two pieces, one on either side of the cut. The gel separates fragments by size, so:

• 0pt
• Plasmid → one band (full length).
• Linear DNA → two bands (one for each side of the cut).

1. Visualise the circular plasmid as a rubber band: a single cut snips it into one piece (a line).
2. Visualise the linear DNA as a stretched string: a single cut splits it at one point into two pieces.
3. The gel therefore shows one band for the plasmid and two bands for the linear DNA.

Topology: one cut on a circle gives one linear piece (one band); one cut on a line gives two pieces (two bands).

Concept used. DNA is colourless and invisible under white light. To see it on an agarose gel:

1. 0pt
2. Stain the gel (during or after the run) with ethidium bromide (EtBr). EtBr intercalates between adjacent base pairs of the DNA.
3. Illuminate the gel with UV light (∼ 300 nm) on a UV transilluminator.
4. EtBr fluoresces orange (∼ 605 nm) at the DNA bands; DNA-free regions stay dark.
5. Photograph through a UV-blocking filter to record the band pattern.

Stain with ethidium bromide, then expose to UV light on a transilluminator — DNA bands fluoresce bright orange against a dark background.

Concept used. A selectable marker (e.g. antibiotic resistance gene like amp^R) lets us distinguish host cells that have taken up the vector from those that have not. Without a marker, we have no way to identify the rare cells that were successfully transformed. After plating the transformation mixture, every colony will grow — both the transformed and the non-transformed — but we cannot tell which are which.

1. Transformation is inherently inefficient: typically only 1 in 10⁴–10⁶ cells takes up a plasmid.
2. Without a marker, the transformed cells are mixed in among ∼ 10⁵× more non-transformed cells.
3. Selecting recombinants becomes impossible because no difference-of-growth is available.

No way to distinguish transformed from non-transformed cells; the rare transformants are lost in the much larger population of untransformed cells, so the cloning effectively fails at the selection step.

Concept used. For DNA to appear as a band on an EtBr-stained gel, several conditions must be met: (i) the DNA must be present at high enough concentration to be detected by EtBr fluorescence, (ii) it must be intact, not degraded into nucleotides, (iii) the EtBr must intercalate into it, and (iv) the gel must be viewed under UV light.

1. DNA concentration too low — below the EtBr detection limit (∼ 1–10 ng per band).
2. DNA fully degraded — small mono/oligonucleotides run off the gel and give no band.
3. Gel illuminated under visible light rather than UV — EtBr fluorescence only shows under UV.
4. EtBr staining step was skipped, ineffective, or used at too low a concentration to intercalate.

Most likely causes: (i) too little DNA, (ii) DNA fully degraded, (iii) UV transilluminator not used, or (iv) faulty/missed EtBr staining.

Concept used. The bacterial cell wall is negatively charged (phosphates on the lipopolysaccharide), and so is DNA (phosphates on the backbone). Two negatives repel — so DNA cannot approach the cell. Divalent Ca^2+ ions neutralise both surfaces, screening the repulsion and letting DNA stick to the cell envelope. A subsequent heat shock at 42 ^∘C creates transient pores through which the bound DNA enters.

1. Soak cells in ice-cold CaCl2 (0.1 M typical). Ca^2+ binds to the cell wall and to phosphate groups on the DNA.
2. Mix cells with the recombinant DNA, keep on ice — DNA-cell complexes form on the wall.
3. Heat shock (42 ^∘C, 90 s) opens transient pores; DNA enters the cell.
4. Cool back, recover in rich medium, then plate on selection.

CaCl2 provides Ca^2+ ions that neutralise the negative charges on the cell wall and on DNA, letting DNA bind to the cell envelope before the heat-shock pores let it in.

Concept used. The recombinant plasmid carries an antibiotic-resistance marker. Carrying the plasmid is a metabolic burden on the cell — replication, transcription and translation of the plasmid and its gene cost energy. In the presence of the antibiotic, only resistant (plasmid-bearing) cells survive, so the population stays 100 % recombinant.

Without the antibiotic the selection pressure vanishes:

1. Cells that spontaneously lose the plasmid (a rare event per division) escape the metabolic cost and grow slightly faster than plasmid-bearing cells.
2. Over many generations, plasmid-free cells out-compete the recombinant cells and gradually dominate the bioreactor.
3. Yield of the recombinant protein drops sharply as the recombinant fraction shrinks.
4. Even worse, the dominant non-recombinant population will continue to consume nutrients without producing the target protein.

Plasmid-free cells (faster-growing because they avoid the metabolic burden) will out-compete recombinant cells over generations; the bioreactor soon produces little to no recombinant protein.

Concept used. Polymerase Chain Reaction (PCR) amplifies a specific DNA segment through repeated cycles of three temperature-defined steps. The figure labels the three steps as A, B and C:

• 0pt
• A — Denaturation. The double-stranded template DNA is heated to ∼ 94–95 ^∘C, which breaks the hydrogen bonds between the two strands. The result is two single strands. Arrow ``Heat'' in the figure.
• B — Annealing. Temperature is dropped to ∼ 50–60 ^∘C so that the two short primers (one for each strand) can base-pair with their complementary sequences flanking the region to be amplified.
• C — Extension (Elongation). Temperature is raised to ∼ 72 ^∘C, the optimum for Taq DNA polymerase. Taq adds deoxynucleotides to the 3'-OH end of each annealed primer, synthesising a new strand complementary to the template.

1. A (Denaturation): heat at 94–95 ^∘C separates the two parental strands.
2. B (Annealing): cool to 50–60 ^∘C, primers bind the single-stranded templates at their target sequences.
3. C (Extension): warm to 72 ^∘C, Taq polymerase extends each primer into a full complementary strand using dNTPs.
4. Repeat A–B–C for ∼ 30 cycles. After n cycles, the number of double-stranded copies of the target region is N = N₀ × 2ⁿ, i.e. ∼ 10⁹-fold amplification.

A = Denaturation; B = Annealing of primers; C = Extension by Taq DNA polymerase.

Concept used. The figure shows pBR322, the classic E. coli cloning vector developed by Bolivar and Rodriguez in 1977. The plasmid is 4361 bp long and carries (i) an origin of replication (ori), (ii) two antibiotic-resistance genes (amp^R for ampicillin and tet^R for tetracycline), and (iii) several unique restriction sites suitable for cloning (EcoRI, ClaI, HindIII, BamHI, SalI, PvuI, PvuII). Reading the labelled and unlabelled regions:

• 0pt
• A — amp^R: the ampicillin-resistance gene (encodes β-lactamase). Cells carrying intact amp^R survive on ampicillin plates.
• B — BamH I (BamHI) restriction site: a unique cloning site inside tet^R. Insertion at this site inactivates tetracycline resistance, providing the basis for insertional-inactivation screening of recombinants.
• C — tet^R: the tetracycline-resistance gene. Intact tet^R confers tetracycline resistance.

1. Recall the standard pBR322 map: two antibiotic markers amp^R (1 o'clock–3 o'clock arc) and tet^R (9 o'clock–12 o'clock arc), plus ori (5 o'clock) and rop (6 o'clock).
2. Map labels to the figure: A on the right belongs to the ampicillin-resistance arc; C on the upper-left belongs to the tet-resistance arc; B on the left is the BamHI restriction site (inside tet^R).

A = amp^R (ampicillin-resistance gene); B = BamHI restriction site (inside tet^R); C = tet^R (tetracycline-resistance gene).

Concept used. In the classical insertional-inactivation scheme using two antibiotic markers (e.g. amp^R and tet^R of pBR322), recombinant selection takes two steps: (i) grow all transformants on ampicillin plates, and (ii) replica-plate onto tetracycline plates to identify the tet^S (recombinant) colonies. This requires duplicate plates and the labour of replica-plating.

The newer blue-white screening replaces one antibiotic marker with a marker gene encoding a chromogenic enzyme — most commonly the lacZα fragment of β-galactosidase. The vector carries a multiple cloning site (MCS) inserted in the middle of lacZα. When plated on a medium containing X-gal (a colourless substrate) and the inducer IPTG:

• 0pt
• Non-recombinant colonies have intact lacZα, make active β-galactosidase, cleave X-gal to a blue product, and form blue colonies.
• Recombinant colonies have a foreign DNA insert disrupting lacZα, cannot cleave X-gal, and form white colonies.

So we identify recombinants on a single plate in a single growth step — much faster and cheaper than two-antibiotic replica plating.

1. Compare workflow lengths: two-marker antibiotic screening requires primary plate + replica plate + comparison. Blue-white screening needs only one plate.
2. Compare resource use: two-antibiotic method needs two antibiotics and replica-plating apparatus. Blue-white needs only one antibiotic plus X-gal/IPTG.
3. Compare ease of scoring: tet-sensitivity needs colony-by-colony comparison between two plates. Blue/white is a direct visual readout on one plate.
4. Compare safety: X-gal is non-toxic to the user. The duplicate antibiotics aren't hazardous either, but the overall reagent burden is lower with blue-white.
5. Practical bonus: blue-white plates can be photographed and counted automatically; replica plates need manual eye-balling.

Blue-white screening using lacZα insertional inactivation needs only one plate (vs two with the antibiotic-only method), gives a direct colour readout (vs growth-vs-no-growth comparison), avoids replica-plating labour, and uses fewer reagents — so it has replaced the older antibiotic-only insertional inactivation.

Concept used. Agrobacterium tumefaciens is a soil bacterium that naturally infects wounded dicot plants and causes crown-gall tumours. The infection is mediated by a large plasmid called the Ti (Tumour-inducing) plasmid. Two regions of Ti are critical:

• 0pt
• T-DNA (transferred DNA) — a 20–25 kb segment delimited by two ``border'' sequences. T-DNA is excised from the Ti plasmid, transferred into the plant cell, and integrated into the plant chromosomal DNA.
• Virulence (vir) region — a cluster of genes whose products process and shuttle T-DNA into the plant.

Step-wise natural infection.

1. A wounded plant releases phenolic compounds (e.g. acetosyringone) that activate the vir genes in nearby Agrobacterium.
2. Agrobacterium attaches to the plant cell wall using surface proteins.
3. vir gene products nick the T-DNA borders, excise the T-DNA as a single-stranded molecule (T-strand), and coat it with VirD2 and VirE2 proteins for protection.
4. A bacterial Type IV secretion system ferries the T-DNA-protein complex into the plant cell.
5. Inside the plant cell, nuclear localisation signals on VirD2/VirE2 drive the T-DNA into the nucleus.
6. The T-DNA is integrated at random sites into the plant chromosomes by the host's own DNA-repair machinery. Genes carried on the T-DNA are now transcribed by plant RNA polymerase.
7. In nature, the T-DNA encodes auxin and cytokinin biosynthesis genes (causing tumour formation) plus opine synthesis genes (which feed Agrobacterium).

Engineering for plant transformation.

1. Engineers disarm the Ti plasmid by deleting the tumour-causing auxin/cytokinin genes from the T-DNA.
2. The gene of interest plus a plant-selectable marker (e.g. nptII for kanamycin resistance) is inserted between the T-DNA borders.
3. The recombinant Ti is reintroduced into Agrobacterium, and plant tissue (leaf disc, callus) is co-cultured with this strain.
4. Agrobacterium transfers the engineered T-DNA into the plant genome via its retained vir machinery — exactly as it would in nature.
5. Transformed plant cells are selected on kanamycin, regenerated into whole plants, and screened for stable integration and expression of the foreign gene.

Agrobacterium tumefaciens acts as a natural plant-transformation vector: its Ti plasmid's T-DNA is excised and transferred (via the vir-encoded type-IV secretion system) into the plant cell nucleus, where it integrates into the chromosomes. Biotechnologists disarm the T-DNA, replace the tumour genes with the gene of interest plus a plant marker, and exploit Agrobacterium's natural delivery to produce transgenic plants.

Concept used. A bioreactor is a closed vessel designed to grow microbial, plant or animal cells under tightly controlled physical and chemical conditions so that they convert raw materials into a target biological product. The most widely used industrial bioreactor is the stirred-tank bioreactor (Fig. alongside).

Schematic design of a stirred-tank bioreactor.

Key components of a stirred-tank bioreactor.

• 0pt
• Agitator (impeller). Driven by a motor; keeps the cells suspended and the broth uniformly mixed.
• Sparger (air inlet). Bubbles sterile air through the broth from the bottom, supplying oxygen to aerobic cells.
• Cooling jacket. Removes the heat released by cellular metabolism to keep temperature constant (typically ∼ 30–37 ^∘C).
• Sensors and probes. Continuously monitor pH, dissolved oxygen and temperature; feedback to controllers that adjust acid/base or air flow.
• Inlets for nutrients, acid/base, antifoam.
• Outlets for harvest, sampling and exhaust gases.
• Foam breaker at the top to prevent foam overflow.

Laboratory flask vs. continuous-culture bioreactor.

1. A lab flask is a closed batch vessel with no flow of medium and only passive aeration. Cells grow to stationary phase in hours to days, then stop producing.
2. A continuous-culture bioreactor is a controlled vessel with steady flow of fresh medium in and used medium out. Cells stay in log phase for days or weeks, producing the target protein continuously.
3. The bioreactor's combined control of temperature, pH, dissolved O2 and agitation maximises growth rate and product yield. A shake flask cannot do this.

A bioreactor is a sterile, agitated, aerated, sensor-controlled vessel with inlets and outlets that lets cells grow continuously in log phase at industrial scale; a lab flask is a small, batch-only, manually-shaken vessel with no environmental control. The bioreactor gives orders-of-magnitude higher and longer-running product yield.