NCERT Exemplar Biotechnology Principle & Processes - Biology Class 12 - NEET

MULTIPLE CHOICE QUESTIONS

Q.1. Rising of dough is due to:
(a) Multiplication of yeast
(b) Production of CO₂
(c) Emulsification
(d) Hydrolysis of wheat flour starch into sugars.
Ans. (b)
Solution.
The puffed appearance of dough used in bread and other leavened products is due to the production of carbon dioxide (CO₂) gas. In bread-making, the fungus Saccharomyces cerevisiae (baker's yeast) ferments sugars and releases CO₂, which becomes trapped in the dough matrix and causes it to rise. Fermentation of batter for idli and dosa is carried out by a community of lactic acid bacteria and some yeasts; CO₂ and organic acids produced contribute to the texture and flavour.

Q.2. Which of the following enzymes catalyse the removal of nucleotides from the ends of DNA?
(a) Endonuclease
(b) Exonuclease
(c) DNA ligase
(d) Hind - II
Ans. (b)
Solution.
Exonucleases remove nucleotides from the free ends of DNA molecules. In contrast, endonucleases cleave phosphodiester bonds at internal sites within DNA. DNA ligase joins DNA fragments by forming phosphodiester bonds. Hind II is an example of a restriction endonuclease (endonuclease), not an exonuclease.

Q.3. The transfer of genetic material from one bacterium to another through the mediation of a viral vector is termed as:
(a) Transduction
(b) Conjugation
(c) Transformation
(d) Translation
Ans. (a)
Solution.
Transduction is the process by which bacteriophages (viruses that infect bacteria) transfer bacterial DNA from one cell to another. Conjugation involves direct cell-to-cell contact and transfer of DNA (often plasmids). Transformation is uptake of naked DNA from the environment. Translation is the process of protein synthesis from mRNA.

Q.4. Which of the given statements is correct in the context of visualizing DNA molecules separated by agarose gel electrophoresis?
(a) DNA can be seen in visible light
(b) DNA can be seen without staining in visible light
(c) Ethidium bromide stained DNA can be seen in visible light
(d) Ethidium bromide stained DNA can be seen under exposure to UV light
Ans. (d)
Solution.
Separated DNA fragments are visualised after staining with a fluorescent intercalating dye such as ethidium bromide and exposing the gel to ultraviolet (UV) light. The dye intercalates between base pairs and fluoresces under UV, producing visible bands. Unstained DNA is not visible under ordinary visible light.

Q.5. 'Restriction' in Restriction enzyme refers to:
(a) Cleaving of phosphodiester bond in DNA by the enzyme
(b) Cutting of DNA at specific position only
(c) Prevention of the multiplication of bacteriophage by the host bacteria
(d) All of the above
Ans. (c)
Solution.
The term restriction originates from the bacterial restriction-modification system, where restriction enzymes (restriction endonucleases) restrict or prevent multiplication of invading bacteriophages by cutting their DNA. The enzymes cleave phosphodiester bonds at specific recognition sequences, which is the biochemical basis of the restriction function; the historical naming emphasises the biological role of restricting phage growth.

Q.6. Which of the following is not required in the preparation of a recombinant DNA molecule?
(a) Restriction endonuclease
(b) DNA ligase
(c) DNA fragments
(d) E.coli
Ans. (d)
Solution.
To construct recombinant DNA the essential tools include: restriction enzymes to cut DNA, DNA ligase to join fragments, suitable DNA fragments (insert and vector) and a vector to carry the insert. A host organism (for propagation and expression) is required for many applications but it need not be Escherichia coli specifically; other hosts (yeast, insect, mammalian cells, plants) or cell-free systems may be used. Thus E. coli is not strictly required for the concept of recombinant DNA construction.

Q.7. In agarose gel electrophoresis, DNA molecules are separated on the basis of their:
(a) Charge only
(b) Size only
(c) Charge to size ratio
(d) All of the above
Ans. (b)
Solution. DNA fragments are separated according to size due to sieving effect of agarose gel. Although DNA has charge, charge-to-size ratio is not used as the criterion.

Q.8. The most important feature in a plasmid to serve as a vector in gene cloning experiment is:
(a) Origin of replication (ori)
(b) Presence of a selectable marker
(c) Presence of sites for restriction endonuclease
(d) Its size
Ans. (a)
Solution.
The origin of replication (ori) is essential because it allows the plasmid to replicate inside the host cell independently of the host chromosome. Without an ori the vector cannot be maintained and amplified. Selectable markers and restriction sites are important supporting features; size influences ease of manipulation and copy number but the ori is the most crucial feature for a functional vector.

Q.9. While isolating DNA from bacteria, which of the following enzymes is not required?
(a) Lysozyme
(b) Ribonuclease
(c) Deoxyribonuclease
(d) Protease
Ans. (c)
Solution.
When isolating DNA we use lysozyme to weaken bacterial cell walls, proteases to remove contaminating proteins and ribonuclease (RNase) to remove RNA. A deoxyribonuclease (DNase) would degrade the target DNA and is therefore not used in DNA isolation.

Q.10. Which of the following contributed in popularising the PCR (polymerase chain reactions) technique?
(a) Easy availability of DNA template
(b) Availability of synthetic primers
(c) Availability of cheap deoxyribonucleotides
(d) Availability of 'Thermostable' DNA polymerase
Ans. (d)
Solution.
The discovery and use of a thermostable DNA polymerase (Taq polymerase) from the thermophilic bacterium Thermus aquaticus made PCR practical and automated. Because the enzyme withstands high-temperature denaturation steps, cycles could be run repeatedly without adding fresh enzyme each cycle, greatly simplifying and accelerating the technique.

Q.11. An antibiotic resistance gene in a vector usually helps in the selection of:
(a) Competent bacterial cells
(b) Transformed bacterial cells
(c) Recombinant bacterial cells
(d) None of the above
Ans. (b)
Solution.
An antibiotic resistance gene on a vector allows selection for transformed cells (cells that have taken up the plasmid). By growing bacteria on medium containing the antibiotic, only cells harbouring the plasmid survive. Further screening may be needed to distinguish recombinant plasmids from non-recombinant plasmids.

Q.12. Significance of 'heat shock' method in bacterial transformation is to facilitate:
(a) Binding of DNA to the cell wall
(b) Uptake of DNA through membrane transport proteins
(c) Uptake of DNA through transient pores in the bacterial cell wall
(d) Expression of antibiotic resistance gene
Ans. (c)
Solution.
The heat shock method briefly raises the temperature of competent cells, creating a thermal imbalance that helps DNA cross the cell membrane through transient pores formed in the cell envelope. This improves the uptake of plasmid DNA into the bacterial cytoplasm.

Q.13. The role of DNA ligase in the construction of a recombinant DNA molecule is:
(a) Formation of phosphodiester bond between two DNA fragments
(b) Formation of hydrogen bonds between sticky ends of DNA fragments
(c) Ligation of all purine and pyrimidine bases
(d) None of the above
Ans. (a)
Solution.
DNA ligase catalyses formation of the phosphodiester bond between adjacent nucleotides, sealing nicks and covalently joining DNA fragments (for example, vector and insert) to produce a continuous DNA molecule suitable for transformation.

Q.14. Which of the following bacteria is not a source of restriction endonuclease?
(a) Haemophilus influenzae
(b) Escherichia coli
(c) Entamoeba coli
(d) Bacillus amyloliquefaciens
Ans. (c)
Solution.
Restriction endonucleases are primarily isolated from bacteria such as Haemophilus influenzae, Escherichia coli and various Bacillus species. Entamoeba coli is an amoeboid eukaryote and is not a typical source of bacterial restriction enzymes.

Q.15. Which of the following steps are catalysed by Taq DNA polymerase in a PCR reaction?
(a) Denaturation of template DNA
(b) Annealing of primers to template DNA
(c) Extension of primer end on the template DNA
(d) All of the above
Ans. (c)
Solution.
Taq DNA polymerase catalyses the extension step in PCR, synthesising new DNA from the primers along the template strand. Denaturation and annealing are temperature-driven physical steps; polymerase acts during the extension (elongation) step at the appropriate temperature.

Q.16. A bacterial cell was transformed with a recombinant DNA molecule that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reasons could be:
(a) Human gene may have intron which bacteria cannot process
(b) Amino acid codons for humans and bacteria are different
(c) Human protein is formed but degraded by bacteria
(d) All of the above
Ans. (a)
Solution.
A common reason is that a eukaryotic gene contains introns which bacteria cannot splice out; consequently the bacterial transcription-translation machinery cannot produce the correct protein unless a cDNA (intron-free copy) is used. Codon usage differences and proteolytic degradation in bacteria can also affect expression, but the presence of introns is a primary barrier.

Q.17. Which of the following should be chosen for best yield if one were to produce a recombinant protein in large amounts?
(a) Laboratory flask of largest capacity
(b) A stirred-tank bioreactor without in-lets and out-lets
(c) A continuous culture system
(d) Any of the above
Ans. (c)
Solution.

• To produce recombinant proteins at industrial scale, small lab flasks are inadequate due to limited volume and poor control of parameters.
• A stirred-tank bioreactor with proper inlets and outlets, aeration and control systems is appropriate, but for sustained high yield the continuous culture system is preferred because it allows steady-state production and continuous harvest of biomass or product.

Q.18. Who among the following was awarded the Nobel Prize for the development of PCR technique?
(a) Herbert Boyer
(b) Hargovind Khurana
(c) Kary Mullis
(d) Arthur Kornberg
Ans. (c)
Solution.
Kary Mullis was awarded the Nobel Prize in Chemistry in 1993 for the invention of the polymerase chain reaction (PCR) technique.

Q.19. Which of the following statements does not hold true for restriction enzyme?
(a) It recognises a palindromic nucleotide sequence
(b) It is an endonuclease
(c) It is isolated from viruses
(d) It can produce the same kind of sticky ends in different DNA molecules
Ans. (c)
Solution.

• Restriction enzymes typically recognise palindromic DNA sequences.
• They are endonucleases because they cleave internal phosphodiester bonds in DNA.
• They are isolated from bacteria as part of the restriction-modification defence system; they are not generally isolated from viruses.
• They can produce compatible sticky ends when two different DNA molecules are cut by the same restriction enzyme, facilitating ligation.

VERY SHORT ANSWER TYPE QUESTIONS

Q.1. How is copy number of the plasmid vector related to yield of recombinant protein?
Ans. Higher plasmid copy number results in more copies of the gene per cell and therefore generally leads to higher production of the encoded protein, provided expression is efficient and cellular resources are not limiting.

Q.2. Would you choose an exonuclease while producing a recombinant DNA molecule?
Ans. No. Exonucleases remove nucleotides from free DNA ends and would degrade linear DNA fragments and could destroy the insert or vector ends needed for cloning. Restriction endonucleases and ligases are used instead.

Q.3. What does H in' 'd' and 'III' refer to in the enzyme Hind III?
Ans. In the name HindIII: 'H' denotes the genus Haemophilus, 'in' denotes the species influenzae, 'd' denotes the particular strain (Rd) and 'III' indicates that it was the third enzyme isolated from that strain.

Q.4. Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Ans. If a restriction enzyme has more than one recognition site within a vector, digestion will fragment the vector into multiple pieces, preventing recovery of an intact vector for cloning. A single unique site is preferred in the cloning region to allow insertion without destroying the vector.

Q.5. What does 'competent' refer to in competent cells used in transformation experiments?
Ans. Competent cells are bacterial cells treated (chemically, e.g., with CaCl₂, or by electroporation) to become capable of taking up foreign DNA from their environment.

Q.6. What is the significance of adding proteases at the time of isolation of genetic material (DNA).
Ans. Proteases degrade proteins that are bound to or contaminate DNA preparations. Removal of proteins prevents interference with downstream applications such as restriction digestion, ligation and sequencing.

Q.7. While doing a PCR, 'denaturation' step is missed. What will be its effect on the process?
Ans. Without the denaturation step, the double-stranded DNA template will not separate into single strands, so primers cannot anneal and polymerase cannot extend; consequently, no amplification will occur.

Q.8. Name a recombinant vaccine that is currently being used in vaccination program.
Ans. The Hepatitis B recombinant vaccine (for example, Engerix) is produced by expressing the viral surface antigen in yeast and is widely used in vaccination programmes.

Q.9. Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?
Ans. No. Biomolecules such as DNA and proteins generally require an aqueous environment to maintain proper conformation and exhibit biological activity; water is essential for the chemistry of life.

Q.10. What modification is done on the Ti plasmid of Agrobacterium tumefaciens to convert it into a cloning vector?
Ans. The tumour-inducing (Ti) plasmid is disarmed by removing genes responsible for pathogenicity while retaining the T-DNA transfer machinery. The T-DNA region is then engineered to carry the gene of interest, converting Ti plasmids into plant cloning vectors that can stably transfer genes into plant genomes.

SHORT ANSWER TYPE QUESTIONS

Q.1. What is meant by gene cloning?
Ans. Gene cloning is the process of creating multiple identical copies of a gene. A gene of interest is inserted into a vector to form recombinant DNA, introduced into a host cell, and propagated so that each daughter cell carries copies of the gene; colonies derived from a single transformed cell therefore contain many identical copies of the inserted gene.

Q.2. Both a wine maker and a molecular biologist who had developed a recombinant vaccine claim to be biotechnologists. Who in your opinion is correct?
Ans. Both are correct. Biotechnology broadly covers the use of living organisms or their components to produce useful products or processes. A winemaker uses microbial fermentation (yeast) to produce wine, while a molecular biologist uses genetic engineering to produce vaccines; both are practising biotechnology, though at different levels of technology and scale.

Q.3. A recombinant DNA molecule was created by ligating a gene to a plasmid vector. By mistake, an exonuclease was added to the tube containing the recombinant DNA. How does this affect the next step in the experiment i.e. bacterial transformation?
Ans. A closed circular recombinant plasmid is generally resistant to exonuclease action because exonucleases require free ends to act. Therefore a correctly closed circular plasmid is unlikely to be degraded by exonuclease and transformation may still succeed. However, if the plasmid preparation contained any linear forms or nicked circles, these could be degraded and reduce transformation efficiency.

Q.4. Restriction enzymes that are used in the construction of recombinant DNA are endonucleases which cut the DNA at 'specific-recognition sequence'. What would be the disadvantage if they do not cut the DNA at specific-recognition sequence?
Ans. If restriction enzymes cut at random sites instead of specific recognition sequences, the resulting DNA fragments would be unpredictable and would not generate defined sticky or blunt ends required for precise ligation. This would complicate cloning, make insertion of specific fragments unreliable and hamper construction of recombinant molecules.

Q.5. A plasmid DNA and a linear DNA (both are of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.
Ans. A circular plasmid with a single restriction site becomes a single linear molecule when cut, producing one band. A linear DNA with one internal restriction site is cut into two separate fragments of different sizes, which migrate to form two bands on the gel. Thus plasmid yields one band and linear DNA yields two bands under these conditions.

Q.6. How does one visualise DNA on an agarose gel?
Ans. DNA is commonly stained with a fluorescent dye such as ethidium bromide which intercalates between base pairs. When the stained gel is exposed to ultraviolet (UV) light, the ethidium bromide fluoresces and DNA fragments appear as orange bands. Safer alternative dyes are also available (e.g., SYBR Safe) that fluoresce under blue or UV light.

Q.7. A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment?
Ans. Without a selectable marker (such as an antibiotic resistance gene), it is difficult to distinguish transformed from non-transformed cells. After transformation, only a fraction of cells take up the plasmid; a selectable marker enables growth or screening for transformed cells and simplifies isolation of clones carrying the vector.

Q.8. A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gel with ethidium bromide, no DNA bands were observed. What could be the reason?
Ans. Possible reasons include:
(i) The DNA sample was degraded by nucleases (exo- or endonucleases).
(ii) The gel was run with reversed polarity so DNA migrated out of the gel (wells towards anode).
(iii) Ethidium bromide was omitted or used at insufficient concentration so bands could not be visualised.
Checking sample integrity, electrode orientation and staining procedure helps identify the cause.

Q.9. Describe the role of CaCl₂ in the preparation of competent cells?
Ans. Calcium chloride (CaCl₂) treatment increases competence by neutralising negative charges on the DNA and cell surface, helping DNA approach the cell membrane. Divalent Ca²⁺ ions also facilitate formation of transient pores in the bacterial envelope during heat shock, enhancing DNA uptake during transformation.

Q.10. What would happen when one grows a recombinant bacterium in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing?
Ans. Without antibiotic selection pressure, cells may lose plasmids carrying recombinant genes (especially high-copy plasmids that impose a metabolic burden). Non-plasmid-bearing cells often grow faster and outcompete plasmid-bearing cells, reducing recombinant yield over time.

Q.11. Identify and explain steps 'A', 'B' and 'C' in the PCR diagram given below.

Ans. 
A-Denaturation
B-Annealing
C-Extension of primers

Q.12. Name the regions marked A, B and C.

Ans. 
A-Bam HI
B-Pst I .
C-Ampicillin resistance gene (amp^R)

LONG ANSWER TYPE QUESTIONS

Q.1. For selection of recombinants, insertional inactivation of antibiotic marker has been superceded by insertional inactivation of a marker gene coding for a chromogenic substrate. Give reasons.
Ans. Selection based on insertional inactivation of antibiotic markers requires multiple antibiotic markers and multiple plates to distinguish recombinants from non-recombinants, making the process laborious and time-consuming. For example, one method requires screening on two different antibiotic plates and replica plating to identify recombinants. Insertional inactivation of a chromogenic marker (a gene that produces a coloured product in the presence of a chromogenic substrate) allows easy visual differentiation on a single plate: colonies carrying non-recombinant plasmids produce colour, while recombinants where the marker is disrupted by an insert remain colourless (or vice versa depending on design). This single-step visual screening saves time and labour and improves throughput, which is why chromogenic marker systems (e.g., blue-white screening using lacZ/ X-gal) are widely used.

Q.2. Describe the role of Agrobacterium tumefaciens in transforming a plant cell.
Ans. Agrobacterium tumefaciens harbours a large plasmid called the Ti (tumour-inducing) plasmid. The Ti plasmid contains a T-DNA region flanked by left and right border sequences. During infection, the T-DNA is excised and transferred from the bacterium into the plant cell, where it integrates into the plant genome; genes carried on the T-DNA can be expressed in plant cells. Molecular biologists have exploited this natural gene transfer system by disarming the Ti plasmid (removing tumour-causing genes) and using the transfer machinery to deliver desired genes into plants, thereby generating stable transgenic plants.

Q.3. Illustrate the design of a bioreactor. Highlight the difference between a flask in your laboratory and a bioreactor which allows cells to grow in a continuous culture system.
Ans. A bioreactor is a vessel designed to support biological reactions and growth of cells at controlled scale. It provides regulated conditions such as temperature, pH, aeration, agitation, nutrient feed and waste removal to maximise yield and maintain product quality. Key features include:

• Controlled stirring/agitation to keep cells and nutrients evenly distributed and to improve gas transfer.
• Aeration system (sparger) to supply oxygen and remove CO₂.
• Sensors and control systems for pH, dissolved oxygen, temperature and foam.
• Inlets and outlets for continuous feed of medium and removal of product or biomass in continuous culture mode.

Compared with a laboratory flask, which is limited in volume and offers poor control of environmental parameters, a bioreactor enables precise control, scale-up to hundreds or thousands of litres and operation in continuous or fed-batch modes for sustained high-yield production. Continuous culture maintains cells in a steady physiological state, enabling constant product formation and higher overall productivity for some processes.

Sparged stirred-tank bioreactor through which sterile air bubbles are sparged