NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 11 Biotechnology: Principles and Processes

Multiple Choice Questions

Question 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.
Answer:
(b)

Question 2.
An enzyme catalysing the removal of nucleotides from the ends of DNA is
(a) endonuclease
(b) exonuclease
(c) DNA ligase
(d) HindII
Answer:
(b) : Exonucleases remove nucleotides from the terminal ends (either 5′ or 3′) of DNA of one strand of duplex. Endonucleases make cut at specific position within the DNA. DNA ligases are known as joining or sealing enzymes. Ligases form phosphodiester bonds between adjacent nucleosides and covalently link two individual fragments of DNA. Hindll is restriction endonuclease, it cuts the DNA and produces blunt end.

Question 3.
The transfer of genetic material from one bacterium to another through the mediation of a vector like virus is termed as
(a) transduction
(b) conjugation
(c) transformation
(d) translation.
Answer:
(a)

Question 4.
Which of the given statements is correct in the context of observing DNA 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.
Answer:
(d) : Electrophoresis is a technique of separation of molecules such as DNA, RNA or protein, under the influence of an electric field, so that they migrate in the direction of electrode bearing the opposite charge, viz, positively charged molecules move towards cathode (-ve electrode) and negatively charged molecules travel towards anode (+ve electrode) through a medium/ matrix. Since DNA fragments are negatively charged molecules, they can be separated by allowing them to move towards the anode (+ve electrode) under an electric field through a matrix of agarose gel.

The DNA fragments separate’according to their size through the agarose gel, with smaller fragments moving farther away as compared to larger ones. The DNA fragments can be visualised by staining them with ethidium bromide followed by exposure to UV radiations. Bright orange coloured bands of DNA can be observed. The separated DNA bands are then cut out from the agarose gel and extracted from the gel piece, this step is known as elution.

Question 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 in bacteria.
(d) all of the above.
Answer:
(c) : A restriction enzyme is an enzyme that cuts DNA by hydrolysing phosphodiester bonds at or near a specific recognition nucleotide sequences known as restriction sites. The term ‘restriction’ refers to the function of an enzyme restricting the propagation of foreign DNA of bacteriophage in the host bacterium. These enzymes are found in bacteria and provide a defense mechanism against invading virus. The host DNA is protected by a modification enzyme that modifies the prokaryotic DNA and blocks cleavage. Together, these processes form restriction modification system.

Question 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
Answer:
(d) : To produce a recombinant DNA following procedure is followed :

• Genetic material is isolated by using the enzyme, lysozyme or cellulase or chitinase for animal, plant or fungal cell respectively.
• Purified DNA is cut at specific sites by restriction enzymes.
• DNA fragments are separated using agarose gel electrophoresis and the gene or fragment of interest is amplified using PCR.
• After the cutting of the source DNA and the vector DNA with specific restriction enzyme, the cut out ‘gene of interest’ from the source DNA and the cut vector with space are mixed and ligase enzyme is added. This results in the formation of a rDNA or hybrid DNA or chimeric DNA.
• The ultimate aim of recombinant DNA technology is to produce a desirable protein. The foreign gene gets expressed under appropriate condition, culturing methods are used to produce higher yields of the desired protein. Therefore, E. coli is not required for preparation of a recombinant DNA molecule. we have to break the cell open to release DNA and other macromolecules

Question 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.
Answer:
(b) : Electrophoresis is a technique of separation of molecules, such as DNA, RNA or proteins on the basis of their size, under the influence of an electric field. So they migrate towards electrode of opposite charge. DNA fragments separate according to the size through pores of agarose gel.

Question 8.
The most important feature in a plasmid to be used as a vector is
(a) origin of replication (or/)
(b) presence of a selectable marker
(c) presence of sites for restriction endonuclease
(d) its size.
Answer:
(a,b,c & d) : Origin of replication (On), a selectable marker, sites for restriction 1 endonuclease and its size, all are importantfeatures required to facilitate cloning into a vector. A good DNA vector should be able to replicate autonomously in the host cell, for which it needs to have an origin of replication site (Ori). This is important for the replication of inserted gene. A prokaryotic DNA has a single Ori while eukaryotic DNA may have more than one Ori. Selectable marker helps in identifying and eliminating non-transformants and selectively permitting the growth of transformants. Cloning sites (recognition sites) are the sites where the DNA is cut by a restriction endonuclease. A vector should be ideally less than 10 kb in size because large DNA molecules can break down during purification procedure.

Question 9.
While isolating DNA from bacteria, which of the following enzymes is not used?
(a) Lysozyme
(b) Ribonuclease
(c) Deoxyribonuclease
(d) Protease
Answer:
(c) : In order to cut the DNA with restriction enzymes, it needs to be in pure form, free from other macromolecules. Since the DNA is enclosed by the membranes,we haive to break the cell open to release DNA and other macromolecules like RNA, proteins, polysaccharides and lipids. It is obtained by treating the bacterial cells/ plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells) and chitinase (fungus). DNA is interwined with proteins like histones. RNA can be removed by treatment with ribonuclease while proteins can be removed by treatment with protease. Other molecules are removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

Question 10.
Which of the following has popularised the PCR (polymerase chain reaction)?
(a) Easy availability of DNA template.
(b) Availability of synthetic primers.
(c) Availability of cheap deoxyribonudeo- tides.
(d) Availability of ‘thermostable’ DNA poly-merase.
Answer:
(d) : The final step of PCR is extension, wherein Taq DNA polymerase (isolated from a thermophilic bacterium Thermits aquaticus) synthesises the DNA region between the primers, using dNTPs (deoxynucleoside triphosphates) and Mg²⁺. The primers are extended towards each other so that the DNA segment lying between the two primers is copied. The optimum temperature for this polymerisation step is 72CC. Taq polymerase remains active during high temperature induced denaturation of double stranded DNA.

Question 11.
An antibiotic resistance gene in a vector usually helps in the selection of
(a) competent cells
(b) transformed cells
(c) recombinant cells
(d) none of the above.
Answer:
(b) :The ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic resistance genes. For example, you can ligate a foreign DNA at the BamHl site of tetracycline resistance gene in the vector pBR322. The recombinant plasmids will lose tetracycline resistance other macromolecules like RNA, proteins, polysaccharides and lipids. It is obtained by treating the bacterial cells/ plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells) and chitinase (fungus). DNA is interwined with proteins like histones. RNA can be removed by treatment with ribonuclease while proteins can be removed by treatment with protease. Other molecules are removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

Question 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.
Answer:
(c) : Transformation is a process by which a cell takes up naked DNA fragment from the environment, incorporates it into its own chromosomal DNA and finally expresses the trait controlled by the incoming DNA. Since DNA is a hydrophilic molecule, it cannot pass through membranes, so the bacterial cells must be made competent to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium (Ca²⁺) which increases the efficiency with which DNA enters the bacterium through pores in its cell wall. Recombinant DNA (rDNA) can then be forced into such cells by incubating the cell with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock), and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Question 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.
Answer:
(a) : DNA ligases (joining or sealing enzymes) are also called genetic gum. They join two individual fragments of double  straqded DNA by forming phosphodiester bonds between them. Thus, they help in sealing gaps in DNA fragments. Therefore, they act as a molecular glue.

Question 14.
Which of the following is not a source of restriction endonuclease?
(a) Haemophilus influenzae
(b) Escherichia coli
(c) Entamoeba coli
(d) Bacillus amyloliquifaciens
Answer:
(c)

Source	Restriction endonuclease
(1) Haemophilus influenzae	Hindll and HmdIII
(2) Escherichia coli	EcoRI and EcoRII
(3)Bacillus amyloliquefaciens	Bam HI

Question 15.
Which of the following steps are catalysed by Taq 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.
Answer:
(c) : Refer answer 10.

Question 16.
A bacterial cell was transformed with a recombinant DNA 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 codonsfor humans and bacteria are different
(c) human protein is formed but degraded by bacteria
(d) all of the above.
Answer:
(a) : Eukaryotic genes do not function properly when transferred into bacterial cell because introns are present in eukaryotic cells but are absent in prokaryotic cells. Hence, when bacterial cell is transformed with recombinant DNA which is generated using human gene, it could not process it. As a result, no desired protein will be produced.

Question 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
Answer:
(c) : The cells having cloned genes of interest can be grown on a small scale in the laboratory. The cultures may be used tor extracting and purifying the desired protein. The cells can also be multiplied in a continuous culture system where the used medium is passed out from one side and fresh medium is added from the other side to maintain the cells in their physiologically most active log/ exponential phase – rapid multiplication of the cells. This type of culturing method produces a larger biomass to get higher yields of desired protein.

Question 18.
Who among the following was awarded the Nobel Prize for the development of PCR technique?
(a) Herbert Boyer
(b) Hargovind Khurana
(c) KaryMullis 
(d) Arthur Kornberg
Answer:
(c)

Question 20.
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 produces the same kind of sticky ends in , different DNA molecules.
Answer:
(c) : More than 900 restriction enzymes have been isolated from over 230 strains of bacteria each of which recognise different recognition sequences. No restriction enzyme has been isolated from viruses.

Very Short Answer Type Questions

Question 1.
How is copy number of the plasmid vector related to yield of recombinant protein?
Answer:
The copy number is the number of copies of plasmids compared with the amount of chromosomal DNA found in single bacterial cell. The copy number influences the stability of plasmid. The rDNA can multiply as many times as the copy number of vector plasmid. So, higher the copy number of plasmid, higher would be the copy number of gene and protein yield would be high.

Question 2.
Would you choose an exonuclease while producing a recombinant DNA molecule?
Answer:
For producing recombinant DNA, exonuclease is not added because it removes nucleotides from terminal ends (either 5′ or 3′) and it acts on single strand of DNA. It does not recognise any specific sequence.

Question 3.
What does ‘H’, ‘in’, ‘d’ and ‘III’ refer to in the enzyme Hindlll?
Answer:
ln naming of restriction endonuclease, the first letter of enzyme is the first letter of bacterium’s, generic name, “H”-Haemophilus, “in” represents two letters of species name- influenzae, “d” is for strain “Rd”, and “HI” indicates the order in which enzyme is synthesised.

Question 4.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
Restriction enzymes should preferably have single recognition site because presence of more than one recognition sites within the vector will generate several fragments, which will complicate the gene cloning.

Question 5.
What does ‘competent’ refer to in competent cells used in transformation experiments?
Answer:
Competent cells are the cells that have altered cell wall so that they can take up foreign DNA easily. Competent cell is the one that is able to undergo genetic transformation.

Question 6.
What is the significance of adding proteases at the time of isolation of genetic material (DNA)?
Answer:
Genes located on the DNA are interwined with proteins such as histones, which are removed by proteases. It is important to add protease during DNA isolation because DNA has to be free from any other .macromolecule such as RNA, proteins to get cut with restriction endonuclease.

Question 7.
While doing a PCR/denaturation’step is missed. What will be its effect on the process?
Answer:
Denaturation is the first step in PCR reaction. If denaturation does not take place, annealing and extension will also not take place. Therefore, there will be no amplification.

Question 8.
Name a recombinant vaccine that is currently being used in vaccination program.
Answer:
HepaH tis-R surface antigen (HBsAg) is currently used as vaccine in vaccination programme against hepatitis virus.

Question 9.
Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?
Answer:
DNA and protein do not show biological activity in anhydrous conditions because in non-aqueous conditions rigidity of biomolecules increases due to weakening of hydrogen bond strength.

Question 10.
What modification is done on the Ti plasmid Agrobacterium tumefaciens to convert it into a cloning vector?
Answer:
Agrobacterium tumefaciens, a pathogen of several dicot plants is able to deliver T-DNA to transform normal cell into tumour and direct tumour cells to produce chemicals required by pathogen. The Ti (tumor inducing) plasmid has been modified (disarmed) by removing gene responsible for causing tumour and inserting gene to be used as selectable marker. Modified plasmid is non-pathogenic to plants and delivers the gene of interest.

Short Answer Type Questions

Question 1.
What is meant by gene cloning?
Answer:
Gene cloning is formation of multiple identical copies of any template DNA. For cloning, an alien DNA is linked with origin of replication of a vector, so that it can replicate and multiply itself in the host organism.

Question 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?
Answer:
Biotechnology is technological employment of biological entities and processes to generate products and services useful to man. In this case a wine maker has utilised a strain of yeast, which is commonly used for making wine by fermentation. The molecular biologist has used cloned gene for an antigen and antigen is used as vaccine. Thus, both can be considered as biotechnologists.

Question 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?
Answer:
Recombinant DNA is formed by ligating an alien DNA with plasmid DNA, a circular, self-replicating, extra-chromosomal, dsDNA. Therefore, rDNA is also circular DNA and exonucleases cleave DNA at terminal ends (either 5′ or 3′) in a linear DNA segment. Thus, addition of exonuclease to a tube containing rDNA would not affect bacterial transformation.

Question 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?
Answer:
If endonucleases used in the recombinant DNA technology do not cut DNA at specific locations, they may cut the DNA at random location. The sticky ends will not be generated to ligate the DNA segments and recombinant DNA would not be formed.

Question 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.
Answer:
A plasmid DNA is a circular DNA. When this circular DNA having one site for restriction endonuclease is cleaved with an enzyme, it would form a strand of linear DN? and would be visible as single band on agarose gel. On the other hand, when linear DNA with one restriction site is cleaved, it would produce two fragments and thus two bands would be seen.

Question 6.
How does one visualise DNA on an agarose gel?
Answer:
The separated DNA fragments on an agarose gel can be visualised only after staining the DNA with ethidium bromide followed by exposure to UV radiations as bright orange coloured bands.

Question 7.
A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment?
Answer:
If a cloning vector does not have a selectable marker, then it would not be possible to distinguish between transformants (host bacterium having rDNA) and non-transformants. Therefore, an ideal cloning vector should have selectable markers for the selection of transformants.

Question 8.
A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gef with ethidium bromide, no DNA bands were observed. What could be the reason?
Answer:
There could be various reasons for DNA bands not visible after electrophoresis, such as:

1. DNA might be impure, i.e., have not been isolated properly and is associated with RNA and proteins.
2. Restriction endonucleases have not been added in appropriate amount.
3. Electrodes might not have been connected correctly.
4. Stained bands have not been visualised under UV rays.
5. Electrodes were put in opposite orientation in the gel assembly, that is anodes towards the well (where DNA is loaded).
6. Ethidium bromide has not been added to stain DNA fragments.

Question 9.
Describe the role of CaCI₂ in the preparation of competent cells.
Answer:
Competent cell is one which can take up naked DNA fragment and incorporate it into its own chromosomal DNA. DNA is a hydrophilic molecule and. cannot pass through membranes, so bacterial cell can be made competent by CaCH method. In this method, bacterial cell is treated with divalent cation Ca²⁺ which increases the efficiency with which DNA enters the bacterium through pores in its cell wall.

Question 10.
What would happen when one grows a recombinant bacterium in a bioreactor but forgets to add antibiotic to the medium in which the recombinant is growing?
Answer:
Antibiotics do not allow other bacteria to grow in the medium. In the absence of antibiotics, the desirable bacteria may not be able to grow to its optimum level. The plasmids may also be lost, since maintaining a high copy number of plasmids is a metabolic burden on the bacteria.

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

Answer:
Polymerase Chain Reaction (PCR) involves amplification of gene of interest.
A single PCR amplification cycle involves three basic steps:

1. A- Denaturation : It is separation of two strands of DNA by heating the DNA strand at a temperature of 94 – 96 °C. Each single strand of target DNA is used as template DNA.
2. B-Annealing : Hybridisation of two oligo-nucleotide primers to each of the ssDNA template. It is carried out at a temperature of 40- 60 °C.
3. C-Extension : It is the final step of PCR. In this, enzyme Taq DNA polymerase synthesises the DNA region between the primers, using dNTPs and Mg²⁺.

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

Answer:
A – BarnHI
B – PtI
C – ampR

Long Answer Type Questions

Question 1.
For selection of recombinants, insertional inactivation of antibiotic marker has been superceded by insertional inactivation of a marker gene coding for a chormogenic substrate. Give reasons.
Answer:
Selection of recombinants by insertional inactivation of an antibiotic has been superceded by insertional inactivation of selectable marker because selection of recombinants due to inactivation of antibiotics is a cumbersome procedure, because it requires simultaneous plating on two plates having different antibiotics (ampicillin and tetracyiine). Therefore, alternative selectable markers have been developed which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme p-galactosidase.

This results in the inactivation of the enzyme (insertional inactivation). The presence of chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. Presence of insert results into insertional inactivation of the (1-galactosidase and the colonies do not produce any colour, and these are identified as recombinant colonies.

Question 2.
Describe the role of Agiobacterium tumefaciens in transforming a plant cell.
Answer:
Aemhactprium tumefaciens is a bacterium that causes tumours in plants. It has ability to transform plant cells. For this reason it has become an important tool in plant improvement by genetic engineering.
The plasmid is disarmed by deletion of the tumour inducing gene. These modified bacteria can still transform plant cells. The part of Ti plasmid transferred into plant cell DNA, is called T-DNA. This T-DNA with desired DNA spliced into it, is inserted into the chromosomes of the host plant where it produces copies of itself. But it no longer produces tumours. Such plant cells are then cultured, induced to multiply and differentiate to form plantlets. Thus the ability of Agrobactcrium to transfer genes to the plants has been exploited in genetic engineering for plant improvement programme.

Question 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.
Answer:
Bioreactors are vessels in which raw materials are biologically converted into specific products by microbes, plant and anirhal cells and their enzymes. They are allowed to synthesise the desired proteins which are finally extracted and purified from cultures.

Small volume of cultures are usually employed in laboratories for research and production of less quantities of products. Large scale production of products is carried out in bioreactors. The most commonly used bioreactors are of stirring type, that have provision for batch culture or continuous culture. In continuous culture, the culture medium is added and the product is taken out continuously.

Flask is used in laboratory for testing and provides batch type of culture. With flask no gadget can be connected. Bioreactors are used commercially and provide continuous culture. With a bioreactor, gadgets can be attached to give better functioning. A bioreactor provides optimal conditions for optimal growth (temperature pH, substrate, salts, vitamins, oxygen).

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 11Biotechnology: Principles and Processes help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 11Biotechnology: Principles and Processes, drop a comment below and we will get back to you at the earliest.