NCERT Exemplar Class 12 Biology Solutions Chapter 11 Biotechnology Principles And Processes

Detailed Answers to the NCERT Exemplar for Class 12 Science Chapter 11 Biotechnology: Principles and Processes(Multiple Choice Questions)

The detailed answers to the MCQ solutions are given below:

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:

The answer is option (b), Production of CO₂
Explanation: Carbon dioxide creates bubbles in the dough and which leads to the rising of the dough.

Question:2

Which of the following enzymes catalyze the removal of nucleotides from the ends of DNA?
a. endonuclease
b. exonuclease
c. DNA ligase
d. Hind – II

Answer:

The answer is option (b), exonuclease
Explanation: These are of two kinds; exonucleases and endonucleases. Exonucleases remove nucleotides from the ends of the DNA whereas endonucleases make cuts at Specific positions within the DNA. DNA ligase facilitates joining of sticky ends of DNA fragments. Hind-II is a type of endonuclease.

Question: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

Answer:

The answer is option (a) Transduction
Explanation: Transfer of genetic materials from a virus or bacterium is called transduction.

Question: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

Answer:

The answer is option (d), Ethidium bromide-stained DNA can be seen under exposure to UV light
Explanation: DNA cannot be seen in visible light, so options a and b are incorrect. Staining with ethidium bromide and the presence of UV light are necessary to see DNA.

Question:5

'Restriction' in Restriction enzyme refers to:
a. Cleaving of the phosphodiester bond in DNA by the enzyme
b. Cutting of DNA at a specific position only
c. Prevention of the multiplication of bacteriophage by the host bacteria
d. All of the above

Answer:

The answer is option (c), Prevention of the multiplication of bacteriophage in bacteria
Explanation: Restriction enzymes belong to a class of enzymes called nuclease. Restriction enzymes prevent the multiplication of bacteriophages in bacteria. They are used for cutting DNA at a specific position, which becomes possible by cleaving of phosphodiester bond in DNA by the enzyme. Option 'c' is the correct answer.

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:

The answer is option (d) E. coli
Explanation: E. coli was used in several experiments involving DNA replication and it has no role in preparation of DNA molecules.

Question:7

In agarose gel electrophoresis, DNA molecules are separated on the basis of their:
a. Charge the only
b. Size only
c. Charge to size ratio
d. All of the above

Answer:

The answer is option (b), Size only
Explanation: (b) Size only

Question:8

The most important feature in a plasmid to serve as a vector in a 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

Answer:

The answer is option (a), Origin of replication (ori)
Explanation: Origin of replication is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA. Therefore, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin support high copy number.

Question:9

While isolating DNA from bacteria, which of the following enzymes is not required?
a. Lysozyme
b. Ribonuclease
c. Deoxyribonuclease
d. Protease

Answer:

The answer is option (c), Deoxyribonuclease
Explanation: Lysozyme is used for breaking the membranes which surround DNA. Ribonuclease is used for removing RNA. Protease is used for removing protein. All these steps are necessary for the isolation of DNA from bacteria.

Question:10

Which of the following contributed to popularising the PCR (polymerase chain reaction) technique?
a. Easy availability of DNA template
b. Availability of synthetic primers
c. Availability of cheap deoxyribonucleotides
d. Availability of 'Thermostable' DNA polymerase

Answer:

The answer is option (d), Availability of 'Thermostable' DNA polymerase
Explanation: Thermostable DNA is isolated from the bacterium Thermus aquaticus. It remains active during high temperature-induced denaturation of DNA. Before the discovery of thermostable polymerase, the polymerase in use suffered denaturation during the heat treatment step of the process.

Question: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

Answer:

The answer is option (b), Transformed cells
Explanation: During culture, only transformed cells survive on agar plate. These cells show resistant to antibiotic and can be easily recognized.

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:

The answer is option (c), Uptake of DNA through transient pores in the bacterial cell wall
Explanation: Putting the cells with recombinant DNA on ice; followed by a heat shock and then putting them back on ice helps in uptake of DNA through transient pores in the bacterial cell wall.

Question:13

The role of DNA ligase in the construction of a recombinant DNA molecule is:
a. Formation of a 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:

The answer is option (a), Formation ofa phosphodiester bond between two DNA fragments
Explanation: DNA ligase facilitates joining of two sticky ends of DNA fragments by formation of a phosphodiester bond between them.

Question: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

Answer:

The answer is option (c), Entamoeba coli
Explanation: Entamoeba coli is not a bacterium and hence is not a source of restriction endonuclease.

Question:15

Which of the following steps are catalyzed 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

Answer:

The answer is option (c), Extension of primer end on the template DNA
Explanation: (c) Extension of primer end on the template DNA

Question: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. A human gene may have an intron that 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

Answer:

The answer is option (a) Human gene may have an intron that bacteria cannot process
Explanation: Any nucleotide sequence in a gene that is removed by RNA splicing during maturation of the final RNA product is called an intron. Intron can be different in different organisms.

Question:17

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

Answer:

The answer is option (c), A continuous culture system
Explanation: A large flask, without a suitable culture system, is not going to yield anything. The same holds for a tank bioreactor. It is the continuous culture system that is a must for producing a recombinant protein.

Question:18

Who among the following was awarded the Nobel Prize for the development of the PCR technique?
a. Herbert Boyer
b. Hargovind Khurana
c. Kary Mullis
d. Arthur Kornberg

Answer:

The answer is option (c), Kary Mullis
Explanation: (c) Kary Mullis

Question:19

Which of the following statements does not hold for a restriction enzyme?
a. It recognizes 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

Answer:

The answer is option (c). It is isolated from viruses
Explanation: It is isolated from bacteria.

Find Answers to Chapter 11 NCERT Exemplar for Class 12 Science Biotechnology: Principles and Processes(Very short answer Questions)

The detailed answers to the very short questions are given below:

Question:1

How is the copy number of the plasmid vector related to the yield of recombinant protein?

Answer:

A higher number of copies of the plasmid vector helps in producing a large quantity of recombinant protein.

Question:2

Would you choose an exonuclease while producing a recombinant DNA molecule?

Answer:

Exonuclease removes nucleotides from the ends of the DNA, and hence it cannot help in producing circular DNA. So, exonuclease cannot be used for making a recombinant DNA molecule.

Question:3

What does H in 'd' and 'III' refer to in the enzyme Hind III?

Answer:

In the enzyme Hind III, 'H in' refers to Haemophilus influenza, D refers to the strain of H. influenza, and III refers to the sequence in which this enzyme was discovered.

Question:4

Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.

Answer:

Presence of more than one recognition site on vector will generate many fragment of DNA; which will complicate the matter. Hence, restriction enzymes should have one recognition site at the vector.

Question:5

What does 'competent' refer to in incompetent cells used in transformation experiments?

Answer:

The ability of the bacterial cell to take up DNA through pores in the cell wall is called competence. DNA is a hydrophilic molecule, and hence, it cannot pass through the cell membrane. So, the cell is made 'competent' by treating with suitable divalent ion; like calcium.

Question:6

What is the significance of adding proteases at the time of isolation of genetic material (DNA)?

Answer:

Protease helps in removing protein during the process of obtaining pure DNA. Other macromolecules are eliminated with the help of suitable enzymes during this process. In the end, pure DNA is isolated.

Question:7

While doing a PCR, 'denaturation' step is missed. What will be its effect on the process?

Answer:

If denaturation is missed, then primers will not be able to join at the template. This will result in no extension, no amplification, and thus a large number of copies of DNA cannot be made.

Question:8

Name a recombinant vaccine that is currently being used in the vaccination program.

Answer:

Hepatitis B vaccine

Question:9

Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?

Answer:

Biomolecules do not exhibit biological activity in anhydrous conditions. DNA may get damaged under anhydrous condition but has the ability to repair later on. Protein molecule may get denatured under anhydrous conditions.

Question:10

What modification is done on the Ti plasmid of Agrobacterium tumefaciens to convert it into a cloning vector?

Answer:

Ti plasmid in Agrobacterium has the ability to induce tumours in plants. This plasmid is 'disarmed' by suitable modification, and then it can be used as a cloning vector for delivering a gene of interest to plants and animals.

Get Answers to the Class 12 Science Chapter 11 NCERT Exemplar of Biotechnology: Principles and Processes( Short answer Questions)

The detailed answers to the short answer questions are given below:

Question:1

What is meant by gene cloning?

Answer:

A set of experimental methods used to assemble recombinant DNA molecules and to use them for cloning in a host organism is called gene cloning. Gene cloning involves the following main steps:

1. Selection of a suitable gene.
2. Treating the gene to obtain small fragments.
3. Transferring the fragment to a suitable host.

Question:2

Both a winemaker and a molecular biologist who had developed a recombinant vaccine claim to be biotechnologists. Who, in your opinion, is correct?

Answer:

Biotechnology can be defined as a set of methods to use live organisms to produce products and processes for the benefit of humankind. Therefore, it is correct to include a winemaker as well as a molecular biologist under the category of biotechnologist. They have developed a recombinant vaccine which will increase the production of a vaccine for human welfare.

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:

When a DNA molecule is created by ligating a gene to a plasmid vector, it becomes a circular DNA which is ready to replicate in the host organism. Addition of exonuclease is not going to affect the process after this stage because the DNA does not have a free end, and hence enzyme exonuclease will not get a substrate to show its action. Therefore, in this experiment, bacterial transformation is not going to be disturbed.

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:

Specific-recognition sequence in a DNA provide sticky ends at which recombination of genes takes place. This further leads to the replication of the selected gene. If endonuclease fails to cut DNA at specific-recognition sequence; then recombination or replication will fail to take place.

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, the plasmid shows one DNA band while the linear DNA shows two fragments. Explain.

Answer:

A circular DNA opens up to resemble a single linear DNA. A Linear DNA is divided into two fragments after cleavage. Hence, circular DNA shows one band, while linear DNA shows two bands.

Question:6

How does one visualize DNA on an agarose gel?

Answer:

DNA fragments separate when they are moved towards the anode in an electric field. The agarose gel provides the matrix through which DNA fragments separate due to a sieving effect. Separated DNA fragments can be visualized only after staining with ethidium bromide and then by exposure to UV radiation. After staining, DNA fragments appear as bright orange bands.

Question:7

A plasmid without a selectable marker was chosen as a vector for cloning a gene. How does this affect the experiment?

Answer:

Selectable marker helps in identifying and eliminating non-transformant DNA and in selectively permitting the growth of transformants. In the absence of a selectable marker, it will not be possible to differentiate between transformants and non-transformants. Thus, carrying the experiment to its logical end would be impossible in the absence of a selectable marker.

Question: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?

Answer:

The following are the possible reasons for the non-observation of DNA bands:
(a) DNA may have been contaminated because of the accidental addition of a nuclease enzyme.
(b) Electrodes may have been put in the wrong orientation with the anode towards the loading well. As DNA is negatively charged, it moves towards the anode. When the anode is near the loading well, separated DNA may move out of the gel.
(c) Quantity of ethidium bromide may not have been sufficient.

Question:9

Describe the role of CaCl₂ in the preparation of competent cells.

Answer:

Divalent ions increase the efficiency of uptake of DNA through the pores in the bacterial cell wall. CaCl₂ provides the divalent ion Ca^2+, which creates transient pores on the bacterial cell wall, which facilitate entry of foreign DNA into the bacterial cells. Loading of divalent ions makes the cell competent.

Question:10

What would happen when one grows a recombinant bacterium in a bioreactor but forgets to add an antibiotic to the medium in which the recombinant is growing?

Answer:

In the absence of an antibiotic, the recombinant bacterium does not need to produce a gene, which can make it resistant to the antibiotic. In other words, there is no pressure on the recombinant bacterium to make the desirable gene. Thus, in the absence of an antibiotic, a gene of interest will not be produced by the recombinant bacterium.

Question:11

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

Answer:

A: Denaturation, B: Annealing, C: Extension

Question:12

Name the regions marked A, B and C.

Answer:

A shows a tetracycline resistance site
B shows restriction site Pst I
C shows an ampicillin resistance site

Comprehensive Answers to the NCERT Exemplar Class 12 Science Chapter 11 of Biotechnology: Principles and Processes (Long Answer Questions)

The detailed answers to the long-answer questions are given below:

Question:1

For the selection of recombinants, insertional inactivation of the antibiotic marker has been superseded by insertional inactivation of a marker gene coding for a chromogenic substrate. Give reasons.

Answer:

A marker gene helps in differentiating between transformant genes and non-transformant genes. This helps in selecting the suitable recombinants. In the case of E. coli, pBR322 is the vector for resistance to the antibiotic tetracycline. The insertional inactivation of pBR322 will result in loss of resistance to tetracycline by E. coli. This can be found out by growing the recombinants on two plates: one containing tetracycline and another containing ampicillin. The recombinant will grow in ampicillin but not in tetracycline.
A marker gene for chromogenic substrate helps in identifying the recombinant DNA on the basis of gain or loss of colour from the chromogenic substrate. Insertional inactivation of a marker gene coding for a chromogenic substrate will result in no blue colour imparted in the colony. These steps are taken for easy identification of recombinants from non-recombinants.

Question:2

Describe the role of Agrobacterium tumefaciens in transforming a plant cell.

Answer:

Agrobacterium tumefaciens infects walnuts, grapevines, sugar beets, horseradish, etc.
The role of Agrobacterium tumefaciens in transforming a plant cell can be described as:

1. Agrobacterium tumefaciens infects plants usually through an open wound.
2. Once it enters its plant host, it injects a section of its DNA called the T-DNA which is derived from its Ti (tumour-inducing) plasmid into its host.
3. The T-DNA first directs the plant cells to make auxins and cytokinins, which causes the cells to become irregularly shaped and form a visible tumour called a gall.
4. The T-DNA then directs the plant cell to start making opines (usually nopaline or atropine), which A. tumefaciens uses as an energy source. Thus, A. tumefaciens creates a special niche for itself inside the gall.

Thus, this property of Ti-plasmid has been exploited for cloning of the gene of interest and stably integrating them in the plant genes. Therefore, by using Ti-plasmid or its derivatives, recombinant plant cells with desired genes of interest stably integrated into the plant genome have been successfully produced.

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:

(a) Simple stirred-tank bioreactor;
(b) Sparged stirred-tank bioreactor through which sterile air bubbles are sparged
Structure of Bioreactor:

1. It is a cylindrical structure with a curved base.
2. A stirrer is present for even mixing and oxygen availability throughout the reactor.
3. There is an agitator system, an oxygen delivery system, a foam control system, a temperature control system, etc.
4. There is a sampling port through which small volumes of culture can be taken out periodically.

A flask in a laboratory cannot be used for producing recombinant DNA on a large scale. Unlike a bioreactor, a flask cannot be used to grow culture continuously.