NCERT Solutions for Class 12 Science Biology Chapter 11 Biotechnology: Principles And Processes are provided here with simple step-by-step explanations. These solutions for Biotechnology: Principles And Processes are extremely popular among Class 12 Science students for Biology Biotechnology: Principles And Processes Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Biology Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 12 Science Biology are prepared by experts and are 100% accurate.

Page No 75:

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:

Rising  of dough is due to the production of COby yeast when is added to the flour and also mixed with lukewarm water and sugar. 
Hence, the correct answer is option (b).

Page No 75:

Question 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

Answer:

Restriction enzyme belongs to a larger class of enzyme called nucleases an it is of two types:-
Endonucleases and exonucleases
Exonucleases removes nucleotides from the ends of the DNA.
Hence, the correct answer is option (b).

Page No 75:

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 process of transfer of DNA from on  bacterium to another by a virusis called transduction.
Hence, the correct answer is option (a).

Page No 75:

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 DNA fragments separated by gel electrophoresis can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation.
Hence, the correct answer is option (d).



Page No 76:

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 by the host bacteria
d. All of the above

Answer:

The restriction enzyme are called 'molecular scissors. They are responsible for cutting DNA. The term 'restriction' refers to the function of these enzymes in restricting the propagation of forign DNA if bacteriphage in host bacterium i.e, cutting of DNA  at a specific position only.
Hence, the correct answer is option (c).

Page No 76:

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:

E.Coli is not required for the preparation of a recombinant DNA molecules as it is required for the expression of rDNA molecule.
Hence, the correct answer is option (d).

Page No 76:

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:

The DNA fragments can be separated by a technique known as gel electrophoresis. As, DNA Fragments are negatively charged molecules they can be separated by forcing them to move towards the anode under electric field through a medium/matrix. They are separated (Resolved) according to their size through sieving effect provided by the agarose gel.
Hence, the correct answer is option (b).

Page No 76:

Question 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

Answer:

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.
Hence, the correct answer is option (a).

Page No 76:

Question 9:

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

Answer:

During DNA isolation from bacteria, the DNA is enclosed within the membrane and there is need to break the cell open in order to release the DNA along with other macromolecules such as RNA, proteins, polysaccharides and also lipids.
It can be done by treating the bacterial cell with an lysozyme enzyme. The RNA can be removed by treatment with ribonuclease while proteins can be removed by the treatment with protease enzyme.
Hence, the correct answer is option (c).
 

Page No 76:

Question 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

Answer:

PCR stands for Polymerase chain reaction. Multiple copies of the gene(or DNA) of interest is synthesised invitro with the help of this reaction. The repeated amplification of DNA is achieved by the use of thermostable enzyme called  tag DNA polymerase. It is isolated from a bacterium thermus aquaticus. This remains active during the high temperature that induced the denaturation of double stranded DNA.
Hence, the correct answer is option (d).



Page No 77:

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 vector requires a selectable marker that helps in the identification and elimination of non-transformants and selectively permit the  growth of the transformants.
Hence, the correct answer is option (b).

Page No 77:

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:

Top force bacteria to take up the plasmid, the bacterial cells made competent to take up DNA. It can be done by treating the cell with a specific concentration of a divalent cation such as calcium, that increases the efficiency with which DNA enters the bacterium through pores in its cell wall. rDNA can then be forced into such cells by incubating the cells with rDNA on ice. followed by placing them briefly at 42ºc(heat shock) and then putting them back on ice.
Hence, the correct answer is option (c).

Page No 77:

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 purime and pyrimidine bases
d. None of the above

Answer:

The DNA ligase enzyme is used in gene cloning to join DNA molecules together to construct rDNA. This enzyme facilitates the joining of DNA strands together by catalyzing the formation of a phosphodiester bond.
Hence, the correct answer is option (a).

Page No 77:

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:

Escherichia coli, Haemophilus influenzae, and  Bacillus amyloliquefaciens are sources of restriction enzymes such as ECORI, Hind III and Bam HI.
Hence, the correct answer is option (c).

Page No 77:

Question 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

Answer:

Taq DNA polymerase is a DNA-dependent DNA polymerase.
The process of DNA polymerisation and extension of the primer end on the template strands of the DNA by using this enzyme in PCR,
Hence, the correct answer is option (c),

Page No 77:

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. 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

Answer:

A nucleotide sequence present in a gene is removed by RNA splicing during maturation of final RNA product is called intron but bacterial system does not possess the ability to splice RNA in order to remove introns to produce mRNA, hence translation could not take place and so the protein is not produced. 
Hence, the correct answer is option (a),



Page No 78:

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:

Recombinant protein is any protein that encode gene is expressed in a heterologous host. The cells harbouring cloned genes of interest are grown on a small scale in the laboratory. To extract the desired protein, the cultures are used and then are purified by using different separation techniques.
Such alls can be multiplied in a continuous culture system while the used medium is drained out from one side and fresh medium is added from the other to maintain the cells in their physiologically active log/exponential phase.
Hence, the correct answer is option (c). 

Page No 78:

Question 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

Answer:

Polymerase chain reaction (PCR) technique was developed by Kary Mullis in 1985. He also received a Nobel prize four this 1993.
Hence, the correct answer is option (c).

Page No 78:

Question 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

Answer:

Restriction enzymes are endonucleases that cleave phosphodiester bonds in the DNA backbone.
They are used in genetic engineering techniques to isolate specific segments of DNA. These enzymes are obtained from bacteria. 
For example:- ECORI, 
They are not isolated from viruses.
Hence, the correct answer is option (c).

Page No 78:

Question 1:

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

Answer:

The copy number of plasmid vector is directly related to the yield of recombinant protein.
More the copy number of plasmid vector, the copy number of genes also increases. This results in the higher yield of the recombinant protein is in higher amounts.

Page No 78:

Question 2:

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

Answer:

No, one cannot choose an exonucleases while producing a recombinant DNA molecule because an exonuclease remove nucleotides from the ends of the DNA while endonucleases make cuts at specific positions within the DNA.

Page No 78:

Question 3:

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

Answer:

The 'H' refers to the genus of an organism Haemophillus from which the enzyme has been isolated. While 'd' and 'III' refers to the strain type and sequence in which the restriction enzyme is isolated.

Page No 78:

Question 4:

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

Answer:

If a restriction enzyme have more than one site of action in the cloning site of a vector then it results in the formation of multiple and fragments and complicates the process of genetic engineering.

Page No 78:

Question 5:

What does ‘competent’ refer to in competent cells used in transformation experiments?

Answer:

A competent in 'competent cell' which is used in transformation experiments means the capability of a bacterial cell wall for uptake of hydrophilic DNA fragments when it is treated with divalent through the cell membrane.

Page No 78:

Question 6:

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

Answer:

During DNA isolation, DNA is released along with some other macromolecules such as RNA, proteins, polysaccharides and lipids. So, the RNA can be removed by treatment with ribonucleases whereas proteins can be removed by treatment with proteases.

Page No 78:

Question 7:

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

Answer:

If the initial step i.e, denaturation is missing during PCR, the dsRNA will not split into two template stands on which the reaction would take place.



Page No 79:

Question 8:

 

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

Answer:

Hepatitis B vaccine is a recombinant vaccine that is currently being used in a vaccination programme.

Page No 79:

Question 9:

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

Answer:

All the biological processes that occurs in the body will takes place in the presence of water or they are mediated by water. So, the biomolecules such as DNA, protein, will not exhibit any biological activity in anhydrous condition. As, they undergo denaturation under anhydrous conditions.

Page No 79:

Question 10:

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

Answer:

The Ti plasmid in Agrobacterium tumefaciens has the ability to transforms normal plant cells into a tumor in plants. This pathogen is now been modified/disarmed into a cloning vector that is no more pathogenic to the plants but is still able to deliver the desired genes of interest into a variety of plants.

Page No 79:

Question 1:

What is meant by gene cloning?

Answer:

The process of gene cloning involves the insertion of a desired of interest in DNA of the vector. They are used to introduce the gene of interest inside the host cell.

Page No 79:

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 deals with the techniques of using live organisms or enzymes from organisms to produce products and processes that are useful to humans. It is correct to say that both winemakers and molecular biologist are biotechnologist. As, the wine makers uses microorganisms such as yeast in order to convert it into wine. Whereas the scientists who devloped an recombinant vaccine is also called biotechnologist because they also used a vector to produce large number of pathogens.

Page No 79:

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:

The addition of an exonuclease enzyme to the tube that contains a recombinant DNA will not affect the further steps of gene transfer. The rDNA is a circular and closed plasmid without free ends. Hence, rDNA will not be acted on by exonucleases, as it removes nucleotides from the ends of the DNA.

Page No 79:

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:

The restriction enzyme cleaves the DNA at the specific sites. If the restriction enzyme is not able to cut the DNA molecule at specific positions, then it would not produce sticky ends and hence, the process of construction of rDNA would not be possible.

Page No 79:

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:

When the restriction endonuclease cuts the specific restriction site of the circular DNA of plasmid, the circular DNA changes into linear DNA. But, when the endonuclease cuts within the linear DNA, it would results in the formation of two linear DNA fragments.

Page No 79:

Question 6:

How does one visualise DNA on an agarose gel?

Answer:

The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by the exposure to ultra violet radiation.

Page No 79:

Question 7:

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

Answer:

Selectable markers helps in the identification and elimination of non-transformation and selectively permit the growth of the transformants. If selectable markers are not present in the plasmid which is chosen as a cloning vector then it will not be possible to differentiate between transformants and non-transformants.

Page No 79:

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 DNA bonds are not observed due to following reasons:-
(i) Insufficient concentration of Ethidium bromide as it is responsible for the intercalation within the bases of the DNA
(ii) Improper concentration of gel
(iii) Gel mixture got contaminated with nucleases.

Page No 79:

Question 9:

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

Answer:

Divalent cation such as CaClincreases the efficiency with which the DNA molecule enters the bacterium through transient pores in its cell wall.

Page No 79:

Question 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?

Answer:

Antibiotics present in the medium ensures that the bacteria retain the plasmid that contains the antibiotic resistant gene. While in the absence of antibiotic, the bacteria will grow successfully and there will be no selection pressure.
Hence, in a bioreactor, the quantity of a desired product will be reduced.



Page No 80:

Question 11:

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

Answer:

Step A:
Denaturation: In this step, heat treatment is provided to the PCR mixture, containing the DNA. This leads to denaturation i.e., breaking of H-bonds between the two strands of the DNA.
(b) Annealing: In this step of PCR, the single strands of DNA are obtained, and are hybridized with primers in the presence of DNA polymerase enzyme.
(c) Extension of Primers: DNA polymerase adds complementary nucleotides to the DNA strand to form a new double-stranded DNA. 
Hence, doubling the initial DNA content. 

Page No 80:

Question 12:

Name the regions marked A, B and C.


 

Answer:

The A, B, C in the E.Coli cloning vector are:-
A- Barn HI
B- Pst I
C- amPR



Page No 81:

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:

Selectable markers are developed to differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of chromogenic substrate. In this process, a recombinant DNA is inserted within the coding sequence of an enzyme, B-galactosidase. This results into inactivation of the gen for synthesis of B-galactosidase enzyme and this process is referred to as insertional inactivation. So, the presence of a chromogenic substrate give blue coloured colonies if the plasmid in the bacteria does not have an insert. Whereas the presence of insert results into the insertional inactivation of the B-galactosidase gene and the colonies do not produce any colour and are identified as recombinant colonies.

Page No 81:

Question 2:

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

Answer:

Agrobacterium tumifaciens is a pathogen of several dicot plants. It is able to deliver a piece of DNA known as 'T-DNA' to transform normal plant cells into a tumor  cell. It direct these tumor cells to produce the chemicals required by the pathogen. The Ti plasmid (Tumor inducing) plasmid of this bacterium has now been modified into a cloning vector that is no more pathogenic to the plants but is able to use the mechanism to deliver desired genes of interest into a variety of plants.

Page No 81:

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:

  • Flask which is used in laboratory is for small scale testing of culture while bioreactor which is used for commercial-scale production.
  • Small volume culture in flask cannot yield appreciable quantities of products while in bioreactor, large quantities are cultured.



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