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

#CBSE Class 12th

Irshad AnwarUpdated on 04 Jun 2025, 01:49 PM IST

NCERT Exemplar Solutions Class 12 Biology Chapter 11, Biotechnology Principles and Processes, is a valuable study material for students. It covers all types of questions, such as MCQs, short answers, and long answers, making it easier for students to grasp the concepts. It becomes easier to understand the fundamentals of biotechnology, like genetic engineering, recombinant DNA technology, and applications in agriculture and medicine. The NCERT Exemplar Solutions of this chapter cover important topics such as restriction enzymes, vectors, and cloning for clarity through the questions and well-explained answers.

This Story also Contains

NCERT Exemplar Class 12 Biology Solutions Chapter 11
Get NCERT Exemplar for Class 12 Biology Chapter 11(MCQs)
Find NCERT Exemplar for Class 12 Biology Chapter 11(Very short answer Questions)
Know NCERT Exemplar for Class 12 Biology Chapter 11(Short Answer Questions)
Access NCERT Exemplar Class 12 Science Chapter 11(Long Answer Questions)
Approach to Solve Questions of Class 12 Biology Chapter 11
Important Topics of Class 12 Biology Chapter 11
Important Question in Class 12 Biology Chapter 11
NCERT Exemplar Class 12 Biology Chapter Wise

NCERT exemplar class 12 Biology solutions chapter 11 Biotechnology Principles And Processes

These NCERT Exemplar Class 12 Solutions help to understand complex topics in easy step-by-step answers that make it less difficult for the students to master techniques such as gene cloning and vectors. It also includes well-structured answers that match the CBSE exam pattern, helping students write the correct and well-prepared answers. These are extremely helpful for board examinations and competitive tests such as NEET, as everything is explained so clearly and is easy to revise. Through studying NCERT Exemplar Class 12 Biology Solutions, students can build a strong understanding of the topics and improve problem-solving skills and accuracy.

NCERT Exemplar Class 12 Biology Solutions Chapter 11

The chapter Biotechnology: Principles and Processes helps students understand how living organisms and biological systems are used to develop useful products. The questions come in various formats such as MCQs, short and long answers. Solving these helps students build a strong understanding of the concepts and prepare better for exams.

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

Apply

Get NCERT Exemplar for Class 12 Biology Chapter 11(MCQs)

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: All DNA molecules have a uniform negative charge, so during electrophoresis, they move based on size smaller fragments travel faster and farther through the gel than larger ones.

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: Kary Mullis was awarded the Nobel Prize in Chemistry in 1993 for developing the Polymerase Chain Reaction (PCR) technique, which revolutionized molecular biology by allowing rapid DNA amplification.

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: Restriction enzymes are actually isolated from bacteria, where they serve as a defense mechanism against invading viral DNA.

Find NCERT Exemplar for Class 12 Biology Chapter 11(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.

Recommended eBooks

Related eBooks & Sample Papers

CBSE Class 12 Sample Paper English Elective 2025-26 with Solution

52+ Downloads

CBSE Class 12 Sample Paper English Core 2025-26 with Solution

27+ Downloads

Know NCERT Exemplar for Class 12 Biology Chapter 11(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:

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

NEET Highest Scoring Chapters & Topics

Most Scoring concepts For NEET (Based On Previous Year Analysis)

Download EBook

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.
PCR

Answer:

A: Denaturation, B: Annealing, C: Extension

Question:12

Name the regions marked A, B and C.
plasmid

Answer:

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

Access NCERT Exemplar Class 12 Science Chapter 11(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:

Agrobacterium tumefaciens infects plants usually through an open wound.
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.
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.
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:

It is a cylindrical structure with a curved base.
A stirrer is present for even mixing and oxygen availability throughout the reactor.
There is an agitator system, an oxygen delivery system, a foam control system, a temperature control system, etc.
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.

NCERT Exemplar Class 12 Solutions Subject wise:

Approach to Solve Questions of Class 12 Biology Chapter 11

To answer Biotechnology Principles and Processes questions well, adopt this easy-to-follow approach:

Understand that the principles of biotechnology are based primarily on two fundamental truths: genetic engineering (altering DNA to incorporate new characteristics) and bioprocess engineering.
Understand the major steps of genetic engineering, which are purifying the desired DNA, placing it into a vector (such as a plasmid), introducing the vector into a host organism, and replicating the recombinant DNA individually to create lots of copies.
Understand that restriction enzymes (molecular scissors) cleave DNA at precise sites, and ligases seal DNA fragments to form recombinant DNA.
Practice diagramming and describing the steps involved in recombinant DNA technology, such as the use of vectors, hosts, and selectable markers.
Revise the NCERT textbook and practice previous years' questions to become familiar with diagrams, terminology, and question types from this chapter

Also, Read the NCERT Solution subject-wise

Important Topics of Class 12 Biology Chapter 11

Biotechnology: Principles and Processes is a chapter that introduces students to the basics of biotechnology, including how biological systems and organisms are used to develop useful products and technologies.

Introduction to Biotechnology
Genetic Engineering
Recombinant DNA Technology
Cloning
Applications in Medicine
Applications in Agriculture
Environmental Biotechnology

Must Read NCERT Notes subject-wise

Important Question in Class 12 Biology Chapter 11

Biotechnology: Principles and Processes chapter covers the fundamental principles and techniques used in biotechnology, including genetic engineering, DNA manipulation, and the processes involved in producing recombinant DNA.

Question. What is biotechnology?
A. Study of fossils
B. Use of living organisms or their products to modify living or non-living materials for human use
C. Chemical synthesis of drugs
D. Study of plants only

Answer:
Biotechnology is the use of living organisms or their products to modify living or non-living materials to produce useful products for human benefit.
Hence, the correct option is B. Use of living organisms or their products to modify living or non-living materials for human use.

Also, check the NCERT Books and the NCERT Syllabus here

NCERT Exemplar Class 12 Biology Chapter Wise

Find all chapter-wise practice questions and solutions in the table below to strengthen your concepts and prepare effectively for exams.

NCERT Exemplar Class 12 Biology Chapter 1 Reproduction In Organisms

NCERT Exemplar Class 12 Biology Chapter 2 Sexual Reproduction In Flowering Plants

NCERT Exemplar Class 12 Biology Chapter 3 Human Reproduction

NCERT Exemplar Class 12 Biology Chapter 4 Reproductive Health

NCERT Exemplar Class 12 Biology Chapter 5 Principles of Inheritance And Variation

NCERT Exemplar Class 12 Biology Chapter 6 Molecular Basis of Inheritance

NCERT Exemplar Class 12 Biology Chapter 7 Evolution

NCERT Exemplar Class 12 Biology Chapter 8 Human Health and Disease

NCERT Exemplar Class 12 Biology Chapter 9 Strategies For Enhancement in Food Reproduction

NCERT Exemplar Class 12 Biology Chapter 10 Microbes in Human Welfare

NCERT Exemplar Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

NCERT Exemplar Class 12 Biology Chapter 12 Biotechnology and Its Applications

NCERT Exemplar Class 12 Biology Chapter 13 Organisms and Populations

NCERT Exemplar Class 12 Biology Chapter 14 Ecosystems

NCERT Exemplar Class 12 Biology Chapter 15 Biodiversity and Conservation

NCERT Exemplar Class 12 Biology Chapter 16 Environmental Issues

Frequently Asked Questions (FAQs)

Q: Why is practicing Exemplar problems important for Chapter 11?

Practising exemplar problems helps students build conceptual clarity, improve problem-solving skills, and prepare for varied question formats found in board exams and competitive tests like NEET. The questions help to understand and encourage the application of concepts in new situations.

Q: What is PCR, and why is it important in biotechnology?

PCR (Polymerase Chain Reaction) is a technique used to amplify specific DNA sequences, making millions of copies from a small initial sample. It is important in genetic testing, disease diagnosis, forensic analysis, and research.

Q: What types of questions are included in the NCERT Exemplar for this chapter?

The Exemplar provides a wide range of question types: multiple-choice questions (MCQs), fill in the blanks, match the following, true/false, and conceptual short and long answer questions. These are designed to test both basic understanding and higher-order thinking skills, making them valuable for board and entrance exam preparation.

Q: What are restriction enzymes and why are they important in biotechnology?

Restriction enzymes are molecular scissors that cut DNA at specific sequences, enabling scientists to isolate and manipulate genes. They are essential tools in genetic engineering, allowing for the creation of recombinant DNA molecules used in cloning, gene therapy, and research.

Q: What is the main focus of Chapter 11: Biotechnology – Principles and Processes in the NCERT Exemplar?

This chapter introduces the foundational principles and techniques of biotechnology, including genetic engineering, cloning, use of restriction enzymes, vectors, and the process of recombinant DNA technology. It emphasizes understanding both the theoretical concepts and practical applications relevant to modern biology and medicine.

Articles

NCERT Solutions for Class 12 Biology Chapter 3 – Reproductive Health

4 days ago

NCERT Resources

NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit

Aug 26 2025

NCERT Resources

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei

Aug 25 2025

NCERT Resources

NCERT Solutions for Class 12 Physics Chapter 12 - Atoms

Aug 25 2025

NCERT Resources

Certifications By Top Providers

People Management with Impact

Via Indian Institute of Management Bangalore

Forensic Science DNA Analysis

Via University of Cambridge, Cambridge

Basic 3D Animation using Blender

Via Indian Institute of Technology Bombay

Preparing for the AP English Language and Composition Exam

Via Tennessee Board of Regents

Consumer Behaviour

Via Indian Institute of Management Bangalore

Certificate Course Animation Technology

Via Manipal Institute of Communication, Manipal

1078 courses

816 courses

394 courses

281 courses

Harvard University, Cambridge

96 courses

University of Michigan, Ann Arbor

83 courses

IT & Software

Teaching and Academics

Programming and Development

Health, Fitness and Medicine

Business and Management

Engineering and Architecture

General Management

Financial Management

Environmental Studies

History

Psychology

Artificial Intelligence

Explore Top Universities Across Globe

University of Essex, Colchester

Wivenhoe Park Colchester CO4 3SQ

University College London, London

Gower Street, London, WC1E 6BT

The University of Edinburgh, Edinburgh

Old College, South Bridge, Edinburgh, Post Code EH8 9YL

University of Delaware, Newark

Newark, DE 19716

University of Nottingham, Nottingham

University Park, Nottingham NG7 2RD

Bristol Baptist College, Bristol

The Promenade, Clifton Down, Bristol BS8 3NJ

PR in Ireland After Study: Eligibility, Steps to Apply, Document Required

12 min ⋆ Sep 01, 2025 05:09 AM IST

Advantages and Disadvantages of Studying MBBS in Abroad 2025-26

9 min ⋆ Sep 01, 2025 05:09 AM IST

IELTS 2025 Exam: Test Dates (September), Registration, Syllabus, Fees

20 min ⋆ Sep 01, 2025 05:09 AM IST

PG in Abroad After MBBS for Indian Students 2025: Fees and Courses

11 min ⋆ Sep 01, 2025 05:09 AM IST

MCAT Syllabus 2025 - Check Section-wise Exam Syllabus & Pattern

8 min ⋆ Sep 01, 2025 05:09 AM IST

Is Studying MBBS Abroad in 2025 Worth It for Indian Students?

12 min ⋆ Sep 01, 2025 05:09 AM IST

Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer

gopi

4 Sep'25

Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

Read Complete Answer

Prachi Kumari

4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

Read Complete Answer

gopi

3 Sep'25

I gave my 12th exam this year in which I got less marks. Then I gave improvement exam in which also I could not score full marks. So can I become a private candidate and appear for the exam again next year or not?

Hello

Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

Read Complete Answer

SEJAL JAIN

20 Aug'25

i gave improvement 2025 as the roll no is different from my previous year marksheet so will i get new migration certificate

Hello Aspirant,

Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

Never miss an CBSE Class 12th update

Get timely CBSE Class 12th updates directly to your inbox. Stay informed!

Student Community: Where Questions Find Answers

Ask and get expert answers on exams, counselling, admissions, careers, and study options.

Download Careers360 App

All this at the convenience of your phone

Regular Exam Updates
Best College Recommendations
College & Rank predictors
Detailed Books and Sample Papers
Question and Answers

Scan and download the app

Applications Open

NEET Test Series

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

NCERT Exemplar Class 12 Biology Solutions Chapter 11

Aakash Repeater Courses

Get NCERT Exemplar for Class 12 Biology Chapter 11(MCQs)

Find NCERT Exemplar for Class 12 Biology Chapter 11(Very short answer Questions)

Related eBooks & Sample Papers

Know NCERT Exemplar for Class 12 Biology Chapter 11(Short Answer Questions)

Access NCERT Exemplar Class 12 Science Chapter 11(Long Answer Questions)

Approach to Solve Questions of Class 12 Biology Chapter 11

Important Topics of Class 12 Biology Chapter 11

Important Question in Class 12 Biology Chapter 11

NCERT Exemplar Class 12 Biology Chapter Wise

Questions related to CBSE Class 12th

Popular CBSE Class 12th Questions

Study Resources, Applications and Opportunities

Aakash Repeater Courses

NEET Most scoring concepts

NEET Previous 10 Year Questions

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

JEE Main high scoring chapters and topics

CBSE Class 12th Notifications

Latest updates

Never miss an CBSE Class 12th update

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

NVS

JNVST

Schools By Medium

By Ownership

By City

By State

Download Careers360 App

All this at the convenience of your phone

Scan and download the app