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NCERT Solutions for Class 12 Biology Chapter 9 Biotechnology Principles and Processes

NCERT Solutions for Class 12 Biology Chapter 9 Biotechnology Principles and Processes

Edited By Irshad Anwar | Updated on Jun 21, 2025 02:16 PM IST | #CBSE Class 12th
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The NCERT Solutions for Class 12 Biology Chapter 9 Biotechnology Principles and Processes is an important chapter that discusses the principles and methods applied to genetic engineering as well as microbial biotechnology. It also includes topics like recombinant DNA technology, restriction enzymes, cloning vectors, PCR, and bioprocess engineering. The NCERT Solutions include answers to all the questions in a clear language that allows students to understand the difficult and complex topics easily, which helps in better learning and improves exam preparation.

This Story also Contains
  1. Download PDF of NCERT Solutions For Class 12 Biology Chapter 9
  2. NCERT Solutions for Class 12 Biology Chapter 9(Exercise Questions)
  3. Approach to Solve Questions of Class 12 Chapter 9
  4. Important Question of Class 12 Biology Chapter 9
  5. What Extra Should Students Study Beyond the NCERT for NEET?
  6. NCERT Solutions for Class 12 Biology- Chapter-wise
NCERT Solutions for Class 12 Biology Chapter 9 Biotechnology Principles and Processes
NCERT Solutions for Class 12 Biology Chapter 9 Biotechnology Principles and Processes
The NCERT Solutions for Class 12 offer step-by-step explanations and practice questions that enable students to build a strong foundation in Biology and perform well in the exams. Biotechnology has transformed medicine, agriculture, and industry by providing creative solutions through genetic engineering and molecular biology methods. The concepts and answers of this chapter serve as the basis for developments such as gene therapy, genetically modified organisms (GMOs), and drug manufacture. The NCERT Solutions for Class 12 Biology help students revise important topics quickly before exams.


Also Read,

NCERT Exemplar for Class 12 Biology Biotechnology Principles and Processes

Download PDF of NCERT Solutions For Class 12 Biology Chapter 9

After going through the solutions of Biotechnology: Principles and Processes, students must be able to understand all the answers to the following questions:

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NCERT Solutions for Class 12 Biology Chapter 9(Exercise Questions)

Given below are the detailed answers to all the questions included in the Biotechnology: Principles and Processes chapter.

Q1. Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).

Answer:

Recombinant proteins are proteins produced as a result of recombinant DNA technology. In this technology, there is the transfer of some specific gene from one organism to another by using molecular tools such as biological vectors, restriction enzymes etc. Some of the proteins produced through RDT and are being used for therapeutic uses are as follows:

S.No
Name of the recombinant protein
Therapeutic use of the recombinant protein
1.
DNAase I
To treat cystic fibrosis
2.
Antithrombin III
To prevent the formation of blood clots
3.
Insulin
To treat type I diabetes mellitus
4.
Interferon α
Used for chronic hepatitis C
5.
Interferon AZA
Used for herpes and virus enteritis
6.
Coagulation factor VIII
To treat haemophilia A
7.
Coagulation factor IX
To treat haemophilia B
8.
Interferon B
To treat multiple sclerosis
9.
Human growth hormone recombinant
To promote growth in humans
10.
Tissue plasminogen activator
To treat the myocardial infection


Q2. Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces

Answer:

The following chart shows the action of the restriction enzyme EcoRI, the substrate DNA on which it acts and the site where it cuts

Action of Restriction enzyme


Q3. What would be the molar concentration of human DNA in a human cell? Consult your teacher.

Answer:

The molar concentration of DNA in a human cell will be the total no. of chromosomes multiplied by 6.023 × 1023

Hence, the molar concentration of DNA in each diploid cell in humans is 46 × 6.023 × 1023 = 2.77×1025 moles

Q4. Do eukaryotic cells have restriction endonucleases? Justify your answer.

Answer:

No, eukaryotic cells do not possess restriction enzymes. All the restriction endonucleases have been developed and isolated from different strains of bacteria. The bacteria possess these restriction endonucleases as a defence mechanism to restrict the growth of viruses. Their own DNA remain safe from these enzymes because it is methylated. The eukaryotic cell has RNA interference as a defence mechanism against foreign DNA. Thus, eukaryotic cells do not have restriction endonucleases.


Q5. Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?

Answer:

The advantages of stirred tank bioreactors over shake flasks are as follows:

1. Stirred tank bioreactors are utilised for large-scale production of biotechnological products, unlike the shake flask method which is used for small-scale production of products.

2. In stirred tank bioreactors, a small sample can be taken out for testing.

3. Stirred tank bioreactors have foam breakers to control the foam.

4. Stirred tank bioreactors have temperature and pH control systems.

Q6. Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following the base-pair rule

Answer:

Palindromic sequences in the DNA molecule refer to groups of bases forming the same sequence when read either backwardly or forwardly. The recognition sites of restriction endonucleases are palindromic sequences. Five examples of palindromic DNA sequences are given below

1. ACTAGT/TGATCA

2. AAGCTT/TTCGAA

3. GGATCC/CCTAGG

4. AGGCCT/TCCGGA

5. ACGCGT/TGCGCA


Q7. Can you recall meiosis and indicate at what stage a recombinant DNA is made?

Answer:

In meiosis, during the pachytene stage of Prophase I, crossing-over takes place and recombinant DNA is formed by combining portions of male and female DNA.

Q8. Can you think and answer how a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker?

Answer:

In recombinant DNA technology, the selection of transformed and non-transformed cells can be done using reporter genes that encode reporter enzymes. During the RDT experiment, the foreign gene is joined with a reporter gene. The reporter gene should be such that it produces a visible expression. For example, the Lac Z gene, which codes for the enzyme beta-galactosidase, is used as a reporter gene. The activity of this gene is not found in transformed cells as the product formed by its catalysis is not formed in transformed cells, and bacterial colonies appear white. In non-transformed cells, this gene shows its activity and the catalysed product is formed; as a result of this, bacterial colonies appear blue. Thus, a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker.

Q9. Describe briefly the following:

(a) Origin of replication

Answer:

Origin of replication- This refers to the DNA sequence, from where replication of DNA starts. By linking a DNA sequence with the origin of replication, it can be allowed to replicate in the host cells. Origin of replication also controls the copy number of linked DNA sequences.

(b) Bioreactor

Answer:

Bioreactors - These are large vessels (100-1000 litres) that are used for large-scale production of biotechnological products such as proteins, enzymes etc. from raw materials. In a bioreactor, optimum conditions such as temperature, pH, vitamins, oxygen, salts etc. are maintained. Stirred bioreactors are the most commonly used bioreactors. Stirred bioreactors can be simply stirred tank bioreactors or sparged tank bioreactors.


(c) Downstream processing

Answer:

Downstream processing- The process of separation and purification of biotechnological products is called downstream processing. The processes in downstream processing vary depending on the quality of the product. Before the release of the product, it undergoes clinical trials and quality control testing.


Q10. Explain briefly :

(a) PCR

Answer:

Polymerase Chain Reaction (PCR)- The molecular technique to amplify a gene and obtain its several copies is referred to as PCR. The process of PCR has certain requirements i.e. a thermostable enzyme called Taq polymerase ( obtained from Thermus aquaticus ), primers ( short stretches of DNA ), dNTPs, a template strand etc. The process of PCR takes place in three steps.

1. Denaturation- The double-stranded DNA helix is opened up by breaking their H-bonds at high temperatures.

2. Annealing- The primers are allowed to hybridise to complementary regions of DNA. This step takes place at 45-55°C temperature.

3. Extension- The primers are extended with the help of the Taq polymerase enzyme and the cycle is repeated several times to obtain the desired number of copies.

(b) Restriction enzymes and DNA

Answer:

Restriction enzymes and DNA- Restriction enzymes are those enzymes which cut DNA at particular places. The restriction enzyme first scans the DNA template and looks for its recognition site. Once it finds the recognition site, it binds at that region of DNA and cuts each of the two strands in its sugar-phosphate backbone. The sites at which restriction enzymes cut DNA are called recognition sites of DNA. These are palindromic sequences i.e. they read similarly from the backward and forward direction.

(c) Chitinase

Answer:

Chitinase - The enzyme that catalyses the breakdown of chitin polysaccharide which is usually found in the cell wall of fungi. Chitinase is mainly used during DNA isolation from fungi.


Q11. Discuss with your teacher and find out how to distinguish between

(a) Plasmid DNA and Chromosomal DNA

Answer:

The differences between plasmid DNA and chromosomal DNA are as follows:

Plasmid DNA
Chromosomal DNA
Circular, extra-chromosomal DNA which is capable of self-replication and is found in bacteria is called plasmid DNA.
The entire DNA (excluding extrachromosomal DNA) present in the cell constitutes chromosomal DNA
It is found only in bacteria
It is found in both bacteria and other eukaryotic cells.

(b) RNA and DNA

Answer:

The differences between RNA and DNA are as follows:

RNA
DNA
RNA contains ribose sugar
DNA contains deoxyribose sugar
In RNA, adenine and uracil are found as pyrimidines
In DNA, adenine and uracil are found as pyrimidines
It has catalytic properties and is less stable than DNA
DNA is non-catalytic and is more stable than RNA

(c) Exonuclease and Endonuclease

Answer:

The differences between exonuclease and endonuclease are as follows:

Exonuclease
Endonuclease
These are nucleases (enzymes) that cut DNA from its ends.
These are nucleases that cut DNA from internal sites on DNA


NCERT Solutions for Class 12- Subject-wise

Approach to Solve Questions of Class 12 Chapter 9

Given below are the steps which students can follow to solve the questions of the Biotechnology: Principles and Processes chapter effectively.

  • Students must start by understanding the basic concepts like genetic engineering, recombinant DNA technology, and cloning vectors such as plasmids.
  • Focus on the steps involved in the process of recombinant DNA technology, such as isolation of DNA, cutting with enzymes, amplification using PCR, and insertion into host cells.
  • Students should learn about the tools used in biotechnology, including restriction enzymes, ligases, and vectors, along with their specific functions and examples.
  • Short notes of important definitions, steps, diagrams (like PCR cycle and gel electrophoresis), and important terms can be prepared for quick revision before exams.
  • Practice previous year questions and diagram-based MCQs to get familiar with the exam pattern and to improve the speed and accuracy.
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Important Question of Class 12 Biology Chapter 9

Question: Among the following statements, carefully identify the ones that contain incorrect information about the palindromic sequences.

A. Palindromic sequences are DNA sequences that read the same in both directions but have an antiparallel orientation.

B. Palindromic sequences are DNA sequences that read the same in both directions but have a parallel orientation.

C. Palindromic sequences are recognised by a specific restriction endonuclease, leading to specific cleavage.

D. Palindromic sequences are recognised by exonucleases, resulting in non-specific cleavage.

Choose the option(s) that contain incorrect statements:

Options:

  1. A and C

  2. A and D

  3. B and C

  4. B and D

Answer: The correct answer is option (4).

Explanation:

Statement B states that the Palindromic sequences are DNA sequences that read the same in both directions but have a parallel orientation. This statement is incorrect. Palindromic sequences have an antiparallel orientation, not a parallel one. Therefore, statement B is incorrect.

Statement D states that the palindromic sequences are recognized by exonucleases, resulting in non-specific cleavage. This statement is incorrect. Exonucleases degrade DNA from the ends, but they do not specifically recognize and cleave palindromic sequences. Therefore, statement D is incorrect.

Hence, the correct statements are B and D.

NCERT Exemplar Class 12 Solutions

What Extra Should Students Study Beyond the NCERT for NEET?

Here is a table for all the important topics from the chapter:

NCERT Solutions for Class 12 Biology- Chapter-wise

Below mentioned are the Chapterwise solutions:


Frequently Asked Questions (FAQs)

1. Why you should use NCERT solutions for class 12 biology chapter 9 biotechnology principles and processes?
  • Solutions for NCERT biotechnology principles and processes will help you to clear your concepts as NCERT is the base of your learning.  
  • You will get all the answers to this chapter and the biotechnology class 12 NCERT PDF will help you to score good marks in the exam.   
  • Biotechnology principles and processes NCERT PDF will also help you with competitive exams like NEET.   
  • NCERT solutions for class 12 biology chapter 9, biotechnology principles and processes will also help you in your 12th board exam.   
  • Biotechnology NCERT will also boost your knowledge.  
2. What are the important topics covered in NCERT Solutions for Class 12 Biology Chapter 9?

The important topics of NCERT Solutions for Class 12 Biology Chapter 9 are

  • Biotechnology: Principles and Processes
  • Biotechnology and its applications
  • Principles of Biotechnology
  • Tools of Recombinant DNA Technology
  • Processes of Recombinant DNA Technology.
3. Are NCERT Solutions for Class 12 Biology Chapter 9 enough for board exam preparation?

Yes, NCERT Solutions for Class 12 Biology Chapter 9 provide solutions for all questions given in NCERT Textbook Biology for Class 12. The majority of the board exam questions are taken from these exercises. Students who master these ideas will perform well in their final exams.

4. What are multiple alleles and co-dominance?

Multiple alleles are the occurrence of more than two alternative forms of a gene at one locus, for example, the ABO blood group system in human beings. Co-dominance is when both alleles in a heterozygous state are expressed to the full without one being dominant to the other, for example, in the AB blood group where A and B alleles are equally expressed.

5. How does recombinant DNA technology work in biotechnology?

Recombinant DNA technology in biotechnology entails the joining of genetic material from two or more sources to form altered DNA. It is achieved through the isolation of a desired gene, placing it into a vector (such as a plasmid), and inserting it into a host cell. The host cell makes copies, which results in the desired protein or characteristic. It finds application in medicine, agriculture, and industry.

6. What are restriction enzymes and their role in genetic engineering?

Restriction enzymes are proteins that cut DNA at precise sequences, working as molecular scissors. They are pivotal in genetic engineering since they facilitate the precise cutting and manipulation of DNA fragments. Researchers utilize them to insert, delete, or alter genes in organisms. This method is important for cloning, gene therapy, and recombinant DNA technology.

7. How do plasmids act as vectors in recombinant DNA technology?

Plasmids act as vectors in recombinant DNA technology by carrying foreign genes into host cells for expression. They are small, self-replicating DNA molecules that can integrate target genes using restriction enzymes and DNA ligase. Once introduced into a host (usually bacteria) via transformation, they replicate independently, ensuring gene propagation. Selectable markers (e.g., antibiotic resistance) help identify successfully modified cells.

8. What is the role of biotechnology in medicine and agriculture?

Biotechnology significantly contributes to medicine by facilitating genetic modification, drug discovery, and targeted treatments. In agriculture, it increases crop yield, resistance to pests, and climate resilience with genetically modified organisms (GMOs). It also supports sustainable agriculture by minimizing chemical application. Biotechnology generally enhances health and food security across the world.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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