CBSE Class 12 Biology Bio Technology Principles And Processes Question Bank

Access the latest CBSE Class 12 Biology Bio Technology Principles And Processes Question Bank. We have provided free printable Class 12 Biology worksheets in PDF format, specifically designed for Chapter 9 Biotechnology Principles and Processes. These practice sets are prepared by expert teachers following the 2025-26 syllabus and exam patterns issued by CBSE, NCERT, and KVS.

Chapter 9 Biotechnology Principles and Processes Biology Practice Worksheet for Class 12

Students should use these Class 12 Biology chapter-wise worksheets for daily practice to improve their conceptual understanding. This detailed test papers include important questions and solutions for Chapter 9 Biotechnology Principles and Processes, to help you prepare for school tests and final examination. Regular practice of these Class 12 Biology questions will help improve your problem-solving speed and exam accuracy for the 2026 session.

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Question. Write the two components of the first artificial recombinant DNA molecule constructed by Cohen and Boyer. 
Answer. Two components of first artificial recombinant DNA molecule constructed are :An antibiotic resistance gene and plasmid of Salmonella typhimurium.

Question. Explain the work carried out by Cohen and Boyer that contributed immensely in biotechnology
Answer. Cohen and Boyer contributed to the field of biotechnology by constructing the first recombinant DNA molecule in 1972. They cut the piece of DNA from a plasmid carrying antibiotic resistance gene, using restriction enzymes. This piece of foreign DNA, was linked with the plasmid DNA, acting as a vector with the help of enzyme DNA ligase. This newly formed DNA molecule is called recombinant DNA.

Question. Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer. Separation of DNA fragments can be done by a technique called agarose gel electrophoresis. In this technique the DNA molecules are separated according to their size, under the influence of an electric field (DNA being negatively charged moves from cathode to anode).

Question. Mention the type of host cells suitable for the gene guns to introduce an alien DNA.
Answer. Undifferentiated plant cells are the most suitable host cells for the gene gun to introduce an alien DNA. It is because plant cells have rigid cell wall which can be easily broken down by bombarding them with high velocity micro-particles of gold or tungsten coated with DNA.

Question. Name the host cells in which micro-injection technique is used to introduce an alien DNA.
Answer. Microinjection technique is used to introduce an alien DNA directly into the nucleus of the animal host cells such as oocytes, eggs and embryo.

Question. Name the material used as matrix in gelelectrophoresis and mention its role.
Answer. Most commonly used matrix in DNA gel electrophoresis is agarose. It provides sieving effect for separation of DNA fragments according to their size.

Question. State what happens when an alien gene is ligated at SalI site of pBR322 plasmid. 
Answer. When an alien gene is ligated at SalI site of pBR322, the gene tetR becomes non functional and plasmid loses its tetracycline resistance. Hence, the cell possessing such recombinant pBR322 will not be able to grow on tetracycline.

Question. How is Agrobacterium tumifaciens able to transform a normal plant cell into a tumor?
Answer. When Agrobacterium tumifaciens infects a plant cell, its Ti plasmid (Ti-tumour inducing plasmid) transfers a piece of its DNA known as ‘T-DNA’ into plant cells and transform normal plant cells into a tumour.

Question. How can retroviruses be used eficiently in biotechnology experiments inspite of them being disease causing? 
Answer. Retroviruses are first disarmed (disease causing gene is removed/inactivated), this disarmed virus do not cause disease and hence are used to transfer desirable genes into host cells. So, inspite of them being disease causing, it (disarmed retrovirus) is used effciently in biotechnology experiments.

Question. State what happens when an alien gene is ligated at PvuI site of pBR322 plasmid.
Answer. When an alien gene is ligated at PvuI site of pBR322 plasmid the transformant cell loses ampicillin-resistance as ampR gene becomes nonfunctional.
Thus, the recombinant will be unable to grow in presence of ampicillin.

Question. Why is ‘plasmid’ an important tool in biotechnology experiments?
Answer. Plasmid have the ability to replicate within bacterial cells independent of the control of chromosomal DNA and have high copying number, therefore any alien DNA ligated to it, also multiplies to equal the copy number of plasmids. So, it is used as a vector in gene cloning experiments and thus plays a role of an important tool in biotechnology.

Question. Name the specific sequence of DNA in a plasmid that the ‘gene of interest’ ligates with, to enable it to replicate. 
Answer. Origin of replication (ori) is the sequence of DNA in a plasmid to which gene of interest ligates in order to get replicated.

Question. State the role of DNA ligase in biotechnology.
Answer. DNA ligases join two individual fragments of double stranded DNA by the formation of phosphodiester bond between them.

Question. Mention the use of cloning vector in biotechnology. 
Answer. Cloning vectors are essential in biotechnological experiment as they help in transferring a fragment of foreign DNA (gene of interest) into suitable host.
They help in cloning, have high copy number and help in easy linking of alien DNA.

Question. Why is it essential to have a ‘selectable marker’ in a cloning vector?
Answer. Selectable marker helps in identifying and eliminating non transformants and selectively permitting the growth of the transformants. Hence, they are considered as essential property in a cloning vector.

Question.Why do DNA fragments move towards the anode during gel-electrophoresis?
Answer. DNA is a negatively charged molecule hence during gel electrophoresis it move towards anode (positive electrode) under the influence of electrical field.

Question. In the year 1963, two enzymes responsible for restricting the growth of bacteriophage in E. coli were isolated. How did the enzymes act to restrict the growth of the bacteriophage?
Answer. In 1963, the two enzymes responsible for restricting the growth of bacteriophage in Escherichia coli were isolated. One of these added methyl groups to DNA while the other cut DNA at specific sequences i.e., restriction endonuclease. To restrict the growth of bacteriophage, the E. coli recognises and cut foreign DNA into pieces by restriction endonucleases. And in order to prevent cleavage of its own DNA by these enzymes, it modifies its DNA by adding methyl groups. And thus remain unrecognised.

Question. How is the action of exonuclease different from that of endonuclease? 
Answer. Differences between action of exonucleases and endonucleases are as follows :

Exonucleases Endonucleases
These nucleases
cleave base pairs
of DNA at their
terminal ends.
They cleave DNA at
any point except the
terminal ends.
They act on single
strand of DNA
or gaps in double
stranded DNA.
They cleave one
strand or both
strands of double
stranded DNA.

Question. Mention the role of ‘molecular scissors’ in recombinant DNA technology.
Answer. Restriction endonuclease such as EcoRI,HindIII, Bam HI acts as molecular scissors or chemical scalpels. They serve as the tools for cutting DNA molecules at specific palindromic sites, which is the basic requirement for gene cloning or recombinant DNA technology.

Question. Name the technique used for separating DNA fragments in the laboratory. 
Answer. Separation of DNA fragments can be done by a technique called agarose gel electrophoresis. In this technique the DNA molecules are separated according to their size, under the influence of an electric field (DNA being negatively charged moves from cathode to anode).

Question. What is the role of ethidium bromide during agarose-gel electrophoresis of DNA fragments?
Answer. DNA fragments can be seen only after staining.Ethidium bromide is used to stain DNA fragments followed by exposure to UV radiation. This gives bright orange colour to DNA fragments which helps to view separated DNA fragments.

Question. What is EcoRI? How does EcoRI differ from an exonuclease? 
Answer. EcoRI is a restriction endonuclease enzyme it recognises base sequences 5′-GAATTC-3′ 3′-CTTAAG-5′ in DNA duplex and cuts each of the two strands between G and A. On the other hand, exonuclease remove nucleotide from the terminal ends of DNA in one strand of duplex. Hence, EcoRI cut each of the two strand of DNA duplex at specific point whereas exonuclease remove nucleotide from the terminal ends (either 5′ or 3′) of DNA in one strand of duplex.

Question. State how has Agrobacterium tumifaciens been made a useful cloning vector to transfer DNA to plant cells. 
Answer. Agrobacterium tumifaciens is a soilinhabiting bacterium that may invade growing plants at the junction of root and stem, where it can cause a cancerous growth known as a crown gall. A. tumifaciens contains Ti plasmid which carries gene for tumour formation. For using Agrobacterium tumifaciens as a cloning vector researchers deleted the genes which governs auxin and cytokinin production (the oncogene) from T-DNA of Ti plasmid. It is known as disarming. After disarming, this T-DNA is inserted into chromosomes of the host plant where it produces copies of itself.

Question. How are ‘sticky ends’ formed on a DNA strand? Why are they so called? 
Answer. When restriction enzymes cut the strand of DNA a little away from the centre of the palindromic sites, between the same two bases on the opposite strands, it leaves single stranded portions at the ends. This forms overhanging stretches called sticky ends on each strand. They are called sticky as they form hydrogen bonds with their complementary cut counterparts. The stickiness of the ends facilitates the action of the enzyme DNA ligase.

Question. Write the role of ori and ‘restriction’ site in a cloning vector pBR322. 
Answer. Origin of replication (ori) site in cloning vector pBR322 is a sequence from where replication starts. Any piece of DNA when linked to this sequence can be made to replicate within host cell. Restriction site within the markers tetR and ampR genes permit an easy selection for cells transformed or the recombinant pBR322.

Question. How does a restriction nuclease function? Explain. 
Answer. Restriction nucleases acts as molecular scissors or chemical scalpels. Each restriction endonuclease functions by ‘inspecting’ the length of a DNA sequence. Once it finds its sepcific recognition sequence, it will bind to the DNA and cut each of the two strands of the double helix at specific points in their sugar-phosphate backbones. Each restriction endonuclease recognises a specific palindromic nucleotide sequence in the DNA.

Question. How is insertional inactivation of an enzyme used as a selectable marker to differentiate recombinants from non-recombinants?
Answer. Insertional inactivation refers to the process where insertion of rDNA within the coding sequence of an enzyme causes its inactivation. The non-recombinants having intact functional gene e.g. β-galactosidase produce blue colour with chromogenic substrate but when rDNA is inserted within the coding sequence of enzyme β-galactosidase, recombinants do not produce any colour. Hence recombinants can be easily differentiated from non-recombinants due to insertional inactivation.

Question. Explain palindromic nucleotide sequence with the help of a suitable example. 
Answer. The palindrome sequence in DNA is a sequence of base pairs that reads same on the two strands when orientation of reading is kept the same.
For example, the following sequences reads the same on the two strands in 5′ → 3′ direction. This is also true if read in the 3′ → 5′ direction.
5′ - GAATTC-3′
3′ - CTTAAG-5′

Question. Why is making cells competent essential for biotechnology experiments? List any two ways by which this can be achieved. 
Answer. Competent host is essential for biotechnology experiment. Since DNA is a hydrophilic molecule, it cannot pass through membranes, so the bacterial cells must be made capable to take up DNA i.e., made competent.
This can be achieved by :
(i) treatment of DNA with divalent cation of CaCl2 or rubidium chloride. Treating them with a specific concentration of a divalent cation, increases the efficiency with which DNA enters the bacterium through pores in its cell wall.
(ii) Heat shock treatment of DNA. Recombinant DNA (rDNA) can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock) and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Question. (a) Explain how to find whether an E. coli bacterium has transformed or not when a recombinant DNA bearing ampicillin resistant gene is transferred into it.
(b) What does the ampicillin resistant gene act as in the above case?
Answer. (a) If the recombinant DNA bearing ampcillin resistant is transferred into E.coli bacteria, it will confer resistance to ampicillin and therefore will grow on ampicillin containing medium. However the non- transformants will not grow on same medium. The recombinants or transformed E.coli can be selected by use of selectable markers. The transformants must be plated on a plate bearing ampicillin containing medium. Hence, transformants can be selected from non-transformants.
(b) Ampicillin resistant gene act as selectable marker and helps in selecting the transformant in the above case.

Question. Why and how bacteria can be made ‘competent’?
Answer. Competent host is essential for biotechnology experiment. Since DNA is a hydrophilic molecule, it cannot pass through membranes, so the bacterial cells must be made capable to take up DNA i.e., made competent.
This can be achieved by :
(i) treatment of DNA with divalent cation of CaCl2 or rubidium chloride. Treating them with a specific concentration of a divalent cation, increases the efficiency with which DNA enters the bacterium through pores in its cell wall.
(ii) Heat shock treatment of DNA. Recombinant DNA (rDNA) can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock) and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Question. Write any four ways used to introduce a desired DNA segment into a bacterial cell in recombinant technology experiments. 
Answer. Four ways to form a desired DNA segment into a host cell are (i) Electroporation, (ii) Chemical mediated gene transfer (iii) Microinjection and (iv) biolistic gun.

Question. Why is this method of selection referred to as “insertional inactivation”?
Answer. In this method, insertion of recombinat DNA in the coding sequence of enzyme b-galactosidase causes its inactivation hence named insertional inactivation.

Question. List the key tools used in recombinant DNA technology.
Answer. Biological or key tools used in recombinant
DNA technology are :
(i) Enzymes : Different kinds of specific enzymes used in recombinant DNA technology are lysing enzymes (Used to open up the cells to get DNA), it includes lysozyme, cellulose and chitinase and cleaving enzymes (enzymes used to break DNA molecules) it includes exonuclease, endonuclease and restriction endouclease.
(ii) Cloning vectors : These are DNA molecules that can carry foreign DNA segment and replicate inside a host cell. It may be plasmids, a bacteriophage, cosmids, Yeast Artifical Chromosomes(YACs), Bacterial Artifical Chromosomes (BACs) and viruses.
(iii) Competent host : A competent host is essential for transformation with recombinant DNA. It includes DNA mediated or vector mediated gene transfer and direct or vectorless gene transfer (microinjection, electroporation, chemical mediated gene transfer, biolistic method or gene gun method).

Question. How are recombinant vectors created? Why is only one type of restriction endonuclease required for creating one recombinant vector?
Answer. Recombinant vectors are created by cutting the vector and source DNA using the same restriction enzyme. This results in production of complementary
‘sticky ends’. DNA ligase help in linking alien DNA
with the plasmid DNA.
Same restriction enzyme is used for creating one recombinant vector because it recognises and cuts the DNA at a particular sequence (recognition site) and create sticky ends.

Question. Explain how the sticky ends are formed and get joined. 
Answer. The sticky ends get joined end to end with the help of enzyme DNA ligase.

Question. Rearrange the following in the correct sequence to accomplish an important biotechnological reaction :
(i) Denaturation of ds-DNA
(ii) Chemically synthesised oligonucleotides
(iii) Primers
(iv) Complementary region of DNA
(v) Thermostable DNA polymerase (from Thermus aquaticus)
(vi) Nucleotides provided
(vii) Genomic DNA template
(viii) In vitro synthesis of copies of DNA of interest
(ix) Enzyme DNA-polymerase. 
Answer. The sequence of steps shows the reaction of PCR.
Denaturation of ds DNA
                ↓
Genomic DNA template
                ↓
Chemically synthesised oligonucleotides (Primers)
                ↓
Complementary regions of DNA
                ↓
Thermostable DNA polymerase (Enzyme polymerase) (from Thermus aquaticus)
                ↓
dNTPs (Nucleotides) provided
                ↓
In vitro synthesis of copies of DNA of interest

Question. Prepare a flow chart in formation of recombinant DNA by the action of restriction endonuclease enzyme EcoRI. 
Answer. 
Isolation of DNA fragments or gene to be cloned
                ↓
The purified DNA is cut by the restriction enzyme EcoRI.
                ↓
Insertion of isolated gene in a suitable plasmid vector
                ↓
Introduction of the recombinant DNA into a suitable cell called host
                ↓
Selection of transformed host cell, and identification of the clone containing the desired gene/DNA fragment
                ↓
Multiplication / expression of the introduced gene in the host.

Question. What is a bioreactor used for? Name a commonly used bioreactor and any two of its components.  
Answer. A bioreactor is a device in which a substrate of low value is utilised by living cells or enzymes to generate a product of higher value. These are used for food processing, fermentation, waste treatment etc. The most commonly used bioreactors are of stirring type. Stirring type bioreactors may be (i) Simple stirred tank bioreactor and (ii) Sparged stirred - tank bioreactor. Two components of a simple stirred tank bioreactor are cooling jacket and stirrer blades.

Question. (a) Name the technique to obtain multiple copies, of a DNA segment of interest,synthesised in vitro. Name two sets of primers that are necessary for reaction to occur.
(b) Mention three diagnostic applications of this technique.
Answer. Applications of PCR :
(i) Diagnosis of pathogens
(ii) Diagnosis of specific mutations
(iii) DNA fingerprinting
(iv)  In prenatal diagnosis
(v)   In gene therapy

Question. Mention the role of vectors in recombinant DNA technology. Give any two examples.
Answer. The cloning vectors are DNA molecules that can carry a foreign DNA segment and replicate inside the host cell. These are plasmids, bacteriophages, cosmids, phagemids, yeast artificial chromosome (YAC), bacterial artificial chromosome (BAC), transposons and virus. Cloning vector carry rDNA and they generally have high copy number, they can produce multiple number of required gene. Vectors help in easy linking of foreign DNA and in selection of recombinants from non recombinants.

Question. (a) Illustrate the recognition sequence of EcoRI and mention what such sequences are called.
(b) How does restriction endonuclease act on a DNA molecule? 
Answer. (a) Recognition sequence of EcoRI is
5′ - GAATTC-3′
3′ - CTTAG-5′
Such sequences are called palindrome sequence.
(b) Restriction endonuclease recognise the base sequence at palindrome site in DNA duplex and cuts its strands.

Question. A and B are the two different cloning vectors in two different bacterial colonies cultured in chromogenic substrate. Bacterial colonies with cloning vector A were colourless whereas those with B were blue coloured. Explain giving reasons the cause of the difference in colour that appeared. 
Answer. Presence of the insert within a gene results in insertional inactivation of the enzyme -galactosidase, hence bacterial colonies do not produce any colour. Therefore, bacterial colonies with cloning vector A are colourless as they are recombinants with the insert and bacterial colonies with cloning vector B are blue coloured as they are non-recombinants.

Question. How is DNA isolated in purified form from a bacterial cell? 
Answer. DNA is isolated in purified form from a bacterial cell in following steps :
(i) Break the cell to release DNA along with other macromolecules. This can be done by treating cells with enzymes like lysozyme in case of bacteria, cellulase in case of plant cells and chitinase for fungus.
(ii) RNA and proteins are also present with DNA. These can be removed by ribonuclease and proteases respectively.
(iii) Similarly other molecules (if any) are removed by appropriate treatments.
(iv) Purified DNA can be precipitated out after treating with chilled ethanol and DNA fragments can be seen as fine threads in the suspension.

Question. Name the natural source of agarose. Mention one role of agarose in biotechnology.
Answer. Agarose is a commonly used as matrix in agarose gel electrophoresis. It is extracted from sea weeds. In recombinant DNA technology, agarose is used to separate DNA fragments according to their size.

Question. What is a plasmid? 
Answer. Plasmids are extra-chromosomal, self replicating, usually circular, double stranded DNA molecules, found naturally in many bacteria and also in some yeast.

Question. How can DNA segments, separated by gel electrophoresis be visualised and isolated?
Answer. The central dye in agarose gel electrophoresis is ethidium bromide. It has the unique property of fluorescence under UV light when intercalated with DNA. By running DNA through an ethidium bromide- treated gel and exposing it to UV light, distinct bands of DNA become visible. The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution.

Question. How do ‘ori’ and ‘cloning sites’ facilitate cloning into a vector?
Answer. Ori sites are the origin of replication sites which control replication of the DNA in which they are present. Cloning of a vector containing rDNA requires its multiplication to produce large number of copies and ori is essential for it. Cloning sites are the specific sites in vector that possess recognition sequences for a particular enzyme. It enables insertion of foreign DNA segment into that particular site.

Question. (a) Name the selectable markers in the cloning vector pBR322? Mention the role they play.
(b) Why is the coding sequence of an enzyme β-galactosidase a preferred selectable marker in comparison to the ones named above?
Answer. (a) Plasmid pBR322 is a most widely used cloning vector. It has two resistance genes ampicillin resistance (ampR) and tetracycline resistance (tetR) which are considered useful as selectable markers. Selectable markers help in identifying and eliminating non-transformants and selectively permitting the growth of transformants. Bacterial cells containing recombinant pBR322 will be unable to grow in the presence of ampicillin, but will grow on tetracycline.
(b) Selection of recombinants due to inactivation of antibiotics is a cumbersome procedure because it requires simultaneous plating on two plates having different antibiotics. Therefore, alternative selectable markers have been developed which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme, -galactosidase. This results into inactivation of the enzyme, which is referred to as insertional inactivation. The presence of a chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. Presence of insert results into insertional inactivation of the -galactosidase and the colonies do not produce any colour, these are identified as recombinant colonies.

Question. State the role of ‘biolistic gun’ in biotechnology experiments. 
Answer. Biolistic gun helps in the process of gene transfer into the host cell without using a vector. In biolistic method or gene gun method, tungsten or gold particles, coated with foreign DNA are bombarded into target cells at a very high velocity. This method is suitable for plants, but is also used to insert genes into animal that promote tissue repair into cells (particularly cancer of mouth)  near wounds. It  has made  great impact  in  the  field of vaccine development.

Question. Name and describe the technique that helps in separating the DNA fragments formed by the use of restriction endonuclease.
Answer.  After the cutting of DNA by restriction enzyme, fragments of DNA are formed. Separation of DNA
rop : rop codes for protein involved in the replication of plasmid ampR: gene for ampicillin resistance which help in selecting transformants.
56.
ClaI HindIII
fragments according to their size or length is done by a technique called agarose gel electrophoresis.
It is a technique of separation of molecules such as DNA, RNA or protein, under the influence of an electrical field, so that they migrate in the direction of electrode bearing the opposite charge, viz., positively charged molecules move towards cathode (–ve electrode) and negatively charged molecules travel towards anode (+ve electrode) through a medium/matrix. Most commonly used matrix is agarose. DNA fragments separate according to size through the pores of agarose gel. Hence the smaller, the fragment size, the farther it moves.
The separated DNA fragments can be seen only after staining the DNA with a compound known as ethidium bromide (EtBr) followed by exposure to UV radiation. The fragments are visible as bright orange coloured bands.

Question. State the functions of the following in the
cloning vector pBR322 :
(a) Ori,
(b) rop, and
(c) HindIII sites 
Answer. (a) Origin of replication (Ori) is a specific sequence of DNA bases which is responsible for initiating replication. It is also responsible for controlling the copy number of the linked DNA.
(b) rop in pBR322 encodes for a protein involved in replication of plasmid.
(c) HindIII is restriction site in pBR322. It is the site where HindIII endonucleaese make a cut so that a foreign DNA segment can be introduced to this vector.

Question. Draw pBR322 cloning vector. Label ‘ori’, ‘rop’ and any one antibiotic resistance site on it and state their functions. 
Answer. Origin of replication (ori) : This 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. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin support high copy number. rop : rop codes for protein involved in the replication of plasmid ampR: gene for ampicillin resistance which help in selecting transformants.

Question. Write any two ways the products obtained through this technique can be utilised.
Answer. The product obtained can be used in the following way :
(i) Fragments of DNA obtained can be used to construct a recombinant DNA molecule by joining them with cloning vector
(ii) The desired DNA fragment can be amplified using polymerase chain reaction (PCR).


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Chapter 9 Biotechnology Principles and Processes CBSE Class 12 Biology Worksheet

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