GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

   

Gujarat Board GSEB Textbook Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes Textbook Questions and Answers, Additional Important Questions, Notes Pdf.

Gujarat Board Textbook Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

GSEB Class 12 Biology Biotechnology: Principles and Processes Text Book Questions and Answers

Question 1.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).
Answer:

  • Human Insulin (Humulin) – Treatment of Diabetes Type 1
  • Human Growth Hormone (HGH) – Replacement of deficient hormone in short stature persons.
  • Calcitonin – Treatment of rickets.
  • Chorionic Gonadotropin – Treatment of infertility.
  • Erythropoietin – stimulates erythrocyte formation in anaemics.
  • Tissue Plasminogen Activator – Dissolves blood clots after stroke and heart attack.
  • Blood Clotting factors VIII and IX – Replacement of clotting factors missing in haemophitra A or B patients.
  • Platelet Growth Factor – Stimulation of wound healing.
  • Interferon (α,β,γ) – Treatment of viral infection and cancer.
  • Interleukins – Enhancing activity of the immune system.
  • Vaccines – Preventing diseases like herpes, hepatitis B, etc.,

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 2.
From what you have learned, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
DNA is bigger in molecular size. DNA can be studied by Electrophoresis

Question 3.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
DNA molecules are the biggest biomolecules containing number of genes. A gene operates through synthesis of the polypeptide. An enzyme is formed of one or few polypeptides. It is always smaller than DNA which contains hundreds of genes.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 4.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
Most restriction endonucleases are prokaryotic in origin. However, there are several found in eukaryotic cells, including our own. In eukaryotes they are not referred to as restriction enzymes, just endonucleases. An example of an endonuclease in eukaryotes is Apnl, isolated from yeast. This enzyme helps to prevent DNA damage from environmental agents. Another common enzyme family called the topoisomerases (DNA Gyrase) has endonuclease activity.

Topoisomerases prevent the supercoiling of DNA at replication forks, by cutting the back bone, relieving the tension and pasting the ends together again – hence the endonuclease activity. In prokaryotes, restriction enzymes actually restrict the proliferation of viruses by cleaving their nucleic acids at specific base-pair sequences. These enzymes cut DNA at the exact same sequence no matter which organism the DNA belongs to – that’s why they’re such powerful tools in genetic engineering. Eukaryotic endonucleases may not all help in restricting invading nucleic acids and in fact perform many distinct “jobs”. That is probably why they are never referred to as restriction enzymes.

Question 5.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors to have over shake flasks?
Answer:
No. Eukaryotic cells do not have restriction endonucleases. Restriction endonucleases are present in bacteria. They protect bacteria from viral attack by disintegrating viral DNA without harming the bacterial genome which has methylation of sensitive sites.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 6.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:
GSEB Solutions Class 12 Biology Chapter 11 Biotechnology Principles and Processes 1

Question 7.
Can you recall meiosis and indicate at what stage recombinant DNA is made?
Answer:
Crossing over

Question 8.
Describe briefly the followings:
a. Origin of replication
b. Bioreactors
c. Downstream processing
Answer:
a. A site of DNA at which replication starts.
b. A large vessel in which microbes or plant cells or animal cells are cultured under aerobic condition to produce a novel product.
c. The extraction and purification of desired products from the cultured broth.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 9.
Explain briefly
a. PCR
b. Restriction enzymes and DNA
c. Chitinase
Answer:
(a) PCR:
PCR stands for Polymerase chain reaction. In this reaction, multiple copies of the gene (DNA) of interest is synthesized in vitro using two sets of primers (small chemically synthesized oligonucleotides that are complementary to the region of DNA) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as a template. Then the process of replication of DNA is repeated many times i.e. one billion copies are made.

(b) Restriction enzyme:
The enzymes which are responsible for restricting the growth of bacteriophage in Escherichia coli are called restriction enzymes. One of these enzymes is added to the methyl groups of DNA, while the other cuts DNA. This enzyme is known as the restriction endonuclease enzyme. The first restriction endonuclease is Hind II. Now more than 900 restriction enzymes have been isolated from over 230 strains of bacteria.

(c) Chitinase:
It is an enzyme obtained from the fungus Trichoderma which is specialized to digest chitin of fungal cell walls. The enzyme is used both as an effective fungicide as well as dissolving fungal cell walls to obtain their protoplasts for separation of DNA is genetic engineering.

Question 10.
Discuss with your teacher and find out how to distinguish between
a. Plasmid DNA and Chromosomal DNA
b. RNA and DNA
c. Exonuclease and Endonuclease
Answer:
a. A circular double stood DNA present in the bacteria that replicate independently of the chromosomal DNA is called plasmid DNA. Chromosomal DNA is the highly coiled and circular DNA present in the cytoplasm of bacteria.

b. DNA is the nucleic acid that carries the genetic information in the cell and it is capable of self-replication and synthesis of RNA. It consists of two long chains of nucleotides twisted into a double helix and joined by hydrogen bonds between the complementary bases adenine and thymine or cytosine and guanine.

RNA is another class of nucleic acid mainly involved in translating into proteins the genetic information i.e., carried in DNA. RNA is a linear polymer of four different nucleotides.

Each nucleotide is composed of three parts: a five-carbon sugar known as ribose, a phosphate group, and one of four bases attached to each ribose, either adenine (A), cytosine (C), guanine (G), or uracil (U). The structure of RNA is basically a repeating chain of ribose and phosphate moieties, with one of the four bases attached to each ribose.

c. Exonucleases are enzymes (found as individual enzymes, or as parts of larger enzyme complexes) that cleave nucleotides one at a time from an end of a polynucleotide chain. These enzymes hydrolyze phosphodiester bonds from either the 3′ or 5′ terminus of a polynucleotide molecule.

Endonucleases are enzymes that cleave the phosphodiester bond within a polynucleotide chain, in contrast to exonucleases, which cleave phosphodiester bonds at the end of a polynucleotide chain. Restriction endonucleases (Restriction Enzymes) cleave DNA at specific sites and these enzymes are often used in genetic engineering to make recombinant DNA for introduction into bacterial, plant, or animal cells.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

GSEB Class 12 Biology Biotechnology: Principles and Processes Additional Important Questions and Answers

Question 1.
Expand
(i) PCR
(ii) Bt
Answer:
(i) PCR – Polymerase chain reaction
(ii) Bt – Bacillus thermogenesis.

Question 2.
If mosquito acts as insect vector to transfer the malarial parasite into human body, what is used to deliver an alien piece of DNA into host cell?
Answer:
Plasmid DNA or bacteriophages

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 3.
What is a plasmid? (Delhi 2008)
Answer:
A plasmid is an autonomously replicating circular extrachromosomal DNA found in the bacterial cells.

Question 4.
Name two important cloning vectors.
Answer:
Plasmid DNA and bacteriophages

Question 5.
Match the words in box ‘A’ with the most accurate answers from box ‘B\
Answer:

A B
i. Lysozyme a. T-DNA
ii. Agrobacterium tumifaciens b. DNA stain
iii. Gel electrophoresis c. Escherichia coli
iv. Ethidium bromide d. Restriction enzymes
v. EcoRI e. Bacterial cell breakage
vi. Molecular scissors f. Separation of DNA fragments

i – e, ii – a, iii – f, iv – b, v – c, vi – d

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 6.
Biotechnology deals with techniques of using live organisms or enzymes from organisms to produce products useful to humans. Explain briefly the two core techniques that enabled the birth of modern biotechnology.
Answer:
i. Genetic engineering: It is the technique used to alter the chemistry of genetic material to introduce it into host organism and thus change the phenotype of the host.

ii. Maintenance of sterile ambiance in chemical engineering processes to enable the growth of only desired cell or microbes.
Question 7.
Explain the merits of genetic engineering when compared to traditional hybridization procedures.
Answer:
Traditional hybridization procedures, very often lead to the inclusion and multiplication of undesirable genes along with the desired genes. But genetic engineering overcomes this limitation and allows to isolate and introduce only one or a set of desirable genes without introducing undesirable genes into the target organism.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 8.
Explain the three basic steps in genetically modifying an organism.
Answer:

  • Identification of DNA with desirable genes.
  • Introduction of identified DNA into the host.
  • Maintenance of introduced DNA in the host and transfer of DNA to its progeny.

Question 9.
Genetic engineering leads to the production of desired products, which can be accomplished with the help of certain tools. Name the five important tools of genetic engineering.
Answer:

  1. Restriction enzymes
  2. Polymerase enzymes
  3. Ligases
  4. Vectors and
  5. Host organisms

Question 10.
EcoRI is the name of a restriction endonuclease. What do ‘E’, ‘co’, R’ and T mean?
Answer:
‘E’ stands for Escherichia (genus)
‘co’ – Coli RY13 (species)
‘R’ is derived from the name the strain.
‘I’ indicates the order in which the enzymes were isolated from that strain of bacteria.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 11.
How do “Ori” and “Cloning site” facilitate cloning into a vector? (AI 2008)
Answer:
“Ori” is a specific sequence of DNA, where replication occurs; the alien DNA must be linked to it if it has to replicate. The restriction site where a foreign DNA is linked is called the “cloning site”.

Question 12.
During genetic engineering, DNA fragments are produced by cutting DNA with endonucleases. These fragments are further used for producing recombinant DNA. Explain the technique and principle that help to separate these DNA fragments.
Answer:
The fragments can be separated by a technique known as gel electrophoresis. Since DNA fragments are negatively charged molecules they can be separated by forcing them to move towards the anode under an electric field through a medium. A natural polymer called agarose is commonly used as the medium. The fragments separate according to their size. The smaller fragments move farther.

Question 13.
Separation and isolation of DNA fragments is a major event in recombinant DNA technology. The separated fragments cannot be seen in a normal way. Is there any method to visualise the separated DNA? Explain.
Answer:
The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation. DNA can be seen as bright orange coloured bands.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 14.
How are the DNA fragments separated through gel electrophoresis extracted?
Answer:
The separated bands of DNA on agarose gel are cut out from the gel piece and extracted out. This step is known as elution. The DNA fragments are purified in this way.

Question 15.
Name two important cloning vectors.
Answer:
Plasmid DNA and bacteriophages

Question 16.
What are the main features that are required to facilitate cloning into a vector?
Answer:
Origin of replication, a selectable marker, cloning sites, and vectors for cloning genes in plants and animals.

Question 17.
Explain the significance of the origin of replication and selectable marker in gene cloning experiments.
Answer:
Origin of replication: 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.
Selectable marker: The sequence of DNA, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 18.
Name any three suitable ‘selectable markers’ for E.coli.
Answer:
The genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline are suitable ‘selectable markers’ for E.coli.

Question 19.
During genetic engineering experiments recombinants are distinguished from nonrecombinants by different methods. One of the methods is known as insertional inactivation. Explain the process.
Answer:
In this process a recombinant DNA is inserted within the coding sequence of an enzyme a -galactosidase. This results into the inactivation of the enzyme, which is referred to as insertional inactivation. Presence of insert results into insertional inactivation of the a -galactosidase and the colonies do not produce any colour, these are identified as recombinant colonies.

Question 20.
DNA is a hydrophilic molecule, so it cannot pass through cell membranes. Then how does a recombinant DNA inserted into a plasmid vector, enter into a bacterial cell?
Answer:
This is done by treating them with a specific concentration of a divalent cation such as calcium, which increases the efficiency with which DNA enters the bacterium through pores in its cell wall. Recombinant DNA can be then forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 21.
The number of recognition sites usually preferred for a vector is
Answer:
One

Question 22.
You are provided with the mixture of transformed bacteria having recombinant plasmids in which a foreign DNA is linked at the site of tetracycline resistance and non-transformed bacteria. How will you select the recombinant plasmids from the mixture?
Answer:
The recombinant plasmids will lose tetracycline resistance due to the insertion of foreign DNA. Normal plasmids will grow both in tetracycline-containing medium and ampicillin medium. The bacterial cells are plated first on an ampicillin-containing medium and then on a tetracycline-containing medium. The transformants do not grow on a tetracycline-containing medium. But the non-transformants will grow on both mediums.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 23.
Agrobacterium tumifaciens act as a natural genetic engineer, who transforms normal plant cells into tumors. Justify the statement.
Answer:
Agrobacterium tumifaciens is a pathogen of several dicot plants. They are able to deliver a piece of DNA known as T-DNA to transform normal plant cells into a tumor. T-DNA is seen in T: – plasmid of Agrobacterium tumifaciens. As they transform a normal cell into a tumor with the help of an inserted DNA, they can be called as natural genetic engineers.

Question 24.
What are restriction enzymes? Describe the naming of the restriction enzyme.
Answer:
Restriction enzymes are that enzyme which are present in bacterial cells as a defense mechanism to restrict the growth of bacteriophage, by cutting the DNA at specific sites.

  • The first letter of the name comes from the genus of the prokaryote.
  • The second and third letters come from the name of the species of the cell from where it is obtained.
  • The fourth letter comes from the strain of the prokaryote.
  • The Roman numbers following these 4 letters indicate the order in which enzymes were isolated from the strain of the bacterize.
  • e.g. E Coli RI is isolated from Escherichia coli strain RY13. The letter ‘R” stands for the strain.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 25.
DNA is usually intertwined with histone proteins and RNA. But in genetic engineering experiments, DNA must be isolated in a veiy pure form. How is this possible?
Answer:
Histone can be removed by treatment with a protease, RNA with ribonuclease, and other molecules by appropriate treatments.

Question 26.
Briefly explain the ‘cutting’ and ligation’ of DNA during recombinant DNA technology or how is recombinant DNA produced?
Answer:
‘Cutting’ of DNA is by restriction enzymes. Purified DNA molecules are incubated with restriction endonucleases to make the ‘cut’. This process is repeated in vectors also.
The ligation’ of DNA involves several processes. The cut-out ‘gene of interest’ from the source DNA and the cut vector with space are mixed and DNA ligase is added. This results in the preparation of recombinant DNA.

Question 27.
One of the methods by which recombinant DNA can be introduced into host cells is by ‘heat shock’ treatment (electro- poration). Write down other three important methods by which we can introduce alien DNA into host cells.
Answer:

  •  Micro-injection: Recombinant DNA is directly injected into the nucleus of animal cell.
  • Biolistics or gene gun: Cells are bombarded with high velocity microparticles of gold or tungsten coated with DNA.
  • By disarmed pathogen vectors: These vectors are allowed to infect the cell, which transfers the recombinant DNA into the host.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 28.
Explain how the gene of interest is amplified invitro during biotechnological experiments.
Answer:
Multiple copies of gene of interest is synthesised invitro using Polymerase Chain Reaction’ (P.C.R). It is done by using a set of primers and enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and genomic DNA as template. This process is repeated several times and DNA can be amplified.

Question 29.
What is the significance of thermostable DNA polymerase during PCR? From where is this DNA polymerase extracted?
Answer:
During PCR, repeated amplification of DNA is achieved using thermostable DNA polymerase which remains active during high temperature induced denaturation of double stranded DNA. The enzyme is isolated from a bacterium Thermus aquaticus.

Question 30.
Distinguish between recombinant DNA and recombinant protein.
Answer:
Recombinant DNA is a genetically modified DNA fragment that is formed by rejoining DNA sequences from two or more different organisms. Recombinant protein: If any protein encoding gene is expressed in a heterologous host, it is called recombinant protein.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 31.
Give the significance of using a bioreactor’ in biotechnological experiments.
Which is the commonly used type of bioreactor?
Answer:
A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions.
The most commonly used bioreactor is stirring type.

Question 32.
Explain the importance of (a) ori, (b) ampR and (c) rop in the E.coli vector shown below
GSEB Solutions Class 12 Biology Chapter 11 Biotechnology Principles and Processes 2
Answer:
a = ori = origin of replication
b = ampR = ampicillin resistance genes
c = rop = proteins involved in replication of plasmid.
ori = origin of replication is a specific portion of plasmid genome that serves as start signal for self replication.
Plasmids are characterised to possess some important genes like the antibiotic resistance gene like ampR, tetR.
The proteins produced by rop genes will help in the replication of plas

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 33.
A plasmid and a DNA sequence in a cell need to be cut for producing recombinant DNA. Name the enzyme which acts as molecular scissors to cut the DNA segments.
Answer:
Restriction enzyme

Question 34.
Name the particular technique in biotechnology whose steps are shown in the figure. Use the figure to summarise the technique in three steps.
Answer:
This technique is called cloning. There are three basic steps to modify an organism genetically.

  • Identification of desired gene in human DNA.
  • Introduction of the identified DNA into a vector namely plasmid.
  • Introduction of this recombinant DNA into a host namely E. coli.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology Principles and Processes 3

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 35.
Name any two cloning vectors. Describe the features required to facilitate cloning into a vector.
Answer:
Plasmid and bacteriophages are the commonly used vectors.
The following features are required to facilitate cloning in a vector.
i. Origin of replication (ori)
ii. Selectable marker
iii. Cloning (recognition) site
iv. Small size of vector

i. 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. The target DNA should be cloned in a vector whose origin supports high copy number.

ii. Selectable marker: In addition to ‘ori’, the vector requires a selectable marker which helps in identifying and eliminating non transformants and selectively permitting the growth of the transformants. Transformation is a procedure through which a piece of DNA is introduced in a host bacterium.

iii. Small size of vector: The mechanisms to deliver genes of our interest into variety of plants are now used to deliver desirable genes into plant cells. So once a gene or a DNA fragment has been ligated into a suitable vector, it is transferred into a bacterial plant or animal host.

Question 36.
State the principle underlying ‘gel electrophoresis’ and mention the application of this technique in biotechnology.
Answer:
The principle underlying gel electrophoresis is that when a biomolecule like DNA is placed in a gel matrix (agarose gel) under an electric field, the molecule separates according to their charge and size. It is widely used in the separation of fragments produced as a result of restriction endonuclease digestion.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 37.
How has Agrobacterium tumifaciens been suitably modified to act as a cloning vector?
Answer:
The tumour inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanisms to deliver genes of our interest into variety of plants.

Question 38.
An interesting property of restriction enzymes is molecular cutting and pasting. Restriction enzymes typically recognize a symmetrical sequence of DNA.
GSEB Solutions Class 12 Biology Chapter 11 Biotechnology Principles and Processes 4
Notice that the top strand is the same as the bottom strand, but reads backward. When the enzyme cuts the strand between G and A, it leaves over-hanging chains
GSEB Solutions Class 12 Biology Chapter 11 Biotechnology Principles and Processes 5
a. What is this symmetrical sequence of DNA known as?
b. What is the significance of these overhanging chains?
c. Name the restriction enzyme that cuts the strand between G and A.
Answer:
a. This symmetrical sequence of DNA is known as ‘palindromic sequence’.
b. These overhanging chains have sticky ends and they can be easily joined by DNA ligase.
c. Restriction endonuclease from E.coli strain (Eco RI)

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 39.
How and why is the bacterium Thermus aquaticus employed in recombinant DNA technology? Explain.
Answer:
The DNA polymerase enzyme isolated from Thermus aquaticus is widely used in PCR for amplification of gene of interest. This enzyme is heat stable and can act under the temperature conditions used in PCR machine.

Question 40.
a. What are molecular scissors ? Give one example.
b. Explain their role in recombinant DNA technology.
Answer:
a. These are restriction endonucleases which cut the DNA molecule into fragments with sticky ends.
e. g. Haemophilus parainfluenzas
b. The enzyme restriction endonuclease cleaves DNA at specific site resulting in the formation of sticky ends. In practice, the digestion by the restriction enzyme is performed by incubation of the purified DNA molecules with the specific enzyme keeping all other conditions at the optimum level and checked by using agarose gel electrophoresis.

Question 41.
Mention the role of vectors in recombinant DNA technology. Give any two examples.
Answer:
Vectors are used for inserting the gene of interest and amplifying it in a suitable host. Eg. Plasmid, bacteriophage.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 42.
Why is Agrobacterium tumefaciens a good cloning vector? Explain.
Answer:
The tumour inducing plasmid of Agrobacterium tumefaciens has now been modified into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanism to deliver genes of our interest into a variety of plants.

Question 43.
Explain the importance of (a) ori, (b) ampR and (c) rop in the E.coli vector shown below.
GSEB Solutions Class 12 Biology Chapter 11 Biotechnology Principles and Processes 6
Asnwer:
a = ori = origin of replication
b = ampR = ampicillin resistance genes c = rop = proteins involved in replication of plasmid.
ori = origin of replication is a specific portion of plasmid genome that serves as start signal for self replication.
Plasmids are characterised to possess some important genes like the antibiotic resistance gene like ampR, tetR.
The proteins produced by rop genes will help in the replication of plasmid.

GSEB Solutions Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 44.
If a certain plant when introduced into new environment neither produces seeds nor it responds to vegetative reproduction (propagation), how can more plants be produced of its kinds from this plant? State the method.
Answer:
At this stage, only tissue culture is the best method to produce more plants of its own kind. For this purpose following steps are to be taken.
i. A tissue from parental is to be sterilised.
ii. A suitable culture medium is provided to the separated tissue.
iii. Inoculation of this tissue and maintenance in aseptic conditions.
iv. Certain growth promoting hormones are also used for stimulating growth and differentiation.

Leave a Comment

Your email address will not be published. Required fields are marked *