Biotechnology: Principles And Processes Class 12th Biology CBSE Solution

Class 12th Biology CBSE Solution

Exercises
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:

The ten recombinant proteins which are used in medical practice are as follows:

Note: Monoclonal antibodies are recombinant antibodies produced by myeloma B – cells and have unique specificity for antigens used to treat diseases caused by specific antigens by stimulating host immune system against the attack.



Question 2.

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 action of restriction enzyme in the production of the recombinant DNA molecule is:



Question 3.

From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?


Answer:

A DNA molecule is always bigger than a protein, in size because a protein is the product of one or more mRNAs whereas these mRNAs are produced from DNA. For example, human DNA houses approximately about 30,000 genes which are used to produce all the different types of proteins required by the cell. Even after this the only transcribed region is the euchromatin region and not the heterochromatic region of the genome. Thus, the DNA molecule is certainly larger than a protein which is only a fraction of the translated DNA.



Question 4.

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


Answer:

The molar concentration of DNA is the molar concentration of the total nucleotides present in one human cell. As the size of cell varies from cell to cell the whole calculation is an assumption.

Assuming the cell is of radius 0.05nm.


The volume of the perfect sphere would be 5.24 × 10-4 or 5.24 × 10-11 L


Total number of nucleotides:


In the nucleus: According to the Human Genome Project the 23 pairs of human chromosomes have about 3 billion base pairs that adds upto 6 billion bases = 6×109 nucleotides


In the mitochondria: There are about 1500 mitochondria per cell. Each mitochondrion contains about 16000 nucleotides. Thus total mitochondrial genome per cell is 16000×1500 = 2.3×107 nucleotides


Thus total DNA in a human cell is = (6 × 109 + 2.3 × 107) = 6 × 109 nucleotides. (Mitochondrial DNA is negligible)


Therefore,


Mole = (6 × 109)/(6.023 × 1023) = 10-14 mol (6.023 × 1023 is the Avogadro number)


Thus,


Molarity = mole of solute/litre of solution


= 10-14/5.24×10-11


= 0.0002 M



Question 5.

Do eukaryotic cells have restriction endonucleases? Justify your answer.


Answer:

No, eukaryotic cells do not have restriction endonucleases. All restriction endonucleases have been isolated from various strains of bacterias till date. Even the restriction endonucleases are named according to the genus, species, strains and isolation number of the bacterias. The first letter of the restriction endonuclease is the genus and the last two is the species of the organism. The subscript is first letter the strain of the organism, if already used the next letter is used. The Roman numbers in the end is the isolation number of the enzyme chronologically when the organism has more than one restriction endonuclease.

Example: EcoRI, where “E” is for the genus of organism Escherichia. “co” is for coli, which is the species. “R” belongs to the strain of E. coli from which EcoRI is isolated, that is, RY13. And the Roman numeral “I” says it is the first isolated endonuclease of the organism E. coli (RY13).



Question 6.

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


Answer:

The advantages of using stirred tank bioreactors over shake flasks are:

i) It has an agitator system for better mixing of oxygen.


ii) It has an oxygen delivery system to deliver fresh oxygen for the cells in the bioreactors.


iii) It has a foam control system to avoid frothing and degeneration of the required product.


iv) It has a temperature and pH control system to maintain optimum growing conditions for the cells.


v) It also has sampling ports so that small volumes of culture can be withdrawn periodically to avoid over population of the cells and overflow of reaction mixtures when new batches are added.



Question 7.

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


Answer:

A palindromic sequence is a sequence of double stranded DNA when read from either side of the double strand it is read the same. For example, “MADAM” that is, when we read it from either side, from forward as well as backward it is read the same. The same happens in case of a DNA double helix only difference is that it is read on two strands and when read from 3’ or 5’ on both strands it reads the same. The five palindromic sequences are:

1. 3’....CGTATACG....5’


5’....GCATATGC....3’


2. 3’....AAGCTT....5’ (restriction site of endonuclease Hind III)


5’....TTCGAA....3’


3. 3’....GAATTC....5’ (restriction site of endonuclease EcoRI)


5’....CTTAAG....3’


4. 3’....AGCT....5’ (restriction site of endonuclease Alu1)


5’....TCGA....3’


5. 3’....TTATTCGAATAA....5’


5’....AATAAGCTTATT....3’



Question 8.

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


Answer:

Recombinant DNA in meiosis is produced by crossing over. The crossing over occurs in the pachytene stage of prophase of the meiosis I. In this stage the sister chromatids are joined at the centromere as well as at the chiasmata. Chiasma is the position where the genetic recombination occurs by strand exchange.



Question 9.

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


Answer:

When a reporter enzyme such as the Lac Z gene is present in the genome of a transformed cell it produces the enzyme β–galactosidase. The substrate on which β–galactosidase acts on is lactose. When lactose is present in the culture medium as the carbon source of the bacteria β – galactosidase breaks down lactose to produce the end product X – gal which gives a characteristic blue colour. This blue colour is visible with naked eye and by observing the culture plates it can be identified which of the host cells are transformed. When a gene of interest is incorporated within this Lac Z gene the gene becomes inactivated as it gets disrupted. This is known as Insertional inactivation. When such a plasmid transforms the host bacterial cell it does not develop any colour and thus can be identified. This process of identification is known as the blue–white screening process.



Question 10.

Describe briefly the following:

Origin of replication


Answer:

Origin of replication is the site on the DNA strand, both genomic and extrachromosomal, which is recognised by the DNA polymerase for the initiation of replication. It is also the site where the unwinding of DNA occurs to facilitate replication. The stretch of DNA replication from one origin in one single go is known as a replicon. The eukaryotic DNA has only one replicon thus has only one origin of replication whereas the eukaryotic DNA is a multiple replicon system consisting of multiple recognisable origins of replication.



Question 11.

Describe briefly the following:

Bioreactors


Answer:

When microorganisms are cultured for desirable products in the laboratories the quantity of produce is fairly low and cannot meet the high market demand. Thus, to meet the market demands when the desired products are produced in an industrial scale in the bioreactors the yield is much higher. In industries these cells are allowed to multiply in a continuous culture system where the used medium is drained out from one side while fresh medium is added from the other to maintain the cells in their exponential phase. Thus bioreactors can be defined as large containers in which raw materials are biologically converted to specific desired products such as enzymes, hormones, antibodies, etc., using microbial, plant, animal or human cells. A bioreactor provides optimum growth conditions including temperature, pH, substrate, salts, vitamins, oxygen, etc., and even mixing if needed to achieve its desired product.


Note: Exponential phase of a cell is the most metabolically active phase of any cell where itutilises high amount of substrate and grows and divides at a higher rate.



Question 12.

Describe briefly the following:

Downstream processing


Answer:

After the industrial synthesis of the required product is complete it is obtained in crude form and requires passing through a series of processes before the product can be marketed. These processes include separation and purification of the desired product which are collectively known as the downstream processing. In some cases, especially drugs, these products have to undergo certain clinical trials. Quality control testing is strictly required and some product has to be formulated with suitable preservatives too. The downstream processing of the obtained product varies with product variety.



Question 13.

Explain briefly

PCR


Answer:

PCR or the polymerase chain reaction is the invitro amplification of DNA for experimental purpose when the sample is in very low quantity. This process was first developed by Kary Mullis in 1985. The process uses template DNA which is first denatured by heating at 96˚C followed by annealing of two primers at the 3’ ends of the template DNA strands at a comparatively lower temperature. Now DNA polymerase is added along with the four deoxyribonucleoside triphosphates (dATP, dTTP, dCTP and dGTP) to elongate the DNA strand. Normal polymerases cannot withstand the high temperature of the PCR thus a special polymerase called the Taq polymerase is added. This Taq polymerase is thermostable in nature as it is isolated from the bacterium Thermus aquaticus which is found in the hot springs and thus is active in very high temperatures. The full process is maintained at a range of 72˚C to 75˚C and is allowed to run for any required number of cycles.



Question 14.

Explain briefly

Restriction enzymes and DNA


Answer:

Restriction enzymes are synthesised by eubacterias as a defence mechanism. These are specifically endonucleases and cleave double stranded DNA where they find their specific restriction sites. Restriction enzymes are of three types: type I, type II and type III. Of which only type II restriction endonucleases restricts DNA within the recognition site and thus are used as molecular scissors for genetic engineering. Sites recognisable by the restriction enzymes are also present in the host genome but they are methylated for protection against their own enzyme, so that they do not undergo self digestion.



Question 15.

Explain briefly

Chitinase


Answer:

Chitinase is an enzyme which is used to degrade the glycosidic bond present in chitin. Chitin makes up the fungal cell membrane exoskeleton. Thus when it is required to break open a fungal cell for experimentation Chitinase is used.

Note: Glycosidic bonds are those that occur between two glucose molecules.



Question 16.

Discuss with your teacher and find out how to distinguish between

Plasmid DNA and Chromosomal DNA


Answer:

The differences between plasmid DNA and chromosomal DNA are



Question 17.

Discuss with your teacher and find out how to distinguish between

RNA and DNA


Answer:

the differences between RNA and DNA are



Question 18.

Discuss with your teacher and find out how to distinguish between

Exonuclease and Endonuclease


Answer:

The differences between exonuclease and endonuclease are


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