NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

NCERT Solutions for Class 12 Biology Chapter 11 Free PDF Download

Please Click on Free PDF Download link to Download the NCERT Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Pocesses

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

Ans.(i) Human insulin – Diabetes
(ii) Human growth hormone – Dwarfism cure
(iii) Blood clotting factor Y1H/IX-Haemophilia
(iv) TPA (tissue plasminogen activator) – Heart attack/strokes.
(v) PDGF (platelet derived growth factor) – Stimulates wound healing.
(vi) Interferon – Treatment of viral infection.
(vii) Interlinking – Enhances immune reaction,
(viii) Hepatitis B vaccine – Prevention of infectious disease.
(ix) Herpes Vaccine – Prevention of infectious disease.
(x) DNase I – Treatment of cystic fibrosis.

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.

Ans. Name of the Restriction enzyme – Bam HI.
The substrate DNA on which it acts:

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

Ans. DNA is bigger in molecular size than enzymes. Because DNA contains genetic information for the development and functioning of all living cells. Only 5-7% of entire DNA contains instructions for the synthesis of protein. On the other hands, enzymes are proteins which are synthesized from a small stretch of DNA known as genes.

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

Ans. Molar concentration = mol/L
The average molecular weight of 1 base pair
= 650D
= 650 × 1.66 × 10^–24 gm
= 1.079 × 10^–21 gm
The number of Base pair in DNA = 3.3 × 10⁹
So the molecular weight of DNA will be = molecular weight of base pair × Number of base pair
= 1.079 × 10^–21 gm × 3.3 × 10⁹
= 3.56 × 10^–12 gm
We know that 1 mole contain 6.023 × 10²³ molecules
So, 3.56 × 10^–12 molecule will be equivalent to
1 mole = 6.023 × 10²³ molecule

$$3.56 × 10^{–12} molecule =\frac{1 Mole×356×10^{-12}}{6.023×10^{-23}}\\= 5.5 × 10^{–15} M\\The\space average\space volume\space of\space Human\space DNA\space is = 5 × 10^{–12}L\\So\space molar\space concentration\space will\space be =\frac{5.5×10^{-15}M}{5×10^{-12}L}\\= 0.0011M/L$$

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

Ans. No, eukaryotic cells do not have restriction endonuclease because DNA molecules of eukaryotes are heavily methylated. Methylation protects the DNA from the activity of restriction enzymes.

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

Ans. Shake flasks are used for growing and mixing the desired materials on a small scale in the laboratory. A large scale production of desired biotechnological product is done by using ‘bioreactors’. Besides better aeration and mixing properties, the bioreactors have following advantages:
(i) Small volumes of cultures are periodically withdrawn from die reactor for sampling.
(ii) It has a foam control system, pH control system and temperature control system.
(iii) Facilitates even mixing and oxygen availability throughout the bioreactor.

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

Ans. Palindrome nucleotide sequences in the DNA molecule are groups of bases that form the same sequence when read both forward and backward. Five examples of palindromic DNA sequences are as follows:
(i) 5′___________GGATCC___________3’
3′___________CCTAGG___________5′
(ii) 5’___________AAGCTT___________3′
3′___________TTCGAA___________5′
(iii) 5′___________ACGCGT___________3′
3′___________TGCGCA___________5′
(iv) 5′___________ACTAGT___________3′
3′___________TGATCA___________5′
(v) 5′___________AGGCCT___________3′
3′___________TCCGGA___________5′

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

Ans.Meiosis is a process in which single cell divides twice to form four haploid daughter cells.
During the pachytene stage of prophase I, crossing over of chromosomes takes place and hence recombinant DNA is made.

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?

Ans. A reporter enzyme can be used to differentiate transformed cells by tracking down the activity of its corresponding genes (receptor gene). For e.g., (3-galactosidase (Lac Z) activity is not found in transformed cells so that they appear white in colour. The others, which appear blue in colour, indicate that cells do not carry foreign DNA.

10. Describe briefly the followings:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing

Ans.(a) Origin of Replication: It is a DNA sequence that serves as a start site for the replication process 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.
(b) Bioreactors: Bioreactors are large vessels used for the large scale production of biotechnology products. A bioreactor provides the optimal conditions for achieving the desired production labels by providing optimum growth conditions of temperature, pH, substrate, oxygen etc.
(c) Downstream processing: It is a technique for separation and purification of foreign gene products after the completion of the biosynthetic state.

11. Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase

Ans.(a) PCR: It is a method of amplifying fragments of DNA using sets of primers and the enzyme Taq DNA polymerase.
It takes place in 3 steps:

Denaturation of target DNA: The two strands of double stranded DNA is separate due to high temperature.
Annealing of primer: Primer binds to the DNA strand.
Polymerization: Extension of primer into complete DNA strand complementary to the temple strand.

(b) Restriction enzymes: These enzymes are used in genetic engineering to cut the large DNA molecules into smaller fragments. When DNA from two different sources are cut the same restriction enzyme, the resultant DNA fragments have the kind of sticky ends.

(c) Chitinase: It is an enzyme that break down chitin, a component of fungal call wall. It is useful to isolate fungal call DNA.

12. 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

Ans.(a) Plasmid DNA and Chromosomal DNA

Plasmid DNA	Chromosomal DNA
Plasmid DNA is an extra-chromosomal DNA molecule in bacteria that is capable of replicating, independent of chromosomal DNA.	Chromosomal DNA is the entire DNA an organism present inside chromosomes.

(b) RNA and DNA

RNA	DNA
RNA is a single stranded molecule.	DNA is a double stranded molecule.
It contains ribose sugar.	It contains deoxyribose sugar.
The pyrimidines in RNA are adenine and uracil.	The pyrimidines in DNA are adenine and thymine.
RNA cannot replicate itself.	DNA molecules have the ability to replicate.
It is a component of the ribosomes.	It is a component of the chromosomes.

Exonuclease	Endonuclease
It is a type of restriction enzyme that removes the nucleotide from 5 or 3 end of the DNA molecule.	It is a type of restriction enzyme that makes a cut within the DNA at a specific site to generate sticky ends.