ISC Biotechnology 2015 Class-12 Previous Year Question Papers Solved

ISC Biotechnology 2015 Class-12 Previous Year Question Paper Solved for practice. Step by step Solutions with Questions of Part-1 and Part-2 (Section A , Section-B). By the practice of Biotechnology 2015 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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ISC Biotechnology 2015 Class-12 Previous Year Question Paper Solved

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Maximum Marks: 80
Time allowed: Three hours

  • Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
  • Answer Question 1 (Compulsory) from Part I and five questions from Part II, choosing two questions from Section A, two questions from Section B and one question from either Section A or Section B.
  • The intended marks for questions or parts of questions are given in brackets [ ].
  • Transactions should be recorded in the answer book.
  • All calculations should be shown clearly.
  • All working, including rough work, should be done on the same page as, and adjacent to the rest of the answer.

Part -1 (20 Marks)
(Answer all questions)

ISC Biotechnology 2015 Class-12 Previous Year Question Paper Solved 

Question 1.
(a) Mention any one significant difference between each of the following : [5] (i) Prokaryotic genome and Eukaryotic genome
(ii) Purine and Pyrimidine
(iii) Centrifugation and Crystallography
(iv) Glucose and Glycogen
(v) Codon and Cosmid

(b) Answer the following questions : [5]
(i) Why are Flavr Savor tomatoes preferred over natural tomatoes ?
(ii) State one gene – one enzyme hypothesis.
(iii) Why is the amino acid glycine said to be optically inactive ?
(iv) State Chargaff’s Law of DNA bases.
(v) What is electroporation ?

(c) Write the full form of each of the following: [5]
(i) PIR
(ii) SSB
(iii) STS
(iv) BAC
(v) GBB

(d) Explain briefly: [5]
(i) Insertional inactivation
(ii) Microprocessor
(iii) Supramolecular assembly
(iv) Callus .
(v) Cybrid
Answer 1:

(i) Prokaryotic genome: In prokaryotic genome, a naked DNA is present, equal to single chromosome.

Eukaryotic genome: In eukaryotic genome, DNA is associated with histone proteins and the number of chromosome is 2 to numerous.

(ii) Purine: They are large size double ring structure e.g., Adenine, Guanine.

Pyrimidine: They are small in size, single ring structures e.g., Thymine, Cytosine.

(iii) Centrifugation: A process to separate small molecules by the action of centrifugal force, simply a physical phenomenon.

Crystallography: It is a technique of studying 3-D structure of macromolecules /atomic arrangement / crystal structure (proteins and nuclear acids) by placing in an intense beam of monochromatic X-rays, producing the regular pattern of reflections.

(iv) Glucose: It is a monosaccharide

Glycogen: It is a polymer formed by condensation of large number of glucose monomers.

(v) Codon: It is a sequence of three nitrogen bases in wRNA which determine the incorpo¬ration of a specific amino acid in a polypeptide chain

Cosmid: They are formed by integration of plasmid with bacterial ori, an antibiotic selection marker, and a cloning site with one or more ‘cos’ sites derived from (λ) lambda bacteriophage.


(i) Flavr Savor tomatoes are genetically modified tomatoes characterised by delayed ripening and longer shelf life.

(ii) One gene-oneenzyme hypothesis is the idea that genes act to the production of enzymes with each gene responsible for producing a single enzyme that intum affects a single step in a metabolic pathway. Concept was proposed by George Beadle and Edward Lawric Tatum in 1941.

(iii) Glycine is optically inactive because it does not affect the plane of polarised light.


(i) Chargaff’s rule states that purines and pyrimidines are equal in amount ;.e.,A+G = C + T
(ii) Molar amount of purine adenine is always equal to molar amount of thymine. Similarly guanine is equal by cytosine
(iii) Sugar deoxyribosome and phosphate occur in equimolar proportion
(iv) The ratio of A + T/ G + Cis constant for a species.

(v) Electroporation: It is the technique of introducing DNA into the cell by a brief exposure to a very high voltage electric pulse.

(c) (i) PIR – Protein Information Resources

(ii) SSB – Single Strand Breaks

(iii) STS – Sequence Tagged Sites

(iv) BAC – Bacterial Artificial Chromosome

(v) GDB-Genomic Data Bank


(i) Insertional inactivation: Harder problem to solve is to determine which of the transformed colonies comprise cells that contain recombinant DNA molecules, and which contain self-ligated vector molecules. Insertional inactivation is the inactivation of a gene by inserting a fragment of DNA into the middle of its coding sequence. Any future products from the inactivated gene will not work because of the extra codes added to it. Recombinants can therefore, be identified because the characteristic coded by the inactivated gene is no longer visible.

pBR322 contains genes which code for ampicillin resistance and tetracycline resistance. BamHI cuts in the middle of the gene which codes for tetracycline resistance. If a gene is inserted here, the plasmid loses its ability to code for tetracycline resistance. Thus, the plasmid containing the recombinant gene is resistant to ampicillin but sensitive to tetracycline. To screen, we use replica plates.

The pUC8 plasmid is ampicillin resistant and contains a gene lac Z’which partially codes for β galactosidase. To make the plasmid capable of coding for the whole protein, we add the missing DNA along with the recombinant gene. The host which contains the plasmid pUC8 is resistant to ampicillin and is also capable of producing β galactosidase.

(ii) Microprocessor: A microprocessor is a programmable digital electronic component that incorporates the functions of a central processing unit (CPU) on a single integrated circuit (IC).

(iii) Supra molecular assembly is based on weak bonds as supposed to covalent bonds. As mentioned weak bond, though harder to control have the advantage of possible reopening. Relying on weak bond in surface organisation, can lead to molecular assemblies and raise the probability of reorganization processes.

(iv) Callus: It is a mass of meristematic undifferentiated unorganized cells produced in a culture,

(v) Cybrids or a cytoplasmic hybrids are cells or plants containing nucleus of one species but cytoplasm from both the parental species.

Part – II (50 Marks)
(Answer any Five questions)

ISC Biotechnology 2015 Class-12 Previous Year Question Paper Solved 

Question 2.
(a) Explain the general structure of mRNA and tRNA. Mention their functions during protein
synthesis. [4] (b) Give one application of each of the following in genetic engineering techniques. [4] (i) Restriction enzymes and DNA ligases
(ii) Shuttle vectors and Expression vectors
(c) What is meant by genetic code ? [2] Answer 2:
(a) Formation of RNA over DNA template is called translation. It is meant for taking the coded information from DNA to the site in cytoplasm where it is required for protein synthesis. Only one of the two strands – sense strand transcribes it. mRNA brings instructions from the DNA for the formation of particular type of polypeptide. The instructions are present in the base sequence of nucleotide’s. Formation of polypeptide occurs over the ribosome, mRNA gets attached to the ribosome.

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