ISC Biotechnology 2019 Class-12 Previous Year Question Papers Solved

ISC Biotechnology 2019 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 2019 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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ISC Biotechnology 2019 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 2019 Class-12 Previous Year Question Paper Solved 

Question 1.
(a) Mention any one significant difference between each of the following : [5] (i) Anti codon and codon
(ii) Intrinsic fluorescence and extrinsic fluorescence
(iii) Introns and Exons
(iv) Genomic DNA library and cDNA library
(v) RAM and ROM

(b) Answer the following questions: [5] (i) Which amino acid is optically inactive and why ?
(ii) What is meant by exponential phase 9
(iii) What are designer oils ?
(iv) What is palindromic sequence 9
(v) Which substance is used in diploidization of haploid plants ?

(c) Write the full form of each of the following : [5] (i) NBRI
(ii) NBTB
(iii) BLAST
(iv) PIR
(v) YAC

(d) Explain briefly the following terms : [5] (i) Callus
(ii) SNPs
(iii) Lyophilisation
(iv) Gene cloning
(v) Cybrids
Answers 1: (ISC Biotechnology 2019 Class-12)
(a) difference between Anti codon and codon

(i) Anti codon is the corresponding triplet sequence on the transfer RNA (tRNA) which brings in the specific amino acid to the ribosome during translation. The anti codon is complementary to the codon, that is, if the codon is AUU, then the anti codon is UAA.
Codon is the triplet sequence in the messenger RNA (mRNA) transcript which specifies a corresponding amino acid (or a start or stop command).

(ii) difference between Intrinsic fluorescence and extrinsic fluorescence

Intrinsic fluorescence Extrinsic fluorescence
The fluorescence shown by natural compounds is known as intrinsic fluorescence. There are certain chemical compounds which do not fluorescence. It can be detected after coupling them with a fluorescence probe or flour. This phenomenon is called extrinsic fluorescence.


(iii) difference between Introns and Exons

Introns Exons
Introns are non-coding DNA base sequences, which are found between exons, but are not transcribed part of mature mRNA. Exons are coding DNA base sequences that are transcribed into w/RNA and finally code for amino acids in the proteins.


(iv)  difference between Genomic DNA Library and cDNA Library

Genomic DNA Library cDNA Library
(1) It is a collection of clones that represent the complete genome of an organism.

(2) It is the mixture of genomic DNA.

(1) The library made from complementary or copy DNA (cDNA) is called cDNA library.

(2) cDNA library consists of cDNA clones prepared by using mRNA.


(v) difference between RAM and ROM

(1) It stores data and programme temporarily.

(2) It contains the data of operating system, software’s and other application programmes.

(1) It provides a non-volatile storage of data.

(2) It loads operating system into the memory and starts the computer when switched on.



(i) Glycine is the only amino acid which is optically inactive. It is the first neutral amino acid. In Glycine, the central atom has 2 hydrogen atoms. It does not contain any chiral carbon that’s why it is optically inactive.
ISC Biotechnology Question Paper 2019 Solved for Class 12 1

(ii) Exponential Phase : It is a growth phase. In the exponential (log) phase, cells divide as fast as possible according to the growth medium, the microorganism itself and environmental conditions. This phase has a limited duration.
ISC Biotechnology Question Paper 2019 Solved for Class 12 2
(iii) Designer Oil : “Designer oil” that reduces LDL ( bad ) blood cholesterol levels in humans and increases energy expenditure, which may prevent people from gaining weight. The oil incorporates a phytosterol-based functional food ingredient Phytrol (TM) from Forbes into oil using proprietary technology.

(iv) Palindromic sequence are the base pair sequences that are the same when read forward (left to right) or backward (right to left) from a central axis of symmetry. The sequences read the same on the two strands in 5 → 3 direction and this is also true when we read in the 3 → 5 directions.

(v) Colchicine is used in diploidization of haploid plants. There are mainly two approaches for diploidization-colchicine treatment and endometriosis.

(c) NBRI: National Botanical Research Institute
NBTB : National Biotechnology Board
BLAST: Basic Local Alignment Search Tool
PIR : Protein Information Resource
YAC : Yeast Artificial Chromosome


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

(ii) SNPs : Single Nucleotide Polymorphisms (SNPs) are the variations in a nucleotide at genomic DNA in different individuals at a population which occur due to change even in a single base (e.g., A, G, T or C). In human genome, SNPs occur at 1 n – 3.2 million sites.

(iii) Lyophilisation : In this method the microbial suspension is put in small ampoules and frozen through drying under vacuum. Vaccum drying results in sublimation of cell water. The lyophilised cultures are stored at 4°C. The culture remain viable for about 10 years. By this method a large number of cultures are maintained without variation in the characteristics.

(iv) Gene Cloning : It is a process of copying a specific desired gene/section on DNA and producing in large number/quantities.

(v) Cybrids : A cytoplasmicallv hybrid cell with organelles from both parental sources (obtained through fusion of cytoplast with a whole cell) and a nucleus of only one cell Nucleus of the other cell denatured.

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

ISC Biotechnology 2019 Class-12 Previous Year Question Paper Solved 

Question 2.
(a) With reference to composition of culture medium, answer the following : [4] (i) Cytokinins
(ii) Auxin’s
(b) Explain the induced fit hypothesis of enzyme action with the help of suitable illustrations. [4] (c) Write a note on Quaternary’ structure of proteins. [2] Answers 3:  (ISC Biotechnology 2019 Class-12)

(i) Cytokinins:

  • It stimulates the cell division process.
  • It helps in the morphogenesis of plant cell, along with auxins.

(ii) Auxin:

  • It helps in the formation of callus and in the development of xylem along with the promotion of cambial activity.
  • It is used for the elongation and enlargement of plant cell and it inhibits the promotion of growths of apical and lateral buds.

(b) The Induced-fit theory:
The key-lock hypothesis does not fully account for enzymatic action; i.e.. certain properties of enzymes cannot be accounted for by the simple relationship between enzyme and substrate proposed by the key-lock hypothesis. A theory called the induced-fit theory retains the key lock idea of a fit of the substrate at the active site but postulates in addition that the substrate must do more than simply fit into the already preformed shape of an active site.

Rather, the theory 7 states, the binding of the substrate to the enzyme must cause a change in the shape of the enzyme that results in the proper alignment of the catalytic groups on its surface. This concept has been likened to the fit of a hand in a glove, the hand (substrate) inducing a change in the shape of the glove (enzyme). Although some enzymes appear to function according to the older key-lock hypothesis, most apparently function according to the induced-fit theory .

Typically, the substrate approaches the enzyme surface and induces a change in its shape that results in the correct alignment of the catalytic groups. In the case of the digestive enzyme carboxypeptidase.
ISC Biotechnology Question Paper 2019 Solved for Class 12 3
The induced-fit theory 7 explains a number of anomalous properties of enzymes. An example is “non-competitive inhibition.” in which a compound inhibits the reaction of an enzyme but does not prevent the binding of the substrate. In this case, the inhibitor compound attracts the binding group so that the catalytic group is too far away from the substrate to react. The site at which the inhibitor binds to the enzyme is not the active site and is called an allosteric site. The inhibitor changes the shape of the active site to prevent catalysis without preventing binding of the substrate. An inhibitor also can distort the active site by affecting the essential binding group: as a result, the enzyme can no longer attract the substrate.

(c) Quaternary Structure of proteins (4° Structure) : It is the last or fourth level of protein organisation found in only oligomeric proteins or multimers. The multimeric proteins are formed of two to several poly peptides. The monomers or polypeptide sub units are also called protomers. Protomers maybe similar, e.g., two similar a polypeptides in enzyme phosphorvlase. It is known as homogeneous Quaternary structure. An oligomeric protein having dissimilar sub units shows heterogeneous Quaternary structure, .e.g., tetrameric hemoglobin with two ot (141 amino acids each) and two p (146 amino acids each) polypeptide chains.

Question 3. (ISC Biotechnology 2019 Class-12)
(a) Explain the important postulates of central dogma. [4] (b) Name and explain the method used to sterilize the following : [4] (i) Vitamins
(ii) Forceps and Scalpels
(iii) Nutrient Media
(iv) Explant
(c) What is the Chargaff’s rule of equivalence ? [2] Answers 4: (ISC Biotechnology 2019 Class-12)
(a) The ‘Central Dogma’ is the process by which the instructions in DNA are converted into a functional product. It was first proposed in 1958 by Francis Crick, discoverer of the structure of DNA.

1. The central dogma of molecular biology explains the flow of genetic information, from DNA to RNA, to make a functional product, a protein.

2. The central dogma suggests that DNA contains the information needed to make all of our proteins, and that RNA is a messenger that carries this information to the ribosomes .

3. The ribosomes serve as factories in the cell where the information is ‘translated’ from a code into the functional product.

4. The process by which the DNA instructions are converted into the functional product is called gene expression.

5. Gene expression has two key stages – transcription and translation .

6. In transcription, the information in the DNA of every cell is converted into small, portable RNA messages.

7. During translation, these messages travel from where the DNA is in the cell nucleus to the ribosomes where they are ‘read’ to make specific proteins.

8. The central dogma states that the pattern of information that occurs most frequently in our cells is :

  • From existing DNA to make new DNA (DNA replication)
  • From DNA to make new RNA (transcription)
  • From RNA to make new proteins (translation)

9. Reverse transcription is the transfer of information from RNA to make new DNA. this occurs in the case of retroviruses, such as HIV . It is the process by which the genetic information from RNA is assembled into new DNA.

Does the ‘Central Dogma’ always apply?

  • With modem research it is becoming clear that some aspects of the central dogma are not entirely accurate.
  • Current research is focusing on investigating the function of non-coding RNA.
  • Although this does not follow the central dogma it still has a functional role in the cell.

(b) Sterilization methods:

(i) Vitamins : Vitamins can be sterilize by the method of autoclaving. For this a concentrated stock solution is required to be prepared followed by fitration and subsequent addition in sterile medium at the temperature of growth (or room temperature).

(ii) Forceps & Scalpels : The metallic instruments like forceps, scalpels, needles, spatulas can be sterilized by flame sterilization method. For this we have to dip them in 25% ethanol ‘ followed by flaming and cooling. It is called incineration.

(iii) Nutrient media : Culture media are properly dispensed in glass container, plugged with cotton or sealed with plastic closures and sterilized by autoclaving (steam sterilization) at 15 psi for 30 minutes. Minimum temperature required for autoclaving of nutrient media is 15 minutes.

(iv) Explant: The explants are sterilized by disinfectants (e g., sodium, hypochlorite. NaOCl, mercuric chloride-HgCl2) and washed aseptically for 6-10 times with sterlized distilled water.

(c) Chargaff’s rule of equivalence : Deoxy ribonucleic Acid (DNA), the genetic material is made up of four types of organic nitrogenous bases : adenine (A), guanine (G), thymine (T) and cytosine (C). Of these, A and G are the purines and T and C are the pyrimidines. Chargaff gave the base pairing rule or the rule of base equivalence which states that only one purine can combine with one pyrimidine. That means A can combine with T and G with C. Two purines or two pyrimidines cannot combine with each other; if they do so, there w ill be a sudden change in the characteristic of an organism. This sudden change is called mutation.

Question 4, (ISC Biotechnology 2019 Class-12)
(a) Differentiate between oils and fats. Discuss hydrolysis, rancidity and hardening shown by lipids. [4] (b) Using tissue culture method one can produce disease free plants. Discuss the method used to produce virus free plants. [4] (c) Write the main objectives of HGP. [2] Answer 4: (ISC Biotechnology 2019 Class-12)
(a) Difference between Fats and Oils :

Fats Oils
1. Solid at room temperature 1. Liquid at room temperature
2. Saturated and trans are its types 2. Unsaturated fats like monounsaturated and polyunsaturated are its types
3. Mostly derived from animal 3. Mostly derived from plants
4. Increases cholesterol levels 4. Improves cholesterol levels
5. Mainly comes from animal food but also through vegetable oil by process called hydrogenation

Example: Butter, beef fat Contains 9 cal/gm

5. Mainly comes from plants or fish

Example:Vegetable oil, fish oil Contains 9 cal/gm

Hydrolysis of lipids: Hydrolysis is the breakdown of a substance by the addition of water. Fats and oils are hydrolyzed by moisture to yield glycerol and 3 fatty acids. This leads to hydrolytic rancidity of food product characterized by unpleasant flavor and aroma thereby making it undesirable for consumers. Chemically fats are esters, so they are liable to hydrolysis. This reaction is catalyzed by lipase enzyme or can occur via non-enzymatic hydrolysis. Partial hydrolysis of triglycerides will yield mono- and di- glycerides and free fatty’ acids.

Rancidity: Rancidity is the natural process of decomposition of fats or oils by either hydrolysis or oxidation or by both. The process of degradation converts fatty acids esters of oils into free fatty acids. This gives rise to an unpleasant odour and taste in food. These lipids degrade to the point of becoming either unpalatable or unhealthy to ingest.

Hydrogenation and hardening: Hydrogenation of unsaturated fatty acids is widely practiced. This treatment affords saturated fatty acids. The extent of hydrogenation is indicated by the iodine number. Hydrogenated fatty- acids are less prone toward rancidification. Since the saturated fatty acids are higher melting than the unsaturated precursors, the process is called hardening.

(b) Production of virus-free plants : Plant viruses are found in nature and propagate inside the cells of living plant. There are many plants that suffer from serious diseases developing a variety of symptoms such as mosaic, chlorosis, necrosis, vein clearing, lead curling, leaf rolling, etc. Sometimes the disease severity is so high that viruses migrate to vascular bundles and also get established inside the seeds.

For the first time G Moral and C. Martin (1952) produced virus-free shoots of Dahlia from virus-infected plants through shoot meristem culture, On 26th November, 2002 a US Patent was granted to H.C. Chaturvedi and co-workers for successfully regenerating and proliferating shoot meristem of Cirus spp.

Using tissue culture method we can produce disease-free plants. Generally, the apical meristem of plants are free from viruses and the other microorganisms. However, if viral particles are present, their number would be quite low.

Methods of virus elimination :
There are two methods that are used to produce virus-free plants.
(i) Heat Treatment of Meristem : Before meristem culture, viruses associated with meristem are eliminated in vitro by heat treatment (thermotherapy) of whole plants. Heat treatment is given through hot water or air at 35-40°C for a varying period (a few minutes to several months). It should be noted that prolonged heat treatment inactivates that host resistance. Percentage of plant survival is also low. Temperature tolerant viruses may survive inside the plant tissue.

(ii) Meristem-tip Culture : Meristem-tip culture is the most widely applicable approach for virus elimination. In most of the cases explant (shoot tip) of 100-1000 pm has been cultured to raise virus-free plants. But the explant of small size (1 mm) i.e., meristem tip is preferred for in-vitro culture. Meristem tip is excised in aseptic conditions and cultured on nutrient medium. If required thermotherapy is given to the mother plant before excision of meristem tip. The inoculated tubes are incubated properly in light and dark regime (Fig.)

There are some chemicals which are used to eliminate viruses from plant tissues and protoplasts. These are malachite green, thiouracil, acetyl salicylic acid, cycloheximide, actinomvcin-D, etc.
ISC Biotechnology Question Paper 2019 Solved for Class 12 4
(c) Objectives of HGP :
(1) The goals of the original HGP were not only determine all 3 billion base pairs in the human genome with a minimal error rate, but also to identify’ all the genes in this vast amount of data. This part of the project is still ongoing although a preliminary count indicates about 30,000 genes in the human genome, which is far fewer than predicated by most scientists.

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