ISC Class 12 Biotechnology Board Exam Question Paper 2010 with Solutions

ISC Class 12 Biotechnology Board Exam Question Paper with Solutions 2010

Part - I (Compulsory)

Question 1.
(a) Mention any one significant difference between each of the following : [5]
(i) Aldoses and ketoses
(ii) Pluripotent and multipotent cell
(iii) Chemostat and turbidostat
(iv) Codon and anticodon
(v) Androgenesis and gynogenesis.

Answer:
(i) Aldose has one aldehyde group (carbonyl group at the end of the carbon chain), whereas a ketose has one ketone group (carbonyl group at any other position along the chain).
(ii) Pluripotent cells can differentiate into cells of all three germ layers (ectoderm, mesoderm, and endoderm), whereas multipotent cells have a more limited differentiation capacity and can only give rise to a restricted family of related cell types.
(iii) In a chemostat, the growth rate is controlled by regulating the concentration of a limiting nutrient added at a constant flow rate, whereas in a turbidostat, the flow rate of fresh medium is adjusted automatically based on maintaining a constant turbidity (cell density).
(iv) A codon is a triplet of nucleotides on mRNA that specifies a particular amino acid or stop signal during translation, whereas an anticodon is the complementary triplet of nucleotides on tRNA that recognizes the mRNA codon.
(v) Androgenesis is the development of an embryo derived exclusively from the male genome (paternal contribution), whereas gynogenesis is the development of an embryo derived exclusively from the female genome (maternal contribution) without participation of the sperm nucleus.

Teacher's Note:
a) Ensure that contrasting pairs are defined with precise biological terminology, focusing on structural or functional distinctions.
b) Students often confuse pluripotent and multipotent stem cells; remember that pluripotent has broader differentiation potential.

(b) Answer the following questions : [5]
(i) What are nonsense codons
(ii) What is a microprocessor?
(iii) What is the role of oligonucleotide primer during DNA replication
(iv) Name two unique bases found in the tRNA structure.
(v) Who is regarded as the father of tissue culture technique ?

Answer:
(i) Nonsense codons (also called stop codons: UAA, UAG, and UGA) do not code for any amino acid and signal the termination of protein synthesis.
(ii) A microprocessor is an integrated circuit that contains all the functions of a central processing unit (CPU) of a computer, responsible for performing arithmetic, logic, and control operations.
(iii) An oligonucleotide primer provides a free 3'-OH group to which DNA polymerase can attach incoming deoxyribonucleotides, as DNA polymerase cannot initiate synthesis de novo.
(iv) Dihydrouridine (DHU) and pseudouridine.
(v) Gottlieb Haberlandt.

Teacher's Note:
a) Note down the three stop codons (UAA, UAG, UGA) accurately as they frequently appear in molecular biology questions.
b) Gottlieb Haberlandt is universally recognized as the pioneer and father of plant tissue culture for his early work on plant cell totipotency.

(c) Write the full form of the following: [5]
(i) NBPGR
(ii) VNTR
(iii) ARS
(iv) PEG
(v) ORF

Answer:
(i) National Bureau of Plant Genetic Resources
(ii) Variable Number Tandem Repeat
(iii) Autonomously Replicating Sequence
(iv) Polyethylene Glycol
(v) Open Reading Frame

Teacher's Note:
a) Memorize standard abbreviations in biotechnology as full-form questions are direct scoring opportunities.
b) Spelling accuracy is critical; minor spelling errors in terms like Polyethylene Glycol or Autonomously Replicating Sequence result in loss of marks.

(d) Explain briefly:
(i) Chargaff's rule of equivalence
(ii) Database
(iii) Biolistic
(iv) Coenzyme
(v) Optical activity

Answer:
(i) Chargaff's rule of equivalence states that in double-stranded DNA, the amount of adenine equals thymine ([A] = [T]), and the amount of guanine equals cytosine ([G] = [C]), meaning total purines equal total pyrimidines ([A] + [G] = [T] + [C]).
(ii) A database is an organized collection of structured information or data stored electronically in a computer system, allowing rapid retrieval, management, and analysis of biological or bibliographic data.
(iii) Biolistic (gene gun method) is a physical method of introducing foreign DNA into target cells or tissues by microprojectiles (gold or tungsten particles) coated with DNA and accelerated at high velocity.
(iv) A coenzyme is a non-protein organic molecule loosely bound to an enzyme, essential for the enzyme's catalytic activity by acting as a carrier for chemical groups or electrons.
(v) Optical activity is the ability of chiral chemical substances to rotate the plane of polarization of plane-polarized light passing through them.

Teacher's Note:
a) Ensure concise definitions are provided covering the core operational principle for each term.
b) For Chargaff's rule, explicitly state the molar equivalence formula.

PART - II
(Answer any five Questions)

Question 2.
(a) What are lipids ? Explain the following properties of lipids: [4]
(i) Amphipathic nature
(ii) Saponification
(iii) Hydrogenation

Answer:
Lipids are a diverse group of naturally occurring hydrophobic or amphiphilic organic molecules, including fats, waxes, sterols, and phospholipids, soluble in non-polar solvents.
(i) Amphipathic nature: Molecules possess both hydrophilic (polar head) and hydrophobic (non-polar tail) regions, allowing them to form lipid bilayers and micelles in aqueous environments.
(ii) Saponification: The alkaline hydrolysis of triglycerides (fats or oils) using a strong base like sodium hydroxide to form glycerol and alkali salts of fatty acids (soap).
(iii) Hydrogenation: The chemical addition of hydrogen across the carbon-carbon double bonds of unsaturated fatty acids in liquid oils to convert them into saturated, semisolid or solid fats.

Teacher's Note:
a) Clearly distinguish between the three biochemical properties with proper context in lipid chemistry.
b) Mentioning the chemical reagents used in saponification (NaOH) and hydrogenation ($H_2$ gas with catalyst) adds precision.

(b) Mention the chief characteristics of embryonic stem cells. Give any two uses of such cells. [4]

Answer:
Chief characteristics of embryonic stem cells:
1. They are unspecialized cells capable of dividing and renewing themselves through long periods of cell division (self-renewal).
2. They are pluripotent, meaning they have the potential to differentiate into derivatives of all three embryonic germ layers.
Uses of embryonic stem cells:
1. Used in regenerative medicine and cell-based therapies to replace damaged tissues (e.g., in Parkinson's disease or diabetes).
2. Used for drug discovery, toxicity testing, and studying early human developmental biology.

Teacher's Note:
a) Emphasize pluripotency and self-renewal as the defining hallmarks of embryonic stem cells.
b) Clearly list two distinct therapeutic or research applications.

(c) Write the names of any four important centres for biotechnology in India. [2]

Answer:
1. Centre for Cellular and Molecular Biology (CCMB), Hyderabad.
2. National Institute of Immunology (NII), New Delhi.
3. Institute of Microbial Technology (IMTECH), Chandigarh.
4. National Centre for Cell Science (NCCS), Pune.

Teacher's Note:
a) List recognized national institutes functioning under the Department of Biotechnology (DBT), India.
b) Provide city locations alongside institute acronyms for complete clarity.

Question 3.
(a) How is DNA sequencing performed using Sanger's dideoxy technique ? Give one limitation of this technique. [4]

Answer:
Sanger's dideoxy chain termination method involves the synthesis of a DNA strand complementary to the template strand using DNA polymerase. The reaction mixture contains single-stranded template DNA, a primer, DNA polymerase, all four normal deoxynucleotides (dNTPs), and a low concentration of fluorescently labeled dideoxynucleotides (ddNTPs). When a ddNTP is incorporated into the growing chain, it lacks the 3'-OH group, terminating further chain elongation. This produces a nested set of fluorescent fragments of varying lengths, which are separated by high-resolution capillary electrophoresis and detected by a laser scanner to read the sequence.
Limitation: The method is limited in read length (typically up to 800-900 base pairs per reaction) and can be affected by secondary structures or homopolymer stretches in the template DNA.

Teacher's Note:
a) Clearly mention the role of ddNTPs in chain termination.
b) Mentioning capillary gel electrophoresis and fluorophore labeling completes the modern explanation of the technique.

(b) Write short notes on : [4]
(i) Shuttle vector
(ii) YAC
(iii) PCR
(iv) Electroporation

Answer:
(i) Shuttle vector: A plasmid or vector constructed to replicate in two different host species (e.g., both Escherichia coli and yeast), allowing genes to be manipulated in one organism and expressed or tested in another.
(ii) YAC (Yeast Artificial Chromosome): An artificially constructed DNA molecule containing yeast telomeres, centromeres, and replication origins, used to clone very large DNA fragments (up to 1000 kb or more) in yeast cells.
(iii) PCR (Polymerase Chain Reaction): An in vitro technique for the enzymatic amplification of a specific DNA segment through repeated thermal cycles of denaturation, primer annealing, and DNA extension.
(iv) Electroporation: A molecular biology technique where a high-voltage electrical pulse is applied across a cell suspension to transiently increase the permeability of the plasma membrane, facilitating the uptake of foreign DNA.

Teacher's Note:
a) Ensure all four sub-notes capture the structural and functional definition of each tool/technique.
b) Mention insert capacity limits for YAC to earn full credit.

(c) Name any two industrial enzymes and give their uses. [2]

Answer:
1. Amylase: Used in the starch processing industry to convert starch into glucose and high-fructose syrups.
2. Protease: Used in the detergent industry for protein stain removal and in the dairy industry for cheese making.

Teacher's Note:
a) State standard industrial enzymes along with their primary commercial applications.
b) Avoid vague descriptions; specify the exact industrial process.

Question 4.
(a) Explain any one biochemical technique based on each of the following principles : [4]
(i) Sedimentation coefficient
(ii) Polarity

Answer:
(i) Sedimentation coefficient: Density Gradient Centrifugation (or Rate Zonal Centrifugation) separates macromolecules based on their sedimentation coefficient (S-value) in a density gradient under centrifugal force, allowing separation of subcellular organelles or nucleic acids of different sizes.
(ii) Polarity: Reversed-phase Chromatography or Adsorption Chromatography separates biomolecules based on differences in their polarity, where non-polar molecules interact more strongly with a non-polar stationary phase, while polar molecules elute faster.

Teacher's Note:
a) Explicitly link the underlying principle (sedimentation rate or molecular polarity) to the chromatographic or centrifugation technique.
b) Keep explanations precise and application-oriented.

(b) What are the steps involved in the construction of an rDNA molecule for the purpose of gene cloning ? [4]

Answer:
1. Isolation of the desired donor DNA containing the gene of interest and selection of a suitable vector (plasmid).
2. Cleavage of both the donor DNA and vector DNA with the same restriction endonuclease to generate compatible sticky or blunt ends.
3. Ligation of the gene of interest into the linearized vector using DNA ligase to form recombinant DNA (rDNA).
4. Introduction of the recombinant DNA into a suitable host organism (transformation) and identification of transformants.

Teacher's Note:
a) List the steps in chronological order: isolation, restriction digestion, ligation, and transformation.
b) Highlight the specific roles of restriction enzymes and DNA ligase.

(c) How is distant hybridization technique useful in plant tissue culture ? [2]

Answer:
Distant hybridization (interspecific or intergeneric crosses) helps in combining desirable agronomic traits from different species (such as disease resistance and high yield) into a single hybrid, and combined with embryo rescue techniques in tissue culture, it helps overcome cross-incompatibility barriers to produce novel crop varieties.

Teacher's Note:
a) Emphasize gene transfer and trait combination across species barriers.
b) Mention embryo rescue as a vital adjunct tissue culture technique.

Question 5.
(a) Discuss the Cell Culture technology under the following headings [4]
(i) Different types of nutrient media used.
(ii) Sterilization of instruments and transfer area.

Answer:
(i) Different types of nutrient media: Culture media contain essential inorganic salts, carbon sources (sucrose/glucose), vitamins, and amino acids. They can be classified as solid media (using gelling agents like agar) or liquid (suspension) media, and natural or synthetic (chemically defined) media.
(ii) Sterilization of instruments and transfer area: Metallic instruments (forceps, scalpels) are sterilized by dipping in 70-95% alcohol followed by flame incineration. Transfer areas (inoculation chambers/laminar air flow hoods) are sterilized by UV irradiation and wiping surfaces with disinfectants like 70% ethanol or sodium hypochlorite.

Teacher's Note:
a) Differentiate between solid and liquid media and state their compositions.
b) Clearly specify autoclaving, flaming, and UV exposure as standard sterilization protocols.

(b) What is a DNA library ? How is a genomic DNA library formed ? [4]

Answer:
A DNA library is a collection of cloned DNA fragments representing the entire genome or transcriptome of an organism.
Formation of a genomic DNA library:
1. Extraction and isolation of total genomic DNA from the organism.
2. Partial digestion of the genomic DNA using restriction endonucleases to generate manageable fragments.
3. Ligation of these DNA fragments into suitable cloning vectors (such as plasmids, phages, or cosmids).
4. Introduction of the recombinant vectors into host cells (bacteria or yeast) to produce a population of clones representing the entire genome.

Teacher's Note:
a) Differentiate clearly between genomic libraries and cDNA libraries.
b) Emphasize partial restriction digestion to ensure overlapping fragments representing the entire genome.

(c) Give two applications of somatic hybridization. [2]

Answer:
1. Production of novel somatic hybrid plants (cybrids or somatic interspecific hybrids) combining disease resistance and stress tolerance traits.
2. Overcoming sexual incompatibility barriers between sterile or sexually incompatible plant species.

Teacher's Note:
a) Focus on protoplast fusion applications in crop improvement.
b) Highlight interspecific crosses and cytoplasmic male sterility management.

Question 6.
(a) Explain the post-transcriptional changes in mRNA and tRNA of a eukaryotic cell. [4]

Answer:
In eukaryotic cells, primary RNA transcripts undergo extensive post-transcriptional processing before becoming functional.
mRNA processing involves:
1. 5' Capping: Addition of a 7-methylguanosine cap at the 5' end to protect against degradation and aid in ribosome binding.
2. 3' Polyadenylation: Cleavage of the transcript and addition of a poly-A tail (about 200 adenine residues) at the 3' end.
3. RNA Splicing: Removal of non-coding introns and ligation of coding exons by spliceosomes.
tRNA processing involves endonucleolytic cleavage of precursor tRNA ends, removal of introns (where present), and specific base modifications (such as methylation or pseudouridylation).

Teacher's Note:
a) Detail all three primary modifications for mRNA: capping, polyadenylation, and splicing.
b) Mention enzymatic cleavage and base modification for tRNA processing.

(b) Classify proteins on the basis of their functions. Also give an example of each type. [4]

Answer:
Proteins are functionally diverse and classified as follows:
1. Enzymes: Catalyze biochemical reactions (e.g., DNA polymerase).
2. Transport proteins: Carry substances across membranes or in body fluids (e.g., Hemoglobin).
3. Structural proteins: Provide mechanical support and shape (e.g., Collagen).
4. Defensive/Antibody proteins: Protect against foreign pathogens (e.g., Immunoglobulins).
5. Hormonal/Regulatory proteins: Coordinate physiological activities (e.g., Insulin).

Teacher's Note:
a) Provide at least four to five major functional classes with standard biological examples.
b) Ensure examples match the functional category.

(c) Mention any two significant applications of bioinformatics. [2]

Answer:
1. Gene prediction and genome annotation to identify open reading frames, promoters, and functional protein domains.
2. Sequence alignment and phylogenetic analysis to study evolutionary relationships and homology between species.

Teacher's Note:
a) Highlight computational tools used in genomics and proteomics.
b) Keep applications distinct and scientifically rigorous.

Question 7.
(a) Discuss the applications of plant tissue culture emphasizing its role in obtaining :
(i) Stress-tolerant plants
(ii) Biodegradable plastic

Answer:
(i) Stress-tolerant plants: Plant tissue culture combined with genetic transformation enables the introduction of stress-tolerance genes (such as those for drought, salinity, or cold tolerance) into crop cells, which are then regenerated into whole plants.
(ii) Biodegradable plastic: Transgenic plants produced via tissue culture can express genes (such as those from bacteria for polyhydroxybutyrate, PHB) that synthesize biodegradable polymers in their plastids or chloroplasts on a commercial scale.

Teacher's Note:
a) Connect plant tissue culture to transgenic crop generation for abiotic stress resistance.
b) Mention PHB production in transgenic chloroplasts for biodegradable plastics.

(b) Briefly explain the principle and application of the technique - Southern blotting similar technique for protein identification.

Answer:
Principle of Southern Blotting: DNA fragments are separated by agarose gel electrophoresis, transferred onto a nitrocellulose or nylon membrane by capillary action, and hybridized with a labeled specific DNA probe to detect complementary sequences.
The similar technique for protein identification is Western Blotting, where proteins are separated by polyacrylamide gel electrophoresis (PAGE), transferred to a membrane, and detected using specific antibodies.

Teacher's Note:
a) Clearly distinguish Southern blotting (for DNA) from Western blotting (for proteins).
b) Mention probe hybridization in Southern blotting versus antibody-antigen interaction in Western blotting.

(c) What is DNA Micro-array technology' ?

Answer:
DNA microarray technology is a high-throughput molecular technique used to monitor gene expression levels or detect single nucleotide polymorphisms across thousands of genes simultaneously. It consists of microscopic spots of DNA oligonucleotides (probes) attached to a solid glass or silicon surface, which hybridize with fluorescently labeled cDNA or cRNA target samples.

Teacher's Note:
a) Emphasize the multiplex nature of microarray analysis for parallel gene expression profiling.
b) Mention probe-target hybridization and fluorescence detection.

Question 8.
(a) Give four differences between an inducible and a repressible operon. [4]

Answer:
1. Inducible operons are normally turned off and switched on in the presence of a specific substrate (inducer), whereas repressible operons are normally active and switched off in the presence of an end product (corepressor).
2. Inducible systems (e.g., lac operon) are typically involved in catabolic pathways, whereas repressible systems (e.g., trp operon) are involved in anabolic pathways.
3. In an inducible operon, the repressor protein synthesized by the regulator gene binds to the operator to block transcription until inactivated by the inducer.
4. In a repressible operon, the aporepressor alone cannot bind to the operator; it requires binding with a corepressor metabolite to form an active repressor complex.

Teacher's Note:
a) Contrast catabolic (inducible) versus anabolic (repressible) pathway roles.
b) Clearly explain the interaction of the inducer versus corepressor with the repressor protein.

(b) Discuss the impact of the following factors on enzyme activity': [4]
(i) Substrate concentration
(ii) Temperature
(iii) Presence of a competitive inhibitor
(iv) Enzyme concentration

Answer:
(i) Substrate concentration: Initially increases the rate of reaction linearly; at high concentrations, active sites become saturated, reaching a maximum velocity (Vmax).
(ii) Temperature: Reaction rate increases with temperature up to an optimum point due to increased kinetic energy; beyond this optimum, enzymes denature, causing a sharp drop in activity.
(iii) Presence of a competitive inhibitor: Competes with the substrate for binding at the active site, decreasing apparent affinity ($K_m$) but overcome by increasing substrate concentration.
(iv) Enzyme concentration: Increases the reaction rate proportionally as long as substrate is available in excess.

Teacher's Note:
a) Describe the graphical trends (saturation curves, bell-shaped temperature optimum curves).
b) Distinguish competitive inhibition kinetics clearly.

(c) Name any two disaccharides and give their structural units.

Answer:
1. Sucrose: Composed of $\alpha$-D-glucose and $\beta$-D-fructose linked by an $\alpha$(1$\rightarrow$2) glycosidic bond.
2. Lactose: Composed of $\beta$-D-galactose and $\beta$-D-glucose linked by a $\beta$(1$\rightarrow$4) glycosidic bond.

Teacher's Note:
a) Specify both monosaccharide subunits and the linkage type for each disaccharide.
b) Note anomeric configurations ($\alpha$ or $\beta$) for complete accuracy.

Question 9.
(a) Explain the following methods of selection of recombinant cell : [4]
(i) Insertional inactivation
(ii) Blue-white selection

Answer:
(i) Insertional inactivation: A method where the insertion of foreign DNA into a marker gene (e.g., tetracycline resistance gene in pBR322) disrupts its function, making the recombinant cell sensitive to that antibiotic while retaining resistance to another.
(ii) Blue-white selection: Uses the lacZ gene encoding $\beta$-galactosidase. Recombinants with foreign DNA interrupt the lacZ gene, failing to produce functional enzyme and forming white colonies on X-gal medium, whereas non-recombinants form blue colonies.

Teacher's Note:
a) Explain antibiotic resistance marker inactivation clearly.
b) Describe the enzymatic basis of blue-white screening using X-gal and $\beta$-galactosidase.

(b) Differentiate between : [4]
(i) Local and global sequence alignment
(ii) RAM and ROM

Answer:
(i) Global alignment forces alignment across the entire length of two sequences (best for closely related sequences of similar length), whereas local alignment finds regions of high similarity within longer sequences (best for divergent sequences).
(ii) RAM (Random Access Memory) is volatile, read-write memory used for temporary data storage, whereas ROM (Read-Only Memory) is non-volatile, permanent memory containing essential startup instructions.

Teacher's Note:
a) Contrast algorithmic goals in sequence alignment.
b) Clearly state volatility and write-capability differences between RAM and ROM.

(c) Name any two protein databases and two genomic databases. [2]

Answer:
Protein databases: UniProt and PIR (Protein Information Resource).
Genomic databases: GenBank and Ensembl.

Teacher's Note:
a) List standard bioinformatics repositories recognized globally.
b) Distinguish protein sequence databases from genomic sequence databases.