ISC Class 12 Biotechnology Board Exam Question Paper 2012 with Solutions

ISC Class 12 Biotechnology Board Exam Question Paper 2012 with Solutions

Part 1

Question 1.

(a) Mention any one significant difference between each of the following: [5]

(i) Hybrid and Cybrid. [1 Mark]

Answer:
A hybrid is produced through the fusion of protoplasts of two different plant species or varieties containing nuclear and cytoplasmic contributions from both parents, whereas a cybrid (cytoplasmic hybrid) contains the nucleus of only one species but cytoplasm from both parental species.

Teacher's Note:
a) Focus on the nuclear and cytoplasmic genetic constitution in both types of somatic fusion products.
b) Students often confuse cybrids as having two nuclei, whereas they possess a single parental nucleus combined with mixed cytoplasm.

(ii) DNA polymerase and Taq DNA polymerase. [1 Mark]

Answer:
Standard DNA polymerase is typically isolated from mesophilic organisms and is heat-sensitive, whereas Taq DNA polymerase is a thermostable enzyme isolated from the thermophilic bacterium Thermus aquaticus that can withstand high temperatures used in polymerase chain reactions.

Teacher's Note:
a) Emphasize the thermal stability property which makes Taq polymerase indispensable for PCR.
b) Mention the source organism correctly to secure full credit in board evaluations.

(iii) Glycosidic bond and Peptide bond. [1 Mark]

Answer:
A glycosidic bond is a covalent bond joining a carbohydrate molecule to another group (which may or may not be another carbohydrate), whereas a peptide bond is a chemical bond formed between the carboxyl group of one amino acid and the amino group of another amino acid.

Teacher's Note:
a) Identify glycosidic bonds with carbohydrates and peptide bonds with proteins (amino acids).
b) State the reacting functional groups clearly for both bonds.

(iv) Oils and Waxes. [1 Mark]

Answer:
Oils are esters of unsaturated fatty acids with glycerol and are liquid at room temperature with low melting points, whereas waxes are esters of long-chain fatty acids with long-chain monohydroxy alcohols other than glycerol.

Teacher's Note:
a) Highlight the alcohol component (glycerol versus long-chain monohydroxy alcohol) as the key distinguishing factor.
b) Note the physical state at room temperature as a supporting distinguishing feature.

(v) Homopolysaccharide and Heteropolysaccharide. [1 Mark]

Answer:
Homopolysaccharides are complex carbohydrates formed by the polymerization of only one type of monosaccharide monomer, whereas heteropolysaccharides are produced by the condensation of two or more different types of monosaccharide monomers or their derivatives.

Teacher's Note:
a) Use monomer composition (single type versus multiple types) as the primary basis of difference.
b) Giving examples such as starch for homo and chitin or agar for hetero adds clarity.

(b) Answer the following questions: [5]

(i) What is a callus? [1 Mark]

Answer:
A callus is a mass of meristematic, undifferentiated cells derived from plant tissue (explants) cultured in vitro.

Teacher's Note:
a) Key terms like meristematic and undifferentiated must be included.
b) Mention that it arises from plant explants under artificial culture conditions.

(ii) Name the method used for the sterilization of plant hormones and vitamins. [1 Mark]

Answer:
Millipore membrane filtration (using a filter paper with a pore size of 0.2 micrometer diameter).

Teacher's Note:
a) Explain that autoclaving denatures vitamins and hormones, necessitating filtration.
b) Mentioning the pore size (0.2 micrometer) demonstrates precise knowledge.

(iii) Why is DNA replication called semi-conservative replication? [1 Mark]

Answer:
DNA replication is called semi-conservative because out of the two strands in the newly formed daughter DNA molecules, one strand is conserved from the parent molecule and the other strand is newly synthesized.

Teacher's Note:
a) Clearly state the retention of one parental strand and synthesis of one new strand.
b) Reference to the Meselson and Stahl experiment can be included conceptually.

(iv) What is a promoter gene? [1 Mark]

Answer:
A promoter gene is a specific segment of DNA that binds RNA polymerase and initiates the transcription of a genetic code.

Teacher's Note:
a) Identify its precise location and role in binding RNA polymerase.
b) Do not confuse it with structural genes or operator genes.

(v) State two uses of stem cells. [1 Mark]

Answer:
1. Bone marrow transplants used to treat leukemia (blood cancer).
2. Treatment and potential cellular therapy for muscular dystrophy.

Teacher's Note:
a) Provide clear biomedical applications.
b) Keep answers concise and direct as required for 1-mark sub-parts.

(c) Write the full form of the following: [5]

(i) HGP [1 Mark]

Answer:
Human Genome Project.

Teacher's Note:
a) Check for accurate spelling of all expanded terms.
b) Avoid common capitalization or spelling errors.

(ii) STS [1 Mark]

Answer:
Sequence Tagged Sites.

Teacher's Note:
a) Ensure exact terminology from genomic mapping is used.
b) Verify abbreviations strictly against standard biotechnology nomenclature.

(iii) CSIR [1 Mark]

Answer:
Council of Scientific and Industrial Research.

Teacher's Note:
a) Standard institutional acronym.

Teacher's Note:
b) Write out the full name without missing any preposition.

(iv) LAF [1 Mark]

Answer:
Laminar Air Flow.

Teacher's Note:
a) Essential laboratory equipment abbreviation in plant tissue culture.
b) Ensure correct spelling of 'Laminar'.

(v) SCP [1 Mark]

Answer:
Single Cell Protein.

Teacher's Note:
a) Refers to protein derived from cultured microbial biomass.
b) Frequently tested acronym in industrial biotechnology.

(d) Explain briefly: [5]

(i) Amphipathic property of lipids. [1 Mark]

Answer:
Most membrane lipids possess both a hydrophilic (polar) head and a hydrophobic (non-polar) tail, allowing them to form bilayer structures like vesicles, liposomes, or cell membranes in an aqueous environment.

Teacher's Note:
a) Mention both polar and non-polar regions.
b) Relate the property to membrane structure formation.

(ii) Replication fork [1 Mark]

Answer:
The replication fork is the Y-shaped region where the two strands of DNA are unwound and separated to allow template-directed replication of each strand.

Teacher's Note:
a) Describe the structural appearance (Y-shaped) and functional significance.
b) Connect it directly to DNA unwinding during replication.

(iii) Androgenesis [1 Mark]

Answer:
Androgenesis is the development of an embryo containing only paternal chromosomes, typically resulting from the failure of the egg nucleus to participate in fertilization.

Teacher's Note:
a) Emphasize the exclusive presence of paternal chromosomes.
b) Distinguish it from normal amphimixis.

(iv) Transamination. [1 Mark]

Answer:
Transamination is the enzymatic process involving the transfer of an alpha-amino group from an amino acid to an alpha-keto acid.

Teacher's Note:
a) Specify the chemical groups involved (alpha-amino group and alpha-keto acid).
b) Highlight that it is a key pathway in amino acid metabolism.

(v) Active site. [1 Mark]

Answer:
The active site is a specific region on an enzyme molecule where substrates bind and undergo a chemical reaction.

Teacher's Note:
a) Define it as the catalytic binding pocket of an enzyme.
b) Essential concept in enzyme kinetics and catalysis.

Part 2

Question 2.

(a) Explain the general structure of an amino acid. What do you understand by essential and non-essential amino acids? [4 Marks]

Answer:
Amino acids are the building blocks of proteins, containing an amino group (-NH2) and a carboxyl group (-COOH) attached to a central alpha-carbon atom, along with a variable side chain (R group) and a hydrogen atom.
Essential amino acids: These are amino acids that cannot be synthesized by the human body and must be obtained through diet (e.g., valine, leucine, lysine).
Non-essential amino acids: These are amino acids that are synthesized within the body through metabolic pathways such as transamination (e.g., serine, alanine).

Teacher's Note:
a) Include structural components (amino, carboxyl, R group on alpha-carbon).
b) Distinguish clearly based on dietary requirement and internal synthesis.

(b) What are cloning vectors ? Write the main characteristics of any three types of cloning vectors. [4 Marks]

Answer:
A cloning vector is a self-replicating DNA molecule used to carry a foreign DNA insert into a host cell for amplification. Three main types of cloning vectors and their characteristics are:
1. Plasmids: Extra-chromosomal circular DNA molecules that replicate autonomously inside bacterial cells; cloning limit is typically 100 to 10,000 base pairs (0.1 to 10 kb).
2. Bacteriophages (e.g., Lambda and M13): Viral DNA molecules whose non-essential regions can be replaced with foreign DNA; cloning limit ranges from 8 to 20 kb.
3. Yeast Artificial Chromosomes (YACs): Artificial chromosomes containing telomeres, an origin of replication, a yeast centromere, and a selectable marker; capable of cloning large fragments up to 1 Mb.

Teacher's Note:
a) Define vector role clearly before listing types.
b) Mention cloning capacities for each vector type to ensure completeness.

(c) What is a Codon ? Name the start codon and any one end codon. [2 Marks]

Answer:
A codon is a unit of genetic code consisting of a sequence of three adjacent nucleotide bases in mRNA that codes for a specific amino acid during protein synthesis.
Start codon: AUG.
End codon: UAG (or UAA, or UGA).

Teacher's Note:
a) Mention triplet nature and mRNA context.
b) State AUG as methionine initiator and any one stop triplet accurately.

Question 3.

(a) Briefly explain the structure of a tRNA molecule. Mention its function during the process of protein synthesis. [4 Marks]

Answer:
Transfer RNA (tRNA) is a small RNA molecule (70-85 nucleotides) that folds into a characteristic clover-leaf structure in two dimensions and an L-shaped tertiary structure due to complementary base pairing. Key structural features include the amino acid binding site (3-prime end with CCA-OH), the anticodon loop for codon recognition, the T-psi-C loop for ribosome binding, and the DHU loop for enzyme binding.
Function: tRNA acts as an adapter molecule that transfers specific amino acids to the ribosomes during polypeptide synthesis, matching its anticodon to the mRNA codon.

Teacher's Note:
a) Describe both structural conformation and specific loop functions.
b) Emphasize the dual adapter role linking nucleic acid sequence to amino acid sequence.

(b) Give the stepwise procedure of sequencing of DNA by Sanger's method. [4 Marks]

Answer:
1. Preparation of single-stranded template DNA.
2. Addition of a mixture of all four normal deoxyribonucleotides (dATP, dGTP, dCTP, dTTP) in ample quantities and four fluorescently labeled dideoxynucleotides (ddATP, ddGTP, ddCTP, ddTTP) in limiting quantities along with DNA polymerase.
3. Chain elongation proceeds normally until a dideoxynucleotide is randomly incorporated, terminating further extension because it lacks a 3-prime OH group.
4. The resulting DNA fragments of varying lengths are separated by high-resolution gel electrophoresis, and an automated laser scanner detects the fluorescent tags to read the sequence.

Teacher's Note:
a) Clearly explain the role of dideoxynucleotides in chain termination.
b) Mention gel separation and fluorescence detection steps.

(c) What is Totipotency ? Give an example of a Totipotent cell. [2 Marks]

Answer:
Totipotency is the ability of a single cell to divide and produce all the differentiated cells in an organism, including extraembryonic tissues, and give rise to a complete new organism.
Example: A plant zygote or fertilized egg cell (or plant explant cells capable of callus regeneration).

Teacher's Note:
a) Define developmental potential comprehensively.
b) Provide a standard biological example like a zygote or spore.

Question 4.

(a) Briefly describe the essential components of the nutrient medium used for the plant tissue culture technique. Also, write the names of any two plant tissue culture media frequently used in the laboratory . [4 Marks]

Answer:
The essential components of a plant tissue culture nutrient medium are:
1. Inorganic nutrients: Essential macro-elements (N, P, K, Ca, S, Mg) and micro-elements (Fe, Zn, Mn, Cu, B, Mo) required for plant growth.
2. Vitamins: Organic supplements like thiamine (essential), inositol, pyridoxine, and nicotinic acid.
3. Carbon source: Sugars, most commonly sucrose (20-50 g/L), to provide energy.
4. Growth regulators: Auxins (e.g., IAA, 2,4-D) and cytokinins (e.g., kinetin, BAP) to stimulate cell division and organogenesis.
Two commonly used media: White's medium and Murashige and Skoog (MS) medium.

Teacher's Note:
a) List all key nutritional and hormonal categories.
b) Name MS medium and White's medium as standard laboratory formulations.

(b) With reference to suspension culture, explain the following : [4 Marks]

(i) Achemostat. [2 Marks]

Answer:
A chemostat is a continuous cell culture system in which a chosen nutrient is maintained at a growth-limiting concentration while fresh medium is added and an equal volume of culture is withdrawn at regular intervals, allowing the study of individual nutrient effects on cell growth.

Teacher's Note:
a) Clarify that nutrient limitation is intentionally controlled.
b) Distinguish from batch culture by mentioning continuous input and harvest.

(ii) A turbidostat. [2 Marks]

Answer:
A turbidostat is a continuous culturing method where the turbidity (cell density) of the culture is kept constant by automatically manipulating the feed rate of fresh medium in response to changes in culture turbidity.

Teacher's Note:
a) Highlight turbidity and cell density control as the regulating parameter.
b) Contrast with chemostat where nutrient concentration is the limiting factor.

(c) What are purines and pyrimidines ? Where are they located in a cell? [2 Marks]

Answer:
Purines and pyrimidines are nitrogenous bases that form the basic building units of nucleic acids (DNA and RNA). Purines are large double-ring structures (adenine and guanine), while pyrimidines are smaller single-ring structures (thymine, cytosine, and uracil).
Location: They are located within the nucleic acids found in the nucleus, mitochondria, chloroplasts, and cytoplasm of a cell.

Teacher's Note:
a) Differentiate structural sizes (double ring versus single ring) and specific base types.
b) Specify cellular locations accurately within DNA and RNA contexts.

Question 5.

(a) Explain giving an example how recombinant DNA technology can be used for the formation of the following: [4 Marks]

(i) A vaccine. [2 Marks]

Answer:
Recombinant DNA technology is used to clone genes encoding immunogenic surface proteins of pathogens into expression hosts like yeast or bacteria. For example, the Hepatitis B vaccine is produced by cloning the surface antigen gene of the hepatitis B virus into transgenic yeast cells, which then express the purified subunit protein used for immunization.

Teacher's Note:
a) Explain gene isolation and host expression principles.
b) Cite Hepatitis B vaccine as the standard textbook example.

(ii) A plant with delayed fruit ripening. [2 Marks]

Answer:
Genetic engineering can suppress enzymes responsible for fruit softening and ripening. For example, in the transgenic tomato FlavrSavr, the gene encoding polygalacturonase (an enzyme that degrades pectin and softens fruit) was inhibited using antisense technology, lowering enzyme activity and delaying ripening.

Teacher's Note:
a) Identify the target enzyme (polygalacturonase) and its role in softening.
b) Mention the FlavrSavr tomato as the classic commercial example.

(b) What is osmotic pressure ? Explain any one biochemical technique based on osmotic pressure. [4 Marks]

Answer:
Osmotic pressure is defined as the minimum pressure required to maintain equilibrium and prevent the net movement of solvent across a semi-permeable membrane. It is a colligative property dependent on solute molar concentration.
Technique (Dialysis): Dialysis is a biochemical technique used to separate molecules in solution based on differential rates of diffusion through a semi-permeable membrane. Small molecules pass through the pores while larger macromolecules (like proteins or DNA) are retained inside the dialysis tubing.

Teacher's Note:
a) Define osmotic pressure as a colligative property.
b) Describe dialysis clearly as a separation technique based on molecular size and membrane permeability.

(c) What are dextro-rotatory and laevo-rotatory substances? [2 Marks]

Answer:
Dextro-rotatory substances are optically active compounds that rotate the plane of polarized light to the right (clockwise), denoted by prefixes (+) or d (e.g., d-glucose).
Laevo-rotatory substances are compounds that rotate the plane of polarized light to the left (counter-clockwise), denoted by prefixes (-) or l (e.g., L-alanine).

Teacher's Note:
a) Define optical rotation direction for both categories.
b) Mention standard prefixes used for identification.

Question 6.

(a) Write short notes on : [4 Marks]

(i) Single nucleotide polymorphism. [2 Marks]

Answer:
Single nucleotide polymorphisms (SNPs, pronounced snips) represent variations at a single nucleotide position in a DNA sequence among individuals. They occur frequently across the human genome (millions of sites) and serve as crucial genetic markers for linkage mapping, disease association studies, and DNA fingerprinting.

Teacher's Note:
a) Define SNPs as single-base variations.
b) Highlight their significance in genomic mapping and personalized medicine.

(ii) Bioinformatics databases. [2 Marks]

Answer:
Bioinformatics databases are structured, computerized repositories of life sciences data gathered from experiments, literature, and high-throughput sequencing. They store nucleotide sequences, protein structures, and macromolecular interactions. Examples include ENA (European Nucleotide Archive), UniProt, and PDB (Protein Data Bank).

Teacher's Note:
a) Explain the purpose of biological databases.
b) Provide recognizable examples like UniProt or PDB.

(b) How are biomolecules separated by the following techniques: [4 Marks]

(i) Chromatography. [2 Marks]

Answer:
Chromatography separates biomolecules based on their differential distribution between a mobile phase (liquid or gas) and a stationary phase (solid or liquid). Components with varying solubilities and adsorption affinities travel through the stationary phase at different rates, leading to effective separation.

Teacher's Note:
a) Describe the roles of mobile and stationary phases.
b) Emphasize differential solubility and affinity as the mechanism of separation.

(ii) Centrifugation. [2 Marks]

Answer:
Centrifugation uses centrifugal force to sediment suspended particles or molecules in a liquid mixture based on their size, shape, and density. Denser components migrate outward and form a pellet at the bottom of the tube, while less dense components remain in the supernatant liquid.

Teacher's Note:
a) Mention centrifugal force and sedimentation principles.
b) Distinguish between pellet and supernatant formation.

(c) Give two differences between enzymes and inorganic catalysts: [2 Marks]

Answer:

Teacher's Note:
a) Present differences clearly in a comparative table.
b) Focus on chemical nature (protein vs inorganic) and reaction specificity.

Question 7.

(a) Differentiate between : [4 Marks]

(i) Prokaryotic genome and Eukaryotic genome. [2 Marks]

Answer:
Prokaryotic genome: Much smaller and simpler, lacks highly repetitive DNA, not bounded by a nuclear membrane, and consists of naked double-stranded DNA without histone proteins.
Eukaryotic genome: Larger and complex, contains highly repetitive DNA sequences, bounded by a nuclear membrane, and consists of DNA associated with histone proteins.

Teacher's Note:
a) Contrast genome size, complexity, and presence of histones.
b) Note the absence or presence of a true nuclear membrane.

(ii) Somatic embryo and Zygotic embryo. [2 Marks]

Answer:
Somatic embryo: Formed from somatic plant cells (other than egg cells) without fertilization, typically lacking endosperm or seed coat.
Zygotic embryo: Formed as a result of sexual double fertilization involving an egg cell, leading to seed development with associated endosperm and seed coat.

Teacher's Note:
a) Emphasize the origin of the cells (somatic vs zygotic after fertilization).
b) Mention the presence or absence of surrounding seed structures.

(b) Explain how a genomic DNA library is formed. How does it differ from cDNA library ? [4 Marks]

Answer:
Genomic library formation involves: (1) Isolating high molecular weight genomic DNA and digesting it with restriction enzymes, (2) Fractionating fragments using agarose gel electrophoresis, (3) Dephosphorylating and ligating fragments into suitable vectors (plasmids, phages, or cosmids), and (4) Introducing the recombinant vectors into host cells for amplification.
Difference from cDNA library: A genomic library contains the entire genomic DNA (including introns and non-coding sequences) of an organism, whereas a cDNA library contains only expressed genes (exons) synthesized from mRNA templates via reverse transcriptase.

Teacher's Note:
a) Outline the step-by-step construction of a genomic library.
b) Highlight the key genetic difference regarding introns and expressed sequences.

(c) Name any two inborn metabolic disorders in human beings. Also, write one main symptom for each of them. [2 Marks]

Answer:
1. Alkaptonuria: Symptom is that the patient's urine turns black upon standing in air due to the oxidation of accumulated homogentisic acid.
2. Phenylketonuria (PKU): Symptom is severe mental retardation resulting from impaired brain development caused by elevated plasma phenylalanine levels.

Teacher's Note:
a) Provide standard inherited metabolic disorders.
b) State distinct clinical or biochemical symptoms for each.

Question 8.

(a) Give the step-wise procedure of Southern blotting technique. Mention any two important applications of this technique. [4 Marks]

Answer:
Procedure of Southern blotting:
1. Genomic DNA is isolated and digested with restriction enzymes into fragments.
2. Fragments are separated by agarose gel electrophoresis according to size.
3. Separated DNA bands are denatured with alkali and transferred (blotted) onto a nylon or nitrocellulose membrane.
4. The membrane is hybridized with a radio-labeled DNA probe complementary to the target sequence.
5. Autoradiography is performed using a photographic film to visualize the hybrid bands.
Applications: (1) Detection of specific DNA sequences in genetic diagnosis, (2) DNA fingerprinting and restriction fragment length polymorphism (RFLP) analysis.

Teacher's Note:
a) List all steps from restriction digestion to autoradiographic detection.
b) Give relevant molecular biology applications.

(b) What are blunt ends and sticky ends? How are they formed? [4 Marks]

Answer:
Sticky ends: Single-stranded complementary overhangs produced when a restriction enzyme (like EcoRI) cuts DNA asymmetrically within a palindromic sequence.
Blunt ends: Flush, double-stranded ends produced when a restriction enzyme (like HaeIII) cuts both strands of DNA symmetrically at the center of the recognition sequence.
Formation: They are formed by the action of restriction endonucleases (Type II restriction enzymes) recognizing specific palindromic sequences on DNA molecules.

Teacher's Note:
a) Define both types of ends structurally.
b) Explain how symmetric versus asymmetric cleavage by restriction enzymes creates them.

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

Answer:
1. Alpha-Amylase: Widely used in the food industry (starch processing) and in laundry detergents.
2. Papain: Used in medicine, food processing (meat tenderizing), and the textile industry.

Teacher's Note:
a) Name common industrial enzymes correctly.
b) Specify their commercial applications clearly.

Question 9.

(a) Enlist the main steps in the regeneration of a complete plant from an explant. [4 Marks]

Answer:
1. Preparation and sterilization of a suitable nutrient medium.
2. Selection and surface sterilization of plant explants (e.g., shoot tips).
3. Inoculation of the explant into the sterile nutrient medium under aseptic conditions.
4. Incubation under controlled physical conditions (light, temperature, humidity) to induce callus formation and organogenesis.
5. Plantlet regeneration, hardening, and subsequent transfer to greenhouse or field conditions.

Teacher's Note:
a) Provide sequential steps from media preparation to plantlet acclimatization.
b) Include crucial aseptic and environmental control steps.

(b) Given below is a list of four bio molecules found in a living cell. For each of them, write the class of biomolecules they belong to and their location in a living cell: [4 Marks]

(i) Histones. [1 Mark]

Answer:
Class: Proteins.
Location: Nucleus (associated with eukaryotic DNA).

Teacher's Note:
a) Identify histones as basic nuclear proteins.
b) Mention correct cellular compartment.

(ii) mRNA. [1 Mark]

Answer:
Class: Nucleic acids (RNA).
Location: Nucleus and cytoplasm.

Teacher's Note:
a) Classify as ribonucleic acid.
b) Note synthesis in nucleus and function in cytoplasm.

(iii) Haemoglobin. [1 Mark]

Answer:
Class: Proteins (conjugated protein / metalloprotein).
Location: Cytoplasm of red blood cells (erythrocytes).

Teacher's Note:
a) Identify as a respiratory protein.
b) Specify localization in red blood cells.

(iv) Glycogen. [1 Mark]

Answer:
Class: Carbohydrates (polysaccharide).
Location: Cytoplasm of liver and muscle cells.

Teacher's Note:
a) Classify as a storage polysaccharide.
b) Mention primary storage organs (liver and muscles).

(c) Write any two uses of transgenic plants. [2 Marks]

Answer:
1. Development of insect-resistant crops (e.g., Bt cotton expressing insecticidal crystal proteins).
2. Enhancement of nutritional quality, such as Golden Rice enriched with beta-carotene.

Teacher's Note:
a) Give practical agricultural or nutritional applications.
b) Standard textbook examples like Bt crops or nutritional fortification are ideal.