ISC Class 12 Biotechnology Sample Paper 2025 with Solutions

SECTION A - 14 MARKS

Question 1

(i) While performing the process of Western Blotting technique, the scientists use specific proteins. What type of proteins do the scientists use during this process? [1 Mark]

Answer:
Antibodies / Immunoglobulins

Teacher's Note:
a) Western blotting is an analytical technique used to detect specific proteins in a given sample of tissue homogenate or extract.
b) Remember that primary and secondary antibodies are specifically used to bind to the target protein immobilized on a membrane.

(ii) A forensic scientist discovered a tiny spot of blood at a crime scene. The sample underwent 10 PCR cycles for 40 minutes. Calculate how many copies of DNA would be present at the end. [1 Mark]

Answer:
1024

Teacher's Note:
a) The number of DNA copies generated after \( n \) cycles is calculated using the formula \( 2^n \), where \( n \) is the number of PCR cycles.
b) For 10 cycles, \( 2^{10} = 1024 \) copies are produced from a single starting template.

(iii) Which nitrogenous base will NOT be present in the genetic material of a eukaryotic cell? [1 Mark]
(a) Cytosine
(b) Guanine
(c) Thymine
(d) Uracil

Answer: (d) Uracil

The genetic material of eukaryotic cells is DNA, which contains thymine instead of uracil. Uracil is found in RNA.

Teacher's Note:
a) DNA contains adenine, guanine, cytosine, and thymine.
b) Uracil replaces thymine exclusively in RNA molecules.

(iv) The following proteins of given molecular weight are subjected to Gel electrophoresis.

Write the order of Sequence in which these proteins are isolated in a gel. [1 Mark]

Answer:
The sequence of proteins obtained from top (cathode) to bottom (anode) in a gel: Myosin > Insulin > Haemoglobin > Ribozyme > Keratin > Albumin.

Teacher's Note:
a) In gel electrophoresis, smaller molecules move faster and travel further through the gel matrix than larger molecules.
b) Arrange the proteins in decreasing order of molecular weight to show the migration path from origin (top) to the front (bottom).

(v) Give a reason for each of the following:

(a) A good vector used in rDNA technology must have atleast one selectable marker. [1 Mark]

Answer:
To select the host cell with recombinant DNA.

Teacher's Note:
a) Selectable markers help in identifying and eliminating non-transformants.
b) They selectively permit the growth of the transformants carrying the vector.

(b) Continuous culture is preferred over batch culture. [1 Mark]

Answer:
To obtain maximum production.

Teacher's Note:
a) In continuous culture, used medium is drained out from one side while fresh medium is added from the other to maintain cells in their physiologically most active exponential phase.
b) This method produces a larger biomass and higher product yield compared to batch culture.

(vi) Base sequence in one of the strands of DNA is 5’ - TAG CAT GAT - 3’.

(a) Write the base sequences of its complementary strand. [1 Mark]

Answer:
3’ - ATC GTA CTA - 5’

Teacher's Note:
a) Adenine pairs with thymine (A-T) and cytosine pairs with guanine (C-G) via hydrogen bonds.
b) Ensure the polarity of the complementary strand is correctly antiparallel (3’ to 5’).

(b) Explain the base complementarity rule. [1 Mark]

Answer:
Chargaff’s equivalence rule states that in double-stranded DNA, the number of adenine residues equals thymine residues, and the number of guanine residues equals cytosine residues.

Teacher's Note:
a) Purines always pair with specific pyrimidines due to structural and spatial constraints.
b) This principle maintains a uniform width of the DNA double helix.

(vii) Observe the following figure carefully and answer the questions that follow:

[Figure: A nucleosome showing histone octamer core labelled as B and DNA wrapped around it labelled as A, with linker DNA connected at the bottom C]

(a) Identify Figure 1. Where is this structure found? [1 Mark]

Answer:
Nucleosome, found in eukaryotic chromosomes.

Teacher's Note:
a) A nucleosome consists of a segment of DNA wound in sequence around eight histone protein cores.
b) It represents the fundamental subunit of chromatin packing in eukaryotes.

(b) What is its importance with respect to a prokaryotic and a eukaryotic organism? [1 Mark]

Answer:
It is found in eukaryotic chromosomes for supercoiling of DNA, and is absent in prokaryotic organisms.

Teacher's Note:
a) Packaging of long DNA molecules into compact structures is essential for nuclear organization in eukaryotes.
b) Prokaryotes lack true histones and organize their nucleoid using different polyamines and proteins.

(viii) Answer the following questions.

(a) Expand EMBL. [1 Mark]

Answer:
European Molecular Biology Laboratory

Teacher's Note:
a) EMBL is a core bioinformatics and molecular biology research institution.
b) Memorize exact standard expansions for major biological databases.

(b) What is the variation in length of DNA fragments due to inherited differences in highly repetitive DNA, known as? [1 Mark]

Answer:
VNTR (Variable Number of Tandem Repeats)

Teacher's Note:
a) VNTR belongs to a class of satellite DNA referred to as mini-satellites.
b) Each individual inherits these differences from their parents, making them useful in DNA fingerprinting.

(ix) Given below are two statements marked Assertion and Reason. Read the two statements carefully and choose the correct option.
Assertion: Trp operon is an inducible operon.
Reason: Tryptophan inhibits the process of formation of Lactose in Trp operon. [1 Mark]
(a) Both Assertion and Reason are true and Reason is correct explanation for assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (d) Both Assertion and Reason are false.

Trp operon is a repressible operon, not an inducible operon, and tryptophan is involved in repressing its own synthesis, not lactose formation.

Teacher's Note:
a) Operons involved in anabolic pathways like tryptophan synthesis are typically repressible.
b) Catabolic pathways like the lac operon are inducible.

(x) Given below are two statements marked Assertion and Reason. Read the two statements carefully and choose the correct option.
Assertion: In gel electrophoresis, molecules get separated according to their charge to mass ratio.
Reason: The more the charge on the molecule, more is its mass. [1 Mark]
(a) Both Assertion and Reason are true and Reason is correct explanation for Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (c) Assertion is true and Reason is false. (The official key shows (c) or Assertion is true but Reason is false.)

Molecules migrate based on their charge-to-mass ratio in certain electrophoretic systems, but having more charge does not inherently mean having more mass.

Teacher's Note:
a) Nucleic acids have a uniform charge-to-mass ratio, separating purely by size through a sieving gel.
b) Proteins separate based on net charge, size, and shape.

SECTION B - 28 MARKS

Question 2

(i) Figure 2 represents an experiment conducted in the laboratory. Observe the following figure carefully and answer the questions that follow: [2 Marks]

[Figure: Four centrifuge tubes labelled Controls, First generation, and Second generation showing different positions of DNA bands containing 14N and 15N isotopes]

(a) What was the objective of the experiment depicted in Figure 2?

Answer:
To prove that replication of DNA is semiconservative.

Teacher's Note:
a) This refers to the classic Meselson and Stahl experiment performed on E. coli.
b) It demonstrated that each newly synthesized DNA molecule consists of one old strand and one new strand.

(b) How was the DNA separated into different layers?

Answer:
Density Gradient Centrifugation using cesium chloride (CsCl).

Teacher's Note:
a) CsCl centrifugation creates a density gradient where DNA molecules settle at positions corresponding to their buoyant density.
b) Heavy 15N DNA, hybrid DNA, and light 14N DNA separate into distinct distinct bands.

(ii) A selectable marker is used in the selection of recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. [2 Marks]

(a) Mention the name of the mechanism involved. Which enzyme is responsible for the production of colour?

Answer:
Insertional Inactivation. Beta-galactosidase.

Teacher's Note:
a) Insertion of a foreign DNA sequence into the coding sequence of the enzyme beta-galactosidase leads to inactivation of the enzyme.
b) Non-recombinants produce blue colonies, while recombinants appear colourless (white).

(b) How is it advantageous to use chromogenic substrate method over using antibiotic resistant gene, to select the recombinants?

Answer:
Selection of recombinants due to inactivation of antibiotics requires simultaneous plating on two plates having different antibiotics.

Teacher's Note:
a) The blue-white screening method allows rapid visual identification in a single plating step.
b) It avoids the cumbersome duplicate plating procedures required for antibiotic resistance markers.

Question 3 [4 Marks]

(i) Answer the following questions:

(a) In general, DNA carries a gene for a particular protein. To produce this protein, DNA synthesises RNA and RNA synthesises protein. Sometimes, few viruses contain only RNA but no DNA, still they synthesise protein. Elaborate the process of formation of DNA in such viruses.

Answer:
1. Retroviruses contain single-stranded RNA as their genetic material.
2. Upon infecting a host cell, the viral enzyme reverse transcriptase uses the single-stranded RNA template to synthesize a complementary DNA (cDNA) strand.
3. The RNA-DNA hybrid is acted upon by the ribonuclease H activity of reverse transcriptase to degrade the RNA strand.
4. A second DNA strand is then synthesized to form a double-stranded DNA molecule that integrates into the host genome.

Teacher's Note:
a) This process violates the original central dogma by showing information flow from RNA to DNA.
b) Mentioning the enzyme reverse transcriptase is essential for full credit.

(b) During plant cell culture, certain growth regulators are required for proper plant culture. Describe the role of such growth regulators.

Answer:
1. Growth regulators such as auxins and cytokinins control cell division, elongation, and differentiation in plant tissue culture.
2. Auxins promote root initiation, callus formation, and cell elongation.
3. Cytokinins stimulate shoot development, cell division, and overcome apical dominance.
4. The precise ratio of auxins to cytokinins determines whether roots, shoots, or an undifferentiated mass of cells (callus) will develop.

Teacher's Note:
a) Plant growth regulators are vital synthetic or natural hormones added to nutrient media.
b) Emphasize the auxin-to-cytokinin balance for successful organogenesis.

OR

(ii) Answer the following questions:

(a) The wavelength that are absorbed by the nucleic acid and the efficiency of its absorption during its estimation, depends both on the structure and concentration of the molecules. Elaborate any one method of nucleic acid estimation based on absorption of light.

Answer:
1. Nucleic acid estimation is commonly performed using UV spectrophotometry based on Beer - Lambert’s Law.
2. DNA and RNA strongly absorb ultraviolet light at a wavelength of \( 260\text{ nm} \) due to the presence of aromatic rings in purine and pyrimidine bases.
3. The amount of light absorbed is directly proportional to the concentration of nucleic acid in the sample solution.
4. An absorbance value of 1.0 at \( 260\text{ nm} \) corresponds to approximately \( 50\,\mu\text{g/mL} \) for double-stranded DNA.

Teacher's Note:
a) Mentioning the peak absorbance wavelength of \( 260\text{ nm} \) is critical.
b) The ratio of absorbance at \( 260\text{ nm} \) to \( 280\text{ nm} \) is used to assess sample purity from protein contamination.

(b) Radhika wishes to find out the sequence of the DNA of a strawberry plant. Suggest the steps she should follow to isolate the DNA from the strawberry plant.

Answer:
1. Maceration: Crush the strawberry plant tissue mechanically to break open the cell walls.
2. Cell Lysis: Treat the homogenate with detergents like SDS (Sodium Dodecyl Sulphate) to solubilize cell membranes and nuclear envelopes.
3. Enzyme Treatment: Add enzymes such as cellulase, pectinase, and protease to degrade cell wall components, polysaccharides, and associated proteins.
4. Precipitation: Add chilled ice-cold ethanol to precipitate purified DNA strands, which appear as white thread-like filaments.

Teacher's Note:
a) Plant DNA extraction requires mechanical disruption combined with enzymatic digestion because of tough cellulose walls.
b) Chilled ethanol is necessary for the precipitation step.

Question 4 [4 Marks]

State any two significant differences between the following:

(i) Purine bases and Pyrimidine bases

Answer:

Teacher's Note:
a) Purines are larger nine-membered heterocyclic compounds.
b) Pyrimidines are smaller six-membered heterocyclic compounds.

(ii) Leading strand and Lagging strand

Answer:

Teacher's Note:
a) DNA polymerase can only synthesize in the 5’ to 3’ direction.
b) This antiparallel nature of DNA strands creates the continuous and discontinuous replication modes.

Question 5 [4 Marks]

(i) A study by few scientists, reports a decrease in non-target insect populations in areas where Bt crops are extensively cultivated.

(a) Explain the mode of action of the Bt toxin that provides resistance to the plants against insect infestations. [2 Marks]

Answer:
1. Bacillus thuringiensis produces insecticidal protein crystals containing inactive protoxins.
2. When ingested by susceptible insect pests, the alkaline pH of the insect midgut solubilizes the crystals.
3. Gut proteases activate the protoxin into active toxin molecules.
4. The active toxin binds to midgut epithelial cells, creating pores that cause cell swelling, lysis, and eventual death of the insect.

Teacher's Note:
a) The toxin is specifically activated in alkaline environments.
b) Emphasize the pore-formation mechanism in the midgut.

(b) What is the source of Bt toxin? [2 Marks]

Answer:
Bacillus thuringiensis

Teacher's Note:
a) It is a soil bacterium that naturally produces crystal proteins.
b) Genes encoding these proteins are isolated and introduced into crop plants like cotton and corn.

OR

(ii) Eli Lily is one of the first pharmaceutical companies to produce human insulin using rDNA technology by cell based fermentation method.

(a) Explain how Eli Lily synthesised the human insulin. [2 Marks]

Answer:
1. Eli Lily prepared two DNA sequences corresponding to human insulin chains A and B.
2. These DNA sequences were introduced into plasmids of Escherichia coli host cells to produce the respective insulin chains separately.
3. The chains A and B were extracted and purified from the bacterial cultures.
4. Finally, chains A and B were combined and linked together by disulfide bonds to form functional human insulin (humulin).

Teacher's Note:
a) Human insulin is composed of two short polypeptide chains linked by disulfide bridges.
b) Separate production prevents folding complications within bacterial systems.

(b) How was insulin obtained before the advent of rDNA technology? [2 Marks]

Answer:
It was obtained from the pancreas of slaughtered animals, specifically from calves (bovine insulin) or pigs (porcine insulin).

Teacher's Note:
a) Animal-sourced insulin frequently caused allergic reactions and immune responses in diabetic patients due to minor structural differences.
b) Recombinant human insulin eliminated these immunological complications.

Question 6 [4 Marks]

Answer the following questions:

(i) A gene was being ligated to the plasmid vector to prepare a recombinant DNA. An exonuclease was added to the tube accidentally. How will it affect the next step of the experiment? [2 Marks]

Answer:
There will be no effect on the experiment. This is because a recombinant DNA is circular and closed with no free ends. Therefore, the exonuclease will not degrade the DNA.

Teacher's Note:
a) Exonucleases require free 5’ or 3’ ends to initiate nucleotide removal.
b) Circular plasmids lack terminal ends, protecting them from exonucleolytic degradation.

(ii) Observe the following DNA sequence carefully.
5’- A T C G A A T T C T A C -3’
3’- T A G C T T A A G A T G -5’ [2 Marks]

(a) Identify the specific sequence that is acted upon by a particular endonuclease enzyme.

Answer:
5’ - G A A T T C - 3’
3’ - C T T A A G - 5’
(Or with cleavage indicator: 5’ - G\( \downarrow \)AATTC - 3’ / 3’ - CTTAA\( \uparrow \)G - 5’)

Teacher's Note:
a) Restriction endonucleases scan DNA for specific recognition sites.
b) Identify the palindromic hexanucleotide sequence within the given strand.

(b) What term is used for the sequence that is acted upon by the endonuclease? What is the name of the endonuclease enzyme that acts on the above sequence?

Answer:
Palindromic sequence. EcoRI.

Teacher's Note:
a) A DNA palindrome reads the same forwards and backwards on opposite strands when read in the 5’ to 3’ direction.
b) EcoRI specifically recognizes and cleaves the GAATTC sequence.

Question 7 [4 Marks]

Falak is starting a new experiment involving the germination of seeds for subsequent tissue culture. Outline a step-by-step protocol for sterilizing the following:

(i) Instruments [1 Mark]

Answer:
Dry heat sterilization in a hot air oven.

Teacher's Note:
a) Surgical instruments and glassware are typically sterilized using dry heat or autoclaving.
b) Prevents microbial contamination of culture vessels.

(ii) Culture medium [1 Mark]

Answer:
Wet heat sterilization in an autoclave.

Teacher's Note:
a) Media are subjected to steam under pressure at \( 121^{\circ}\text{C} \) for 15-20 minutes.
b) Ensures complete destruction of heat-resistant bacterial spores.

(iii) Explants [1 Mark]

Answer:
Surface sterilization or chemical sterilization using NaOCl or HgCl₂.

Teacher's Note:
a) Plant tissues carry natural surface microflora that must be eliminated without killing plant cells.
b) Followed by thorough rinsing with sterile distilled water.

(iv) Vitamins [1 Mark]

Answer:
Membrane filtration.

Teacher's Note:
a) Vitamins and heat-labile organic supplements are destroyed by autoclaving.
b) They are sterilized by passing through bacterial-proof membrane filters (pore size \( 0.22\,\mu\text{m} \)).

Question 8 [4 Marks]

Richard, a young marine biotechnologist, is designing an oil-eating bacteria for bioremediation using genetic engineering. He has chosen a particular bacterial strain and introduced specific genes to enhance its oil-digesting capabilities. Enumerate the steps Richard took while designing the recombinant bacteria for bioremediation. [4 Marks]

Answer:
1. Selection of bacterial strain: Pseudomonas putida was chosen as the host strain.
2. Identification of catabolic plasmids: Different strains of Pseudomonas contain individual plasmids carrying genes encoding enzymes that digest components of petroleum such as octane, naphthalene, camphor, and xylene.
3. Gene isolation and combination: Specific catabolic genes from different plasmids were isolated and introduced into a single bacterial strain.
4. Recombinant creation: Using genetic engineering techniques, Richard successfully introduced these multiple plasmids into one single bacterial strain, creating a superbug capable of simultaneously digesting all major petroleum components.

Teacher's Note:
a) Mentioning Pseudomonas putida and its multi-plasmid nature is crucial.
b) Bioremediation utilizes genetically engineered microorganisms to clean up environmental pollutants.

SECTION C - 28 MARKS

Question 9

(i) What is a monolayer culture? Describe how feeding and nutrient distribution are managed in large-scale monolayer cultures using roller bottles. [4 Marks]

Answer:
1. Definition: When the bottom of a culture vessel is covered with a continuous layer of cells one cell thick, they are called monolayer cultures.
2. Anchorage dependence: Monolayer cultures are anchorage-dependent, requiring a scale-up process by increasing the surface area of the substrate in proportion to cell number and medium volume.
3. Roller bottle setup: A round bottle or tube is rolled slowly around its horizontal axis by motorized rollers.
4. Nutrient distribution: As the bottle rotates, the liquid nutrient medium washes continuously over the inner surface, bathing the attached cells, ensuring efficient gaseous exchange, continuous feeding, and uniform nutrient distribution.

Teacher's Note:
a) Explain anchorage dependence clearly.
b) Highlight the mechanical rotation of roller bottles for large-scale production.

(ii) Mention any three types of alignments developed by BLAST. [3 Marks]

Answer:
1. Local alignment
2. Global alignment
3. Pairwise sequence alignment (or Multiple sequence alignment)

Teacher's Note:
a) BLAST (Basic Local Alignment Search Tool) is a fundamental bioinformatics algorithm.
b) Mentioning any three recognized alignment modes is sufficient.

OR

(iii) What is the purpose of synchronizing cells in a suspension culture? Discuss any three chemical methods commonly used for synchronizing suspension cultures. [4 Marks]

Answer:
1. Purpose: A synchronous culture is a culture in which cell cycles or specific phases for the majority of cultured cells occur simultaneously so that the cell culture grows uniformly and can be analyzed at specific cell cycle stages.
2. Chemical Method 1 - Starvation: Depriving cells of essential nutrients or growth factors halts cell cycle progression at a specific checkpoint.
3. Chemical Method 2 - Mitotic arrest: Using chemical inhibitors like colchicine or colcemid to arrest cells at metaphase by blocking spindle formation.
4. Chemical Method 3 - Metabolic inhibition: Applying specific metabolic blockers (such as hydroxyurea or thymidine block) to reversibly inhibit DNA synthesis and synchronize cells at the G1/S boundary.

Teacher's Note:
a) Define synchronization clearly before listing methods.
b) Chemical agents like inhibitors and starvation protocols are standard laboratory techniques.

(iv) Mention any three major database sources. [3 Marks]

Answer:
1. NCBI (National Center for Biotechnology Information)
2. EMBL (European Molecular Biology Laboratory)
3. DDBJ (DNA Data Bank of Japan)
(Other acceptable answers: SWISS-PROT, GenBank, GENSCAN)

Teacher's Note:
a) These three primary international nucleotide sequence database collaborators exchange data daily.
b) Memorize standard abbreviations for biological repositories.

Question 10

(i) Restriction enzymes are also called molecular scissors. They cut the DNA at specific sites. They are of different types. Discuss different types of restriction enzymes used in rDNA technology. Who discovered these enzymes? [4 Marks]

Answer:
1. Discovery: Restriction enzymes were discovered by Werner Arber, and later isolated and characterized by Hamilton Smith and Daniel Nathans.
2. Type I Restriction Enzymes: These are complex enzymes that possess both endonuclease and methylase activities, cleaving DNA at random sites far (\( > 1000\text{ bp} \)) from their recognition sequences.
3. Type II Restriction Enzymes: These are most commonly used in recombinant DNA technology because they recognize specific palindromic sequences and cleave DNA within or near the recognition site without requiring ATP.
4. Type III Restriction Enzymes: These enzymes recognize specific sequences and cleave DNA at a short distance (\( \sim 24\text{ - }26\text{ bp} \)) downstream of the recognition site, requiring ATP hydrolysis.

Teacher's Note:
a) Type II enzymes are the standard tools used in genetic engineering due to their precise cutting behavior.
b) Credit Werner Arber for the discovery.

(ii) Explain the 3 - D structure of DNA. [3 Marks]

Answer:
1. Double helix model: DNA exists as a 3-D double-stranded helix winding around a central axis in a right-handed direction, proposed by Watson and Crick.
2. Dimensions and grooves: Each turn of the helix is \( 34\text{ Angstroms} \) (\( 3.4\text{ nm} \)) long and contains approximately 10 base pairs, with a constant diameter of \( 20\text{ Angstroms} \). The backbone coiling creates alternating major and minor grooves.
3. Backbone and pairing: The hydrophilic sugar-phosphate backbones form the outer edges, while nitrogenous base pairs stack horizontally inside perpendicular to the axis, linked by hydrogen bonds.

[Figure: Diagram of B-DNA double helix showing 34 Angstrom pitch, 10 base pairs per turn, 3.4 Angstrom distance between adjacent base pairs, 20 Angstrom diameter, major and minor grooves, and base pairing of A-T and G-C]

Teacher's Note:
a) Mention key structural dimensions (\( 34\,\mathring{\text{A}} \), \( 20\,\mathring{\text{A}} \)).
b) Describing major and minor grooves completes the 3-D description.

Question 11

(i) Figure 3 shows an important event that occurs during normal cellular metabolism. Study the figure given below and answer the questions that follow: [4 Marks]

[Figure: Transcription bubble showing DNA double helix, RNA polymerase enzyme labelled E, growing mRNA transcript, coding strand A, and template strand]

(a) Which process is being depicted in Figure 3?

Answer:
Transcription

Teacher's Note:
a) Transcription is the synthesis of an RNA molecule from a DNA template strand.
b) Recognized by the formation of an RNA transcript from a DNA template bubble.

(b) Explain the role of part labelled ‘E’.

Answer:
Part E is the enzyme called RNA polymerase involved in transcription.

Teacher's Note:
a) RNA polymerase catalyzes the polymerization of ribonucleotides.
b) It moves along the template strand in the 3’ to 5’ direction.

(c) Which factor helps the part ‘E’ to initiate the process shown above? What is importance of strand A?

Answer:
Sigma factor. Strand labelled A is the coding strand.

Teacher's Note:
a) The sigma factor (\( \sigma \)) recognizes promoter sites to initiate transcription.
b) The coding strand has the same sequence as the newly synthesized mRNA (except thymine is replaced by uracil).

(d) If the sequence of strand A is 5’ - A T G C A C T A G C T A C G - 3’, then what should be the sequence on the newly formed strand?

Answer:
5’ - A U G C A C U A G C U A C G - 3’

Teacher's Note:
a) The newly formed mRNA transcript has the same sequence as the coding strand, with uracil replacing thymine.
b) Ensure proper 5’ to 3’ polarity.

(ii) Figure 4 shows an important process used in forensics. Study the figure given below and answer the questions that follow. [3 Marks]

[Figure: DNA fingerprinting flowchart showing DNA samples from Person A and Person B, restriction digestion, fragment generation, gel electrophoresis separation, and banding patterns]

(a) Which material is collected from Person A and Person B to start the process depicted in Figure 4? Briefly discuss the role of this material in the body of an individual?

Answer:
DNA, genetic material in the human body. It carries hereditary information responsible for directing protein synthesis, regulating cellular activities, and transmitting genetic traits across generations.

Teacher's Note:
a) Somatic cells from blood, hair follicles, or saliva provide genomic DNA samples.
b) Mention genetic transmission and protein control.

(b) Explain the method of obtaining the fragments collected from sample A and sample B.

Answer:
Using restriction enzymes (restriction endonuclease digestion).

Teacher's Note:
a) Genomic DNA is digested with specific restriction enzymes to cut DNA at specific sites.
b) Produces various fragments of differing lengths known as RFLPs.

(c) Describe the process of obtaining the bands from sample A and sample B?

Answer:
Gel electrophoresis.

Teacher's Note:
a) DNA fragments are loaded onto an agarose gel and subjected to an electric field.
b) Fragments separate by size, yielding distinct banding patterns after staining.

Question 12

Khalid, David and Anuradha were working on yeast genome. They performed a process to digest the yeast genome and plasmid vector containing Lac Z gene with EcoRI restriction enzyme. The genomic fragment that we obtained was mixed with the cut vectors using DNA ligase. The yeast genome was inserted within the Lac Z gene of the plasmid E. coli cells were transformed with the ligation mix containing the rDNA and plated on solid agar medium containing X-Gal. They observed that some plasmid vectors self -ligated and did not carry the yeast genome.

(i) Name the enzyme that was used to ligate the yeast genome with the plasmid. [1 Mark]

Answer:
DNA ligase

Teacher's Note:
a) DNA ligase seals phosphodiester backbones between DNA fragments.
b) Essential for joining inserts into vector backbones.

(ii) Describe the technique they can use to distinguish the non-recombinant vector from the recombinant ones. [1 Mark]

Answer:
Blue white method of screening the recombinants (Blue-white screening).

Teacher's Note:
a) Based on insertional inactivation of the beta-galactosidase gene.
b) Recombinants produce white colonies while non-recombinants form blue colonies.

(iii) Why is this method of selection referred to insertional inactivation of antibiotic resistance gene? [3 Marks]
(Note: The question wording refers to antibiotic resistance gene while contextually describing Lac Z, following the official marking scheme.)

Answer:
1. Insertion of foreign DNA occurs within the coding sequence of an enzyme marker gene located on the plasmid vector.
2. This disrupts the reading frame and inactivates the gene function.
3. Consequently, the resulting enzyme cannot hydrolyze the chromogenic substrate X-Gal, enabling visual colorimetric identification as it can help in identification based on colour with high sensitivity and visual detection.

Teacher's Note:
a) Explain how gene disruption causes loss of functional enzyme synthesis.
b) Highlight visual sensitivity advantages.

(iv) How could they have prevented self-ligation of the plasmid? [2 Marks]

Answer:
By using Alkaline phosphatase enzyme.

Teacher's Note:
a) Alkaline phosphatase removes terminal 5’ phosphate groups from the linearized vector.
b) This prevents self-ligation because DNA ligase requires a 5’ phosphate group to form phosphodiester bonds.