CBSE Class 12 Biotechnology Question Paper 2026 Solved Code 99

SECTION A

1. The formation of inclusion bodies in microbial cell culture is associated with : [1 Mark]
(A) Incorrect folding and intracellular accumulation of recombinant proteins.
(B) Incorrect folding and extracellular secretion of recombinant proteins.
(C) Enhanced cell growth.
(D) Lack of nutrients in the medium

Answer: (A) Incorrect folding and intracellular accumulation of recombinant proteins.

Teacher's Note:
a) Inclusion bodies are dense clumps of wrongly folded recombinant protein that stay inside the host cell.
b) They are common when a foreign protein is over-expressed in E. coli.

2. The incubators used to culture mammalian cells are maintained at : [1 Mark]
(A) 30% CO₂
(B) 2% CO₂
(C) 5-10% CO₂
(D) 15-20% CO₂

Answer: (C) 5-10% CO₂

Teacher's Note:
a) CO₂ works with the bicarbonate buffer in the medium to keep the pH near 7.4.
b) Remember the standard conditions for animal cells: \( 37^{\circ}C \) and 5-10% CO₂.

3. The most common molecule used to spot on a DNA microarray is : [1 Mark]
(A) cDNA
(B) Peptide
(C) mRNA
(D) Antibody

Answer: (A) cDNA

Teacher's Note:
a) cDNA or oligonucleotides are fixed on the chip as probes, since DNA is stable.
b) mRNA from the sample is converted to labelled cDNA and hybridised to these spots.

4. Plant organ culture involves the in vitro growth of isolated _________. [1 Mark]
(A) Protoplast
(B) Plant cell
(C) Anther
(D) Callus

Answer: (C) Anther

Teacher's Note:
a) Organ culture means growing a whole plant organ such as an anther, ovary, root or shoot tip.
b) Protoplast, single cell and callus cultures are not organ cultures.

5. For E. coli, the dry cell weight of one billion cells is approximately : [1 Mark]
(A) 250 mg
(B) 150 mg
(C) 350 mg
(D) 450 mg

Answer: (B) 150 mg

Teacher's Note:
a) This is a direct textbook fact from the chapter on microbial cell culture.
b) Learn such standard values carefully, as options are often close to each other.

6. The natural defence mechanism of bacteria occurs for : [1 Mark]
(A) Destruction of bacteria's own DNA during replication.
(B) Facilitating efficient cloning process.
(C) Destruction of invading viral DNA
(D) Destruction of invading bacterial DNA

Answer: (C) Destruction of invading viral DNA

Teacher's Note:
a) Restriction enzymes cut the DNA of invading bacteriophages; this is called restriction.
b) The bacterium protects its own DNA by methylation (modification), so option (A) is wrong.

7. The technique of Nick Translation was developed by : [1 Mark]
(A) Rigby and Paul Berg in 1877
(B) Rigby and Paul Berg in 1977
(C) Linus Pauling in 1985
(D) Paul Berg in 1973

Answer: (B) Rigby and Paul Berg in 1977

Teacher's Note:
a) Nick translation is used to make labelled DNA probes for techniques like FISH.
b) Option (A) has the same names but a wrong year (1877), so read the dates carefully.

8. Locus link was primarily developed to : [1 Mark]
(A) Provide three dimensional structures of proteins.
(B) Store taxonomic information of species.
(C) Provide information on official gene names.
(D) Provide information on metabolic pathways.

Answer: (C) Provide information on official gene names.

Teacher's Note:
a) LocusLink (NCBI) gave a single query interface for gene names, symbols and locations.
b) Protein 3D structures are stored in PDB, not in LocusLink.

9. Which of the following is a function of collagen in human body ? [1 Mark]
(A) Providing strength to our bones.
(B) Production of hormones.
(C) Transport of oxygen to tissues
(D) Secretion of digestive enzymes

Answer: (A) Providing strength to our bones.

Teacher's Note:
a) Collagen is a fibrous structural protein found in bones, tendons, cartilage and skin.
b) Oxygen transport is the function of haemoglobin, not collagen.

10. Which of the following statements is correct for use in diagnostics and therapeutics ? [1 Mark]
(A) Monoclonal antibodies are preferred over polyclonal antibodies.
(B) Antibiotics are preferred over antibodies.
(C) Antibodies are not used at all.
(D) Only polyclonal antibodies are used.

Answer: (A) Monoclonal antibodies are preferred over polyclonal antibodies.

Teacher's Note:
a) Monoclonal antibodies come from a single clone, so they bind only one epitope with high specificity.
b) They can be produced in unlimited amounts from hybridoma cells.

11. Single Nucleotide Polymorphisms can occur : [1 Mark]
(A) only in the coding regions of a genome.
(B) only in the non-coding regions of a genome.
(C) in both coding and non-coding regions of a genome.
(D) in mitochondrial DNA only

Answer: (C) in both coding and non-coding regions of a genome.

Teacher's Note:
a) A SNP is a single base difference between individuals and can occur anywhere in the genome.
b) Words like "only" in options are often a sign of a wrong choice.

12. SCID (Severe Combined Immuno Deficiency) is caused due to deficiency of which enzyme ? [1 Mark]
(A) Adenosine deaminase
(B) Amylase
(C) Lipase
(D) Reverse Transcriptase

Answer: (A) Adenosine deaminase

Teacher's Note:
a) ADA deficiency leads to loss of T and B lymphocytes, so the immune system fails.
b) ADA-SCID was the first disease treated by gene therapy.

For Questions 13 to 16, two statements are given - one labelled Assertion (A) and other labelled Reason (R). Select the correct answer to these questions from the options (A), (B), (C) and (D) as given below :
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

13. Assertion (A) : The human embryonic stem cells can be derived from the inner cell mass of blastocyst and cultured in a suitable nutrient medium.
Reason (R) : It is possible to selectively remove a gene and make precise genetic modification in mouse embryonic stem cells. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) The Assertion is about the source of human ES cells; the Reason is about gene knock-out in mouse ES cells.
b) Both facts are correct, but one does not explain the other.

14. Assertion (A) : Mutation selection technique identifies micro organisms that have been improved by induced mutations.
Reason (R) : Mutations cannot be induced by irradiations in micro-organisms. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) UV rays and X-rays are commonly used to induce mutations in microbes, so the Reason is false.
b) Improved mutant strains are then picked out by screening and selection.

15. Assertion (A) : M13 bacteriophage uses rolling circle mechanism for DNA replication.
Reason (R) : During rolling circle mechanism, double stranded replicative form of DNA is formed which replicates until there are about its hundred copies in the host bacterial cell. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) M13 has single stranded DNA which is converted to a double stranded replicative form (RF) inside E. coli.
b) The RF copies itself to about a hundred copies and then single strands are made by rolling circle replication.

16. Assertion (A) : The gene that encodes for Barnase is used to induce male sterility in transgenic plants.
Reason (R) : Barnase is a RNA hydrolyzing enzyme that inhibits pollen formation in the tapetal cells of anther. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) The barnase gene is placed under a tapetum-specific promoter, so only the tapetal cells are destroyed.
b) Without a working tapetum, pollen does not develop and the plant becomes male sterile.

SECTION B

17. (a) On which plasmid are Bacterial Artificial Chromosome (BAC) vectors based ? Enlist the important sequences present in these vectors. [2 Marks]

Answer:
1. BAC vectors are based on the Fertility (F) plasmid of E. coli.
2. Important sequences present are the replication and maintenance genes of the F plasmid, and selectable markers (along with cloning sites).

Teacher's Note:
a) "F plasmid of E. coli" is the key term for the first mark.
b) Name any two sequences for the second mark: replication/maintenance genes, selectable marker, cloning sites.

OR

(b) Given below is a single stranded sequence of DNA :
5' ACTAGAATTCGCCA 3'
Which restriction enzyme would cleave this sequence in double stranded form ? Illustrate the fragments that will be generated upon digestion by this restriction enzyme. [2 Marks]

Answer:
1. The sequence contains the site GAATTC, so it is cut by EcoRI (between G and A on each strand).
Double stranded form:
5' ACTAGAATTCGCCA 3'
3' TGATCTTAAGCGGT 5'
2. Fragments formed after digestion (with sticky ends):
Fragment 1: 5' ACTAG 3' / 3' TGATCTTAA 5'
Fragment 2: 5' AATTCGCCA 3' / 3' GCGGT 5'

Teacher's Note:
a) Always write the complementary strand first, then locate the palindrome 5' GAATTC 3'.
b) EcoRI makes staggered cuts, giving 5' AATT overhangs (sticky ends) on both fragments.

18. Although not required for cell growth, antibiotics are one of the most essential components of animal cell culture medium. Why ? Give examples of two such antibiotics often used in animal cell culture medium. [2 Marks]

Answer:
1. Antibiotics are added to control the growth of contaminating bacteria and fungi in the rich culture medium, which would otherwise overgrow the slow-growing animal cells.
2. Examples: Penicillin and Streptomycin.

Teacher's Note:
a) The reason carries 1 mark and each antibiotic carries half a mark.
b) Mention "prevent microbial contamination" clearly in your reason.

19. What is meant by 'Gene knock out' ? How is it useful in developing mouse models for studying human diseases ? [2 Marks]

Answer:
1. Gene knock out means selectively removing (inactivating) a particular gene from mouse embryonic stem (ES) cells.
2. Knock-out mice lacking that gene show the disease condition, so they help to understand the genetic basis of a human disease (and to search for new diagnostic and therapeutic methods).

Teacher's Note:
a) Use the words "selectively removing a gene" and "mouse ES cells" in the definition.
b) Only one use is needed for the second mark.

For Visually Impaired Candidates (in lieu of Q. 19)

19. Why is an inverted microscope required to observe animal cell cultures ? [2 Marks]

Answer:
1. Animal cells attach and grow at the bottom of the culture flask or dish, and an inverted microscope lets us see these cells clearly.
2. This is because in an inverted microscope the optical system (objective) is at the bottom and the light source is on the top.

Teacher's Note:
a) Link the design (objective below, light above) with the position of cells (at the bottom).
b) It also allows cells to be seen through the base of the flask without opening it.

20. How is chymotrypsinogen converted to chymotrypsin in the duodenum ? What is the name of this process ? [2 Marks]

Answer:
1. In the duodenum, trypsin makes a proteolytic cut in the inactive chymotrypsinogen; this causes a conformational change that exposes the active site, forming active chymotrypsin.
2. This process is called in-situ activation.

Teacher's Note:
a) Chymotrypsinogen is a zymogen; it is made inactive to protect the pancreas from self-digestion.
b) Write the keywords "proteolytic cut", "active site exposed" and "in-situ activation".

21. Even if we know where the genes are present in a genome, it is not entirely clear how to count them. Give two reasons for this problem. [2 Marks]

Answer:
1. Overlapping genes: some genes overlap with each other, sharing the same stretch of DNA, so it is hard to decide where one gene ends and the next begins.
2. Splice variants: one gene can give many different mRNAs and proteins by alternative splicing, so it is unclear whether to count it as one gene or many.

Teacher's Note:
a) The two value points are "overlapping genes" and "splice variants", 1 mark each.
b) Add one short line of explanation for each term.

SECTION C

22. Briefly explain the process of sickling of Red Blood Cells (RBCs) during sickle cell anaemia. Why does sickle celled RBCs lead to anaemic conditions in patients ? [3 Marks]

Answer:
1. In sickle cell anaemia, glutamic acid at the 6th position of the beta chain of haemoglobin is replaced by valine.
2. This altered haemoglobin (HbS) sticks together to form long fibres inside the RBCs, which distort the cells into a sickle shape.
3. Sickle RBCs get stuck in narrow capillaries and are destroyed, and sickle haemoglobin has impaired oxygen carrying capacity, so the patient becomes anaemic.

Teacher's Note:
a) State the mutation exactly: valine replaces glutamic acid at position 6 of the beta chain.
b) Do not reverse the two amino acids; this is a very common mistake.

23. How does the metagenomics approach helps to identify novel genes present in an environment ? Explain the process. [3 Marks]

Answer:
1. Metagenomics studies DNA taken directly from an environmental sample, so it covers the genomes of both culturable and non-culturable microbes; this allows novel genes to be found even from bacteria that cannot be grown in the lab.
2. Process: Environmental sample collection and isolation of DNA, followed by DNA manipulation (cutting into fragments).
3. Ligation of fragments into vectors, cloning to build a metagenomic library, and screening by protein expression to identify novel genes and products.

Teacher's Note:
a) Most soil and water microbes cannot be cultured; this is why metagenomics is useful.
b) Write the steps in the correct order: isolation, manipulation, ligation, cloning/library, expression.

24. What is the use of introducing the property of delayed fruit ripening in some plants ? Explain the methods used for delaying fruit ripening. [3 Marks]

Answer:
1. Delayed ripening gives fruits a longer shelf life, so they can be transported over long distances without spoilage.
2. Ripening is slowed down by blocking or reducing the production of ethylene, the ripening hormone, in the fruit.
3. Ethylene-producing genes are introduced in a way (for example, in antisense orientation) that suppresses the plant's own ethylene production.

Teacher's Note:
a) Ethylene is the key word; delaying ripening means controlling ethylene.
b) Flavr Savr tomato is a well-known example of a delayed-ripening transgenic crop.

25. Mention the catalytic traid (along with diagram) which contributes to the enzymatic activity of native subtilisin. Why was this enzyme improved by protein engineering ? Write the method used in this process. [3 Marks]

Answer:
1. The catalytic triad of subtilisin is Serine 221, Histidine 64 and Aspartic acid 32 (Ser 221, His 64, Asp 32), which lie close together in the folded enzyme at the active site.
2. Native subtilisin is inactivated by bleach in detergents, because bleach oxidises methionine 222 next to the active serine.
3. By site directed mutagenesis (protein engineering), Met 222 was substituted by other amino acids such as alanine, so the enzyme keeps its activity and stability in the presence of bleach.

Teacher's Note:
a) In the diagram, show the three residues Ser 221, His 64 and Asp 32 grouped together at the active site.
b) Remember the problem residue: Met 222, which is oxidised by bleach.

For Visually Impaired Candidates (in lieu of Q. 25)

25. Since many centuries, whey has been used for the treatment of numerous ailments. Briefly elaborate upon the scientific relevance of it. [3 Marks]

Answer:
1. Whey proteins raise the level of glutathione in the body.
2. Glutathione detoxifies xenobiotics (foreign harmful chemicals).
3. It also protects cellular components from reactive oxygen intermediates and free radicals.

Teacher's Note:
a) Glutathione is the key word; each of the three points carries 1 mark.
b) Whey is the watery by-product of cheese making and is rich in proteins.

26. Selection methods are based on the expression or non-expression of certain traits in recombinant microbial cells. Enlist such any three traits. [3 Marks]

Answer:
1. Antibiotic resistance: only cells carrying the plasmid with the resistance gene (for example, for ampicillin or tetracycline) grow on a medium containing that antibiotic.
2. Expression of an enzyme such as beta-galactosidase, which changes the colour of colonies on a suitable substrate.
3. Expression of a protein such as GFP (green fluorescent protein), which makes the recombinant cells glow green under UV light.

Teacher's Note:
a) Dependence or independence on a nutrient such as leucine is also accepted as a trait.
b) Write exactly three traits, as asked.

27. (a) Explain briefly the procedures of the following techniques used in recombinant DNA Technology :
(i) Transfection
(ii) Electroporation
(iii) Biolistics [3 Marks]

Answer:
1. Transfection: foreign DNA is mixed with a charged substance such as calcium phosphate, cationic liposomes or DEAE dextran and overlaid on the recipient host cells, which take up the DNA.
2. Electroporation: a short pulse of electric current is used to create temporary microscopic pores in the host cell membrane, through which the rDNA enters.
3. Biolistics: a gene gun (particle gun) is used to shoot microscopic gold or tungsten particles coated with DNA into the host cells.

Teacher's Note:
a) Biolistics is mainly used for plant cells, which have a tough cell wall.
b) Name the chemical (calcium phosphate) and the metal particles (gold/tungsten) for full marks.

OR

(b) Describe the Blue-White selection method of screening for the presence of recombinant plasmids. [3 Marks]

Answer:
1. The foreign DNA is inserted within the lacZ gene (coding for beta-galactosidase) of the pUC19 plasmid, which inactivates this gene (insertional inactivation).
2. After transformation, E. coli host cells are plated on solid medium containing X-gal.
3. Colonies with non-recombinant plasmids make active beta-galactosidase and turn blue; colonies with recombinant plasmids (rDNA) cannot, so they stay white.

Teacher's Note:
a) Remember: white colonies = recombinant, blue colonies = non-recombinant.
b) The key term "insertional inactivation of lacZ" carries a full mark.

28. How can you differentiate between cancerous and non-cancerous cells in culture by microscopic examination ? Give one example of each of such cell types. [3 Marks]

Answer:
1. Contact inhibition: non-cancerous (normal) cells stop dividing when they touch each other and form a single layer (monolayer); cancerous cells lack contact inhibition and pile up in multiple layers.
2. Anchorage dependence: normal cells must attach to a surface to grow, while cancerous cells can grow in suspension without attachment.
3. Examples: non-cancerous cell line - CHO (Chinese Hamster Ovary) cells; cancerous cell line - HeLa cells.

Teacher's Note:
a) Other accepted differences: limited life span, density limitation, growth rate, change in ploidy or cell shape.
b) Any two differences carry 2 marks and the examples carry 1 mark.

SECTION D

Instructions : Q. Nos. 29 and 30 are case-based questions. Each of these questions have sub-parts [(i), (ii) and (iii)] with an internal choice in one sub-part.

29. In human beings, Chronic Mylogenous Leukemia (CML) is caused due to the reciprocal translocation between chromosome 9 and chromosome 22. This translocation is detected by Fluorescence in-situ hybridisation (FISH) technique in such patients, and severity of the disease is observed.

(i) Name the genes which fuse during translocation in Philadelphia chromosome. [1 Mark]

Answer: The abl gene (from chromosome 9) and the bcr gene (from chromosome 22) fuse to form the bcr-abl fusion gene.

Teacher's Note:
a) Each gene name carries half a mark.
b) The Philadelphia chromosome is the shortened chromosome 22 carrying the bcr-abl fusion.

(ii) During FISH technique, is any fluorescence detected in the der 9 chromosome ? Why ? [1 Mark]

Answer: No. The part of chromosome 9 carrying the abl gene is translocated to chromosome 22, so the fluorescent-labelled gene is absent from der 9 and no signal is seen there.

Teacher's Note:
a) "No" carries half a mark and the reason carries half a mark.
b) The probes bind only where their target genes (abl, bcr) are present.

(iii) (a) Briefly describe Nick Translation technique. [2 Marks]

Answer:
1. The enzyme DNase I creates single-stranded nicks (breaks) at random places in the double stranded DNA.
2. DNA polymerase I then starts at each nick, removes old nucleotides ahead of it and synthesises new DNA, incorporating labelled (fluorescent or radioactive) nucleotides, giving a labelled probe.

Teacher's Note:
a) Name both enzymes: DNase I (makes nicks) and DNA polymerase I (fills in).
b) The purpose is to make a labelled DNA probe.

OR

(b) How is CML detected by FISH techniques ? [2 Marks]

Answer:
1. Fluorescent DNA probes for the abl and bcr genes (in two different colours) are hybridised in situ with a smear of lymphocyte cells from the CML patient.
2. In CML cells, the two probes bind to the fused bcr-abl gene and show a merged (yellow) fluorescence colour under the fluorescence microscope, which confirms the translocation.

Teacher's Note:
a) Usually a red and a green probe are used; their merging gives a yellow signal.
b) Mention "in-situ hybridisation" and "fluorescence microscope" as keywords.

30. Microbial culture works are classified based on how nutrients are supplied and culture is maintained during the fermentation process as :
(A) Batch Culture : This is a closed system with fixed amount of nutrient medium.
(B) Fed-Batch culture : In this system, fresh nutrients are added during growth, but culture is not removed.
(C) Continuous culture : This is a culture system where fresh medium containing a limited nutrient is added when required and an equal volume of the culture is removed.

(i) Enlist any two limitations of batch culture. [1 Mark]

Answer: Nutrients get depleted over time, and toxic waste products accumulate in the medium, which limits growth.

Teacher's Note:
a) Other accepted points: small scale, cells face a continually changing environment, harvesting and adding medium cannot be done together.
b) Each limitation carries half a mark.

(ii) What does the term 'fed' in fed-batch culture refer to ? [1 Mark]

Answer: "Fed" refers to the addition of nutrients during cultivation, that is, the culture is continuously or sequentially fed with fresh medium.

Teacher's Note:
a) In fed-batch culture nutrients are added but no culture is removed.
b) Do not confuse it with continuous culture, where culture is also removed.

(iii) (a) Write the significance of using a turbidostat and a chemostat in continuous culture system. [2 Marks]

Answer:
1. A chemostat maintains a constant chemical environment (constant nutrient concentration) in the culture.
2. A turbidostat maintains a constant turbidity of the culture medium, that is, a constant cell concentration.

Teacher's Note:
a) Link "chemo" with chemical environment and "turbido" with turbidity (cell density).
b) Both devices keep the culture in a steady state.

OR

(b) Draw labelled graphical representations of Batch and continuous culture systems. [2 Marks]

[Figure: Two graphs with time on the x-axis. Batch culture: cell density [X] starts low and rises to a plateau; substrate concentration [S] falls with time; cell specific substrate turnover rate [QS] also falls. Continuous culture: [X], [QS] and [S] are all horizontal straight lines, staying constant with time.]

Answer:
1. Batch culture graph: plot against time; cell density [X] increases and then levels off, while substrate concentration [S] and cell specific substrate turnover rate [QS] decrease with time.
2. Continuous culture graph: cell density [X], [QS] and substrate concentration [S] all remain constant with time, shown as horizontal lines.
Where [X] = cell density, [S] = concentration of substrate, [QS] = cell specific substrate turnover rate.

Teacher's Note:
a) Each correctly labelled graph carries 1 mark; always label the axes and all curves.
b) The key contrast: changing values in batch culture versus steady values in continuous culture.

SECTION E

31. (a) State the principle of mass spectrometry. Describe the procedure of Matrix Assisted Laser Desorption Ionisation (MALDI), explaining the significance of the matrix in this technique. [5 Marks]

Answer:
1. Principle: mass spectrometry separates charged molecular ions according to their mass to charge (m/z) ratio.
2. Ionisation: the protein or peptide sample is mixed with a matrix and dried on a plate; a laser beam hits it, and the sample molecules are ionised.
3. Ion separation: the ions are accelerated in an electric field and separated according to their m/z ratio (for example, in a time-of-flight analyser, lighter ions reach first).
4. Detection: the ions reach the detector and the result is analysed as a mass spectrum.
5. Significance of matrix: the matrix absorbs the laser light energy and vaporises, transferring the sample from the condensed phase to the gas phase; it helps to volatilise and protonate peptides and proteins without breaking them.

Teacher's Note:
a) Write the three steps in order: ionisation, ion separation, detection.
b) The term "m/z ratio" is essential for the principle mark.
c) The matrix protects large molecules from direct laser damage.

OR

(b) Why do cereals and legumes have limited nutritional quality ? Describe two genetic engineering approaches used to improve protein quality in seeds. [5 Marks]

Answer:
1. Cereals and legumes have limited nutritional quality because their seed proteins are deficient in certain essential amino acids (cereals lack lysine, legumes lack sulphur-containing amino acids such as methionine).
2. Approach I: the genes that encode seed storage proteins are engineered to contain more nutritionally desirable amino acids.
3. This is done by inserting codons for additional essential amino acids, and by substituting existing amino acids with new ones in the storage protein.
4. Approach II: genes of entirely novel proteins that are highly enriched in specific essential amino acids are introduced into the plant.
5. Endogenous genes can also be modified so that the seed makes more of the limiting amino acid.

Teacher's Note:
a) The reason carries 1 mark and each approach carries 2 marks.
b) Give examples: lysine for cereals, methionine for legumes.
c) Use the phrase "seed storage proteins" in your answer.

32. (a) (i) How are virus-free plants produced from virus-infected plants ? State any two advantages of virus-free plants.
(ii) Elaborate on the production of artificial seed and illustrate its structure with a well-labelled diagram. [5 Marks]

[Figure: Structure of an artificial seed - a round bead showing an outer artificial seed coat (A), a somatic embryo at torpedo stage inside it (B), and the artificial endosperm filling the space around the embryo (C).]

Answer:
1. (i) Virus-free plants are produced by meristem culture: the apical or axillary meristem of the infected plant, which is usually free of virus, is cultured to regenerate whole plants.
2. Advantages: increased yield and better quality of produce (virus-free plants are obtained even from infected crop plants).
3. (ii) Artificial seeds are produced by encapsulating somatic embryos in a protective coating such as calcium alginate beads, or by desiccating the somatic embryos with or without coating.
4. Diagram: a round bead with the somatic embryo in the centre.
5. Labels: A - artificial seed coat; B - somatic embryo at torpedo stage; C - artificial endosperm.

Teacher's Note:
a) Meristems are virus-free because they have no vascular tissue and cells divide very fast.
b) Calcium alginate is the key encapsulating material to name.
c) The diagram needs all three labels for full marks.

OR

(b) Explain the steps involved in the development of Golden Rice through genetic engineering and mention its advantages in providing nutrition. How was Golden Rice further engineered ? [5 Marks]

Answer:
1. Three genes involved in the biosynthetic pathway were introduced into rice.
2. These genes code for enzymes needed for beta-carotene (provitamin A) production.
3. The genes were placed under the control of an endosperm-specific promoter, so the enzymes are made in the rice endosperm.
4. Advantage: provitamin A is produced in the grain, which helps prevent vitamin A deficiency and night blindness.
5. Golden Rice was further engineered by introducing three more genes from different organisms to increase the iron content and its absorption.

Teacher's Note:
a) Normal rice endosperm lacks beta-carotene; that is why these genes were added.
b) "Endosperm-specific promoter" is an important value point.
c) The grain looks yellow (golden) because of beta-carotene.

33. (a) (i) Describe the steps involved in a Polymerase Chain Reaction (PCR) technique. [3 Marks]
(ii) If a researcher starts with three double stranded DNA molecules, how many such molecules will be produced after 10 PCR cycles ? Explain briefly. [2 Marks]

Answer:
(i)
1. Denaturation: the reaction mixture is heated to about \( 94^{\circ}C \) so that the double stranded DNA separates into two single strands.
2. Annealing: the mixture is cooled to about \( 50\text{-}60^{\circ}C \) so that the two primers bind to their complementary sequences on the single strands.
3. Extension: at about \( 72^{\circ}C \), Taq DNA polymerase adds nucleotides to the primers and makes new complementary strands; the cycle is repeated many times.
(ii)
4. Each PCR cycle doubles the number of DNA molecules, so one molecule gives \( 2^{n} \) molecules after n cycles.
5. Number of molecules \( = 3 \times 2^{10} = 3 \times 1024 = 3072 \) double stranded DNA molecules.

Teacher's Note:
a) Mention the temperature with each step to earn full marks.
b) Do not forget to multiply by the starting number of molecules (3).
c) Taq polymerase is used because it is heat stable.

OR

(b) What is the full form of RFLP ? State its basic principle and mention any two important applications. [5 Marks]

Answer:
1. RFLP stands for Restriction Fragment Length Polymorphism.
2. Principle: there is variation in the size (length) of DNA fragments produced by a restriction enzyme between individuals of the same species.
3. After digestion with restriction enzymes, these fragments move differently during gel electrophoresis, giving different band patterns for different individuals.
4. Application 1: forensic science, to identify individuals or criminals (DNA fingerprinting).
5. Application 2: paternity or parentage testing.

Teacher's Note:
a) The variation arises because mutations create or remove restriction sites.
b) Write the full form correctly, as it alone carries 1 mark.

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