CBSE Class 12 Biotechnology Sample Paper 2026 27 with Solutions PDF Download

SECTION A

1. An organism has 20% of its genome as genes and the genome is divided into four chromosomes. Identify this organism. [1 Mark]
A. Arabidopsis thaliana
B. Humans
C. Drosophila
D. Yeast

Answer: C. Drosophila

Teacher's Note:
a) Drosophila (fruit fly) has only four pairs of chromosomes, which makes it a classic model organism.
b) Link the clue "four chromosomes" directly to Drosophila to save time in the exam.

2. Subtilisin is inactivated due to oxidation of amino acid [1 Mark]
A. histidine
B. methionine
C. aspartate
D. serine

Answer: B. methionine

Teacher's Note:
a) Oxidation of a methionine residue near the active site inactivates subtilisin.
b) This is a standard example of protein engineering: replacing this methionine makes the enzyme resistant to oxidation.

3. The charge relay system in serine proteases helps in [1 Mark]
A. stabilising the substrate by forming disulfide bonds.
B. activating the serine residue for nucleophilic attack.
C. lowering the pH of the active site.
D. transporting electrons through the electron transport chain.

Answer: B. activating the serine residue for nucleophilic attack.

Teacher's Note:
a) The charge relay system makes the serine residue reactive so that it can attack the substrate.
b) Do not confuse it with electron transport, which happens in mitochondria.

4. What is the main characteristic of a turbidostat in a continuous microbal culture system? [1 Mark]
A. Maintains a constant rate of nutrient feed
B. Promotes maximum biomass accumulation
C. Increases nutrient supply over time
D. Ensures constant cell concentration

Answer: D. Ensures constant cell concentration

Teacher's Note:
a) A turbidostat keeps the cell concentration (turbidity) constant.
b) Remember: turbidostat - constant turbidity; chemostat - constant nutrient feed rate.

5. Meristem is used as an explant for : [1 Mark]
A. Obtaining virus resistant plants
B. Promoting somaclonal variations
C. Producing commercially important secondary metabolites
D. Producing Virus free plants

Answer: D. Producing Virus free plants

Teacher's Note:
a) Meristem culture gives virus free plants, not virus resistant plants.
b) Read options A and D carefully, as "free" and "resistant" mean different things.

6. A student is interested in amplifying a specific region of genome of an organism. Which of the following would actually help her in targeted amplification? [1 Mark]
A. Probes
B. Primers
C. rNTPs
D. ddNTPs

Answer: B. Primers

Teacher's Note:
a) In PCR, primers bind to the two ends of the target region, so only that region is amplified.
b) Probes are used for detection, and ddNTPs are used in DNA sequencing, not in amplification.

7. There is no increase in the cell concentration in the lag phase due to the following reason - [1 Mark]
A. Exhaustion of the medium.
B. Space constraint
C. Accumulation of toxic products
D. Acclimatization to the new environment

Answer: D. Acclimatization to the new environment

Teacher's Note:
a) In the lag phase, cells are adjusting to the new medium, so they do not divide yet.
b) Exhaustion of medium and toxic products are reasons for the stationary phase, not the lag phase.

8. How does the inclusion of phenol red in many cell culture media demonstrate a practical application of understanding the medium's properties? [1 Mark]
A. Phenol red directly contributes to cell growth by providing essential nutrients.
B. Phenol red acts as a visual indicator, simplifying pH monitoring and maintenance of optimal cell culture conditions.
C. Phenol red is essential for maintaining the correct osmolality of the medium.
D. Phenol red is a necessary cofactor for many enzymatic reactions within the cell.

Answer: B. Phenol red acts as a visual indicator, simplifying pH monitoring and maintenance of optimal cell culture conditions.

Teacher's Note:
a) Phenol red is a pH indicator; its colour change shows the pH of the medium.
b) It is not a nutrient or a cofactor.

9. Somatic embryos are [1 Mark]
A. identical with zygotic embryos and without seed coats
B. identical with zygotic embryos and with seed coats
C. non-identical with zygotic embryos and without seed coats
D. non-identical with zygotic embryos and with seed coats

Answer: A. identical with zygotic embryos and without seed coats

Teacher's Note:
a) Somatic embryos look like zygotic embryos but have no seed coat.
b) This is why they are encapsulated to make artificial seeds.

10. Which of the following protein pharmaceutical is not produced by CHO cell line? [1 Mark]
A. Erythropoietin
B. Factor IX
C. Interleukin 2
D. Monoclonal antibody

Answer: D. Monoclonal antibody

Teacher's Note:
a) Erythropoietin, Factor IX and Interleukin 2 are listed as products of the CHO (Chinese Hamster Ovary) cell line.
b) Monoclonal antibodies are linked with hybridoma cells, so option D is the answer as per the marking scheme.

11. The correct sequence of steps in the technique of Microarray is:
(i) Choosing a cell population
(ii) Conversion to cDNA molecules
(iii) Extraction of mRNA molecules
(iv) Fluorescent labelling
(v) Scanning and interpretation [1 Mark]
A. (i), (ii), (iv), (iii), (v)
B. (i), (ii), (iii), (v), (vi)
C. (i), (iii), (iv), (ii), (v)
D. (i), (iii), (ii), (iv), (v)

Answer: D. (i), (iii), (ii), (iv), (v)

Teacher's Note:
a) mRNA must be extracted first before it can be converted into cDNA.
b) The cDNA is then labelled with fluorescent dye, and the chip is scanned last.

12. The vectors suitable for transformation of animal cells are: [1 Mark]
A. SV 40 vectors and papilloma virus vectors
B. Cauliflower mosaic virus (CMV) vector and Gemini vectors
C. Lambda phage and M13 phage vectors
D. T4 phage vectors

Answer: A. SV 40 vectors and papilloma virus vectors

Teacher's Note:
a) SV 40 and papilloma viruses are animal viruses, so they are used as vectors for animal cells.
b) Cauliflower mosaic virus and Gemini viruses are plant vectors, and phages infect bacteria.

For Questions number 13 to 16, two statements are given one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.

13. Assertion: Proteins have been engineered to show minimum deleterious effects.
Reason: Proteins are the main molecules which provide stimulus for immunity. [1 Mark]
A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
B. Both Assertion (A) and Reason (R) are true, but 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 Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) Proteins can trigger immune responses, so therapeutic proteins are engineered to reduce harmful effects.
b) The Reason explains why the engineering is needed, so option A is correct.

14. Assertion: The Somatic cell hybridization offers an excellent alternative for obtaining distant hybrids.
Reason: Somatic hybrids are created by fusing intact cells of distantly related plants. [1 Mark]
A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
B. Both Assertion (A) and Reason (R) are true, but 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) Somatic hybrids are made by fusing protoplasts (cells without cell wall), not intact cells.
b) Watch for the word "intact" - it makes the Reason false.

15. Assertion: Auxins are added to the culture medium if callus induction is desired.
Reason: Nature and quantity of auxin added depends on source of explant. [1 Mark]
A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
B. Both Assertion (A) and Reason (R) are true, but 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 Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) Both statements are true, but the Reason does not explain why auxins induce callus.
b) Always ask: does the Reason answer "why" for the Assertion?

16. Assertion: Genetic variation between individuals is exploited in DNA fingerprinting.
Reason: Not all genetic variations are useful in DNA fingerprinting. [1 Mark]
A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
B. Both Assertion (A) and Reason (R) are true, but 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 Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) DNA fingerprinting is based on differences in DNA between individuals.
b) The Reason is true but does not explain the Assertion.

SECTION B

17. What is specific growth rate in microbial growth kinetics? Write the relation between specific growth rate and doubling time. [2 Marks]

Answer:
1. Specific growth rate \( (\mu) \) is an index of the rate of growth of cells in a particular environment. It is a characteristic of the microorganism and depends on the growth environment, such as temperature, pH, medium composition and dissolved oxygen.
2. A strain with a higher specific growth rate has a lower doubling time, and vice versa: \( t_d = \frac{0.693}{\mu} \)

Teacher's Note:
a) Write the formula \( t_d = \frac{0.693}{\mu} \) clearly; it carries 1 mark.
b) Mention that the relation is inverse: higher \( \mu \) means shorter doubling time.

18. Attempt either option A or B. [2 Marks]

A. Give any four requirements to carry out the polymerase chain reaction. [2 Marks]

Answer:
1. DNA template that is to be amplified.
2. Two primers - short oligonucleotides (usually 10-18 nucleotides long) that hybridise to the target region, one on each strand.
3. A heat stable DNA polymerase (stable above \( 80^{\circ}C \)), such as Taq polymerase from a thermostable bacterium.
4. Deoxynucleotide triphosphates (dNTPs) and buffer.

Teacher's Note:
a) Each requirement carries half a mark, so list exactly four.
b) Mention that the DNA polymerase must be heat stable, as PCR uses high temperatures.

OR

B. Describe any two methods to directly introduce foreign DNA into the host cells. [2 Marks]

Answer:
1. Microinjection: Foreign DNA is injected directly into animal or plant cells using a fine syringe under a phase contrast microscope.
2. Biolistics: Microscopic particles of gold or tungsten are coated with the DNA of interest and bombarded onto cells with a particle gun.

Teacher's Note:
a) Electroporation (a high voltage pulse makes temporary pores in the membrane) is also accepted as a method.
b) Name the method and give one line on how it works for full marks.

19. A researcher is using baffle flasks for small scale microbial culture in the lab. How are baffle flasks different from ordinary flasks? [2 Marks]

Answer:
1. Baffle flasks have V-shaped notches or indentations in the sides of the flask.
2. These increase the turbulence of the agitated culture medium, which improves oxygen transfer and hence the growth of microbes.

Teacher's Note:
a) One mark is for the structure (notches) and one for the benefit (better oxygen transfer).
b) Use the keywords "turbulence" and "oxygen transfer".

20. What is protoplast culture? Write any two of its uses. [2 Marks]

Answer:
1. Protoplasts are plant cells without cell wall. They are isolated by enzymes (cellulases, hemicellulases and pectinases) from leaf, seedling, calli etc. and grown in culture.
2. Uses: (a) Fusion of two somatic cells to create somatic hybrids. (b) Various biochemical and metabolic studies.

Teacher's Note:
a) Other accepted uses are creating cybrids and genetic manipulation.
b) Name the enzymes used for removing the cell wall to make the definition complete.

21. Explain the key difference between polyclonal and monoclonal antibodies in terms of their binding specificity to antigens. How does this difference arise from their cellular origins? [2 Marks]

Answer:
1. Polyclonal antibodies are a mixture that binds to many different epitopes on an antigen, because they are made by different B lymphocytes, each recognising a different part of the antigen. Monoclonal antibodies bind to only a single epitope.
2. Monoclonal antibodies are highly specific because they are produced by a single clone of immortalised B lymphocytes (hybridoma), derived from one antigen-activated B cell that recognises that one epitope.

Teacher's Note:
a) Key words: "multiple epitopes" for polyclonal and "single epitope" for monoclonal.
b) Mention "single clone" and "hybridoma" to explain the origin.

SECTION C

22. A kidney transplant patient is prescribed OKT3 to prevent acute rejection. Explain how OKT3 modifies T-cell activity? [3 Marks]

Answer:
1. OKT3 is an immunosuppressant that binds to the CD3 molecule on the surface of T cells, which are the key cells in acute graft rejection.
2. This binding blocks the normal function of T cells, so they cannot attack the transplanted kidney.
3. When OKT3 therapy is over, the OKT3 molecules are slowly cleared from the body, and the T cells resume their normal immune function within a week.

Teacher's Note:
a) The keywords "CD3 molecule" and "T cells" earn the first mark.
b) Do not forget the third point: the effect is reversible after the therapy ends.

23. Write about any three ways that allow us to perform gene analysis using Bioinformatics. [3 Marks]

Answer: Bioinformatics processes raw sequence information in these ways:
1. DNA sequence information is processed to identify genes.
2. The proteins encoded by the genes can be inferred, and gene function can be determined.
3. Regulatory sequences can be identified, and phylogenetic relationships can be inferred.

Teacher's Note:
a) Any three of these are accepted: finding genes, inferring proteins, finding gene function, identifying regulatory sequences, inferring phylogeny.
b) Write each way as a separate point so the examiner can count three.

24. A. Identify the parts of fermenter which are labelled as X, Y and Z.
B. "A transgene may be added, which encodes an enzyme to modify a metabolite produced by the organism to yield a new product of interest". Will this be considered as strain improvement? Justify giving an example. [3 Marks]

[Figure: Diagram of a stirred tank fermenter with a motor on top driving a central shaft with three sets of blades inside the culture broth. Labels shown: Motor, pH controller, Acid/base reset and pump, Steam, Sterile seal, Exhaust, Cooling water out, Culture broth, Cooling water in, Steam (at the bottom outlet), Sterile air (entering at the bottom). Label X is on the left side at the level of the upper blades, label Y is on the left side at the level of the middle blades, and label Z is on the right side near the bottom of the vessel, just above the sterile air inlet.]

Answer:
A. X - Impeller; Y - Cooling jacket; Z - Sparger (air bubbles).
B. Yes, this is considered strain improvement. The transgene (foreign gene) is introduced by recombinant DNA technology, a method of genetic engineering that gives the strain improved characters.

Teacher's Note:
a) Each correctly named part in A carries half a mark.
b) In B, first say "Yes", then justify with "recombinant DNA technology / genetic engineering".

For Visually Impaired Candidates (in lieu of Q. 24)

A. What is the capacity of a small scale biofermentor used in laboratories? Mention any two uses of such bioreactors.
B. "A transgene may be added, which encodes an enzyme to modify a metabolite produced by the organism to yield a new product of interest". Will this be considered as strain improvement? Justify giving an example. [3 Marks]

Answer:
A. Small scale fermentors used in research laboratories have a capacity of 10-100 litres. Uses: (1) to optimise the various parameters for the growth of microbes; (2) to produce enough quantity of metabolites from microbes for research.
B. Yes, this is considered strain improvement. The transgene (foreign gene) is introduced by recombinant DNA technology, a method of genetic engineering that gives the strain improved characters.

Teacher's Note:
a) Write the capacity with its unit (10-100 litres).
b) In B, the answer "Yes" carries half a mark and the justification carries one mark.

25. Attempt either option A or B. [3 Marks]

A. A group of Indian scientists observed that group of patients feeding on whey are responding better to retro-viral therapy as compared to those who are not taking it. What could be the possible scientific explanation for this therapeutic effect? [3 Marks]

Answer:
1. Whey proteins raise the level of the tripeptide glutathione (gamma-glutamyl cysteinyl glycine) in cells.
2. Glutathione is a reducing compound with many functions, including detoxification of xenobiotics.
3. It also protects cell components from the effects of oxygen intermediates and free radicals.

Teacher's Note:
a) The key word is "glutathione" - write its full form as a tripeptide.
b) Mention both roles: detoxification and protection from free radicals.

OR

B. Write about the three parameters that define the dietary value of proteins. [3 Marks]

Answer:
1. PER (Protein Efficiency Ratio): the weight gained by consuming 1 g of food protein.
2. BV (Biological Value): the amount of protein nitrogen retained from a given amount of protein nitrogen consumed.
3. BCAA (Branched Chain Amino Acid) profile: important for the biosynthesis of muscle proteins.

Teacher's Note:
a) Write the full form of each abbreviation along with its meaning.
b) Remember the three as "PER, BV, BCAA".

26. A. Illustrate the steps involved in the generation of RFLP.
B. What is the RFLP pattern expected in case of identical twins. [3 Marks]

Answer:
A. Steps:
1. Isolation of genomic DNA.
2. Digestion of the DNA into fragments by a specific restriction enzyme.
3. Separation of the fragments.
4. Performing agarose gel electrophoresis to see the fragment pattern.
B. The RFLP pattern of identical twins would be exactly the same.

Teacher's Note:
a) Each step in A carries half a mark, so write all four in the correct order.
b) Identical twins have the same DNA, so their restriction fragment patterns match.

27. Look at the diagram and answer the following questions
A. Derive at the original DNA sequence by reading above given autoradiogram.
B. Why are ddNTPs used along with dNTPs in the enzymatic method of DNA sequencing?
C. How is a dNTP chemically different from a ddNTP? [3 Marks]

[Figure: Sequencing gel with four lanes labelled ddA, ddT, ddG and ddC. An arrow on the left points downward from "larger strands" (top) to "Smaller strands" (bottom). From top to bottom, the bands are in lanes: ddG, ddT, ddA, ddG, ddC, ddG, ddA, ddT, ddC, ddA.]

Answer:
A. Reading the gel from bottom to top gives the sequenced strand: 5' ACTAGCGATG 3'. The original DNA is its complement: 3' TGATCGCTAC 5'.
B. ddNTPs are used to terminate the growing chain at specific bases.
C. A ddNTP lacks the 3'-OH group that is present in a dNTP.

Teacher's Note:
a) Always read the gel from the smallest band (bottom) to the largest (top); this gives the 5' to 3' sequence.
b) Without the 3'-OH group, the next nucleotide cannot be joined, so the chain stops.

For Visually Impaired Candidates (in lieu of Q. 27)

The sequence of the strand of DNA complementary to the sequenced strand is
5' ACGCCCGAGTAGCCCAGA 3'
A. Write the sequence of the original strand.
B. Which method is commonly used to sequence DNA?
C. Why is automated DNA sequencing considered safer? [3 Marks]

Answer:
A. 5' TCTGGGCTACTCGGGCGT 3'
B. Dideoxy chain termination method.
C. Automated DNA sequencers do not use radioactive labels, so they are considered safer.

Teacher's Note:
a) Write the complementary bases and reverse the order so that the strand reads 5' to 3'.
b) Always mark the 5' and 3' ends in a DNA sequence answer.

28. A. Based on the visual cues, Identify the technique which is depicted here?
B. Expand ICM. From which developmental stage is it obtained, and why is this stage ideal for pluripotency research? [3 Marks]

[Figure: A vector carrying a gene is introduced into cells in a culture dish by electroporation (labelled "Electroporation"). A side panel labelled "Gene of interest" shows the vector lining up with a chromosome and exchanging the gene ("Homologous recombination"). The cells are then grown on a dish labelled "Selection with resistance marker", and the selected cells are injected into a hollow ball-shaped embryo.]

Answer:
A. Homologous recombination / creation of a gene knockout.
B. ICM - Inner Cell Mass. It is obtained from the blastocyst stage. The ICM has pluripotent embryonic stem cells that can form all tissues, so it is ideal for research on regeneration and genetic disorders.

Teacher's Note:
a) The labels "Homologous recombination" and "Selection with resistance marker" in the figure are the visual cues.
b) B has three parts: full form, stage and reason - answer all three.

For Visually Impaired Candidates (in lieu of Q. 28)

A. Differentiate between finite and continuous cell lines based on their lifespan in culture.
B. Herceptin is a monoclonal antibody used to treat HER2 - positive breast cancer. Based on the information provided, what is the role of the HER2 receptor in cancer cell growth, and how does Herceptin interfere with this process to inhibit cancer progression? [3 Marks]

Answer:
A. Finite cell lines have a limited lifespan and stop dividing after some time. Continuous cell lines are immortal and can grow indefinitely in culture.
B. 1. HER2 is a cell surface receptor that receives growth signals and promotes cell growth and division. In HER2-positive breast cancer, it is overexpressed, causing excess growth signals and uncontrolled growth of cancer cells.
2. Herceptin binds specifically to HER2 receptors on cancer cells. This blocks the growth signals, disrupts the signalling pathway and stops the cancer from progressing.

Teacher's Note:
a) Use the words "limited lifespan" and "immortal" for A.
b) In B, the word "overexpressed" and the idea of "blocking the receptor" carry the marks.

SECTION D

29. Chronic Myelogenous Leukemia is caused due to Reciprocal translocation mutation between chromosome 9 and 22 that forms an extra-long chromosome 9 and extra small chromosome 22 containing the fused gene regions. Refer to the schematic view representing metaphase chromosome and answer the question that follow: [4 Marks]

[Figure: Three panels showing chromosomes 9 (green) and 22 (purple). Panel 1: normal chromosome 9 with a red band labelled P on its lower arm, and chromosome 22 with a band labelled Q. Panel 2: the end pieces of both chromosomes break off. Panel 3: arrows show the pieces exchanged, giving a longer chromosome 9 carrying a pink piece and a small chromosome 22 carrying a green piece, with the red bands joined.]

A. Identify the gene region P and Q in the given diagram.

Answer: P is the abl gene and Q is the bcr gene.

Teacher's Note:
a) abl is on chromosome 9 and bcr is on chromosome 22.
b) Do not swap the two gene names.

B. What is the extra small resultant chromosome 22 called?

Answer: Philadelphia chromosome.

Teacher's Note:
a) The Philadelphia chromosome carries the fused bcr-abl gene.
b) Spell "Philadelphia" correctly.

Attempt either subpart C or D.

C. Apart from Karyotyping, which other technique can be used to detect CML in patients? State the principle of the technique.

Answer:
1. FISH - Fluorescence in Situ Hybridization.
2. Principle: fluorescently labelled probes hybridise to the specific DNA regions on the chromosomes, and the fluorescent signal shows where the target sequence lies.

Teacher's Note:
a) Write the full form of FISH.
b) The phrase "fluorescently labelled probes" is the key point for the principle.

OR

D. Why is Karyotyping not, a preferred method to find out the severity of CML disease?

Answer: Karyotyping needs chromosomes arrested at metaphase, and it is also not an easy procedure. So it is not preferred for finding the severity of CML.

Teacher's Note:
a) Mention both points: metaphase arrested chromosomes are needed, and the procedure is difficult.
b) This is why probe based methods like FISH are preferred.

For Visually Impaired Candidates (in lieu of Q. 29)

Chronic Myelogenius Leukemia is caused due to reciprocal translocation between chromosome 9 and 22 that forms an extra-long chromosome 9 and an extra small chromosome 22 containing the fused gene regions. Both the chromosomes are metaphase arrested and visible during Karyotyping. Answer the questions that follow: [4 Marks]

A. What are the names given to both the gene regions of chromosome 9 and 22 that fuse?

Answer: The gene region on chromosome 9 is abl, and the gene region on chromosome 22 is bcr.

Teacher's Note:
a) Link each gene to its chromosome: abl - 9, bcr - 22.
b) Together they form the fused bcr-abl gene.

B. What is the extra small chromosome 22 known as?

Answer: Philadelphia chromosome.

Teacher's Note:
a) This is a one word answer worth 1 mark.
b) Spell "Philadelphia" correctly.

Attempt either subpart C or D.

C. Apart from Karyotyping, which other technique can be used to detect CML in patients? What is its principle?

Answer:
1. FISH - Fluorescence in Situ Hybridization.
2. Principle: fluorescently labelled probes hybridise to the specific DNA regions on the chromosomes, and the fluorescent signal shows where the target sequence lies.

Teacher's Note:
a) Write the full form of FISH.
b) The phrase "fluorescently labelled probes" is the key point for the principle.

OR

D. Why is Karyotyping not, a preferred method to find out the severity of CML disease?

Answer: Karyotyping needs chromosomes arrested at metaphase, and it is also not an easy procedure. So it is not preferred for finding the severity of CML.

Teacher's Note:
a) Mention both points: metaphase arrested chromosomes are needed, and the procedure is difficult.
b) This is why probe based methods like FISH are preferred.

30. Mass spectrometry (MS) has emerged as an important tool for characterisation. A protein with a molecular weight of 10,000 dalton generates five different peaks with the ions containing 5, 4, 3, 2, and 1 charges, respectively, as shown below. [4 Marks]

[Figure: A horizontal m/z axis from 0 to 10,000. Five protein ions are drawn above it with charges 5+, 4+, 3+, 2+ and 1+ from left to right. The 1+ ion is at 10,000, and ions with more charges lie at smaller m/z values, closer to 0.]

A. What happens if there is a loss of charge from a biomolecule?

Answer: Molecular ions are generated. Ions are formed by a loss or gain of charge, for example by electron ejection, protonation or deprotonation.

Teacher's Note:
a) Use the term "molecular ions" in your answer.
b) Give the examples: electron ejection, protonation, deprotonation.

B. Mass spectrometry is an analytical tool. Justify the statement.

Answer: Mass spectrometry is used to obtain protein structural information such as peptide mass or amino acid sequence. It is also used to identify the type and location of amino acid modifications in proteins.

Teacher's Note:
a) Any one use is enough for 1 mark.
b) Connect the use to "analysis" of protein structure.

Attempt either subpart C or D.

C. Calculate the m/z ratio each for protein ions containing 4 and 2 charges.

Answer:
Formula: \( m/z = \frac{M + nH}{n} \), where M = 10,000 and each added proton adds 1 to the mass.
For n = 4: \( m/z = \frac{10000 + 4}{4} = 2501 \)
For n = 2: \( m/z = \frac{10000 + 2}{2} = 5001 \)

Teacher's Note:
a) Write the formula first; it carries marks along with the substitution.
b) Remember to add the mass of the n protons to M before dividing by n.

OR

D. A protein has a molecular weight of 20,000 daltons and it forms two protein ions containing 6 and 7 charges, what will be its mass/charge ratio?

Answer:
Formula: \( m/z = \frac{M + nH}{n} \), where M = 20,000.
For n = 6: \( m/z = \frac{20000 + 6}{6} = 3334.33 \)
For n = 7: \( m/z = \frac{20000 + 7}{7} = 2858.14 \)

Teacher's Note:
a) Show the substitution for both charges separately.
b) Check: the ion with more charges always has the smaller m/z value.

For Visually Impaired Candidates (in lieu of Q. 30)

Mass spectrometry (MS) has emerged as an important tool in biotechnology. It is extremely useful in obtaining protein structural information such as peptide mass or amino acid sequences. The molecular ions are generated either by a loss or gain of a charge (e.g. electron ejection, protonation or deprotonation). After the ions are formed, they can be separated according to their m/z ratio and finally detected. [4 Marks]

A. What happens if there is a loss of charge from a biomolecule? Give two reasons.

Answer: Molecular ions are generated. Ions are formed by a loss or gain of charge, for example by electron ejection, protonation or deprotonation.

Teacher's Note:
a) Use the term "molecular ions" in your answer.
b) Give the examples: electron ejection, protonation, deprotonation.

B. Mass spectrometry is an analytical tool. Justify the statement.

Answer: Mass spectrometry is used to obtain protein structural information such as peptide mass or amino acid sequence. It is also used to identify the type and location of amino acid modifications in proteins.

Teacher's Note:
a) Any one use is enough for 1 mark.
b) Connect the use to "analysis" of protein structure.

Attempt either subpart C or D.

C. Name the major components of a mass spectrometer.

Answer:
1. Ionisation chamber / Ion source
2. Mass analyser
3. Detector
4. Vacuum system

Teacher's Note:
a) Each component carries half a mark; electromagnet and amplifier are also accepted.
b) Remember the order in which the ions travel: source, analyser, detector.

OR

D. How is MALDI used to volatilize and protonate proteins?

Answer:
1. The sample is moved from a condensed phase to a gas phase with the help of a solid matrix.
2. A pulsed laser beam is directed onto the sample suspended or dissolved in the matrix.
3. The matrix absorbs the laser energy and vaporises; in the gas phase, the matrix helps to ionise the sample.
4. The charged molecules are directed by electrostatic lenses from the ion source to the mass analyser.

Teacher's Note:
a) Key words: "matrix", "pulsed laser beam" and "gas phase".
b) MALDI stands for Matrix Assisted Laser Desorption Ionisation.

SECTION E

31. Attempt either option A or B. [5 Marks]

A. Few restriction enzymes break the phosphodiester bond in such a manner that single stranded overhang ends are generated in the DNA strand. EcoRI is one such a restriction enzyme.
I. Write the sequence for restriction site for enzyme EcoRI. Give a name to the kind of ends generated here.
II. If the following DNA strand is digested by EcoRI, how many DNA fragments will be generated. Write the sequence of these double stranded DNA fragments with their respective polarity.
5' GTATTAGCTGAATTCTATAAAACG 3'
3' CATAATCGACTTAAGATATTTTGC 5'
III. Why is EcoRI named so? [5 Marks]

Answer:
I. The EcoRI restriction site is:
5' GAATTC 3'
3' CTTAAG 5'
The ends generated are called sticky ends.
II. EcoRI cuts between G and A, so two DNA fragments are formed:
Fragment 1: 5' GTATTAGCTG 3' / 3' CATAATCGACTTAA 5'
Fragment 2: 5' AATTCTATAAAACG 3' / 3' GATATTTTGC 5'
III. In EcoRI, E refers to the genus (Escherichia), co to the species (coli), the capital R to the strain (RY13), and the Roman numeral I shows that it was the first enzyme isolated from this strain.

Teacher's Note:
a) Always write the 5' and 3' ends on every strand; polarity carries marks in II.
b) Check that each fragment has a 4-base single stranded overhang (AATT) - this is what makes the ends sticky.
c) In III, explain every letter of the name, including the Roman numeral.

OR

B. A gene X was introduced into cloning vector pUC19 at LacZ gene site. What will be the impact on the recombinant plasmids? Also suggest a possible way by which we can differentiate recombinants from non-recombinant plasmids. [5 Marks]

Answer:
1. Insertional inactivation will occur in the recombinant plasmid: the LacZ gene will be inactivated.
2. Recombinants can be told apart from non-recombinants by using a chromogenic medium, that is, blue-white selection.
3. pUC19 carries the LacZ gene, which codes for the enzyme \( \beta \)-galactosidase. In E. coli cells transformed with pUC19, this enzyme cleaves the colourless substrate X-gal into a blue product, so the colonies are blue.
4. The LacZ region also contains the polylinker. Insertion of foreign DNA at any restriction site here gives an altered, non-functional enzyme.
5. This enzyme cannot cleave X-gal, so colonies with recombinant plasmids appear white.

Teacher's Note:
a) Write "insertional inactivation" as the impact; it carries 1 mark.
b) Remember: blue colonies are non-recombinants, white colonies are recombinants.

32. Attempt either option A or B. [5 Marks]

A.
I. A company plans to release a new insect-resistant GM crop. Evaluate three potential bioethical concerns in plant genetic engineering.
II. What was the need to develop "Flavr Savr"? [5 Marks]

Answer:
I. Three major concerns about GM crops:
1. GM plants may cause allergic or toxic reactions in the long run.
2. The transgene may escape through pollen grains and contaminate the genes of natural populations.
3. Non-target and beneficial organisms may also be affected.
II. Flavr Savr (a GM tomato) was developed for:
1. Delayed fruit ripening.
2. Longer shelf life.

Teacher's Note:
a) Other accepted concerns: new microbial and insect strains, transgenic plants becoming weeds, change in evolutionary pattern, and spread of antibiotic resistance marker genes to microbes.
b) Since the question says "evaluate", explain briefly why each concern matters.

OR

B.
I. How can the transgenic plants be used as factories to produce biodegradable plastics?
II. What are artificial seeds and mention any one application? Explain diagrammatically. [5 Marks]

Answer:
I. 1. Genetically engineered Arabidopsis plants with three genes from a bacterium produce PHB (poly hydroxy butyrate) globules only in their chloroplasts, without affecting plant growth and development.
2. Large-scale production of PHB is easily achieved by extraction from the leaves of genetically engineered Populus.
II. 1. Artificial seeds are somatic embryos raised in tissue culture and encapsulated in protective chemicals like calcium alginate, which form an artificial seed coat and prevent them from drying out.
2. Application: rapid and mass propagation of elite plant species and hybrid varieties.
3. Diagram: a round bead showing the artificial seed coat (A) on the outside, the somatic embryo at torpedo stage (B) inside, and the artificial endosperm (C) filling the space around the embryo.

Teacher's Note:
a) Name the biodegradable plastic (PHB) and the plants used (Arabidopsis, Populus).
b) Label all three parts of the artificial seed in the diagram: seed coat, somatic embryo and artificial endosperm.

33. Attempt either option A or B. [5 Marks]

A.
I. Describe the technique developed by O'farrell to study proteomics.
II. Name any two properties that can be manipulated using protein Engineering. [5 Marks]

Answer:
I. The technique is 2-D gel electrophoresis.
1. First dimension: isoelectric focusing (IEF). Proteins are separated on the basis of their isoelectric pH (pI). Ampholytes (polyamino-polycarboxylic acids) are used to create the pH gradient.
2. Second dimension: SDS-PAGE. Proteins are separated on the basis of their molecular mass.
3. The two runs are carried out at right angles to each other, which increases the resolution.
4. Proteins are separated into 2-D patterns with high resolution because two properties, charge and size, are used.
II. Thermal and pH stability; solubility.

Teacher's Note:
a) State the basis of separation for each dimension: charge (pI) for IEF and mass for SDS-PAGE.
b) Other accepted properties in II: solvent tolerance and catalytic potency.

OR

B.
I. Why is sickle cell anaemia called a molecular disease?
II. Who developed peptide mapping? Describe the technique. [5 Marks]

Answer:
I. In sickle cell haemoglobin, a single valine replaces glutamic acid at the 6th position of the beta chain. This changes the structure of the haemoglobin so that it forms fibres inside the RBC and deforms the cell (sickling). As the disease is caused by this molecular change, it is called a molecular disease.
II. Peptide mapping was developed by V.M. Ingram. The principle is to make peptide maps after cutting the protein and separating the peptides.
1. The protein is cut into peptides using trypsin, a proteolytic enzyme.
2. Paper electrophoresis is performed.
3. Paper chromatography is then done at right angles (\( 90^{\circ} \)) to the direction of electrophoresis, separating the peptides by their partition coefficient in the solvent.
4. The peptides are made visible with ninhydrin. Sequencing the changed peptide shows valine in place of glutamic acid.

Teacher's Note:
a) In I, mention "valine replaces glutamic acid at position 6 of the beta chain".
b) In II, the order electrophoresis then chromatography at right angles is the key step.
c) Peptide mapping is also called protein fingerprinting.