CUET UG Biology Booster Test 2-Recombinant DNA Processes -

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CUET UG Biology Booster Test 2-Recombinant DNA Processes

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Biology Booster Test 2-Recombinant DNA Processes

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QUESTION 1 OF 20

Arrange the enzymatic treatments in the logical order required to isolate pure DNA from a bacterial cell:
1. Addition of chilled ethanol
2. Treatment with lysozyme
3. Treatment with ribonuclease and protease

Answer Review

1 Arrange the enzymatic treatments in the logical order required to isolate pure DNA from a bacterial cell:
1. Addition of chilled ethanol
2. Treatment with lysozyme
3. Treatment with ribonuclease and protease

A 2, 3, 1

B 1, 2, 3

C 3, 2, 1

D 2, 1, 3

💡 Short Explanation

First, the cell wall must be broken by enzymes (Lysozyme). Second, the macromolecules (RNA/proteins) are removed by their respective enzymes. Finally, the pure DNA is precipitated by chilled ethanol.

✅ Correct Answer Explanation

Isolation of DNA follows a strict sequence: (2) Cell lysis using lysozyme (for bacteria) to release the cellular content, (3) Purification by adding ribonuclease (removes RNA) and protease (removes proteins), and finally (1) Precipitation using chilled ethanol to make the DNA visible/isolatable as fine threads.

❓ Why Other Options Are Incorrect

Option B → Ethanol precipitation cannot occur before cell lysis or purification.
Option C → Purification enzymes cannot be added before the cell wall is broken.
Option D → Ethanol cannot be added between enzyme treatments.

🎯 Strategy

Used Contextual/Tonal Matching

Application: Defining the logic: Lysis -> Purification -> Precipitation.

Final Logic: The biological order of operations requires breaking the cell, cleaning the sample, then precipitating.

🧠 Memory Trick

"Break (Lysozyme), Clean (RNase/Protease), Precipitate (Ethanol)."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Isolation of Genetic Material –

2 Match the extraction events with their corresponding visual or physical outcome:

Event	Outcome
I. Addition of chilled ethanol	P. Breaks down bacterial cell wall
II. Treatment with lysozyme	Q. Removes RNA
III. Treatment with protease	R. Precipitates pure DNA as fine threads
IV. Treatment with ribonuclease	S. Removes histone proteins

A I-R, II-P, III-S, IV-Q

B I-P, II-R, III-Q, IV-S

C I-R, II-Q, III-S, IV-P

D I-Q, II-S, III-P, IV-R

💡 Short Explanation

I-R: Ethanol precipitates DNA. II-P: Lysozyme digests cell walls. III-S: Protease removes histones/proteins. IV-Q: Ribonuclease removes RNA.

✅ Correct Answer Explanation

Chilled ethanol is strictly used for the precipitation of DNA (I-R). Lysozyme is a specific enzyme used to break bacterial cell walls (II-P). Proteases are used to degrade proteins, including histones associated with DNA (III-S). Ribonucleases specifically degrade RNA (IV-Q).

❓ Why Other Options Are Incorrect

Option B → Incorrectly pairs ethanol with wall breaking (P).
Option C → Incorrectly pairs ribonuclease with wall breaking (P).
Option D → Incorrectly pairs ethanol with RNA removal (Q).

🎯 Strategy

Used Option Grouping

Application: Aligning the specific tool/reagent with its biochemical function.

Final Logic: A is the only choice where all pairs are scientifically accurate.

🧠 Memory Trick

"Ethanol-Precipitate; Lysozyme-Lysis."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Isolation of Genetic Material –

3 Why would a researcher choose to use cellulase rather than lysozyme when attempting to extract genetic material from a leaf sample?

A Leaf samples contain chitin, which cellulase targets.

B Leaf samples are from plants, which possess a cellulose-based cell wall.

C Lysozyme destroys plant DNA immediately upon contact.

D Cellulase is the only enzyme that can break down proteins.

💡 Short Explanation

Plant cell walls are made of cellulose. Cellulase is the enzyme that breaks down cellulose. Lysozyme is for bacterial cell walls (peptidoglycan).

✅ Correct Answer Explanation

Different organisms have different cell wall compositions. Plants have cell walls made of cellulose. Therefore, to effectively break the plant cell wall and release the DNA, one must use cellulase. Lysozyme is specific to bacterial peptidoglycan walls.

❓ Why Other Options Are Incorrect

Option A → Fungi, not plants, contain chitin.
Option C → Lysozyme does not destroy DNA.
Option D → Protease is the enzyme that breaks down proteins.

🎯 Strategy

Used Elimination

Application: Eliminate based on cell wall composition: Plants = Cellulose = Cellulase.

Final Logic: Substrate specificity matches Enzyme specificity.

🧠 Memory Trick

"Cellulose = Cellulase."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Isolation of Genetic Material –

4 Which of the following enzymes should NOT be added to the cellular extract if the goal is to purify and maintain intact DNA?

A Ribonuclease

B Protease

C Deoxyribonuclease

D Lysozyme

💡 Short Explanation

DNA is the target molecule. Deoxyribonuclease (DNase) breaks down DNA. The goal is to keep DNA intact.

✅ Correct Answer Explanation

Deoxyribonuclease is an enzyme that degrades DNA. Adding it to an extract would destroy the very molecule one is trying to isolate. The other enzymes (Ribonuclease, Protease, Lysozyme) are helpful for purification by removing unwanted contaminants (RNA, proteins, or breaking cell walls).

❓ Why Other Options Are Incorrect

Option A → Ribonuclease removes RNA, aiding purity.
Option B → Protease removes proteins, aiding purity.
Option D → Lysozyme helps release DNA from bacterial cells.

🎯 Strategy

Used Odd One Out

Application: All other enzymes help purify/isolate DNA; DNase destroys the target.

Final Logic: You never add an enzyme that destroys your target product.

🧠 Memory Trick

"DNase destroys DNA."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Isolation of Genetic Material –

5 Based on the provided gel image, how does the visual migration pattern in lanes 2 to 4 prove that the restriction enzyme digestion was successful?

A The DNA remains clumped in a single band at the loading well.

B The DNA resolves into multiple distinct bands of smaller fragment sizes that move further through the gel.

C The DNA completely disappears from the gel.

D The DNA moves towards the cathode instead of the anode.

💡 Short Explanation

Undigested DNA is one large band. Digested DNA is many small fragments. Small fragments move faster through the gel.

✅ Correct Answer Explanation

In agarose gel electrophoresis, larger molecules move slowly and stay near the top, while smaller fragments resulting from successful restriction digestion move faster and further through the gel. Multiple distinct bands indicate that the enzyme cut the DNA at specific sites.

❓ Why Other Options Are Incorrect

Option A → This indicates the DNA was not cut (or the gel failed).
Option C → DNA should not disappear; it should be visible as bands.
Option D → DNA always moves towards the anode (positive electrode) because it is negatively charged.

🎯 Strategy

Used Contextual/Tonal Matching

Application: Matching the visual "success" with the principle of gel migration.

Final Logic: Digestion = Fragmentation = Faster migration/multiple bands.

🧠 Memory Trick

"Small bands = Successful cutting."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: DNA Fragmentation/Gel Electrophoresis – -200

6 What fundamental property of DNA forces it to migrate towards the anode during the progression check shown in the gel?

A DNA is a positively charged molecule.

B DNA is a neutral molecule driven by heat.

C DNA is a negatively charged molecule.

D DNA is repelled by the agarose matrix.

💡 Short Explanation

DNA backbone contains phosphates. Phosphates are negatively charged. Opposite charges attract; DNA moves to positive (Anode).

✅ Correct Answer Explanation

The phosphate groups in the sugar-phosphate backbone of DNA give the molecule an overall negative charge at neutral pH. In an electric field, negatively charged molecules are attracted to the positive electrode (anode).

❓ Why Other Options Are Incorrect

Option A → DNA is negative, not positive.
Option B → DNA is not neutral; it is strongly charged.
Option D → Agarose is a porous matrix, not a repulsive agent.

🎯 Strategy

Used Elimination

Application: Recall the chemical nature of DNA.

Final Logic: DNA = Negatively charged backbone = moves to Anode.

🧠 Memory Trick

"DNA is negative; Anode is positive."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: DNA Electrophoresis –

7 Which of the following components is NOT required in the reaction mixture for the in vitro synthesis of a gene using PCR?

A Genomic DNA as a template

B Two sets of oligonucleotide primers

C Restriction endonucleases

D Thermostable DNA polymerase

💡 Short Explanation

PCR is for synthesis/amplification. Restriction enzymes are for cutting. You do not cut the DNA during PCR.

✅ Correct Answer Explanation

PCR is a replication technique. It requires a template (genomic DNA), building blocks (nucleotides), primers to start synthesis, and a polymerase enzyme to build the strands. Restriction endonucleases are used in the cloning or isolation phases, not in the PCR amplification phase itself.

❓ Why Other Options Are Incorrect

Option A → Template is essential for replication.
Option B → Primers are essential to initiate replication.
Option D → Thermostable polymerase is essential to build the new strands.

🎯 Strategy

Used Elimination

Application: Eliminate the tools needed for other phases of rDNA technology.

Final Logic: PCR = Synthesis; Restriction Enzymes = Cutting.

🧠 Memory Trick

"PCR doesn't need scissors (Restriction Enzymes)."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

8 Arrange the events involving primers during a single PCR cycle:
1. The enzyme extends the primers using provided nucleotides.
2. The double-stranded DNA template is denatured into single strands.
3. Primers anneal to complementary regions on the single-stranded DNA.

A 2, 3, 1

B 3, 2, 1

C 1, 2, 3

D 2, 1, 3

💡 Short Explanation

First: Denature DNA. Second: Anneal Primers. Third: Extend chain.

✅ Correct Answer Explanation

PCR cycling occurs in a set sequence: (2) High heat denatures the template, (3) lowering temperature allows primers to anneal to the template, and (1) then the temperature is optimized for the polymerase to extend the chain using nucleotides.

❓ Why Other Options Are Incorrect

Option B, C, D → Incorrect order of physical events needed for synthesis.

🎯 Strategy

Used Contextual/Tonal Matching

Application: DAE (Denature-Anneal-Extend) protocol.

Final Logic: Template must be single-stranded (2) before primers can bind (3), before extension (1) can occur.

🧠 Memory Trick

"DAE."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

9 Match the PCR steps/tools to their underlying mechanisms:

PCR Step/Tool	Mechanism
I. Denaturation	P. Extends primers along the template
II. Thermostable Polymerase	Q. Separates double-stranded DNA via high temperature
III. Annealing	R. Binds chemically synthesized oligonucleotides
IV. Extension	S. Withstands heat-induced degradation

A I-Q, II-S, III-R, IV-P

B I-Q, II-P, III-R, IV-S

C I-S, II-Q, III-R, IV-P

D I-R, II-S, III-Q, IV-P

💡 Short Explanation

I (Denaturation) = Q (Heat separation). II (Polymerase) = S (Heat stability). III (Annealing) = R (Primer binding). IV (Extension) = P (Primer extension).

✅ Correct Answer Explanation

Denaturation (I) involves separating strands via high heat (Q). Thermostable polymerase (II) is unique because it withstands this heat (S). Annealing (III) is where primers bind (R). Extension (IV) is where the polymerase builds the strand (P).

❓ Why Other Options Are Incorrect

Option B → Misaligns Polymerase mechanism with P.
Option C → Misaligns Denaturation with S.
Option D → Misaligns Denaturation with R.

🎯 Strategy

Used Option Grouping

Application: Direct mapping of steps to biological mechanisms.

Final Logic: Option A provides the only set of perfectly aligned definitions.

🧠 Memory Trick

"I-Q, II-S, III-R, IV-P."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

10 During the extension phase of PCR, what serves as the template for the thermostable DNA polymerase to add nucleotides?

A The RNA strand

B The genomic DNA

C The restriction enzyme

D The ampicillin resistance gene

💡 Short Explanation

PCR amplifies a target DNA segment. The original target (genomic) is the template for the new copy.

✅ Correct Answer Explanation

The polymerase uses the single-stranded template DNA—which originates from the genomic DNA—to determine the order of nucleotides added to the growing strand. The primer binds to this template, and the enzyme extends it based on the template's sequence.

❓ Why Other Options Are Incorrect

Option A → PCR uses DNA templates, not RNA.
Option C → Enzymes are not templates.
Option D → An ampicillin gene is a marker, not the template for PCR replication.

🎯 Strategy

Used Substitution

Application: Substituting "Template" with "DNA strand to be copied."

Final Logic: Extension phase follows the sequence of the DNA template (genomic DNA).

🧠 Memory Trick

"Template = The original document."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

11 Why is the origin of Taq polymerase from Thermus aquaticus biologically significant for the PCR process?

A It ensures the enzyme remains active during the high temperature induced denaturation step.

B It allows the enzyme to act as a restriction endonuclease.

C It prevents the DNA from becoming negatively charged.

D It helps in breaking down the bacterial cell wall during DNA isolation.

💡 Short Explanation

PCR cycles involve high heat (94-95°C). Normal enzymes would denature and fail. Taq polymerase is evolutionarily adapted to heat, maintaining activity.

✅ Correct Answer Explanation

The PCR process requires repeated cycles of high-temperature denaturation to separate DNA strands. The significance of using Taq polymerase from Thermus aquaticus is its inherent thermostability, allowing it to function at high temperatures without being denatured, which is essential for repeating the amplification process automatically.

❓ Why Other Options Are Incorrect

Option B → Taq polymerase is a DNA synthesizer, not a cutter.
Option C → The negative charge of DNA is a chemical property of the phosphate backbone, not enzyme-related.
Option D → Cell wall breakdown is handled by lysozyme, not Taq.

🎯 Strategy

Used Substitution

Application: Substitute the biological necessity ("heat survival") with the enzyme's property ("Thermostability").

Final Logic: High-temperature PCR needs a heat-stable enzyme.

🧠 Memory Trick

"Thermus = Thermostable."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

12 Evaluate the following statements concerning Taq polymerase activity:
Statement I: It actively synthesises new DNA strands during the high-temperature denaturation phase.
Statement II: It requires oligonucleotide primers to initiate the extension process.

A Both statements are true.

B Both statements are false.

C Statement I is false, but Statement II is true.

D Statement I is true, but Statement II is false.

💡 Short Explanation

Synthesis occurs during the extension phase, not denaturation. DNA polymerase always requires a primer 3'-OH end.

✅ Correct Answer Explanation

Statement I is false because denaturation is the phase for separating strands, not synthesis; synthesis occurs during the extension phase at a lower temperature (approx. 72°C). Statement II is true because DNA polymerases cannot start synthesis de novo; they must have a primer to provide a starting point.

❓ Why Other Options Are Incorrect

Option A → Statement I is incorrect.
Option B → Statement II is correct.
Option D → Statement I is incorrect.

🎯 Strategy

Used Elimination

Application: Validate both statements independently against PCR mechanics.

Final Logic: Denaturation = separation; Extension = synthesis; Polymerase = needs primer.

🧠 Memory Trick

"Denature = Separate; Extend = Synthesis."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

📄 Passage

"If the process of replication of DNA is repeated many times, the segment of DNA can be amplified to approximately billion times, i.e., 1 billion copies are made. Such repeated amplification is achieved by the use of a thermostable DNA polymerase..."

According to the passage, what specific technological factor enables these amplification cycles to be repeated many times without the process breaking down?

A The use of chilled ethanol

B The use of a thermostable DNA polymerase

C The use of RNA primers

D The use of lysozyme

💡 Short Explanation

Repeating the cycle requires high heat. Only a heat-stable enzyme survives this. Thus, the enzyme enables the repetition.

✅ Correct Answer Explanation

As mentioned in the passage, the repeated amplification is achieved specifically by using a thermostable DNA polymerase. This enzyme does not degrade at the high temperatures used in the denaturation step, allowing it to survive multiple cycles.

❓ Why Other Options Are Incorrect

Option A → Used for final DNA precipitation, not for cycling.
Option C → Primers are DNA, not RNA in PCR, and not the factor for cycling.
Option D → Used for initial bacterial cell wall breakdown.

🎯 Strategy

Used Contextual/Tonal Matching

Application: Extracting information directly from the provided passage.

Final Logic: The passage explicitly attributes repeated amplification to "thermostable DNA polymerase."

🧠 Memory Trick

"Repeated Cycles = Heat Stable Enzyme."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

📄 Passage

Based on the passage, what is the mathematical outcome of subjecting a segment of DNA to repeated cycles of this replication process?

A It degrades into single nucleotides.

B It is amplified to approximately 1 billion copies.

C It transforms into RNA.

D It converts exactly 1 million copies into pure protein.

💡 Short Explanation

PCR is an exponential amplification method. The passage explicitly mentions "1 billion copies."

✅ Correct Answer Explanation

The passage clearly states that repeating the DNA replication process many times results in the segment being amplified to approximately 1 billion copies.

❓ Why Other Options Are Incorrect

Option A → Degradation is not the goal.
Option C → DNA is not converted into RNA in PCR.
Option D → The number mentioned is 1 billion, not 1 million, and it produces DNA, not protein.

🎯 Strategy

Used Contextual/Tonal Matching

Application: Direct retrieval of the numerical fact from the passage.

Final Logic: Passage text: "1 billion copies are made."

🧠 Memory Trick

"PCR = Billion-fold."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: PCR –

15 Why is it crucial that both the source DNA and the vector DNA are cut with the exact same specific restriction enzyme before mixing them?

A So that they develop resistance to the same antibiotics.

B So they generate compatible cut spaces (sticky ends) that can be seamlessly mixed and ligated.

C So the vector can be digested into individual nucleotides.

D Because ligase will only work on uncut DNA.

💡 Short Explanation

Same enzyme = same recognition sequence. This results in identical "sticky" overhangs. Complementary overhangs allow them to join together.

✅ Correct Answer Explanation

Restriction enzymes cut DNA at specific recognition sequences, leaving "sticky ends" (overhangs). By using the same enzyme for both the gene of interest and the vector, they are left with complementary sticky ends. This allows them to base-pair together, enabling ligase to seal them into a single recombinant molecule.

❓ Why Other Options Are Incorrect

Option A → Antibiotic resistance is a marker property, not dependent on the cutting enzyme.
Option C → Digestion into nucleotides is not the goal of cloning.
Option D → Ligase works on cut (nicked) DNA, not uncut.

🎯 Strategy

Used Substitution

Application: Substituting "Compatible cut" with "Complementary sticky ends."

Final Logic: Same enzyme = Same cut ends = Compatible base pairing.

🧠 Memory Trick

"Same Cut = Same Sticky Ends."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Ligation –

16 Which of the following events is NOT a part of creating new DNA combinations in vitro during recombinant DNA technology?

A Cutting source DNA with a restriction enzyme.

B Mixing cut vector with the 'gene of interest'.

C Adding DNA ligase to the mixture.

D Translating the DNA immediately into proteins within the same test tube.

💡 Short Explanation

rDNA technology builds the DNA construct. Translation is a cellular process requiring ribosomes, tRNA, etc. It does not happen "in the tube" with just ligation reagents.

✅ Correct Answer Explanation

Recombinant DNA technology focuses on joining DNA segments. Translating DNA into protein is a complex cellular process (Protein Synthesis) that requires living cells, not just reagents in a tube.

❓ Why Other Options Are Incorrect

Option A → Cutting is the first essential step.
Option B → Mixing is required to bring fragments together.
Option C → Ligation is required to join fragments.

🎯 Strategy

Used Elimination

Application: Eliminate the standard cloning steps to find the process that occurs only in vivo.

Final Logic: Protein synthesis requires ribosomes and cells, not just the tube containing DNA ligation reagents.

🧠 Memory Trick

"Recombinant DNA = Building the Molecule, not the Protein."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Recombinant DNA Technology –

17 When selecting for transformants, the ampicillin resistance gene introduced via the recombinant vector acts as a selectable marker because it:

A allows the host cell to produce ampicillin.

B kills the transformed cells leaving only pure untransformed cells.

C confers the ability for only the transformed cells to grow on ampicillin-containing agar.

D denatures at high temperatures.

💡 Short Explanation

Marker = Resistance. Resistance = Growth in presence of antibiotic. Ampicillin-sensitive cells die; resistant ones survive.

✅ Correct Answer Explanation

A selectable marker is a gene used to identify which cells have successfully taken up the foreign DNA (transformants). By growing the cells on media containing ampicillin, only those that express the ampicillin resistance gene (i.e., the transformants) will survive and grow, allowing for selection.

❓ Why Other Options Are Incorrect

Option A → It gives resistance to the antibiotic, it does not produce it.
Option B → It allows transformed cells to grow, it does not kill them.
Option D → This is irrelevant to the function of a selectable marker.

🎯 Strategy

Used Substitution

Application: Substituting "Selectable Marker" with its functional role in cell culture.

Final Logic: Marker = Antibiotic resistance = Growth survival.

🧠 Memory Trick

"Selectable Marker = Survival Tag."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Transformant Selection –

18 If a population of E. coli cells is subjected to the transformation process but fails to take up the recombinant DNA, which of the following statements will NOT be true?

A They will remain untransformed recipient cells.

B They will lack the ampicillin resistance gene.

C They will die when spread on agar plates containing ampicillin.

D They will act as a selectable marker and grow profusely.

💡 Short Explanation

Untransformed cells are sensitive to ampicillin. They die; they do not grow. They are not markers; they are the failed product.

✅ Correct Answer Explanation

If cells fail to take up the recombinant DNA, they remain "untransformed." Because they do not have the resistance gene, they remain sensitive to ampicillin and will die when plated on antibiotic media. They cannot grow profusely, nor do they function as a marker.

❓ Why Other Options Are Incorrect

Option A → Correct; they are indeed untransformed.
Option B → Correct; they did not get the DNA.
Option C → Correct; they are sensitive to ampicillin.

🎯 Strategy

Used Extreme Word Filter

Application: The statement "grow profusely" is impossible for antibiotic-sensitive cells on an antibiotic plate.

Final Logic: Untransformed = Sensitive = Death.

🧠 Memory Trick

"Fail to take up DNA = Fail to survive."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Transformant Selection –

19 If a protein-encoding gene is successfully expressed in a heterologous host, the resulting product is technically referred to as a:

A Recombinant protein

B Restriction endonuclease

C Selectable marker

D Native plasmid

💡 Short Explanation

Protein encoded by a foreign (recombinant) DNA is called "recombinant protein." Restriction enzymes cut DNA. Markers indicate success.

✅ Correct Answer Explanation

When a foreign gene is placed into a host (the heterologous host) and expressed, the protein product is created using the rDNA construct. Therefore, the product itself is called a "recombinant protein."

❓ Why Other Options Are Incorrect

Option B → This is a tool, not a product.
Option C → This is a marker gene, not the protein product.
Option D → This is a DNA molecule, not a protein.

🎯 Strategy

Used Substitution

Application: Substituting "Expressed recombinant gene product" with the industry standard term.

Final Logic: rDNA-produced protein = Recombinant protein.

🧠 Memory Trick

"Recombinant gene = Recombinant protein."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Desirable Protein Production – -201

20 Consider the following statements about producing desirable proteins:
Statement I: Small volume cultures in the laboratory are sufficient to yield large, commercial quantities of products.
Statement II: To produce in large quantities, bioreactors processing 100-1000 litres of culture are required.

A Both statements are correct.

B Both statements are incorrect.

C Statement I is correct, but Statement II is incorrect.

D Statement I is incorrect, but Statement II is true.

💡 Short Explanation

Lab scales are too small for commercial quantities. Commercial quantities require large bioreactors (100-1000L).

✅ Correct Answer Explanation

Statement I is incorrect because lab-scale experiments are insufficient for mass-producing products like insulin or vaccines; they are for research and development. Statement II is true because commercial quantities require industrial-sized bioreactors (vessels) to control environmental parameters over 100–1000 liters.

❓ Why Other Options Are Incorrect

Option A → Statement I is incorrect.
Option B → Statement II is correct.
Option C → Statement II is correct.

🎯 Strategy

Used Substitution

Application: Contrast "Laboratory scale" vs. "Commercial scale."

Final Logic: Lab volume (small) $\neq$ Commercial demand (large volume requires large bioreactors).

🧠 Memory Trick

"Lab = Test; Bioreactor = Production."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Large Scale Production –

CUET UG Biology Booster Test 2-Recombinant DNA Processes

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