CUET UG Biology Booster Test 2- Tools of Recombinant DNA Technology -

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CUET UG Biology Booster Test 2- Tools of Recombinant DNA Technology

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Biology Booster Test 2- Tools of Recombinant DNA Technology

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QUESTION 1 OF 20

📄 Passage

In the year 1963, the two enzymes responsible for restricting the growth of bacteriophage in Escherichia coli were isolated. One of these added methyl groups to DNA, while the other cut DNA. The later was called restriction endonuclease. The first restriction endonuclease–Hind II, whose functioning depended on a specific DNA nucleotide sequence was isolated and characterised five years later.

What was the primary biological role of the two enzymes isolated from E. coli in 1963?

Answer Review

📄 Passage

What was the primary biological role of the two enzymes isolated from E. coli in 1963?

A To replicate the bacterial genome

B To restrict the growth of bacteriophage

C To synthesize RNA from DNA

D To join fragments of host DNA

💡 Short Explanation

The enzymes were part of the bacterial defense mechanism. They protected the bacteria by destroying invading viral (bacteriophage) DNA. This "restriction" ability is the foundation of restriction endonucleases.

✅ Correct Answer Explanation

The correct answer is ( A). The passage explicitly states that the two enzymes isolated in 1963 from Escherichia coli were responsible for "restricting the growth of bacteriophage." One enzyme methylated the DNA (protecting host DN A), and the other (restriction endonuclease) cut the foreign viral DNA.

❓ Why Other Options Are Incorrect

Option A → DNA replication is performed by DNA polymerases, not restriction enzymes.
Option C → RNA synthesis is performed by RNA polymerase.
Option D → Joining DNA fragments is the role of DNA ligase.

🎯 Strategy

Used Contextual Matching.

Application: The answer is directly derived from the opening sentence of the provided passage.

Final Logic: Restriction enzymes = Bacteriophage defense = Growth restriction.

🧠 Memory Trick

"Restrict = Stop Virus."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

📄 Passage

Why was the discovery of Hind II significantly different from the methyl-adding enzyme isolated earlier?

A Because Hind II was found in animals instead of bacteria.

B Because its functioning depended on recognizing a specific DNA nucleotide sequence.

C Because it joined DNA rather than cutting it.

D Because it removed nucleotides only from the ends.

💡 Short Explanation

Hind II was the first restriction endonuclease. Its key breakthrough was sequence-specific cutting. This precision allowed it to become a tool for biotechnology.

✅ Correct Answer Explanation

The correct answer is ( A). While the methyl-adding enzyme modified DNA to protect it, Hind II was revolutionary because it recognized a specific nucleotide sequence and cut the DNA at that location. This sequence specificity is the basis for its utility in recombinant DNA technology.

❓ Why Other Options Are Incorrect

Option A → Hind II was found in bacteria (Haemophilus influenzae), not animals.
Option C → Hind II is an endonuclease (a cutter), not a ligase (a joiner).
Option D → Endonucleases cut internally, whereas exonucleases cut from the ends.

🎯 Strategy

Used Contextual Matching.

Application: The passage directly states: "Hind II, whose functioning depended on a specific DNA nucleotide sequence."

Final Logic: Hind II = Sequence-specific endonuclease.

🧠 Memory Trick

"Hind II = First Sequence Cutter."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

3 Match the enzyme to its source organism format logic based on the naming convention:

List I	List II
1. First letter	P. Order in which enzymes were isolated
2. Second and third letters	Q. Genus of prokaryotic cell
3. Fourth letter	R. Species of prokaryotic cell
4. Roman numeral	S. Name of the strain

A 1-Q, 2-R, 3-S, 4-P

B 1-Q, 2-S, 3-R, 4-P

C 1-R, 2-Q, 3-S, 4-P

D 1-P, 2-Q, 3-R, 4-S

💡 Short Explanation

1st Letter = Genus (Q). 2nd/3rd Letters = Species (R). 4th Letter = Strain (S). Roman Numeral = Order (P).

✅ Correct Answer Explanation

The correct answer is ( A). The nomenclature of restriction enzymes follows a strict protocol: (1) First letter from the genus (Q), (2) Next two letters from the species (R), (3) Fourth letter from the strain (S), and (4) Roman numeral indicating the order of isolation (P).

❓ Why Other Options Are Incorrect

Options B, C, and D incorrectly map the components of the naming convention.

🎯 Strategy

Used Option Grouping.

Application: Map the known definitions to their placeholders (1-Q, 4-P) to eliminate incorrect sequences.

Final Logic: Genus-Species-Strain-Order = Q-R-S-P.

🧠 Memory Trick

"G-S-S-O (Genus-Species-Strain-Order)."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

4 Consider the following regarding the Roman numeral in an enzyme's name:
I. It indicates the optimal pH at which the enzyme functions.
II. It denotes the order in which the enzyme was isolated from a strain.
III. It signifies the number of cuts the enzyme makes on the DNA.

A I only

B II only

C I and III only

D II and III only

💡 Short Explanation

Roman numerals in enzyme names (e.g., EcoRI) indicate isolation order. "I" means it was the first enzyme discovered in that specific strain. It has nothing to do with pH or the number of cuts.

✅ Correct Answer Explanation

The correct answer is ( A). The Roman numeral in a restriction enzyme's name, such as the "I" in EcoRI, denotes the order in which the enzyme was isolated from that particular bacterial strain. Statements I and III are scientifically incorrect.

❓ Why Other Options Are Incorrect

Options A, C, and D are incorrect because they include false statements about pH or cutting frequency.

🎯 Strategy

Used Elimination.

Application: The order of isolation is the standard definition found in NCERT for the Roman numeral suffix. Eliminate all other claims.

Final Logic: Roman numeral = Order of isolation.

🧠 Memory Trick

"Roman = Rank (Order)."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

5 If a scientist wishes to create sticky ends in the middle of a long linear DNA fragment, which of the following is NOT an appropriate tool, and why?

A Exonuclease, because it removes nucleotides from the ends of the DNA.

B Endonuclease, because it cuts at specific positions within the DNA.

C Exonuclease, because it recognizes palindromes exclusively.

D Endonuclease, because it modifies the sugar-phosphate backbone.

💡 Short Explanation

Exonucleases act at the ends of DNA molecules. To cut "in the middle" (internally), you must use an endonuclease. Option (A) correctly identifies that an exonuclease is inappropriate for internal cutting.

✅ Correct Answer Explanation

(A). Exonucleases remove nucleotides sequentially from the ends of DNA. Therefore, they are incapable of making an internal cut ("in the middle") to create sticky ends. Endonucleases, conversely, are the enzymes that cut internally at specific sites.

❓ Why Other Options Are Incorrect

Option B → Endonuclease is an appropriate tool for internal cutting.
Option C → Exonuclease does not recognize palindromes.
Option D → Endonucleases cut the backbone, which is part of their function.

🎯 Strategy

Used Contextual Matching.

Application: Distinguish the "what" (Internal vs. En

D) from the "why" (Mechanism). Exonucleases are inappropriate because they cut at ends.

Final Logic: Internal cut = Endonuclease; End cut = Exonuclease.

🧠 Memory Trick

"Exo = End; Endo = Inside."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

6 The action of an endonuclease is heavily dependent on "inspecting" the DNA. What specifically is it inspecting to make an internal cut?

A The methylation status of the ends

B The length of a specific palindromic recognition sequence

C The total length of the chromosome

D The presence of a selectable marker

💡 Short Explanation

Restriction endonucleases scan DNA for their specific target site. These sites are palindromic sequences. The enzyme cuts only when it detects the exact palindromic pattern.

✅ Correct Answer Explanation

The correct answer is (A). Restriction endonucleases "inspect" the DNA for a specific palindromic recognition sequence (e.g., GAATTC for EcoRI). Once the enzyme finds the exact base sequence matching its target, it binds and cuts the DNA at a precise location.

❓ Why Other Options Are Incorrect

Option A → Methylation is involved in host defense, but the enzyme's cut site is determined by the palindrome.
Option C → Chromosome length is irrelevant to the specific cut site.
Option D → Selectable markers are used later in the cloning process, not for the cutting mechanism.

🎯 Strategy

Used Substitution.

Application: Replace the word "inspecting" with its biological meaning in the context of restriction endonucleases (Sequence recognition).

Final Logic: Inspection = Palindromic sequence identification.

🧠 Memory Trick

"Restriction = Sequence Reader."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

7 Sequence the steps an endonuclease takes to utilize DNA sequence symmetry:
1. It cuts each of the two strands at specific points.
2. It inspects the length of a DNA sequence.
3. It binds to the DNA.
4. It finds its specific palindromic recognition sequence.

A 2, 4, 3, 1

B 4, 2, 3, 1

C 2, 3, 4, 1

D 4, 3, 2, 1

💡 Short Explanation

1st: Inspect sequence (2). 2nd: Find/identify palindrome (4). 3rd: Bind to the site (3). 4th: Cut (1).

✅ Correct Answer Explanation

The correct answer is ( A). The enzyme first inspects the DNA length (2), identifies the specific palindromic sequence (4), binds to that DNA site (3), and finally cuts the strands (1).

❓ Why Other Options Are Incorrect

Options B, C, and D violate the logical biological order of interaction between the enzyme and the DNA substrate.

🎯 Strategy

Used Logical Sequencing.

Application: Think of the enzyme's action: Scan → Locate → Dock (Bin

D) → Cut. This corresponds to 2 → 4 → 3 → 1.

Final Logic: Scan-Locate-Bind-Cut.

🧠 Memory Trick

"Scan, Find, Bind, Cut."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

8 To confirm a DNA palindrome, the sequence must read the same on both strands. What critical condition must be maintained while reading?

A Both strands must be read exclusively from 3' to 5'.

B The orientation of reading must be kept the same for both strands.

C The reading must start from the exact center of the sequence.

D Only the coding strand should be read.

💡 Short Explanation

DNA palindromes are defined by 5' to 3' symmetry. You must read both strands in the same direction to see the palindrome. Reading in opposite directions (one 5'-3', the other 3'-5') would not show the identical sequence.

✅ Correct Answer Explanation

T (A). A DNA palindrome is only evident if you read both strands in the same direction (e.g., both 5' to 3'). If you read in opposite directions (one 5'-3' and the other 3'-5'), the strands would be complementary, not identical (palindromiA).

❓ Why Other Options Are Incorrect

Option A → 3' to 5' is not the standard convention for identifying palindromic symmetry.
Option C → The center is where the enzyme might cut, but it's not the reference point for reading the sequence.
Option D → Both strands must be considered for symmetry.

🎯 Strategy

Used Contextual Matching.

Application: The passage specifically states "when orientation of reading is kept the same."

Final Logic: Palindrome = Consistent orientation reading.

🧠 Memory Trick

"Same Direction = Same Sequence."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

9 In the provided figure showing the action of EcoRI, where does the restriction enzyme cut relative to the center of the palindrome site to create overhanging stretches?

A Exactly at the center of the palindrome site.

B A little away from the center, but between the same two bases on opposite strands.

C Outside the palindromic sequence altogether.

D It cuts one strand in the center and leaves the other uncut.

💡 Short Explanation

Sticky ends require a "staggered" cut. Staggered means the cut is off-center. The cut occurs at the same relative position on both strands to create the overhang.

✅ Correct Answer Explanation

(A). Restriction enzymes like EcoRI produce sticky ends by cutting the phosphodiester backbone of each strand a few base pairs away from the center of the recognition site. Because they cut at the same relative position on both strands, the resulting DNA ends have complementary, single-stranded "overhangs."

❓ Why Other Options Are Incorrect

Option A → A center cut produces blunt ends.
Option C → The enzyme must cut within the recognition site.
Option D → Cutting only one strand would be a "nick," not an end-creation.

🎯 Strategy

Used Elimination.

Application: Distinguish blunt (center) vs. sticky (off-center). Only (A) correctly describes the staggered cut mechanism.

Final Logic: Staggered cut = Sticky ends.

🧠 Memory Trick

"Staggered = Sticky."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

10 According to the visual sequence of forming recombinant DNA, what specifically allows the sticky ends of the vector DNA and foreign DNA to unite before ligase acts?

A The covalent bonds formed between identical sugar bases.

B The hydrogen bonds formed with their complementary cut counterparts.

C The ionic interactions between the DNA and the enzyme.

D The immediate action of DNA polymerase.

💡 Short Explanation

Sticky ends match because of complementary base pairing. Base pairing is held by hydrogen bonds. This alignment acts like temporary "Velcro" until ligation.

✅ Correct Answer Explanation

The correct answer is (A). The sticky ends of the vector and foreign DNA are complementary to each other. When they align, they spontaneously form hydrogen bonds between the complementary bases (e.g., A-T, G-A). This creates a temporary stable association.

❓ Why Other Options Are Incorrect

Option A → Covalent bonds are formed later, by the ligase, not before it acts.
Option C → Ionic interactions are not the specific mechanism for base-pair complementarity.
Option D → DNA polymerase has no role in joining sticky ends.

🎯 Strategy

Used Contextual Matching.

Application: Base pairing = Hydrogen bonds.

Final Logic: Complementary sticky ends align through Hydrogen bonding.

🧠 Memory Trick

"Sticky ends = Hydrogen bond Velcro."

📚 Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology

11 Match the process to the enzyme responsible:

List I	List II
1. Cutting DNA at a specific site	P. DNA Ligase
2. Connecting sticky ends end-to-end	Q. DNA Polymerase
3. Adding methyl groups to DNA	R. Restriction Endonuclease
4. Synthesizing multiple copies in PCR	S. Methylase (Methyl-adding enzyme)

A 1-R, 2-P, 3-S, 4-Q

B 1-P, 2-R, 3-Q, 4-S

C 1-S, 2-P, 3-R, 4-Q

D 1-R, 2-S, 3-P, 4-Q

💡 Short Explanation

Cut = Restriction Endonuclease (R). Connect = Ligase (P). Methylate = Methylase (S). Synthesize = Polymerase (Q).

✅ Correct Answer Explanation

The correct answer is (A). Restriction endonucleases (R) are the "molecular scissors" that cut DNA. DNA ligase (P) functions as "molecular glue" to join DNA ends. Methylase (S) modifies DNA by adding methyl groups. DNA polymerase (Q) is essential for synthesizing or multiplying DNA copies, notably in processes like PCR.

❓ Why Other Options Are Incorrect

Options B, C, and D misalign the enzymes with their primary biological functions as defined in recombinant DNA technology.

🎯 Strategy

Used Substitution.

Application: Identify the two most certain pairs (e.g., 1-R and 2-P). Only option (A) contains this correct pairing.

Final Logic: Proper matching confirms option (A).

🧠 Memory Trick

"R-Cut, P-Join, S-Methyl, Q-Copy."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Tools of Recombinant DNA Technology – –204

12 Which of the following conditions is NOT required for DNA ligase to successfully create a recombinant DNA molecule from two sources?

A The fragments must have been cut by the same restriction enzyme.

B The fragments must have complementary sticky ends.

C The fragments must be from the exact same biological species.

D The ends of the fragments must be brought together in proximity.

💡 Short Explanation

Recombinant DNA technology is designed to join DNA from different sources (even different species). Requirement of same species is false. The other options (same enzyme, complementary ends, proximity) are physical/chemical requirements.

✅ Correct Answer Explanation

The correct answer is (C). Recombinant DNA technology specifically involves creating combinations of DNA from different biological sources. The fragments do not need to be from the same species; they only need to have complementary ends (produced by the same restriction enzyme) to align and be joined.

❓ Why Other Options Are Incorrect

Option A → Required; using the same enzyme ensures complementary ends.
Option B → Required; complementary ends are necessary for H-bonding.
Option D → Required; proximity allows the enzyme to access the break.

🎯 Strategy

Used Extreme Word Filter.

Application: The phrase "exact same biological species" contradicts the fundamental definition of "recombinant DNA" (which implies joining foreign/different DNA).

Final Logic: Recombinant = Different sources.

🧠 Memory Trick

"Recombinant = Diverse origins."

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

13 Regarding gel electrophoresis, evaluate these statements:
I. DNA moves towards the anode because it is negatively charged.
II. The electric field is the driving force for the movement of fragments.
III. DNA moves towards the cathode because it is positively charged.

A I and II only

B II and III only

C I and III only

D I only

💡 Short Explanation

DNA is negative. Anode is positive. DNA moves to Anode (Statement I true). Electricity moves them (Statement II true). DNA is not positive (Statement III false).

✅ Correct Answer Explanation

The correct answer is (A). DNA is negatively charged due to its phosphate backbone, so it moves toward the positive electrode (Anode). The applied electric field is the force that pulls the DNA through the gel. Statement III is false because DNA is never positively charged.

❓ Why Other Options Are Incorrect

Options B, C, and D are incorrect because they include the false statement III or exclude the true statement I or II.

🎯 Strategy

Used Elimination.

Application: Identify III as false (DNA is never positive). Eliminate all options containing III.

Final Logic: Negative DNA → Positive Anode.

🧠 Memory Trick

"Negative = Anode."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Separation and Isolation of DNA Fragments –

14 Arrange the following DNA fragments in the order they would appear from the loading well to the farthest end of the gel due to the sieving effect:
1. 500 base pairs
2. 2500 base pairs
3. 100 base pairs
4. 1500 base pairs

A 2, 4, 1, 3

B 3, 1, 4, 2

C 2, 1, 4, 3

D 3, 4, 1, 2

💡 Short Explanation

Smallest moves farthest (closest to anode). Largest stays nearest to the well. Order from well: Largest to Smallest (2500, 1500, 500, 100).

✅ Correct Answer Explanation

The correct answer is (A). In agarose gel electrophoresis, DNA fragments separate by size. The largest fragments (2500 bp) are most impeded by the gel matrix and move the least, staying nearest to the loading well. The smallest fragments (100 bp) move the fastest and reach the farthest end of the gel. Thus, the order from well to end is 2500 → 1500 → 500 → 100.

❓ Why Other Options Are Incorrect

Options B, C, and D provide incorrect size-migration sequences.

🎯 Strategy

Used Substitution.

Application: Rule: Largest = Nearest; Smallest = Farthest. Sequence the fragments based on this rule.

Final Logic: Size $\uparrow$ → Speed $\downarrow$ → Distance $\downarrow$.

🧠 Memory Trick

"Small = Sprint."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Separation and Isolation of DNA Fragments –

15 Why is ethidium bromide highly essential in the downstream process of gel electrophoresis?

A It cuts the DNA fragments into smaller pieces.

B It pushes the DNA faster through the agarose matrix.

C It allows the invisible DNA fragments to emit visible color upon UV exposure.

D It protects the DNA from being damaged by the electric current.

💡 Short Explanation

DNA is invisible. EtBr is a fluorescent dye. EtBr + UV = Visible bands.

✅ Correct Answer Explanation

The correct answer is (C). Pure DNA is transparent. Ethidium bromide (EtBr) is an intercalating dye that binds to DNA. When the gel is exposed to UV light, the EtBr-DNA complex emits a bright orange fluorescence, rendering the invisible DNA bands visible.

❓ Why Other Options Are Incorrect

Option A → EtBr is a stain, not a cutting enzyme.
Option B → EtBr does not affect DNA migration speed.
Option D → EtBr is for visualization, not protection from electricity.

🎯 Strategy

Used Contextual Matching.

Application: EtBr = Stain = Visualization under UV.

Final Logic: Stain + UV = Visible.

🧠 Memory Trick

"Stain = See."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Separation and Isolation of DNA Fragments –

16 Which of the following combinations will NOT result in visible bright orange coloured bands of DNA?

A Ethidium bromide stained gel + UV light

B Ethidium bromide stained gel + Visible light

C Pure DNA without staining + Visible light

D Both B and C

💡 Short Explanation

Condition A works (EtBr + UV). Condition B fails (EtBr needs UV to glow). Condition C fails (Pure DNA is invisible). Therefore, both B and C are correct "failures."

✅ Correct Answer Explanation

The correct answer is (D). Visibility of the bands depends on two factors: the presence of the stain (EtBr) and UV radiation to excite the dye. In option (B), there is no UV light; in option (C), there is no stain. Both conditions result in no visible bands.

❓ Why Other Options Are Incorrect

Option A → Results in visible orange bands.

🎯 Strategy

Used Elimination.

Application: Identify the two failures. Since both B and C fail to produce orange bands, the combined option (D) is the correct choice.

Final Logic: Missing Stain OR Missing UV = No visibility.

🧠 Memory Trick

"No Stain/No UV = No Sight."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Separation and Isolation of DNA Fragments –

17 When visually identifying bands to extract, a researcher looks for:

A Blue colonies on the gel.

B Bright orange bands under UV light.

C Transparent bands in the matrix.

D A continuous smear under visible light.

💡 Short Explanation

After staining with EtBr and applying UV, DNA bands fluoresce bright orange. These are the bands selected for extraction.

✅ Correct Answer Explanation

The correct answer is (B). Ethidium bromide is chosen for gel electrophoresis specifically because it renders DNA bands bright orange under UV illumination. These specific bands are then identified and excised for further use (elution).

❓ Why Other Options Are Incorrect

Option A → Blue colonies are used in transformation/screening, not gel electrophoresis bands.
Option C → DNA is transparent, so they are not "visible" as transparent bands.
Option D → A smear indicates degraded or poorly separated DNA, not isolated bands.

🎯 Strategy

Used Contextual Matching.

Application: Recall the specific visual appearance described in NCERT: "bright orange coloured bands."

Final Logic: EtBr + UV = Bright orange.

🧠 Memory Trick

"UV = Bright Orange."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Separation and Isolation of DNA Fragments –

18 Which of the following actions is NOT a part of the elution process?

A Identifying the orange band under UV light.

B Cutting out the separated bands from the agarose gel.

C Ligation of the DNA fragment into a cloning vector inside the gel.

D Extracting the purified DNA fragment from the gel piece.

💡 Short Explanation

Elution is the recovery of the DNA band from the gel. Ligation occurs after the DNA is extracted and purified, not inside the gel. Therefore, ligation in the gel is incorrect.

✅ Correct Answer Explanation

The correct answer is (C). Elution is strictly the process of extracting the DNA band from the excised gel slice. Ligation into a vector is a separate, downstream process that happens after the DNA has been successfully recovered and purified from the gel.

❓ Why Other Options Are Incorrect

Options A, B, and D describe the steps of identifying, excising, and recovering the DNA, which are all part of the elution/recovery workflow.

🎯 Strategy

Used Elimination.

Application: Identify the step that doesn't fit the "recovery" definition. Ligation is a "joining" step, not an extraction step.

Final Logic: Elution = Extraction (Recovery).

🧠 Memory Trick

"Elution = Take out, not link up."

📚 Class XII – Biology – Chapter: Biotechnology: Principles and Processes – Topic: Separation and Isolation of DNA Fragments –

19 What critical role does the host's DNA polymerase enzyme play once the recombinant DNA (like a modified plasmid) is transferred into an E. coli cell?

A It cuts the plasmid to allow gene expression.

B It makes multiple copies of the recombinant DNA.

C It adds methyl groups to protect the host DNA.

D It helps in visualising the DNA.

💡 Short Explanation

Plasmids have an "origin of replication." Host DNA polymerase recognizes this and replicates the plasmid. This multiplies the inserted recombinant DNA.

✅ Correct Answer Explanation

The correct answer is (B). Once a recombinant plasmid is inside the host cell (E. coli), it utilizes the host's cellular machinery—specifically the DNA polymerase—to replicate itself. This multiplication ensures that the alien/foreign DNA is copied many times as the host cell divides.

❓ Why Other Options Are Incorrect

Option A → DNA polymerase is a synthesizer, not a cutter.
Option C → While host enzymes methylate DNA, this is not the role of polymerase in the context of recombinant replication.
Option D → Polymerase is not involved in visualization.

🎯 Strategy

Used Substitution.

Application: Polymerase function = Synthesis/Replication. Therefore, "makes multiple copies" is the correct functional mapping.

Final Logic: DNA Polymerase = Replication/Multiplication.

🧠 Memory Trick

"Polymerase = Polymer-maker (copy maker)."

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

20 Why are vectors like plasmids absolutely necessary for cloning an alien piece of DNA in a host organism?

A Alien DNA lacks an 'origin of replication' and needs a vector's specific sequences to multiply.

B Alien DNA is too large to enter the host cell on its own.

C Plasmids act as molecular scissors to integrate the DNA.

D Host organisms reject any DNA that is not circular.

💡 Short Explanation

Foreign DNA cannot replicate by itself in a host. Vectors contain an Origin of Replication (ori). The host enzyme binds to the 'ori' on the vector to initiate copying.

✅ Correct Answer Explanation

The correct answer is (A). A crucial requirement for any piece of DNA to be cloned in a host is that it must be able to replicate. Foreign (alien) DNA lacks the necessary control sequences, called the "origin of replication" (ori). Vectors provide this 'ori,' which signals the host's DNA polymerase to initiate replication, allowing the foreign DNA to be maintained and multiplied.

❓ Why Other Options Are Incorrect

Option B → Size is not the primary reason; replication capability is.
Option C → Plasmids do not act as scissors; enzymes (restriction endonucleases) do.
Option D → Host organisms can take up linear DNA under certain conditions; circularity is not a strict rejection criteria, but rather a stability requirement for cloning.

🎯 Strategy

Used Contextual Matching.

Application: Key concept of cloning vectors: The presence of an 'ori' (origin of replication) is the defining requirement for multiplication.

Final Logic: Alien DNA + Vector(ori) = Replication.

🧠 Memory Trick

"Vector = 'ori' Provider."

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

CUET UG Biology Booster Test 2- Tools of Recombinant DNA Technology

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