CUET UG Biology Booster Unit 5 Test (M3) -

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CUET Booster Biology Unit 5 Test (M3)

⏱️ 30 Minutes

❓ 20 Questions

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

CUET Booster Biology Unit 5 Test (M3)

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QUESTION 1 OF 20

Consider the following statements regarding the Watson-Crick double helix and its replication:
I. The specific base pairing immediately suggests a possible copying mechanism.
II. The two strands act as a template for synthesizing entirely distinct, non-complementary strands.
III. Upon completion of replication, each DNA molecule possesses one parental strand. Which of the statements is/are correct?

Answer Review

1 Consider the following statements regarding the Watson-Crick double helix and its replication:
I. The specific base pairing immediately suggests a possible copying mechanism.
II. The two strands act as a template for synthesizing entirely distinct, non-complementary strands.
III. Upon completion of replication, each DNA molecule possesses one parental strand. Which of the statements is/are correct?

A I only

B I and III

C II and III

D I, II, and III

💡 Short Explanation

Statement I is the fundamental insight of the Watson-Crick model. Statement II is false because strands synthesize complementary, not distinct/non-complementary strands. Statement III describes the semiconservative nature.

✅ Correct Answer Explanation

The double helix structure, based on specific base pairing (A=T, G=C), inherently implies that one strand can dictate the sequence of the other (I). The semiconservative model dictates that each new helix retains one old (parental) strand and gains one new strand (III). Statement II is incorrect because the newly synthesized strand must be complementary to the template to preserve genetic information.

❓ Why Other Options Are Incorrect

Option A → Incomplete, as Statement III is also true.
Option C → Incorrect because Statement II is false.
Option D → Incorrect because Statement II is false.

🎯 Strategy

Used

Elimination: Evaluate each statement; discard any option containing Statement II.

Final Logic: Base pairing and semiconservative preservation are core tenets; non-complementary synthesis would destroy genetic integrity.

🧠 Memory Trick

"Watson-Crick = Complementary Copying."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Experimental Proof – .

2 What is the direct consequence of the two strands of DNA separating during the semiconservative replication process?

A They fold into complex ribozyme structures.

B They act as templates for synthesizing new complementary strands.

C They immediately degrade to provide nucleotides for RNA synthesis.

D They form a triplex structure with a newly synthesized strand.

💡 Short Explanation

Strand separation exposes base sequences. This exposure allows DNA polymerase to match new nucleotides. This is the definition of a template.

✅ Correct Answer Explanation

The separation of the parental double helix exposes the nitrogenous bases on each strand. According to the principle of complementarity, these exposed bases direct the assembly of a new, matching strand. This process is the heart of semiconservative replication.

❓ Why Other Options Are Incorrect

Option A → DNA does not form ribozymes (that is RNA).
Option C → Parental strands are preserved, not degraded.
Option D → Replication produces duplex DNA, not triplex structures.

🎯 Strategy

Used

Contextual/Tonal Matching: Choosing the standard biological function of a DNA template.

Final Logic: Separation = Template exposure = Complementary synthesis.

🧠 Memory Trick

"Separate to Replicate."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Replication Scheme – .

3 Which of the following properties is NOT true regarding the 15N isotope used by Meselson and Stahl?

A It is the heavy isotope of nitrogen.

B It can be incorporated into newly synthesized DNA.

C It emits radioactive decay that is measured during centrifugation.

D It can be separated from 14N based on differences in densities.

💡 Short Explanation

15N is a stable isotope. It is not radioactive. Separation occurs by density, not radiation.

✅ Correct Answer Explanation

15N is a stable (non-radioactive) isotope. The experiment relied on the fact that 15N-labeled DNA is denser than 14N-DNA, allowing separation via density gradient centrifugation (using CsCl). There is no radioactive decay measured in this experiment.

❓ Why Other Options Are Incorrect

Option A → True; it is the heavier isotope.
Option B → True; nitrogen is a component of DNA bases.
Option D → True; that is the basis of the technique.

🎯 Strategy

Used

Extreme Word Filter: Identify "radioactive decay" as the false concept.

Final Logic: 15N is stable/non-radioactive; the experiment uses density, not radiation.

🧠 Memory Trick

"15N = Heavy, NOT Radioactive."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Meselson-Stahl Experiment – .

4 Match the experimental step/component to its exact function in the Meselson-Stahl experiment:

Column I	Column II
1. ¹⁵NH₄Cl medium	P. One generation time
2. ¹⁴NH₄Cl medium	Q. Heavy DNA generation
3. Cesium chloride	R. Light/Normal DNA generation
4. 20 minute interval	S. Density gradient

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

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

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

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

💡 Short Explanation

15NH4Cl (Q) = Heavy DNA generation. 14NH4Cl (R) = Light/Normal DNA generation. Cesium chloride (S) = Density gradient. 20 minutes (P) = One generation time.

✅ Correct Answer Explanation

15NH4Cl acts as the heavy nitrogen source (Q) to label parent DNA. 14NH4Cl is the medium (R) into which cells are transferred. Cesium chloride (S) creates the density gradient for centrifugation. 20 minutes (P) is the generation time of E. coli, used to capture specific replication stages.

❓ Why Other Options Are Incorrect

The other options provide incorrect mappings of these standard biological experimental parameters.

🎯 Strategy

Used

Option Grouping: Matching experimental components to their known functional roles.

Final Logic: Mapping the specific reagents and conditions to their experimental purpose.

🧠 Memory Trick

"15N = Q (heavy), 14N = R (light/replacement), CsCl = S (gradient), 20m = P (period)."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Meselson-Stahl Experiment – .

5 If E. coli grown in 15N is transferred to a 14N medium and left for exactly 20 minutes, why does the extracted DNA show an intermediate density?

A Because E. coli completes two divisions in 20 minutes.

B Because one parental 15N strand pairs with one newly synthesized 14N strand.

C Because the 15N atoms decay into 14N atoms in 20 minutes.

D Because both strands are newly synthesized with 14N.

💡 Short Explanation

20 minutes = One replication cycle. Each new DNA molecule is a hybrid of one heavy and one light strand. Intermediate density = Hybrid.

✅ Correct Answer Explanation

After 20 minutes, each parent 15N-DNA molecule has replicated once. The resulting DNA molecules each consist of one original 15N strand and one newly synthesized 14N strand. This "hybrid" DNA has a density halfway between heavy and light, confirming the semiconservative model.

❓ Why Other Options Are Incorrect

Option A → E. coli generation time is 20 minutes, so only one division occurs.
Option C → Isotopes are stable, they do not decay into each other in this experiment.
Option D → This would only be true after many generations.

🎯 Strategy

Used

Substitution: Apply the semiconservative model definition to the 20-minute timeline.

Final Logic: One cycle = Hybrid = Intermediate density.

🧠 Memory Trick

"Hybrid = 1 Heavy + 1 Light."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Meselson-Stahl Experiment – .

6 Arrange the proportions of hybrid DNA expected across successive generations of E. coli shifted from 15N to 14N medium (Generation 1, 2, 3, 4 respectively):
1. 12.5% hybrid;
2. 50% hybrid;
3. 25% hybrid;
4. 100% hybrid

A 4, 2, 3, 1

B 4, 3, 2, 1

C 2, 4, 1, 3

D 1, 3, 2, 4

💡 Short Explanation

Gen 1: 100% Hybrid (4). Gen 2: 50% Hybrid (2). Gen 3: 25% Hybrid (3). Gen 4: 12.5% Hybrid (1).

✅ Correct Answer Explanation

In Gen 1, all DNA is hybrid (100%). In Gen 2, two hybrid and two light molecules exist (50% hybrid). In Gen 3, only two hybrid molecules remain among eight total strands (25% hybrid). In Gen 4, this halves again to 12.5%. The sequence follows the decay of the "hybrid" signal by half each generation.

❓ Why Other Options Are Incorrect

They miscalculate the halving progression of the hybrid DNA.

🎯 Strategy

Used

Dimensional/Unit Analysis: Calculate the ratio of hybrid molecules per generation: 1, 1/2, 1/4, 1/8.

Final Logic: The proportion of hybrid DNA follows the geometric progression: 100\% \rightarrow 50\% \rightarrow 25\% \rightarrow 12.5\%.

🧠 Memory Trick

"Hybrid fraction halves each time."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Generation Ratios – .

7 Which of the following is NOT a characteristic of deoxyribonucleoside triphosphates in DNA replication?

A They act as substrates for the polymerization reaction.

B They provide energy required for the highly expensive polymerization.

C They possess two terminal high-energy phosphates similar to ATP.

D They act primarily to unwind the DNA double helix.

💡 Short Explanation

dNTPs are building blocks (substrates). Helicase unwinds the helix, not dNTPs.

✅ Correct Answer Explanation

Deoxyribonucleoside triphosphates (dNTPs) serve two roles: they act as the actual monomers (substrates) to be added to the growing chain, and their bond cleavage provides the energy for the polymerization. Unwinding the helix is the specific job of enzymes like DNA helicase, not the dNTPs themselves.

❓ Why Other Options Are Incorrect

Option A, B, and C are all standard characteristics of dNTPs in the replication process.

🎯 Strategy

Used

Substitution: Replace "dNTP function" with the actual function of helicase.

Final Logic: dNTPs build the chain; helicase unwinds it.

🧠 Memory Trick

"dNTP = Building blocks (not unzippers)."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Replication Enzymes – .

8 Why must the DNA-dependent DNA polymerases be highly efficient during E. coli replication?

A Because they have to synthesize an RNA primer before DNA synthesis.

B Because they have to catalyze the polymerization of a large number of nucleotides in a very short time (18 minutes).

C Because they actively repair double-strand breaks during cell division.

D Because they must constantly switch between 5'→3' and 3'→5' directions.

💡 Short Explanation

E. coli genome is 4.6 million bp. Division time is ~20 mins. Polymerase must be extremely fast to complete this.

✅ Correct Answer Explanation

E. coli has a genome of about 4.6 \times 10^6 base pairs. To complete replication within the cell's division time (~18-20 minutes), the DNA polymerases must catalyze polymerization at an incredible speed (approx. 2000 bp/sec), requiring high efficiency and accuracy.

❓ Why Other Options Are Incorrect

Option A → Primers are synthesized by Primase.
Option C → Repair enzymes handle breaks, not replication polymerase.
Option D → Polymerase only works in 5'→3' direction.

🎯 Strategy

Used

Contextual/Tonal Matching: Aligning enzyme efficiency with the physiological constraints of the cell.

Final Logic: High speed is needed to replicate millions of base pairs in minutes.

🧠 Memory Trick

"Millions of bases / Short time = High speed needed."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Replication Machinery – .

9 Read the following regarding the origin of replication (ori):
I. It is a definite region in the DNA where replication originates.
II. It allows random initiation of replication anywhere along the DNA.
III. A vector must provide an ori for a piece of DNA to be propagated during recombinant DNA procedures. IV. DNA polymerases can naturally initiate replication without an ori. Which are correct?

A I and III

B I, II, and III

C II and IV

D I and IV

💡 Short Explanation

Statement I is correct (definition). Statement II is false (not random). Statement III is correct (vector requirement). Statement IV is false (polymerase needs a primer/template site).

✅ Correct Answer Explanation

Replication is not random; it requires a defined "ori" site (I). Vectors used in cloning must contain an ori to allow the insert to be replicated by the host cell machinery (III). Statement II is incorrect (it is site-specific), and Statement IV is incorrect (DNA polymerases cannot initiate synthesis from scratch).

❓ Why Other Options Are Incorrect

Options containing II or IV are disqualified.

🎯 Strategy

Used

Elimination: Identify the incorrect statements (II and IV) to filter the correct options.

Final Logic: Ori is specific and required for cloning; it is not random and is required for polymerase activity.

🧠 Memory Trick

"Ori = Origin = Specific Start Point."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Replication Fork – .

10 Match the strand synthesis feature with its mechanism:

Column I	Column II
1. Discontinuous synthesis	P. Energy required for helicase/unwinding
2. Continuous synthesis	Q. Direction is restricted (5′→3′ polymerization)
3. 5′→3′ direction	R. Continuous on 3′→5′ template
4. High energy requirement	S. Discontinuous on 5′→3′ template

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

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

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

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

💡 Short Explanation

Discontinuous on 5'-3' template (S). Continuous on 3'-5' template (R). Direction is restricted (Q). Energy required for helicase/unwinding (P).

✅ Correct Answer Explanation

Discontinuous synthesis (Okazaki fragments) occurs on the template with 5'→3' polarity (S). Continuous synthesis occurs on the template with 3'→5' polarity (R). DNA polymerase is limited by the 5'→3' direction (Q). The separation of the DNA helix is energy-intensive (P).

❓ Why Other Options Are Incorrect

The other combinations do not reflect the correct molecular geometry of the replication fork.

🎯 Strategy

Used

Option Grouping: Matching replication mechanics to their fork-geometry descriptions.

Final Logic: Linking template polarity to the type of synthesis (Okazaki vs. leading).

🧠 Memory Trick

"5-3 Template = Fragmented (S); 3-5 Template = Smooth/Continuous (R)."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Replication Fork

11 Which foundational principle distinguishes the process of transcription from DNA replication in terms of base pairing?

A Guanine pairs with Adenine instead of Cytosine.

B Adenosine pairs with Uracil instead of Thymine.

C Thymine pairs with Guanine instead of Adenine.

D Cytosine pairs with Uracil instead of Guanine.

💡 Short Explanation

DNA replication uses Thymine (T) to pair with Adenine (A). Transcription uses Uracil (U) to pair with Adenine (A) in RNA.

✅ Correct Answer Explanation

In DNA replication, Adenine (A) on the template strand pairs with Thymine (T). In transcription, the RNA polymerase synthesizes an RNA strand where Adenine (A) on the DNA template is paired with Uracil (U) instead of Thymine. This is the hallmark chemical difference between the two processes.

❓ Why Other Options Are Incorrect

Option A, C, and D describe incorrect base-pairing rules that do not exist in standard biological transcription.

🎯 Strategy

Used

Substitution: Replace the "DNA rule" (A-T) with the "RNA rule" (A-U).

Final Logic: Transcription uses Uracil to complement Adenine.

🧠 Memory Trick

"Transcription = A to U."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Transcription Principles – .

12 Which of the following is NOT a reason why only one strand of DNA is copied into RNA during transcription?

A Copying both would produce RNA molecules with different sequences.

B Simultaneous transcription of both strands would yield two identical RNA molecules.

C Simultaneous transcription would produce complementary RNA strands that form double-stranded RNA.

D A double-stranded RNA would prevent translation into protein.

💡 Short Explanation

Copying both strands would yield complementary (not identical) RNA. Statement B is therefore false (it is NOT a valid reason).

✅ Correct Answer Explanation

If both strands were transcribed, they would produce two complementary RNA strands because the template strands themselves are complementary. These would bind to form double-stranded RNA (dsRNA), which cells generally treat as a danger signal (often used in RNA interference), effectively preventing protein translation. Statement B is false because the RNAs would be complementary, not identical.

❓ Why Other Options Are Incorrect

Options A, C, and D are all scientifically valid reasons why only one strand is transcribed.

🎯 Strategy

Used

Extreme Word Filter: Identifying "identical" as the incorrect term in this context.

Final Logic: Complementary strands form dsRNA; they are never identical.

🧠 Memory Trick

"Complementary \neq Identical."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Transcription Principles – .

13 Arrange the structural regions of a transcription unit as they are located relative to the coding strand (from 5' upstream to 3' downstream):
1. Terminator;
2. Structural gene;
3. Promoter;
4. Additional downstream regulatory sequences

A 3, 2, 1, 4

B 1, 2, 3, 4

C 3, 1, 2, 4

D 2, 3, 1, 4

💡 Short Explanation

Upstream (5') is the Promoter (3). The middle is the Structural Gene (2). Downstream (3') is the Terminator (1). Further regulatory sequences (4) follow.

✅ Correct Answer Explanation

In a transcription unit, the promoter is located at the 5'-end (upstream) of the coding strand. This is followed by the structural gene that is to be transcribed, and the transcription terminates at the terminator sequence located at the 3'-end (downstream). Additional regulatory sequences often follow the terminator.

❓ Why Other Options Are Incorrect

They place the promoter, gene, or terminator in the wrong order relative to the 5' to 3' flow.

🎯 Strategy

Used

Option Grouping: Mapping the linear geography of a gene.

Final Logic: Promoter > Gene > Terminator > Downstream.

🧠 Memory Trick

"P-G-T (Promoter-Gene-Terminator)."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Transcription Unit – .

14 By convention, if the promoter switches position with the terminator in a transcription unit, what is the immediate consequence?

A The RNA polymerase is destroyed.

B The definition of the coding and template strands is reversed.

C Both strands simultaneously begin to act as coding strands.

D The structural gene is spliced out of the sequence.

💡 Short Explanation

Promoter location defines the template direction. If the promoter flips, the 3'-5' strand flips.

✅ Correct Answer Explanation

The polarity of the coding strand (5'→3') and the template strand (3'→5') is determined by the location of the promoter. If the promoter were to switch to the other end, the strand that was formerly the template would now become the coding strand, and vice-versa.

❓ Why Other Options Are Incorrect

Option A, C, and D are physically impossible or biologically irrelevant outcomes.

🎯 Strategy

Used

Substitution: Replace "flipped promoter" with "flipped template definition."

Final Logic: The Promoter location defines the Template; move the Promoter, move the Template definition.

🧠 Memory Trick

"Promoter flips = Template flips."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Transcription Unit – .

15 Which of the following characteristics is NOT associated with the split-gene arrangement in eukaryotes?

A The coding sequences are called exons.

B Exons appear in mature or processed RNA.

C Introns are defined as expressed sequences.

D Intervening sequences do not appear in mature RNA.

💡 Short Explanation

Exons = Expressed. Introns = Intervening (Non-expressed).

✅ Correct Answer Explanation

In eukaryotes, "Exons" are the coding, expressed sequences, while "Introns" are the intervening, non-coding sequences. Statement C incorrectly calls introns "expressed sequences," which is the exact opposite of their role.

❓ Why Other Options Are Incorrect

Options A, B, and D are all accurate descriptions of the eukaryotic split-gene structure.

🎯 Strategy

Used

Odd One Out: Identify the statement that contradicts the definition of exons vs. introns.

Final Logic: Exons are expressed; Introns are intervening.

🧠 Memory Trick

"Exon = Expressed; Intron = Intervening."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Gene Definitions – .

16 In bacteria, the structural gene in a transcription unit is generally described as:

A Monocistronic

B Polycistronic

C Non-cistronic

D Spliced

💡 Short Explanation

Bacterial genes are typically organized in operons. One promoter can transcribe multiple genes (polycistronic).

✅ Correct Answer Explanation

Bacterial transcription units are characterized by "polycistronic" genes, meaning a single promoter regulates a cluster of genes that are transcribed into a single mRNA molecule. Eukaryotes, in contrast, are typically "monocistronic."

❓ Why Other Options Are Incorrect

Option A → Characteristic of Eukaryotes.
Option C → Not a biological classification.
Option D → Bacteria do not perform splicing.

🎯 Strategy

Used

Substitution: Recall the standard genetic organization of prokaryotes.

Final Logic: Bacteria = Polycistronic; Eukaryotes = Monocistronic.

🧠 Memory Trick

"Bacteria = Many (Poly) at once."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Gene Definitions – .

📄 Passage

"In eukaryotes, there are two additional complexities – (i) There are at least three RNA polymerases in the nucleus... The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA). (ii) The second complexity is that the primary transcripts contain both the exons and the introns and are non-functional. Hence, it is subjected to a process called splicing..."

According to the passage, the primary role of RNA polymerase II in eukaryotes is the transcription of:

A Transfer RNA (tRNA)

B Ribosomal RNA (rRNA)

C Heterogeneous nuclear RNA (hnRNA)

D Small nuclear RNAs (snRNAs)

💡 Short Explanation

Passage explicitly states: "The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA)."

✅ Correct Answer Explanation

The text specifies that RNA polymerase II is dedicated to transcribing the hnRNA, which is the precursor for messenger RNA (mRNA) in eukaryotic cells.

❓ Why Other Options Are Incorrect

The passage defines the specific role of Pol II; the other RNA types are transcribed by Pol I or III (though not explicitly listed in the passage, they are not the answer to this specific question).

🎯 Strategy

Used

Contextual/Tonal Matching: Reading and extracting data directly from the provided text.

Final Logic: The passage names hnRNA as the target of Pol II.

🧠 Memory Trick

"Pol II = hnRNA."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Eukaryotic Complexity – .

📄 Passage

What is the specific nature of the primary transcripts before splicing, as highlighted in the text?

A They consist entirely of active exons.

B They are fully functional mRNAs ready for translation.

C They contain both exons and introns and are non-functional.

D They lack adenylate residues but are fully spliced.

💡 Short Explanation

Passage states: "...primary transcripts contain both the exons and the introns and are non-functional."

✅ Correct Answer Explanation

The passage explains that eukaryotic primary transcripts are "non-functional" because they contain intervening sequences (introns) that must be removed through splicing before the mRNA can be used for protein synthesis.

❓ Why Other Options Are Incorrect

Options A, B, and D contradict the information provided in the passage.

🎯 Strategy

Used

Contextual/Tonal Matching: Extracting the definition from the provided text.

Final Logic: Passage explicitly calls them "non-functional" and lists both components.

🧠 Memory Trick

"Primary Transcript = Mixed/Non-functional."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Eukaryotic Complexity – .

19 Why can translation begin much before the mRNA is fully transcribed in bacterial cells?

A Because bacterial mRNA possesses a dense protective cap.

B Because bacteria lack RNA polymerase completely.

C Because mRNA requires no processing and both processes occur in the same compartment.

D Because transcription in bacteria is extremely slow compared to eukaryotes.

💡 Short Explanation

No nucleus = Same space. No splicing/processing = mRNA is ready immediately.

✅ Correct Answer Explanation

Bacterial cells lack a nucleus, so transcription and translation occur in the same cytoplasmic compartment. Because bacterial mRNA does not require splicing or extensive post-transcriptional processing (like capping/tailing), ribosomes can begin binding to the mRNA even while it is still being transcribed.

❓ Why Other Options Are Incorrect

Option A → Bacteria do not use caps in the eukaryotic sense.
Option B → Bacteria rely entirely on RNA polymerase.
Option D → Transcription is quite fast in bacteria.

🎯 Strategy

Used

Substitution: Combine cellular biology knowledge (no nucleus) with genetic knowledge (no splicing).

Final Logic: No nucleus + No processing = Simultaneous transcription and translation.

🧠 Memory Trick

"Prokaryote = One space, one process."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Prokaryotic Transcription – .

20 What is the consequence of RNA polymerase transiently associating with the initiation-factor (σ) and termination-factor (ρ) in prokaryotes?

A It alters the specificity of the RNA polymerase to either initiate or terminate transcription.

B It permanently binds the polymerase to the promoter region.

C It prevents the elongation phase from occurring.

D It causes the mRNA to be spliced immediately.

💡 Short Explanation

Sigma factor (σ) = Initiation (promoter binding). Rho factor (ρ) = Termination. They direct the polymerase's start and end points.

✅ Correct Answer Explanation

RNA polymerase is a core enzyme. By binding transiently to the sigma (σ) factor, it gains the specificity to recognize and initiate at the promoter. By binding to the rho (ρ) factor, it recognizes the terminator sequence to stop transcription. Thus, these factors modulate the "specificity" of the enzyme's activity.

❓ Why Other Options Are Incorrect

Option B → The binding is transient, not permanent.
Option C → The association allows the process, it does not prevent it.
Option D → Bacteria do not splice mRNA.

🎯 Strategy

Used

Substitution: Replace "association" with "control/specificity."

Final Logic: Sigma starts, Rho stops = Specificity control.

🧠 Memory Trick

"Sigma = Start (Initiation); Rho = Stop (Termination)."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Prokaryotic Transcription

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