Cuet Ug Biology Booster Unit 5 Test (D3) -

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CUET Booster Biology Unit 5 Test (D3)

⏱️ 30 Minutes

❓ 20 Questions

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

CUET Booster Biology Unit 5 Test (D3)

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QUESTION 1 OF 20

The semiconservative model of DNA replication inherently relies on which specific characteristic of the DNA double helix?

Answer Review

1 The semiconservative model of DNA replication inherently relies on which specific characteristic of the DNA double helix?

A The anti-parallel polarity of the two strands

B The uniform distance generated by purine-pyrimidine pairing

C The complementary hydrogen bonding between the specific base pairs.

D The right-handed coiling of the polynucleotide chains

💡 Short Explanation

Replication copies information. Information is encoded in base sequences. Complementarity allows one strand to dictate the synthesis of its partner.

✅ Correct Answer Explanation

The semiconservative model depends on the fact that each DNA strand is complementary to its partner. The hydrogen bonding (A=T, G≡C) allows each parental strand to act as a template. Without this specific chemical complementarity, the parental strands could not guide the exact sequence assembly of new strands.

❓ Why Other Options Are Incorrect

Option A → Polarity relates to the direction of synthesis (5'→3'), not the information-copying ability.
Option B → This provides structural stability, not copying capability.
Option D → This is a geometric property of the helix, not the basis for replication.

🎯 Strategy

Used

Contextual/Tonal Matching: Identifying the functional basis of information transfer.

Final Logic: Complementary base pairing is the physical mechanism that enables accurate duplication.

🧠 Memory Trick

"Complementary = Copying."

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

2 Match the replication concept to its core mechanistic outcome:

Column I	Column II
1. Semiconservative replication	P. Base pairing determines sequence
2. Template strand function	Q. Replication fork = Small opening
3. Replication fork	R. One parent strand and one newly synthesized strand
4. Complementary base pairing	S. Directs nucleotide addition

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

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

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

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

💡 Short Explanation

1-R: Semiconservative = One parent/one new strand. 2-S: Template = Directs nucleotide addition. 3-Q: Replication fork = Small opening. 4-P: Base pairing = Determines sequence.

✅ Correct Answer Explanation

Semiconservative replication results in hybrid molecules (R). The template strand directs the nucleotide addition (S). The replication fork allows access to the DNA in small, energy-efficient openings (Q). Complementary base pairing determines the exact sequence of the new strand (P).

❓ Why Other Options Are Incorrect

Other mappings fail to correctly correlate the biological mechanisms to their definitive outcomes.

🎯 Strategy

Used

Option Grouping: Matching biochemical processes to their specific roles.

Final Logic: Aligning definitions: Semiconservative (Hybrid), Template (Directs Addition), Fork (Opening), Base pairing (Sequence).

🧠 Memory Trick

"Template-Directs; Pairing-Determines."

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

3 Which of the following parameters does NOT contribute to the separation of DNA molecules in the Meselson-Stahl centrifugation method?

A The mass difference between 15N and 14N.

B The application of centrifugal force.

C The density gradient formed by cesium chloride (CsCl).

D The radioactive decay emissions of the nitrogen isotopes.

💡 Short Explanation

15N is a stable isotope. It is not radioactive. Separation relies entirely on mass/density.

✅ Correct Answer Explanation

The Meselson-Stahl experiment used density gradient centrifugation to separate heavy (15N) DNA from light (14N) DNA. 15N is a stable isotope, not a radioisotope, so radioactive decay plays no role in the experiment.

❓ Why Other Options Are Incorrect

Option A, B, and C are all fundamental requirements for the density-based separation of the DNA molecules.

🎯 Strategy

Used

Extreme Word Filter: Identifying "radioactive decay" as a non-applicable concept.

Final Logic: 15N is stable, so radioactivity is irrelevant.

🧠 Memory Trick

"15N = Stable/Heavy."

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

4 Regarding the Taylor and colleagues experiment in 1958, which of the following statements are correct?
I. It involved the use of radioactive thymidine.
II. It was performed on Vicia faba (faba beans).
III. It proved that DNA in chromosomes replicates conservatively.
IV. It demonstrated semiconservative replication in chromosomal DNA.

A I, II, and III

B I, II, and IV

C II, III, and IV

D I and IV

💡 Short Explanation

Used radioactive thymidine (I). Used Vicia faba (II). Proved semiconservative (IV), NOT conservative (III is wrong).

✅ Correct Answer Explanation

Taylor and colleagues (1958) extended the semiconservative evidence to chromosomes using radioactive thymidine to detect DNA distribution in Vicia faba (faba beans). This proved that chromosomal DNA replication is also semiconservative (IV), not conservative (III).

❓ Why Other Options Are Incorrect

Options containing Statement III are incorrect as the experiment disproved conservative replication.

🎯 Strategy

Used

Elimination: Eliminate options containing III (conservative).

Final Logic: Taylor's experiment is the classic proof of semiconservative replication in eukaryotes (chromosomes).

🧠 Memory Trick

"Taylor = Faba Beans = Semiconservative."

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

5 Analyze the expected results if E. coli cells from a 15N medium are cultured in a 14N medium for 80 minutes:
I. The cells will have completed exactly four generations of division.
II. The proportion of light density DNA will be significantly higher than hybrid DNA.
III. Heavy density DNA (fully 15N) will reappear in the 4th generation.
IV. The amount of hybrid DNA remains numerically constant but drops in proportion relative to the total DNA pool.

A I, II, and IV

B I, II, and III

C II, III, and IV

D I and IV

💡 Short Explanation

80 min / 20 min = 4 generations (I). Hybrid decreases; Light increases (II). Heavy DNA is gone forever after Gen 1 (III is wrong). Hybrid molecules number stays constant (IV).

✅ Correct Answer Explanation

In 80 minutes, E. coli completes 4 generations (I). Hybrid molecules are formed only in Gen 1; thereafter, replication produces more light DNA, making light DNA dominant (II). Heavy DNA never reappears because the 15N strands are diluted (III is false). The number of hybrid molecules remains fixed at two, while total DNA increases, so their proportion drops (IV).

❓ Why Other Options Are Incorrect

Options containing III are incorrect because heavy DNA cannot be re-synthesized from 14N.

🎯 Strategy

Used

Elimination: Reject any option containing Statement III.

Final Logic: Gen count is 4; Hybrid molecules stay at 2; Light DNA increases exponentially.

🧠 Memory Trick

"Heavy = Gone; Hybrid = Constant number; Light = Increasing."

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

6 If Meselson and Stahl had allowed the E. coli to grow for 60 minutes instead of 40 in the 14N medium, which outcome would NOT be observed?

A Three replication cycles would have been completed.

B The amount of hybrid DNA would exceed the amount of light DNA.

C The total number of DNA molecules would be eight times the original number.

D Only hybrid and light density DNA bands would be visible in the gradient.

💡 Short Explanation

60 mins = 3 generations (Gen 3). Hybrid = 2 molecules. Light = 6 molecules. Light > Hybrid.

✅ Correct Answer Explanation

At 60 minutes (Gen 3), there are 8 total DNA molecules (2 hybrids and 6 lights). Therefore, light DNA is much more abundant than hybrid DNA. Statement B is false as it claims hybrid > light.

❓ Why Other Options Are Incorrect

A, C, and D are correct: 60/20 = 3 cycles; 2^3 = 8 molecules; only hybrid/light exist.

🎯 Strategy

Used

Dimensional/Unit Analysis: Calculate molecule count for Gen 3.

Final Logic: Gen 3 has 8 molecules (2 Hybrid, 6 Light), so hybrid cannot exceed light.

🧠 Memory Trick

"Gen 3 = 8 molecules (2 Hybrid, 6 Light)."

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

7 Match the replication enzymes/substrates to their energetic or catalytic traits:

Column I	Column II
1. DNA-dependent DNA polymerase	P. dNTPs = Substrate + Energy
2. DNA ligase	Q. Efficient, 5′→3′ catalyst
3. Deoxyribonucleoside triphosphates (dNTPs)	R. Origin of replication (Ori) = Initiation sequence
4. Origin of replication (ori)	S. Ligase = Joins fragments

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

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

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

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

💡 Short Explanation

1-Q: DNA Pol = Efficient, 5'-3' catalyst. 2-S: Ligase = Joins fragments. 3-P: dNTPs = Substrate + Energy. 4-R: Ori = Initiation sequence.

✅ Correct Answer Explanation

DNA Pol is a highly efficient catalyst (Q). Ligase joins fragments (S). dNTPs provide substrate and energy (P). Ori is the sequence for initiation (R).

❓ Why Other Options Are Incorrect

Other mappings fail to match the specific biochemical definitions.

🎯 Strategy

Used

Option Grouping: Aligning the biochemical components to their functional definitions.

Final Logic: Accurate matching based on textbook function definitions.

🧠 Memory Trick

"Ligase = Joins (S); dNTP = Substrate + Energy (P)."

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

8 Arrange the following molecular steps based on their progressive occurrence during discontinuous DNA replication:
1. DNA ligase joins the newly synthesized fragments.
2. DNA polymerase adds deoxynucleotides in the 5'→3' direction.
3. The replication fork opens a small segment of the DNA helix.
4. The template with 5'→3' polarity is exposed for pairing.

A 3, 4, 2, 1

B 4, 3, 2, 1

C 3, 2, 4, 1

D 2, 3, 4, 1

💡 Short Explanation

Fork opens (3). Template exposed (4). Synthesis by Pol (2). Ligase joins (1).

✅ Correct Answer Explanation

The process begins with fork opening (3), which exposes the template (4). DNA polymerase then polymerizes the nucleotides (2). Finally, DNA ligase joins the resulting Okazaki fragments (1).

❓ Why Other Options Are Incorrect

These options reverse the logical order of events required for DNA synthesis.

🎯 Strategy

Used

Contextual/Tonal Matching: Sequencing the biological events of replication.

Final Logic: Unwinding > Exposure > Synthesis > Ligation.

🧠 Memory Trick

"Open, Expose, Add, Join."

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

9 A failure in cellular coordination right after the replication fork completes duplicating the genome at the S-phase results in:

A Uncontrolled initiation at random ori sites

B Polyploidy, a chromosomal anomaly

C Immediate reversal of the DNA polymerase activity

D Cleavage of the newly formed hybrid DNA

💡 Short Explanation

S-phase ensures exactly 2n to 4n. Failure leads to uneven/multiple chromosome sets. This is polyploidy.

✅ Correct Answer Explanation

DNA replication must be strictly controlled to occur exactly once per cell cycle. Failure to coordinate replication with cell division can result in cells containing multiple sets of chromosomes, a condition known as polyploidy.

❓ Why Other Options Are Incorrect

Option A, C, and D are not the standard phenotypic outcomes of replication coordination failure.

🎯 Strategy

Used

Substitution: Connect cell-cycle failure to genetic consequences.

Final Logic: Replication failure in S-phase disrupts normal chromosome count (polyploidy).

🧠 Memory Trick

"S-phase failure = Ploidy disorder."

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

10 Which characteristic is NOT true concerning the synthesis of the lagging strand at the replication fork?

A It is synthesized discontinuously.

B It requires DNA ligase to join the fragments later.

C It is copied from the template strand with a 5'→3' polarity.

D The DNA polymerase on this strand synthesizes DNA in the 3'→5' direction.

💡 Short Explanation

DNA polymerase always synthesizes in 5'→3' direction. Lagging strand is made of fragments, but each fragment is still 5'→3'.

✅ Correct Answer Explanation

DNA polymerase is unidirectional; it can only catalyze polymerization in the 5'→3' direction. The lagging strand is synthesized discontinuously to accommodate the fork geometry, but each individual fragment is still synthesized 5'→3'. Statement D is false.

❓ Why Other Options Are Incorrect

Option A, B, and C are correct descriptions of lagging strand synthesis.

🎯 Strategy

Used

Extreme Word Filter: Identifying the directional error (3'-5') as the false statement.

Final Logic: Polymerase direction is always 5'-3'.

🧠 Memory Trick

"Pol = 5 to 3 always."

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

11 Arrange the logical reasoning steps indicating why RNA was the first genetic material before DNA evolved:
1. DNA evolved from RNA with chemical modifications for greater stability.
2. Essential life processes like metabolism and translation evolved around RNA.
3. RNA functioned as both a genetic material and a reactive catalyst.
4. The reactive nature of the RNA catalyst made it structurally unstable.

A 2, 3, 4, 1

B 3, 2, 4, 1

C 4, 3, 2, 1

D 2, 4, 3, 1

💡 Short Explanation

Life started with RNA (2, 3). Its instability (4) forced the evolution of DNA (1).

✅ Correct Answer Explanation

RNA-based life (metabolism/translation) evolved first (2). RNA acted as both genotype and catalyst (3). Because RNA's reactive nature led to instability (4), life evolved DNA as a stable storage molecule (1).

❓ Why Other Options Are Incorrect

These options incorrectly sequence the evolutionary requirements of stability and functionality.

🎯 Strategy

Used

Option Grouping: Mapping evolutionary progression from RNA-world to DNA-storage.

Final Logic: Function (2,3) -> Instability (4) -> Solution/Stability (1).

🧠 Memory Trick

"Function first, then stability."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: RNA World – .

12 If both DNA strands were transcribed simultaneously, the resulting double-stranded RNA would prevent translation. This specific prevention implies that:

A The genetic information transfer machinery operates exclusively through double-stranded intermediates.

B The exercise of transcription would become futile as protein synthesis is hindered.

C The RNA polymerases would instantly degrade the double-stranded structure.

D The complementary RNA strands would undergo reverse transcription.

💡 Short Explanation

dsRNA stops translation. If you stop translation, the whole process of protein making fails. Transcription becomes a wasted effort.

✅ Correct Answer Explanation

If both strands were transcribed, they would form dsRNA. Since dsRNA cannot be translated into proteins, the cell would have wasted metabolic energy (transcription) for no functional protein product, rendering the process futile.

❓ Why Other Options Are Incorrect

Option A, C, and D are not the direct biological implications of translation prevention by dsRNA.

🎯 Strategy

Used

Contextual/Tonal Matching: Choosing the most logical outcome of inhibiting protein production.

Final Logic: Futile transcription = Energy waste with no protein output.

🧠 Memory Trick

"dsRNA = Protein synthesis blocked = Futile."

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

📄 Passage

"There is a convention in defining the two strands of the DNA in the structural gene of a transcription unit. Since the two strands have opposite polarity and the DNA-dependent RNA polymerase also catalyse the polymerisation in only one direction, that is, 5'→3', the strand that has the polarity 3'→5' acts as a template... The other strand which has the polarity (5'→3') and the sequence same as RNA (except thymine at the place of uracil), is displaced during transcription. Strangely, this strand (which does not code for anything) is referred to as coding strand."

Based on the defined conventions in the passage, if the promoter shifts to the 3'-end of the coding strand, it effectively becomes the:

A Terminator

B Origin of replication

C Enhancer

D Intron

💡 Short Explanation

Promoter defines the start (5' end). If you move the start mechanism to the end, it becomes the terminator or changes the entire gene direction.

✅ Correct Answer Explanation

The promoter defines the polarity of the transcription unit. If the functional element that signals "start" (promoter) is placed at the 3' end, the transcription unit's orientation is reversed, making that site effectively the terminator for the previous orientation.

❓ Why Other Options Are Incorrect

Option B, C, and D are unrelated to the directional control of transcription units.

🎯 Strategy

Used

Substitution: Replace "shifted promoter" with "definition reversal."

Final Logic: Promoter defines direction; move it, change the definition.

🧠 Memory Trick

"Start moves to End = Stop."

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

📄 Passage

According to the passage, the primary functional reason the coding strand is displaced during transcription is because:

A It contains multiple uracil residues that repel RNA polymerase.

B RNA polymerase only catalyzes polymerization in the 5'→3' direction, necessitating the 3'→5' strand as the template.

C The coding strand is structurally too heavy to be transcribed.

D The coding strand lacks a promoter region completely.

💡 Short Explanation

Polymerase is 5'-3'. Must read the 3'-5' strand. Therefore, the 5'-3' strand is "displaced."

✅ Correct Answer Explanation

Transcription is constrained by the strict 5'→3' catalytic directionality of RNA polymerase. Because it must read the template 3'→5', the complementary strand (coding strand) must be moved aside (displaced) to allow the enzyme access to the template.

❓ Why Other Options Are Incorrect

Option A, C, and D are not biological reasons for strand displacement.

🎯 Strategy

Used

Substitution: Replace "displaced" with "polymerase directionality requirement."

Final Logic: Directionality = 3'-5' Template requirement = Coding strand displacement.

🧠 Memory Trick

"5-3 synthesis needs 3-5 template."

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

15 Why does the split-gene arrangement in eukaryotes complicate the definition of a gene in terms of a DNA segment?

A Because overlapping sequences code for identical polypeptides simultaneously.

B Because intervening sequences (introns) do not appear in the mature RNA and do not code for the final product.

C Because exons are entirely removed before the mRNA leaves the nucleus.

D Because the promoter region is considered the actual coding segment.

💡 Short Explanation

Genes are traditionally "segments that code." Split-genes contain segments (introns) that don't code.

✅ Correct Answer Explanation

The traditional definition of a gene is a functional DNA segment coding for a protein. Split-genes contain introns, which interrupt the code and are removed, making the "gene" not a simple, continuous coding segment, but a collection of exons interspersed with non-coding introns.

❓ Why Other Options Are Incorrect

Option A, C, and D are factually inaccurate regarding the structure of eukaryotic genes.

🎯 Strategy

Used

Contextual/Tonal Matching: Matching the complexity of gene definition to the presence of non-coding introns.

Final Logic: Split-genes contain non-coding segments (introns), complicating the simple "DNA segment = Protein" definition.

🧠 Memory Trick

"Split-gene = Introns complicate."

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

16 Regulatory sequences are sometimes loosely defined as regulatory genes. What justifies this classification despite them not coding for RNA or proteins?

A They contain multiple uninterrupted exons.

B They act as physical anchors for ribosomes during translation.

C They significantly affect the inheritance of a character by controlling a structural gene.

D They are transcribed into heavily modified transfer RNAs.

💡 Short Explanation

Genes are "units of inheritance." Regulatory sequences control the expression of traits. Therefore, they fit the definition of a gene in an inheritance context.

✅ Correct Answer Explanation

A gene is a functional unit of inheritance. Even though regulatory sequences do not code for polypeptides, they control when and how structural genes are expressed. By controlling the expression of a trait, they are effectively part of the genetic unit of inheritance.

❓ Why Other Options Are Incorrect

Option A, B, and D describe incorrect functions or characteristics for regulatory sequences.

🎯 Strategy

Used

Contextual/Tonal Matching: Using the broadest definition of a gene ("unit of inheritance").

Final Logic: Inheritance control justifies the "regulatory gene" title.

🧠 Memory Trick

"Control expression = Inheritance control = Gene."

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

17 In a bacterial cell, transcription and translation can be coupled. Analytically, what physical condition uniquely permits this phenomenon?

A The extensive splicing mechanism that accelerates mRNA transport.

B The absence of a nuclear membrane, meaning there is no spatial separation of cytosol and nucleus.

C The utilization of RNA Polymerase II specifically for rapid translation.

D The presence of multiple origins of replication functioning as promoters.

💡 Short Explanation

Transcription and translation require the same space (ribosomes and mRNA). Prokaryotes have no nucleus.

✅ Correct Answer Explanation

Eukaryotes separate transcription (nucleus) and translation (cytosol). Bacteria lack this nuclear barrier, allowing ribosomes to access the mRNA transcript while the RNA polymerase is still creating it.

❓ Why Other Options Are Incorrect

Option A, C, and D are factually false for bacteria.

🎯 Strategy

Used

Substitution: Connect lack of nucleus to lack of spatial separation.

Final Logic: No nucleus = Spatial coupling.

🧠 Memory Trick

"Prokaryote = No Nucleus = Coupled."

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

18 Which of the following is NOT a true representation of the role of RNA polymerase and its associated factors in bacteria?

A The core RNA polymerase is only capable of catalyzing the process of elongation.

B Transient association with the rho (ρ) factor terminates transcription.

C Association with the sigma (σ) factor permanently modifies the active site for replication.

D Interaction with initiation and termination factors alters the enzyme's specific functional state.

💡 Short Explanation

Association is transient (temporary), not permanent. It affects transcription, not replication.

✅ Correct Answer Explanation

The sigma factor (σ) binds transiently to the core enzyme to facilitate initiation of transcription. It does not perform "permanent" modification, nor does it have any function related to DNA replication. Statement C is entirely false.

❓ Why Other Options Are Incorrect

Options A, B, and D are true biological facts about bacterial transcription.

🎯 Strategy

Used

Elimination: Identify the false statement about "permanence" and "replication."

Final Logic: Factor binding is temporary; transcription factors don't modify replication.

🧠 Memory Trick

"Sigma = Transient (temporary) = Transcription."

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

19 The clear-cut division of labor among the three RNA polymerases in the eukaryotic nucleus implies that:

A A mutation disrupting RNA Polymerase III would primarily halt the synthesis of tRNAs and snRNAs.

B RNA Polymerase I is the exclusive synthesizer of messenger RNAs.

C All three polymerases must physically bind to a single promoter simultaneously to transcribe a gene.

D Heterogeneous nuclear RNA (hnRNA) is generated collaboratively by Pol I and Pol III.

💡 Short Explanation

Pol I = rRNA. Pol II = hnRNA (mRNA precursor). Pol III = tRNA, snRNA, 5S rRNA.

✅ Correct Answer Explanation

In eukaryotes, RNA Polymerase III is specifically responsible for transcribing tRNA, 5S rRNA, and snRNAs. If it were mutated, these specific RNA classes would not be synthesized, directly impacting protein synthesis and splicing machinery.

❓ Why Other Options Are Incorrect

Option B, C, and D are factually incorrect regarding the specific functions and collaboration of the polymerases.

🎯 Strategy

Used

Substitution: Map the correct Polymerase (III) to its specific products (tRNA, snRNA).

Final Logic: Mutation in Pol III = tRNA/snRNA failure.

🧠 Memory Trick

"Pol I (rRNA), Pol II (mRNA/hnRNA), Pol III (tRNA/snRNA)."

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

20 Evolutionary biologists suggest that the presence of introns and the process of splicing in eukaryotic genomes is reminiscent of antiquity. What does this arrangement likely represent?

A The emergence of a completely protein-dominated cellular factory.

B The dominance of the ancient RNA-world where RNA-dependent processes were central.

C A recent evolutionary mutation restricting transcription speed.

D The structural degradation of DNA from its original continuous format.

💡 Short Explanation

Introns/Splicing are RNA-heavy, complex processes. Evolutionary theory links RNA complexity to early RNA-based life.

✅ Correct Answer Explanation

The existence of complex splicing machinery is thought to be an evolutionary remnant of the "RNA World," where RNA was both the genetic material and the catalyst for essential biochemical processes, predating the current DNA-based storage/protein-based machinery.

❓ Why Other Options Are Incorrect

Options A, C, and D are not the standard evolutionary justifications for the existence of introns.

🎯 Strategy

Used

Contextual/Tonal Matching: Aligning splicing complexity with RNA World theory.

Final Logic: Splicing is an RNA-dependent process, pointing to RNA-based early life.

🧠 Memory Trick

"Splicing = Remnant of RNA World."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Eukaryotic Complexity/RNA World

CUET Booster Biology Unit 5 Test (D3)

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