CUET UG Biology Booster Unit 5 Test (D2) -

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CUET Booster Biology Unit 5 Test (D2)

⏱️ 30 Minutes

❓ 20 Questions

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

CUET Booster Biology Unit 5 Test (D2)

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QUESTION 1 OF 20

📄 Passage

\"In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous transformation in the bacteria. When Streptococcus pneumoniae are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R). Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria.\"

Biochemically, the smoothness of the S strain colonies mentioned in the passage is specifically attributed to:

Answer Review

📄 Passage

Biochemically, the smoothness of the S strain colonies mentioned in the passage is specifically attributed to:

A The absence of RNA inside the S strain

B The presence of a tough protein coat

C The presence of a mucous polysaccharide coat

D The mutation of its structural genes via heat

💡 Short Explanation

S strain bacteria are virulent due to a protective coating. This coat is composed of polysaccharides (mucous). It protects the bacteria from the host\\\\\\\'s immune system.

✅ Correct Answer Explanation

In Streptococcus pneumoniae, the \\\\\\\"Smooth\\\\\\\" (S) strain is characterized by the presence of a mucous polysaccharide coat (capsule). This capsule is essential for its virulence as it prevents phagocytosis by the host\\\\\\\'s immune cells. Options A, B, and D do not describe the biochemical basis for the colony morphology; the capsule is specifically a polysaccharide structure.

❓ Why Other Options Are Incorrect

Option A → Absence of RNA does not determine colony texture.
Option B → The coat is polysaccharide-based, not primarily a tough protein coat.
Option D → Heat kills the bacteria; it does not induce the mutation responsible for the smooth phenotype.

🎯 Strategy

Used: Elimination

Application: Identifying that the \\\\\\\"smooth\\\\\\\" phenotype is a classic biological reference to a polysaccharide capsule allows for immediate elimination of protein or RNA-based options.

Final Logic: The capsule that gives S-strain its \\\\\\\"smooth\\\\\\\" appearance is composed of polysaccharides.

🧠 Memory Trick

Smooth = Sugar (Polysaccharide).

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Transforming Principle –

📄 Passage

\"In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumoni
A), witnessed a miraculous transformation in the bacteria. When Streptococcus pneumoniae are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R). Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria.\"

In the passage, Griffith observed that heat-killed S strain enabled the R strain to synthesize a smooth coat. What did he term the substance responsible for this phenomenon?

A Central Dogma

B The Transforming Principle

C Polycistronic mRNA

D Nuclein

💡 Short Explanation

Griffith performed experiments on S. pneumoniae. He noted a change in R-strain phenotype. He coined the term for the unknown factor causing this change.

✅ Correct Answer Explanation

Frederick Griffith conducted experiments in 1928 and observed that the R strain bacteria were transformed into S strain bacteria when mixed with heat-killed S strain. Since he did not know the chemical nature of the substance causing this, he labeled it the \\\\\\\"Transforming Principle.\\\\\\\" Central Dogma refers to the flow of information (DNA→RNA→Protein), Polycistronic mRNA is a feature of prokaryotic transcription, and Nuclein was Miescher\\\\\\\'s term for DNA.

❓ Why Other Options Are Incorrect

Option A → Central Dogma describes information flow, not the phenomenon of bacterial conversion.
Option C → This is a genetic term not used by Griffith to describe the unknown factor.
Option D → Nuclein was the early name for DNA discovered by Miescher, not the name Griffith gave to his observation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Matching the historical terminology used by Griffith in the described experiment.

Final Logic: Griffith historically coined \\\\\\\"Transforming Principle\\\\\\\" to describe the unidentified factor that changed R-strain to S-strain.

🧠 Memory Trick

Transformed bacteria = Transforming Principle.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Transforming Principle –

3 Which of the following is NOT a correct inference drawn from the Avery, MacLeod, and McCarty experiments?

A The transforming substance was not a protein or RNA.

B DNA alone from S bacteria caused R bacteria to become transformed.

C Proteases successfully degraded the genetic material to halt transformation.

D The experiments determined the biochemical nature of Griffith\\\\\\\'s transforming principle.

💡 Short Explanation

Avery, MacLeod, and McCarty purified biochemicals. Proteases and RNases did not stop transformation. Only DNase stopped transformation.

✅ Correct Answer Explanation

Avery, MacLeod, and McCarty discovered that digestion with Proteases and RNases did not affect the transformation process. This proved that neither protein nor RNA was the genetic material. Transformation was only inhibited when DNase was added. Therefore, stating that proteases degraded the genetic material to halt transformation is factually incorrect.

❓ Why Other Options Are Incorrect

Option A → This is a correct inference, not the answer to \\\\\\\"NOT a correct inference.\\\\\\\"
Option B → This is the main conclusion of their work.
Option D → This is the overarching goal and result of their study.

🎯 Strategy

Used: Elimination

Application: Finding the statement that contradicts the experimental results provided in standard biological texts regarding the digestion of genetic material.

Final Logic: Proteases do not inhibit transformation because proteins are not the genetic material.

🧠 Memory Trick

Protease/RNase = Pass (Transformation continues); DNase = Dead (Transformation stops).

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Biochemical Characterisation –

4 Match the specific enzyme treatment (Column I) to its specific biological effect on transformation (Column II):

Column I	Column II
1. Protease	P. Breaks down RNA; transformation continues
2. RNase	Q. Breaks down DNA; transformation is inhibited
3. DNase	R. Breaks down proteins; transformation continues
4. Lipase	S. Breaks down lipids; transformation continues (no effect)

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

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

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

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

💡 Short Explanation

Protease → digests Protein (Transformation continues). RNase → digests RNA (Transformation continues). DNase → digests DNA (Transformation stops).

✅ Correct Answer Explanation

The experiment showed that transformation only stops when DNA is removed. Thus, Protease (1) and RNase (2) allow transformation to continue. DNase (3) breaks down the genetic material, inhibiting transformation (Q). Lipase (4) has no effect on the genetic material (S). Option A is the only choice that correctly maps these biological impacts.

❓ Why Other Options Are Incorrect

Option B → Incorrectly assigns Q (inhibition) to RNase.
Option C → Incorrectly assigns P (RNA digestion) to Protease.
Option D → Incorrectly assigns Q to Protease.

🎯 Strategy

Used: Option Grouping

Application: Establishing the key result that DNase is the only enzyme that inhibits transformation, then narrowing down the options.

Final Logic: Only DNase (3) matches Q (inhibition); only Option A matches this condition.

🧠 Memory Trick

DNase = Death of transformation.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Biochemical Characterisation –

5 If Avery, MacLeod, and McCarty had discovered that RNase inhibited transformation instead of DNase, what would have been the logical scientific conclusion?

A That proteins are the genetic material

B That RNA is the hereditary material in Streptococcus pneumoniae

C That DNA relies on RNA to transform cells

D That the R strain intrinsically possessed the S strain genome

💡 Short Explanation

Enzymes degrade specific substrates. Inhibition indicates the substrate is the genetic material. RNase targets RNA.

✅ Correct Answer Explanation

The principle of the experiment was that removing the genetic material would halt transformation. If RNase (which degrades RN A) had been the enzyme that stopped transformation, it would imply that RNA was the molecule carrying the genetic information necessary for the transformation of R strain to S strain.

❓ Why Other Options Are Incorrect

Option A → RNase acts on RNA, not protein.
Option C → The experiment was designed to identify the substance, not a complex interaction.
Option D → The experiment specifically sought to find what external substance transforms the bacteria.

🎯 Strategy

Used: Substitution

Application: Replacing the actual result (DN

A) with the hypothetical result (RN
A) to see what the conclusion would logically become.

Final Logic: If substance X is destroyed and transformation stops, then X must be the genetic material.

🧠 Memory Trick

RNase acts on RNA.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Biochemical Characterisation –

6 Evaluate the following statements regarding the proof of DNA as genetic material:
I. Avery et al. proved DNA was the transforming principle, but Hershey and Chase provided the unequivocal proof.
II. Biologists universally accepted DNA as the genetic material immediately after Avery\\\\\\\'s work.
III. Hershey and Chase proved DNA is the genetic material by tracking radioactive phosphorus from viruses into bacteria. Which of the statements are correct?

A I and II only

B I and III only

C II and III only

D I, II, and III

💡 Short Explanation

Avery et al. provided strong biochemical evidence. Hershey and Chase provided final, unequivocal proof. The scientific community remained skeptical for some time.

✅ Correct Answer Explanation

Statement I is correct as Avery\\\\\\\'s work provided the biochemical evidence, while Hershey and Chase\\\\\\\'s experiment used bacteriophages for definitive proof. Statement II is incorrect because there was significant skepticism in the scientific community regarding the findings of Avery, MacLeod, and McCarty. Statement III is correct; the use of P-32 allowed for the tracking of viral DNA into the host cell.

❓ Why Other Options Are Incorrect

Option A → Includes statement II, which is incorrect.
Option C → Includes statement II, which is incorrect.
Option D → Includes statement II, which is incorrect.

🎯 Strategy

Used: Elimination

Application: Identify that statement II is false, eliminating all options containing it.

Final Logic: Since the scientific community did not immediately accept the findings (statement II), only options containing I and III remain.

🧠 Memory Trick

H&C = Hell-yeah Conclusive (Unequivocal).

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Genetic Material is DNA –

7 What fundamental biological principle allowed Hershey and Chase to utilize bacteriophages to prove DNA is the genetic material?

A Bacteriophages naturally contain both host and viral DNA.

B The bacterial cell treats the entering viral genetic material as if it was its own, manufacturing more viruses.

C Bacteriophages only possess RNA genomes which mimic bacterial DNA.

D Bacterial cells actively consume viral protein coats for energy.

💡 Short Explanation

Bacteriophages are viruses infecting bacteria. They inject genetic material into the host. Host machinery reads and replicates viral material.

✅ Correct Answer Explanation

The Hershey-Chase experiment relied on the ability of the bacteriophage to \\\\\\\"hijack\\\\\\\" the bacterial cell\\\\\\\'s biosynthetic machinery. By proving that only the viral DNA enters the cell and is then used by the host to create new viral progeny, they established DNA as the carrier of genetic instructions.

❓ Why Other Options Are Incorrect

Option A → Bacteriophages only contain their own genetic material.
Option C → T2 bacteriophages are DNA-based, not RNA-based.
Option D → The protein coat stays outside and is removed; it is not consumed by the bacteria.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Matching the core premise of viral infection logic.

Final Logic: If the host manufactures virus particles, the substance that entered must be the genetic blueprint.

🧠 Memory Trick

Hershey-Chase = Hijack (viral infection principle).

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Genetic Material is DNA –

8 If Hershey and Chase had used radioactive Carbon-14 instead of P-32 and S-35, why would the experiment have failed to distinguish between DNA and protein?

A Carbon-14 is too weak to be detected by autoradiography.

B Carbon is completely absent in viral structures.

C Carbon is present in both DNA and proteins, so both would become radioactive.

D Carbon-14 would immediately degrade the bacteriophage.

💡 Short Explanation

P-32 is unique to DNA (in viruses). S-35 is unique to protein. Carbon is universal to all organic life.

✅ Correct Answer Explanation

The success of the Hershey-Chase experiment depended on \\\\\\\"labeling\\\\\\\" DNA and protein specifically. Phosphorus is present in DNA but not in most proteins, while Sulfur is present in proteins but not in DNA. Carbon, however, is a fundamental building block of all organic macromolecules, including both DNA and proteins. If used, both components would be labeled, making it impossible to distinguish which entered the host.

❓ Why Other Options Are Incorrect

Option A → Carbon-14 is detectable by autoradiography.
Option B → Carbon is the backbone of organic chemistry; it is present in viruses.
Option D → Carbon-14 is not destructive in this context.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Analyzing the elemental composition of the macromolecules.

Final Logic: Since Carbon is found in both targets, it cannot serve as a differential tracer.

🧠 Memory Trick

Carbon = Common to both.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Genetic Material is DNA –

9 During the Hershey-Chase experiment, what would have been the likely outcome if the blending step was completely skipped before centrifugation?

A Both radioactive S-35 and P-32 would be found in the bacterial pellet.

B The viral DNA would fail to enter the bacterial cells.

C Centrifugation would successfully separate DNA from protein anyway.

D The bacteria would fail to manufacture more virus particles.

💡 Short Explanation

Blending removes viral coats from cell surfaces. Without blending, coats remain attached to cells. Centrifugation pellets everything that is attached to bacteria.

✅ Correct Answer Explanation

The blending step is crucial for agitating the viral protein coats away from the bacterial cell surface. If this step were skipped, the radioactive viral coats (labeled with S-35) would still be attached to the bacteria. When centrifuged, the heavy bacteria and the attached protein coats would settle together in the pellet, causing both labels to appear there and invalidating the conclusion.

❓ Why Other Options Are Incorrect

Option B → The DNA has already entered; skipping blending doesn\\\\\\\'t change entry.
Option C → Centrifugation separates based on density/size; it cannot distinguish between attached coats and the cell itself.
Option D → Viruses would still be manufactured, but the experiment results would be unreadable.

🎯 Strategy

Used: Elimination

Application: Visualizing the physical separation process of the experiment.

Final Logic: If you don\\\\\\\'t detach the label, it travels with the cell, leading to false-positive results.

🧠 Memory Trick

Blender = Break apart (separates coats).

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Genetic Material is DNA –

10 Arrange the sequence of deductive steps in the Hershey-Chase experiment:
I. Agitation in a blender removes viral coats from bacteria.
II. Viruses grown in P-32 and S-35 mediums infect E. coli.
III. Centrifugation separates heavier bacteria from lighter viral coats.
IV. Radioactive DNA (P-32) is detected inside the bacterial cells, proving DNA is the genetic material.

A I, II, III, IV

B II, I, III, IV

C II, III, I, IV

D IV, II, I, III

💡 Short Explanation

Step 1: Infection (II). Step 2: Blending (I). Step 3: Centrifugation (III). Step 4: Analysis/Detection (IV).

✅ Correct Answer Explanation

The chronological order of the Hershey-Chase experiment starts with infecting E. coli with labeled phages (II). Once infection occurs, a blender is used to remove empty viral coats (I). Then, centrifugation (III) separates the components. Finally, the analysis (IV) shows that P-32 entered the cells, confirming DNA as the genetic material.

❓ Why Other Options Are Incorrect

Option A → Starts with blending before infection, which is impossible.
Option C → Attempts centrifugation before removing the coats via blending.
Option D → Begins with the final conclusion/detection.

🎯 Strategy

Used: Substitution

Application: Ordering events based on the standard biological workflow of a viral infection assay.

Final Logic: Infection must precede any mechanical separation or observation.

🧠 Memory Trick

Infect → Blend → Centrifuge → Detect (IBCD).

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: The Genetic Material is DNA –

11 Which of the following molecules fundamentally fails to fulfill the first essential criterion of a genetic material (the ability to direct its own duplication)?

A DNA

B RNA

C Proteins

D Both A and B

💡 Short Explanation

Genetic material must store information and replicate. DNA and RNA have complementary base pairing for replication. Proteins lack a mechanism to template their own replication.

✅ Correct Answer Explanation

A primary requirement for genetic material is the ability to generate a replica of itself (replication). DNA and RNA utilize complementary base pairing (A=T/U, G≡ C) to act as templates for synthesis. Proteins consist of amino acids, which lack a direct complementary template mechanism to replicate their own sequence, thus failing this fundamental criterion.

❓ Why Other Options Are Incorrect

Option A → DNA is the primary genetic material and replicates accurately.
Option B → RNA can replicate (e.g., in RNA viruses or the RNA World hypothesis).
Option D → Both DNA and RNA successfully meet the replication criterion.

🎯 Strategy

Used: Elimination

Application: Identifying which macromolecule lacks the structural basis (complementary pairing) for copying.

Final Logic: Proteins are products of translation, not templates for their own synthesis.

🧠 Memory Trick

Protein = Product, not Procreator.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Properties of Genetic Material –

12 How did Griffith\\\\\\\'s experiment inadvertently demonstrate the structural stability of the genetic material long before its chemical nature was known?

A By showing that heat, which killed the bacteria, did not destroy the properties of the transforming principle.

B By showing that mice lived when injected with the R strain.

C By proving that DNA is less reactive than RNA.

D By showing that S strain mutations are reversible.

💡 Short Explanation

Heat kills cells by denaturing proteins. The \\\\\\\"Transforming Principle\\\\\\\" survived this heat. Survival implies high structural/chemical stability.

✅ Correct Answer Explanation

Griffith used heat to kill the virulent S-strain. While the heat was sufficient to destroy the bacterial cell structure and protein machinery, the \\\\\\\"Transforming Principle\\\\\\\" remained intact and functional, as evidenced by its ability to still transform the live R-strain. This demonstrated that the genetic material is structurally stable even under harsh conditions.

❓ Why Other Options Are Incorrect

Option B → This observation relates to the non-virulence of R-strain, not genetic stability.
Option C → Griffith\\\\\\\'s experiment did not identify the chemical nature as DNA or RNA.
Option D → The experiment did not demonstrate reversible mutations.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Matching the observation (surviving heat) to the property (stability).

Final Logic: If the factor works after heat, the molecule must be stable.

🧠 Memory Trick

Heat-killed = Hardened/Stable material.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Properties of Genetic Material –

13 Why do certain viruses, such as Tobacco Mosaic Virus, mutate and evolve at a significantly faster rate than organisms with DNA genomes?

A Because their genome contains extra thymine residues.

B Because they possess an RNA genome, which is inherently unstable and mutates faster.

C Because they lack any protein coats to protect their genome.

D Because they do not rely on an adapter molecule for translation.

💡 Short Explanation

RNA is chemically more reactive than DNA. RNA-dependent replication (like in many viruses) lacks efficient proofreading. Higher reactivity leads to higher mutation rates.

✅ Correct Answer Explanation

RNA is inherently unstable due to the 2\\\\\\\'-OH group on the ribose sugar. Viruses that use RNA as their genetic material have higher mutation rates because RNA replication is generally more error-prone and the molecule itself is more susceptible to chemical degradation and modification compared to the stable double-stranded DNA.

❓ Why Other Options Are Incorrect

Option A → RNA contains Uracil, not Thymine.
Option C → TMV does have a protein coat (capsi

Option D → This is not the cause of rapid mutation rates.

🎯 Strategy

Used: Elimination

Application: Identifying the specific characteristic of RNA that leads to instability.

Final Logic: RNA\\\\\\\'s structural properties inherently make it more prone to change than DNA.

🧠 Memory Trick

RNA = Rapidly changing.

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Properties of Genetic Material –

14 Consider the following statements regarding the expression of Mendelian characters:
I. RNA can directly code for the synthesis of proteins.
II. DNA directly synthesizes proteins without any intermediates.
III. The cellular protein-synthesizing machinery has evolved around RNA. Which of the statements are correct?

A I and II only

B II and III only

C I and III only

D I, II, and III

💡 Short Explanation

RNA (mRN A) is the direct code for proteins. DNA requires transcription into RNA first (DNA does not directly code). Ribosomes (protein machinery) are essentially ribozymes.

✅ Correct Answer Explanation

Statement I is correct: mRNA directly codes for amino acid sequences. Statement III is correct: ribosome structure and function are dependent on rRNA (ribozymes). Statement II is incorrect: DNA requires the intermediate step of transcription (mRNA formation) before translation can occur.

❓ Why Other Options Are Incorrect

Option A → Incorrect because of statement II.
Option B → Incorrect because of statement II.
Option D → Incorrect because of statement II.

🎯 Strategy

Used: Elimination

Application: Eliminate any option that includes Statement II.

Final Logic: Since DNA is not the direct template for translation, statement II is false.

🧠 Memory Trick

Intermediate = Involved (Transcription).

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

15 Which of the following is NOT cited as biological evidence suggesting that essential life processes initially evolved around the RNA world?

A Metabolism relies on RNA.

B Translation processes rely heavily on RNA.

C Splicing represents the dominance of the RNA-world.

D The stability of the double-helix repair mechanism directly evolved from RNA.

💡 Short Explanation

RNA World evidence includes ribozymes and translation components. DNA repair mechanisms are specific to DNA\\\\\\\'s double-stranded structure. DNA repair evolved to protect the evolved DNA, not the other way around.

✅ Correct Answer Explanation

The RNA world hypothesis posits that life began with RNA. Metabolism, splicing (introns), and translation (ribosomes) are all RNA-based processes. Option D is incorrect because DNA repair mechanisms are associated with the evolved stability of DNA, not the original RNA-based world.

❓ Why Other Options Are Incorrect

Option A, B, and C → These are standard arguments supporting the RNA World hypothesis.

🎯 Strategy

Used: Odd One Out

Application: Identify which statement is fundamentally linked to DNA, not RNA.

Final Logic: DNA repair is a feature of DNA, not a relic of the RNA world.

🧠 Memory Trick

RNA = Ribozyme (Metabolism/Translation).

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

16 Match the specific characteristic of early RNA (Column I) to its biological implication (Column II):
Column I Column II
1. 2′-OH group at every nucleotide P. Easily expresses Mendelian characters
2. Catalytic properties Q. Acts as an enzyme (ribozyme) but makes it reactive
3. Directly codes for proteins R. Makes the molecule labile and easily degradable
4. Lacks thymine S. Lowers its chemical stability relative to DNA

A 1–R, 2–Q, 3–P, 4–S

B 1–Q, 2–R, 3–P, 4–S

C 1–R, 2–P, 3–Q, 4–S

D 1–S, 2–Q, 3–P, 4–R

💡 Short Explanation

2\\\\\\\'-OH = Reactive/Labile (R).
Catalytic = Ribozyme (Q).
Coding = Expression (P).
Lack of Thymine = Less stable than DNA (S).

✅ Correct Answer Explanation

The presence of the 2\\\\\\\'-OH group makes RNA reactive and degradable (1-R). RNA\\\\\\\'s ability to act as an enzyme is the definition of a ribozyme (2-Q). RNA can code for proteins, facilitating the expression of traits (3-P). The absence of Thymine (which is present in DNA) makes RNA less stable (4-S).

❓ Why Other Options Are Incorrect

Options B, C, and D → These incorrectly pair the chemical or structural properties of RNA with their biological consequences.

🎯 Strategy

Application: Verify the most direct links first (e.g., 2-Q is a standard definition) to narrow choices.

Final Logic: 1-R and 2-Q are strongly established scientific associations.

🧠 Memory Trick

OH = Open to reaction (labile).

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

17 The presence of thymine at the place of uracil confers additional stability to DNA. According to the text, the detailed discussion of this stability is strongly associated with the understanding of which complex cellular process?

A The process of transcription initiation

B The process of repair in DNA

C The splicing of hnRNA

D The process of translation at the ribosome

💡 Short Explanation

Thymine is methylated Uracil. Uracil can naturally occur by cytosine deamination. Using Thymine allows DNA repair systems to distinguish accidental Uracil.

✅ Correct Answer Explanation

DNA evolved thymine (5-methyl uracil) to increase stability. The evolutionary advantage is that since cytosine can spontaneously deaminate to form uracil, the presence of thymine allows the DNA repair machinery to recognize and replace the \\\\\\\"incorrect\\\\\\\" uracil, ensuring genome integrity.

❓ Why Other Options Are Incorrect

Option A, C, and D → These processes do not rely on the Thymine vs. Uracil distinction for structural stability/repair.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match the concept of \\\\\\\"stability\\\\\\\" in DNA to the specific evolutionary mechanism (repair).

Final Logic: DNA stability = Genome protection = Repair.

🧠 Memory Trick

Thymine = Tougher/Reliable.

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

18 The lability and easy degradability of RNA is chemically attributed to the reactivity of which specific structural component?

A The N-glycosidic linkage

B The 5\\\\\\\'-end phosphate moiety

C The 2\\\\\\\'-OH group present at every nucleotide

D The 3\\\\\\\'-5\\\\\\\' phosphodiester linkage

💡 Short Explanation

Ribose has a 2\\\\\\\'-OH group; Deoxyribose does not. The hydroxyl group is a nucleophilic site. This site facilitates self-hydrolysis in RNA.

✅ Correct Answer Explanation

In the pentose sugar of RNA (ribose), the 2\\\\\\\' carbon holds a hydroxyl (-OH) group. This group is chemically reactive and can perform a nucleophilic attack on the phosphodiester backbone, leading to cleavage or \\\\\\\"lability\\\\\\\" of the RNA molecule. DNA, lacking this oxygen (hence \\\\\\\"deoxy\\\\\\\"), is significantly more stable.

❓ Why Other Options Are Incorrect

Option A, B, and D → These components are shared by both DNA and RNA and do not account for the specific instability of RNA relative to DNA.

🎯 Strategy

Used: Elimination

Application: Distinguish the structural difference between ribose and deoxyribose.

Final Logic: The \\\\\\\"deoxy\\\\\\\" in DNA refers to the removal of the very group that makes RNA unstable.

🧠 Memory Trick

OH = Open to attack.

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

19 Arrange the conceptual events detailing the evolution of genetic storage:
I. Essential life processes (metabolism, translation) evolve around RNA.
II. RNA acts as a genetic material as well as a reactive catalyst.
III. Chemical modifications of RNA yield a chemically less reactive molecule.
IV. Double-stranded DNA evolves a repair process, becoming preferred for genetic storage.

A I, II, III, IV

B II, I, III, IV

C II, III, I, IV

D I, III, II, IV

💡 Short Explanation

Initial state: RNA as engine (I) and carrier (II). Evolution: Chemical refinement (III). Final state: DNA as superior storage (IV).

✅ Correct Answer Explanation

The evolutionary sequence begins with life processes centering on RNA (I). Because RNA is both catalyst and genetic material (II), it was initially used for both. To solve the instability issue, chemical modifications created DNA (III). Finally, DNA\\\\\\\'s double-stranded structure and repair mechanisms made it the ideal long-term storage medium (IV).

❓ Why Other Options Are Incorrect

Options B, C, and D fail to follow the logical order of evolutionary progression (from reactive RNA to stable DN
A).

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Sequence the evolutionary timeline from primitive RNA functionality to complex DNA stability.

Final Logic: Functionality → Refinement → Stability.

🧠 Memory Trick

RNA (Reactive) → DNA (Durable).

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

20 Which of the following is NOT a feature that makes DNA preferred over RNA for the long-term storage of genetic information?

A It is structurally more stable due to its double-stranded nature.

B It is chemically less reactive because it lacks a 2\\\\\\\'-OH group.

C It has the ability to directly act as an enzyme and catalyze its own storage.

D It has evolved a process of repair leveraging its complementary strands.

💡 Short Explanation

RNA is a catalyst (ribozyme). DNA is not an enzyme. DNA\\\\\\\'s advantage is stability, not catalytic activity.

✅ Correct Answer Explanation

DNA is preferred for storage because it is stable, less reactive, and repairable. It does not function as an enzyme (ribozyme). Catalytic activity is a hallmark of RNA, which is why RNA is considered the precursor \\\\\\\"engine\\\\\\\" of life, while DNA is the \\\\\\\"safe\\\\\\\" storage unit.

❓ Why Other Options Are Incorrect

Option A, B, and D → These are the very reasons why DNA is superior for genetic storage.

🎯 Strategy

Used: Elimination

Application: Identify the option that describes a property of RNA (catalysis) rather than DNA (storage).

Final Logic: DNA is a \\\\\\\"passive\\\\\\\" information carrier; RNA is the \\\\\\\"active\\\\\\\" molecule.

🧠 Memory Trick

DNA = Database (Storage), RNA = Reactor (Catalyst).

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

CUET Booster Biology Unit 5 Test (D2)

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