CUET UG Biology Booster Unit 5 Test (D4) -

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CUET Booster Biology Unit 5 Test (D4)

⏱️ 30 Minutes

❓ 20 Questions

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

CUET Booster Biology Unit 5 Test (D4)

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QUESTION 1 OF 20

George Gamow's triplet hypothesis was a bold postulation derived strictly from logical deduction. What was the core mathematical necessity driving this hypothesis?

Answer Review

1 George Gamow's triplet hypothesis was a bold postulation derived strictly from logical deduction. What was the core mathematical necessity driving this hypothesis?

A RNA molecules are chemically unstable unless bases are arranged in triplets.

B Only 4 bases must account for 20 amino acids, requiring a 4^3 permutation to yield enough distinct combinations.

C The cell-free system could only process nucleotides in distinct groups of three.

D The ribosome structure possesses three distinct slots for mRNA reading.

💡 Short Explanation

Genetic code requires 20 amino acids. 4^1=4, 4^2=16 (not enough). 4^3=64 (sufficient).

✅ Correct Answer Explanation

Gamow utilized combinatorial mathematics to show that a triplet code (4^3=64) is the minimum necessary to encode 20 amino acids. Options A, C, and D are incorrect because they cite structural or experimental claims not related to Gamow's initial mathematical deduction.

❓ Why Other Options Are Incorrect

Option A → Relates to chemistry, not the mathematical logic Gamow used.
Option C → Relates to Nirenberg's later experimental work.
Option D → Describes the ribosome's later-discovered mechanism, not Gamow's initial hypothesis.

🎯 Strategy

Used

Substitution: Replace "triplet hypothesis" with "mathematical requirement (4^3)."

Final Logic: 4^3=64 is the only permutation that meets the 20-amino-acid requirement.

🧠 Memory Trick

"Gamow-Math: 4 bases, 20 AAs, 64 combinations."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Genetic Code – .

2 In the context of deciphering the genetic code, which methodology was NOT utilized by the mentioned scientists?

A Using polynucleotide phosphorylase for template-independent RNA synthesis.

B Synthesizing RNA copolymers with defined combinations of bases.

C Utilizing reverse transcriptase to generate DNA from unpunctuated RNA.

D Employing a cell-free protein synthesis system.

💡 Short Explanation

Reverse transcriptase was not used in the deciphering of the code. Ochoa, Khorana, and Nirenberg used the other three methods.

✅ Correct Answer Explanation

Deciphering the code relied on cell-free systems (Nirenberg), synthetic RNA polymers (Khorana), and polynucleotide phosphorylase (Ochoa). Reverse transcriptase (Option C) was not a technique used in the initial code-breaking period.

❓ Why Other Options Are Incorrect

Option A → Ochoa utilized this enzyme.
Option B → Khorana synthesized these defined polymers.
Option D → Nirenberg established the cell-free system.

🎯 Strategy

Used

Odd One Out: Identify the modern molecular tool that wasn't part of the classic code-breaking experiments.

Final Logic: Reverse transcriptase is not a core method for the initial genetic code deciphering.

🧠 Memory Trick

"No Reverse-T in classic code-breaking."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Genetic Code – –111.

3 Analyze the following statements about the universal and non-overlapping nature of the genetic code:
I. The property of the code being read in a contiguous fashion directly defines its unambiguous nature.
II. Universality implies that a specific codon, like UUU, codes for the same amino acid in both bacteria and humans.
III. Rare deviations from the universal nature of the code are observed in some protozoans and mitochondrial DNA. Which statements are accurate?

A I and II

B II and III

C I and III

D I, II, and III

💡 Short Explanation

I is false: Contiguity defines "no punctuation," not "unambiguous" nature. II is true: Universality definition. III is true: Known exceptions exist.

✅ Correct Answer Explanation

Statement II correctly defines universality. Statement III acknowledges accepted scientific exceptions in organelles and protozoans. Statement I is incorrect because "contiguous" refers to reading without gaps, while "unambiguous" means one codon = one specific amino acid.

❓ Why Other Options Are Incorrect

Options A, C, and D contain Statement I, which misidentifies the definition of the genetic code properties.

🎯 Strategy

Used

Elimination: Recognize that Statement I conflates two different code properties.

Final Logic: Universality and exceptions are correct; contiguity is not ambiguity.

🧠 Memory Trick

"Contiguous=Continuous, Unambiguous=Specific."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Salient Features – –112.

4 Arrange the numerical breakdown of the genetic code's distinct permutations from broad availability to functional application:
1. Codons that function exclusively as stop/terminator signals.
2. Total mathematical permutations generated by a triplet code of 4 bases.
3. Total number of amino acids required to be coded by the genetic material.
4. Total number of codons that actively specify amino acids.

A 2 (64), 4 (61), 3 (20), 1 (3)

B 2 (64), 3 (20), 4 (61), 1 (3)

C 4 (61), 1 (3), 3 (20), 2 (64)

D 2 (64), 4 (61), 1 (3), 3 (20)

💡 Short Explanation

64 total codons. 61 code for amino acids. 3 code for termination. 20 amino acids are coded.

✅ Correct Answer Explanation

The logical breakdown is: 64 total triplet permutations (2), of which 61 code for amino acids (4), leaving 3 to act as stops (1), which specify the 20 required amino acids (3).

❓ Why Other Options Are Incorrect

They fail to place the total permutations (64) at the start or misorder the coding capacity.

🎯 Strategy

Used

Dimensional/Unit Analysis: Categorizing numbers (Total, Coding, Stop, AA count).

Final Logic: 64 Total > 61 Coding > 3 Stop > 20 Amino Acids.

🧠 Memory Trick

"64-61-3-20."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Genetic Code – .

5 The dual function of the AUG codon is biologically critical because it establishes the reading frame. Based on genetic code properties, what does this indicate about the translation process?

A The reading frame initiated by AUG operates with clear physical punctuations between subsequent codons.

B AUG defines the starting point from which the mRNA is read in an unbroken, contiguous triplet manner.

C AUG simultaneously terminates the previous translation unit while initiating a new one.

D AUG must be skipped if it appears in the middle of a structural sequence.

💡 Short Explanation

Translation is "comma-less." AUG sets the frame to be read as triplets.

✅ Correct Answer Explanation

The genetic code is "comma-less" or contiguous. AUG marks the beginning; from that point on, the ribosome reads the mRNA as consecutive three-nucleotide triplets.

❓ Why Other Options Are Incorrect

Option A → Incorrect, as the code is non-punctuated.
Option C → Incorrect, AUG initiates, it doesn't terminate.
Option D → Incorrect, AUG codes for Methionine wherever it is found.

🎯 Strategy

Used

Contextual/Tonal Matching: Defining "reading frame" in the context of a contiguous code.

Final Logic: AUG sets the start, contiguity ensures the reading continues without spaces.

🧠 Memory Trick

"AUG = Frame-setter."

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

6 Match the specific codon sequences to their functional consequences during translation:

Column I	Column II
1. AUG	P. Initiator codon / Codes for Methionine (Met)
2. UAA	Q. Stop (Termination) codon
3. UAG	R. Stop (Termination) codon
4. UGA	S. Stop (Termination) codon

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

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

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

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

💡 Short Explanation

AUG = Start/Met. UAA, UAG, UGA = Terminators.

✅ Correct Answer Explanation

AUG (1-P) is the initiator/Met. UAA (2-Q), UAG (3-R), and UGA (4-S) are all functionally identical stop codons that halt translation.

❓ Why Other Options Are Incorrect

These options incorrectly assign initiator functions to stop codons or vice-versa.

🎯 Strategy

Used

Option Grouping: Aligning the initiator/stop codon list.

Final Logic: AUG is unique; UAA/UAG/UGA are all terminators.

🧠 Memory Trick

"AUG = Start, Three Stop-ers."

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

7 Which characteristic is NOT true regarding the point mutation responsible for Sickle Cell Anemia?

A It involves a change in a single base pair of the DNA.

B It affects the structural gene for the beta globin chain.

C It alters the reading frame of the entire subsequent sequence.

D It results in the substitution of glutamate with valine.

💡 Short Explanation

Point mutation is a substitution, not a frameshift.

✅ Correct Answer Explanation

Sickle cell anemia is a point mutation (substitution of a single base), which replaces one amino acid (Glutamate to Valine) without shifting the reading frame of the remaining protein. Option C describes a frameshift mutation, which is different.

❓ Why Other Options Are Incorrect

A, B, and D are factually correct characteristics of the sickle cell mutation.

🎯 Strategy

Used

Elimination: Point mutation \neq Frameshift.

Final Logic: Point mutation is a substitution; frameshift requires insertion/deletion.

🧠 Memory Trick

"Point mutation = Single change, no shift."

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

8 Based on the frameshift mutation analogy (RAM HAS RED CAP), analyze the following:
I. Insertion of one or two letters alters the reading frame from the point of insertion.
II. Insertion of three letters alters the reading frame of the entire remaining sentence.
III. Deletion of three contiguous letters removes a single "amino acid" equivalent, leaving the rest of the reading frame unaltered.

A I and II

B I and III

C II and III

D I, II, and III

💡 Short Explanation

I is true: Frameshift occurs. II is false: Three letters (one codon) restore the frame. III is true: Deleting one codon removes one amino acid, leaving the frame intact.

✅ Correct Answer Explanation

Insertion/Deletion of 1 or 2 bases causes a frameshift (I). Insertion/Deletion of 3 bases (a whole codon) maintains the reading frame (III). Statement II is false because 3-letter insertions do not shift the entire remaining frame.

❓ Why Other Options Are Incorrect

Any option containing II is incorrect.

🎯 Strategy

Used

Substitution: Apply the "3-base rule" to distinguish between frameshift and frame-retention.

Final Logic: 1-2 bases = Frameshift; 3 bases = In-frame.

🧠 Memory Trick

"3 = Frame Safe."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Frame-shift Mutation – .

9 Francis Crick's adapter hypothesis resolved a major conceptual hurdle in molecular biology. Without an adapter molecule, what fundamental difficulty prevents direct translation?

A Amino acids lack the complex tertiary structures necessary for self-polymerization.

B Amino acids possess no structural specialities enabling them to directly read or uniquely interact with mRNA codon sequences.

C The ribozyme would uncontrollably degrade any unprotected amino acids.

D mRNA nucleotides repel the charged side chains of all 20 basic amino acids.

💡 Short Explanation

No direct affinity between amino acid and codon. tRNA acts as a physical adapter.

✅ Correct Answer Explanation

Crick proposed the adapter (tRNA) because amino acids cannot chemically recognize mRNA bases. The adapter is needed to match the anticodon to the codon and provide the specific amino acid.

❓ Why Other Options Are Incorrect

A, C, and D are incorrect explanations that do not capture the chemical necessity of the adapter.

🎯 Strategy

Used

Contextual/Tonal Matching: Matching Crick's hypothesis to the chemical logic of non-affinity.

Final Logic: Amino acids and mRNA don't "speak the same language," so they need an interpreter (tRNA).

🧠 Memory Trick

"No affinity = Adapter needed."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Adapter Hypothesis – .

10 While the secondary structure of tRNA is depicted as a clover-leaf, its actual compact three-dimensional structure resembles:

A A double helix

B A continuous hairpin loop

C An inverted L

D A spherical globular protein

💡 Short Explanation

Cloverleaf = Secondary structure. Inverted L = Tertiary (3D) structure.

✅ Correct Answer Explanation

Although the clover-leaf model illustrates the base-pairing and loops, the real 3D shape is a compact "Inverted L" structure, which allows it to fit properly into the ribosome.

❓ Why Other Options Are Incorrect

A, B, and D are incorrect shapes for tRNA.

🎯 Strategy

Used

Substitution: Match 3D structure to "Inverted L."

Final Logic: 2D = Clover; 3D = Inverted L.

🧠 Memory Trick

"3D-L."

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

11 Which of the following is NOT an accurate representation of the amino acid activation process?

A It is alternatively referred to as aminoacylation of tRNA.

B The process strictly requires the expenditure of ATP.

C It involves linking an amino acid to the anticodon loop of the tRNA.

D It generates an energetically favored state for subsequent peptide bond formation.

💡 Short Explanation

Amino acid binds to the acceptor end, not the anticodon loop.

✅ Correct Answer Explanation

The amino acid binds to the 3' end (acceptor end) of the tRNA. The anticodon loop is on the opposite side of the molecule to bind with mRNA, not the amino acid.

❓ Why Other Options Are Incorrect

A, B, and D are correct descriptions of aminoacylation.

🎯 Strategy

Used

Extreme Word Filter: Identifying "anticodon loop" as the incorrect binding site.

Final Logic: AA acceptor = Acceptor end; mRNA reader = Anticodon.

🧠 Memory Trick

"Acceptor end = Amino acid."

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

12 Arrange the following molecular assembly steps that constitute the initiation of the translation machinery:
1. The small ribosomal subunit encounters and binds to the mRNA.
2. The complete ribosome forms, presenting two sites in the large subunit.
3. The initiator tRNA recognizes the start codon.
4. Amino acids undergo activation in the presence of ATP.

A 4, 1, 3, 2

B 1, 3, 4, 2

C 4, 3, 1, 2

D 1, 4, 2, 3

💡 Short Explanation

Activation (4) > Small subunit binds (1) > Initiator tRNA binds (3) > Ribosome assembles (2).

✅ Correct Answer Explanation

Translation begins with activating amino acids (4). Then, the small subunit finds the mRNA (1). Next, the initiator tRNA recognizes the start codon (3). Finally, the complete ribosome assembles to start elongation (2).

❓ Why Other Options Are Incorrect

They shuffle the order of the pre-initiation and assembly steps.

🎯 Strategy

Used

Contextual/Tonal Matching: Chronological order of translation initiation steps.

Final Logic: Activate -> Bind mRNA -> tRNA recognition -> Full assembly.

🧠 Memory Trick

"Activate-SmallSub-tRNA-LargeSub."

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

13 During translation initiation, the precise binding of the ribosome to the mRNA ensures that:

A The structural RNAs act as immediate templates for elongation.

B The translation machinery aligns exactly with the AUG start codon to set the correct contiguous reading frame.

C The UTRs are translated first to provide an anchor peptide.

D The release factor prepares the complex for early termination.

💡 Short Explanation

AUG sets the frame.

✅ Correct Answer Explanation

The ribosome's binding to mRNA is specific; it anchors onto the start codon (AUG) to ensure the subsequent triplets are read in the correct frame.

❓ Why Other Options Are Incorrect

A, C, and D are incorrect biological claims.

🎯 Strategy

Used

Substitution: "Ensures that" = "Aligns/Sets frame."

Final Logic: Start codon alignment = Frame setting.

🧠 Memory Trick

"Start-Frame = Correct reading."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Translation Initiation – .

14 Match the components of translation initiation to their defined roles:

Column I	Column II
1. mRNA	P. Start signal (AUG codon)
2. Small ribosomal subunit	Q. Platform for translation initiation
3. Initiator tRNA	R. Carries the complementary anticodon to the mRNA sequence
4. AUG codon	S. Adapter molecule carrying methionine

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

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

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

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

💡 Short Explanation

mRNA: Sequence (R). Small subunit: Platform (Q). tRNA: Adapter (S). AUG: Start signal (P).

✅ Correct Answer Explanation

mRNA contains the sequence (R). The small subunit is the initial platform (Q). The initiator tRNA is the specific adapter (S). AUG is the start signal (P).

❓ Why Other Options Are Incorrect

These mismatch the functional roles with the components.

🎯 Strategy

Used

Option Grouping: Matching initiation components to their functional role.

Final Logic: Direct component-role mapping.

🧠 Memory Trick

"mRNA-Seq, Subunit-Platform, tRNA-Adapter, AUG-Start."

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

15 The mechanism of peptide bond formation is highly energy-dependent. How is this cellular hurdle biologically bypassed during elongation?

A The ribosome continuously hydrolyzes the mRNA backbone for energy.

B The energy is pre-invested during the ATP-dependent charging (aminoacylation) of the tRNAs.

C The 23S rRNA synthesizes ATP molecules natively during each translocation step.

D The release factor donates its intrinsic kinetic energy.

💡 Short Explanation

"Pre-invested" = Amino acid activation.

✅ Correct Answer Explanation

The energy needed for peptide bond formation is stored in the bond between the tRNA and the amino acid, which was created during the ATP-dependent aminoacylation (charging) process.

❓ Why Other Options Are Incorrect

A, C, and D are incorrect biological mechanisms.

🎯 Strategy

Used

Substitution: Replace "energy source" with "pre-invested energy."

Final Logic: ATP used in activation = Energy source for peptide bond.

🧠 Memory Trick

"Charging = Pre-investing energy."

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Polypeptide Elongation – .

16 As translocation occurs, complexes of amino acid-linked tRNAs sequentially bind to the mRNA. What molecular rule governs this critical binding?

A The template-independent polymerization activity of ribozymes.

B The principle of complementary base pairing between the mRNA codon and tRNA anticodon.

C The degenerate nature of the start codon overlapping adjacent triplets.

D The uniform distance created by purine-pyrimidine structural rigidity.

💡 Short Explanation

Base pairing = Binding rule.

✅ Correct Answer Explanation

Translation specificity relies entirely on the complementary base pairing between the codon on the mRNA and the anticodon on the tRNA.

❓ Why Other Options Are Incorrect

A, C, and D describe wrong principles for translation binding.

🎯 Strategy

Used

Substitution: "Molecular rule" = "Base pairing."

Final Logic: Codon-Anticodon pairing = Correct AA delivery.

🧠 Memory Trick

"Codon-Anticodon = Complementary."

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

17 Which of the following scenarios does NOT occur during the binding of the release factor?

A A specific terminator tRNA binds simultaneously to assist the release factor.

B The process of translation is formally halted.

C The complete polypeptide chain is released.

D The interaction occurs specifically at a stop codon sequence (e.g., UAA).

💡 Short Explanation

Terminator tRNAs do not exist.

✅ Correct Answer Explanation

There is no terminator tRNA. The release factor alone binds to the stop codon to trigger the release of the polypeptide chain and end translation. Option A is false.

❓ Why Other Options Are Incorrect

B, C, and D are the actual events of termination.

🎯 Strategy

Used

Extreme Word Filter: Identifying "terminator tRNA" as the contradiction.

Final Logic: Stop codon = Release factor; No terminator tRNA.

🧠 Memory Trick

"Stop = Release factor only."

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

18 The final dissociation of the polypeptide chain from the translational machinery fundamentally relies on the fact that:

A Ribozymes inherently degrade protein chains upon reaching 50 amino acids in length.

B Stop codons lack complementary tRNAs, requiring a release factor to step in.

C The UTRs block the physical path of the translocating ribosome.

D Initiator tRNAs re-bind to force the existing chain out.

💡 Short Explanation

Absence of tRNA = Stop signal.

✅ Correct Answer Explanation

Because there are no tRNA molecules that can base-pair with stop codons, the empty site allows the release factor protein to bind, which triggers the hydrolysis of the polypeptide.

❓ Why Other Options Are Incorrect

A, C, and D describe incorrect mechanisms for polypeptide dissociation.

🎯 Strategy

Used

Substitution: "Fundamentally relies on" = "Absence of tRNA."

Final Logic: Stop codon = No tRNA = Release Factor bound.

🧠 Memory Trick

"No tRNA = Stop."

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

📄 Passage

"An mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR). The UTRs are present at both 5'-end (before start codon) and at 3'-end (after stop codon). They are required for efficient translation process. For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA.",

Based on the text, if an mRNA lacked a 3'-end UTR, what would be the most direct theoretical consequence?

A The start codon would be completely degraded.

B The mRNA would likely suffer from a less efficient translation process.

C The ribosome would initiate translation backwards.

D The initiator tRNA would fail to recognize the AUG codon.

💡 Short Explanation

UTRs are required for efficiency.

✅ Correct Answer Explanation

The passage states: "UTRs are required for efficient translation process." Therefore, lacking one would logically lead to decreased efficiency.

❓ Why Other Options Are Incorrect

A, C, and D are not logical consequences of missing a 3'-UTR.

🎯 Strategy

Used

Contextual/Tonal Matching: Extracting the functional importance of UTRs from the passage.

Final Logic: UTRs = Efficiency; No UTR = Less efficiency.

🧠 Memory Trick

"UTR = Efficiency."

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

📄 Passage

According to the passage, the physical position of the 5'-UTR is explicitly defined relative to which critical functional sequence?

A It is positioned directly after the stop codon.

B It is positioned physically inside the large ribosomal subunit.

C It is positioned before the start codon (AUG).

D It is intermingled within the primary structural coding sequence.

💡 Short Explanation

5' UTR = Before start codon.

✅ Correct Answer Explanation

The passage states: "The UTRs are present at both 5'-end (before start codon)..."

❓ Why Other Options Are Incorrect

These misidentify the location based on the provided passage.

🎯 Strategy

Used

Contextual/Tonal Matching: Locating the text-based definition.

Final Logic: Passage defines 5'-UTR location as "before start codon."

🧠 Memory Trick

"5' UTR = Before Start."

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

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