CUET UG Biology Booster Unit 5 Test (M4) -

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CUET Booster Biology Unit 5 Test (M4)

⏱️ 30 Minutes

❓ 20 Questions

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

CUET Booster Biology Unit 5 Test (M4)

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QUESTION 1 OF 20

Arrange the logical sequence of discoveries and postulations that led to the deciphering of the genetic code:
1. Har Gobind Khorana synthesized RNA molecules with defined base combinations.
2. George Gamow proposed that the code should be a combination of three nucleotides.
3. A complete checker-board for the genetic code was prepared.
4. Marshall Nirenberg's cell-free system aided in decoding the code.

Answer Review

1 Arrange the logical sequence of discoveries and postulations that led to the deciphering of the genetic code:
1. Har Gobind Khorana synthesized RNA molecules with defined base combinations.
2. George Gamow proposed that the code should be a combination of three nucleotides.
3. A complete checker-board for the genetic code was prepared.
4. Marshall Nirenberg's cell-free system aided in decoding the code.

A 2, 1, 4, 3

B 2, 4, 1, 3

C 1, 2, 4, 3

D 4, 2, 1, 3

💡 Short Explanation

Theory (Gamow) came first. Cell-free experiments (Nirenberg) allowed initial decoding. Chemical synthesis (Khorana) refined specific sequences. The complete checkerboard was the final culmination.

✅ Correct Answer Explanation

Gamow (2) provided the triplet hypothesis. Nirenberg (4) created the experimental breakthrough (cell-free system). Khorana (1) utilized chemical methods to synthesize specific defined polymers. These combined efforts led to the final genetic code dictionary/checker-board (3).

❓ Why Other Options Are Incorrect

Option A puts Khorana before Nirenberg's cell-free breakthrough.
Options C and D place the theoretical proposal (Gamow) after experimental breakthroughs, which is historically inaccurate.

🎯 Strategy

Used

Contextual/Tonal Matching: Ordering historical scientific progression.

Final Logic: Theoretical postulate precedes biochemical experimentation and final synthesis.

🧠 Memory Trick

"Gamow Theory > Nirenberg Cell-free > Khorana Synthesis > Checkerboard."

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

2 Which of the following statements about the biochemical methods used to decipher the genetic code are accurate?
I. George Gamow synthesized RNA copolymers to prove his triplet hypothesis.
II. Polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-independent manner.
III. Marshall Nirenberg developed a cell-free system for protein synthesis.

A I and II

B II and III

C I and III

D I, II, and III

💡 Short Explanation

I is False: Gamow was a physicist who used math, not chemical synthesis. II is True: Severo Ochoa discovered the enzyme. III is True: Nirenberg pioneered the cell-free system.

✅ Correct Answer Explanation

Statement II describes the role of Ochoa's enzyme, and III accurately reflects Nirenberg's contribution. Statement I is false because Gamow did not synthesize RNA; he used mathematical reasoning based on the number of amino acids and bases.

❓ Why Other Options Are Incorrect

Option A, C, and D are incorrect as they include statement I, which attributes experimental synthesis to Gamow.

🎯 Strategy

Used

Elimination: Identify statement I as false to eliminate options A, C, and D.

Final Logic: Gamow provided the concept, not the bench-side synthesis.

🧠 Memory Trick

"Gamow-Math; Nirenberg-System; Ochoa-Enzyme."

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

3 Match the salient feature of the genetic code to its corresponding description:

Column I	Column II
1. Degenerate	P. Conserved across almost all living organisms
2. Universal	Q. Three nucleotides specify one amino acid
3. Contiguous	R. Multiple codons can specify the same amino acid
4. Triplet	S. Read continuously without commas or gaps

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

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

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

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

💡 Short Explanation

Degenerate: More than one codon per amino acid. Universal: Same in bacteria/humans. Contiguous: No punctuation. Triplet: 61 codons for amino acids, 3 for stop.

✅ Correct Answer Explanation

Degeneracy (1-R) means redundancy. Universality (2-P) means the code is conserved across life. Contiguity (3-S) means no commas. The Triplet nature (4-Q) defines the coding capacity (64 total, 61 code, 3 stop).

❓ Why Other Options Are Incorrect

They mismatch the terms with their definitions (e.g., matching Triplet to P or S).

🎯 Strategy

Used

Option Grouping: Linking definitions to the correct salient feature.

Final Logic: Accurate mapping of terminology to biological description.

🧠 Memory Trick

"Degenerate = Redundant; Universal = Shared; Contiguous = Continuous; Triplet = 64/61/3."

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

4 Which statement does NOT accurately reflect the degeneracy or specificity of the genetic code?

A A single amino acid can be specified by multiple codons.

B The code consists of 64 distinct codons generated by a permutation of 4 bases.

C Three specific codons out of the 64 code for highly specialized terminator amino acids.

D The genetic code is read in an unpunctuated, contiguous fashion.

💡 Short Explanation

Stop codons do not code for amino acids. The term "terminator amino acids" is biologically incorrect.

✅ Correct Answer Explanation

Stop (terminator) codons like UAA, UAG, and UGA signal the end of translation. They do not encode any amino acid. Option C is therefore the incorrect statement.

❓ Why Other Options Are Incorrect

A, B, and D are accurate definitions regarding the degeneracy, number of codons, and the reading process of the genetic code.

🎯 Strategy

Used

Extreme Word Filter: Identifying "terminator amino acids" as an internal contradiction.

Final Logic: Terminator codons = No amino acids.

🧠 Memory Trick

"Stop codon = Empty codon."

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

5 Regarding the dual functional role of AUG, which statements are correct?
I. It is responsible for initiating the process of translation.
II. It codes for the amino acid Methionine.
III. It acts as a terminator codon in mitochondrial genomes.

A I and II

B II and III

C I and III

D I, II, and III

💡 Short Explanation

AUG is Start/Met (I, II True). AUG does not function as a terminator (III False).

✅ Correct Answer Explanation

AUG is the universal start codon that codes for Methionine. It does not act as a stop codon in mitochondrial genomes; mitochondrial exceptions exist, but AUG is not one of them.

❓ Why Other Options Are Incorrect

Any option including III is false.

🎯 Strategy

Used

Elimination: Exclude III to find the correct pair.

Final Logic: AUG = Start + Met; not a terminator.

🧠 Memory Trick

"AUG = Methionine/Start."

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

6 Which of the following is NOT a characteristic of the terminators UAA, UAG, and UGA?

A They define the end of the translated polypeptide sequence.

B They bind directly to specific terminator tRNAs to halt the ribosome.

C They do not code for any amino acids.

D A release factor binds to them to terminate translation.

💡 Short Explanation

There are no "terminator tRNAs." Release factors perform the termination.

✅ Correct Answer Explanation

Stop codons are recognized by release factors, not tRNAs. There are no tRNAs that correspond to UAA, UAG, or UGA.

❓ Why Other Options Are Incorrect

A, C, and D are accurate descriptions of stop codon function.

🎯 Strategy

Used

Substitution: Replace "terminator tRNAs" with "release factors."

Final Logic: Terminators bind release factors, not tRNAs.

🧠 Memory Trick

"Stop = Release factor."

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

7 If a single base pair changes in the structural gene for the beta globin chain, altering one amino acid residue, what specific type of mutation has occurred?

A Frameshift insertion

B Point mutation

C Frameshift deletion

D Chromosomal rearrangement

💡 Short Explanation

Single base pair change = Point mutation. Sickle cell example.

✅ Correct Answer Explanation

A change in a single nucleotide base pair (substitution) is the definition of a point mutation. This specifically happens in the sickle cell gene (GAG to GUG).

❓ Why Other Options Are Incorrect

A and C involve changing the reading frame (not just one amino acid). D involves large-scale structural change.

🎯 Strategy

Used

Substitution: Match the definition "single base pair change" to "point mutation."

Final Logic: Single base change = Point mutation.

🧠 Memory Trick

"Point mutation = Single point change."

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

8 When insertion or deletion of three or its multiple bases occurs in a sequence, what is the consequence on the reading frame according to the genetic code rules?

A The entire reading frame shifts completely from the beginning of the sequence.

B One or multiple amino acids are inserted or deleted, but the reading frame remains unaltered from that point onwards.

C The ribosome stalls permanently, preventing any further translation.

D The sequence undergoes a point mutation that changes a single amino acid.

💡 Short Explanation

Triplets maintain the frame. Insertion/deletion of 3 restores the frame.

✅ Correct Answer Explanation

Since the genetic code is read in triplets, inserting or deleting multiples of three leaves the downstream sequence "in frame." One or more amino acids will be added/removed, but the rest of the protein remains the same.

❓ Why Other Options Are Incorrect

A describes a frame-shift (which happens with 1 or 2 bases). C and D are biologically incorrect.

🎯 Strategy

Used

Dimensional/Unit Analysis: Multiples of 3 = Full triplets = No frame shift.

Final Logic: Triplets are the unit; adding/removing units keeps the frame.

🧠 Memory Trick

"Multiple of 3 = Frame saved."

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

9 Why was an adapter molecule theoretically necessary for the process of translation, according to Francis Crick?

A Because amino acids have no structural specialities to read the code uniquely.

B Because mRNA degrades rapidly without a protective adapter molecule.

C Because ribosomes require an adapter to bind to the rough endoplasmic reticulum.

D Because DNA cannot leave the nucleus without being bound to an adapter.

💡 Short Explanation

Amino acids cannot bind to nucleotides directly. Adapter (tRNA) acts as the bridge.

✅ Correct Answer Explanation

Crick realized that amino acids and the genetic code (nucleotides) have no intrinsic affinity. Therefore, a chemical bridge ("adapter") was needed to connect the two.

❓ Why Other Options Are Incorrect

B, C, and D are incorrect biological explanations for the necessity of tRNA.

🎯 Strategy

Used

Contextual/Tonal Matching: Recall Crick's specific reasoning for the adapter hypothesis.

Final Logic: No structural affinity = Need an adapter.

🧠 Memory Trick

"No affinity = Need adapter."

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

10 Arrange the physical orientations and actions associated with the tRNA adapter molecule during translation:
1. The anticodon loop forms complementary base pairs with the mRNA code.
2. The amino acid binds to the amino acid acceptor end of the tRNA.
3. The charged tRNA is brought into close proximity within the large ribosomal subunit.
4. The tRNA, initially known as sRNA, is activated by ATP.

A 4, 2, 1, 3

B 2, 4, 3, 1

C 4, 1, 2, 3

D 1, 2, 4, 3

💡 Short Explanation

Activation (4) > Charging (2) > Base pairing (1) > Ribosomal proximity (3).

✅ Correct Answer Explanation

Step 4: Activation by ATP. Step 2: Amino acid links to tRNA (Charging). Step 1: Anticodon finds its match on mRNA. Step 3: Ribosome facilitates the interaction of charged tRNAs.

❓ Why Other Options Are Incorrect

Other orders jumble the physical logic of tRNA charging and translation.

🎯 Strategy

Used

Contextual/Tonal Matching: Chronological order of translation mechanics.

Final Logic: Activate > Charge > Match > Assemble.

🧠 Memory Trick

"Activate-Charge-Match-Assemble."

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

11 During amino acid activation, why is it vital that ATP is present for the aminoacylation of tRNA?

A Because it acts as an enzyme that degrades uncharged tRNAs.

B Because the formation of a peptide bond between charged tRNAs requires energy favored by this activation.

C Because ATP directly base-pairs with the mRNA start codon.

D Because ATP constitutes the structural backbone of the ribosome.

💡 Short Explanation

Peptide bond formation is endergonic. Activation (ATP) provides the energy.

✅ Correct Answer Explanation

The formation of a peptide bond is thermodynamically unfavorable. By using ATP to charge the tRNA (aminoacylation), the cell "primes" the amino acid, providing the necessary energy for the subsequent bond formation.

❓ Why Other Options Are Incorrect

A, C, and D are factually incorrect regarding the role of ATP in translation.

🎯 Strategy

Used

Substitution: "Activation" = "Energy priming."

Final Logic: Activation = Energy for peptide bond.

🧠 Memory Trick

"Activation = Energy boost."

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

12 Match the translation machinery component with its functional state or role:

Column I	Column II
1. Small ribosomal subunit	P. Has binding sites for tRNA
2. Large ribosomal subunit	Q. Initiates translation by binding mRNA
3. Aminoacylated tRNA	R. Catalytic RNA (Ribozyme)
4. Ribozyme	S. Carries the amino acid to the ribosome

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

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

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

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

💡 Short Explanation

1 (Small): Initiates (Q). 2 (Large): Has binding sites (P). 3 (tRNA): Carries AA (S). 4 (Ribozyme): Catalyst (R).

✅ Correct Answer Explanation

The small subunit (1-Q) encounters mRNA first. The large subunit (2-P) has the binding sites. Aminoacylated tRNA (3-S) is the "charged" carrier. The ribozyme (4-R) is the catalyst for the peptide bond.

❓ Why Other Options Are Incorrect

These mismatch the machinery with their biological functions.

🎯 Strategy

Used

Option Grouping: Aligning the components with their NCERT-defined roles.

Final Logic: Direct mapping of component to function.

🧠 Memory Trick

"Small-Init; Large-Sites; Charged-AA; Ribozyme-Catalyst."

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

📄 Passage

"The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits; a large subunit and a small subunit. When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins. There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond.",

Based on the passage, what specific event marks the absolute beginning of the translation process of mRNA into protein?

A The binding of an amino acid to the large subunit.

B The encounter of the small ribosomal subunit with an mRNA.

C The release of the 80 different structural proteins.

D The dissociation of the large and small subunits.

📄 Passage

Which of the following is an INCORRECT interpretation of the ribosome's structural role during initiation, based on the text?

A It operates as a complex cellular factory.

B It relies on an inactive state composed of two distinct subunits.

C The small subunit provides the catalytic sites for peptide bond formation.

D The large subunit houses sites that bring amino acids close together.

💡 Short Explanation

The large subunit provides the catalytic/binding sites.

✅ Correct Answer Explanation

The passage states that the two sites for amino acid binding and peptide bond formation are in the large subunit, not the small one. Therefore, Option C is incorrect.

❓ Why Other Options Are Incorrect

A, B, and D are supported by the passage.

🎯 Strategy

Used

Extreme Word Filter: Identifying "small subunit" as the error.

Final Logic: Large subunit = Catalyst; Small subunit = mRNA encounter.

🧠 Memory Trick

"Large = Linker/Catalyst."

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

15 What energetically favors the formation of a peptide bond between two amino acids during the elongation phase?

A The removal of introns from the mRNA sequence.

B Bringing two aminoacylated (charged) tRNAs close enough together in the large subunit.

C The temporary binding of a release factor to the ribosome.

D The rapid hydrolysis of the structural RNAs in the small subunit.

💡 Short Explanation

Proximity within the ribosome allows the reaction to occur. The "charged" nature provides the energy.

✅ Correct Answer Explanation

The large ribosomal subunit holds the charged (aminoacylated) tRNAs in close physical proximity, which is necessary for the ribozyme to facilitate the peptide bond formation.

❓ Why Other Options Are Incorrect

A, C, and D are either unrelated or physically incorrect processes.

🎯 Strategy

Used

Contextual/Tonal Matching: Recall the elongation mechanism.

Final Logic: Proximity of charged tRNAs = Reaction site.

🧠 Memory Trick

"Close proximity = Bond formation."

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

16 Translocation involves the ribosome moving sequentially along the mRNA. What guides the specific addition of amino acids during this movement?

A The random diffusion of free amino acids into the ribosomal core.

B The sequence of bases in the mRNA Dictating complementary base pairing with tRNA anticodons.

C The independent enzymatic activity of the UTRs.

D The sequential breakdown of the 5' methyl guanosine cap.

💡 Short Explanation

mRNA sequence is the template. Anticodons provide specificity.

✅ Correct Answer Explanation

Translation is governed by base pairing. The codon on the mRNA matches the anticodon on the incoming tRNA, ensuring the correct amino acid is added in the sequence specified by the gene.

❓ Why Other Options Are Incorrect

A, C, and D are incorrect mechanisms for translational specificity.

🎯 Strategy

Used

Substitution: "Guidance" = "Complementary base pairing."

Final Logic: mRNA template dictates the tRNA order via anticodons.

🧠 Memory Trick

"Base pairing = Specificity."

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

17 Which critical interaction occurs to finally halt the elongation of the polypeptide chain?

A The start codon loops back to block the ribosome.

B A release factor binds to the stop codon.

C The ribozyme degrades the remaining mRNA.

D ATP depletion forcibly stops the translocation.

💡 Short Explanation

Stop codon + Release factor = Stop.

✅ Correct Answer Explanation

Termination occurs when the ribosome encounters a stop codon. A release factor (a protein) recognizes the stop codon and binds, signaling the process to finish.

❓ Why Other Options Are Incorrect

A, C, and D are biologically wrong descriptions of translation termination.

🎯 Strategy

Used

Substitution: "Halt" = "Release factor binding."

Final Logic: Stop signal recognized by release factor.

🧠 Memory Trick

"Stop codon = Release Factor."

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

18 Which of the following is NOT an event associated with the dissociation of the polypeptide chain?

A The complete polypeptide is released from the cellular factory.

B Translation is formally terminated.

C A specific tRNA adapter molecule binds to the UAA stop codon.

D The interaction occurs specifically at a termination sequence dictated by DNA.

💡 Short Explanation

No tRNA binds to stop codons (UAA). This is the incorrect (NOT) statement.

✅ Correct Answer Explanation

Option C is the incorrect statement because there is no tRNA for the stop codon UAA. Termination involves release factors, not tRNA.

❓ Why Other Options Are Incorrect

A, B, and D are true aspects of the termination process.

🎯 Strategy

Used

Elimination: Identify the statement that contradicts the rule of no-tRNA-for-stops.

Final Logic: No tRNA for stop codons.

🧠 Memory Trick

"Stop = No tRNA."

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

19 Relative to the structural sequence of the mRNA, where are the Untranslated Regions (UTRs) precisely positioned to ensure efficient translation?

A Sandwiched between multiple consecutive AUG codons.

B At both the 5'-end (before start codon) and the 3'-end (after stop codon).

C Embedded within the introns before splicing occurs.

D Attached to the anticodon loop of the initiator tRNA.

💡 Short Explanation

5' UTR: Before start. 3' UTR: After stop.

✅ Correct Answer Explanation

UTRs flank the coding region of the mRNA. They are located at the 5'-end (upstream of the start codon) and the 3'-end (downstream of the stop codon).

❓ Why Other Options Are Incorrect

These suggest incorrect locations (introns, tRNA, or between codons).

🎯 Strategy

Used

Substitution: Replace "UTRs" with "flanking regions."

Final Logic: Flanking = 5' and 3' ends.

🧠 Memory Trick

"UTR = 5' and 3' Flanks."

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

20 What is the primary biological importance of the UTRs located on the mRNA transcript?

A They serve as the direct templates for ribozyme synthesis.

B They are essential for ensuring an efficient translation process.

C They act as physical anchors for the release factor.

D They code for the 80 different proteins that assemble the ribosome.

💡 Short Explanation

They are "required for efficient translation."

✅ Correct Answer Explanation

UTRs contain regulatory signals that are critical for the efficiency of translation, ensuring the ribosome binds and protein synthesis proceeds correctly.

❓ Why Other Options Are Incorrect

A, C, and D are not the functions of UTRs.

🎯 Strategy

Used

Substitution: "Importance" = "Efficiency."

Final Logic: UTRs = Translation efficiency.

🧠 Memory Trick

"UTR = Efficiency."

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

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