CUET UG Categorised PYQ Biology Unit 5 -

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CUET UG Categorised PYQ Biology Unit 5

BIOLOGY

⏱️ 30 Minutes

❓ 27 Questions

📅 2022, 2023, 2024 & 2025

✍️ Long Answers

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

CUET UG Categorised PYQ Biology Unit 5

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QUESTION 1 OF 27

Which one of the following enzyme brings about hydrolysis of lactose to glucose and galactose? (PYQ 2022 Shift 1)

Answer Review

1 Which one of the following enzyme brings about hydrolysis of lactose to glucose and galactose? (PYQ 2022 Shift 1)

A Transacetylase

B Amylase

C Permease

D β-galactosidase

💡 Short Explanation

Encoded by lacZ gene. Breaks lactose into monosaccharides. Important in lac operon.

✅ Correct Answer Explanation

β-galactosidase catalyzes the hydrolysis of lactose into glucose and galactose. It is encoded by the lacZ gene of the lac operon in E. coli.

❓ Why Other Options Are Incorrect

Option A) Transacetylase → Product of lacA gene.
Option B) Amylase → Digests starch.
Option C) Permease → Helps lactose entry into cell.

🎯 Strategy

Used: NCERT Recall

Option A → Not hydrolysis enzyme.
Option B → Starch digestion.
Option C → Transport protein.
Option D → Correct.

Final Answer → β-galactosidase.

🧠 Memory Trick

📚 → Section/Theme: Lac Operon

2 Match List – I with List – II (PYQ 2022 Shift 1)

List – I	List – II
(A) Initiation factor	(I) Tailing
(B) Introns	(II) Rho(ρ)
(C) Termination factor	(III) Sigma(σ)
(D) Adenylate residue	(IV) Splicing

A (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

B (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

C (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

D (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

💡 Short Explanation

Sigma factor initiates transcription. Introns are removed by splicing. Rho factor terminates transcription. Adenylate residues form poly-A tail.

✅ Correct Answer Explanation

Correct matching: Initiation factor → Sigma (σ) Introns → Splicing Termination factor → Rho (ρ) Adenylate residue → Tailing Thus, Option D is correct.

❓ Why Other Options Are Incorrect

Option A) Initiation factor and introns mismatched.
Option B) Introns incorrectly matched with Rho.
Option C) Multiple incorrect pairings.

🎯 Strategy

Used: Option Grouping

Option A → Incorrect.
Option B → Incorrect.
Option C → Incorrect.
Option D → Correct.

Final Answer → Option D.

🧠 Memory Trick

📚 → Section/Theme: Transcription and RNA Processing

3 Replication of DNA is characterised by: (PYQ 2022 Shift 1)
(A) The direction of replication is 5' → 3'
(B) Only template with 5' → 3' polarity is replicated
(C) Replication is initiated at ori
(D) DNA polymerase catalyses the process
(E) The daughter molecule formed has one parental strand

A (A), (B), (D), (E)

B (A), (B), (C), (D)

C (A), (C), (D), (E)

D (B), (C), (D), (E)

💡 Short Explanation

DNA synthesis occurs 5'→3'. Replication begins at ori. Semiconservative replication occurs.

✅ Correct Answer Explanation

Statements A, C, D and E are correct: DNA synthesis proceeds 5'→3'. Replication starts at origin (ori). DNA polymerase catalyses replication. Each daughter DNA contains one parental strand. Statement B is incorrect because both template strands participate in replication.

❓ Why Other Options Are Incorrect

Option A) Includes incorrect B.
Option B) Includes incorrect B.
Option D) Includes incorrect B.

🎯 Strategy

Used: Elimination

Option A → B incorrect.
Option B → B incorrect.
Option C → Correct set.
Option D → B incorrect.

Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: DNA Replication

4 A students was repeating the experiments of Alfred Hershey and Martha Chase (1952). Results obtained by him are shown in the given figure. Select the biomolecule that was radio actively labelled by the students for his experiment. (PYQ 2022 Shift 1)

A DNA of bacteriophage with radioactive Phosphorous

B Plasmid DNA of E. coli with radioactive Sulphur

C Proteins of bacteriophage with radioactive Sulphur

D Chromosomal DNA of E. coli with radioactive Sulphur

💡 Short Explanation

Sulphur labels proteins. DNA lacks sulphur. Hershey-Chase used S³⁵ for protein coat.

✅ Correct Answer Explanation

In the Hershey-Chase experiment, radioactive sulphur (³⁵S) was used to label bacteriophage proteins because proteins contain sulphur-containing amino acids while DNA does not.

❓ Why Other Options Are Incorrect

Option A) DNA was labelled with radioactive phosphorus.
Option B) DNA does not contain sulphur.
Option D) DNA does not contain sulphur.

🎯 Strategy

Used: NCERT Recall

Option A → Phosphorus label.
Option B → Impossible.
Option C → Correct.
Option D → Impossible.

Final Answer → Proteins of bacteriophage.

🧠 Memory Trick

📚 → Section/Theme: Hershey-Chase Experiment

5 The sequence of nitrogen bases of the template strand of DNA in a transcription unit is 3'-ATTGAACG-5'. The sequence of nitrogen bases in its mRNA transcript would be: (PYQ 2023 Shift 1)

A 3'-AUUGAACAG-5'

B 5'-UAACUUGC-3'

C 3'-CAGUUCAAU-5'

D 5'-GUUAGUCCA-3'

💡 Short Explanation

mRNA is complementary to template strand. RNA contains U instead of T. mRNA is synthesized 5'→3'.

✅ Correct Answer Explanation

Template DNA = 3'-ATTGAACG-5' Complementary RNA bases: A → U T → A T → A G → C A → U A → U C → G G → C Therefore mRNA sequence = 5'-UAACUUGC-3'.

❓ Why Other Options Are Incorrect

Option A → Wrong orientation and extra base.
Option C → Reverse complementary sequence.
Option D → Incorrect base pairing.

🎯 Strategy

Used: Substitution

Option A → Wrong polarity.
Option B → Correct complementary sequence.
Option C → Reverse order.
Option D → Pairing errors.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Transcription

6 Which of the following nitrogenous base is NOT present in RNA? (PYQ 2023 Shift 1)

A Adenine

B Guanine

C Cytosine

D Thymine

💡 Short Explanation

RNA contains Uracil. DNA contains Thymine. Thymine is absent in RNA.

✅ Correct Answer Explanation

RNA consists of Adenine, Guanine, Cytosine and Uracil. Thymine is replaced by Uracil in RNA molecules.

❓ Why Other Options Are Incorrect

Option A → Present in RNA.
Option B → Present in RNA.
Option C → Present in RNA.

🎯 Strategy

Used: Odd One Out

Option A → RNA base.
Option B → RNA base.
Option C → RNA base.
Option D → DNA-specific base.

Final Answer → Thymine.

🧠 Memory Trick

📚 → Section/Theme: Structure of RNA

7 Identify the regions of the transcription unit: (PYQ 2023 Shift 1)

A A, B, and D only

B A, B, and C only

C A, C, and D only

D B, C, and D only

💡 Short Explanation

Promoter present. Structural gene present. Terminator present. Sigma factor is not a region.

✅ Correct Answer Explanation

A transcription unit consists of promoter, structural gene and terminator. Sigma factor is a protein associated with RNA polymerase in prokaryotes and is not a structural region of the transcription unit.

❓ Why Other Options Are Incorrect

Option A → Sigma factor incorrectly included.
Option B → Sigma factor included.
Option D → Promoter omitted.

🎯 Strategy

Used: Elimination

Option A → Sigma factor error.
Option B → Sigma factor error.
Option C → Correct.
Option D → Missing promoter.

Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: Transcription Unit

8 Observe the given figure and answer the question. What does it show? (PYQ 2023 Shift 1)

A Post transcriptional processing in Prokaryotes

B Termination of translation

C Process of transcription in Prokaryotes

D Process of transcription in Eukaryotes

💡 Short Explanation

Eukaryotes undergo RNA processing. Involves nucleus-based transcription. hnRNA formation is characteristic.

✅ Correct Answer Explanation

The figure represents transcription in eukaryotes where primary transcript (hnRNA) is formed and later processed through capping, tailing and splicing before becoming mature mRNA.

❓ Why Other Options Are Incorrect

Option A) Prokaryotes lack extensive post-transcriptional processing.
Option B) Figure is not related to translation.
Option C) Prokaryotic transcription lacks hnRNA processing.

🎯 Strategy

Used: Contextual/Tonal Matching

Option A → Wrong organism group.
Option B → Different process.
Option C → No extensive RNA processing.
Option D → Correct.

Final Answer → Eukaryotic transcription.

🧠 Memory Trick

📚 → Section/Theme: Transcription in Eukaryotes

9 Observe the given figure and answer the question.
Match the following (RNA processing diagram): (PYQ 2023 Shift 1)

List I	List II
A. Methyl Guanosine Triphosphate	I. F
B. Intron	II. I
C. Poly A tail	III. C
D. Fully processed hn RNA	IV. H

A (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

B (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

C (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

D (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

💡 Short Explanation

5' cap = Methyl guanosine triphosphate. Introns are removed during splicing. Poly-A tail added at 3' end.

✅ Correct Answer Explanation

During RNA processing, methyl guanosine triphosphate forms the 5' cap, introns are removed through splicing, Poly-A tail is added at the 3' end and mature processed RNA is formed after completion of all modifications.

❓ Why Other Options Are Incorrect

Option A) Incorrect mapping of cap and intron.
Option B) Poly-A tail incorrectly assigned.
Option C) Mature RNA incorrectly matched.

🎯 Strategy

Used: Option Grouping

Option A → Incorrect.
Option B → Incorrect.
Option C → Incorrect.
Option D → Correct.

Final Answer → Option D.

🧠 Memory Trick

📚 → Section/Theme: RNA Processing

10 Observe the given figure and answer the question.
RNA polymerase III is responsible for transcription of: (PYQ 2023 Shift 1)

A tRNA, rRNA, mRNA

B tRNA, 5SrRNA, snRNA

C 5SrRNA, tRNA, mRNA

D snRNA, rRNA, 5SrRNA

💡 Short Explanation

RNA Pol I → rRNA. RNA Pol II → mRNA. RNA Pol III → tRNA, 5S rRNA, snRNA.

✅ Correct Answer Explanation

In eukaryotes, RNA polymerase III transcribes tRNA, 5S rRNA and certain small nuclear RNAs (snRNA). This division of labor among polymerases is a standard NCERT concept.

❓ Why Other Options Are Incorrect

Option A) mRNA is synthesized by RNA polymerase II.
Option C) Includes mRNA incorrectly.
Option D) Major rRNA is synthesized by RNA polymerase I.

🎯 Strategy

Used: NCERT Recall

Option A → mRNA error.
Option B → Correct.
Option C → mRNA error.
Option D → rRNA error.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Eukaryotic Transcription

11 Observe the given figure and answer the question.
Identify sequence of events: (PYQ 2023 Shift 1)

A (B), (A), (C), (D)

B (B), (A), (D), (C)

C (C), (D), (A), (B)

D (A), (D), (B), (C)

💡 Short Explanation

Transcription occurs first. Capping follows. Splicing and tailing occur next. Mature mRNA exits nucleus.

✅ Correct Answer Explanation

RNA polymerase II first transcribes hnRNA. The transcript undergoes capping, followed by tailing and splicing. Mature mRNA is then transported out of the nucleus for translation.

❓ Why Other Options Are Incorrect

Option A) Transport cannot occur first.
Option B) Transcription cannot be last.
Option D) Processing cannot precede transcription.

🎯 Strategy

Used: Contextual/Tonal Matching

Option A → Wrong chronology.
Option B → Wrong chronology.
Option C → Correct order.
Option D → Processing before transcription.

Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: Post-Transcriptional Processing

12 Transcription can be coupled with translation in bacteria because: (PYQ 2023 Shift 1)

A No separation of cytosol and nucleus

B Post transcriptional processing is simple

C Transcription is faster than translation

D RNA molecule is complex

💡 Short Explanation

Prokaryotes lack a true nucleus. Transcription and translation occur in same compartment. Both processes can occur simultaneously.

✅ Correct Answer Explanation

Bacteria are prokaryotes and do not possess a nuclear membrane. Since transcription occurs directly in the cytoplasm where ribosomes are present, translation can begin even before transcription is completed, leading to coupling of the two processes.

❓ Why Other Options Are Incorrect

Option B) Simplicity of processing is not the main reason.
Option C) Relative speed is not responsible for coupling.
Option D) RNA complexity is unrelated.

🎯 Strategy

Used: Elimination

Option A → Fundamental reason.
Option B → Secondary factor.
Option C → Incorrect explanation.
Option D → Unrelated.

Final Answer → Option A.

🧠 Memory Trick

📚 → Section/Theme: Prokaryotic Transcription

13 Match the following: (PYQ 2023 Shift 2)

List I	List II
A. RNA Polymerase II	I. 28S rRNA
B. RNA Polymerase I	II. hnRNA
C. Ribosome	III. 5S rRNA
D. RNA Polymerase III	IV. 23S rRNA in bacteria

A (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

B (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

C (A)-(I), (B)-(I), (C)-(IV), (D)-(III)

D (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

💡 Short Explanation

Pol I → 28S rRNA. Pol II → hnRNA. Pol III → 5S rRNA. Ribosome contains 23S rRNA in bacteria.

✅ Correct Answer Explanation

RNA Polymerase I synthesizes large rRNAs (28S, 18S, 5.8S), RNA Polymerase II synthesizes hnRNA (precursor of mRNA), RNA Polymerase III synthesizes 5S rRNA and tRNA, while bacterial ribosomes contain 23S rRNA.

❓ Why Other Options Are Incorrect

Option A) Incorrect polymerase assignments.
Option C) Multiple mismatches.
Option D) Incorrect matching of all major components.

🎯 Strategy

Used: Option Grouping

Option A → Incorrect.
Option B → Correct.
Option C → Incorrect.
Option D → Incorrect.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Eukaryotic Transcription

14 The size of VNTR varies from: (PYQ 2023 Shift 2, 2024 Shift 1)

A 0.1 to 20 kb

B 0.2 to 10 bp

C 0.1 to 10 kb

D 0.2 to 20 kb

💡 Short Explanation

VNTR = Variable Number Tandem Repeats. Used in DNA fingerprinting. Length ranges from 0.1–20 kb.

✅ Correct Answer Explanation

VNTRs are repetitive DNA sequences whose lengths vary among individuals. NCERT specifies that their size ranges approximately from 0.1 kb to 20 kb.

❓ Why Other Options Are Incorrect

Option B) Too small.
Option C) Upper limit incorrect.
Option D) Lower limit incorrect.

🎯 Strategy

Used: NCERT Recall

Option A → Correct range.
Option B → Too short.
Option C → Incomplete range.
Option D → Incorrect lower value.

Final Answer → Option A.

🧠 Memory Trick

📚 → Section/Theme: DNA Fingerprinting

15 Which is NOT true for a molecule that can act as a genetic material? (PYQ 2023 Shift 3)

A It should be able to generate its replica

B It should be chemically and structurally unstable

C It should provide scope for slow changes that are required for evolution

D It should be able to express itself

💡 Short Explanation

Genetic material must remain stable. Stability ensures faithful inheritance. Instability would cause excessive mutations.

✅ Correct Answer Explanation

A genetic material should: Replicate itself. Be chemically and structurally stable. Express itself in the form of characters. Permit variations required for evolution. Therefore, being chemically and structurally unstable is not a desirable property of genetic material.

❓ Why Other Options Are Incorrect

Option A) Essential property of genetic material.
Option C) Necessary for evolution.
Option D) Must express information through proteins.

🎯 Strategy

Used: Elimination

Option A → Required property.
Option B → Opposite of required property.
Option C → Required property.
Option D → Required property.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Criteria for Genetic Material

16 Identify the statements which hold true for DNA in E. coli. (PYQ 2023 Shift 3)

A A, B and C only

B B and D only

C A, B, C and D

D A, C and D only

💡 Short Explanation

E. coli DNA occurs in nucleoid. DNA is arranged in loops. DNA length is about 1.36 mm, not metres.

✅ Correct Answer Explanation

E. coli DNA: Is localized in the nucleoid region. Is not scattered throughout the cell. Is organized into large loops held by proteins. The DNA length is approximately 1.36 mm (not 1.36 metres). Therefore Statement B is incorrect.

❓ Why Other Options Are Incorrect

Option A) Includes incorrect Statement B.
Option B) Statement B is incorrect.
Option C) Includes incorrect Statement B.

🎯 Strategy

Used: Elimination

Option A → Contains B.
Option B → B incorrect.
Option C → B incorrect.
Option D → All correct statements.

Final Answer → Option D.

🧠 Memory Trick

📚 → Section/Theme: Packaging of DNA

17 Who performed experiments on Vicia faba to prove that the DNA replicates semi-conservatively? (PYQ 2024 Shift 1, 2024 Shift 2)

A Taylor and colleagues

B Matthew Meselson

C Stahl

D Hershey and Chase

💡 Short Explanation

Experiment performed on Vicia faba root tip cells. Demonstrated semi-conservative DNA replication in eukaryotes. Used radioactive thymidine labeling.

✅ Correct Answer Explanation

Taylor, Woods and Hughes performed experiments on Vicia faba (broad bean) root tip cells using radioactive thymidine. Their observations supported the semi-conservative mode of DNA replication in eukaryotic chromosomes.

❓ Why Other Options Are Incorrect

Option B) Meselson worked on E. coli.
Option C) Stahl worked with Meselson in E. coli experiments.
Option D) Hershey and Chase proved DNA is genetic material.

🎯 Strategy

Used: Elimination

Option A → Vicia faba experiment.
Option B → E. coli experiment.
Option C → E. coli experiment.
Option D → Different objective.

Final Answer → Option A.

🧠 Memory Trick

📚 → Section/Theme: Experimental Proof of DNA Replication

18 In Hershey and Chase experiment, some viruses grew on medium that contained: (PYQ 2024 Shift 1, 2024 Shift 2)

A ³⁵S, ³²P

B ³⁶S, ³⁴P

C ³²S, ³⁶P

D ³⁴S, ³⁶P

💡 Short Explanation

Protein labeled with ³⁵S. DNA labeled with ³²P. Proved DNA is genetic material.

✅ Correct Answer Explanation

Sulfur is present in proteins but absent in DNA, so proteins were labeled with radioactive ³⁵S. Phosphorus is present in DNA, so DNA was labeled with radioactive ³²P.

❓ Why Other Options Are Incorrect

Option B) Incorrect isotopes.
Option C) Incorrect isotopes.
Option D) Incorrect isotopes.

🎯 Strategy

Used: Elimination

Option A → Standard experimental isotopes.
Option B → Wrong labels.
Option C → Wrong labels.
Option D → Wrong labels.

Final Answer → Option A.

🧠 Memory Trick

📚 → Section/Theme: Hershey-Chase Experiment

19 Nucleosome is: (PYQ 2024 Shift 1, 2024 Shift 2)

A Positively charged DNA wrapped around negatively charged histone octamer

B Negatively charged DNA wrapped around positively charged histone octamer

C Positively charged DNA wrapped around positively charged histone octamer

D Negatively charged DNA wrapped around negatively charged histone octamer

💡 Short Explanation

DNA contains negatively charged phosphate groups. Histones are rich in lysine and arginine. Histones are positively charged.

✅ Correct Answer Explanation

A nucleosome consists of negatively charged DNA wrapped around a positively charged histone octamer. The positive charge of histones helps neutralize DNA's negative charge and facilitates DNA packaging.

❓ Why Other Options Are Incorrect

Option A) DNA is not positively charged.
Option C) DNA is not positively charged.
Option D) Histones are not negatively charged.

🎯 Strategy

Used: Elimination

Option A → DNA charge incorrect.
Option B → Both charges correct.
Option C → DNA charge incorrect.
Option D → Histone charge incorrect.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Packaging of DNA Helix

20 "Transforming Principle" was given by: (PYQ 2024 Shift 1, 2024 Shift 2)

A Maclyn McCarty

B Frederick Griffith

C Alfred Hershey

D Watson and Crick

💡 Short Explanation

Conducted bacterial transformation experiment. Worked with Streptococcus pneumoniae. Proposed transforming principle.

✅ Correct Answer Explanation

Frederick Griffith demonstrated that a substance from heat-killed virulent bacteria could transform non-virulent bacteria into virulent forms. He called this substance the "transforming principle."

❓ Why Other Options Are Incorrect

Option A) McCarty later identified DNA as transforming material.
Option C) Hershey proved DNA is genetic material.
Option D) Watson and Crick proposed DNA structure.

🎯 Strategy

Used: Elimination

Option A → DNA identification.
Option B → Transformation experiment.
Option C → Phage experiment.
Option D → DNA model.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Transforming Principle

21 Select the incorrect statement: (PYQ 2024 Shift 1, 2024 Shift 2)

A Chromosome 1 has the most genes and chromosome Y has the fewest.

B Chromosome 21 has the most genes and chromosome Y has the fewest.

C Less than 2 percent of the genome codes for proteins.

D The functions of over 50 percent of discovered genes are unknown.

💡 Short Explanation

Chromosome 1 contains highest number of genes. Chromosome Y contains the fewest. Statement B reverses the fact.

✅ Correct Answer Explanation

Human Genome Project findings show that Chromosome 1 has the largest number of genes while Chromosome Y has the fewest. Therefore Statement B is incorrect.

❓ Why Other Options Are Incorrect

Option A) Correct HGP finding.
Option C) Correct; less than 2% codes for proteins.
Option D) Correct; many gene functions remain unknown.

🎯 Strategy

Used: Elimination

Option A → Correct.
Option B → Factually incorrect.
Option C → Correct.
Option D → Correct.

Final Answer → Option B.

🧠 Memory Trick

📚 → Chapter: Human Genome Project

22 Match the following: (PYQ 2024 Shift 1, 2024 Shift 2, 2025 Shift 1)

List – I	List – II
(A) i	(I) permease
(B) a	(II) β-galactosidase
(C) y	(III) transacetylase
(D) z	(IV) repressor

A (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

B (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

C (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

D (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

💡 Short Explanation

i gene → Repressor.
z gene → β-galactosidase.
y gene → Permease.
a gene → Transacetylase.

✅ Correct Answer Explanation

In the lac operon:

i → Repressor
z → β-galactosidase
y → Permease
a → Transacetylase

Thus:

A → IV
B → III
C → I
D → II

❓ Why Other Options Are Incorrect

All other options contain incorrect gene-product pairings.

🎯 Strategy

i → Repressor
z → β-galactosidase
y → Permease
a → Transacetylase
Final Answer → Option B

🧠 Memory Trick

📚 → Class: XII → Book: Biology → Chapter: Molecular Basis of Inheritance → Section/Theme: Lac Operon

23 Amino acid is attached to which site of tRNA? (PYQ 2024 Shift 1, 2024 Shift 2)

A Anticodon loop

B 3′ end

C 5′ end

D D-loop

💡 Short Explanation

Amino acid attaches at the CCA sequence. Located at the 3′ end. Forms aminoacyl-tRNA.

✅ Correct Answer Explanation

The amino acid attachment site of tRNA is the 3′ terminal CCA sequence. Aminoacyl-tRNA synthetase attaches the correct amino acid to this site before translation.

❓ Why Other Options Are Incorrect

Option A) Anticodon recognizes mRNA codon.
Option C) Not the amino acid attachment site.
Option D) Involved in enzyme recognition.

🎯 Strategy

Used: Elimination

Option A → Codon recognition.
Option B → Amino acid attachment.
Option C → Incorrect end.
Option D → Recognition loop.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Translation

24 Given below is the DNA coding sequence. Its complementary strand would read as: (PYQ 2024 Shift 1, 2024 Shift 2)
5′ – GTATTACG – 3′

A 5′ – CUTUUTGC – 3′

B 3′ – CATAATGC – 5′

C 3′ – CUTUUTGC – 5′

D 5′ – CTUTTUGC – 3′

💡 Short Explanation

DNA base pairing: G-C and A-T. Complementary strands are antiparallel. No uracil in DNA.

✅ Correct Answer Explanation

Complementary base pairing: G ↔ C T ↔ A A ↔ T Therefore: 5′–GTATTACG–3′ 3′–CATAATGC–5′

❓ Why Other Options Are Incorrect

Option A) Contains uracil (U), found in RNA.
Option C) Contains uracil (U), found in RNA.
Option D) Contains uracil (U), found in RNA.

🎯 Strategy

Used: Substitution

Option A → RNA bases.
Option B → Correct antiparallel complement.
Option C → RNA bases.
Option D → RNA bases.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Structure of DNA

25 The process of copying genetic information from one strand of DNA into RNA is termed as: (PYQ 2025 Shift 1)

A Replication

B Translation

C Transcription

D Regulation

💡 Short Explanation

DNA copied into RNA. First step of gene expression. Catalyzed by RNA polymerase.

✅ Correct Answer Explanation

Transcription is the process by which genetic information from DNA is copied into RNA. RNA polymerase reads the DNA template strand and synthesizes complementary RNA.

❓ Why Other Options Are Incorrect

Option A) DNA → DNA copying.
Option B) RNA → Protein synthesis.
Option D) Controls gene expression but is not RNA synthesis.

🎯 Strategy

Used: Elimination

Option A → DNA duplication.
Option B → Protein formation.
Option C → DNA to RNA.

Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: Transcription

26 Which one of the following is not associated with the process of transcription in bacteria? (PYQ 2025 Shift 1)

A Rho factor

B Methyl guanosine triphosphate

C Sigma factor

D DNA dependent RNA polymerase

💡 Short Explanation

Involved in mRNA capping. Found in eukaryotes. Not involved in bacterial transcription.

✅ Correct Answer Explanation

Methyl guanosine triphosphate forms the 5′ cap of eukaryotic mRNA during post-transcriptional modification. Bacteria generally do not possess this capping process. Rho factor, Sigma factor and DNA-dependent RNA polymerase are directly associated with bacterial transcription.

❓ Why Other Options Are Incorrect

Option A) Participates in transcription termination.
Option C) Helps RNA polymerase recognize promoters.
Option D) Synthesizes RNA from DNA template.

🎯 Strategy

Used: Odd One Out

Option A → Bacterial transcription component.
Option B → Eukaryotic RNA processing.
Option C → Bacterial transcription factor.
Option D → Bacterial transcription enzyme.

Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Transcription

27 If a double-stranded DNA has 15% of adenine, find out the percent of cytosine in the DNA? (PYQ 2025 Shift 1)

A 15%

B 30%

C 35%

D 85%

💡 Short Explanation

A = T in DNA. A + T = 30%. Remaining 70% is G + C.

✅ Correct Answer Explanation

According to Chargaff's rule: Adenine (A) = Thymine (T) Guanine (G) = Cytosine (C) Given: A = 15% Therefore: T = 15% So, A + T = 30% Remaining: G + C = 70% Since G = C, C = 70 ÷ 2 = 35%

❓ Why Other Options Are Incorrect

Option A) Confuses cytosine with adenine percentage.
Option B) Represents combined A + T percentage.
Option D) Exceeds possible value for a single base.

🎯 Strategy

Used: Substitution

Option A → Incorrect application of Chargaff's rule.
Option B → Represents A + T.
Option C → Correct calculation.

Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: Structure of DNA

CUET UG Categorised PYQ Biology Unit 5

Test Complete!

Answer Review