Molecular Basis of Inheritance

Topic: Gene Expression and Regulation;  Subtopic: Termination of Transcription

Keyword Definitions:

• Transcription: The process of synthesizing RNA from a DNA template by RNA polymerase.

• Rho factor: A helicase enzyme in prokaryotes that terminates transcription at specific sites.

• RNA polymerase: The enzyme that catalyzes RNA synthesis during transcription.

• Termination: The final step of transcription where the RNA transcript is released.

• Template strand: The DNA strand that serves as a guide for RNA synthesis.

Lead Question – 2025

Which factor is important for termination of transcription?

(1) α
(2) σ
(3) ρ
(4) γ

Explanation: The correct answer is (3) ρ. In prokaryotes, termination of transcription occurs by Rho-dependent or Rho-independent mechanisms. The Rho factor is a helicase that unwinds the RNA-DNA hybrid, causing release of the nascent RNA. In Rho-independent termination, a hairpin loop forms followed by uracil residues. Hence, Rho factor ensures efficient transcription termination in bacteria.

1. Which of the following is NOT required for Rho-independent termination?

(1) GC-rich hairpin
(2) Rho protein
(3) Poly-U tail
(4) RNA polymerase

Explanation: The correct answer is (2) Rho protein. Rho-independent termination functions without Rho. It depends on the formation of a GC-rich hairpin followed by a poly-U sequence, causing RNA polymerase to pause and detach. Hence, Rho factor is absent in this mechanism.

2. In prokaryotes, which signal stops RNA synthesis during transcription?

(1) Stop codon
(2) Terminator sequence
(3) Promoter sequence
(4) Operator sequence

Explanation: The correct answer is (2) Terminator sequence. The terminator region of DNA contains specific signals that mark the end of transcription, leading to RNA polymerase detachment. Stop codons function during translation, not transcription.

3. The Rho factor possesses which type of enzymatic activity?

(1) Ligase
(2) Helicase
(3) Polymerase
(4) Nuclease

Explanation: The correct answer is (2) Helicase. Rho factor is an ATP-dependent RNA helicase that unwinds RNA-DNA hybrids during transcription termination. It moves along the RNA strand toward RNA polymerase to release the transcript.

4. Assertion-Reason Type
Assertion (A): Rho-dependent termination needs ATP.
Reason (R): Rho factor moves along RNA using energy from ATP hydrolysis.

(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true

Explanation: The correct answer is (1). Rho factor uses ATP to translocate along RNA toward RNA polymerase. The energy from ATP hydrolysis helps unwind the RNA-DNA hybrid, leading to transcription termination.

5. Match the following:
A. Promoter
B. Rho factor
C. Terminator
D. RNA polymerase

I. Initiation
II. Elongation
III. Termination
IV. Regulation

(1) A-I, B-III, C-III, D-II
(2) A-I, B-III, C-II, D-II
(3) A-I, B-II, C-III, D-II
(4) A-I, B-III, C-III, D-IV

Explanation: The correct answer is (3). Promoter initiates transcription, RNA polymerase performs elongation, and Rho factor terminates transcription. Thus, each component plays a specific stage role in the transcription cycle.

6. The termination signal is located on which strand of DNA?

(1) Coding strand
(2) Template strand
(3) Both strands
(4) mRNA

Explanation: The correct answer is (2). The terminator sequence resides on the template strand, where transcription ends. RNA polymerase recognizes this signal and releases the RNA transcript, marking termination.

7. Fill in the blanks:
Transcription termination in prokaryotes occurs either by _______ mechanism or by _______ factor.

(1) Rho-independent, Rho-dependent
(2) Sigma, Omega
(3) Alpha, Beta
(4) Enhancer, Repressor

Explanation: The correct answer is (1). Two mechanisms exist: Rho-dependent (using Rho helicase) and Rho-independent (using hairpin loop and U-rich tail). Both ensure precise RNA transcript release at gene end.

8. Choose the correct statements:
Statement I: Termination requires stop codons.
Statement II: Termination can occur via Rho or hairpin structure.

(1) Both are true
(2) Only Statement I true
(3) Only Statement II true
(4) Both are false

Explanation: The correct answer is (3). Stop codons act during translation, not transcription. Termination in transcription occurs through either Rho-dependent or Rho-independent mechanisms, involving specific structural or enzymatic cues.

9. Which subunit of RNA polymerase is mainly responsible for promoter recognition?

(1) σ
(2) β
(3) α
(4) ρ

Explanation: The correct answer is (1). Sigma (σ) subunit recognizes the promoter region, initiating transcription. Once initiation occurs, sigma dissociates, allowing elongation by the core enzyme.

10. Which of the following steps immediately follows transcription termination?

(1) RNA capping
(2) Translation initiation
(3) RNA splicing
(4) DNA replication

Explanation: The correct answer is (2). In prokaryotes, transcription and translation are coupled; as soon as RNA is released, ribosomes begin translating it into protein. This efficiency aids rapid gene expression response.

Topic: Genetic Code; Subtopic: Triplet Code Hypothesis

Keyword Definitions:

• Genetic code: The sequence of nucleotides in DNA or RNA that determines the sequence of amino acids in proteins.

• Nucleotides: The building blocks of nucleic acids, consisting of a sugar, phosphate group, and nitrogen base.

• Triplet code: A set of three nucleotides that codes for one amino acid.

• Codon: A specific sequence of three nucleotides on mRNA that corresponds to a particular amino acid.

• Amino acids: Organic compounds that combine to form proteins, vital for cellular structure and function.

Lead Question - 2025

Who proposed that the genetic code for amino acids should be made up to three nucleotides:

(1) George Gamow
(2) Francis Crick
(3) Jacque Monod
(4) Franklin Stahl

Explanation: George Gamow proposed the triplet code hypothesis suggesting that three nucleotides together could code for one amino acid. Since there are four nitrogen bases, combinations of three (4³ = 64) provide enough codons to encode 20 amino acids. His model explained genetic information storage and translation, forming the basis for later discoveries. Answer: (1) George Gamow

1. Which scientist experimentally proved that the genetic code is a triplet and non-overlapping?
(1) Nirenberg and Matthaei
(2) Watson and Crick
(3) Hershey and Chase
(4) Jacob and Monod
Explanation: Nirenberg and Matthaei provided experimental evidence supporting the triplet and non-overlapping nature of the genetic code by synthesizing RNA sequences and observing corresponding polypeptide formation. This confirmed the coding relationship between RNA triplets and specific amino acids. Answer: (1) Nirenberg and Matthaei

2. How many codons are possible if the genetic code is made of three bases?
(1) 16
(2) 32
(3) 64
(4) 128
Explanation: A triplet code means 4 possible bases can arrange in 3 positions, resulting in 4³ = 64 codons. Of these, 61 code for amino acids and 3 act as stop codons, ending translation. Thus, the total number of possible codons is 64. Answer: (3) 64

3. Which of the following acts as a stop codon during protein synthesis?
(1) UAA, UAG, UGA
(2) AUG, GUA, CUA
(3) GCU, ACC, UCA
(4) UUU, GGG, CCC
Explanation: UAA, UAG, and UGA are stop codons that signal the termination of translation. These codons do not code for any amino acid, ensuring the proper end of protein synthesis on the ribosome. Answer: (1) UAA, UAG, UGA

4. Which of the following codons initiates protein synthesis in most organisms?
(1) AUG
(2) UAA
(3) UAG
(4) UGA
Explanation: AUG acts as the initiation codon for translation and codes for methionine in eukaryotes and formylmethionine in prokaryotes. It sets the reading frame and signals the start of protein synthesis. Answer: (1) AUG

5. Which property of genetic code ensures that one codon codes for only one amino acid?
(1) Degeneracy
(2) Unambiguous nature
(3) Universality
(4) Non-overlapping
Explanation: The genetic code is unambiguous, meaning each codon specifies only one amino acid. This precision prevents errors in protein synthesis and ensures fidelity in the translation of genetic information. Answer: (2) Unambiguous nature

6. Which term describes the ability of multiple codons to code for the same amino acid?
(1) Redundancy
(2) Universality
(3) Degeneracy
(4) Non-ambiguity
Explanation: Degeneracy refers to the phenomenon where more than one codon codes for the same amino acid, such as GAA and GAG both coding for glutamic acid. This property provides protection against mutations. Answer: (3) Degeneracy

7. Assertion-Reason Type:
Assertion (A): Genetic code is universal.
Reason (R): All organisms use the same codon for each amino acid.
(1) Both A and R are true, and R is the correct explanation of A
(2) Both A and R are true, but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Explanation: The genetic code is universal because nearly all living organisms interpret each codon identically, meaning the same codon codes for the same amino acid across species. This demonstrates a shared evolutionary origin. Answer: (1) Both A and R are true, and R is the correct explanation of A

8. Matching Type:
Match List-I with List-II:
List-I (Feature) – List-II (Description)
A. Start Codon – I. UAA
B. Stop Codon – II. AUG
C. Degenerate Code – III. Multiple codons for one amino acid
(1) A-II, B-I, C-III
(2) A-I, B-II, C-III
(3) A-III, B-I, C-II
(4) A-II, B-III, C-I
Explanation: AUG serves as the start codon, UAA as a stop codon, and degeneracy refers to the presence of multiple codons for a single amino acid. These are fundamental features ensuring proper translation of mRNA. Answer: (1) A-II, B-I, C-III

9. Fill in the Blanks:
The codon ______ codes for Methionine and also acts as a start codon.
(1) UAA
(2) AUG
(3) UGA
(4) UAG
Explanation: AUG codes for Methionine and functions as the initiation codon during translation. It signals the start of polypeptide synthesis on the ribosome and establishes the correct reading frame for mRNA decoding. Answer: (2) AUG

10. Choose the Correct Statements:
Statement I: Genetic code is overlapping.
Statement II: Genetic code is degenerate.
(1) Both statements are correct
(2) Only Statement I is correct
(3) Only Statement II is correct
(4) Both statements are incorrect
Explanation: The genetic code is non-overlapping because each base is part of only one codon, and it is degenerate since multiple codons can specify the same amino acid. Therefore, only Statement II is correct. Answer: (3) Only Statement II is correct

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Topic: Structure of Chromatin; Subtopic: Histones and Nucleosomes

Keyword Definitions:

• Histones: Small, positively charged proteins that help in DNA packaging within the nucleus by forming nucleosomes.

• Nucleosome: Structural unit of chromatin consisting of DNA wrapped around histone proteins.

• Chromatin: The complex of DNA and proteins that forms chromosomes within the nucleus.

• Lysine & Arginine: Basic amino acids that carry positive charges, allowing binding to negatively charged DNA.

• DNA Packaging: The process of coiling DNA with histones to fit inside the cell nucleus.

Lead Question – 2025

Histones are enriched with:

(1) Lysine & Arginine
(2) Leucine & Lysine
(3) Phenylalanine & Leucine
(4) Phenylalanine & Arginine

Explanation:
Histones are positively charged proteins rich in basic amino acids like lysine and arginine. Their positive charge allows strong electrostatic attraction to the negatively charged phosphate groups of DNA, facilitating tight DNA binding and nucleosome formation. These proteins play a key role in DNA packaging, gene regulation, and chromatin organization. Hence, the correct answer is (1) Lysine & Arginine.

Guessed Questions:

1. Which of the following is the structural unit of chromatin?
(1) Chromosome
(2) Gene
(3) Nucleosome
(4) Histone
Explanation: A nucleosome is the repeating structural unit of chromatin, composed of DNA wrapped around a histone octamer. It helps compact long DNA molecules into the nucleus while maintaining accessibility for transcription. Hence, the correct answer is (3) Nucleosome.

2. How many types of histone proteins form the nucleosome core?
(1) Three
(2) Four
(3) Five
(4) Eight
Explanation: The nucleosome core consists of an octamer made up of four types of histone proteins — H2A, H2B, H3, and H4 — present in pairs. DNA winds around this core 1¾ times, forming the nucleosome structure. Thus, the correct answer is (2) Four.

3. Which histone is not part of the nucleosome core particle?
(1) H2A
(2) H2B
(3) H3
(4) H1
Explanation: Histone H1 is known as the linker histone. It is not part of the nucleosome core but binds to the DNA between nucleosomes, aiding in further chromatin compaction. Hence, the correct answer is (4) H1.

4. Which statement best explains why histones bind to DNA?
(1) Both are negatively charged
(2) Both are positively charged
(3) DNA is negatively charged, histones are positively charged
(4) DNA is neutral
Explanation: DNA carries a negative charge due to its phosphate backbone, while histones are rich in positively charged amino acids like lysine and arginine. This opposite charge interaction allows histones to bind tightly to DNA. Hence, the correct answer is (3) DNA is negatively charged, histones are positively charged.

5. Which of the following amino acids is basic in nature?
(1) Alanine
(2) Glycine
(3) Arginine
(4) Valine
Explanation: Arginine is a basic amino acid due to its guanidino group, which remains positively charged at physiological pH. This positive charge allows interaction with negatively charged DNA, making it a major component of histones. Thus, the correct answer is (3) Arginine.

6. Which of the following statements about histones is correct?
(1) They are acidic proteins
(2) They are basic proteins
(3) They are neutral proteins
(4) They are hydrophobic proteins
Explanation: Histones are basic proteins because they contain lysine and arginine residues, which carry positive charges. This basicity allows them to neutralize DNA’s negative charge and compact it into nucleosomes. Hence, the correct answer is (2) They are basic proteins.

7. Assertion–Reason Type:
Assertion (A): Histones help in DNA packaging.
Reason (R): They are rich in basic amino acids that interact with negatively charged DNA.
(1) Both A and R are true, and R explains A
(2) Both A and R are true, but R does not explain A
(3) A is true, R is false
(4) A is false, R is true
Explanation: Histones are positively charged proteins rich in lysine and arginine, which bind electrostatically with DNA’s phosphate backbone. This interaction helps in DNA condensation and nucleosome formation. Hence, both A and R are true, and R correctly explains A. The answer is (1).

8. Matching Type:
Match the following:
List I (Histone type) – List II (Function)
A. H1 – 1. Linker histone
B. H2A & H2B – 2. Core histones
C. H3 & H4 – 3. Core histones
(1) A–1, B–2, C–3
(2) A–2, B–3, C–1
(3) A–1, B–3, C–2
(4) A–3, B–1, C–2
Explanation: Histone H1 acts as a linker histone, stabilizing DNA between nucleosomes. H2A, H2B, H3, and H4 form the nucleosome core. Thus, the correct matching is A–1, B–2, C–3.

9. Fill in the Blanks:
Histones are ________ proteins that bind to DNA.
(1) Acidic
(2) Neutral
(3) Basic
(4) Hydrophobic
Explanation: Histones are basic proteins enriched with lysine and arginine, allowing them to bind tightly with DNA through ionic interactions. Their basicity neutralizes DNA’s negative charge and aids chromatin compaction. Hence, the correct answer is (3) Basic.

10. Choose the Correct Statements:
Statement I: Histones help in packaging of DNA into nucleosomes.
Statement II: Histones are rich in acidic amino acids.
(1) Both statements are correct
(2) Both statements are incorrect
(3) Statement I correct, Statement II incorrect
(4) Statement I incorrect, Statement II correct
Explanation: Histones play an essential role in DNA packaging by forming nucleosomes. However, they are basic, not acidic, due to lysine and arginine residues. Therefore, Statement I is correct, Statement II is incorrect. The correct answer is (3).

Topic: DNA as Genetic Material; Subtopic: Experimental Proofs and Chromatin Structure

Keyword Definitions:

• DNA: Deoxyribonucleic acid, the hereditary material carrying genetic instructions.

• Chromatin: The complex of DNA and proteins forming chromosomes within the nucleus.

• Euchromatin: Lightly packed chromatin region active in gene transcription.

• Heterochromatin: Densely packed chromatin region inactive in transcription.

• Transformation: Process of genetic material transfer between cells demonstrated by Griffith.

• Bacteriophage: Virus that infects bacteria, used by Hershey and Chase in their DNA experiments.

Lead Question - 2025

Match List I with List II
List - I    List - II
A. Alfred Hershey and Martha Chase    I. Streptococcus pneumoniae
B. Euchromatin    II. Densely packed and dark-stained
C. Frederich Griffith    III. Loosely packed and light-stained
D. Heterochromatin    IV. DNA as genetic material confirmation

Choose the correct answer from the options given below
(1) A-II, B-IV, C-I, D-III
(2) A-IV, B-II, C-I, D-III
(3) A-IV, B-III, C-I, D-II
(4) A-III, B-II, C-IV, D-I

Explanation: The Hershey–Chase experiment confirmed that DNA, not protein, is the genetic material (A–IV). Euchromatin is lightly packed and transcriptionally active (B–III). Griffith worked with Streptococcus pneumoniae to demonstrate transformation (C–I). Heterochromatin is densely packed and transcriptionally inactive (D–II). Thus, the correct answer is (3) A–IV, B–III, C–I, D–II. These findings collectively formed the foundation of molecular genetics and chromatin classification.

Guessed Questions:

1. Who discovered the phenomenon of bacterial transformation?
(1) Watson
(2) Griffith
(3) Hershey
(4) Avery

Explanation: Frederick Griffith in 1928 discovered bacterial transformation while working on Streptococcus pneumoniae. He showed that a “transforming principle” from heat-killed virulent bacteria could genetically alter non-virulent bacteria, paving the way for identifying DNA as the hereditary material. The correct answer is (2) Griffith.

2. Hershey and Chase used radioactive isotopes to label DNA and proteins. Which isotopes were used?
(1) 35S and 32P
(2) 14C and 35S
(3) 32P and 15N
(4) 3H and 14C

Explanation: Hershey and Chase labeled viral DNA with 32P and protein coat with 35S to trace which material entered bacterial cells. They found only DNA entered, confirming DNA as the genetic material. Hence, the correct answer is (1) 35S and 32P.

3. Which type of chromatin remains transcriptionally active?
(1) Euchromatin
(2) Heterochromatin
(3) Satellite DNA
(4) All chromatin

Explanation: Euchromatin is the less condensed region of chromatin where genes are transcriptionally active and DNA is accessible for RNA synthesis. It stains lightly and appears dispersed under a microscope. Hence, the correct answer is (1) Euchromatin.

4. Which scientist provided biochemical evidence that DNA is the genetic material?
(1) Griffith
(2) Avery, MacLeod, and McCarty
(3) Hershey and Chase
(4) Watson and Crick

Explanation: Avery, MacLeod, and McCarty, in 1944, provided biochemical proof that DNA is the transforming principle. They treated Griffith’s bacterial extracts with enzymes, showing only DNA destruction stopped transformation. Therefore, the correct answer is (2) Avery, MacLeod, and McCarty.

5. Which of the following is true about heterochromatin?
(1) It is transcriptionally active
(2) It stains lightly
(3) It remains condensed throughout the cell cycle
(4) It forms genes

Explanation: Heterochromatin remains condensed during the cell cycle, is transcriptionally inactive, and stains darkly due to tight packing. It helps maintain chromosome structure and gene regulation. Therefore, the correct answer is (3) It remains condensed throughout the cell cycle.

6. What conclusion did Hershey and Chase draw from their experiment?
(1) Proteins are genetic material
(2) RNA is the genetic material
(3) DNA is the genetic material
(4) Lipids are the genetic material

Explanation: The Hershey and Chase experiment with T2 bacteriophage concluded that DNA, not protein, enters bacteria during infection and directs viral replication. This established DNA as the molecule of inheritance. Hence, the correct answer is (3) DNA is the genetic material.

7. Assertion-Reason Type Question:
Assertion (A): Euchromatin is transcriptionally active.
Reason (R): It is densely packed and stains darkly.
(1) Both A and R are true, and R is the correct explanation.
(2) Both A and R are true, but R is not the correct explanation.
(3) A is true, R is false.
(4) A is false, R is true.

Explanation: Euchromatin is transcriptionally active due to its loose structure, which allows access to enzymes involved in transcription. It stains lightly, not darkly. Thus, Assertion is true but Reason is false. The correct answer is (3) A is true, R is false.

8. Matching Type Question:
Match the scientist with their contribution.
A. Watson and Crick — I. X-ray crystallography
B. Rosalind Franklin — II. DNA double helix model
C. Hershey and Chase — III. DNA as genetic material
(1) A-II, B-I, C-III
(2) A-III, B-II, C-I
(3) A-I, B-II, C-III
(4) A-II, B-III, C-I

Explanation: Watson and Crick proposed the double helix structure (A–II), Rosalind Franklin used X-ray crystallography to reveal helical DNA (B–I), and Hershey–Chase confirmed DNA as genetic material (C–III). Thus, the correct match is (1) A–II, B–I, C–III.

9. Fill in the Blanks / Completion Type:
The transforming principle identified by Avery, MacLeod, and McCarty was ________.
(1) Protein
(2) DNA
(3) RNA
(4) Lipid

Explanation: Avery, MacLeod, and McCarty demonstrated that DNA is the transforming principle responsible for heredity. Enzymes that degraded DNA prevented transformation, proving that DNA carries genetic information. Thus, the correct answer is (2) DNA.

10. Choose the Correct Statements:
Statement I: Heterochromatin is transcriptionally active.
Statement II: Euchromatin is lightly stained and active.
(1) Both statements are true.
(2) Both statements are false.
(3) Statement I is true, Statement II is false.
(4) Statement I is false, Statement II is true.

Explanation: Heterochromatin is transcriptionally inactive and darkly stained, while euchromatin is lightly stained and active in transcription. Hence, Statement I is false, and Statement II is true. The correct answer is (4) Statement I is false, Statement II is true.

Topic: Gene Expression and Regulation; Subtopic: Post-Transcriptional Modifications in Eukaryotes

Keyword Definitions:

• Transcription: The process of synthesizing RNA from a DNA template.

• Pre-mRNA (hnRNA): The primary transcript produced in eukaryotes before modification.

• Introns and Exons: Introns are non-coding sequences removed during RNA splicing, while exons are coding sequences joined together.

• Capping: Addition of a methylated guanosine cap at the 5′ end of hnRNA for protection and stability.

• Polyadenylation: Addition of adenine residues at the 3′ end of hnRNA forming the poly-A tail.

Lead Question – 2025

Which of the following are the post-transcriptional events in an eukaryotic cell?

A. Transport of pre-mRNA to cytoplasm prior to splicing.

B. Removal of introns and joining of exons.

C. Addition of methyl group at 5′ end of hnRNA.

D. Addition of adenine residues at 3′ end of hnRNA.

E. Base pairing of two complementary RNAs.

(1) A, B, C only

(2) B, C, D only

(3) B, C, E only

(4) C, D, E only

Explanation:

Post-transcriptional modifications in eukaryotic cells include three main processes: capping, splicing, and polyadenylation. Capping adds a methylated guanosine to the 5′ end, splicing removes introns and joins exons, and polyadenylation adds adenine residues to the 3′ end. Transport occurs after these events. Hence, the correct answer is (2) B, C, D only.

Guessed Questions:

1. The addition of a 5′ cap and 3′ poly-A tail occurs in

(1) Prokaryotic mRNA

(2) Eukaryotic hnRNA

(3) tRNA only

(4) Ribosomal RNA

2. Which enzyme catalyzes removal of introns from hnRNA?

(1) DNA polymerase

(2) RNA polymerase II

(3) Spliceosome complex

(4) Ligase

3. The 3′ end of eukaryotic hnRNA is modified by addition of

(1) Poly-A tail

(2) 5′ cap

(3) tRNA sequence

(4) Ribosomal subunit

4. In eukaryotes, capping of mRNA occurs

(1) After splicing

(2) Before transcription begins

(3) During elongation stage of transcription

(4) After mRNA reaches cytoplasm

5. The process of removing non-coding sequences from pre-mRNA is known as

(1) Transcription

(2) Translation

(3) Splicing

(4) Replication

6. Which one of the following is NOT a post-transcriptional modification?

(1) Polyadenylation

(2) Splicing

(3) Capping

(4) DNA replication

Assertion – Reason Question

7. Assertion (A): In eukaryotic cells, introns are removed from hnRNA by splicing.

Reason (R): Introns are non-coding sequences that interfere with translation.

(1) Both A and R are true and R is the correct explanation of A

(2) Both A and R are true but R is not the correct explanation of A

(3) A is true but R is false

(4) A is false but R is true

Matching Type Question

8. Match the following:

List I (Process) — List II (Function)

A. Capping — 1. Removal of introns

B. Splicing — 2. Stability of mRNA

C. Polyadenylation — 3. Export of mRNA

D. hnRNA — 4. Precursor of mRNA

(1) A–2, B–1, C–3, D–4

(2) A–3, B–2, C–4, D–1

(3) A–4, B–3, C–2, D–1

(4) A–1, B–2, C–3, D–4

Fill in the Blanks / Completion Question

9. The non-coding sequences present in the primary transcript of eukaryotic genes are called __________.

(1) Introns

(2) Exons

(3) Codons

(4) Cistrons

Choose the Correct Statements Question

10. Statement I: Capping protects mRNA from degradation.

Statement II: Splicing joins coding sequences (exons) together.

(1) Both statements are true

(2) Both statements are false

(3) Statement I is true, Statement II is false

(4) Statement I is false, Statement II is true

Explanation:

Capping and splicing are essential post-transcriptional modifications. The 5′ methyl cap prevents degradation by exonucleases and facilitates ribosome binding. Splicing removes introns and links exons into a continuous coding sequence for translation. Therefore, both statements I and II are true, and the correct option is (1). 

Topic: Nucleic Acids; Subtopic: RNA World Hypothesis and Evolution of DNA

Keyword Definitions:
• RNA World: The hypothesis that RNA was the first genetic material capable of storing information and catalyzing reactions.
• Genetic Material: Molecules responsible for storing and transmitting hereditary information (DNA or RNA).
• Catalyst: A substance that increases the rate of a biochemical reaction without being consumed.
• DNA: Deoxyribonucleic acid, a stable double-stranded molecule storing genetic information in all living organisms except some viruses.
• Double Helix: The twisted-ladder structure of DNA formed by complementary base pairing.
• Repair Mechanism: Cellular process that corrects errors or damages in DNA to maintain genetic stability.

Lead Question – 2025
Given below are two statements:
Statement – I: In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.
Statement – II: DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Both statements I and statement II are correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct

Explanation:
Both statements I and II are correct. The RNA world hypothesis suggests that RNA acted as the first genetic material capable of both storing genetic information and catalyzing reactions. However, due to its instability, DNA evolved as a more stable molecule. DNA’s complementary strands and repair systems ensure greater genetic fidelity and evolutionary advantage.

1. Which among the following proves that RNA could have been the first genetic material?
(1) RNA can store information and act as an enzyme
(2) RNA is double-stranded and stable
(3) RNA cannot act as an enzyme
(4) RNA is chemically inert
Explanation: RNA can act both as a genetic material and as a catalyst (ribozyme). This dual function supports the RNA world hypothesis, suggesting that RNA preceded DNA and proteins in evolution. RNA’s catalytic nature enabled early biochemical reactions essential for life formation, even before DNA or enzymes evolved.

2. Which type of RNA acts as a template during protein synthesis?
(1) mRNA
(2) tRNA
(3) rRNA
(4) snRNA
Explanation: Messenger RNA (mRNA) carries genetic information from DNA to the ribosome, where it serves as a template for protein synthesis. The sequence of nucleotides in mRNA determines the sequence of amino acids, ensuring proper translation of genetic information into functional proteins.

3. Which enzyme is responsible for catalyzing DNA synthesis from an RNA template?
(1) DNA polymerase
(2) RNA polymerase
(3) Reverse transcriptase
(4) Ligase
Explanation: Reverse transcriptase synthesizes DNA from an RNA template. It is found in retroviruses such as HIV. This enzyme supports the idea that RNA could have preceded DNA, as it demonstrates that RNA can serve as a template for the evolution of DNA, leading to more stable genetic storage systems.

4. Which of the following gives DNA its greater stability compared to RNA?
(1) Presence of uracil
(2) Presence of thymine and deoxyribose
(3) Presence of ribose sugar
(4) Lack of phosphodiester bonds
Explanation: DNA’s stability arises from the presence of thymine instead of uracil and deoxyribose instead of ribose. Deoxyribose lacks an oxygen atom at the 2′ position, making DNA less reactive and more stable. This chemical property allows DNA to serve as the long-term genetic material in living organisms.

5. Which of the following statements about DNA and RNA is incorrect?
(1) DNA contains thymine, RNA contains uracil
(2) DNA is single-stranded, RNA is double-stranded
(3) Both DNA and RNA contain adenine and guanine
(4) DNA has deoxyribose sugar while RNA has ribose sugar
Explanation: Option (2) is incorrect. DNA is typically double-stranded, while RNA is usually single-stranded. Their difference in sugars and bases makes RNA more reactive but less stable, and DNA more suited for hereditary information storage due to its stable double-helical structure and repair mechanisms.

6. In DNA, adenine pairs with thymine through:
(1) One hydrogen bond
(2) Two hydrogen bonds
(3) Three hydrogen bonds
(4) Phosphodiester bonds
Explanation: In DNA, adenine pairs with thymine through two hydrogen bonds, while cytosine pairs with guanine via three hydrogen bonds. These complementary base pairings ensure genetic fidelity during replication and transcription processes, making DNA a reliable molecule for long-term genetic information storage.

7. Assertion-Reason Type:
Assertion (A): DNA is more stable than RNA.
Reason (R): DNA lacks the 2'-hydroxyl group present in RNA.
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Explanation: Both A and R are true and R correctly explains A. DNA’s deoxyribose lacks a hydroxyl group at the 2′ position, reducing chemical reactivity and enhancing stability. This chemical difference ensures DNA’s resistance to hydrolysis and allows it to serve as a stable genetic material.

8. Matching Type:
Match the following:
A. mRNA    1. Carries amino acids
B. tRNA    2. Template for protein synthesis
C. rRNA    3. Forms ribosomal structure
(1) A–2, B–1, C–3
(2) A–3, B–2, C–1
(3) A–1, B–3, C–2
(4) A–2, B–3, C–1
Explanation: The correct match is A–2, B–1, C–3. mRNA serves as the template for protein synthesis, tRNA brings amino acids to the ribosome, and rRNA forms the structural and catalytic part of ribosomes. Together, they coordinate the translation process of gene expression efficiently.

9. Fill in the Blank:
DNA replication proceeds in the ______ direction.
(1) 3′ → 5′
(2) 5′ → 3′
(3) Random
(4) Both directions equally
Explanation: DNA replication occurs in the 5′ → 3′ direction. DNA polymerase adds nucleotides only to the 3′ end of the growing strand. The leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously as Okazaki fragments, ensuring accurate and semiconservative DNA replication.

10. Choose the Correct Statements (Statement I & II):
Statement I: RNA can catalyze certain biochemical reactions.
Statement II: DNA can act as an enzyme in metabolic reactions.
(1) Both statements are correct
(2) Statement I is correct, Statement II is incorrect
(3) Both statements are incorrect
(4) Statement I is incorrect, Statement II is correct
Explanation: Statement I is correct while Statement II is incorrect. Some RNA molecules (ribozymes) possess catalytic activity, playing roles in RNA splicing and peptide bond formation. DNA, however, is chemically stable and does not exhibit catalytic activity. This further supports RNA’s ancient biological significance.

Topic: Gene Expression and Regulation; Subtopic: RNA Types and RNA Interference (RNAi)

Keyword Definitions:

• mRNA: Messenger RNA that carries genetic information from DNA to ribosomes for protein synthesis.

• tRNA: Transfer RNA that brings specific amino acids to the ribosome during translation by pairing with mRNA codons.

• rRNA: Ribosomal RNA forming structural and catalytic parts of the ribosome, helping in peptide bond formation.

• RNA interference (RNAi): A gene-silencing mechanism in eukaryotes that uses small RNA molecules to inhibit gene expression by degrading mRNA.

• siRNA: Small interfering RNA responsible for degrading target mRNA molecules during RNAi.

• Translation: The process of synthesizing proteins from mRNA using tRNA and rRNA.

Lead Question – 2025

Given below are two statements:
Statement I: Transfer RNAs and ribosomal RNA do not interact with mRNA.
Statement II: RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Both Statement I and Statement II are correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Statement I is incorrect but Statement II is correct

Explanation: During translation, both tRNA and rRNA directly interact with mRNA to ensure proper decoding and peptide synthesis, so Statement I is incorrect. RNA interference occurs in eukaryotes as a defense mechanism against viral genomes or transposons by degrading complementary mRNA using siRNA, so Statement II is correct. Hence, the correct answer is (4).

Guessed Questions:

1. Which of the following RNAs carries amino acids to the ribosome during protein synthesis?
(1) mRNA
(2) tRNA
(3) rRNA
(4) snRNA

Explanation: Transfer RNA (tRNA) has an anticodon loop complementary to the codon on mRNA. It carries specific amino acids to the ribosome during translation, facilitating the sequential addition of amino acids into a growing polypeptide chain. Therefore, tRNA is called an adaptor molecule.

2. Which of the following statements about RNAi is correct?
(1) It promotes mRNA translation.
(2) It degrades specific mRNA molecules.
(3) It enhances transcription.
(4) It prevents DNA replication.

Explanation: RNA interference (RNAi) silences gene expression by degrading complementary mRNA molecules through small interfering RNA (siRNA) or microRNA (miRNA). It acts as a natural defense mechanism and a tool for gene regulation and therapeutic applications. Hence, option (2) is correct.

3. Match the following types of RNA with their functions:
A. mRNA → I. Carries genetic code
B. tRNA → II. Transfers amino acids
C. rRNA → III. Forms ribosome
Choose the correct answer:
(1) A-I, B-II, C-III
(2) A-II, B-I, C-III
(3) A-III, B-II, C-I
(4) A-I, B-III, C-II

Explanation: Messenger RNA (mRNA) carries the genetic message from DNA to ribosomes, transfer RNA (tRNA) carries amino acids, and ribosomal RNA (rRNA) makes up the structural framework of ribosomes. Therefore, option (1) is correct.

4. Assertion (A): RNAi is used in biotechnology to silence specific genes.
Reason (R): It allows degradation of mRNA complementary to double-stranded RNA molecules.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, R is false.
(4) A is false, R is true.

Explanation: RNA interference is a valuable tool for gene silencing. The introduction of double-stranded RNA leads to the degradation of complementary mRNA, preventing protein synthesis. Therefore, both statements are true, and R correctly explains A.

5. Fill in the Blanks:
RNA interference uses ________ molecules to degrade target mRNA.
(1) miRNA
(2) tRNA
(3) rRNA
(4) snRNA

Explanation: RNA interference utilizes small interfering RNA (siRNA) and microRNA (miRNA) to silence genes. These short RNA molecules bind complementary mRNA and promote its degradation, thereby suppressing gene expression at the post-transcriptional level. Therefore, the answer is miRNA.

6. Choose the correct statements:
Statement I: mRNA is synthesized from the DNA template.
Statement II: tRNA and rRNA are involved in protein synthesis.
(1) Both statements are true.
(2) Both statements are false.
(3) Statement I is true, Statement II is false.
(4) Statement I is false, Statement II is true.

Explanation: mRNA transcription occurs using the DNA template strand, and both tRNA and rRNA play essential roles in translation. Therefore, both statements are true, making option (1) correct.

7. Which of the following enzymes is responsible for synthesizing tRNA?
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) DNA polymerase

Explanation: RNA polymerase III synthesizes transfer RNA (tRNA) and some small nuclear RNAs in eukaryotic cells. RNA polymerase I synthesizes rRNA, and RNA polymerase II synthesizes mRNA. Thus, option (3) is correct.

8. Which of the following RNAs acts as a template during translation?
(1) mRNA
(2) rRNA
(3) tRNA
(4) miRNA

Explanation: Messenger RNA (mRNA) acts as the direct template for translation, carrying codons that determine the sequence of amino acids in the polypeptide chain. Ribosomes read these codons and assemble amino acids accordingly. Hence, option (1) is correct.

9. Which of the following statements is true regarding tRNA structure?
(1) It is double-stranded.
(2) It has an anticodon loop and an amino acid acceptor end.
(3) It has no secondary structure.
(4) It carries codons for protein synthesis.

Explanation: tRNA is a single-stranded RNA molecule with cloverleaf secondary structure containing an anticodon loop for codon recognition and a 3'-CCA end that binds amino acids. Hence, option (2) is correct.

10. Which of the following statements about RNA interference is incorrect?
(1) It is a post-transcriptional process.
(2) It prevents mRNA translation.
(3) It occurs only in prokaryotes.
(4) It uses small RNA molecules for silencing.

Explanation: RNA interference occurs in eukaryotic cells, not prokaryotes. It prevents mRNA translation or promotes degradation using siRNA or miRNA. Hence, the incorrect statement is option (3).

Topic: Chromosomal Gene Distribution; Subtopic: Gene Mapping and Chromosome Structure

Keyword Definitions:

• Chromosome: A thread-like structure made of DNA and proteins carrying genetic information.

• Gene: A sequence of DNA that codes for a specific protein or trait.

• Genome: The complete set of genes or genetic material in an organism.

• Gene density: The number of genes present per unit length of DNA on a chromosome.

• Human Genome Project (HGP): An international scientific research project aimed at mapping all human genes.

Lead Question – (2025)

Which chromosome in the human genome has the highest number of genes?

(1) Chromosome X

(2) Chromosome Y

(3) Chromosome 1

(4) Chromosome 10

Explanation:

Among all human chromosomes, Chromosome 1 contains the highest number of genes, approximately 2,968 functional genes. It is also the largest human chromosome, containing about 249 million base pairs. In contrast, the Y chromosome has the least number of genes, primarily involved in male sex determination. Hence, the correct answer is (3) Chromosome 1.

1. Which human chromosome carries the smallest number of genes?

(1) Chromosome 21

(2) Chromosome Y

(3) Chromosome X

(4) Chromosome 18

Explanation:

The Y chromosome contains the smallest number of genes, around 70–80, most of which are associated with male sex determination and spermatogenesis. In contrast, Chromosome 1 contains the largest number of genes. The Y chromosome is thus genetically poor but functionally significant in male-specific traits.

2. Which of the following statements about the human genome is correct?

(1) It contains about 23,000 functional genes.

(2) It contains more than 100,000 functional genes.

(3) It contains only 5,000 functional genes.

(4) It contains 1,000,000 functional genes.

Explanation:

The human genome contains approximately 23,000 functional genes distributed across 23 pairs of chromosomes. Earlier predictions suggested over 100,000 genes, but genome sequencing revealed fewer. These genes account for only about 1.5% of total DNA; the rest is non-coding regions with regulatory or unknown functions. Correct answer is (1).

3. Which of the following chromosomes is associated with Down’s syndrome?

(1) Chromosome 13

(2) Chromosome 18

(3) Chromosome 21

(4) Chromosome 22

Explanation:

Down’s syndrome results from trisomy of chromosome 21, where an individual possesses three copies of chromosome 21 instead of two. This leads to developmental and intellectual disabilities, characteristic facial features, and other systemic abnormalities. It is one of the most common chromosomal disorders. Hence, the answer is (3) Chromosome 21.

4. The X chromosome in humans contains genes responsible for:

(1) Only sex determination

(2) Sex determination and other traits

(3) Only hair color

(4) Only muscle formation

Explanation:

The X chromosome carries genes not only related to sex determination but also essential for other physiological and metabolic processes like vision, muscle function, and immune regulation. Males have one X and one Y chromosome, whereas females have two X chromosomes. Hence, correct answer is (2).

5. Which human chromosome carries the gene for SRY (Sex-determining Region Y)?

(1) Chromosome 1

(2) Chromosome 22

(3) Chromosome Y

(4) Chromosome X

Explanation:

The SRY gene (Sex-determining Region of Y) is located on the Y chromosome and is crucial for male sex determination. It triggers the development of testes and male phenotypic characteristics during embryonic growth. Its absence or mutation can lead to sex reversal. Thus, the correct answer is (3) Chromosome Y.

6. The smallest human chromosome is:

(1) Chromosome 21

(2) Chromosome 22

(3) Chromosome 20

(4) Chromosome Y

Explanation:

The smallest human autosome is Chromosome 21, containing about 48 million base pairs and roughly 200–300 genes. It is smaller than Chromosome 22 in terms of base pairs but not in gene density. The Y chromosome, though shorter, is a sex chromosome, not autosomal. Hence, the correct answer is (1).

7. Assertion–Reason Type Question

Assertion (A): Chromosome 1 has the highest number of genes in humans.

Reason (R): Chromosome 1 is the largest human chromosome with about 249 million base pairs.

(1) Both A and R are true, and R is the correct explanation of A.

(2) Both A and R are true, but R is not the correct explanation of A.

(3) A is true but R is false.

(4) A is false but R is true.

Explanation:

Both statements are correct. Chromosome 1 indeed has the highest number of genes and is also the largest in size with about 249 million base pairs, directly explaining why it carries the most genes. Hence, the correct answer is (1).

8. Matching Type Question

Match List I with List II:

List I – Chromosome

A. 1

B. 21

C. Y

D. X

List II – Characteristics

I. Causes Down’s syndrome when trisomic

II. Smallest number of genes

III. Largest chromosome, most genes

IV. Contains both sex-linked and non-sex-linked genes

Options:

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

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

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

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

Explanation:

Chromosome 1 has the largest number of genes, Chromosome 21 causes Down’s syndrome when trisomic, Chromosome Y has the fewest genes, and Chromosome X contains both sex-linked and non-sex-linked genes. Therefore, the correct match is (1) A-III, B-I, C-II, D-IV.

9. Fill in the Blanks

The human chromosome with the greatest number of genes is ________.

(1) Chromosome X

(2) Chromosome 1

(3) Chromosome 22

(4) Chromosome Y

Explanation:

The Chromosome 1 has the highest number of genes, containing approximately 2,968 genes and being the largest chromosome. Its vast size and gene content make it vital for numerous cellular and developmental functions in humans. Hence, the correct answer is (2) Chromosome 1.

10. Choose the Correct Statements

Statement I: Chromosome 1 has the largest size and maximum gene number.

Statement II: Chromosome Y has the smallest size and highest gene number.

(1) Both statements are correct

(2) Both statements are incorrect

(3) Statement I is correct but Statement II is incorrect

(4) Statement I is incorrect but Statement II is correct

Explanation:

Statement I is correct as Chromosome 1 is the largest and richest in gene content, while Statement II is incorrect because Chromosome Y is the smallest and poorest in gene number. Therefore, the correct answer is (3).

Topic: Genetic Code; Subtopic: Features and Universality of Codons

Keyword Definitions:

• Genetic Code: The sequence of nucleotides in mRNA that determines the sequence of amino acids in a protein.

• Codon: A triplet of nucleotides that codes for a specific amino acid or stop signal during translation.

• Initiator Codon: A codon that signals the start of translation, usually AUG coding for Methionine.

• Terminator Codon: Codons (UAA, UAG, UGA) that signal the end of protein synthesis.

• Universality: Property of the genetic code being nearly the same across most organisms.

• Degeneracy: More than one codon can code for the same amino acid, providing tolerance to mutations.

Lead Question - 2024 (Jhajjhar)
Identify the correct statements about Genetic codons.
A. AUG is initiator codon and codes for Glycine
B. UAA codes for Tyrosine
C. The codons are mostly universal
D. UAG is terminator codon
E. More than one codons code for single amino acid
1. C, B, D
2. A, D, E
3. C, D, E
4. A, B, C

Explanation: The genetic code is universal, nearly the same in all organisms. AUG is the initiator codon coding for Methionine, not Glycine. UAG acts as a stop codon. The code is degenerate since multiple codons can specify the same amino acid. Therefore, the correct answer is Option 3 (C, D, E).

1. Which of the following codons acts as the start signal during translation?
1. UAA
2. AUG
3. UGA
4. UAG

The start codon in almost all organisms is AUG, which codes for Methionine. It signals the ribosome to begin translation. Other codons like UAA, UAG, and UGA are stop codons that terminate translation. Hence, the correct answer is Option 2 (AUG).

2. Which of the following is not a stop codon?
1. UAA
2. UGA
3. UAG
4. AUG

There are three stop codons in the genetic code: UAA, UAG, and UGA. These signal the end of translation. AUG is a start codon, not a stop codon, and codes for Methionine. Thus, the correct answer is Option 4 (AUG).

3. The genetic code is termed ‘degenerate’ because:
1. One codon codes for more than one amino acid
2. More than one codon codes for a single amino acid
3. All codons code for the same amino acid
4. Codons are not specific

The degeneracy of the genetic code means that multiple codons can specify a single amino acid. For example, both UUU and UUC code for Phenylalanine. This property helps minimize the effect of mutations. Hence, the correct answer is Option 2.

4. Assertion-Reason Type:
Assertion (A): AUG acts as both initiator and coding codon.
Reason (R): It codes for Methionine and marks the start of translation.
1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

AUG serves dual functions—it is the initiator codon and codes for Methionine during translation. Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion. The correct answer is Option 1.

5. Match the following:
List-I (Codon) — List-II (Amino Acid)
A. UUU — I. Valine
B. GCU — II. Alanine
C. UGG — III. Phenylalanine
D. AUG — IV. Methionine
1. A-III, B-II, C-I, D-IV
2. A-II, B-I, C-IV, D-III
3. A-III, B-II, C-IV, D-I
4. A-I, B-III, C-II, D-IV

UUU codes for Phenylalanine, GCU for Alanine, UGG for Tryptophan, and AUG for Methionine. Correct pairing is A-III, B-II, C-IV, D-I. Thus, the correct answer is Option 3.

6. Fill in the Blank:
The codon UGA is usually a stop codon but in some cases, it codes for the amino acid ______.

UGA generally serves as a stop codon. However, in certain organisms like mitochondria, it codes for Tryptophan. This is one of the few exceptions to the universal nature of the genetic code. Therefore, the correct word to fill is Tryptophan.

7. Choose the correct statements:
Statement I: Genetic code is overlapping.
Statement II: Genetic code is non-ambiguous.
1. Both statements are correct.
2. Both are incorrect.
3. Only Statement I is correct.
4. Only Statement II is correct.

The genetic code is non-overlapping, meaning each base is part of only one codon. It is also non-ambiguous because each codon specifies only one amino acid. Therefore, only Statement II is correct. The correct answer is Option 4.

8. Which property of the genetic code allows silent mutations?
1. Overlapping nature
2. Universality
3. Degeneracy
4. Ambiguity

Due to the degeneracy of the genetic code, a change in the third base of a codon may not alter the amino acid sequence. This permits silent mutations that do not affect protein function. Thus, the correct answer is Option 3 (Degeneracy).

9. Which of the following statements best describes the universality of genetic code?
1. Codons are unique for each organism.
2. Codons code the same amino acids in almost all organisms.
3. Codons code differently in bacteria.
4. Codons vary within the same organism.

The universality of the genetic code means that codons specify the same amino acids across most species, from bacteria to humans. This strongly supports the theory of a common origin of life. The correct answer is Option 2.

10. Which of the following amino acids is coded by only one codon?
1. Methionine
2. Leucine
3. Serine
4. Valine

Methionine (AUG) and Tryptophan (UGG) are unique because each is coded by a single codon, while others have multiple codons. Hence, the correct answer is Option 1 (Methionine).

Topic: DNA Fingerprinting and Experiments on Genetic Material; Subtopic: Experimental Proof and Key Discoveries

Keyword Definitions:

• DNA Fingerprinting: Technique used to identify individuals based on unique DNA patterns.

• Pneumococcus: Bacterium used by Griffith in transformation experiments.

• E.coli: Model organism used in molecular biology and genetics.

• Bacteriophage: Virus that infects bacteria, crucial in proving DNA as genetic material.

• Meselson and Stahl Experiment: Proved the semi-conservative replication of DNA.

• Alec Jeffreys: Developed the technique of DNA fingerprinting.

Lead Question – (September 2024, Jhajjhar)

Match List-I with List-II

List-I          List-II

1. DNA fingerprinting          I. M. Meselson and F. Stahl

2. Pneumococcus          II. A. Hershey and M. Chase

3. E.coli          III. F. Griffith

4. Bacteriophage          IV. Alec Jeffreys

Choose the correct answer from the options given below:

1. A-IV, B-III, C-II, D-I

2. A-IV, B-III, C-I, D-II

3. A-II, B-III, C-I, D-IV

4. A-III, B-II, C-I, D-IV

Explanation:

The correct matching is A-IV, B-III, C-I, D-II. Alec Jeffreys discovered DNA fingerprinting, Griffith used Pneumococcus to show transformation, Meselson and Stahl demonstrated DNA replication in E.coli, and Hershey-Chase proved DNA as genetic material using bacteriophage. Together, these discoveries form the foundation of molecular genetics. Answer: Option 2

1. Which scientist demonstrated the semi-conservative mode of DNA replication?

1. Watson and Crick

2. Hershey and Chase

3. Meselson and Stahl

4. Avery and MacLeod

Explanation:

Meselson and Stahl demonstrated that DNA replicates in a semi-conservative manner using nitrogen isotope labeling in E.coli. One parental strand is conserved in each daughter molecule. This experiment confirmed Watson and Crick’s model of DNA replication. Answer: Option 3

2. Transformation principle was discovered by:

1. Avery

2. Griffith

3. Hershey

4. Chargaff

Explanation:

Frederick Griffith in 1928 conducted experiments with Pneumococcus bacteria and mice, discovering the “transforming principle” where genetic information passed from heat-killed virulent to non-virulent cells. This established the foundation of molecular genetics. Answer: Option 2

3. Assertion-Reason Type:

Assertion (A): Hershey and Chase used bacteriophages labeled with radioactive isotopes.

Reason (R): They proved that DNA is the genetic material entering bacterial cells.

1. Both A and R are true, and R is the correct explanation.

2. Both A and R are true, but R is not the correct explanation.

3. A is true but R is false.

4. A is false but R is true.

Explanation:

Hershey and Chase labeled protein with S³⁵ and DNA with P³², infecting E.coli. Only DNA entered bacteria, proving it as genetic material. Both assertion and reason are true, and R correctly explains A. Answer: Option 1

4. Which of the following is not correctly matched?

1. DNA fingerprinting – Alec Jeffreys

2. Pneumococcus – Griffith

3. E.coli – Hershey and Chase

4. Bacteriophage – Meselson and Stahl

Explanation:

Bacteriophage was used by Hershey and Chase, not Meselson and Stahl. The latter worked with E.coli on DNA replication. Hence, option 4 is incorrectly matched. Answer: Option 4

5. Fill in the Blanks:

DNA replication follows the ________ model proposed by Watson and Crick.

1. Conservative

2. Dispersive

3. Semi-conservative

4. Random

Explanation:

DNA replication follows the semi-conservative model, meaning one parental strand acts as a template for a new complementary strand. This ensures genetic stability. Meselson and Stahl experimentally confirmed this in E.coli using nitrogen isotopes. Answer: Option 3

6. Choose the correct statements:

Statement I: Hershey and Chase used radioactive sulfur to label DNA.

Statement II: Meselson and Stahl used nitrogen isotopes to prove replication.

1. Only Statement I is correct

2. Only Statement II is correct

3. Both Statements are correct

4. Both Statements are incorrect

Explanation:

Hershey and Chase labeled DNA with P³² and protein with S³⁵, so Statement I is false. Meselson and Stahl used N¹⁵ and N¹⁴ to prove semi-conservative replication, hence Statement II is true. Answer: Option 2

7. The transforming principle was identified as DNA by:

1. Griffith

2. Avery, MacLeod and McCarty

3. Hershey and Chase

4. Chargaff

Explanation:

Avery, MacLeod and McCarty demonstrated that DNA was the transforming principle responsible for heredity. Their work extended Griffith’s observations, confirming that DNA carries genetic information. Answer: Option 2

8. The ratio of purines to pyrimidines in DNA was given by:

1. Watson

2. Chargaff

3. Pauling

4. Franklin

Explanation:

Erwin Chargaff formulated base pairing rules, stating that in DNA, the amount of adenine equals thymine and guanine equals cytosine, maintaining a 1:1 ratio of purines to pyrimidines. Answer: Option 2

9. The double helical model of DNA was proposed by:

1. Watson and Crick

2. Franklin and Wilkins

3. Meselson and Stahl

4. Avery and MacLeod

Explanation:

James Watson and Francis Crick proposed the double helix structure of DNA in 1953 using X-ray diffraction data from Rosalind Franklin. The model revealed antiparallel strands joined by hydrogen bonds. Answer: Option 1

10. In Griffith’s experiment, mice injected with heat-killed virulent and live non-virulent strains died because:

1. Heat-killed bacteria became active

2. Non-virulent bacteria became virulent by transformation

3. Mice had immunity issues

4. DNA destroyed virulence

Explanation:

Griffith’s transformation experiment showed that live non-virulent bacteria acquired genes from heat-killed virulent bacteria, becoming virulent. This established that genetic material could be transferred between cells. Answer: Option 2 

Topic: DNA Replication; Subtopic: Semiconservative Mode of DNA Replication

Keyword Definitions:

• DNA Replication: The process by which DNA makes an identical copy of itself before cell division.

• Semiconservative Replication: Each new DNA molecule contains one parental strand and one newly synthesized strand.

• Autoradiography: A technique that uses radioactive materials to visualize molecules or fragments.

• Vicia faba: The broad bean plant used in experiments by Taylor and colleagues to demonstrate semiconservative DNA replication.

Lead Question – 2024 (Jhajjhar)
Which experimental material was used by Taylor and colleagues to prove that DNA in chromosomes replicates semiconservatively?
1. Vicia faba
2. Pisum sativum
3. Solanum tuberosum
4. Oryza sativa

Explanation: Taylor, Woods, and Hughes demonstrated semiconservative replication in eukaryotes using root tip cells of Vicia faba (broad bean). They incorporated radioactive thymidine and observed its distribution using autoradiography. After one replication, only one chromatid was labeled, and after the second, both chromatids were labeled. Hence, DNA replicates semiconservatively. Correct answer: Option (1).

1. Who proposed the semiconservative model of DNA replication?
1. Hershey and Chase
2. Watson and Crick
3. Meselson and Stahl
4. Taylor and Hughes

Explanation: The semiconservative model of DNA replication was proposed by Watson and Crick in 1953 after discovering the double-helical structure of DNA. It was later experimentally confirmed by Meselson and Stahl in 1958 using E. coli and isotopes of nitrogen (¹⁴N and ¹⁵N). Correct answer: Option (2).

2. Meselson and Stahl used which organism in their experiment to prove DNA replication is semiconservative?
1. Escherichia coli
2. Vicia faba
3. Drosophila melanogaster
4. Saccharomyces cerevisiae

Explanation: Meselson and Stahl used Escherichia coli grown in a medium containing heavy isotope ¹⁵N, then transferred to a ¹⁴N medium. The DNA showed hybrid density after one generation and light density after two, proving semiconservative replication. Correct answer: Option (1).

3. Which enzyme joins Okazaki fragments on the lagging strand during DNA replication?
1. DNA polymerase I
2. DNA ligase
3. Helicase
4. Primase

Explanation: DNA ligase is the enzyme responsible for joining Okazaki fragments by forming phosphodiester bonds between adjacent nucleotides. DNA polymerase I removes RNA primers, helicase unwinds the DNA, and primase synthesizes RNA primers. Correct answer: Option (2).

4. In semiconservative DNA replication, each new DNA molecule consists of:
1. Two new strands
2. Two old strands
3. One old and one new strand
4. One RNA and one DNA strand

Explanation: In semiconservative replication, the parental DNA strands separate and serve as templates for new strand synthesis. Each daughter DNA molecule thus has one parental and one newly synthesized strand, ensuring genetic stability. Correct answer: Option (3).

5. The enzyme responsible for unwinding the DNA helix during replication is:
1. DNA polymerase
2. Helicase
3. Ligase
4. Gyrase

Explanation: Helicase unwinds the DNA double helix by breaking hydrogen bonds between complementary bases. DNA gyrase (a topoisomerase) relieves torsional strain, DNA ligase joins fragments, and DNA polymerase adds nucleotides. Correct answer: Option (2).

6. Which of the following statements about DNA polymerase is correct?
1. It synthesizes DNA in 3′ → 5′ direction.
2. It synthesizes DNA in 5′ → 3′ direction.
3. It joins RNA fragments.
4. It is involved only in transcription.

Explanation: DNA polymerase synthesizes new DNA strands in the 5′ → 3′ direction using a DNA template. It adds nucleotides complementary to the template strand. DNA synthesis proceeds continuously on the leading strand and discontinuously on the lagging strand. Correct answer: Option (2).

7. Assertion-Reason Question:
Assertion (A): DNA replication is bidirectional in eukaryotes.
Reason (R): Both strands act as templates, and replication proceeds from multiple origins of replication.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Explanation: Eukaryotic DNA replication is bidirectional and initiates from multiple origins to ensure rapid duplication of long DNA molecules. Both strands act as templates, forming leading and lagging strands. Therefore, both A and R are true, and R correctly explains A. Correct answer: Option (1).

8. Matching Type Question:
Match the enzymes with their respective functions:
A. Helicase – I. Unwinds DNA helix
B. Ligase – II. Joins Okazaki fragments
C. Primase – III. Synthesizes RNA primers
D. DNA polymerase – IV. Synthesizes new DNA strand

1. A-I, B-II, C-III, D-IV
2. A-II, B-I, C-IV, D-III
3. A-III, B-IV, C-I, D-II
4. A-IV, B-II, C-I, D-III

Explanation: Helicase unwinds DNA, ligase joins Okazaki fragments, primase synthesizes RNA primers, and DNA polymerase extends new DNA strands. Thus, A-I, B-II, C-III, D-IV is the correct matching sequence. Correct answer: Option (1).

9. Fill in the Blanks:
The short DNA fragments synthesized on the lagging strand are called __________.
1. Okazaki fragments
2. DNA primers
3. Exons
4. Replicons

Explanation: The lagging strand is synthesized discontinuously in small segments called Okazaki fragments, later joined by DNA ligase. These fragments are named after Reiji Okazaki, who discovered them during DNA replication studies. Correct answer: Option (1).

10. Choose the Correct Statements:
Statement I: DNA replication is a semiconservative process.
Statement II: In each new DNA molecule, both strands are newly synthesized.
1. Both statements are true.
2. Both statements are false.
3. Statement I is true but Statement II is false.
4. Statement I is false but Statement II is true.

Explanation: DNA replication is semiconservative, meaning one parental strand and one new strand are present in each daughter molecule. Statement II is false because both strands are not new. Hence, Statement I is true, Statement II is false. Correct answer: Option (3).

Topic: Gene Expression and Regulation; Subtopic: Lac Operon Mechanism

Keyword Definitions:

• Lac Operon: A cluster of genes in E. coli responsible for lactose metabolism, regulated by an operator and repressor system.

• i Gene: Regulatory gene that codes for a repressor protein which can bind to the operator region to block transcription.

• Repressor: A protein that inhibits gene expression by binding to the operator sequence.

• Inducer: A molecule (like allolactose) that binds to the repressor and prevents it from binding to the operator, enabling gene transcription.

Lead Question – 2024 (Jhajjhar)
In the lac operon the i gene codes for:
1. Inducer
2. Repressor
3. β-galactosidase
4. Permease

Explanation: The correct answer is Repressor. The i gene of the lac operon encodes a repressor protein that binds to the operator region, preventing transcription of structural genes. When lactose (inducer) is present, it inactivates the repressor, allowing RNA polymerase to transcribe genes responsible for lactose utilization.

1. Single Correct Answer MCQ:
Which of the following enzymes is coded by the lacZ gene?
1. Lactase
2. β-galactosidase
3. Transacetylase
4. Permease

Explanation: The correct answer is β-galactosidase. The lacZ gene of the lac operon encodes this enzyme, which hydrolyzes lactose into glucose and galactose. It plays a crucial role in lactose metabolism in E. coli, acting only when lactose is available as an energy source.

2. Single Correct Answer MCQ:
Which component of the lac operon functions as a switch to turn transcription on or off?
1. Promoter
2. Operator
3. Structural genes
4. i gene

Explanation: The correct answer is Operator. The operator site serves as a regulatory switch where the repressor binds to block RNA polymerase activity. When the repressor is removed by an inducer, the operator becomes accessible, enabling transcription of structural genes necessary for lactose metabolism.

3. Single Correct Answer MCQ:
What acts as an inducer in the lac operon system?
1. Lactose
2. Allolactose
3. Glucose
4. Galactose

Explanation: The correct answer is Allolactose. It is an isomer of lactose formed in small quantities when lactose enters the cell. Allolactose binds to the lac repressor, inactivating it and allowing transcription of genes responsible for lactose metabolism in E. coli.

4. Single Correct Answer MCQ:
Which of the following statements about the lac operon is incorrect?
1. It is an inducible system.
2. It functions only in the presence of glucose.
3. It consists of regulatory and structural genes.
4. It controls lactose metabolism in bacteria.

Explanation: The correct answer is It functions only in the presence of glucose. The lac operon is repressed when glucose is present because bacteria prefer glucose as a carbon source. It is induced only when glucose is absent and lactose is available, making it an inducible system.

5. Assertion–Reason Type:
Assertion (A): Lac operon is switched on in the presence of lactose.
Reason (R): Lactose acts as an inducer and binds to the repressor, preventing its attachment to the operator region.
1. Both A and R are true and R explains A
2. Both A and R are true but R does not explain A
3. A is true but R is false
4. A is false but R is true

Explanation: The correct answer is Both A and R are true and R explains A. Lactose (in its allolactose form) binds to the repressor, causing a conformational change that prevents it from binding to the operator. This derepresses the operon, allowing transcription of genes for lactose metabolism.

6. Matching Type MCQ:
Match the genes of the lac operon with their respective enzymes:
A. lacZ – I. β-galactosidase
B. lacY – II. Permease
C. lacA – III. Transacetylase
Options:
1. A-I, B-II, C-III
2. A-II, B-I, C-III
3. A-III, B-I, C-II
4. A-I, B-III, C-II

Explanation: The correct match is A-I, B-II, C-III. The lacZ gene codes for β-galactosidase, lacY for permease that facilitates lactose entry, and lacA for transacetylase, which detoxifies byproducts. Together, they coordinate lactose utilization when the operon is induced.

7. Fill in the Blanks Type:
In lac operon, the structural genes are transcribed only when _________ binds to the repressor.
1. Allolactose
2. Glucose
3. RNA polymerase
4. Lactase

Explanation: The correct answer is Allolactose. It binds to the repressor protein, altering its shape so that it can no longer attach to the operator region. This allows RNA polymerase to transcribe the operon, initiating synthesis of lactose-metabolizing enzymes.

8. Single Correct Answer MCQ:
Which of the following enzymes is responsible for lactose entry into the bacterial cell?
1. Permease
2. Transacetylase
3. β-galactosidase
4. RNA polymerase

Explanation: The correct answer is Permease. The lacY gene encodes permease, which facilitates lactose transport across the bacterial membrane. Without permease, lactose cannot enter the cell to induce the lac operon and initiate the transcription of metabolic enzymes.

9. Choose the Correct Statements (Statement I & II):
Statement I: The repressor binds to the operator in the absence of an inducer.
Statement II: The repressor binds to the promoter in the presence of an inducer.
1. Both statements are true
2. Both statements are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is true

Explanation: The correct answer is Statement I is true but Statement II is false. The repressor binds to the operator (not promoter) when no inducer is present, blocking transcription. When the inducer binds to the repressor, it detaches from the operator, enabling transcription initiation.

10. Single Correct Answer MCQ:
What happens to the lac operon when both glucose and lactose are present in the medium?
1. Operon is fully active
2. Operon is repressed by catabolite repression
3. Operon is permanently off
4. Only lacY gene is transcribed

Explanation: The correct answer is Operon is repressed by catabolite repression. When glucose is present, cyclic AMP levels fall, preventing CAP binding near the promoter. This suppresses transcription even in the presence of lactose, demonstrating glucose preference over lactose as an energy source.

Topic: Structure of DNA; Subtopic: Experimental Techniques Used in DNA Discovery

Keyword Definitions:

• DNA (Deoxyribonucleic Acid): The molecule that carries genetic information in living organisms.

• Double Helix: The structural arrangement of DNA with two strands coiled around each other.

• X-ray Diffraction: A technique used to determine molecular structures by analyzing X-ray patterns.

• Nucleotides: Building blocks of DNA, composed of a sugar, phosphate group, and nitrogenous base.

• Watson and Crick Model: Proposed the double helix structure of DNA based on X-ray diffraction data.

Lead Question - 2024 (Jhajjhar)
Which of the following technique was used to elucidate the double helix model of DNA?
1. β-radiation
2. Electromagnetic radiation
3. UV-vis spectroscopy
4. X-ray diffraction

Explanation: The double helix structure of DNA was revealed through X-ray diffraction studies conducted by Rosalind Franklin and Maurice Wilkins. The diffraction pattern provided critical insights into DNA’s helical structure, enabling Watson and Crick to propose the correct model in 1953. This technique measures how X-rays scatter when they hit crystallized molecules, revealing atomic arrangement.

1. Which scientist provided X-ray diffraction images crucial for DNA model discovery?
1. Rosalind Franklin
2. Gregor Mendel
3. Alfred Hershey
4. Erwin Chargaff

Explanation: Rosalind Franklin used X-ray diffraction to capture Photo 51, which showed the helical pattern of DNA. This evidence was crucial for Watson and Crick’s double helix model. Franklin’s precision in analyzing DNA fibers made her contribution indispensable to understanding DNA’s three-dimensional structure and molecular organization.

2. The two strands of DNA are held together by:
1. Covalent bonds
2. Ionic bonds
3. Hydrogen bonds
4. Peptide bonds

Explanation: DNA strands are held together by hydrogen bonds between complementary nitrogenous bases. Adenine pairs with thymine via two hydrogen bonds, and guanine pairs with cytosine via three hydrogen bonds. These weak bonds enable the double helix to unwind easily during replication and transcription while maintaining stability of the DNA molecule.

3. Which of the following forms of DNA is most commonly found in cells?
1. A-DNA
2. B-DNA
3. Z-DNA
4. C-DNA

Explanation: The B-DNA form is the most common and biologically relevant conformation of DNA found in living cells. It is a right-handed double helix with about 10 base pairs per turn. B-DNA’s structure is ideal for replication and transcription processes, ensuring accurate genetic information transfer between generations.

4. Assertion-Reason Type Question:
Assertion (A): DNA replication is semi-conservative.
Reason (R): Each daughter DNA molecule has one newly synthesized and one parental strand.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Explanation: The correct answer is option 1. DNA replication is semi-conservative, as demonstrated by the Meselson-Stahl experiment. Each new DNA molecule contains one original (parental) strand and one new strand. This ensures genetic fidelity during cell division and precise transmission of hereditary information in all living organisms.

5. Match the following:
A. Watson and Crick — (i) Semi-conservative replication
B. Meselson and Stahl — (ii) Double helix model
C. Franklin — (iii) X-ray diffraction
D. Chargaff — (iv) Base pairing rule
1. A-(ii), B-(i), C-(iii), D-(iv)
2. A-(iii), B-(ii), C-(i), D-(iv)
3. A-(iv), B-(iii), C-(ii), D-(i)
4. A-(ii), B-(iv), C-(iii), D-(i)

Explanation: The correct match is option 1. Watson and Crick proposed the double helix model; Meselson and Stahl proved semi-conservative replication; Franklin discovered DNA’s helical nature through X-ray diffraction; and Chargaff established base pairing rules. Together, their findings built the foundation of modern molecular genetics.

6. Which nitrogenous base pairs with cytosine in DNA?
1. Thymine
2. Adenine
3. Guanine
4. Uracil

Explanation: In DNA, cytosine pairs with guanine through three hydrogen bonds, ensuring stability and accurate base pairing. Adenine pairs with thymine via two hydrogen bonds. This complementary pairing follows Chargaff’s rule and is essential for replication, maintaining the precise genetic code within the double helix structure of DNA.

7. Fill in the Blank:
The sugar present in DNA is _______.
1. Ribose
2. Deoxyribose
3. Fructose
4. Glucose

Explanation: The correct answer is deoxyribose. DNA’s backbone consists of deoxyribose sugars linked to phosphate groups. Each deoxyribose connects to a nitrogenous base forming a nucleotide. The absence of an oxygen atom at the 2′ position differentiates DNA from RNA, contributing to DNA’s chemical stability and suitability for genetic storage.

8. Which of the following statements about DNA is correct?
1. It contains uracil instead of thymine.
2. It is single-stranded in most organisms.
3. Its two strands are antiparallel.
4. Bases face outward.

Explanation: The correct answer is option 3. DNA strands are antiparallel, meaning one runs 5′→3′ and the other 3′→5′. This orientation facilitates replication and enzyme binding. Bases lie inward forming pairs, while the sugar-phosphate backbone faces outward, giving DNA stability and its characteristic right-handed helical structure.

9. Choose the Correct Statements (Statement I & Statement II):
Statement I: DNA replication occurs in the 5′→3′ direction.
Statement II: DNA polymerase catalyzes the addition of nucleotides at the 5′ end.
1. Both statements are true.
2. Statement I true, II false.
3. Statement I false, II true.
4. Both statements false.

Explanation: The correct answer is option 2. Statement I is true because DNA synthesis occurs in the 5′→3′ direction. Statement II is false because DNA polymerase adds nucleotides only at the 3′ end of the growing strand. This ensures directionality and accuracy in replication.

10. Which scientist discovered that A=T and G≡C in DNA?
1. Rosalind Franklin
2. James Watson
3. Erwin Chargaff
4. Francis Crick

Explanation: Erwin Chargaff discovered that in DNA, adenine equals thymine, and guanine equals cytosine in quantity. These Chargaff’s rules were crucial for Watson and Crick’s double helix model, explaining complementary base pairing that allows faithful DNA replication and transmission of genetic information between generations.

Topic: Genetic Code; Subtopic: Properties of Genetic Code

Keyword Definitions:

• Genetic Code: The sequence of nucleotides in mRNA that determines the amino acid sequence of a protein.

• Codon: A triplet of bases in mRNA that codes for a specific amino acid.

• Degeneracy: A property of the genetic code where more than one codon codes for the same amino acid.

• Universality: The same codons specify the same amino acids across almost all organisms.

• Punctuation: The genetic code has no commas or punctuation marks; it is continuous.

Lead Question – (Jhajjar 2024)
Which of the following is not the characteristic feature of genetic code?
1. The codon is triplet
2. The code is nearly universal
3. The code has punctuations
4. Some amino acids are coded by more than one codon, hence the code is degenerate

Answer: The code has punctuations
Explanation: The genetic code is continuous and comma-less, meaning there are no punctuations or gaps between codons. Each codon is a triplet of bases that specifies one amino acid. The code is nearly universal and degenerate since several codons can code for the same amino acid. Hence, the code has punctuations is not a correct feature.

1. How many codons in total specify amino acids and stop signals?
1. 61 sense codons and 3 stop codons
2. 60 sense codons and 4 stop codons
3. 64 sense codons and no stop codons
4. 62 sense codons and 2 stop codons

Answer: 61 sense codons and 3 stop codons
Explanation: Out of 64 possible codons in the genetic code, 61 specify amino acids and are called sense codons, while 3 (UAA, UAG, and UGA) are stop or termination codons. These stop codons signal the end of translation, preventing addition of further amino acids to the polypeptide chain during protein synthesis.

2. Which codon acts as the start codon during translation?
1. AUG
2. UAA
3. UGA
4. UAG

Answer: AUG
Explanation: AUG functions as the initiation or start codon for protein synthesis. It codes for methionine in eukaryotes and formyl methionine (fMet) in prokaryotes. Translation begins when the ribosome recognizes AUG on mRNA, ensuring correct reading frame and initiation of the polypeptide chain with the proper amino acid sequence.

3. Which of the following codons is not a stop codon?
1. UGA
2. UAG
3. UAA
4. AUG

Answer: AUG
Explanation: The three termination codons are UAA, UAG, and UGA, which stop the translation process. AUG, on the other hand, is a start codon that initiates translation. It ensures the ribosome starts reading mRNA from the correct position and codes for methionine, marking the beginning of the polypeptide sequence.

4. Which of the following best describes the degeneracy of the genetic code?
1. One codon codes for one amino acid
2. Many amino acids are coded by the same codon
3. Many codons may code for the same amino acid
4. Codons overlap each other

Answer: Many codons may code for the same amino acid
Explanation: Degeneracy of the genetic code means that more than one codon can specify the same amino acid. For example, leucine is coded by six different codons. This feature provides genetic robustness, minimizing the effects of mutations by allowing substitutions in the third base without changing the amino acid outcome.

5. Who experimentally confirmed that the genetic code is triplet in nature?
1. Watson and Crick
2. Nirenberg and Khorana
3. Crick, Brenner, and colleagues
4. Beadle and Tatum

Answer: Crick, Brenner, and colleagues
Explanation: Francis Crick, Sydney Brenner, and colleagues demonstrated through frame-shift mutation experiments that the genetic code is read in triplets. Addition or deletion of one or two bases altered the reading frame, whereas insertion or deletion of three bases restored it, proving that codons consist of three nucleotide bases specifying one amino acid.

6. Which of the following statements about the genetic code is correct?
1. The code is overlapping
2. The code is ambiguous
3. The code is commaless
4. The code is species-specific

Answer: The code is commaless
Explanation: The genetic code is continuous and commaless, meaning codons are read sequentially without gaps or punctuation marks. It is also non-overlapping and nearly universal, with each codon specifying a single amino acid. This feature ensures correct reading during translation, maintaining the integrity of protein synthesis in all living organisms.

7. Assertion–Reason Type:
Assertion (A): The genetic code is unambiguous.
Reason (R): Each codon specifies only one amino acid.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, but R is false.
4. A is false, but R is true.

Answer: Both A and R are true, and R is the correct explanation of A
Explanation: The genetic code is unambiguous because every codon specifies only one amino acid, leaving no confusion in translation. For example, UUU always codes for phenylalanine. This precise correspondence ensures accurate protein synthesis. Hence, both assertion and reason are true, and the reason correctly explains the assertion.

8. Matching Type:
Match the following codons with their functions:
A. AUG → 1. Stop codon
B. UGA → 2. Start codon
C. UAG → 3. Termination codon
D. UAA → 4. Termination codon

1. A-2, B-1, C-3, D-4
2. A-2, B-3, C-3, D-4
3. A-2, B-3, C-4, D-1
4. A-1, B-2, C-3, D-4

Answer: A-2, B-3, C-3, D-4
Explanation: AUG is the initiation codon, signaling the start of translation, while UGA, UAG, and UAA are stop or termination codons that signal the end of polypeptide synthesis. These codons ensure accurate translation initiation and termination, preventing formation of incomplete or incorrect protein chains within the cell.

9. Fill in the Blank Type:
The genetic code is said to be ________ because the same codon always codes for the same amino acid in most organisms.

Answer: Universal
Explanation: The genetic code is almost universal, meaning it is shared by nearly all organisms, from bacteria to humans. For instance, the codon UUU codes for phenylalanine in all known species. This universality supports the theory of common ancestry and is crucial in biotechnology for expressing genes across species.

10. Choose the Correct Statements (Statement I & II):
Statement I: The genetic code is overlapping.
Statement II: The genetic code is non-ambiguous.
1. Both statements are correct.
2. Statement I is correct, Statement II is incorrect.
3. Statement I is incorrect, Statement II is correct.
4. Both statements are incorrect.

Answer: Statement I is incorrect, Statement II is correct
Explanation: The genetic code is non-overlapping, meaning each base is part of only one codon. It is also unambiguous, as each codon codes for only one amino acid. This ensures clarity during translation and prevents confusion in protein synthesis, maintaining the fidelity of genetic information expression across organisms.

Topic: Transcription in Eukaryotes; Subtopic: RNA Polymerases and Transcriptional Control

Keyword Definitions:
• RNA Polymerase: Enzyme responsible for synthesizing RNA from DNA template.
• TATA Box: DNA sequence acting as a promoter for transcription initiation.
• Rho Factor: Protein involved in terminating transcription in prokaryotes.
• snRNPs: Small nuclear ribonucleoproteins that assist in splicing of pre-mRNA.
• Promoter: Specific DNA region where RNA polymerase binds to initiate transcription.
• Splicing: Process of removing introns and joining exons to form mature mRNA.

Lead Question (2024):
Match List I with List II
A. RNA polymerase III I. snRNPs
B. Termination of transcription II. Promotor
C. Splicing of Exons III. Rho factor
D. TATA box IV. SnRNAs, tRNA
Choose the correct answer from the options given below:
(1) A-III, B-II, C-IV, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-III, C-I, D-II
(4) A-I, B-III, C-IV, D-II

Explanation: The correct answer is (3) A-IV, B-III, C-I, D-II. RNA polymerase III synthesizes small RNAs like tRNA and snRNA. Rho factor assists in termination of transcription in prokaryotes. snRNPs help in splicing of pre-mRNA exons, while the TATA box serves as a promoter site for transcription initiation by RNA polymerase II.

1. Which RNA polymerase synthesizes mRNA in eukaryotes?
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) DNA polymerase

The correct answer is RNA polymerase II. In eukaryotes, RNA polymerase II is responsible for transcribing DNA into messenger RNA (mRNA), which carries genetic information from the nucleus to the cytoplasm for protein synthesis. It recognizes promoters like the TATA box for transcription initiation.

2. What is the role of RNA polymerase I in eukaryotic cells?
(1) Synthesis of mRNA
(2) Synthesis of tRNA
(3) Synthesis of rRNA
(4) Synthesis of snRNA

The correct answer is Synthesis of rRNA. RNA polymerase I transcribes the genes coding for ribosomal RNA (except 5S rRNA), which forms the structural and functional components of ribosomes essential for translation in eukaryotic cells.

3. The rho factor in transcription is involved in:
(1) Initiation of transcription
(2) Elongation of mRNA chain
(3) Termination of transcription
(4) Splicing of RNA

The correct answer is Termination of transcription. The rho factor is a protein in prokaryotes that binds to the RNA transcript and moves along it until it reaches the RNA polymerase, causing dissociation and termination of transcription efficiently.

4. What is the function of the TATA box in transcription?
(1) DNA replication initiation
(2) Transcription termination
(3) Promoter recognition by RNA polymerase II
(4) RNA splicing

The correct answer is Promoter recognition by RNA polymerase II. The TATA box, found around 25 base pairs upstream of the transcription start site, helps position RNA polymerase II correctly for transcription initiation by attracting transcription factors like TATA-binding protein (TBP).

5. The splicing of pre-mRNA involves removal of:
(1) Exons
(2) Introns
(3) Both exons and introns
(4) DNA sequences

The correct answer is Introns. Splicing removes non-coding intron sequences from pre-mRNA and joins the coding exons together to produce mature mRNA. This process is catalyzed by a complex known as the spliceosome, which includes snRNPs.

6. Which RNA polymerase is responsible for transcription of 5S rRNA in eukaryotes?
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) RNA polymerase IV

The correct answer is RNA polymerase III. It transcribes genes for tRNA, 5S rRNA, and certain small nuclear RNAs. It operates in the nucleus and recognizes internal promoter sequences distinct from those recognized by RNA polymerase II.

7. Assertion-Reason Question:
Assertion (A): Splicing is necessary to form functional mRNA in eukaryotes.
Reason (R): Introns must be removed and exons joined to produce a continuous coding sequence.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.

Both statements are true, and R correctly explains A. Splicing removes introns from pre-mRNA and connects exons, ensuring a continuous coding sequence for translation. Without splicing, mRNA cannot produce functional proteins, demonstrating the precision of eukaryotic gene expression mechanisms.

8. Matching Type Question:
Match the following transcriptional components with their roles:
A. Promoter — (i) Site for RNA polymerase binding
B. Introns — (ii) Non-coding regions removed
C. Exons — (iii) Coding regions retained
D. Terminator — (iv) Stops RNA synthesis
(1) A-i, B-ii, C-iii, D-iv
(2) A-ii, B-iv, C-i, D-iii
(3) A-iv, B-iii, C-ii, D-i
(4) A-iii, B-i, C-iv, D-ii

The correct match is (1) A-i, B-ii, C-iii, D-iv. The promoter provides binding sites for RNA polymerase, introns are removed during splicing, exons are retained as coding regions, and the terminator sequence ends transcription, ensuring precise RNA synthesis control.

9. Fill in the Blanks:
The enzyme responsible for the synthesis of tRNA in eukaryotes is ________.
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) DNA polymerase

The correct answer is RNA polymerase III. This enzyme synthesizes tRNA molecules essential for translation. It also transcribes small RNAs like 5S rRNA and snRNA, showing its diverse role in non-coding RNA formation within the nucleus.

10. Choose the Correct Statements (Statement I & Statement II):
Statement I: In eukaryotes, RNA polymerase II transcribes genes for mRNA.
Statement II: RNA polymerase II is also responsible for rRNA synthesis.
(1) Both statements are true.
(2) Both statements are false.
(3) Statement I is true, but Statement II is false.
(4) Statement I is false, but Statement II is true.

The correct answer is (3) Statement I is true, but Statement II is false. RNA polymerase II synthesizes mRNA and some snRNAs, whereas rRNA synthesis is primarily the function of RNA polymerase I, showing specialized roles of RNA polymerases in eukaryotic transcription.

Topic: Transcription; Subtopic: DNA Dependent RNA Polymerase and RNA Synthesis

Keyword Definitions:

• DNA dependent RNA polymerase: Enzyme that synthesizes RNA using DNA as a template strand.

• Template strand: The DNA strand that directs the sequence of nucleotides in RNA synthesis.

• Transcription: The process of synthesizing RNA from DNA.

• Codon: A sequence of three nucleotides that codes for an amino acid.

• Complementary base pairing: A-T and G-C in DNA; A-U and G-C in RNA.

Lead Question – 2024
Which one is the correct product of DNA dependent RNA polymerase to the given template?
3’TACATGGCAAATATCCATTCA5’
(1) 5’AUGUAAAGUUUAUAGGUAAGU3’
(2) 5’AUGUACCGUUUAUAGGGAAGU3’
(3) 5’ATGTACCGTTTATAGGTAAGT3’
(4) 5’AUGUACCGUUUAUAGGUAAGU3’
Answer and Explanation:
The correct answer is (4). RNA is synthesized complementary and antiparallel to the DNA template. Here, A pairs with U, T with A, C with G, and G with C. Thus, the RNA transcript formed from 3’TACATGGCAAATATCCATTCA5’ is 5’AUGUACCGUUUAUAGGUAAGU3’. This RNA can be later translated into protein during translation.

1. Which of the following processes occurs in the nucleus of eukaryotic cells?
(1) Translation
(2) Transcription
(3) Replication of ribosomes
(4) Protein modification
Answer and Explanation:
The correct answer is (2). Transcription occurs inside the nucleus where DNA serves as a template to synthesize mRNA using RNA polymerase. The newly formed pre-mRNA undergoes splicing and modification before leaving the nucleus for translation in the cytoplasm. This ensures gene expression in a regulated manner.

2. The enzyme responsible for removing RNA primers during replication is:
(1) DNA polymerase I
(2) DNA polymerase III
(3) DNA ligase
(4) Primase
Answer and Explanation:
The correct answer is (1). DNA polymerase I removes RNA primers using its 5′→3′ exonuclease activity and replaces them with DNA nucleotides. DNA ligase then seals the remaining gaps. This activity is crucial for producing a continuous DNA strand during replication and ensuring genome stability.

3. The direction of RNA synthesis is always:
(1) 3′→5′
(2) 5′→3′
(3) Both directions
(4) Random
Answer and Explanation:
The correct answer is (2). RNA synthesis always proceeds in the 5′→3′ direction because nucleotides are added to the 3′-OH group of the growing RNA chain. The DNA template is read in the 3′→5′ direction, ensuring that the RNA transcript is complementary and antiparallel to the DNA strand.

4. Which of the following statements is true about RNA polymerase in prokaryotes?
(1) It has no σ factor
(2) It synthesizes proteins directly
(3) It synthesizes all types of RNA
(4) It needs a primer
Answer and Explanation:
The correct answer is (3). In prokaryotes, a single type of RNA polymerase synthesizes all RNA types—mRNA, tRNA, and rRNA. It recognizes the promoter region with the help of σ factor and initiates transcription without requiring a primer. This enzyme is essential for bacterial gene expression.

5. Which of the following is not a component of a transcription unit?
(1) Promoter
(2) Terminator
(3) Enhancer
(4) Structural gene
Answer and Explanation:
The correct answer is (3). A transcription unit consists of a promoter, structural gene, and terminator. Enhancers are regulatory DNA sequences that enhance transcription efficiency but are not a part of the transcription unit itself. They function by interacting with activator proteins to increase transcription rate.

6. Which strand of DNA has the same sequence as the RNA transcript (except for T replaced by U)?
(1) Coding strand
(2) Template strand
(3) Complementary strand
(4) Antisense strand
Answer and Explanation:
The correct answer is (1). The coding strand has the same nucleotide sequence as the RNA transcript except that thymine (T) in DNA is replaced with uracil (U) in RNA. The template strand serves as the actual template during transcription and is complementary to the coding strand.

7. Assertion-Reason Type:
Assertion (A): In transcription, RNA polymerase binds to the promoter region of DNA.
Reason (R): The promoter region codes for proteins.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, R is false.
(4) A is false, R is true.
Answer and Explanation:
The correct answer is (3). The promoter region does not code for proteins but acts as a binding site for RNA polymerase to initiate transcription. It contains specific sequences like TATA box that signal the start point, ensuring accurate transcription initiation.

8. Matching Type:
Match the following:
A. mRNA → (i) Carries amino acids
B. tRNA → (ii) Template for protein synthesis
C. rRNA → (iii) Forms ribosomal structure
(1) A-(ii), B-(i), C-(iii)
(2) A-(iii), B-(ii), C-(i)
(3) A-(i), B-(iii), C-(ii)
(4) A-(ii), B-(iii), C-(i)
Answer and Explanation:
The correct answer is (1). mRNA carries genetic information from DNA to ribosomes, tRNA carries amino acids to the ribosome, and rRNA forms the core structural framework of ribosomes. Together, they ensure accurate translation and protein synthesis in cells.

9. Fill in the Blanks Type:
The RNA strand synthesized during transcription is ______ to the DNA template strand.
(1) Similar
(2) Identical
(3) Complementary
(4) Random
Answer and Explanation:
The correct answer is (3). The RNA strand is complementary and antiparallel to the DNA template strand. This ensures that the genetic code is correctly transcribed from DNA to RNA. Complementary base pairing maintains fidelity during transcription, ensuring proper protein synthesis downstream.

10. Choose the Correct Statements:
Statement I: RNA polymerase requires a primer to start transcription.
Statement II: RNA polymerase can start RNA synthesis de novo.
(1) Both statements are true
(2) Both statements are false
(3) Statement I true, Statement II false
(4) Statement I false, Statement II true
Answer and Explanation:
The correct answer is (4). RNA polymerase does not require a primer to begin transcription; it can initiate RNA synthesis de novo by recognizing the promoter region. This is a major difference from DNA polymerase, which needs a primer for replication.

Topic: DNA Replication; Subtopic: Enzymes and Mechanism of Replication in Prokaryotes

Keyword Definitions:

• DNA Replication: The process by which a DNA molecule produces two identical copies of itself before cell division.

• DNA Polymerase: An enzyme that synthesizes new DNA strands by adding nucleotides to the 3′-end, always in the 5′ → 3′ direction.

• Okazaki Fragments: Short DNA segments synthesized discontinuously on the lagging strand during replication.

• Leading and Lagging Strands: The leading strand is synthesized continuously, while the lagging strand is synthesized in fragments.

• Replication Fork: The Y-shaped structure formed by unwinding DNA where replication proceeds.

Lead Question (September 2024)
Which of the following statements is correct regarding the process of replication in E. coli?
(1) The DNA-dependent RNA polymerase catalyses polymerization in one direction, that is 3′ → 5′.
(2) The DNA-dependent DNA polymerase catalyses polymerization in both 5′ → 3′ and 3′ → 5′ directions.
(3) The DNA-dependent DNA polymerase catalyses polymerization in 5′ → 3′ direction.
(4) The DNA-dependent DNA polymerase catalyses polymerization in one direction, that is 3′ → 5′.

Explanation: The correct answer is (3). DNA-dependent DNA polymerase catalyses the synthesis of new DNA strands only in the 5′ → 3′ direction. The enzyme adds deoxyribonucleotides to the free 3′-OH end of the growing chain, using the parent strand as a template. This unidirectional synthesis ensures accurate replication in E. coli and other organisms.

1. Which enzyme is responsible for joining Okazaki fragments during DNA replication?
(1) DNA Polymerase I
(2) DNA Ligase
(3) DNA Gyrase
(4) Helicase

Explanation: The correct answer is (2). DNA ligase seals the gaps between Okazaki fragments on the lagging strand by forming phosphodiester bonds. This enzyme ensures that discontinuous DNA segments become a continuous strand, completing replication smoothly and maintaining genetic integrity across generations of cells.

2. Which enzyme unwinds the DNA double helix at the replication fork?
(1) Helicase
(2) DNA Ligase
(3) RNA Primase
(4) DNA Polymerase III

Explanation: The correct answer is (1). Helicase breaks hydrogen bonds between complementary bases of DNA, creating a replication fork. This unwinding allows DNA polymerase to access single-stranded templates for synthesis, ensuring accurate and efficient replication initiation in prokaryotic and eukaryotic systems.

3. Assertion–Reason Question:
Assertion (A): DNA replication in E. coli is bidirectional.
Reason (R): Two replication forks proceed in opposite directions from the origin of replication.
(1) Both A and R are true, and R is the correct explanation of A
(2) Both A and R are true, but R is not the correct explanation of A
(3) A is true, but R is false
(4) A is false, but R is true

Explanation: The correct answer is (1). In E. coli, replication starts at the origin (OriC) and proceeds bidirectionally with two replication forks moving in opposite directions. This mechanism allows faster and efficient duplication of the circular chromosome within the bacterial cell cycle duration.

4. Fill in the Blanks:
Okazaki fragments are formed on the __________ strand during DNA replication.
(1) Leading
(2) Lagging
(3) Template
(4) Continuous

Explanation: The correct answer is (2). The lagging strand is synthesized discontinuously as Okazaki fragments, each initiated by RNA primers. DNA ligase later joins these fragments into a continuous strand. This occurs because DNA polymerase can only extend in the 5′ → 3′ direction, opposite to the fork’s movement.

5. Which of the following statements about DNA polymerase is true?
(1) It can initiate synthesis of new strands independently.
(2) It requires an RNA primer with a free 3′-OH group.
(3) It adds nucleotides at the 5′-end.
(4) It functions only in transcription.

Explanation: The correct answer is (2). DNA polymerase cannot initiate DNA synthesis de novo. It requires an RNA primer that provides a free 3′-OH group for nucleotide addition. Primase synthesizes these primers, enabling DNA polymerase III to extend the chain during replication initiation in prokaryotes.

6. Choose the Correct Statements:
Statement I: The leading strand is synthesized continuously.
Statement II: The lagging strand is synthesized discontinuously.
(1) Both statements are correct
(2) Only Statement I is correct
(3) Only Statement II is correct
(4) Both statements are incorrect

Explanation: The correct answer is (1). The leading strand is synthesized continuously toward the replication fork, while the lagging strand forms short Okazaki fragments synthesized away from the fork. This difference arises because DNA polymerase acts only in the 5′ → 3′ direction, making replication semi-discontinuous.

7. Which enzyme removes RNA primers during replication in E. coli?
(1) DNA Polymerase I
(2) DNA Polymerase III
(3) DNA Ligase
(4) Primase

Explanation: The correct answer is (1). DNA Polymerase I removes RNA primers using its 5′ → 3′ exonuclease activity and fills the resulting gaps with DNA nucleotides. This enzyme ensures that the replicated DNA strand contains only deoxyribonucleotides, maintaining accuracy and continuity of the genome.

8. Matching Type Question:
Match the enzymes involved in replication with their functions:
A. Helicase  I. Joins Okazaki fragments
B. Primase  II. Synthesizes RNA primer
C. DNA Ligase  III. Unwinds DNA helix
D. DNA Polymerase I  IV. Removes RNA primers
(1) A–III, B–II, C–I, D–IV
(2) A–I, B–IV, C–III, D–II
(3) A–IV, B–I, C–II, D–III
(4) A–III, B–I, C–IV, D–II

Explanation: The correct answer is (1). Helicase unwinds the DNA double helix, Primase synthesizes RNA primers, DNA Ligase joins Okazaki fragments, and DNA Polymerase I removes primers and fills gaps. Each enzyme performs a specific function vital for accurate DNA replication and strand completion.

9. Which of the following enzymes relieves supercoiling tension during replication?
(1) DNA Ligase
(2) Helicase
(3) Topoisomerase
(4) DNA Polymerase III

Explanation: The correct answer is (3). Topoisomerase (DNA gyrase in prokaryotes) relieves torsional stress created ahead of the replication fork by introducing temporary breaks in the DNA strand. It prevents supercoiling, ensuring smooth progression of the replication machinery along the DNA template.

10. DNA replication is called semiconservative because:
(1) Each daughter molecule contains both parental strands.
(2) Each daughter molecule has one old and one new strand.
(3) Both strands are newly synthesized.
(4) Only one strand acts as a template.

Explanation: The correct answer is (2). Meselson and Stahl’s experiment demonstrated that DNA replication is semiconservative. Each daughter DNA molecule consists of one parental and one newly synthesized strand, preserving the original sequence information for genetic stability across generations.

Topic: Experiments in Molecular Genetics; Subtopic: Key Discoveries in DNA and Gene Regulation

Keyword Definitions:

• Transformation: The process by which genetic material from one organism is transferred to another, changing its genotype and phenotype.

• Lac Operon: A gene regulatory system in E. coli that controls lactose metabolism, discovered by Jacob and Monod.

• Genetic Code: The set of nucleotide triplets (codons) that determine amino acid sequences in protein synthesis.

• Semi-conservative replication: The mechanism of DNA replication where each new molecule has one parental and one new strand.

• DNA experiments: Pioneering studies that revealed DNA’s role as the hereditary material and its mode of expression and replication.

Lead Question – 2024

Match List I with List II
List I      List II
A. Frederick Griffith   I. Genetic code
B. Francois Jacob & Jacques Monod   II. Semi–conservative mode of DNA replication
C. Har Gobind Khorana   III. Transformation
D. Meselson & Stahl   IV. Lac operon
Choose the correct answer from the options given below:
(1) A–III, B–IV, C–I, D–II
(2) A–II, B–III, C–IV, D–I
(3) A–IV, B–I, C–II, D–III
(4) A–III, B–II, C–I, D–IV

Explanation: Frederick Griffith discovered transformation in Streptococcus pneumoniae. Jacob and Monod proposed the lac operon model explaining gene regulation. Har Gobind Khorana decoded the genetic code, and Meselson & Stahl proved the semi-conservative mode of DNA replication using 15N isotope labeling. Hence, the correct answer is option (1) A–III, B–IV, C–I, D–II.

1. Who demonstrated that DNA replication is semi-conservative?
(1) Hershey and Chase
(2) Meselson and Stahl
(3) Griffith and Avery
(4) Watson and Crick

Meselson and Stahl proved DNA replicates semi-conservatively by using 15N-labeled DNA in E. coli. After replication in 14N medium, they found hybrid DNA in the first generation and light and hybrid DNA in the second, confirming each new molecule carries one parental and one new strand. Hence, option (2) is correct.

2. Griffith’s experiment demonstrated the phenomenon of:
(1) Transduction
(2) Transformation
(3) Translation
(4) Transcription

Frederick Griffith’s experiment on Streptococcus pneumoniae showed transformation, where non-virulent R strain bacteria became virulent when mixed with heat-killed S strain. This indicated a “transforming principle” that later proved to be DNA. Therefore, the correct answer is option (2) Transformation.

3. The lac operon regulates:
(1) DNA replication
(2) Protein degradation
(3) Lactose metabolism in bacteria
(4) RNA transport

The lac operon in E. coli controls lactose metabolism through inducible enzyme synthesis. When lactose is absent, the repressor binds to the operator region. When lactose is present, it binds to the repressor, enabling transcription of structural genes. Hence, the correct answer is option (3) Lactose metabolism in bacteria.

4. The scientists who deciphered the genetic code were:
(1) Hershey and Chase
(2) Watson and Crick
(3) Har Gobind Khorana and Marshall Nirenberg
(4) Jacob and Monod

Har Gobind Khorana and Marshall Nirenberg elucidated the genetic code by synthesizing artificial RNA sequences to determine codon–amino acid relationships. Their work explained how triplet codons in mRNA specify amino acids during translation. Hence, option (3) Har Gobind Khorana and Marshall Nirenberg is correct.

5. Which experiment confirmed DNA as the genetic material?
(1) Griffith’s experiment
(2) Avery, MacLeod, and McCarty experiment
(3) Meselson-Stahl experiment
(4) Hershey-Chase experiment

Avery, MacLeod, and McCarty proved DNA is the genetic material by showing only DNA from virulent bacteria could transform non-virulent strains. Treatment with DNase destroyed transformation, confirming DNA’s role in heredity. Hence, the correct answer is option (2) Avery, MacLeod, and McCarty experiment.

6. Which of the following scientists proposed the operon model?
(1) Watson and Crick
(2) Beadle and Tatum
(3) Jacob and Monod
(4) Meselson and Stahl

Francois Jacob and Jacques Monod proposed the operon model to explain gene regulation in prokaryotes. The lac operon describes how structural genes are controlled by regulatory genes, an operator, and a promoter. Hence, option (3) Jacob and Monod is correct.

Assertion–Reason Type Question

7. Assertion (A): Lac operon is an inducible system.
Reason (R): The presence of lactose induces expression of genes responsible for its metabolism.

(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true

The lac operon is an inducible system, activated only in the presence of lactose. Lactose acts as an inducer by binding to the repressor, allowing transcription of lac genes. Hence, both Assertion and Reason are true, and R correctly explains A. Correct answer: (1).

Matching Type Question

Match the scientist with the discovery:
A. Watson & Crick – I. Double helix model
B. Hershey & Chase – II. DNA as genetic material
C. Griffith – III. Transformation
D. Khorana – IV. Genetic code
(1) A–I, B–II, C–III, D–IV
(2) A–II, B–III, C–IV, D–I
(3) A–III, B–IV, C–I, D–II
(4) A–IV, B–I, C–II, D–III

Watson & Crick proposed the double helix model, Hershey & Chase confirmed DNA as the genetic material, Griffith discovered transformation, and Khorana deciphered the genetic code. Hence, the correct answer is option (1) A–I, B–II, C–III, D–IV.

Fill in the Blanks Question

The genetic code is ______ and ______ in nature.

(1) Universal and degenerate
(2) Ambiguous and overlapping
(3) Species-specific and linear
(4) Unique and reversible

The genetic code is universal and degenerate. Universal means it is shared by nearly all organisms, while degeneracy means multiple codons can code for the same amino acid, ensuring mutation tolerance. Hence, option (1) is correct.

Choose the Correct Statements (Statement I & II)

Statement I: Transformation demonstrates that DNA is the genetic material.
Statement II: Lac operon model explains gene regulation in eukaryotes.

(1) Both statements are true
(2) Statement I true, Statement II false
(3) Statement I false, Statement II true
(4) Both statements are false

Transformation experiments demonstrated DNA as the genetic material. However, the lac operon model explains gene regulation in prokaryotes like E. coli, not in eukaryotes. Hence, Statement I is true and Statement II is false. The correct option is (2).

Topic: Gene Structure and Function; Subtopic: Transcription Unit

Keyword Definitions:

• Transcription Unit: A segment of DNA that is transcribed into RNA and includes promoter, structural gene, and terminator.

• Promoter: A DNA sequence where RNA polymerase binds to initiate transcription.

• Structural Gene: A gene that codes for a functional RNA or protein.

• Terminator: A DNA sequence that signals the end of transcription.

• Operator Gene: A regulatory sequence where repressors bind to control gene expression.

• Inducer: A molecule that initiates transcription by inactivating repressors.

• Repressor: A protein that inhibits transcription by binding to the operator.

Lead Question – 2024

A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and downstream end:

(1) Structural gene, Transposons, Operator gene
(2) Inducer, Repressor, Structural gene
(3) Promoter, Structural gene, Terminator
(4) Repressor, Operator gene, Structural gene

Explanation: A transcription unit is the basic DNA segment transcribed into RNA. It consists of a promoter, which recruits RNA polymerase, a structural gene encoding the RNA or protein, and a terminator signaling the end of transcription. Promoter is upstream, terminator downstream, and the structural gene lies in between. Operator genes, inducers, repressors, and transposons are regulatory or mobile elements, not part of the canonical transcription unit. Understanding this layout is crucial for studying gene expression, regulation, and transcription mechanisms. (Answer: 3)

1. Single Correct Answer:
Which region of a transcription unit initiates transcription?
(1) Structural gene
(2) Promoter
(3) Terminator
(4) Operator

Explanation: The promoter is the DNA sequence where RNA polymerase binds to begin transcription. Structural gene codes for RNA or protein, terminator signals transcription end, and operator regulates expression. Accurate identification of the promoter is essential for understanding transcription initiation and gene regulation processes. (Answer: 2)

2. Single Correct Answer:
Which part of the transcription unit signals the end of RNA synthesis?
(1) Promoter
(2) Terminator
(3) Structural gene
(4) Inducer

Explanation: The terminator is the DNA sequence that signals RNA polymerase to stop transcription. The promoter initiates transcription, structural gene encodes RNA or protein, and inducers regulate expression. Correct identification of the terminator ensures understanding of transcription termination in molecular biology. (Answer: 2)

3. Single Correct Answer:
The structural gene in a transcription unit:
(1) Recruits RNA polymerase
(2) Encodes RNA or protein
(3) Signals transcription termination
(4) Binds repressor protein

Explanation: The structural gene is the segment transcribed into RNA and translated into a protein if applicable. The promoter recruits RNA polymerase, terminator ends transcription, and repressors bind to operator sequences. Understanding the structural gene's function is essential for analyzing gene expression and protein synthesis. (Answer: 2)

4. Single Correct Answer:
Which regulatory element can bind a repressor to control transcription?
(1) Promoter
(2) Terminator
(3) Operator
(4) Structural gene

Explanation: The operator is a DNA sequence that binds repressors to regulate transcription. Promoter initiates RNA synthesis, terminator ends transcription, and structural gene codes for RNA/protein. Operator-repressor interactions are central to gene expression control in prokaryotic operons. (Answer: 3)

5. Single Correct Answer:
Which molecule can inactivate a repressor to allow transcription?
(1) Promoter
(2) Inducer
(3) Terminator
(4) Operator

Explanation: Inducers are small molecules that inactivate repressors, enabling RNA polymerase to transcribe the structural gene. Promoters initiate transcription, terminators stop it, and operators are binding sites for repressors. Inducers play a critical role in gene regulation by modulating transcriptional activity. (Answer: 2)

6. Assertion-Reason:
Assertion (A): Promoter is always upstream of the structural gene.
Reason (R): Promoter provides binding site for RNA polymerase to initiate transcription.
(1) Both A and R are true, R explains A
(2) Both A and R are true, R does not explain A
(3) A true, R false
(4) A false, R true

Explanation: The promoter lies upstream of the structural gene and recruits RNA polymerase to initiate transcription. Both the assertion and reason are true, and the reason explains the assertion accurately. Proper understanding of promoter location and function is essential for studying transcription regulation. (Answer: 1)

7. Matching Type:
Match List I (DNA region) with List II (Function):
A. Promoter – (i) Encodes RNA/protein
B. Structural gene – (ii) Signals transcription end
C. Terminator – (iii) Initiates transcription
Options:
(1) A-iii, B-i, C-ii
(2) A-i, B-iii, C-ii
(3) A-ii, B-iii, C-i
(4) A-iii, B-ii, C-i

Explanation: The promoter initiates transcription, the structural gene encodes RNA or protein, and the terminator signals transcription end. Correct matching is crucial for understanding transcription units and their roles in gene expression. (Answer: 1)

8. Fill in the Blanks:
A transcription unit begins at the __________ and ends at the __________.
(1) Promoter, Terminator
(2) Structural gene, Promoter
(3) Terminator, Operator
(4) Operator, Structural gene

Explanation: Transcription starts at the promoter and ends at the terminator. The structural gene lies between them and is transcribed into RNA. Operators and repressors are regulatory elements, not part of the canonical transcription unit. Understanding start and end sites is key in molecular biology. (Answer: 1)

9. Single Correct Answer:
Which region is not part of the standard transcription unit?
(1) Promoter
(2) Structural gene
(3) Terminator
(4) Transposon

Explanation: Transposons are mobile DNA elements and not part of the canonical transcription unit. The standard unit includes promoter, structural gene, and terminator. Promoters and terminators define transcription boundaries, while structural genes encode RNA/protein. Understanding distinctions between functional and mobile elements is important for gene regulation studies. (Answer: 4)

10. Choose the Correct Statements:
Statement I: Promoter recruits RNA polymerase to start transcription.
Statement II: Terminator signals the end of transcription.
(1) Both statements are true
(2) Both statements are false
(3) Statement I true, Statement II false
(4) Statement I false, Statement II true

Explanation: Promoter and terminator are essential regions of a transcription unit. The promoter recruits RNA polymerase for initiation, and the terminator signals transcription termination. Both statements accurately describe transcription processes. Correct identification of these regions is vital for understanding gene expression mechanisms. (Answer: 1)

Topic: Gene Regulation and Lac Operon; Subtopic: Transport and Metabolism of Lactose

Keyword Definitions:

• Lactose: A disaccharide sugar found in milk, composed of glucose and galactose units.

• Permease: A membrane protein that facilitates the transport of specific molecules like lactose into bacterial cells.

• Beta-galactosidase: Enzyme that hydrolyzes lactose into glucose and galactose.

• Polymerase: Enzyme that synthesizes DNA or RNA strands from nucleotides.

• Acetylase: Enzyme that transfers acetyl groups to substrates, not involved in lactose transport.

Lead Question – 2024

The lactose present in the growth medium of bacteria is transported to the cell by the action of:

(1) Acetylase
(2) Permease
(3) Polymerase
(4) Beta-galactosidase

Explanation: Lactose transport into bacterial cells is mediated by the enzyme permease, a membrane-bound protein that facilitates the entry of lactose from the medium into the cytoplasm. Once inside, beta-galactosidase hydrolyzes lactose into glucose and galactose for metabolism. Acetylase and polymerase are unrelated to lactose transport. Permease is crucial for lac operon function, ensuring lactose availability for enzymatic processing and gene regulation. (Answer: 2)

1. Which enzyme cleaves lactose into glucose and galactose?
(1) Permease
(2) Polymerase
(3) Beta-galactosidase
(4) Acetylase

Explanation: Beta-galactosidase hydrolyzes lactose into glucose and galactose inside bacterial cells, enabling metabolism and energy production. Permease only transports lactose, polymerase synthesizes nucleic acids, and acetylase transfers acetyl groups to molecules. Beta-galactosidase is essential for lac operon expression and bacterial utilization of lactose. (Answer: 3)

2. Which gene encodes lactose permease in E. coli?
(1) lacZ
(2) lacY
(3) lacA
(4) lacI

Explanation: The lacY gene encodes lactose permease in E. coli, responsible for transporting lactose into the cytoplasm. lacZ encodes beta-galactosidase, lacA encodes transacetylase, and lacI encodes the repressor protein. lacY is crucial for lac operon regulation and bacterial lactose utilization. (Answer: 2)

3. Which of the following is a function of beta-galactosidase?
(1) Transport lactose
(2) Hydrolyze lactose
(3) Synthesize DNA
(4) Transfer acetyl group

Explanation: Beta-galactosidase catalyzes hydrolysis of lactose into glucose and galactose for metabolism in bacteria. It does not transport lactose, synthesize DNA, or transfer acetyl groups. Its activity is regulated by the lac operon, ensuring enzyme production only when lactose is present. (Answer: 2)

4. What is the role of permease in lac operon?
(1) Hydrolyzes lactose
(2) Represses transcription
(3) Transports lactose into the cell
(4) Acetylates lactose

Explanation: Permease is a membrane protein that transports lactose into the bacterial cell, enabling subsequent metabolism. It is encoded by lacY and is critical for lac operon function. Hydrolysis is performed by beta-galactosidase, repression by lacI, and acetylation by lacA. Permease ensures lactose availability for enzymatic breakdown. (Answer: 3)

5. Which enzyme is not involved in lac operon function?
(1) Beta-galactosidase
(2) Permease
(3) Polymerase
(4) Acetylase

Explanation: Polymerase synthesizes RNA or DNA and is not directly involved in lac operon enzymatic activity. Beta-galactosidase, permease, and acetylase are encoded by lacZ, lacY, and lacA respectively. Polymerase does transcribe lac operon genes but is not part of the lactose metabolism pathway. (Answer: 3)

6. Single Correct Answer:
Which molecule induces the lac operon in E. coli?
(1) Glucose
(2) Lactose
(3) Beta-galactosidase
(4) Permease

Explanation: Lactose acts as the inducer of the lac operon by binding to the lac repressor, releasing it from the operator site, allowing transcription of lacZ, lacY, and lacA. Glucose represses the operon through catabolite repression. Beta-galactosidase and permease are products, not inducers. (Answer: 2)

7. Assertion-Reason:
Assertion (A): Permease is essential for lactose utilization in bacteria.
Reason (R): It transports lactose across the cytoplasmic membrane into the cell.
(1) Both A and R are true, R explains A
(2) Both A and R are true, R does not explain A
(3) A true, R false
(4) A false, R true

Explanation: Permease transports lactose into the cell, enabling hydrolysis and metabolism by beta-galactosidase. Without permease, lactose cannot enter the cytoplasm efficiently, making it essential for utilization. Both Assertion and Reason are correct, and Reason correctly explains Assertion. (Answer: 1)

8. Matching Type:
Match the lac operon components with their function:
A. lacZ – (i) Transport lactose
B. lacY – (ii) Hydrolyze lactose
C. lacA – (iii) Acetylation
D. lacI – (iv) Repressor protein
Options:
(1) A–ii, B–i, C–iii, D–iv
(2) A–i, B–ii, C–iv, D–iii
(3) A–iii, B–iv, C–i, D–ii
(4) A–iv, B–iii, C–ii, D–i

Explanation: lacZ encodes beta-galactosidase to hydrolyze lactose, lacY encodes permease to transport lactose, lacA encodes transacetylase for acetylation, and lacI encodes the repressor protein regulating transcription. Correct matching is A–ii, B–i, C–iii, D–iv. (Answer: 1)

9. Fill in the Blanks:
The enzyme that facilitates the entry of lactose into the bacterial cell is __________.
(1) Beta-galactosidase
(2) Permease
(3) Polymerase
(4) Acetylase

Explanation: Permease is a membrane protein that enables lactose transport from the medium into the bacterial cytoplasm, a critical step in lac operon function. Beta-galactosidase metabolizes lactose after entry. Polymerase synthesizes RNA/DNA, and acetylase modifies substrates, neither involved in transport. (Answer: 2)

10. Choose the Correct Statements:
Statement I: Beta-galactosidase hydrolyzes lactose into glucose and galactose.
Statement II: Permease transports lactose into bacterial cells.
(1) Both statements are true
(2) Both statements are false
(3) Statement I true, Statement II false
(4) Statement I false, Statement II true

Explanation: Beta-galactosidase hydrolyzes lactose for metabolism, while permease transports lactose into bacterial cells. Both functions are essential for lactose utilization under lac operon regulation, making both statements correct. (Answer: 1)

Subtopic: Genetic Code

Keyword Definitions:

• Genetic code: The set of rules by which information encoded in DNA or RNA is translated into proteins by living cells.

• Codon: A sequence of three nucleotides in mRNA that specifies a particular amino acid.

• Initiator codon: A codon that signals the start of translation, usually AUG.

• Stop codon: A codon that signals the end of translation, e.g., UAA, UAG, UGA.

• Unambiguous: Each codon codes for only one amino acid.

• Degenerate: More than one codon can code for the same amino acid.

• Palindromic: Sequence reads the same forward and backward; not typical of genetic code.

• Universal: Codons are conserved across most organisms.

• mRNA: Messenger RNA that carries genetic information from DNA to ribosomes.

• Translation: The process of synthesizing proteins from mRNA templates.

• Amino acid: Building block of proteins.

Lead Question - 2023 (Manipur)

The salient features of genetic code are:

(A) The code is palindromic

(B) UGA act as initiator codon

(C) The code is unambiguous and specific

(D) The code is nearly universal

Choose the most appropriate answer from the options given below:

1. (A) and (D) only

2. (B) and (C) only

3. (A) and (B) only

4. (C) and (D) only

Explanation:

The genetic code has several key features. It is unambiguous, meaning each codon specifies only one amino acid, and it is nearly universal, being conserved across most organisms. The code is not palindromic, and UGA is a stop codon, not an initiator codon. Therefore, statements (C) and (D) correctly describe the genetic code. The unambiguous and universal nature ensures accurate protein synthesis and evolutionary conservation. Correct answer is 4.

1. Single Correct Answer MCQ:

Which codon serves as the initiator of translation in most organisms?

1. UGA

2. UAA

3. AUG

4. UAG

Explanation:

The codon AUG codes for methionine and serves as the initiator codon for translation in most organisms. Stop codons like UGA, UAA, and UAG terminate translation. Correct answer is 3.

2. Single Correct Answer MCQ:

Which feature of genetic code allows multiple codons to code for the same amino acid?

1. Universal

2. Unambiguous

3. Degenerate

4. Palindromic

Explanation:

Degeneracy of the genetic code means that several codons can code for the same amino acid, providing flexibility and error tolerance during protein synthesis. Correct answer is 3.

3. Single Correct Answer MCQ:

A codon that signals the end of protein synthesis is called:

1. Initiator codon

2. Stop codon

3. Start codon

4. Degenerate codon

Explanation:

Stop codons (UAA, UAG, UGA) signal termination of translation, preventing further elongation of the polypeptide chain. Correct answer is 2.

4. Single Correct Answer MCQ:

Which statement about the genetic code is correct?

1. It is palindromic

2. It is universal

3. UGA is an initiator codon

4. It is ambiguous

Explanation:

The genetic code is nearly universal across organisms, ensuring consistency in protein synthesis. It is not palindromic, UGA is a stop codon, and it is unambiguous. Correct answer is 2.

5. Single Correct Answer MCQ:

The codon UUU specifies which amino acid?

1. Phenylalanine

2. Leucine

3. Methionine

4. Valine

Explanation:

The codon UUU codes specifically for phenylalanine. The unambiguous nature of genetic code ensures that UUU cannot code for any other amino acid. Correct answer is 1.

6. Single Correct Answer MCQ:

Which feature of genetic code minimizes the effect of point mutations?

1. Degeneracy

2. Universality

3. Unambiguity

4. Palindromicity

Explanation:

Degeneracy allows multiple codons to code for the same amino acid, reducing the effect of single-nucleotide changes, thereby stabilizing protein sequences. Correct answer is 1.

7. Assertion-Reason MCQ:

Assertion (A): The genetic code is nearly universal.

Reason (R): Most organisms use the same codon assignments for amino acids.

1. Both A and R are true and R explains A

2. Both A and R are true but R does not explain A

3. A is true but R is false

4. Both A and R are false

Explanation:

The genetic code is highly conserved across organisms. The Reason explains this universality by noting that most organisms use the same codons for amino acids. Correct answer is 1.

8. Matching Type MCQ:

Match the feature of genetic code with its description:

A. Degenerate — (i) Each codon codes for one amino acid

B. Unambiguous — (ii) Same codon may code for multiple amino acids

C. Universal — (iii) Codons are conserved across species

D. Stop codon — (iv) Signals termination of translation

1. A-(ii), B-(i), C-(iii), D-(iv)

2. A-(i), B-(ii), C-(iii), D-(iv)

3. A-(ii), B-(iii), C-(i), D-(iv)

4. A-(i), B-(i), C-(ii), D-(iii)

Explanation:

Degenerate allows multiple codons for same amino acid, unambiguous ensures one codon codes for only one amino acid, universal codons are conserved, and stop codons terminate translation. Correct answer is 1.

9. Fill in the Blanks MCQ:

The codon ______ acts as a stop codon, while ______ acts as an initiator codon.

1. AUG, UGA

2. UGA, AUG

3. UAA, UAG

4. UAG, UUU

Explanation:

UGA is one of the stop codons that signals the end of translation, whereas AUG codes for methionine and serves as the initiator codon. Correct answer is 2.

10. Choose the correct statements MCQ:

Statement I: Each codon in the genetic code is specific to one amino acid.

Statement II: The genetic code is palindromic.

1. Statement I only

2. Statement II only

3. Both statements are true

4. Both statements are false

Explanation:

The genetic code is unambiguous, with each codon specifying a single amino acid. It is not palindromic. Therefore, only Statement I is correct. Correct answer is 1.

Topic: DNA as Genetic Material; Subtopic: Hershey-Chase Experiment

Keyword Definitions:

• Hershey-Chase experiment: An experiment using bacteriophages to prove DNA is the genetic material.

• Virus: Infectious agent that requires a host cell for replication.

• Bacteriophage: Virus that infects bacteria.

• DNA: Deoxyribonucleic acid, carries genetic information.

• Protein: Macromolecule made of amino acids, forms viral coat.

• Radioactive isotope: Unstable element emitting radiation, used to label molecules.

• Phosphorus-32: Radioactive isotope used to label DNA.

• Sulfur-35: Radioactive isotope used to label proteins.

• Genetic material: Molecule responsible for inheritance.

• Bacterial infection: Entry of bacteriophage into bacterial host.

• Blender experiment: Step in Hershey-Chase method to separate viral coat from bacteria.

Lead Question - 2023 (Manipur)

With reference to Hershey and Chase experiments, select the correct statements:

A: Viruses grown in the presence of radioactive phosphorus contained radioactive DNA.

B: Viruses grown on radioactive sulphur contained radioactive proteins.

C: Viruses grown on radioactive phosphorus contained radioactive protein.

D: Viruses grown on radioactive sulphur contained radioactive DNA.

E: Viruses grown on radioactive protein contained radioactive DNA.

Choose the most appropriate answer from the options given below:

1. (D) and (E) only

2. (A) and (B) only

3. (A) and (C) only

4. (B) and (D) only

Explanation:

Hershey and Chase labeled DNA with radioactive phosphorus (³²P) and protein with radioactive sulphur (³⁵S). After infection of bacteria, radioactive DNA entered bacterial cells, proving DNA carries genetic information, while radioactive protein remained outside. Statements A and B correctly describe the experimental design and observations. C, D, and E are incorrect interpretations. Correct answer is 2.

1. Single Correct Answer MCQ:

Which component of bacteriophage entered bacterial cells during Hershey-Chase experiment?

1. Protein

2. DNA

3. Lipid

4. Carbohydrate

Explanation:

Hershey and Chase demonstrated that DNA, not protein, enters bacterial cells during infection, carrying genetic information. The viral protein coat remained outside. This experiment provided strong evidence that DNA is the hereditary material. Correct answer is 2.

2. Single Correct Answer MCQ:

Which radioactive isotope was used to label DNA in Hershey-Chase experiment?

1. Sulphur-35

2. Carbon-14

3. Phosphorus-32

4. Nitrogen-15

Explanation:

Phosphorus-32 (³²P) was incorporated into DNA of bacteriophages because DNA contains phosphorus, whereas protein lacks phosphorus. This allowed tracking of DNA during bacterial infection. Correct answer is 3.

3. Single Correct Answer MCQ:

Which radioactive isotope was used to label viral protein in Hershey-Chase experiment?

1. Sulphur-35

2. Phosphorus-32

3. Carbon-14

4. Iodine-131

Explanation:

Sulphur-35 (³⁵S) was used to label proteins because sulfur is present in amino acids like cysteine and methionine, but absent in DNA. This distinction allowed separation of DNA and protein during the experiment. Correct answer is 1.

4. Single Correct Answer MCQ:

The main conclusion of the Hershey-Chase experiment was:

1. Proteins carry genetic information

2. DNA carries genetic information

3. Lipids carry genetic information

4. Carbohydrates carry genetic information

Explanation:

The experiment showed that only DNA enters bacterial cells and directs phage replication, whereas protein remains outside. This conclusively proved DNA is the genetic material, not protein, lipids, or carbohydrates. Correct answer is 2.

5. Single Correct Answer MCQ:

Which step was critical for separating viral coat from bacterial cells in the experiment?

1. Centrifugation

2. Heating

3. Blending

4. Filtration

Explanation:

Blending was used to shear off the viral protein coat from bacterial cells, allowing separation of viral DNA (inside bacteria) from viral protein (outside). This step was crucial for determining which molecule carried genetic information. Correct answer is 3.

6. Single Correct Answer MCQ:

What type of virus was used in the Hershey-Chase experiment?

1. Influenza virus

2. T2 bacteriophage

3. Adenovirus

4. HIV

Explanation:

T2 bacteriophage, a virus that infects E. coli, was used. Its simple structure allowed labeling DNA and protein separately, making it ideal for studying genetic material transfer. Correct answer is 2.

7. Assertion-Reason MCQ:

Assertion (A): DNA, not protein, carries hereditary information.

Reason (R): During phage infection, only radioactive DNA enters bacteria, whereas protein remains outside.

1. Both A and R are true and R explains A

2. Both A and R are true but R does not explain A

3. A is true but R is false

4. Both A and R are false

Explanation:

The experiment proved that DNA enters bacterial cells and directs phage replication, while protein remains outside. This observation demonstrates DNA is the genetic material. Both statements are true, and R explains A. Correct answer is 1.

8. Matching Type MCQ:

Match the label with its component:

A. ³²P — (i) DNA

B. ³⁵S — (ii) Protein

C. Viral DNA enters bacteria — (iii) Hereditary material

D. Viral protein remains outside — (iv) Structural coat

1. A-(i), B-(ii), C-(iii), D-(iv)

2. A-(ii), B-(i), C-(iv), D-(iii)

3. A-(i), B-(iv), C-(iii), D-(ii)

4. A-(iv), B-(iii), C-(i), D-(ii)

Explanation:

³²P labeled DNA, ³⁵S labeled protein. Viral DNA enters bacteria, serving as hereditary material. Viral protein remains outside as structural coat. Correct matching is 1.

9. Fill in the Blanks MCQ:

In Hershey-Chase experiment, _______ labeled DNA, and _______ labeled protein.

1. Sulphur-35, Phosphorus-32

2. Phosphorus-32, Sulphur-35

3. Carbon-14, Sulphur-35

4. Nitrogen-15, Carbon-14

Explanation:

Phosphorus-32 labels DNA because DNA contains phosphorus, while Sulphur-35 labels protein due to sulfur in amino acids. This distinction allowed identification of DNA as genetic material. Correct answer is 2.

10. Choose the correct statements MCQ:

Statement I: Protein remained outside bacterial cells during phage infection.

Statement II: DNA entered bacterial cells and directed replication.

1. Statement I only

2. Statement II only

3. Both statements are true

4. Both statements are false

Explanation:

During Hershey-Chase experiment, radioactive protein stayed outside the bacterial cells, while DNA entered and directed phage replication. Both statements correctly describe the observations and conclusions. Correct answer is 3.

Topic: Gene Regulation; Subtopic: Lac Operon in Prokaryotes

Keyword Definitions:

• Lac operon: Cluster of genes in E. coli responsible for lactose metabolism.

• Inducer: Molecule that activates gene expression by inactivating the repressor.

• Repressor: Protein that binds operator to prevent transcription.

• Operator: DNA sequence where repressor binds.

• Promoter: DNA region where RNA polymerase binds to initiate transcription.

• Lactose: Disaccharide sugar composed of glucose and galactose, acts as inducer.

• Glucose: Monosaccharide sugar; its presence can repress lac operon (catabolite repression).

• Allolactose: Isomer of lactose that actually binds repressor and induces lac operon.

• Gene expression: Process by which information from a gene is used to synthesize a functional product.

• Prokaryotes: Organisms without nucleus, such as bacteria.

• Enzyme induction: Mechanism where enzymes are synthesized only when their substrate is present.

Lead Question - 2023 (Manipur)

Which one of the following acts as an inducer for lac operon?

1. Sucrose

2. Lactose

3. Glucose

4. Galactose

Explanation:

The lac operon is a set of genes in E. coli that code for enzymes to metabolize lactose. Lactose or its isomer allolactose binds to the lac repressor, inactivating it and allowing transcription. Glucose presence inhibits the operon via catabolite repression. Sucrose and galactose do not induce the lac operon. Therefore, lactose is the correct inducer. Correct answer is 2.

1. Single Correct Answer MCQ:

Which protein binds to the operator to block transcription in lac operon?

1. RNA polymerase

2. Repressor

3. Lactase

4. Promoter

Explanation:

The repressor protein binds to the operator sequence of the lac operon, preventing RNA polymerase from transcribing structural genes. Lactase is the enzyme produced, and promoter is where RNA polymerase binds. Correct answer is 2.

2. Single Correct Answer MCQ:

Which sugar can suppress lac operon expression even in presence of lactose?

1. Glucose

2. Sucrose

3. Galactose

4. Fructose

Explanation:

Glucose inhibits lac operon expression via catabolite repression even if lactose is present. Other sugars like sucrose and galactose do not cause this suppression. Correct answer is 1.

3. Single Correct Answer MCQ:

Allolactose is:

1. A repressor

2. An inducer

3. An enzyme

4. A promoter

Explanation:

Allolactose, an isomer of lactose, functions as an inducer by binding the lac repressor and inactivating it, allowing transcription of structural genes. It is not an enzyme or promoter. Correct answer is 2.

4. Single Correct Answer MCQ:

Which enzyme is produced by lac operon to cleave lactose?

1. RNA polymerase

2. Lactase (β-galactosidase)

3. DNA polymerase

4. Ligase

Explanation:

The lac operon produces β-galactosidase, commonly called lactase, which cleaves lactose into glucose and galactose. RNA polymerase transcribes genes, DNA polymerase replicates DNA, and ligase joins DNA fragments. Correct answer is 2.

5. Single Correct Answer MCQ:

Which DNA sequence does the repressor bind to?

1. Promoter

2. Operator

3. Structural gene

4. Enhancer

Explanation:

The operator is the DNA sequence where the lac repressor binds to block transcription of the structural genes. Promoter is for RNA polymerase, structural genes code enzymes, enhancer is regulatory in eukaryotes. Correct answer is 2.

6. Single Correct Answer MCQ:

In absence of lactose, lac operon is:

1. Induced

2. Repressed

3. Constitutively expressed

4. Mutated

Explanation:

Without lactose, the lac repressor binds the operator and prevents transcription, keeping the operon in a repressed state. Induction occurs only when lactose/allolactose is present. Correct answer is 2.

7. Assertion-Reason MCQ:

Assertion (A): Lactose induces the lac operon.

Reason (R): Lactose binds to the repressor and inactivates it.

1. Both A and R are true and R explains A

2. Both A and R are true but R does not explain A

3. A is true but R is false

4. Both A and R are false

Explanation:

Lactose acts as an inducer by binding to the lac repressor and inactivating it, thereby allowing transcription of the operon. Both statements are true, and the reason correctly explains the assertion. Correct answer is 1.

8. Matching Type MCQ:

Match lac operon components with function:

A. Operator — (i) Binding site for repressor

B. Promoter — (ii) RNA polymerase binding site

C. Structural genes — (iii) Code enzymes for lactose metabolism

D. Repressor — (iv) Protein that inhibits transcription

1. A-(i), B-(ii), C-(iii), D-(iv)

2. A-(ii), B-(i), C-(iv), D-(iii)

3. A-(iii), B-(iv), C-(i), D-(ii)

4. A-(iv), B-(iii), C-(ii), D-(i)

Explanation:

Operator is where the repressor binds, promoter is RNA polymerase binding site, structural genes code for enzymes, and repressor protein inhibits transcription. Correct answer is 1.

9. Fill in the Blanks MCQ:

The lac operon in E. coli is _______ in absence of lactose.

1. Induced

2. Repressed

3. Constitutive

4. Mutated

Explanation:

In the absence of lactose, the lac operon is repressed because the repressor protein binds to the operator and prevents transcription of structural genes. Correct answer is 2.

10. Choose the correct statements MCQ:

Statement I: Lactose acts as an inducer for lac operon.

Statement II: Glucose presence inhibits lac operon via catabolite repression.

1. Statement I only

2. Statement II only

3. Both statements are true

4. Both statements are false

Explanation:

Both statements are correct. Lactose induces the lac operon by inactivating the repressor, and glucose represses the operon through catabolite repression, preventing unnecessary lactose metabolism. Correct answer is 3.

Topic: Genetic Material – DNA and RNA; Subtopic: Properties and Function of RNA

Keyword Definitions:

• RNA (Ribonucleic acid): A single-stranded nucleic acid that carries genetic information and helps in protein synthesis.

• Mutation: A sudden change in the nucleotide sequence of DNA or RNA causing genetic variation.

• Protein synthesis: The process by which RNA directs the formation of proteins using amino acids.

• Genetic expression: The process through which information from a gene is used to create a functional product, often a protein.

• Stability of genetic material: Refers to the resistance of nucleic acids to degradation or mutation over time.

Lead Question – 2023 (Manipur)

Given below are two statements:
Statement I: RNA being unstable, it mutates at a faster rate.
Statement II: RNA can directly code for synthesis of proteins hence can easily express the characters.
In the light of the above statements, choose the correct answer from the options given below:
1. Statement I is correct but Statement II is incorrect
2. Statement I is incorrect but Statement II is correct
3. Both Statement I and Statement II are correct
4. Both Statement I and Statement II are incorrect

Explanation: Both statements are correct. RNA is an unstable molecule due to the presence of the 2'-hydroxyl group in ribose, making it more prone to hydrolysis and mutations. RNA viruses mutate faster, enhancing adaptability. Moreover, RNA can directly code for protein synthesis without DNA intermediates, allowing faster gene expression. (Answer: 3)

Guessed Questions:

1. Which of the following is the most stable genetic material?
1. DNA
2. RNA
3. mRNA
4. tRNA

Explanation: DNA is the most stable genetic material due to the absence of a 2'-hydroxyl group in deoxyribose sugar, making it less prone to hydrolysis. Its double-stranded structure and complementary base pairing ensure genetic fidelity across generations, while RNA is unstable and mutates faster. Hence, the correct answer is DNA. (Answer: 1)

2. Which RNA carries amino acids to the ribosome during translation?
1. mRNA
2. tRNA
3. rRNA
4. snRNA

Explanation: tRNA (transfer RNA) carries specific amino acids to the ribosome during protein synthesis. It recognizes codons on mRNA through its anticodon loop and ensures the correct sequence of amino acids in the polypeptide chain. Hence, the correct answer is tRNA. (Answer: 2)

3. Which of the following statements about RNA viruses is correct?
1. They mutate slowly.
2. They have DNA as genetic material.
3. They mutate rapidly due to instability of RNA.
4. They cannot infect humans.

Explanation: RNA viruses mutate rapidly because RNA lacks the proofreading mechanism of DNA polymerases and is chemically unstable. This rapid mutation rate allows them to adapt quickly to changing environments and host defenses. Hence, the correct answer is option 3. (Answer: 3)

4. Assertion (A): RNA viruses evolve faster than DNA viruses.
Reason (R): RNA-dependent RNA polymerases lack proofreading activity.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, R is false.
4. A is false, R is true.

Explanation: RNA viruses mutate faster because RNA-dependent RNA polymerases, responsible for their replication, lack proofreading ability, causing frequent errors. This results in higher mutation rates and faster evolution. Hence, both Assertion and Reason are true, and R correctly explains A. (Answer: 1)

5. Match List-I with List-II:
List-I: (A) mRNA (B) tRNA (C) rRNA (D) snRNA
List-II: (I) Catalyzes peptide bond formation (II) Template for translation (III) Participates in splicing (IV) Brings amino acids
Options:
1. A–II, B–IV, C–I, D–III
2. A–IV, B–II, C–III, D–I
3. A–I, B–III, C–II, D–IV
4. A–III, B–I, C–IV, D–II

Explanation: The correct matching is A–II (mRNA – template for translation), B–IV (tRNA – brings amino acids), C–I (rRNA – catalyzes peptide bond formation), and D–III (snRNA – participates in splicing). Each RNA type plays a distinct role in protein synthesis and gene regulation. (Answer: 1)

6. Which of the following is not a function of RNA?
1. Genetic material in some viruses
2. Catalytic activity in ribozymes
3. Template for DNA synthesis in eukaryotes
4. Template for protein synthesis

Explanation: In eukaryotes, RNA does not act as a template for DNA synthesis; rather, DNA serves as a template for RNA during transcription. Reverse transcription occurs in retroviruses only. RNA serves other roles like catalysis and protein synthesis. Hence, the correct answer is option 3. (Answer: 3)

7. Fill in the blanks:
Retroviruses use _______ to synthesize DNA from RNA.
1. RNA polymerase
2. Reverse transcriptase
3. Ligase
4. Helicase

Explanation: Retroviruses, such as HIV, possess the enzyme reverse transcriptase which synthesizes complementary DNA (cDNA) from their RNA genome. This allows integration of viral DNA into the host genome, facilitating infection and replication. Hence, the correct answer is reverse transcriptase. (Answer: 2)

8. Choose the correct statements:
Statement I: RNA acts as both genetic material and catalyst.
Statement II: DNA can perform catalytic functions.
1. Both statements are correct.
2. Statement I is correct, Statement II is incorrect.
3. Both statements are incorrect.
4. Statement I is incorrect, Statement II is correct.

Explanation: RNA can act as both genetic material (in RNA viruses) and as a catalyst (ribozymes), demonstrating its versatility. DNA, however, is not catalytic due to its stable double-stranded nature. Hence, Statement I is correct and Statement II is incorrect. The correct answer is option 2. (Answer: 2)

9. Which type of RNA forms the major part of the ribosome?
1. mRNA
2. tRNA
3. rRNA
4. snRNA

Explanation: rRNA (ribosomal RNA) forms the structural and catalytic core of ribosomes, enabling protein synthesis. It binds with ribosomal proteins to form the 40S and 60S subunits in eukaryotes. Hence, the correct answer is rRNA. (Answer: 3)

10. Which RNA serves as a template for protein synthesis?
1. tRNA
2. rRNA
3. mRNA
4. miRNA

Explanation: mRNA (messenger RNA) serves as the template for protein synthesis. It carries the genetic code transcribed from DNA and determines the sequence of amino acids in a protein. Ribosomes translate this code into polypeptides. Hence, the correct answer is mRNA. (Answer: 3)

Topic: DNA as Genetic Material; Subtopic: Hershey-Chase Experiment

Keyword Definitions:

• Phage: A virus that infects bacteria, also called bacteriophage.

• 32P and 35S labeling: Radioactive isotopes used to label DNA (32P) and protein (35S) to trace genetic material.

• Genetic material: Substance that carries genetic information, either DNA or RNA.

• Hershey-Chase experiment: Demonstrated that DNA, not protein, is the genetic material using radioactive labeling in bacteriophages.

• Bacteriophage: Virus that infects bacteria and injects genetic material into the host.

Lead Question – 2023 (Manipur)

Which scientist conducted an experiment with 32P and 35S labelled phages for demonstrating that DNA is the genetic material?
1. James D. Watson and F.H.C Crick
2. A.D. Hershey and M.J. Chase
3. F. Griffith
4. O.T. Avery, C.M. MacLeod and M. McCarty

Explanation: A.D. Hershey and M.J. Chase used bacteriophages labeled with radioactive isotopes 32P for DNA and 35S for protein to determine which component enters bacterial cells during infection. They observed that 32P-DNA, not 35S-protein, entered the host, confirming DNA as the genetic material. This experiment conclusively demonstrated DNA carries hereditary information. Correct answer: 2. A.D. Hershey and M.J. Chase.

1. Single Correct Answer MCQ:
Which isotope was used to label DNA in Hershey-Chase experiment?
1. 35S
2. 32P
3. 14C
4. 3H
Explanation: DNA contains phosphorus but not sulfur, hence 32P was used to label DNA in the experiment. Proteins, containing sulfur, were labeled with 35S. This distinction allowed tracing of DNA vs protein entry into bacteria. Correct answer is 2. 32P.

2. Single Correct Answer MCQ:
Which component of bacteriophage enters bacterial cells during infection?
1. Protein
2. DNA
3. Both protein and DNA
4. Neither
Explanation: Hershey-Chase experiment showed that only DNA enters bacterial cells and directs viral replication, whereas protein remains outside. This established DNA as the genetic material. Correct answer: 2. DNA.

3. Single Correct Answer MCQ:
Which of the following was not a part of Hershey-Chase experiment?
1. Bacteriophage T2
2. 32P labeling
3. 35S labeling
4. X-ray crystallography
Explanation: X-ray crystallography was used by Watson and Crick for DNA structure, not Hershey-Chase. Hershey-Chase used T2 phage with 32P-DNA and 35S-protein labeling. Correct answer: 4. X-ray crystallography.

4. Single Correct Answer MCQ:
The radioactive sulfur (35S) labeled which part of phage?
1. DNA
2. RNA
3. Protein
4. Lipid
Explanation: Proteins contain sulfur in amino acids like methionine and cysteine. Hershey-Chase labeled phage proteins with 35S to track their entry into bacterial cells. DNA has no sulfur. Correct answer: 3. Protein.

5. Single Correct Answer MCQ:
The Hershey-Chase experiment was important because it:
1. Determined DNA structure
2. Showed RNA is genetic material
3. Demonstrated DNA is genetic material
4. Identified proteins as hereditary molecules
Explanation: Hershey-Chase experiment conclusively demonstrated that DNA, not protein, is the hereditary material responsible for transferring genetic information in viruses. Correct answer: 3. Demonstrated DNA is genetic material.

6. Single Correct Answer MCQ:
Which virus was used in Hershey-Chase experiment?
1. T2 bacteriophage
2. Lambda phage
3. Influenza virus
4. Tobacco mosaic virus
Explanation: T2 bacteriophage infects E. coli and was used in Hershey-Chase experiment to trace DNA vs protein during infection. Correct answer: 1. T2 bacteriophage.

7. Assertion-Reason MCQ:
Assertion (A): Hershey-Chase experiment used two radioactive isotopes to differentiate DNA and protein.
Reason (R): 32P labels DNA and 35S labels protein, enabling tracking during phage infection.
1. Both A and R true, R correctly explains A
2. Both A and R true, R does not explain A
3. A true, R false
4. A false, R true
Explanation: Hershey-Chase used 32P and 35S to label DNA and protein, respectively, to determine which molecule enters bacterial cells. This design allowed direct observation, proving DNA as genetic material. Correct answer: 1.

8. Matching Type MCQ:
Match the labels with their target:
A. 32P – (i) DNA
B. 35S – (ii) Protein
C. Phage T2 – (iii) Virus used
D. E. coli – (iv) Host bacteria
1. A–i, B–ii, C–iii, D–iv
2. A–ii, B–i, C–iv, D–iii
3. A–iii, B–iv, C–i, D–ii
4. A–iv, B–iii, C–ii, D–i
Explanation: 32P labeled DNA, 35S labeled protein, T2 phage was the virus, and E. coli was the host bacterium. Correct answer: 1.

9. Fill in the Blanks MCQ:
The ______ isotope labels protein in the Hershey-Chase experiment.
1. 32P
2. 35S
3. 14C
4. 3H
Explanation: 35S labels sulfur-containing amino acids in proteins, allowing differentiation from DNA. DNA was labeled with 32P. Correct answer: 2. 35S.

10. Choose the Correct Statements MCQ:
Statement I: DNA of phage enters bacterial cell during infection.
Statement II: Protein of phage enters bacterial cell to carry genetic information.
1. Both true
2. Both false
3. I true, II false
4. I false, II true
Explanation: Hershey-Chase showed DNA, not protein, enters bacterial cells to transmit genetic information. Hence, Statement I is true, Statement II is false. Correct answer: 3.

Topic: Transcription in Prokaryotes and Eukaryotes; Subtopic: Structure of Transcription Unit

Keyword Definitions:

• Transcription: The process of synthesizing RNA from a DNA template, copying genetic information from DNA to RNA.

• Promoter: A DNA sequence where RNA polymerase binds to initiate transcription.

• Structural gene: DNA segment that codes for RNA or protein.

• Terminator: DNA sequence signaling the end of transcription.

• Transcription unit: A segment of DNA that includes a promoter, structural gene, and terminator, producing a single RNA molecule.

Lead Question – 2023 (Manipur)

Given below are two statements:
I: The process of copying genetic information from one strand of the DNA into RNA is termed as transcription.
II: A transcription unit in DNA is defined primarily by the three regions in the DNA i.e., a promoter, the structural gene and a terminator.
In the light of the above statements, choose the correct answer from the options given below:
1. Statement I is true but Statement II is false
2. Statement I is false but Statement II is true
3. Both Statement I and Statement II are true
4. Both Statement I and Statement II are false

Explanation: Transcription is the synthesis of RNA from a DNA template, accurately copying genetic information. A transcription unit consists of three main regions: a promoter to initiate transcription, the structural gene that encodes the RNA product, and a terminator signaling transcription termination. Both statements correctly describe transcription and the definition of a transcription unit. Therefore, both Statement I and Statement II are true. Correct answer: 3.

1. Single Correct Answer MCQ:
Which enzyme synthesizes RNA using DNA as a template?
1. DNA polymerase
2. RNA polymerase
3. Ligase
4. Reverse transcriptase
Explanation: RNA polymerase reads the DNA template strand and synthesizes a complementary RNA strand. It does not require a primer and catalyzes the formation of phosphodiester bonds. DNA polymerase synthesizes DNA, ligase joins DNA fragments, and reverse transcriptase synthesizes DNA from RNA. Hence, correct answer is 2. RNA polymerase.

2. Single Correct Answer MCQ:
Which region of a transcription unit signals the start of transcription?
1. Terminator
2. Structural gene
3. Promoter
4. Enhancer
Explanation: The promoter is a DNA sequence where RNA polymerase binds to initiate transcription. Structural gene codes for RNA, terminator signals the end, and enhancers increase transcription efficiency. Hence, correct answer is 3. Promoter.

3. Single Correct Answer MCQ:
Which of the following marks the end of transcription?
1. Promoter
2. Operator
3. Terminator
4. Enhancer
Explanation: The terminator is a DNA sequence that instructs RNA polymerase to stop transcription, releasing the newly synthesized RNA. Promoter starts transcription, operator regulates gene expression, and enhancer boosts transcription efficiency. Hence, correct answer is 3. Terminator.

4. Single Correct Answer MCQ:
In prokaryotic transcription, which strand serves as the template?
1. Coding strand
2. Template strand
3. Non-template strand
4. Both strands equally
Explanation: RNA polymerase uses the template strand (non-coding strand) to synthesize RNA complementary to it. The coding strand has the same sequence as RNA except T is replaced by U. Hence, correct answer is 2. Template strand.

5. Single Correct Answer MCQ:
Which of the following is not part of a typical transcription unit?
1. Promoter
2. Structural gene
3. Terminator
4. Ribosome
Explanation: Ribosomes are cellular structures for protein synthesis, not part of a transcription unit. The transcription unit includes promoter, structural gene, and terminator. Hence, correct answer is 4. Ribosome.

6. Single Correct Answer MCQ:
Which RNA is synthesized directly from DNA?
1. mRNA
2. tRNA
3. rRNA
4. All of the above
Explanation: mRNA, tRNA, and rRNA are all transcribed from DNA by RNA polymerase. They differ in function, but the synthesis mechanism is similar. Hence, correct answer is 4. All of the above.

7. Assertion-Reason MCQ:
Assertion (A): RNA polymerase does not require a primer to initiate transcription.
Reason (R): It binds directly to the promoter and starts RNA synthesis.
1. Both A and R are true, and R correctly explains A.
2. Both A and R are true, but R does not explain A.
3. A true, R false
4. A false, R true
Explanation: RNA polymerase can initiate RNA synthesis de novo at the promoter without a primer. Its binding to the promoter enables accurate initiation of transcription. Hence, both A and R are true, and R correctly explains A. Correct answer: 1.

8. Matching Type MCQ:
Match the transcription-related terms with their functions:
A. Promoter – (i) Start site
B. Structural gene – (ii) Codes for RNA
C. Terminator – (iii) Stops transcription
D. Enhancer – (iv) Increases transcription efficiency
1. A–i, B–ii, C–iii, D–iv
2. A–ii, B–i, C–iv, D–iii
3. A–iii, B–iv, C–i, D–ii
4. A–iv, B–iii, C–ii, D–i
Explanation: Promoter initiates transcription, structural gene encodes RNA, terminator ends transcription, and enhancer boosts transcription efficiency. Hence, correct answer is 1. A–i, B–ii, C–iii, D–iv.

9. Fill in the Blanks MCQ:
The ______ strand of DNA has the same sequence as the RNA produced during transcription, except T is replaced by U.
1. Template
2. Coding
3. Non-template
4. Complementary
Explanation: The coding strand (sense strand) has the same sequence as RNA except thymine (T) is replaced by uracil (U). The template strand is complementary. Hence, correct answer is 2. Coding.

10. Choose the Correct Statements MCQ:
Statement I: A transcription unit produces a single RNA molecule.
Statement II: Enhancers are part of the structural gene.
1. Both statements true
2. Both statements false
3. Statement I true, Statement II false
4. Statement I false, Statement II true
Explanation: A transcription unit generates a single RNA molecule, but enhancers are regulatory elements outside structural genes that increase transcription efficiency. Hence, Statement I is true, Statement II is false. Correct answer: 3.

Topic: Gene Regulation; Subtopic: Operon Model

• Operon: A cluster of genes under the control of a single promoter and operator, functioning together in transcription regulation.

• Operator: A DNA segment within an operon where regulatory proteins bind to control transcription.

• Repressor protein: A protein that binds to the operator to block RNA polymerase and prevent gene transcription.

• Promoter: A DNA sequence where RNA polymerase binds to initiate transcription.

• Inducer: A molecule that binds to repressor protein, inactivating it and allowing transcription.

• Regulator protein: A protein coded by a regulatory gene, controlling operon activity; can be a repressor or activator.

• RNA polymerase: Enzyme responsible for synthesizing RNA from a DNA template.

• Lac operon: A classic inducible operon in E. coli regulated by repressor and inducer.

• Gene transcription: The process of copying DNA into RNA by RNA polymerase.

• Inducible operon: An operon that is normally off and can be activated by an inducer.

• Constitutive gene: Gene that is continuously expressed regardless of regulatory mechanisms.

Lead Question - 2023 (Manipur)

Name the component that binds to the operator region of an operon and prevents RNA polymerase from transcribing the operon.

• 1. Promoter

• 2. Regulator protein

• 3. Repressor protein

• 4. Inducer

Explanation: The repressor protein binds to the operator region of an operon, physically blocking RNA polymerase from transcribing downstream structural genes. This prevents unnecessary protein synthesis in the absence of an inducer. Promoter is the RNA polymerase binding site, regulator protein may code for repressor, and inducer deactivates the repressor. Correct answer: 3.

1. In the lac operon, allolactose acts as:

• a) Repressor

• b) Inducer

• c) Promoter

• d) Operator

Explanation: Allolactose binds to the repressor protein in the lac operon, inactivating it and allowing transcription of structural genes. It does not directly act as a repressor, promoter, or operator. This regulatory mechanism ensures lactose metabolism occurs only when needed. Correct answer: b.

2. The operator in an operon is:

• a) Where RNA polymerase initiates transcription

• b) Where the repressor binds

• c) A regulatory gene coding site

• d) A sequence that enhances translation

Explanation: The operator is a DNA segment within the operon where repressor proteins bind, controlling access of RNA polymerase to structural genes. It does not code proteins, initiate transcription, or enhance translation. Correct answer: b.

3. A regulatory gene codes for:

• a) Structural proteins

• b) RNA polymerase

• c) Repressor protein

• d) Ribosomal RNA

Explanation: The regulatory gene produces the repressor protein, which binds the operator to control transcription. Structural genes code for functional proteins, RNA polymerase synthesizes RNA, and rRNA is part of ribosomes. Correct answer: c.

4. Inducible operons are typically:

• a) Always on

• b) Normally off

• c) Constitutively expressed

• d) Repressed permanently

Explanation: Inducible operons, like the lac operon, are normally off. They are activated by an inducer molecule that inactivates the repressor, enabling transcription when needed. Constitutive expression or permanent repression does not allow regulation. Correct answer: b.

5. In the trp operon, tryptophan acts as:

• a) Inducer

• b) Corepressor

• c) Repressor

• d) Promoter

Explanation: In the trp operon, tryptophan serves as a corepressor by binding the repressor protein, allowing it to attach to the operator and inhibit transcription of tryptophan synthesis genes. It is not an inducer, the repressor is a separate protein, and promoter is DNA. Correct answer: b.

6. Which part of an operon directly interacts with RNA polymerase?

• a) Operator

• b) Promoter

• c) Repressor

• d) Inducer

Explanation: The promoter is the DNA sequence where RNA polymerase binds to initiate transcription. The operator is for repressor binding, the repressor protein blocks transcription, and the inducer modulates repressor activity. Correct answer: b.

7. Assertion (A): Repressor proteins prevent transcription of operon genes.
Reason (R): Repressors bind to operator regions and block RNA polymerase binding.

• a) Both A and R are true, R explains A

• b) Both A and R are true, R does not explain A

• c) A is true, R is false

• d) A is false, R is true

Explanation: Repressor proteins bind the operator to block RNA polymerase, thus preventing transcription of structural genes in the operon. Both the assertion and reason are correct, and the reason explains why transcription is inhibited. Correct answer: a.

8. Match the operon with its regulation type:

• A) Lac operon

• B) Trp operon

• C) Arabinose operon

• D) Beta-galactosidase gene

(I) Inducible
(II) Repressible
(III) Catabolite-activated
(IV) Constitutive

• a) A-I, B-II, C-III, D-IV

• b) A-II, B-I, C-IV, D-III

• c) A-III, B-IV, C-II, D-I

• d) A-IV, B-III, C-I, D-II

Explanation: Lac operon is inducible (activated by lactose), Trp operon is repressible (inhibited by tryptophan), Arabinose operon is catabolite-activated, and Beta-galactosidase gene can be constitutively expressed under artificial constructs. Correct answer: a.

9. Fill in the blank: The molecule that inactivates the repressor in an inducible operon is called __________.

• a) Inducer

• b) Repressor

• c) Promoter

• d) Operator

Explanation: An inducer molecule binds to the repressor protein, causing conformational change that prevents it from attaching to the operator, allowing RNA polymerase to transcribe genes. Repressor, promoter, and operator are involved in regulation but do not inactivate the repressor. Correct answer: a.

10. Choose the correct statements:

Statement I: Repressor binds to operator to inhibit transcription.
Statement II: Promoter is the binding site for repressor protein.

• a) Both I and II are correct

• b) Only I is correct

• c) Only II is correct

• d) Both I and II are incorrect

Explanation: Repressor proteins bind to the operator to block transcription, making Statement I correct. Promoter is the site for RNA polymerase, not repressor, so Statement II is incorrect. Correct answer: b.

Topic: Human Genome Project; Subtopic: Chromosome Sequencing

• Human Genome Project (HGP): International research project aimed to map and sequence all human genes.

• Chromosome: DNA molecule with part or all of the genetic material of an organism.

• Sequencing: Determining the exact order of nucleotides in a DNA molecule.

• Gene: Functional unit of heredity that codes for proteins or RNA.

• Exons: Coding regions of a gene that remain in mature mRNA.

• Introns: Non-coding regions of a gene removed during RNA splicing.

• Genome: Complete set of DNA including all genes of an organism.

• Bioinformatics: Application of computational tools to manage and analyze biological data.

• Chromosome 1: Largest human chromosome in terms of DNA content.

• Chromosome 22: One of the smallest human autosomes sequenced early in HGP.

• Chromosome 6: Last chromosome fully sequenced in the Human Genome Project.

Lead Question - 2023 (Manipur)

The last chromosome sequenced in Human Genome Project was:

• 1. Chromosome 6

• 2. Chromosome 1

• 3. Chromosome 22

• 4. Chromosome 14

Explanation: Chromosome 6, which contains the major histocompatibility complex (MHC) crucial for immune response, was the last human chromosome fully sequenced in the Human Genome Project. Chromosome 1, being the largest, and Chromosome 22, one of the smallest, were sequenced earlier. Sequencing chromosome 6 completed the mapping of all human autosomes, providing comprehensive information for studying genetic disorders, human evolution, and biomedical research. Correct answer: 1. Chromosome 6.

1. Which chromosome is the largest in human genome?

• a) Chromosome 1

• b) Chromosome 6

• c) Chromosome 22

• d) Chromosome X

Explanation: Chromosome 1 is the largest human chromosome containing approximately 249 million base pairs. Chromosome 6 is smaller, though significant due to MHC genes. Chromosome 22 is one of the smallest, and X chromosome is smaller than chromosome 1 but carries important sex-linked genes. Correct answer: a.

2. The Human Genome Project officially started in:

• a) 1990

• b) 1980

• c) 2000

• d) 1995

Explanation: The Human Genome Project began in 1990 as an international collaborative effort to sequence and map all human genes. It used sequencing technologies and bioinformatics to compile the human genome, completed in 2003. Correct answer: a.

3. The function of major histocompatibility complex (MHC) is:

• a) Immune response regulation

• b) Protein synthesis

• c) Cell division

• d) DNA replication