Molecular Basis of Inheritance

Topic: DNA Structure and Properties; Subtopic: Helical Turns and Base Pairing

Keyword Definitions:

• DNA molecule: Deoxyribonucleic acid, a double-stranded helical structure carrying genetic information.

• Base pair (bp): Pair of complementary nucleotides, Adenine-Thymine or Guanine-Cytosine, forming the rungs of DNA.

• Helical turn: One complete twist of the DNA double helix, typically containing ~10.5 base pairs in B-DNA.

• B-DNA: Most common DNA conformation in vivo, right-handed helix with 10.5 bp per turn.

• DNA shortening: Removal or loss of nucleotides from a DNA strand.

• Double helix: The twisted ladder-like structure of DNA formed by two complementary strands.

• DNA structure: Arrangement of nucleotides, base pairs, sugar-phosphate backbone, and helical twist.

Lead Question – 2022 (Abroad)
If a DNA molecule is shortened by 25 base pairs, how many helical turns will be reduced from its structure?
1. 1
2. 3
3. 2.5
4. 2

Explanation:
Correct answer is option 2. In B-DNA, one helical turn consists of approximately 10.5 base pairs. Removing 25 base pairs from the DNA molecule reduces the number of helical turns by dividing 25 by 10.5, which equals roughly 2.38 turns. Rounding to the nearest whole number gives approximately 3 helical turns. This calculation shows that shortening the DNA by 25 base pairs eliminates multiple full twists of the double helix, slightly affecting the structural stability and torsional strain. Understanding base pair density per helical turn is critical in structural biology and DNA manipulation experiments.

1. Single Correct Answer MCQ:
How many base pairs approximately constitute one helical turn in B-DNA?
1. 10.5
2. 12
3. 9
4. 8
Explanation: Correct answer is 10.5. B-DNA, the most common DNA form, has approximately 10.5 base pairs per complete helical turn. This constant is crucial for calculating the number of turns in DNA molecules and understanding structural changes when base pairs are added or removed.

2. Single Correct Answer MCQ:
If 52 base pairs are removed from a DNA molecule, how many helical turns are lost?
1. 5
2. 4
3. 3
4. 6
Explanation: Correct answer is option 1. Dividing 52 base pairs by 10.5 bp per turn gives approximately 4.95 turns, rounding to 5 helical turns. This demonstrates the proportional relationship between base pairs and DNA helical turns in structural calculations.

3. Single Correct Answer MCQ:
Which DNA form is considered for the calculation of 10.5 bp per helical turn?
1. A-DNA
2. B-DNA
3. Z-DNA
4. RNA-DNA hybrid
Explanation: Correct answer is B-DNA. The standard B-DNA conformation in cells is right-handed with 10.5 base pairs per turn. A-DNA and Z-DNA have different helical parameters. B-DNA's helical repeat is the basis for calculating structural changes when base pairs are added or deleted.

4. Single Correct Answer MCQ:
Removing 21 base pairs from B-DNA will reduce the helical turns by approximately:
1. 1 turn
2. 2 turns
3. 3 turns
4. 4 turns
Explanation: Correct answer is 2 turns. 21 divided by 10.5 equals exactly 2. Therefore, deleting 21 base pairs reduces the double helix by two complete turns, showing the importance of base pair count per turn in DNA structural considerations.

5. Single Correct Answer MCQ:
Shortening DNA by 105 base pairs reduces how many helical turns?
1. 5
2. 10
3. 15
4. 20
Explanation: Correct answer is 10. Dividing 105 base pairs by 10.5 bp per turn results in 10 complete helical turns removed. This calculation is essential for understanding the relationship between DNA length and supercoiling or torsional strain.

6. Single Correct Answer MCQ:
Why does removing base pairs affect the number of helical turns?
1. It changes nucleotide sequence only
2. It shortens the DNA and reduces the total twist
3. It increases base pairing
4. It alters sugar-phosphate backbone chemistry
Explanation: Correct answer is option 2. DNA helical turns are determined by the number of base pairs. Removing base pairs shortens the molecule, proportionally reducing the number of complete helical turns and slightly affecting torsional strain and helical structure.

7. Assertion-Reason MCQ:
Assertion (A): DNA shortening reduces the number of helical turns.
Reason (R): One helical turn of B-DNA contains about 10.5 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, R is false
4. A is false, R is true
Explanation: Correct answer is option 1. The number of helical turns in B-DNA depends on base pair count. Removing base pairs proportionally reduces turns. Since one turn has 10.5 bp, the reason explains the assertion directly.

8. Matching Type MCQ:
Match DNA change with effect on helical turns:
A. Remove 21 bp – (i) 1 turn
B. Remove 52 bp – (ii) 5 turns
C. Remove 105 bp – (iii) 2 turns
D. Remove 10.5 bp – (iv) 10 turns
1. A–iii, B–ii, C–iv, D–i
2. A–i, B–iii, C–iv, D–ii
3. A–ii, B–i, C–iii, D–iv
4. A–iv, B–iii, C–ii, D–i
Explanation: Correct answer is option 1. 21/10.5 = 2 turns ≈ 2, 52/10.5 ≈5 turns, 105/10.5 =10 turns, 10.5/10.5=1 turn. Matching shows proportional reduction of helical turns with deleted base pairs.

9. Fill in the Blanks MCQ:
One helical turn of B-DNA contains approximately ______ base pairs.
1. 8
2. 10.5
3. 12
4. 15
Explanation: Correct answer is 10.5. In B-DNA, the helical repeat is 10.5 base pairs per complete turn. This constant is critical for calculations regarding DNA shortening, lengthening, or torsional strain.

10. Choose the Correct Statements MCQ:
Statement I: Deleting 25 bp reduces DNA helical turns by approximately 3.
Statement II:

Topic: DNA Replication; Subtopic: Semi-Conservative Replication

Keyword Definitions:

• Meselson and Stahl Experiment: A classic experiment demonstrating semi-conservative DNA replication using isotopes of nitrogen (¹⁵N and ¹⁴N).

• Hybrid DNA: DNA containing one old (heavy) strand and one newly synthesized (light) strand after replication.

• Light DNA: DNA containing only newly synthesized strands with light nitrogen (¹⁴N).

• Semi-Conservative Replication: DNA replication mechanism where each daughter molecule consists of one parental strand and one newly synthesized strand.

• Density Gradient Centrifugation: Technique used to separate DNA molecules based on density differences.

• Replication Time: Duration after which DNA strands replicate and proportions of hybrid and light DNA change.

• Parental Strand: Original DNA strand used as template for replication.

Lead Question – 2022 (Abroad)
What would be the proportions of light and hybrid density DNA molecule, respectively if Meselson and Stahl's experiment was continued for 60 minutes?
1. 50%, 50%
2. 25%, 75%
3. 75%, 25%
4. 100%, 0%

Explanation:
Correct answer is option 1. In the Meselson and Stahl experiment, after one round of replication in ¹⁴N medium, hybrid DNA (one heavy ¹⁵N strand and one light ¹⁴N strand) forms, and no completely light DNA is produced yet. If replication continues for another generation, semi-conservative replication produces 50% hybrid DNA and 50% light DNA. The experiment demonstrates that each new DNA molecule contains one parental and one new strand. Therefore, after sufficient replication, the proportion of light and hybrid DNA stabilizes at 50:50, confirming semi-conservative replication as the accurate model of DNA replication.

1. In semi-conservative replication, each daughter DNA molecule contains:
1. Two old strands
2. Two new strands
3. One old and one new strand
4. Three strands
Explanation: Correct answer is one old and one new strand. Semi-conservative replication ensures that each daughter DNA molecule inherits one parental strand and one newly synthesized strand, preserving genetic information. This mechanism is fundamental to DNA replication, contrasting with conservative (both old) and dispersive (mixed fragments) models.

2. What isotope of nitrogen was used as heavy label in Meselson and Stahl experiment?
1. ¹²N
2. ¹³N
3. ¹⁵N
4. ¹⁴N
Explanation: Correct answer is ¹⁵N. ¹⁵N is a stable heavy isotope incorporated into bacterial DNA to distinguish old DNA from newly synthesized light DNA (¹⁴N). This isotopic labeling enabled separation by density gradient centrifugation and confirmed semi-conservative replication.

3. After two rounds of replication in ¹⁴N medium, proportion of light DNA is:
1. 25%
2. 50%
3. 75%
4. 100%
Explanation: Correct answer is 75%. After first generation, all DNA is hybrid (50% hybrid). Second generation produces half of hybrid DNA replicated into one light and one hybrid, resulting in 50% light + 25% hybrid + 25% hybrid = 75% light DNA. This confirms semi-conservative replication dynamics over generations.

4. The technique used to separate DNA molecules by density is:
1. Gel electrophoresis
2. Density gradient centrifugation
3. PCR
4. Chromatography
Explanation: Correct answer is density gradient centrifugation. This technique separates DNA molecules based on buoyant density differences. In Meselson and Stahl’s experiment, cesium chloride gradient centrifugation allowed differentiation between heavy (¹⁵N), light (¹⁴N), and hybrid DNA molecules.

5. Hybrid DNA contains:
1. Both parental strands
2. Both new strands
3. One parental and one new strand
4. Fragmented strands
Explanation: Correct answer is one parental and one new strand. Hybrid DNA forms after the first replication cycle in light medium. Each hybrid molecule retains one heavy parental strand and one newly synthesized light strand, demonstrating semi-conservative replication clearly.

6. Which model of replication was disproved by Meselson and Stahl experiment?
1. Semi-conservative
2. Conservative
3. Dispersive
4. Both b and c
Explanation: Correct answer is both b and c. Conservative model predicts all old DNA remains together, while dispersive predicts mixed old and new fragments. Density gradient results showed hybrid DNA formation, disproving conservative and dispersive models and supporting semi-conservative replication.

7. Assertion-Reason:
Assertion (A): Semi-conservative replication ensures one parental strand is retained in daughter DNA.
Reason (R): Each parental strand acts as a template for new strand synthesis.
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: Correct answer is option 1. Each parental strand serves as template for synthesis of a complementary strand. This guarantees that one parental strand is conserved in each daughter DNA, confirming the semi-conservative mechanism and maintaining genetic fidelity across replication cycles.

8. Matching Type:
Match the type of DNA molecule with its description:
A. Light DNA – (i) DNA with only new strands
B. Heavy DNA – (ii) DNA with only old strands
C. Hybrid DNA – (iii) DNA with one old and one new strand
1. A–i, B–ii, C–iii
2. A–iii, B–i, C–ii
3. A–ii, B–i, C–iii
4. A–i, B–iii, C–ii
Explanation: Correct answer is option 1. Light DNA contains only newly synthesized strands, heavy DNA contains only old strands, and hybrid DNA contains one old (¹⁵N) and one new (¹⁴N) strand. This classification explains the density gradient results observed in the Meselson and Stahl experiment.

9. Fill in the Blanks:
The Meselson and Stahl experiment provided evidence for ______ replication of DNA.
1. Conservative
2. Semi-conservative
3. Dispersive
4. Random
Explanation: Correct answer is semi-conservative. The experiment showed that after DNA replication, each daughter molecule contains one parental and one new strand, confirming the semi-conservative model, and ruling out both conservative and dispersive models of DNA replication.

10. Choose the correct statements:
Statement I: Hybrid DNA forms after first replication cycle in ¹⁴N medium.
Statement II: After one generation, proportion of light DNA is zero.
1. Both I and II are correct
2. Only I is correct
3. Only II is correct
4. Both I and II are incorrect
Explanation: Correct answer is option 2. Hybrid DNA forms after the first replication in light medium, containing one old ¹⁵N strand and one new ¹⁴N strand. Since all molecules are hybrid, proportion of purely light DNA is zero at this stage, confirming the semi-conservative replication model.

Topic: Molecular Genetics; Subtopic: DNA Structure and Base Pairing

Keyword Definitions:

• DNA: Deoxyribonucleic acid, the hereditary material in organisms.

• Complementary strand: The DNA strand whose bases pair specifically with the original strand: A-T, G-C.

• Adenine (A): Purine base that pairs with thymine via two hydrogen bonds.

• Thymine (T): Pyrimidine base that pairs with adenine via two hydrogen bonds.

• Guanine (G): Purine base that pairs with cytosine via three hydrogen bonds.

• Cytosine (C): Pyrimidine base that pairs with guanine via three hydrogen bonds.

• Chargaff’s rule: In double-stranded DNA, amount of A = T and G = C.

• Base composition: Percentage of each nucleotide in DNA.

• Double stranded DNA: DNA composed of two complementary strands forming a double helix.

• Hydrogen bonding: Specific pairing between nucleotide bases via H-bonds.

• Genetic stability: Maintained by correct base pairing and complementary strands.

Lead Question - 2022 (Abroad)

One of the strands of double stranded DNA has base composition as follows: 15% A, 15% T, 40% G and 30% C. What will be the percentage of these bases in the complementary strand?

1. 15% A, 15% T, 30% G, 40% C

2. 15% A, 30% T, 40% G, 15% C

3. 15% A, 15% T, 40% G, 30% C

4. 15% A, 40% T, 15% G, 30% C

Explanation: According to base pairing rules in double stranded DNA, adenine pairs with thymine and guanine pairs with cytosine. Therefore, the complementary strand must have equal amounts of A and T, and equal amounts of G and C as in the original strand. Given 15% A, 15% T, 40% G, 30% C, the complementary strand will have 15% T, 15% A, 40% C, 30% G. Correct answer: 3

1. SINGLE CORRECT ANSWER MCQ

If a DNA strand contains 20% G, what percentage of C will the complementary strand have?

1. 20%

2. 30%

3. 40%

4. 50%

Explanation: In double stranded DNA, guanine pairs with cytosine. Therefore, the percentage of cytosine in the complementary strand will match the guanine percentage in the original strand. Given 20% G, the complementary strand will have 20% C, maintaining the A-T and G-C ratios. Correct answer: 1

2. SINGLE CORRECT ANSWER MCQ

Which base pairs via three hydrogen bonds?

1. Adenine-Thymine

2. Guanine-Cytosine

3. Adenine-Cytosine

4. Thymine-Guanine

Explanation: Guanine pairs with cytosine via three hydrogen bonds, providing stability to the DNA double helix. Adenine pairs with thymine through two hydrogen bonds. Mismatched pairs like A-C or T-G are unstable. Correct answer: 2

3. SINGLE CORRECT ANSWER MCQ

Chargaff’s rule states:

1. A + G = T + C

2. A = T and G = C

3. A + T = G + C

4. A + C = G + T

Explanation: Chargaff’s rule specifies that in double stranded DNA, the amount of adenine equals thymine, and guanine equals cytosine, ensuring complementarity and stability. This explains the matching percentages in complementary strands. Correct answer: 2

4. SINGLE CORRECT ANSWER MCQ

If the original strand has 25% T, what will be the percentage of A in the complementary strand?

1. 20%

2. 25%

3. 30%

4. 50%

Explanation: Adenine pairs with thymine. Therefore, in the complementary strand, A percentage equals T percentage in the original strand. Given 25% T, complementary strand will have 25% A, maintaining base pairing. Correct answer: 2

5. SINGLE CORRECT ANSWER MCQ

If one DNA strand has 35% A, what is the total percentage of G and C?

1. 15%

2. 30%

3. 65%

4. 35%

Explanation: Total A + T = 35% + 35% = 70%. Remaining 30% are G + C. DNA is complementary; hence percentage of G and C is 30%. Correct answer: 2

6. SINGLE CORRECT ANSWER MCQ

Which of the following is not a purine base?

1. Adenine

2. Guanine

3. Cytosine

4. Both A and B

Explanation: Purines are adenine and guanine. Cytosine is a pyrimidine base. Hence, Cytosine is not a purine. Correct answer: 3

7. ASSERTION-REASON MCQ

Assertion (A): Complementary strand has equal percentages of A and T.

Reason (R): Adenine always pairs with thymine in double-stranded 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 but R is false

4. A is false but R is true

Explanation: A pairs specifically with T. This base pairing ensures equal percentage of A and T in complementary strand. Both assertion and reason are correct and the reason explains the assertion. Correct answer: 1

8. MATCHING TYPE MCQ

Match base with complementary base:

Column A:

1. Adenine

2. Thymine

3. Guanine

4. Cytosine

Column B:

1. Guanine

2. Adenine

3. Thymine

4. Cytosine

Explanation: In double stranded DNA: A pairs with T, T with A, G with C, and C with G. These hydrogen bonding rules maintain complementary base pairing and DNA stability. Correct matching: 1-B, 2-C, 3-D, 4-A

9. FILL IN THE BLANKS / COMPLETION MCQ

Percentage of cytosine in the complementary strand will be ________ if the original strand has 30% cytosine.

1. 30%

2. 40%

3. 15%

4. 60%

Explanation: Cytosine pairs with guanine. The complementary strand will have guanine percentage equal to cytosine percentage in the original strand. Given 30% C, complementary strand has 30% G. Correct answer: 1

10. CHOOSE THE CORRECT STATEMENTS MCQ

Statement I: In double-stranded DNA, A = T and G = C.
Statement II: Complementary strand has same sequence as original strand.

1. Only Statement I is correct

2. Only Statement II is correct

3. Both Statements I and II are correct

4. Both Statements I and II are incorrect

Explanation: A equals T and G equals C in double-stranded DNA, maintaining base pairing. Complementary strand does not have same sequence; it has complementary bases. Therefore, only Statement I is correct. Correct answer: 1

Topic: Chromatin Structure; Subtopic: Nucleosome Organization

Keyword Definitions:

• Nucleosome: Basic structural unit of chromatin, consisting of DNA wrapped around histone proteins.

• Histone Octamer: Core of eight histone proteins around which DNA wraps to form nucleosome.

• Beads on String: Appearance of nucleosomes under electron microscope, with DNA connecting them.

• Chromatin: DNA-protein complex that packages eukaryotic DNA into nucleus, consisting of nucleosomes as repeating units.

• Base Pair (bp): Unit of measurement for DNA length, one pair of complementary nucleotides.

• Electron Microscope: High-resolution microscope used to visualize nucleosomes and chromatin structure.

• DNA Wrapping: DNA is negatively charged and wraps around positively charged histones for compaction.

Lead Question - 2022 (Abroad)
With respect to nucleosome, which of the following statements is incorrect ?
1. Nucleosome contains 120 bp of DNA helix
2. Nucleosomes are seen as 'beads on string' under Electron Microscope
3. DNA is wrapped around positively charged histone octamer to form nucleosome
4. Nucleosome is the repeating unit of chromatin
Explanation: The correct answer is Nucleosome contains 120 bp of DNA helix. Each nucleosome consists of approximately 147 base pairs of DNA wrapped around a histone octamer, not 120 bp. Nucleosomes appear as 'beads on a string' under electron microscopy, and DNA wraps around positively charged histones due to electrostatic interactions, forming the repeating structural unit of chromatin. This organization compacts DNA efficiently while allowing accessibility for transcription, replication, and repair. The nucleosome structure is fundamental to understanding eukaryotic chromatin dynamics and gene regulation, and deviations in base pair count can alter nucleosome stability and chromatin function.

1. Single Correct Answer Type:
How many histone proteins form the nucleosome core?
1. 4
2. 6
3. 8
4. 10
Explanation: The correct answer is 8. A nucleosome core consists of eight histone proteins: two each of H2A, H2B, H3, and H4. DNA wraps around this octamer to form the nucleosome. This arrangement compacts DNA into chromatin while preserving accessibility for regulatory proteins. Proper histone assembly is critical for nucleosome stability, chromatin organization, and gene expression control. Misassembly can disrupt chromatin structure and affect transcription and replication, demonstrating the central role of the histone octamer in eukaryotic DNA packaging and nucleosome function.

2. Single Correct Answer Type:
Nucleosomes are primarily composed of:
1. DNA and RNA
2. DNA and histone proteins
3. RNA and ribosomal proteins
4. DNA and transcription factors
Explanation: The correct answer is DNA and histone proteins. Nucleosomes consist of DNA wrapped around a histone octamer, forming the repeating unit of chromatin. RNA and ribosomal proteins are involved in translation, not chromatin packaging. Transcription factors regulate gene expression but do not form nucleosome cores. DNA-histone interactions provide structural stability, compact the genome, and enable regulated access to genetic information. This organization is essential for cellular processes including transcription, replication, repair, and epigenetic modifications in eukaryotic cells.

3. Single Correct Answer Type:
Under electron microscopy, nucleosomes appear as:
1. Loops
2. Fibers
3. Beads on string
4. Rings
Explanation: The correct answer is Beads on string. Nucleosomes, with DNA wrapped around histone octamers, appear as bead-like structures connected by linker DNA, giving a 'beads on a string' morphology under electron microscopy. This structure reflects the basic unit of chromatin compaction. Loops, fibers, and rings represent higher-order chromatin organization, not individual nucleosomes. Visualization of nucleosomes provides insight into chromatin dynamics, DNA accessibility, and regulation of transcription, replication, and repair processes in eukaryotic cells, demonstrating the hierarchical organization of DNA within the nucleus.

4. Single Correct Answer Type:
DNA wraps around histones due to:
1. Covalent bonds
2. Hydrogen bonds
3. Ionic interactions
4. Disulfide bonds
Explanation: The correct answer is Ionic interactions. DNA is negatively charged due to phosphate groups, and histones are positively charged. Electrostatic (ionic) interactions facilitate the wrapping of DNA around the histone octamer to form nucleosomes. Hydrogen bonds stabilize base pairing within DNA, while covalent or disulfide bonds do not mediate DNA-histone binding. These ionic interactions are essential for chromatin compaction, nucleosome stability, and regulation of gene accessibility, allowing dynamic chromatin remodeling during transcription, replication, and DNA repair in eukaryotic cells.

5. Single Correct Answer Type:
The repeating unit of chromatin is:
1. Gene
2. Nucleosome
3. Chromosome
4. Centromere
Explanation: The correct answer is Nucleosome. Nucleosomes, composed of DNA wrapped around histone octamers, form the repeating structural units of chromatin. This modular organization compacts DNA while maintaining accessibility for transcription, replication, and repair. Genes are functional units of DNA, chromosomes are higher-order structures, and centromeres are specialized regions of chromosomes. Nucleosomes provide the foundation for chromatin folding, epigenetic regulation, and genome stability, and are essential for orchestrating cellular processes within eukaryotic nuclei.

6. Single Correct Answer Type:
Approximate number of DNA base pairs in one nucleosome?
1. 100 bp
2. 120 bp
3. 147 bp
4. 200 bp
Explanation: The correct answer is 147 bp. Each nucleosome contains about 147 base pairs of DNA wrapped around a histone octamer. This precise length allows optimal DNA compaction and accessibility. 120 bp is incorrect and underestimates DNA content, while 100 or 200 bp does not reflect canonical nucleosome structure. This arrangement is fundamental for chromatin packaging, nucleosome spacing, and epigenetic regulation, ensuring proper genome organization, transcriptional control, and efficient replication in eukaryotic cells.

7. Assertion-Reason Type:
Assertion (A): Nucleosomes compact DNA into chromatin.
Reason (R): Histone proteins provide a positively charged scaffold for negatively charged DNA.
1. Both A and R are correct, and R is the correct explanation of A
2. Both A and R are correct, but R is not the correct explanation of A
3. A is correct, R is false
4. A is false, R is true
Explanation: Correct answer is Both A and R are correct, and R is the correct explanation of A. DNA is negatively charged due to phosphate groups, and histone octamers are positively charged. This electrostatic interaction enables DNA to wrap around histones, forming nucleosomes, the repeating unit of chromatin. This compacts DNA efficiently while allowing controlled access for transcription, replication, and repair. Nucleosomes organize DNA within the nucleus, and histone-DNA interactions are crucial for chromatin stability and regulation of genetic information in eukaryotic cells.

8. Matching Type:
Match the structure with its function:
A. Nucleosome → (i) Gene expression regulation
B. Histone octamer → (ii) DNA packaging
C. Linker DNA → (iii) Connect nucleosomes
D. Chromatin → (iv) Higher-order DNA organization
1. A-(i), B-(ii), C-(iii), D-(iv)
2. A-(

Topic: DNA Replication; Subtopic: Meselson and Stahl Experiment

• Semiconservative replication: Each new DNA molecule consists of one old (parental) strand and one newly synthesized strand after replication.

• Meselson and Stahl experiment: Conducted in 1958 using 15N-labeled E. coli to prove DNA replication is semiconservative.

• Equilibrium density gradient centrifugation: Technique using cesium chloride gradient to separate DNA based on density differences.

• Isotope labeling: Use of radioactive or heavy isotopes to trace biological molecules during experiments.

• DNA replication: Process of producing two identical copies of DNA from one original DNA molecule.

• Parental strand: Original DNA strand serving as a template during replication.

• Template mechanism: Each old strand guides synthesis of a new complementary strand in DNA replication.

• Replication fork: Y-shaped structure where DNA strands unwind for replication.

• DNA polymerase: Enzyme that synthesizes new DNA strands complementary to the template strand.

• Cesium chloride (CsCl): Compound used to create a density gradient in centrifugation to separate molecules by density.

• Isopycnic centrifugation: Separation of molecules solely based on their buoyant densities in a gradient medium.

Lead Question - 2022 (Abroad)

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Semiconservative replication was experimentally proved by Mathew Meselson and Franklin Stahl (1958).

Reason (R): Meselson and Stahl used radioactive isotope and equilibrium density gradient centrifugation technique.

In the light of the above statements, choose the correct answer from the options given below:

1. (A) is correct but (R) is not correct

2. (A) is not correct but (R) is correct

3. Both (A) and (R) are correct and (R) is the correct explanation of (A)

4. Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Explanation: Both statements are correct, but the reason is slightly inaccurate as they used non-radioactive heavy isotope 15N, not radioactive isotope. They employed equilibrium density gradient centrifugation in CsCl to demonstrate semiconservative replication. Hence, (A) is correct but (R) is not correct. Answer: 1

Q1: Which isotope was used by Meselson and Stahl in their experiment?

1. 32P

2. 14N

3. 15N

4. 3H

Explanation: Meselson and Stahl used heavy isotope 15N to label DNA in E. coli. The labeled DNA showed intermediate density after one generation in 14N medium, proving semiconservative replication. Radioactive isotopes were not used in the experiment. Answer: 15N. Answer: 3

Q2: What is the density of hybrid DNA after one generation in 14N medium?

1. Same as 15N DNA

2. Same as 14N DNA

3. Intermediate between 15N and 14N DNA

4. Higher than 15N DNA

Explanation: After one generation, DNA molecules consist of one old (15N) and one new (14N) strand, resulting in intermediate density. This hybrid band supported semiconservative replication. Answer: Intermediate between 15N and 14N DNA. Answer: 3

Q3: Which enzyme unwinds DNA strands during replication?

1. DNA polymerase

2. Helicase

3. Ligase

4. Primase

Explanation: DNA helicase unwinds the double helix at the replication fork, separating strands to allow complementary synthesis. DNA polymerase synthesizes new strands, ligase joins fragments, and primase adds RNA primers. Answer: Helicase. Answer: 2

Q4: DNA replication is called semiconservative because:

1. Both strands are newly synthesized

2. One strand is parental and one is new

3. Both strands are parental

4. DNA synthesis is incomplete

Explanation: Semiconservative replication means each daughter DNA molecule contains one parental (old) and one newly synthesized strand. This mechanism was confirmed by Meselson and Stahl’s experiment. Answer: One strand is parental and one is new. Answer: 2

Q5: Which technique separated DNA molecules of different densities in Meselson-Stahl experiment?

1. Chromatography

2. Gel electrophoresis

3. Equilibrium density gradient centrifugation

4. Ultrafiltration

Explanation: They used equilibrium density gradient centrifugation in cesium chloride (CsCl) solution, allowing separation of DNA based on density. This technique visually confirmed semiconservative replication pattern. Answer: Equilibrium density gradient centrifugation. Answer: 3

Q6: What type of gradient was used in the Meselson-Stahl experiment?

1. Sucrose gradient

2. Cesium chloride gradient

3. Potassium chloride gradient

4. Sodium chloride gradient

Explanation: Cesium chloride (CsCl) gradient was used in ultracentrifugation to separate DNA of different densities. Sucrose gradient is used for protein separation, not DNA. Answer: Cesium chloride gradient. Answer: 2

Q7 (Assertion-Reason): Assertion (A): DNA replication requires a primer. Reason (R): DNA polymerase can initiate synthesis de novo.

1. (A) is correct but (R) is not correct

2. (A) is not correct but (R) is correct

3. Both (A) and (R) are correct and (R) explains (A)

4. Both (A) and (R) are correct but (R) does not explain (A)

Explanation: DNA polymerase cannot start synthesis de novo; it requires a primer with a free 3’-OH group. Therefore, (A) is correct but (R) is not correct. Answer: 1

Q8 (Matching Type): Match the enzyme with its function:

1. Helicase    A. Joins Okazaki fragments
2. Primase    B. Synthesizes RNA primer
3. Ligase    C. Unwinds DNA helix
4. DNA polymerase    D. Synthesizes new DNA strand

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

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

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

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

Explanation: Helicase unwinds DNA (C), Primase synthesizes RNA primer (B), Ligase joins Okazaki fragments (A), and DNA polymerase adds nucleotides (D). Correct match is 1-C, 2-B, 3-A, 4-D. Answer: 1

Q9 (Fill in the Blank): Okazaki fragments are joined by the enzyme ______ during DNA replication.

1. Helicase

2. Primase

3. Ligase

4. Topoisomerase

Explanation: DNA ligase joins Okazaki fragments on the lagging strand by forming phosphodiester bonds, ensuring continuity of the strand. Helicase unwinds, primase adds primer, and topoisomerase reduces tension. Answer: Ligase. Answer: 3

Q10 (Choose Correct Statements): Choose the correct statements:

1. Replication proceeds bidirectionally in E. coli

2. Okazaki fragments form on the lagging strand

3. DNA synthesis occurs in 5’→3’ direction

4. DNA polymerase initiates synthesis without a primer

Explanation: Statements 1, 2, and 3 are correct. Replication in E. coli is bidirectional, Okazaki fragments occur on the lagging strand, and synthesis always proceeds 5’→3’. DNA polymerase requires a primer, so statement 4 is incorrect. Answer: 1, 2, and 3

Topic: Bacterial Transformation; Subtopic: Griffith Experiment

• Bacterial transformation: Process by which bacteria acquire new genetic traits from external DNA.

• Griffith experiment: Classic experiment demonstrating transformation in Streptococcus pneumoniae.

• S-strain: Smooth strain of bacteria, virulent due to polysaccharide capsule.

• R-strain: Rough strain of bacteria, non-virulent lacking capsule.

• Heat-killed bacteria: Bacteria killed by heat, unable to reproduce but may transfer genetic material.

• Capsule: Protective polysaccharide layer surrounding some bacteria, contributing to virulence.

• Virulence: Ability of a microorganism to cause disease.

• Genetic material: DNA responsible for heredity and inheritance of traits.

• Mice model: Laboratory mice used to study bacterial infections and transformations.

• Recombination: Incorporation of external genetic material into bacterial genome.

• Transformation principle: Discovery that genetic traits can be transferred from dead to living bacteria.

Lead Question - 2022 (Abroad)

Which one of the following experiments of Frederick Griffith resulted in the discovery of bacterial transformation?

1. S-strain (heat-killed) → injected into Mice → Mice lived

2. S-strain (heat-killed) + R-strain (live) → injected into Mice → Mice died

3. S-strain → injected into Mice → Mice died

4. R-strain → injected into Mice → Mice lived

Explanation: Griffith’s key discovery was that non-virulent R-strain bacteria acquired virulence when mixed with heat-killed virulent S-strain. The mixture injected into mice caused death, proving genetic material from dead S-strain transformed R-strain. This experiment demonstrated bacterial transformation. Answer: S-strain (heat-killed) + R-strain (live) → injected into Mice → Mice died. Answer: 2

Q1: Which component of S-strain contributes to its virulence?

1. Ribosome

2. Polysaccharide capsule

3. Flagella

4. Cell wall peptidoglycan

Explanation: The polysaccharide capsule surrounding S-strain bacteria protects it from host immune response, making it virulent. R-strain lacks this capsule and is non-virulent. Capsules play a critical role in pathogenicity and in the process of transformation where genetic material encoding capsule is transferred. Answer: Polysaccharide capsule. Answer: 2

Q2: Which bacterium was used in Griffith’s transformation experiment?

1. Escherichia coli

2. Streptococcus pneumoniae

3. Bacillus subtilis

4. Salmonella typhi

Explanation: Griffith used Streptococcus pneumoniae to study transformation. S-strain was virulent, R-strain non-virulent. Mixing heat-killed S-strain with live R-strain transformed R-strain to virulent form. This experiment identified the principle of bacterial transformation, paving the way for discovering DNA as genetic material. Answer: Streptococcus pneumoniae. Answer: 2

Q3: What did the live R-strain acquire from heat-killed S-strain?

1. Virulence gene

2. Antibiotic resistance

3. Flagella

4. Capsule synthesis inhibitor

Explanation: R-strain acquired the virulence gene from heat-killed S-strain, enabling capsule synthesis and pathogenicity. This transfer of genetic material caused previously harmless R-strain to kill mice. Griffith’s experiment demonstrated horizontal gene transfer in bacteria. Answer: Virulence gene. Answer: 1

Q4: Which type of experiment did Griffith perform?

1. In vitro protein synthesis

2. In vivo bacterial infection

3. In vitro DNA replication

4. In vivo plant transformation

Explanation: Griffith conducted in vivo experiments injecting bacteria into mice. He observed effects of different bacterial strains and their combinations on survival, discovering transformation. The in vivo model allowed demonstration of horizontal gene transfer under physiological conditions, unlike purely in vitro studies. Answer: In vivo bacterial infection. Answer: 2

Q5: Which R-strain feature changed after transformation?

1. Cell shape

2. Capsule production

3. Flagella type

4. Growth rate

Explanation: After transformation, R-strain acquired genes for capsule production from heat-killed S-strain. Capsule production changed its phenotype from non-virulent to virulent. Other features like shape, flagella, or growth rate were unchanged. Capsule formation was crucial for pathogenicity and identification of transformation principle. Answer: Capsule production. Answer: 2

Q6: What principle did Griffith’s experiment demonstrate?

1. DNA replication

2. Bacterial transformation

3. Protein synthesis

4. Mutation repair

Explanation: Griffith’s experiment revealed bacterial transformation, where genetic material from dead S-strain transformed live R-strain into virulent bacteria. This discovery established that genetic information could be transferred horizontally between bacteria, laying the foundation for DNA as the hereditary material. Answer: Bacterial transformation. Answer: 2

Q7: Assertion (A): Heat-killed S-strain alone cannot kill mice.
Reason (R): Virulence requires live bacterial activity and capsule production.

1. (A) is correct but R is not correct

2. (A) is not correct but R is correct

3. Both A and R are correct and R explains A

4. Both A and R are correct but R does not explain A

Explanation: Heat-killed S-strain lacks metabolic activity and cannot cause disease. Virulence depends on live bacterial processes including capsule synthesis. Therefore, mice injected with heat-killed S-strain survive. Both assertion and reason are correct, and reason explains the assertion by describing why live bacteria are necessary. Answer: Both A and R are correct and R explains A. Answer: 3

Q8: Match the bacterial strains with their characteristics:

A. S-strain    1. Non-virulent
B. R-strain    2. Virulent
C. Heat-killed S-strain    3. Cannot replicate

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

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

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

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

Explanation: S-strain is virulent, R-strain is non-virulent, and heat-killed S-strain cannot replicate. Matching identifies characteristics correctly and explains the basis for transformation, demonstrating transfer of virulence genes to R-strain. Answer: A-2, B-1, C-3. Answer: 1

Q9: The discovery that genetic traits can be transferred from dead to living bacteria is called ______.

1. Mutation

2. Transformation

3. Transduction

4. Conjugation

Explanation: Griffith’s experiment demonstrated transformation, the process by which genetic traits from dead bacteria are transferred to living bacteria, changing their phenotype. This principle laid the foundation for understanding DNA as the genetic material. Other processes like transduction or conjugation involve different mechanisms. Answer: Transformation. Answer: 2

Q10: Which of the following statements about Griffith’s experiment are correct?

1. Heat-killed S-strain alone did not kill mice

2. Live R-strain alone did not kill mice

3. Mixture of heat-killed S-strain and live R-strain killed mice

4. Live S-strain killed mice

Explanation: Heat-killed S-strain alone and live R-strain alone did not kill mice. Live S-strain killed mice, and the combination of heat-killed S-strain with live R-strain killed mice due to transformation. Correct statements are 1, 2, 3, and 4 depending on context. Answer: 1, 2, 3, 4

Topic: DNA Replication

Subtopic: Semi-conservative Replication

• DNA replication: Process of producing two identical copies of DNA from one original molecule.

• Semi-conservative: Each new DNA molecule consists of one original strand and one newly synthesized strand.

• Eukaryotes: Organisms with membrane-bound nuclei containing genetic material.

• Meselson-Stahl experiment: Classic experiment proving semi-conservative DNA replication in prokaryotes.

• Chromosome: DNA molecule carrying genes, structural unit of inheritance.

• Nucleotide: Building block of DNA consisting of a sugar, phosphate, and nitrogenous base.

• DNA polymerase: Enzyme catalyzing the synthesis of new DNA strands.

• Parent strand: Original DNA strand used as template during replication.

• Complementary base pairing: Adenine pairs with thymine and guanine pairs with cytosine in DNA.

• Hershey and Chase: Experiment demonstrating DNA as genetic material using bacteriophages.

• Taylor and colleagues: Experimental proof of semi-conservative DNA replication in eukaryotes.

Lead Question - 2022 (Abroad)

DNA replication is semi-conservative in nature was experimentally proved in eukaryotes by:

1. Hershey and Chase

2. Macleod and McCarty

3. Meselson and Stahl

4. Talyor and his colleagues

Explanation: The semi-conservative nature of DNA replication in eukaryotes was confirmed by Taylor and colleagues using radioactive labeling of chromosomes in Vicia faba root cells. Each daughter DNA molecule contained one old strand and one newly synthesized strand, proving semi-conservative replication. Other experiments focused on prokaryotes or DNA as genetic material. Answer: Talyor and his colleagues. Answer: 4

Q1: Which enzyme is responsible for unwinding the DNA double helix during replication?

1. DNA polymerase

2. Helicase

3. Ligase

4. Primase

Explanation: Helicase unwinds the DNA double helix by breaking hydrogen bonds between complementary bases, creating replication forks. DNA polymerase synthesizes new strands, primase provides RNA primers, and ligase joins Okazaki fragments. Helicase is essential for initiating replication, enabling polymerases to access template strands. Answer: Helicase. Answer: 2

Q2: In DNA replication, the strand synthesized continuously is called:

1. Lagging strand

2. Leading strand

3. Template strand

4. Okazaki strand

Explanation: The leading strand is synthesized continuously in the 5' to 3' direction toward the replication fork. The lagging strand is synthesized discontinuously as Okazaki fragments. Accurate synthesis ensures faithful replication of genetic material. Answer: Leading strand. Answer: 2

Q3: Which type of bond joins nucleotides in a DNA strand?

1. Hydrogen bond

2. Phosphodiester bond

3. Peptide bond

4. Ionic bond

Explanation: Nucleotides in a DNA strand are linked via phosphodiester bonds between the phosphate of one nucleotide and the sugar of the next. Hydrogen bonds connect complementary bases between strands. Phosphodiester bonds maintain the DNA backbone’s structural integrity. Answer: Phosphodiester bond. Answer: 2

Q4: Which nitrogenous base pairs with guanine in DNA?

1. Adenine

2. Thymine

3. Cytosine

4. Uracil

Explanation: In DNA, guanine pairs with cytosine via three hydrogen bonds, ensuring complementary base pairing. Adenine pairs with thymine, and uracil replaces thymine in RNA. Complementary base pairing is critical for semi-conservative replication and accurate transmission of genetic information. Answer: Cytosine. Answer: 3

Q5: Okazaki fragments are found on which DNA strand during replication?

1. Leading strand

2. Lagging strand

3. Template strand

4. RNA strand

Explanation: The lagging strand is synthesized discontinuously away from the replication fork in short segments called Okazaki fragments. These fragments are later joined by DNA ligase to form a continuous strand. The leading strand is synthesized continuously. Answer: Lagging strand. Answer: 2

Q6: Which enzyme joins Okazaki fragments on the lagging strand?

1. DNA polymerase

2. Helicase

3. Ligase

4. Primase

Explanation: DNA ligase seals the nicks between Okazaki fragments on the lagging strand, forming a continuous DNA molecule. DNA polymerase synthesizes new DNA, helicase unwinds the helix, and primase provides RNA primers. Ligase ensures structural continuity and stability of the replicated DNA. Answer: Ligase. Answer: 3

Q7: Assertion (A): DNA replication is semi-conservative.
Reason (R): Each daughter DNA molecule contains one parental and one newly synthesized strand.

1. (A) is correct but R is not correct

2. (A) is not correct but R is correct

3. Both A and R are correct and R explains A

4. Both A and R are correct but R does not explain A

Explanation: Semi-conservative replication means each daughter DNA has one old and one new strand. Taylor’s experiment in eukaryotes confirmed this. Both assertion and reason are correct, and the reason directly explains the assertion by describing the molecular outcome of semi-conservative replication. Answer: Both A and R are correct and R explains A. Answer: 3

Q8: Match the DNA replication components with their functions:

A. DNA polymerase    1. Unwinds helix
B. Helicase    2. Synthesizes DNA
C. Ligase    3. Joins fragments
D. Primase    4. Synthesizes RNA primer

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

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

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

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

Explanation: DNA polymerase synthesizes DNA, helicase unwinds the double helix, ligase joins Okazaki fragments, and primase synthesizes RNA primers. Correctly matching enzymes to their functions is essential to understand the replication process. Answer: A-2, B-1, C-3, D-4. Answer: 1

Q9: The process of making two identical DNA molecules from one original molecule is called ______.

1. Transcription

2. Replication

3. Translation

4. Mutation

Explanation: DNA replication produces two identical molecules from one parental DNA. It involves unwinding, complementary base pairing, and synthesis of new strands. This process ensures accurate transmission of genetic information across generations. Other options relate to protein synthesis or changes in DNA sequence. Answer: Replication. Answer: 2

Q10: Which statements are correct about DNA replication?

1. It is semi-conservative

2. Replication occurs in 5' to 3' direction

3. Leading strand is synthesized discontinuously

4. Okazaki fragments occur on lagging strand

Explanation: DNA replication is semi-conservative, and synthesis proceeds in 5' to 3' direction. The leading strand is synthesized continuously, not discontinuously. Okazaki fragments occur on the lagging strand. Therefore, correct statements are 1, 2, and 4. Answer: 1, 2, 4

Topic: Genetic Material

Subtopic: Criteria of Genetic Material

• Genetic Material: Molecule carrying hereditary information of an organism.

• DNA: Deoxyribonucleic acid, primary genetic material in most organisms.

• RNA: Ribonucleic acid, can also carry genetic information in some viruses.

• Mendelian Character: Trait that follows Mendel’s laws of inheritance.

• Replication: Process of making identical copies of genetic material.

• Chemical Stability: Ability of genetic material to remain chemically unchanged under normal conditions.

• Mutation: Change in nucleotide sequence of genetic material.

• Evolution: Change in gene frequency in a population over generations.

• Allele: Alternative form of a gene at the same locus.

• Chromosome: DNA molecule carrying genes, structural unit of inheritance.

• Phenotype: Observable characteristics of an organism determined by genotype.

Lead Question - 2022 (Abroad)

Which one of the following is not a criterion of genetic material?

1. Should not provide the scope for changes for evolution

2. Should be able to express itself in the form of Mendelian character

3. Should be able to generate its replica

4. Should be stable chemically and structurally

Explanation: Genetic material must allow variation for evolution, express Mendelian characters, replicate accurately, and remain chemically stable. Option 1 contradicts evolutionary requirement; genetic material must provide scope for changes. Therefore, it is not a valid criterion. Other options correctly describe essential properties of genetic material. Answer: Should not provide the scope for changes for evolution. Answer: 1

Q1: Which nucleic acid is the primary genetic material in most organisms?

1. RNA

2. DNA

3. ATP

4. tRNA

Explanation: DNA carries hereditary information in most organisms, encoding genes responsible for phenotype. RNA mainly functions in protein synthesis, but in some viruses it acts as genetic material. DNA is chemically stable and capable of accurate replication, fulfilling criteria for genetic material. Answer: DNA. Answer: 2

Q2: Which of the following features ensures faithful transmission of genetic material?

1. Replication

2. Mutation

3. Transcription

4. Translation

Explanation: Replication allows genetic material to produce exact copies, ensuring continuity of hereditary information across generations. Mutations introduce variation, and transcription/translation convert genetic information into functional proteins. Accurate replication is a defining criterion of genetic material. Answer: Replication. Answer: 1

Q3: Which statement is true about RNA as genetic material?

1. It is chemically unstable and cannot mutate

2. It can serve as genetic material in some viruses

3. It cannot encode proteins

4. It is found only in the nucleus

Explanation: RNA serves as genetic material in RNA viruses like retroviruses. It is less chemically stable than DNA but can mutate, providing evolution potential. RNA encodes proteins via transcription and translation. Answer: It can serve as genetic material in some viruses. Answer: 2

Q4: Which property of genetic material allows evolutionary change?

1. Replication accuracy

2. Mutability

3. Chemical stability

4. Mendelian expression

Explanation: Mutability of genetic material allows variation, which is essential for evolution. Without mutations, no new traits arise. Other properties like replication accuracy, chemical stability, and Mendelian expression ensure heredity, but variation for evolution requires mutability. Answer: Mutability. Answer: 2

Q5: Which of the following correctly expresses a Mendelian character?

1. Height determined by multiple genes

2. Blood group controlled by a single gene

3. Skin colour influenced by environment

4. Eye colour varying continuously

Explanation: Mendelian characters are controlled by single genes showing dominant/recessive inheritance. Blood group fits this definition. Polygenic traits like height or skin colour do not strictly follow Mendelian ratios. Therefore, blood group correctly expresses a Mendelian character. Answer: Blood group controlled by a single gene. Answer: 2

Q6: Chemical stability of genetic material is important because:

1. It prevents mutation completely

2. It allows long-term storage of information

3. It ensures expression in phenotype

4. It increases mutation rate

Explanation: Stability ensures that DNA remains intact to store genetic information reliably across generations. While mutations occur occasionally, chemical stability prevents frequent degradation. Stability also allows accurate replication and transmission of hereditary information. Other options misrepresent the role of chemical stability. Answer: It allows long-term storage of information. Answer: 2

Q7: Assertion (A): Genetic material must be able to replicate accurately.
Reason (R): Accurate replication ensures continuity of hereditary information across generations.

1. (A) is correct but R is not correct

2. (A) is not correct but R is correct

3. Both A and R are correct and R explains A

4. Both A and R are correct but R does not explain A

Explanation: Accurate replication is essential for genetic material to maintain hereditary information. Any errors can lead to mutations. The reason correctly explains the assertion because faithful replication is the mechanism that preserves gene sequences across generations. Answer: Both A and R are correct and R explains A. Answer: 3

Q8: Match the molecules with their genetic roles:

A. DNA    1. Protein synthesis
B. RNA    2. Genetic information storage
C. tRNA    3. Amino acid transport
D. Ribosome    4. Translation site

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

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

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

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

Explanation: DNA stores genetic information, RNA carries messages for protein synthesis, tRNA transports amino acids, and ribosomes are the translation sites. Correct matching identifies functions accurately for understanding molecular genetics. Answer: A-2, B-1, C-3, D-4. Answer: 1

Q9: The molecule responsible for hereditary information in most organisms is ______.

1. RNA

2. DNA

3. ATP

4. Protein

Explanation: DNA is the primary genetic material in most organisms, encoding all hereditary information. It is chemically stable, can replicate, express Mendelian characters, and allows variation. RNA carries information only in some viruses, and proteins or ATP do not store genetic information. Answer: DNA. Answer: 2

Q10: Choose the correct statements about genetic material:

1. It must replicate accurately

2. It must allow mutations for evolution

3. It must encode Mendelian characters

4. It must be chemically unstable

Explanation: Genetic material must replicate accurately, allow mutations to drive evolution, and encode Mendelian characters. Chemical instability is undesirable. Therefore, statements 1, 2, and 3 are correct, while 4 is wrong. Answer: 1, 2, 3

Topic: Gene Expression

Subtopic: Pleiotropy

Keyword Definitions:

• Pleiotropy: A condition in which a single gene affects multiple phenotypic traits in an organism.

• Gene: A segment of DNA that codes for a specific protein or functional RNA, influencing traits.

• Phenotype: Observable characteristics or traits of an organism resulting from gene expression and environment.

• Phenylketonuria (PKU): A genetic disorder caused by a defective gene affecting multiple traits, such as mental retardation and skin pigmentation.

Lead Question - 2022 (Abroad)

Select the correct statements with respect to pleiotropism:
(a) A gene is said to be pleiotropic if it affects more than one trait
(b) Phenylketonuria is an example of pleiotropy
(c) A condition where one gene has several alleles is referred to as pleiotropism
(d) A trait is said to be pleiotropic if several genes control it
Choose the correct answer from the options given below:
1. (a) and (b) only
2. (a) and (d) only
3. (a), (b) and (c) only
4. (b), (c) and (d) only

Explanation: Pleiotropy occurs when a single gene influences multiple traits. Phenylketonuria is a classic example, where one defective gene affects mental ability and skin color. Hence, the correct answer is (a) and (b) only.

1. Which of the following disorders shows pleiotropy?

1. Sickle cell anemia
2. Color blindness
3. Albinism
4. Haemophilia

Explanation: Sickle cell anemia shows pleiotropy, as one gene mutation affects hemoglobin shape, oxygen transport, and resistance to malaria. Thus, the correct answer is sickle cell anemia.

2. Which statement correctly defines pleiotropy?

1. A single gene controls multiple traits
2. Multiple genes control one trait
3. A gene has many alleles
4. Genes are located on the same chromosome

Explanation: Pleiotropy occurs when a single gene affects multiple traits. This contrasts with polygenic inheritance, where many genes affect one trait. Therefore, the correct answer is a single gene controls multiple traits.

3. Which of the following is NOT an example of pleiotropy?

1. Sickle cell anemia
2. Phenylketonuria
3. Marfan syndrome
4. ABO blood groups

Explanation: ABO blood groups involve multiple alleles but not pleiotropy, as each gene affects only one trait—blood type. Therefore, the correct answer is ABO blood groups.

4. Pleiotropy differs from polygenic inheritance in that:

1. Pleiotropy involves one gene, polygenic involves many
2. Pleiotropy involves many genes, polygenic involves one
3. Both involve the same gene
4. Both involve multiple alleles only

Explanation: Pleiotropy involves a single gene influencing many traits, whereas polygenic inheritance involves many genes influencing a single trait. Hence, the correct answer is pleiotropy involves one gene, polygenic involves many.

5. In humans, phenylketonuria affects:

1. Only nervous system
2. Only skin pigmentation
3. Both nervous system and skin pigmentation
4. Only liver function

Explanation: In phenylketonuria, a single gene defect affects both the nervous system and skin pigmentation due to accumulation of phenylalanine. Thus, the correct answer is both nervous system and skin pigmentation.

6. Which of the following describes pleiotropy best?

1. One gene → one trait
2. One gene → multiple traits
3. Multiple genes → one trait
4. Multiple genes → multiple traits

Explanation: Pleiotropy describes a single gene influencing multiple traits, such as in Marfan syndrome. Hence, the correct answer is one gene → multiple traits.

7. (Assertion–Reason Type)
Assertion: Pleiotropy means a single gene affects several traits.
Reason: It occurs because one gene may control a product used in different metabolic pathways.

1. Both Assertion and Reason are true, and Reason is the correct explanation.
2. Both are true, but Reason is not the correct explanation.
3. Assertion is true, but Reason is false.
4. Both are false.

Explanation: A pleiotropic gene affects several traits because the gene product may be utilized in multiple pathways. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion. The correct answer is both true and Reason is correct explanation.

8. (Matching Type)
Match the examples with their pleiotropic effects:

A. Marfan syndrome — (i) Connective tissue disorder
B. Phenylketonuria — (ii) Affects mental ability and pigmentation
C. Sickle cell anemia — (iii) Affects RBC shape and oxygen transport
D. Albinism — (iv) Lack of melanin only
1. A-i, B-ii, C-iii, D-iv
2. A-ii, B-i, C-iv, D-iii
3. A-iii, B-ii, C-i, D-iv
4. A-i, B-iii, C-ii, D-iv

Explanation: Marfan syndrome affects connective tissue, PKU affects brain and skin, sickle cell anemia affects RBCs, and albinism affects pigmentation only. Thus, the correct match is A-i, B-ii, C-iii, D-iv.

9. (Fill in the Blanks)
When a single gene affects multiple traits, the condition is called ___________.

1. polygenic inheritance
2. pleiotropy
3. codominance
4. multiple allelism

Explanation: The phenomenon where a single gene affects multiple phenotypic traits is known as pleiotropy. Hence, the correct answer is pleiotropy.

10. (Choose the Correct Statements)
Choose the correct statements about pleiotropy:
(a) One gene affects several traits
(b) It occurs in all genetic disorders
(c) Marfan syndrome is pleiotropic
(d) It involves many genes

1. (a) and (c) only
2. (b) and (d) only
3. (a), (b), and (c) only
4. (a), (c), and (d) only

Explanation: Pleiotropy refers to one gene affecting multiple traits. Marfan syndrome is a known pleiotropic condition. It does not occur in all genetic disorders. Hence, the correct answer is (a) and (c) only.

Topic: DNA Replication
Subtopic: Semi-conservative Replication in Prokaryotes

Keyword Definitions:

• E. coli: A rod-shaped, Gram-negative bacterium commonly used in molecular biology studies.

• 15N-dsDNA: DNA labelled with heavy nitrogen isotope 15N to track replication.

• 14N nucleotide: Normal nitrogen isotope nucleotide used in growth medium for new DNA synthesis.

• Semi-conservative replication: DNA replication method where each daughter DNA molecule contains one old and one newly synthesized strand.

• Prokaryotes: Unicellular organisms lacking a nucleus and membrane-bound organelles.

• DNA replication: Process of copying the DNA molecule to pass genetic information to daughter cells.

• Daughter cells: Cells produced after cell division, containing newly synthesized DNA.

Lead Question (2022):
Ten E. coli cells with 15N-dsDNA are incubated in medium containing 14N nucleotide. After 60 minutes, how many E. coli cells will have DNA totally free from 15N?
(1) 40 cells
(2) 60 cells
(3) 80 cells
(4) 20 cells

Explanation: E. coli divides roughly every 20–30 minutes under optimal conditions. After 60 minutes, approximately two generations occur. Starting with 10 cells, two doublings yield 40 cells. Semi-conservative replication produces hybrid DNA in first generation, but after two divisions, all 40 cells contain DNA synthesized only with 14N. Hence, option (1) is correct.

1. Semi-conservative replication means:
(1) Both strands of DNA are old
(2) Both strands of DNA are newly synthesized
(3) Each DNA molecule has one old and one new strand
(4) DNA does not replicate

Explanation: In semi-conservative replication, each daughter DNA contains one parental strand and one newly synthesized strand. This ensures genetic continuity. Therefore, the correct answer is option (3).

2. If 20 E. coli cells with 15N-dsDNA are grown in 14N medium for one generation, the DNA composition will be:
(1) 20 cells with only 15N DNA
(2) 20 cells with hybrid DNA
(3) 10 hybrid, 10 14N DNA
(4) All 14N DNA

Explanation: After one generation, each parental 15N strand pairs with a new 14N strand forming hybrid DNA. All 20 cells contain hybrid DNA. Hence, the correct answer is option (2).

3. The medium containing 14N nucleotides ensures:
(1) No DNA replication
(2) Only 15N incorporation
(3) Incorporation of 14N into new DNA strands
(4) DNA degradation

Explanation: Growing E. coli in a medium containing 14N ensures that new DNA strands incorporate 14N, replacing 15N over generations. This is used to demonstrate semi-conservative replication. Hence, option (3) is correct.

4. After two generations in 14N medium, the fraction of DNA free from 15N is:
(1) 0%
(2) 25%
(3) 50%
(4) 100%

Explanation: Semi-conservative replication yields hybrid DNA in the first generation. By the second generation, half of the DNA is fully 14N. Starting with 10 cells, this results in 40 cells with only 14N DNA. Hence, the correct answer is option (3).

5. E. coli DNA replication occurs in which manner?
(1) Conservative
(2) Semi-conservative
(3) Dispersive
(4) Random

Explanation: E. coli replicates DNA semi-conservatively, where each daughter DNA has one parental strand and one newly synthesized strand. This was confirmed by the Meselson-Stahl experiment using 15N and 14N. Hence, the correct answer is option (2).

6. In the Meselson-Stahl experiment, heavy DNA is labelled with:
(1) 12C
(2) 14N
(3) 15N
(4) 13C

Explanation: The Meselson-Stahl experiment used 15N, a heavy nitrogen isotope, to label parental DNA. It allowed observation of semi-conservative replication when E. coli were transferred to 14N medium. Hence, the correct answer is option (3).

7. Assertion-Reason:
Assertion (A): All daughter DNA molecules are hybrid after one generation in 14N medium.
Reason (R): Each parental 15N strand pairs with a newly synthesized 14N strand.
(1) Both A and R are true and R is correct explanation
(2) Both A and R are true but R is not correct explanation
(3) A is true, R is false
(4) A is false, R is true

Explanation: After one generation in 14N medium, each parental 15N strand pairs with 14N, forming hybrid DNA in all daughter cells. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion. Correct answer is option (1).

8. Matching Type: Match DNA replication type with description:
A. Conservative – 1. Parental strands separate and combine with new strands
B. Semi-conservative – 2. Entire parental DNA remains intact
C. Dispersive – 3. DNA segments are interspersed with old and new
(1) A–2, B–1, C–3
(2) A–1, B–2, C–3
(3) A–3, B–1, C–2
(4) A–2, B–3, C–1

Explanation: Conservative replication keeps parental DNA intact (A–2), semi-conservative produces one old and one new strand per DNA (B–1), and dispersive replication produces interspersed old and new DNA segments (C–3). Hence, option (1) is correct.

9. Fill in the blank:
The experiment demonstrating semi-conservative DNA replication was performed by ________.

(1) Watson and Crick
(2) Meselson and Stahl
(3) Avery and MacLeod
(4) Hershey and Chase

Explanation: The Meselson-Stahl experiment using 15N and 14N isotopes demonstrated that DNA replication is semi-conservative. Therefore, the correct answer is option (2).

10. Choose the correct statements:
(a) E. coli DNA replication is semi-conservative
(b) 15N DNA becomes hybrid after one generation in 14N
(c) Two generations are required to produce fully 14N DNA
(d) 14N DNA cannot replicate

(1) a, b, c
(2) a, d
(3) b, d
(4) a, c

Explanation: Statements (a), (b), and (c) correctly describe semi-conservative replication and the replacement of 15N with 14N over two generations. Statement (d) is false. Hence, the correct answer is option (1).

Topic: DNA Structure
Subtopic: DNA Length and Base Pair Relationship

Keyword Definitions:

• DNA: Deoxyribonucleic acid, a molecule that carries genetic information in living organisms.

• Base Pair (bp): A pair of nitrogenous bases connected by hydrogen bonds in a DNA molecule, such as A–T and G–C.

• Nucleotide: Basic unit of DNA containing a sugar, phosphate group, and nitrogen base.

• Double Helix: The coiled structure of DNA with two complementary strands wound around each other.

• Genomic DNA: Total genetic material contained within an organism’s chromosomes.

Lead Question – 2022
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs:
(1) 6.6×109 bp
(2) 3.3×106 bp
(3) 6.6×106 bp
(4) 3.3×109 bp

Explanation: The distance between two consecutive base pairs is 0.34 nanometres. Therefore, 1.1 m DNA equals 1.1×109 nm. Dividing by 0.34 nm per bp gives approximately 3.3×109 base pairs. Hence, the correct answer is (4).

1. In a DNA double helix, the two strands are:
(1) Parallel and identical
(2) Antiparallel and complementary
(3) Antiparallel and identical
(4) Parallel and complementary
Explanation: In DNA, one strand runs 5′ to 3′ and the other 3′ to 5′. Their base sequences are complementary (A–T, G–C). Thus, the two strands are antiparallel and complementary. Hence, the correct answer is (2).

2. The distance between two consecutive base pairs in DNA is:
(1) 0.34 nm
(2) 3.4 nm
(3) 34 nm
(4) 0.034 nm
Explanation: Each base pair in the DNA double helix is separated by a distance of 0.34 nanometres. Thus, ten base pairs together form one complete turn measuring 3.4 nm. Hence, the correct answer is (1).

3. The number of base pairs in one complete turn of B-DNA is:
(1) 5
(2) 10
(3) 15
(4) 20
Explanation: In B-form DNA, there are 10 base pairs per complete turn of the helix, and each turn measures 3.4 nm. Hence, the correct answer is (2).

4. The length of a human diploid cell DNA is approximately:
(1) 1.1 m
(2) 2.2 m
(3) 3.3 m
(4) 6.6 m
Explanation: A haploid human genome (one set of chromosomes) contains about 3.3×109 base pairs, corresponding to 1.1 m DNA. Hence, a diploid cell with two sets has approximately 2.2 m DNA. Correct answer is (2).

5. If 200 base pairs of DNA make one turn, what is its total length?
(1) 68 nm
(2) 6.8 nm
(3) 0.68 μm
(4) 0.068 μm
Explanation: Each base pair contributes 0.34 nm to the DNA length. Thus, 200 × 0.34 = 68 nm. Therefore, total length = 68 nm. Hence, the correct answer is (1).

6. Which of the following statements about DNA is correct?
(1) It contains uracil instead of thymine
(2) Its sugar component is ribose
(3) Its backbone is made of phosphate and deoxyribose sugar
(4) It is single-stranded in all organisms
Explanation: DNA is made up of deoxyribose sugar and phosphate forming its backbone, while bases (A, T, G, C) are attached. Uracil is found in RNA, not DNA. Hence, the correct answer is (3).

7. Assertion (A): DNA strands are antiparallel.
Reason (R): The hydrogen bonds form only when one strand runs 5′→3′ and the other 3′→5′.
(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: The hydrogen bonding between bases is possible only when strands are oriented oppositely (antiparallel). Thus, both A and R are true, and R correctly explains A. Hence, the correct answer is (1).

8. Match the following:
A. Hydrogen bond — (i) Adenine–Thymine
B. Phosphodiester bond — (ii) Between sugars and phosphates
C. Complementary base pairing — (iii) DNA stability
D. Double helix — (iv) Watson and Crick
Options:
(1) A-(i), B-(ii), C-(iii), D-(iv)
(2) A-(ii), B-(i), C-(iv), D-(iii)
(3) A-(iv), B-(iii), C-(ii), D-(i)
(4) A-(iii), B-(i), C-(iv), D-(ii)
Explanation: Hydrogen bonds link complementary bases (A–T, G–C), phosphodiester bonds link sugar-phosphate backbone, and the double helix structure was proposed by Watson and Crick. Hence, the correct answer is (1).

9. Fill in the blank:
Each complete turn of B-DNA measures _______ nm.
(1) 0.34
(2) 3.4
(3) 34
(4) 0.034
Explanation: In the B-form of DNA, there are 10 base pairs per turn, and the distance between adjacent base pairs is 0.34 nm. Hence, one complete turn measures 3.4 nm. Correct answer is (2).

10. Choose the correct statements:
(1) A–T pairs have three hydrogen bonds.
(2) G–C pairs have two hydrogen bonds.
(3) Base pairing follows Chargaff’s rule.
(4) A–T pairs have two hydrogen bonds.
Explanation: According to Chargaff’s rule, A pairs with T via two hydrogen bonds, and G pairs with C via three hydrogen bonds. This complementary pairing ensures structural stability of DNA. Hence, statements (3) and (4) are correct.

Topic: Gene Expression and Regulation
Subtopic: Lac Operon Model in E. coli

Keyword Definitions:

• i gene: A regulatory gene in lac operon that codes for the repressor protein controlling transcription.

• Inducer: A molecule, such as lactose or allolactose, that binds to the repressor to initiate gene expression.

• RNA polymerase: Enzyme responsible for transcribing DNA into RNA.

• z, y, a genes: Structural genes of lac operon coding for β-galactosidase, permease, and transacetylase respectively.

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

Lead Question – 2022
In an E. coli strain, i gene mutated and its product cannot bind the inducer molecule. If the growth medium is provided with lactose, what will be the outcome:
(1) z, y, a genes will be transcribed
(2) z, y, a genes will not be translated
(3) RNA polymerase will bind the promoter region
(4) Only z gene will get transcribed

Explanation: A mutated i gene produces a defective repressor that cannot bind the inducer. Thus, the repressor remains active and continues to block transcription of structural genes. Hence, z, y, a genes are not transcribed. The correct answer is (2) z, y, a genes will not be translated.

1. The lac operon is switched on in the presence of:
(1) Glucose
(2) Maltose
(3) Lactose
(4) Galactose
Explanation: Lactose acts as an inducer in the lac operon model, inactivating the repressor and initiating transcription of z, y, a genes. The correct answer is (3) Lactose.

2. Which of the following acts as a repressor in the lac operon?
(1) z gene
(2) y gene
(3) i gene product
(4) a gene product
Explanation: The i gene codes for the repressor protein that binds to the operator region to prevent transcription. The correct answer is (3) i gene product.

3. The lac operon controls the metabolism of:
(1) Lactose
(2) Sucrose
(3) Fructose
(4) Maltose
Explanation: The lac operon enables E. coli to metabolize lactose by producing enzymes β-galactosidase, permease, and transacetylase. The correct answer is (1) Lactose.

4. The function of β-galactosidase is to:
(1) Convert lactose into glucose and galactose
(2) Transport lactose into cell
(3) Convert glucose into lactose
(4) Bind repressor to operator
Explanation: β-galactosidase catalyzes the hydrolysis of lactose into glucose and galactose. The correct answer is (1) Convert lactose into glucose and galactose.

5. In the lac operon, the operator gene functions as:
(1) Promoter for RNA polymerase
(2) Binding site for repressor
(3) Structural gene for enzyme
(4) Site for inducer binding
Explanation: The operator gene serves as the binding site for the repressor protein, controlling transcription of structural genes. The correct answer is (2) Binding site for repressor.

6. If both glucose and lactose are present in E. coli medium:
(1) Only lactose is utilized
(2) Only glucose is utilized
(3) Both are utilized equally
(4) Neither is utilized
Explanation: Due to catabolite repression, E. coli first utilizes glucose as the preferred carbon source before lactose. The correct answer is (2) Only glucose is utilized.

7. Assertion (A): In lac operon, the presence of lactose induces gene expression.
Reason (R): Lactose binds to the repressor, making it inactive.
(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) Both A and R are false
Explanation: Lactose acts as an inducer by binding to the repressor, inactivating it, allowing transcription of structural genes. Both statements are true, and R correctly explains A. Correct answer is (1).

8. Match the following:
A. z gene — (i) β-galactosidase
B. y gene — (ii) Permease
C. a gene — (iii) Transacetylase
Options:
(1) A-(ii), B-(i), C-(iii)
(2) A-(i), B-(ii), C-(iii)
(3) A-(iii), B-(ii), C-(i)
(4) A-(ii), B-(iii), C-(i)
Explanation: In lac operon, z codes for β-galactosidase, y for permease, and a for transacetylase. The correct answer is (2) A-(i), B-(ii), C-(iii).

9. Fill in the blank:
The structural genes of lac operon are transcribed when ______ is present.
(1) Glucose
(2) Maltose
(3) Lactose
(4) Fructose
Explanation: Lactose acts as an inducer that inactivates the repressor protein, enabling transcription of structural genes. The correct answer is (3) Lactose.

10. Choose the correct statements:
(1) Lac operon is an example of inducible operon
(2) Repressor binds to promoter region
(3) Inducer increases repression
(4) Structural genes code for repressors
Explanation: Lac operon is an inducible operon because its transcription is activated by the presence of lactose, the inducer. The correct answer is (1) Lac operon is an example of inducible operon.

Topic: Genome Sequencing
Subtopic: Functional Annotation of Genes

Keyword Definitions:
• Blind Approach: Sequencing strategy without prior knowledge of gene order.
• Genome Sequencing: Determining the complete DNA sequence of an organism’s genome.
• Gene Mapping: Determining the physical or genetic location of genes on chromosomes.
• Expressed Sequence Tags (ESTs): Short DNA sequences derived from expressed genes used for identifying gene transcripts.
• Bioinformatics: Computational analysis of biological data, including DNA, RNA, and protein sequences.
• Sequence Annotation: Assigning function or features to specific DNA segments after sequencing.
• Function Assignment: Predicting the role of a DNA segment based on sequence analysis.
• Contigs: Overlapping DNA sequences assembled to reconstruct longer genomic regions.
• ORF: Open reading frame, a DNA segment potentially coding for a protein.
• Comparative Genomics: Using sequence similarity to infer function or evolutionary relationships.

Lead Question (2022):
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as:
Options:
1. Gene mapping
2. Expressed sequence tags
3. Bioinformatics
4. Sequence annotation

Explanation: The correct answer is 4. Sequence annotation. In the blind approach, the genome is sequenced without prior knowledge, and specific DNA segments are later analyzed to assign functions. Gene mapping determines positions, ESTs identify transcripts, and bioinformatics involves computational analysis, but functional annotation defines roles of sequenced regions.

Guessed MCQs:

1. Which technique identifies short sequences of expressed genes?
Options:
(a) ESTs
(b) Gene mapping
(c) Sequence annotation
(d) Recombinant DNA
Explanation: The correct answer is (a) ESTs. Expressed Sequence Tags are short DNA sequences derived from cDNA that represent transcribed genes. They are used for discovering new genes, analyzing expression, and aiding genome annotation, unlike mapping or recombinant DNA approaches.

2. Which process reconstructs genome sequences from overlapping DNA fragments?
Options:
(a) Contig assembly
(b) Gene mapping
(c) PCR
(d) Transformation
Explanation: The correct answer is (a) Contig assembly. Overlapping DNA fragments are computationally aligned to form continuous sequences, facilitating whole genome reconstruction. Gene mapping locates genes, PCR amplifies DNA, and transformation introduces DNA into organisms, not genome assembly.

3. Assertion-Reason MCQ:
Assertion (A): Bioinformatics is essential in genome sequencing.
Reason (R): It allows analysis of large datasets and prediction of gene function.
Options:
(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: The correct answer is (a). Bioinformatics handles massive genomic datasets, enabling sequence assembly, gene prediction, and functional annotation. Without computational tools, sequencing projects would be inefficient and error-prone, making bioinformatics crucial for analyzing and interpreting genome information.

4. Matching Type MCQ:
Match the methodology with its description:
List - I List - II
(a) Gene mapping (i) Assigns positions on chromosomes
(b) ESTs (ii) Short sequences from expressed genes
(c) Sequence annotation (iii) Assigns function to DNA segments
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 answer is 1. Gene mapping determines gene positions, ESTs are short sequences from transcribed genes, and sequence annotation assigns functions to genomic segments, forming a coordinated framework in genome analysis and functional genomics.

5. Which approach sequences an entire genome without prior gene order knowledge?
Options:
(a) Blind sequencing
(b) Targeted sequencing
(c) PCR amplification
(d) Microarray analysis
Explanation: The correct answer is (a) Blind sequencing. Also called shotgun sequencing, it fragments the genome randomly and sequences all fragments. Targeted sequencing focuses on specific regions, PCR amplifies segments, and microarrays detect gene expression, not whole genome sequences.

6. Single Correct Answer:
Which computational method predicts coding regions in DNA?
Options:
(a) ORF prediction
(b) Gene mapping
(c) PCR
(d) Restriction digestion
Explanation: The correct answer is (a) ORF prediction. Open Reading Frame analysis identifies sequences potentially coding for proteins, aiding annotation. Gene mapping locates genes, PCR amplifies DNA, and restriction digestion cuts DNA but does not predict coding regions.

7. Fill in the Blanks:
Functional assignment of sequenced DNA segments is known as __________.
Options:
(a) Sequence annotation
(b) Gene mapping
(c) EST analysis
(d) Cloning
Explanation: The correct answer is (a) Sequence annotation. After sequencing, genomic segments are analyzed to assign functions using computational tools, databases, and comparative genomics. Mapping, ESTs, and cloning serve different purposes.

8. Which term describes using computational tools to analyze genomic data?
Options:
(a) Bioinformatics
(b) Gene mapping
(c) PCR
(d) Cloning
Explanation: The correct answer is (a) Bioinformatics. It encompasses computational techniques for sequence alignment, annotation, functional prediction, and comparative genomics. Gene mapping locates genes physically or genetically, while PCR and cloning are experimental wet-lab techniques.

9. Single Correct Answer:
Which method identifies all expressed genes in a tissue?
Options:
(a) EST sequencing
(b) Shotgun genome sequencing
(c) Gene mapping
(d) Southern blotting
Explanation: The correct answer is (a) EST sequencing. Expressed Sequence Tags are derived from cDNA and represent active genes in a tissue. Shotgun sequencing sequences the entire genome, gene mapping locates genes, and Southern blotting detects specific DNA sequences.

10. Choose the correct statements:
(i) Blind sequencing fragments the genome randomly
(ii) Sequence annotation assigns function
(iii) Gene mapping predicts coding sequences
(iv) Bioinformatics supports computational analysis
Options:
(a) i, ii, iv
(b) i, iii
(c) ii, iii, iv
(d) i, ii, iii, iv
Explanation: The correct answer is (a) i, ii, iv. Blind sequencing randomly fragments genomes, sequence annotation assigns functions, and bioinformatics analyzes large datasets. Gene mapping locates genes but does not predict coding sequences directly.

Topic: Molecular Genetics
Subtopic: DNA Polymorphism and Applications

Keyword Definitions:

• DNA Polymorphism: Variations in DNA sequences among individuals of a species, useful in genetic studies.

• DNA Fingerprinting: Technique to identify individuals using unique DNA patterns.

• Genetic Mapping: Process of determining the location of genes on chromosomes.

• Translation: Process of synthesizing proteins from mRNA template.

• Chromosome: DNA-protein complex carrying genes in eukaryotic cells.

• Restriction Enzymes: Proteins that cut DNA at specific sequences, used in fingerprinting.

• Polymorphic Markers: DNA sequences that vary between individuals, aiding identification.

• Genome: Complete set of genetic material in an organism.

• Allele: Variant form of a gene at a specific locus.

• Marker: DNA sequence used to track inheritance or identify individuals.

Lead Question (2022)
DNA polymorphism forms the basis of:
(1) DNA fingerprinting
(2) Both genetic mapping and DNA fingerprinting
(3) Translation
(4) Genetic mapping

Explanation:
DNA polymorphisms create unique patterns among individuals, which are utilized in DNA fingerprinting for identification and in genetic mapping to locate genes on chromosomes. Translation does not involve DNA polymorphism. Therefore, the correct answer is (2).

1. Single Correct Answer MCQ:
Which technique identifies individuals using unique DNA sequences?
(1) DNA fingerprinting
(2) Translation
(3) Genetic mapping
(4) PCR only

Explanation:
DNA fingerprinting analyzes polymorphic DNA sequences to uniquely identify individuals. Translation synthesizes proteins, genetic mapping locates genes, and PCR amplifies DNA but does not identify individuals alone. Correct answer is (1).

2. Single Correct Answer MCQ:
Polymorphic markers are used for:
(1) DNA fingerprinting
(2) Protein synthesis
(3) RNA splicing
(4) Photosynthesis

Explanation:
Polymorphic markers vary among individuals and are used to generate DNA fingerprints for identification and in genetic mapping. They are not involved in protein synthesis, RNA splicing, or photosynthesis. Correct answer is (1).

3. Single Correct Answer MCQ:
Genetic mapping determines:
(1) Protein sequence
(2) Location of genes on chromosomes
(3) Translation efficiency
(4) DNA replication rate

Explanation:
Genetic mapping identifies positions of genes on chromosomes using markers, including DNA polymorphisms. Protein sequence, translation, and replication rates are unrelated to mapping. Correct answer is (2).

4. Single Correct Answer MCQ:
DNA polymorphism is NOT used for:
(1) DNA fingerprinting
(2) Genetic mapping
(3) Translation
(4) Forensic analysis

Explanation:
DNA polymorphisms are used for fingerprinting, genetic mapping, and forensic analysis. They are unrelated to translation, which synthesizes proteins from mRNA. Correct answer is (3).

5. Single Correct Answer MCQ:
Which enzyme is used to cut DNA for fingerprinting?
(1) RNA polymerase
(2) Restriction enzyme
(3) DNA ligase
(4) DNA helicase

Explanation:
Restriction enzymes cut DNA at specific sequences to generate fragments for fingerprinting. RNA polymerase synthesizes RNA, DNA ligase joins fragments, and helicase unwinds DNA. Correct answer is (2).

6. Single Correct Answer MCQ:
Which term describes different forms of a gene?
(1) Chromosome
(2) Allele
(3) Marker
(4) Exon

Explanation:
Alleles are variant forms of a gene at a specific locus. Chromosome carries genes, markers help track inheritance, and exons code for proteins. Correct answer is (2).

7. Assertion-Reason MCQ:
Assertion (A): DNA polymorphism is used in forensic science
Reason (R): Individuals have identical DNA sequences
Options:
(1) Both A and R correct and R explains A
(2) A correct, R incorrect
(3) A incorrect, R correct
(4) Both incorrect

Explanation:
DNA polymorphism is used in forensic science to distinguish individuals because their DNA sequences vary. The reason is incorrect because sequences are not identical. Correct answer is (2).

8. Matching Type MCQ:
Match technique with application:
A. DNA fingerprinting — 1. Protein synthesis
B. Genetic mapping — 2. Locating genes
C. Translation — 3. Individual identification
Options:
(1) A–3, B–2, C–1
(2) A–2, B–3, C–1
(3) A–1, B–2, C–3
(4) A–3, B–1, C–2

Explanation:
DNA fingerprinting identifies individuals (A–3), genetic mapping locates genes on chromosomes (B–2), and translation synthesizes proteins (C–1). Correct answer is (1).

9. Fill in the Blanks:
Technique that uses DNA polymorphism to identify individuals is ________.
(1) Genetic mapping
(2) DNA fingerprinting
(3) Translation
(4) PCR only

Explanation:
DNA polymorphism generates unique patterns among individuals, which is utilized in DNA fingerprinting for identification in forensic science and paternity tests. Genetic mapping locates genes, translation synthesizes proteins, and PCR alone does not identify individuals. Correct answer is (2).

10. Choose the correct statements MCQ:
(a) DNA polymorphism is used in genetic mapping
(b) DNA polymorphism is used in DNA fingerprinting
(c) Translation uses DNA polymorphism
(d) Polymorphic markers help identify individuals
Options:
(1) a, b, d only
(2) a, c only
(3) b, c only
(4) a, b, c, d

Explanation:
DNA polymorphisms are used in genetic mapping, DNA fingerprinting, and for identification using polymorphic markers. Translation does not involve DNA polymorphism. Correct statements are (a), (b), and (d). Therefore, the correct answer is (1).

Topic: Molecular Genetics
Subtopic: Chromatin Structure and Nucleosome

Keyword Definitions:

• Euchromatin: Loosely packed chromatin that is transcriptionally active and accessible for RNA synthesis.

• Heterochromatin: Densely packed chromatin that is usually transcriptionally inactive.

• Histone octamer: Protein complex of eight histone molecules around which DNA wraps to form a nucleosome.

• DNA: Negatively charged nucleic acid that wraps around histones to form chromatin.

• Histones: Positively charged proteins rich in lysine and arginine that help package DNA.

• Nucleosome: Basic structural unit of chromatin, consisting of DNA wrapped around a histone octamer.

• Base pairs (bp): Pair of complementary nucleotides in DNA; 1 bp ≈ 0.34 nm in length.

• Transcriptionally active: Regions of DNA being actively transcribed into RNA.

• Transcriptionally inactive: DNA regions that are tightly packed and not transcribed.

• Chromatin: Complex of DNA and protein found in eukaryotic cells.

Lead Question (2022)
Read the following statements and choose the set of correct statements:
(a) Euchromatin is loosely packed chromatin
(b) Heterochromatin is transcriptionally active
(c) Histone octomer is wrapped by negatively charged DNA in nucleosome
(d) Histones are rich in lysine and arginine
(e) A typical nucleosome contains 400 bp of DNA helix
(1) (a), (c), (d) only
(2) (b), (e) only
(3) (a), (c), (e) only
(4) (b), (d), (e) only

Explanation:
Euchromatin is loosely packed and transcriptionally active, heterochromatin is inactive. DNA wraps around histone octamer, and histones are rich in lysine and arginine. A typical nucleosome contains approximately 146 bp, not 400 bp. Correct statements are (a), (c), and (d). Therefore, the correct answer is (1).

1. Single Correct Answer MCQ:
Which chromatin type is transcriptionally inactive?
(1) Euchromatin
(2) Heterochromatin
(3) Nucleosome
(4) Histone octamer

Explanation:
Heterochromatin is densely packed and transcriptionally inactive, whereas euchromatin is active. Nucleosome is the structural unit of chromatin, and histone octamer is a protein complex, not a chromatin type. Correct answer is (2).

2. Single Correct Answer MCQ:
The histone proteins are rich in which amino acids?
(1) Lysine and Arginine
(2) Glycine and Serine
(3) Alanine and Valine
(4) Aspartate and Glutamate

Explanation:
Histones are basic proteins rich in positively charged lysine and arginine residues, which interact with negatively charged DNA to form nucleosomes. Other amino acids like glycine, alanine, or acidic residues do not predominate in histones. Correct answer is (1).

3. Single Correct Answer MCQ:
Approximately how many base pairs of DNA wrap around a histone octamer in a nucleosome?
(1) 146 bp
(2) 400 bp
(3) 50 bp
(4) 1000 bp

Explanation:
Each nucleosome contains DNA wrapped around histone octamer, about 146 base pairs long. Option 400 bp is incorrect. This wrapping helps compact DNA and regulate gene expression. Correct answer is (1).

4. Single Correct Answer MCQ:
Which statement about euchromatin is correct?
(1) It is transcriptionally inactive
(2) It is loosely packed
(3) It is rich in heterochromatin
(4) It contains 400 bp DNA

Explanation:
Euchromatin is loosely packed and transcriptionally active, allowing gene expression. It is not heterochromatin, nor does it contain exactly 400 bp DNA. Correct answer is (2).

5. Single Correct Answer MCQ:
DNA wraps around histone octamer due to:
(1) DNA being positively charged
(2) Histones being negatively charged
(3) Histones being positively charged
(4) Hydrophobic interactions only

Explanation:
DNA is negatively charged due to phosphate backbone. Histones are positively charged, rich in lysine and arginine, enabling electrostatic interactions that wrap DNA around histone octamers to form nucleosomes. Correct answer is (3).

6. Single Correct Answer MCQ:
Heterochromatin is generally found:
(1) In actively transcribed regions
(2) At centromeres and telomeres
(3) Loosely packed
(4) Wrapped around histone H1 only

Explanation:
Heterochromatin is densely packed, transcriptionally inactive, and commonly located at centromeres and telomeres, contributing to chromosome stability. Euchromatin is active and loosely packed. Correct answer is (2).

7. Assertion-Reason MCQ:
Assertion (A): Nucleosome formation compacts DNA
Reason (R): DNA is negatively charged and wraps around positively charged histones
Options:
(1) Both A and R correct and R explains A
(2) A correct, R incorrect
(3) A incorrect, R correct
(4) Both incorrect

Explanation:
Nucleosome formation compacts DNA in chromatin, and the negative DNA wraps around positively charged histones due to electrostatic interactions. Both statements are correct, and the reason explains the assertion. Correct answer is (1).

8. Matching Type MCQ:
Match component with property:
A. Histone octamer — 1. 146 bp DNA wrapped
B. Nucleosome — 2. Eight histone proteins
C. Euchromatin — 3. Loosely packed, transcriptionally active
D. Heterochromatin — 4. Densely packed, transcriptionally inactive
Options:
(1) A–2, B–1, C–3, D–4
(2) A–1, B–2, C–4, D–3
(3) A–3, B–4, C–1, D–2
(4) A–2, B–1, C–4, D–3

Explanation:
Histone octamer: eight histone proteins (A–2), Nucleosome: 146 bp DNA wrapped (B–1), Euchromatin: loosely packed active (C–3), Heterochromatin: densely packed inactive (D–4). Correct answer is (1).

9. Fill in the Blanks:
Transcriptionally inactive chromatin is called ________.
(1) Euchromatin
(2) Heterochromatin
(3) Nucle

Topic: Gene Expression
Subtopic: Translation

Keyword Definitions:

• mRNA (Messenger RNA): RNA molecule that carries genetic code from DNA to ribosome for protein synthesis.

• Translation: Process by which ribosomes synthesize proteins using mRNA as a template.

• Ribosome: Cellular machinery composed of small and large subunits that facilitates translation.

• tRNA (Transfer RNA): RNA molecule that delivers amino acids to the ribosome during protein synthesis.

• Subunit: Either small or large component of the ribosome that assembles during translation.

• Protein Synthesis: The process of building polypeptides according to the genetic code carried by mRNA.

• Initiation: The first phase of translation where ribosomal subunits, mRNA, and initiator tRNA assemble.

Lead Question (2022)
The process of translation of mRNA to proteins begins as soon as :
(1) The larger subunit of ribosome encounters mRNA
(2) Both the subunits join together to bind with mRNA
(3) The tRNA is activated and the larger subunit of ribosome encounters mRNA
(4) The small subunit of ribosome encounters mRNA

Explanation:
Translation starts when the small ribosomal subunit binds to mRNA and locates the start codon. Only after this, initiator tRNA pairs with start codon, followed by joining of the large subunit to form a functional ribosome. Correct answer is (4).

1. Single Correct Answer MCQ:
Which RNA carries amino acids to the ribosome during translation?
(1) mRNA
(2) rRNA
(3) tRNA
(4) snRNA

Explanation:
tRNA molecules carry specific amino acids to the ribosome by matching their anticodon with mRNA codons, enabling polypeptide chain formation. mRNA carries the code, rRNA forms ribosome structure, and snRNA participates in splicing. Correct answer is (3).

2. Single Correct Answer MCQ:
Which codon signals the start of translation?
(1) UAA
(2) AUG
(3) UAG
(4) UGA

Explanation:
AUG codon serves as the start codon for translation, encoding methionine and signaling the small ribosomal subunit to assemble with mRNA and initiator tRNA. UAA, UAG, and UGA are stop codons signaling termination. Correct answer is (2).

3. Single Correct Answer MCQ:
Which ribosomal subunit recognizes the mRNA first during initiation?
(1) Large subunit
(2) Small subunit
(3) Both simultaneously
(4) None

Explanation:
The small ribosomal subunit binds first to the mRNA, scans for the start codon, and then allows the initiator tRNA to pair before joining with the large subunit. The large subunit alone cannot initiate translation. Correct answer is (2).

4. Single Correct Answer MCQ:
Which RNA type is a structural component of ribosomes?
(1) mRNA
(2) tRNA
(3) rRNA
(4) miRNA

Explanation:
rRNA forms the core structure of ribosomal subunits and catalyzes peptide bond formation. mRNA is the template, tRNA delivers amino acids, and miRNA regulates gene expression. Correct answer is (3).

5. Single Correct Answer MCQ:
What is the role of initiator tRNA in translation?
(1) Carries ribosomal proteins
(2) Brings first amino acid to start codon
(3) Terminates translation
(4) Splices introns from mRNA

Explanation:
The initiator tRNA carries methionine and pairs with the AUG start codon on mRNA, beginning translation. Other tRNAs bring subsequent amino acids. Splicing is unrelated, and termination requires stop codons. Correct answer is (2).

6. Single Correct Answer MCQ:
Which phase of translation involves ribosome assembling around mRNA and initiator tRNA?
(1) Elongation
(2) Initiation
(3) Termination
(4) Splicing

Explanation:
Initiation is the first phase where the small ribosomal subunit binds mRNA, the initiator tRNA pairs with the start codon, and the large subunit assembles to form a functional ribosome. Elongation extends the chain, termination ends it, and splicing occurs pre-translation. Correct answer is (2).

7. Assertion-Reason MCQ:
Assertion (A): Translation starts when the small ribosomal subunit binds mRNA.
Reason (R): Large ribosomal subunit alone can read codons and initiate protein synthesis.
Options:
(1) Both A and R are correct and R explains A
(2) A correct, R incorrect
(3) A incorrect, R correct
(4) Both A and R incorrect

Explanation:
Translation begins with the small subunit binding mRNA and locating the start codon. The large subunit alone cannot initiate translation. Therefore, the Assertion is correct but the Reason is incorrect. Correct answer is (2).

8. Matching Type MCQ:
Match molecule with its function:
A. mRNA — 1. Carries genetic code
B. tRNA — 2. Transfers amino acids
C. rRNA — 3. Ribosome structure and catalysis
D. Initiator tRNA — 4. Brings first amino acid
Options:
(1) A–1, B–2, C–3, D–4
(2) A–2, B–1, C–3, D–4
(3) A–1, B–3, C–2, D–4
(4) A–3, B–2, C–1, D–4

Explanation:
mRNA carries the genetic code, tRNA transfers amino acids, rRNA forms ribosome structure and catalyzes peptide bonds, and initiator tRNA brings the first methionine. Correct matching is A–1, B–2, C–3, D–4. Correct answer is (1).

9. Fill in the Blanks:
The ________ subunit of ribosome binds first to mRNA during translation initiation.
(1) Large
(2) Small
(3) Both
(4) None

Explanation:
The small ribosomal subunit binds to mRNA and scans for the start codon, allowing initiator tRNA binding before large subunit joins. Correct answer is (2).

10. Choose the Correct Statements:
(a) Small ribosomal subunit binds mRNA first
(b) Initiator tRNA pairs with start codon
(c) Large subunit joins later
(d) Translation starts with large subunit alone
Options:
(1) a, b, c
(2) a and d only
(3) b and c only
(4) all of the above

Explanation:
Translation initiation involves small subunit binding mRNA (a), initiator tRNA pairing with start codon (b), and subsequent joining of large subunit (c). Translation cannot start with large subunit alone (d is false). Correct answer is (1).

Topic: Chromatin Structure
Subtopic: Histones

Keyword Definitions:

• Histones: Proteins around which DNA winds to form nucleosomes in chromatin.

• Lysine and Arginine: Positively charged amino acids abundant in histones.

• Nucleosome: Structural unit of chromatin consisting of DNA wrapped around 8 histone proteins.

• Chromatin: Complex of DNA and proteins in the nucleus.

• Positive charge: Enables histones to bind negatively charged DNA phosphate backbone.

• Acidic pH: Incorrect for histones; they are basic due to positive amino acids.

Lead Question - 2021

Which one of the following statements about Histones is wrong?

(1) The pH of histones is slightly acidic
(2) Histones are rich in amino acids Lysine and Arginine
(3) Histones carry positive charge in the side chain
(4) Histones are organized to form a unit of 8 molecules

Explanation: Histones are basic proteins, rich in lysine and arginine, carrying positive charges, and organized as octamers (8 molecules) in nucleosomes. The statement about acidic pH is incorrect. Correct answer is option (1) The pH of histones is slightly acidic.

1. Which histone is not part of the core octamer in nucleosomes?
(1) H2A
(2) H2B
(3) H3
(4) H1
Explanation: Histone H1 is not part of the nucleosome core octamer; it acts as a linker histone stabilizing DNA between nucleosomes. H2A, H2B, H3, and H4 form the octamer core. Correct answer is option (4) H1.

2. Histones bind to DNA primarily because:
(1) DNA is positively charged
(2) DNA is negatively charged
(3) Histones are hydrophobic
(4) Histones are acidic
Explanation: DNA has a negatively charged phosphate backbone, which attracts positively charged histone proteins rich in lysine and arginine. This electrostatic interaction facilitates nucleosome formation. Correct answer is option (2) DNA is negatively charged.

3. The basic amino acids in histones are:
(1) Glutamate and Aspartate
(2) Lysine and Arginine
(3) Phenylalanine and Tyrosine
(4) Serine and Threonine
Explanation: Lysine and Arginine are basic amino acids present in histones, providing positive charges that bind negatively charged DNA. Glutamate and aspartate are acidic, phenylalanine/tyrosine are aromatic, and serine/threonine are polar uncharged. Correct answer is option (2) Lysine and Arginine.

4. Nucleosomes are formed by wrapping DNA around:
(1) H1 only
(2) H2A-H2B and H3-H4 octamer
(3) H1-H3 only
(4) H2B-H4 dimer
Explanation: Nucleosomes are DNA wrapped around an octamer of histones: two molecules each of H2A, H2B, H3, and H4. H1 is a linker histone not in the core octamer. Correct answer is option (2) H2A-H2B and H3-H4 octamer.

5. Histone modification that relaxes chromatin is:
(1) Methylation
(2) Acetylation
(3) Phosphorylation
(4) Ubiquitination
Explanation: Acetylation of lysine residues on histone tails neutralizes positive charges, reducing DNA binding and relaxing chromatin structure, facilitating transcription. Methylation can repress or activate, phosphorylation and ubiquitination have regulatory roles. Correct answer is option (2) Acetylation.

6. Which histone is involved in higher-order chromatin folding?
(1) H2A
(2) H2B
(3) H3
(4) H1
Explanation: Histone H1 is the linker histone that binds DNA between nucleosomes and is essential for higher-order chromatin folding into 30 nm fibers. Core histones (H2A, H2B, H3, H4) form the nucleosome. Correct answer is option (4) H1.

7. Assertion-Reason Question:
Assertion (A): Histones are rich in lysine and arginine.
Reason (R): These amino acids give histones a positive charge enabling DNA binding.
(1) Both A and R are true, R is correct explanation of A
(2) Both A and R are true, R is not correct explanation of A
(3) A true, R false
(4) A false, R true
Explanation: Histones are indeed rich in lysine and arginine. These basic amino acids provide positive charges which allow electrostatic interaction with negatively charged DNA. Both statements are true and the reason correctly explains the assertion. Correct answer is option (1).

8. Matching Type Question:
Match histone types with function:
(a) H2A/H2B - 1. Linker histone
(b) H3/H4 - 2. Nucleosome core
(c) H1 - 3. Stabilizes linker DNA
(1) a-2, b-2, c-3
(2) a-1, b-2, c-3
(3) a-2, b-1, c-3
(4) a-3, b-2, c-1
Explanation: H2A/H2B and H3/H4 form nucleosome core (a-2, b-2). H1 is linker histone stabilizing DNA between nucleosomes (c-3). Correct answer is option (1).

9. Fill in the Blanks:
The ________ histone binds between nucleosomes and helps in higher-order chromatin structure.
(1) H2A
(2) H2B
(3) H3
(4) H1
Explanation: Histone H1 is the linker histone that binds between nucleosomes, facilitating higher-order chromatin folding. H2A, H2B, H3 are core nucleosome histones. Correct answer is option (4) H1.

10. Choose the correct statements:
(a) Histones are basic proteins.
(b) DNA wraps around histone octamers.
(c) Histones are acidic in nature.
(d) H1 is a linker histone.
(1) a, b, d
(2) a, c, d
(3) b, c, d
(4) a, b, c
Explanation: Histones are basic proteins (a), DNA wraps around octamers (b), and H1 is a linker histone (d). They are not acidic. Correct answer is option (1) a, b, d.

Topic: Genetic Code
Subtopic: Codons and Translation

Keyword Definitions:

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

• AUG: Start codon that codes for methionine and initiates translation.

• AAA and AAG: Codons that code for the amino acid lysine.

• Methionine: Amino acid coded by start codon AUG.

• Phenylalanine: Amino acid coded by UUU or UUC codons, not AUG.

• Lysine: Essential amino acid coded by AAA and AAG codons.

Lead Question - 2021

Statement I : The codon 'AUG' codes for methionine and phenylalanine.
Statement II : 'AAA' and 'AAG' both codons code for the amino acid lysine.
In the light of the above statements, choose the correct answer from the options given below.

(1) Both Statement I and Statement II are false
(2) Statement I is correct but Statement II is false
(3) Statement I is incorrect but Statement II is true
(4) Both Statement I and Statement II are true

Explanation: AUG is the start codon coding only for methionine, not phenylalanine, making Statement I incorrect. AAA and AAG codons both code for lysine, so Statement II is correct. Therefore, the correct answer is option (3) Statement I is incorrect but Statement II is true.

1. Which codon is recognized as the start codon in mRNA?
(1) UAA
(2) AUG
(3) UAG
(4) UGA
Explanation: AUG serves as the start codon in mRNA, signaling the initiation of translation and coding for methionine. UAA, UAG, and UGA are stop codons that terminate translation. Correct answer is option (2) AUG.

2. The codons UUU and UUC code for:
(1) Methionine
(2) Phenylalanine
(3) Lysine
(4) Leucine
Explanation: UUU and UUC codons specifically code for phenylalanine. Methionine is coded by AUG, lysine by AAA and AAG, and leucine by UUA, UUG, CUU-CUC-CUA-CUG. Correct answer is option (2) Phenylalanine.

3. Which of the following is a stop codon?
(1) AUG
(2) AAA
(3) UGA
(4) AAG
Explanation: UGA is one of the three stop codons (UAA, UAG, UGA) that terminate protein synthesis. AUG is start codon, AAA and AAG code for lysine. Correct answer is option (3) UGA.

4. Codons that specify the same amino acid are called:
(1) Degenerate codons
(2) Start codons
(3) Stop codons
(4) Anticodons
Explanation: Degenerate codons refer to multiple codons coding for the same amino acid, providing redundancy in genetic code. Start codon initiates translation, stop codons terminate, and anticodons are in tRNA. Correct answer is option (1) Degenerate codons.

5. Which amino acid is coded by both AAA and AAG?
(1) Lysine
(2) Methionine
(3) Phenylalanine
(4) Arginine
Explanation: Lysine is encoded by codons AAA and AAG. Methionine is AUG, phenylalanine is UUU/UUC, arginine is CGU-CGC-CGA-CGG. Correct answer is option (1) Lysine.

6. In mRNA, the codon for methionine is:
(1) AUG
(2) UUU
(3) AAA
(4) UAA
Explanation: AUG codon is both start signal and codes for methionine. UUU codes phenylalanine, AAA codes lysine, UAA is a stop codon. Correct answer is option (1) AUG.

7. Assertion-Reason Question:
Assertion (A): AUG codon codes for methionine.
Reason (R): AUG is recognized by tRNA with anticodon UAC.
(1) Both A and R are true, R is correct explanation of A
(2) Both A and R are true, R is not correct explanation of A
(3) A true, R false
(4) A false, R true
Explanation: AUG codon codes methionine, and the corresponding tRNA anticodon is UAC, which pairs with AUG. Both statements are true and the reason explains the assertion. Correct answer is option (1).

8. Matching Type Question:
Match codons with amino acids:
A. UUU - 1. Lysine
B. AAA - 2. Phenylalanine
C. AAG - 3. Lysine
D. AUG - 4. Methionine
(1) A-2, B-1, C-3, D-4
(2) A-1, B-2, C-3, D-4
(3) A-2, B-3, C-1, D-4
(4) A-4, B-2, C-3, D-1
Explanation: UUU codes phenylalanine (A-2), AAA codes lysine (B-1), AAG codes lysine (C-3), AUG codes methionine (D-4). Correct answer is option (1).

9. Fill in the Blanks:
The codon ________ serves as the initiation codon and codes for methionine.
(1) UUU
(2) AUG
(3) AAA
(4) UGA
Explanation: AUG is the start codon initiating translation and codes for methionine. UUU codes phenylalanine, AAA lysine, and UGA is stop codon. Correct answer is option (2) AUG.

10. Choose the correct statements:
(a) UUU codes phenylalanine.
(b) AUG is start codon.
(c) AAA and AAG code lysine.
(d) UGA codes methionine.
(1) a, b, c
(2) a, b, d
(3) b, c, d
(4) a, c, d
Explanation: UUU codes phenylalanine, AUG is start codon, AAA and AAG code lysine. UGA is stop codon, not methionine. Correct answer is option (1) a, b, c.

Topic: Protein Synthesis

Subtopic: Role of RNA Types

Keyword Definitions:

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

• tRNA Transfer RNA brings specific amino acids to ribosomes during protein synthesis.

• rRNA Ribosomal RNA forms the structural and catalytic core of ribosomes.

• siRNA Small interfering RNA regulates gene expression by silencing specific mRNAs.

• Protein synthesis Biological process where ribosomes assemble amino acids into proteins using genetic instructions.

Lead Question - 2021

Which of the following RNAs is not required for the synthesis of protein?
(1) tRNA
(2) rRNA
(3) siRNA
(4) mRNA

Explanation: Protein synthesis requires mRNA, tRNA, and rRNA for accurate translation. siRNA, however, is not directly involved in protein synthesis but functions in RNA interference and silencing of specific genes. Thus, the correct answer is siRNA, as it does not take part in translation machinery.

1. Which enzyme synthesizes mRNA during transcription?
(1) DNA polymerase
(2) RNA polymerase II
(3) Reverse transcriptase
(4) Ligase

Explanation: mRNA is synthesized from a DNA template by RNA polymerase II in eukaryotes. This enzyme catalyzes complementary base pairing and elongation of the RNA strand. DNA polymerase functions in replication, not transcription. Reverse transcriptase synthesizes DNA from RNA. Therefore, the correct answer is RNA polymerase II.

2. Ribosomes are composed mainly of:
(1) Proteins and rRNA
(2) DNA and proteins
(3) Only proteins
(4) Only RNA

Explanation: Ribosomes are complex structures made up of ribosomal RNA (rRNA) and ribosomal proteins. rRNA provides catalytic function for peptide bond formation, while proteins stabilize structure. DNA does not form ribosomes. Hence, the correct answer is proteins and rRNA, which together drive translation.

3. During protein synthesis, tRNA carries:
(1) DNA
(2) Nucleotides
(3) Amino acids
(4) Ribosomes

Explanation: Transfer RNA (tRNA) has anticodon loops complementary to mRNA codons and attaches specific amino acids. This ensures correct placement of amino acids during polypeptide elongation. It does not carry DNA, nucleotides, or ribosomes. The correct answer is amino acids.

4. Assertion (A): mRNA carries codons for protein synthesis.
Reason (R): Codons are three-nucleotide sequences that specify 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, R is false
(4) A is false, R is true

Explanation: Both statements are true. Codons in mRNA consist of three nucleotide bases, each representing a specific amino acid during translation. mRNA indeed carries codons and provides the template for protein synthesis. Therefore, the correct option is (1).

5. Match the following:
A. mRNA - i. Structural role in ribosome
B. tRNA - ii. Template for protein
C. rRNA - iii. Brings amino acids
Options:
(1) A-ii, B-iii, C-i
(2) A-iii, B-ii, C-i
(3) A-i, B-ii, C-iii
(4) A-ii, B-i, C-iii

Explanation: mRNA acts as a template carrying genetic code, tRNA brings specific amino acids, and rRNA forms structural and catalytic core of ribosomes. Thus, the correct matching is A-ii, B-iii, C-i, making option (1) correct.

6. The first amino acid in protein synthesis in eukaryotes is ________.
(1) Valine
(2) Methionine
(3) Glycine
(4) Alanine

Explanation: In eukaryotic cells, protein synthesis initiates with methionine coded by start codon AUG. This is brought by a special initiator tRNA. In prokaryotes, N-formyl methionine serves the same role. Hence, the correct answer is methionine.

7. Choose the correct statements about genetic code:
(1) It is triplet in nature
(2) It is degenerate
(3) It is universal
(4) All of these

Explanation: Genetic code has universal features. It is a triplet code where three nucleotides form a codon, it is degenerate as multiple codons code for the same amino acid, and it is nearly universal across organisms. Thus, the correct option is all of these.

8. The anticodon is located on:
(1) mRNA
(2) rRNA
(3) tRNA
(4) DNA

Explanation: Anticodon is a set of three nucleotide bases complementary to mRNA codon and is present in tRNA. It ensures specific binding and incorporation of amino acids into the polypeptide chain. Therefore, the correct answer is tRNA.

9. The ribosomal site where the first tRNA binds during initiation is:
(1) A-site
(2) P-site
(3) E-site
(4) None of these

Explanation: During initiation of translation, the initiator tRNA carrying methionine binds to the P-site of the ribosome. A-site receives new tRNAs, and E-site releases empty tRNAs. Hence, the correct answer is P-site.

10. Which RNA has catalytic activity in peptide bond formation?
(1) tRNA
(2) rRNA
(3) mRNA
(4) siRNA

Explanation: Ribosomal RNA (rRNA), specifically 23S rRNA in prokaryotes and 28S rRNA in eukaryotes, acts as a ribozyme catalyzing peptide bond formation during protein synthesis. Neither tRNA nor mRNA catalyzes bond formation. Thus, the correct answer is rRNA.

Topic: Transcription in Prokaryotes

Subtopic: Enzymes Involved in Transcription

• DNA dependent RNA polymerase: Enzyme that synthesizes RNA using DNA as a template, capable of initiation, elongation, and termination in prokaryotes.

• DNA Ligase: Enzyme joining DNA fragments during replication, not involved in transcription.

• DNase: Enzyme degrading DNA, unrelated to RNA synthesis.

• DNA dependent DNA polymerase: Enzyme synthesizing DNA from DNA template during replication.

• Initiation: Starting phase of transcription where RNA polymerase binds to promoter.

• Elongation: Addition of nucleotides to growing RNA strand.

• Termination: Ending phase of transcription where RNA polymerase releases RNA transcript.

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

• Prokaryotes: Single-celled organisms lacking nucleus.

• Transcription: Process of synthesizing RNA from DNA template.

• RNA transcript: Newly synthesized RNA strand from transcription.

Lead Question - 2021

Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes?

(1) DNA dependent RNA polymerase
(2) DNA Ligase
(3) DNase
(4) DNA dependent DNA polymerase

Explanation: DNA dependent RNA polymerase is the only enzyme that can catalyze all three stages of transcription—initiation, elongation, and termination—in prokaryotes. DNA ligase, DNase, and DNA polymerase do not participate in RNA synthesis. Answer: DNA dependent RNA polymerase.

1. Single Correct Answer MCQ: Which enzyme synthesizes RNA using DNA template in prokaryotes?
Options:
A. DNA dependent RNA polymerase
B. DNA Ligase
C. DNase
D. DNA polymerase

Explanation: DNA dependent RNA polymerase synthesizes RNA from DNA in prokaryotes. DNA ligase joins DNA fragments, DNase degrades DNA, and DNA polymerase synthesizes DNA. Answer: DNA dependent RNA polymerase.

2. Single Correct Answer MCQ: The initiation phase of transcription involves:
Options:
A. RNA polymerase binding to promoter
B. RNA degradation
C. DNA replication
D. DNA repair

Explanation: During initiation, RNA polymerase binds to the promoter to start RNA synthesis. DNA replication and repair are unrelated, and RNA degradation occurs after transcription. Answer: RNA polymerase binding to promoter.

3. Single Correct Answer MCQ: Which enzyme is responsible for elongation of RNA transcript?
Options:
A. DNA Ligase
B. DNase
C. DNA dependent RNA polymerase
D. DNA polymerase

Explanation: DNA dependent RNA polymerase adds ribonucleotides to the growing RNA chain during elongation. DNA ligase, DNase, and DNA polymerase do not participate in RNA synthesis. Answer: DNA dependent RNA polymerase.

4. Single Correct Answer MCQ: Termination of transcription is catalyzed by:
Options:
A. DNA polymerase
B. DNA Ligase
C. DNA dependent RNA polymerase
D. DNase

Explanation: RNA polymerase recognizes termination signals and releases the RNA transcript. DNA dependent RNA polymerase is responsible for termination, not DNA polymerase, ligase, or DNase. Answer: DNA dependent RNA polymerase.

5. Single Correct Answer MCQ: Which enzyme is absent in RNA synthesis but present in DNA replication?
Options:
A. DNA dependent RNA polymerase
B. DNA Ligase
C. DNase
D. Sigma factor

Explanation: DNA ligase is needed for joining DNA fragments in replication but does not function in RNA synthesis. RNA polymerase catalyzes transcription. Answer: DNA Ligase.

6. Single Correct Answer MCQ: Sigma factor is associated with which phase of transcription?
Options:
A. Initiation
B. Elongation
C. Termination
D. DNA repair

Explanation: Sigma factor guides RNA polymerase to promoter regions during initiation. It is released during elongation and not involved in termination or DNA repair. Answer: Initiation.

7. Assertion-Reason MCQ:
Assertion (A): DNA dependent RNA polymerase can perform initiation, elongation, and termination.
Reason (R): DNA polymerase synthesizes RNA in prokaryotes.
Options:
A. Both A and R true, R correct explanation
B. Both A and R true, R not correct explanation
C. A true, R false
D. A false, R true

Explanation: DNA dependent RNA polymerase performs all transcription stages, but DNA polymerase synthesizes DNA, not RNA. Assertion true, reason false. Answer: A true, R false.

8. Matching Type MCQ:
List I: a. Initiation b. Elongation c. Termination d. RNA polymerase
List II: i. Adds nucleotides ii. Binds promoter iii. Releases RNA iv. Catalyzes all stages
Options:
A. a-ii, b-i, c-iii, d-iv
B. a-i, b-ii, c-iii, d-iv
C. a-iii, b-ii, c-i, d-iv
D. a-iv, b-i, c-ii, d-iii

Explanation: Initiation = promoter binding, Elongation = nucleotide addition, Termination = RNA release, RNA polymerase catalyzes all stages. Answer: a-ii, b-i, c-iii, d-iv.

9. Fill in the Blanks MCQ: The enzyme responsible for transcription in prokaryotes is _______.
Options:
A. DNA Ligase
B. DNA dependent RNA polymerase
C. DNase
D. DNA polymerase

Explanation: DNA dependent RNA polymerase synthesizes RNA from DNA template, performing initiation, elongation, and termination. Answer: DNA dependent RNA polymerase.

10. Choose the correct statements MCQ:
Options:
A. DNA dependent RNA polymerase catalyzes all transcription stages
B. DNA polymerase synthesizes RNA
C. DNase degrades DNA
D. Sigma factor assists in initiation
Select:
1. A, B
2. A, C, D
3. B, C
4. All of the above

Explanation: DNA dependent RNA polymerase catalyzes all transcription stages (A), DNase degrades DNA (C), and sigma factor assists initiation (D). DNA polymerase does not synthesize RNA. Correct statements: A, C, D. Answer: A, C, D.

Topic: Transcription

Subtopic: Gene Expression and RNA Processing

• RNA Polymerase: Enzyme that synthesizes RNA using DNA template.

• Rho Factor: Protein involved in termination of transcription in prokaryotes.

• Coding Strand: DNA strand with same sequence as mRNA (except T→U).

• Transcription Unit: Segment of DNA transcribed into RNA.

• Split Gene: Gene containing exons and introns, typical of eukaryotes.

• Prokaryotes: Organisms lacking nucleus, with continuous genes and no introns.

• Capping: Addition of 7-methylguanosine to 5' end of hnRNA.

• hnRNA: Heterogeneous nuclear RNA, precursor to mRNA.

• Exons: Coding sequences retained in mature mRNA.

• Introns: Non-coding sequences removed during RNA processing.

• Termination of Transcription: Process ending RNA synthesis at specific sequences.

Lead Question - 2021

Identify the correct statement:

1. RNA polymerase binds with Rho factor to terminate transcription in bacteria.
Options:
A. True
B. False
C. Only in eukaryotes
D. Cannot say

Explanation: In prokaryotes, Rho factor binds to RNA and moves toward RNA polymerase to terminate transcription at specific sites. This process is known as rho-dependent termination. It ensures proper release of RNA transcript. Answer: True.

2. The coding strand in a transcription unit is copied to an mRNA.
Options:
A. True
B. False
C. Only partially
D. Only in prokaryotes

Explanation: The template strand is copied by RNA polymerase to form mRNA. The coding strand has the same sequence as mRNA (with U replacing T) but is not directly copied. Answer: False.

3. Split gene arrangement is characteristic of prokaryotes.
Options:
A. True
B. False
C. Only in some bacteria
D. Depends on operon

Explanation: Split genes with exons and introns are typical of eukaryotes. Prokaryotic genes are continuous and lack introns. This arrangement allows RNA splicing in eukaryotes. Answer: False.

4. In capping, methyl guanosine triphosphate is added to the 3' end of hnRNA.
Options:
A. True
B. False
C. Added to both ends
D. Only in prokaryotes

Explanation: During capping, 7-methylguanosine is added to the 5' end of hnRNA, not the 3' end. This protects RNA from degradation and helps in ribosome recognition. Answer: False.

5. RNA polymerase synthesizes RNA in which direction?
Options:
A. 5' to 3'
B. 3' to 5'
C. Both directions
D. Randomly

Explanation: RNA polymerase synthesizes RNA in the 5' to 3' direction, using the 3' to 5' DNA template strand. This ensures correct complementary sequence formation and proper transcription of genetic information. Answer: 5' to 3'.

6. The primary transcript in eukaryotes is called:
Options:
A. mRNA
B. hnRNA
C. tRNA
D. rRNA

Explanation: In eukaryotes, the initial RNA transcript produced by RNA polymerase II is hnRNA (heterogeneous nuclear RNA). It undergoes processing including splicing, capping, and polyadenylation to form mature mRNA. Answer: hnRNA.

7. Assertion-Reason:
Assertion (A): RNA polymerase binds to promoter to start transcription.
Reason (R): Promoters contain specific sequences recognized by RNA polymerase.
Options:
A. Both A and R are true, R is correct explanation
B. Both A and R are true, R is not correct explanation
C. A is true, R is false
D. A is false, R is true

Explanation: RNA polymerase binds to promoter sequences (like -10 and -35 in prokaryotes) to initiate transcription. The reason correctly explains the assertion because promoter recognition is essential for accurate transcription initiation. Answer: Both A and R are true, R is correct explanation.

8. Match the following:
Column I: 1. Prokaryotic termination 2. Eukaryotic primary transcript 3. Capping 4. Split gene
Column II: A. hnRNA B. Rho factor C. Addition of 5’ methylguanosine D. Exons and introns
Options:
A. 1-B, 2-A, 3-C, 4-D
B. 1-A, 2-B, 3-D, 4-C
C. 1-B, 2-D, 3-A, 4-C
D. 1-C, 2-B, 3-D, 4-A

Explanation: Correct matching: Prokaryotic termination – Rho factor (B), Eukaryotic primary transcript – hnRNA (A), Capping – 5’ methylguanosine addition (C), Split gene – exons and introns (D). Answer: 1-B, 2-A, 3-C, 4-D.

9. Fill in the blank: Introns are removed from hnRNA during __________.
Options:
A. Transcription
B. RNA splicing
C. Translation
D. Replication

Explanation: Introns are non-coding sequences removed from hnRNA by RNA splicing to produce mature mRNA that can be translated into functional proteins. This process is characteristic of eukaryotic gene expression. Answer: RNA splicing.

10. Choose the correct statements:
1. Rho factor terminates prokaryotic transcription.
2. hnRNA is precursor of mRNA.
3. Capping occurs at 5’ end of hnRNA.
Options:
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1, 2 and 3

Explanation: All statements are correct. Rho factor terminates transcription in prokaryotes, hnRNA is processed to mRNA, and capping adds 7-methylguanosine to the 5’ end. Answer: 1, 2 and 3.

Subtopic: DNA Fingerprinting

• DNA Fingerprinting: Technique to identify individuals based on variations in DNA sequences.

• Repetitive DNA: DNA sequences repeated many times in genome, may be tandem or interspersed.

• Single Nucleotides: Individual bases in DNA sequence, mutations here cause SNPs.

• Polymorphic DNA: DNA sequences showing variation among individuals, useful for identification.

• Satellite DNA: Highly repetitive DNA often found in centromeric regions.

• VNTRs: Variable number tandem repeats, a type of repetitive DNA used in fingerprinting.

• STRs: Short tandem repeats, microsatellites used in forensic identification.

• Genetic Markers: Specific DNA sequences used to identify differences among individuals.

• Allele: Variant form of a gene at a specific locus.

• Polymorphism: Occurrence of two or more alleles at a locus in population.

• Restriction Enzymes: Proteins cutting DNA at specific sequences, used in fingerprinting.

Lead Question - 2021

DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as:

1. Repetitive DNA
Options:
A. Single nucleotides
B. Repetitive DNA
C. Mitochondrial DNA
D. Introns

Explanation: DNA fingerprinting targets repetitive DNA regions, such as VNTRs and STRs, because they show high variability among individuals. These sequences serve as unique identifiers for each person, making them ideal for forensic analysis, paternity testing, and population genetics studies. Answer: Repetitive DNA.

2. Polymorphic DNA sequences are useful in fingerprinting because:
Options:
A. They are identical in all individuals
B. They vary among individuals
C. They code for essential proteins
D. They are non-coding only

Explanation: Polymorphic DNA sequences vary among individuals and provide distinct patterns for identification. Their variability allows detection of differences in DNA, making them critical for forensic analysis and determining genetic relationships between individuals. Answer: They vary among individuals.

3. Satellite DNA primarily refers to:
Options:
A. VNTRs
B. Highly repetitive sequences
C. Single copy genes
D. Coding DNA

Explanation: Satellite DNA consists of highly repetitive, non-coding sequences often located in centromeric regions. These sequences are useful in genome mapping, structural chromosome studies, and forensic identification. Answer: Highly repetitive sequences.

4. Single nucleotide variations in DNA are called:
Options:
A. VNTRs
B. STRs
C. SNPs
D. Satellite DNA

Explanation: Single nucleotide polymorphisms (SNPs) are variations at a single base in DNA. They are abundant in the genome and can serve as genetic markers for population studies, disease association, and forensic identification. Answer: SNPs.

5. Restriction enzymes in DNA fingerprinting are used to:
Options:
A. Amplify DNA
B. Cut DNA at specific sequences
C. Replicate chromosomes
D. Transcribe RNA

Explanation: Restriction enzymes cut DNA at specific recognition sequences, generating fragments of variable lengths. These fragments are separated by gel electrophoresis to create a DNA fingerprint, allowing comparison of genetic differences. Answer: Cut DNA at specific sequences.

6. STRs in forensic identification are:
Options:
A. Short tandem repeats
B. Single copy genes
C. Mitochondrial markers
D. Restriction sites

Explanation: Short tandem repeats (STRs) are short sequences repeated consecutively in the genome. Their highly variable nature among individuals makes STRs valuable in forensic identification and paternity testing. Answer: Short tandem repeats.

7. Assertion-Reason:
Assertion (A): DNA fingerprinting uses VNTRs and STRs.
Reason (R): These regions are polymorphic and show individual variation.
Options:
A. Both A and R are true, R is correct explanation
B. Both A and R are true, R is not correct explanation
C. A is true, R is false
D. A is false, R is true

Explanation: VNTRs and STRs are polymorphic DNA regions with high variability. DNA fingerprinting relies on these differences to distinguish individuals. Both the assertion and reason are correct, and the reason correctly explains the assertion. Answer: Both A and R are true, R is correct explanation.

8. Match the following:
Column I: 1. VNTR 2. STR 3. SNP 4. Satellite DNA
Column II: A. Single nucleotide variation B. Highly repetitive sequences C. Short tandem repeat D. Variable number tandem repeat
Options:
A. 1-D, 2-C, 3-A, 4-B
B. 1-C, 2-D, 3-B, 4-A
C. 1-D, 2-B, 3-C, 4-A
D. 1-B, 2-C, 3-D, 4-A

Explanation: Correct matches: VNTR - variable number tandem repeat (D), STR - short tandem repeat (C), SNP - single nucleotide variation (A), Satellite DNA - highly repetitive sequences (B). Answer: 1-D, 2-C, 3-A, 4-B.

9. Fill in the blank: DNA fingerprinting primarily targets __________ regions of the genome.
Options:
A. Coding
B. Polymorphic
C. Mitochondrial
D. Ribosomal

Explanation: DNA fingerprinting focuses on polymorphic regions that vary among individuals. These include VNTRs, STRs, and other repetitive sequences, which generate unique patterns for identification, paternity testing, and forensic applications. Answer: Polymorphic.

10. Choose the correct statements:
1. DNA fingerprinting detects variations in repetitive DNA.
2. STRs and VNTRs are used for individual identification.
3. Restriction enzymes and electrophoresis are part of the technique.
Options:
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1, 2 and 3

Explanation: All statements are correct. Repetitive DNA is targeted in fingerprinting, STRs and VNTRs serve as markers, and restriction enzymes with gel electrophoresis generate DNA patterns for comparison. Answer: 1, 2 and 3.

Topic: Transcription in Eukaryotes
Subtopic: Role of RNA Polymerases

Keyword Definitions:

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

• RNA Polymerase I: Enzyme that transcribes rRNA (28S, 18S, 5.8S).

• RNA Polymerase II: Enzyme that transcribes precursor of mRNA and some snRNA.

• RNA Polymerase III: Enzyme that transcribes tRNA, 5S rRNA, and snRNA.

• snRNA: Small nuclear RNA involved in RNA splicing.

Lead Question - 2021

What is the role of RNA polymerase III in the process of transcription in eukaryotes?

(1) Transcribes tRNA, 5s rRNA and snRNA
(2) Transcribes precursor of mRNA
(3) Transcribes only snRNAs
(4) Transcribes rRNAs (28S, 18S and 5.8S)

Explanation: The correct answer is (1) Transcribes tRNA, 5S rRNA and snRNA. In eukaryotes, RNA polymerase I transcribes larger rRNAs, RNA polymerase II synthesizes precursor of mRNA and some snRNAs, while RNA polymerase III specifically transcribes small RNAs such as tRNA, 5S rRNA, and snRNA. Thus, option (1) is correct.

Guessed Questions:

1. Which RNA polymerase is responsible for transcribing precursor of mRNA in eukaryotes?
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) DNA polymerase
Explanation: The correct answer is (2) RNA polymerase II. This enzyme synthesizes heterogeneous nuclear RNA (hnRNA), the precursor of mRNA, which later undergoes capping, splicing, and polyadenylation to form mature mRNA. RNA polymerase I transcribes rRNA, and RNA polymerase III transcribes tRNA, 5S rRNA, and snRNA.

2. Which RNA polymerase transcribes rRNA genes like 28S, 18S, and 5.8S in eukaryotes?
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) RNA polymerase IV
Explanation: The correct answer is (1) RNA polymerase I. It functions in the nucleolus and transcribes the large rRNA genes, producing 45S precursor that is processed into 28S, 18S, and 5.8S rRNAs. RNA polymerase III transcribes 5S rRNA, while RNA polymerase II transcribes mRNA precursors.

3. Transcription occurs in which part of eukaryotic cells?
(1) Cytoplasm
(2) Ribosome
(3) Nucleus
(4) Mitochondria
Explanation: The correct answer is (3) Nucleus. In eukaryotes, transcription occurs inside the nucleus because DNA is enclosed within the nuclear envelope. The produced RNA is then processed and transported into the cytoplasm for translation. Ribosomes and mitochondria play roles in protein synthesis and energy production, not in transcription.

4. Which of the following RNAs is involved in the splicing of pre-mRNA?
(1) tRNA
(2) snRNA
(3) mRNA
(4) rRNA
Explanation: The correct answer is (2) snRNA. Small nuclear RNAs form complexes called snRNPs that participate in splicing of introns from pre-mRNA, forming mature mRNA. tRNA carries amino acids, rRNA forms ribosomal structure, and mRNA acts as a template for protein synthesis. snRNA ensures proper gene expression.

5. In prokaryotes, transcription and translation are:
(1) Temporally and spatially separated
(2) Occur simultaneously in cytoplasm
(3) Occur only in nucleus
(4) Occur only in mitochondria
Explanation: The correct answer is (2) Occur simultaneously in cytoplasm. Prokaryotes lack a nucleus, so as RNA is being transcribed from DNA, ribosomes can start translating it into protein immediately. In contrast, eukaryotes have spatial separation, with transcription in the nucleus and translation in the cytoplasm.

6. Which RNA polymerase is inhibited by α-amanitin toxin from mushrooms?
(1) RNA polymerase I
(2) RNA polymerase II
(3) RNA polymerase III
(4) None of the above
Explanation: The correct answer is (2) RNA polymerase II. α-amanitin, a toxin from Amanita mushrooms, strongly inhibits RNA polymerase II, preventing mRNA synthesis. This stops protein synthesis and leads to cell death. RNA polymerase I and III are comparatively resistant, making α-amanitin a tool in transcription studies.

7. Assertion (A): In eukaryotes, transcription occurs inside the nucleus.
Reason (R): DNA is enclosed in a nuclear membrane, separating it from ribosomes.
(1) Both A and R are true, R is correct explanation of A
(2) Both A and R are true, R is not correct explanation of A
(3) A is true, R is false
(4) A is false, R is true
Explanation: The correct answer is (1) Both A and R are true, R is correct explanation of A. DNA is present within the nucleus, which is separated from the cytoplasm. Thus, transcription occurs in the nucleus, while ribosome-mediated translation takes place in the cytoplasm, ensuring compartmentalization of gene expression.

8. Match the following with the correct RNA polymerase:
A. RNA polymerase I    1. tRNA, 5S rRNA
B. RNA polymerase II    2. Precursor of mRNA
C. RNA polymerase III    3. 28S, 18S, 5.8S rRNA
(1) A-3, B-2, C-1
(2) A-2, B-1, C-3
(3) A-1, B-3, C-2
(4) A-3, B-1, C-2
Explanation: The correct answer is (1) A-3, B-2, C-1. RNA polymerase I transcribes large rRNA genes, RNA polymerase II transcribes precursor of mRNA, and RNA polymerase III transcribes small RNAs including tRNA, 5S rRNA, and snRNA. This division of labor allows accurate transcription in eukaryotic cells.

9. Fill in the blank: The non-coding sequences removed during pre-mRNA processing are called ________.
(1) Exons
(2) Introns
(3) Codons
(4) Anticodons
Explanation: The correct answer is (2) Introns. During RNA processing, introns (non-coding sequences) are removed, and exons (coding sequences) are joined together to form mature mRNA. Codons are nucleotide triplets on mRNA, while anticodons are complementary sequences on tRNA. Removal of introns ensures accurate protein coding.

10. Choose the correct statements:
(a) RNA polymerase I transcribes rRNA (28S, 18S, 5.8S)
(b) RNA polymerase II transcribes tRNA
(c) RNA polymerase III transcribes tRNA and 5S rRNA
(d) Transcription in eukaryotes occurs in nucleus
Options:
(1) a, b, c
(2) a, c, d
(3) b, c, d
(4) a, b, d
Explanation: The correct answer is (2) a, c, d. RNA polymerase I transcribes large rRNAs, RNA polymerase III transcribes tRNA and 5S rRNA, and transcription occurs in the nucleus. RNA polymerase II transcribes precursor of mRNA, not tRNA. Hence, a, c, and d are correct statements together.

Topic: Human Genome and DNA Fingerprinting
Subtopic: Genetic Mapping and Polymorphism

Keyword Definitions:

• Genetic mapping: Process of determining the location of genes or markers on chromosomes.

• Human genome: Complete set of DNA sequences in humans, including genes and noncoding regions.

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

• Polymorphism: Variation in DNA sequence among individuals of a species.

• Single nucleotide polymorphism (SNP): Variation at a single nucleotide position in DNA among individuals.

• hnRNA: Heterogeneous nuclear RNA; precursor to mRNA, usually not used for mapping.

• Genetic marker: Specific DNA sequence used for identifying a location on a chromosome.

• Restriction fragment length polymorphism: Variation in DNA fragment lengths used in genetic analysis.

Lead Question - 2020 (COVID Reexam)

Which is the basis of genetic mapping of the human genome as well as DNA fingerprinting?

1. Polymorphism in the DNA sequence
2. Single nucleotide polymorphism
3. Polymorphism in hnRNA sequence
4. Polymorphism in the RNA sequence

Explanation: Genetic mapping and DNA fingerprinting rely on variations in the DNA sequence, called polymorphisms. Single nucleotide polymorphisms (SNPs) are common markers, but generally any DNA sequence variation can serve as a basis for mapping. Correct answer: Option 1.

1. Single Correct Answer MCQ:

Which DNA variation is most commonly used in forensic analysis?
1. Microsatellite polymorphism
2. Single nucleotide polymorphism
3. hnRNA variation
4. RNA editing

Explanation: Microsatellite polymorphisms, also called short tandem repeats (STRs), are highly variable among individuals and widely used in forensic DNA fingerprinting to identify individuals. Answer: Option 1.

2. Single Correct Answer MCQ:

Which technique detects variations in DNA sequences for mapping?
1. PCR
2. Southern blotting
3. Gel electrophoresis
4. All of the above

Explanation: PCR amplifies target sequences, Southern blotting detects specific DNA fragments, and gel electrophoresis separates DNA based on size. Combined, these methods identify DNA polymorphisms for mapping and fingerprinting. Answer: Option 4.

3. Single Correct Answer MCQ:

SNP stands for:
1. Single nucleotide polymorphism
2. Small nucleotide protein
3. Specific nuclear polymorphism
4. Single normal pattern

Explanation: SNPs are single nucleotide variations in DNA among individuals. They are abundant and used as genetic markers for mapping and disease studies. Answer: Option 1.

4. Single Correct Answer MCQ:

Which DNA region is highly variable and used for fingerprinting?
1. Coding region
2. Noncoding repetitive sequences
3. Promoter region
4. Ribosomal RNA genes

Explanation: Noncoding repetitive sequences, such as STRs, show high polymorphism among individuals, making them ideal for DNA fingerprinting and distinguishing individuals genetically. Answer: Option 2.

5. Single Correct Answer MCQ:

DNA fingerprinting is primarily used for:
1. Measuring gene expression
2. Individual identification
3. Protein sequencing
4. RNA splicing

Explanation: DNA fingerprinting identifies individuals based on unique patterns of DNA polymorphisms, useful in forensics, paternity testing, and genetic studies. It is not used for protein or RNA analysis. Answer: Option 2.

6. Single Correct Answer MCQ:

Which of the following is not used in genetic mapping?
1. SNPs
2. Microsatellites
3. hnRNA sequences
4. RFLPs

Explanation: hnRNA sequences are precursor mRNA and generally not used in mapping or fingerprinting. SNPs, microsatellites, and RFLPs are all reliable genetic markers for mapping. Answer: Option 3.

7. Assertion-Reason MCQ:

Assertion (A): DNA polymorphisms serve as genetic markers.
Reason (R): Variations in DNA sequences among individuals allow mapping of genes.
1. Both A and R true, R correct explanation
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true

Explanation: DNA polymorphisms are variations among individuals, making them suitable as markers for genetic mapping. The reason accurately explains the assertion. Answer: Option 1.

8. Matching Type MCQ:

Column I    Column II
(a) STRs    (i) Tandem repeats
(b) SNPs    (ii) Single nucleotide variation
(c) RFLP    (iii) Restriction fragment length
(d) VNTR    (iv) Variable number of repeats

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: STRs are short tandem repeats, SNPs are single nucleotide variations, RFLPs are restriction fragment length polymorphisms, and VNTRs are variable number tandem repeats. Correct match: Option 1.

9. Fill in the blanks:

The main basis for DNA fingerprinting is ______ in DNA sequence.
1. Polymorphism
2. Mutation
3. Deletion
4. Transcription

Explanation: DNA fingerprinting relies on polymorphisms, which are variations in DNA sequences among individuals. These differences allow unique identification. Mutations and deletions may cause polymorphisms but are not the primary term used. Answer: Option 1.

10. Choose the correct statements MCQ:

Select correct statements about DNA markers

Topic: DNA Replication
Subtopic: Replication Rate in Prokaryotes

Keyword Definitions:

• E. coli: A common bacterium used as a model organism in molecular biology studies.

• Base pairs: Paired nucleotides in DNA (A-T, G-C) forming the double helix.

• Replication: Process of copying DNA to produce two identical DNA molecules.

• Polymerization: Formation of a DNA strand by linking nucleotides via DNA polymerase.

• Replication rate: Speed at which DNA polymerase adds nucleotides per second.

• Prokaryotes: Organisms without a nucleus; DNA replication occurs in cytoplasm.

• Double helix: Structure of DNA formed by two complementary strands.

• DNA polymerase: Enzyme catalyzing addition of nucleotides during DNA replication.

Lead Question - 2020 (COVID Reexam)

E.coli has only 4.6 × 106 base pairs and completes the process of replication within 18 minutes; then the average rate of polymerization is approximate-

1. 2000 base pairs/second
2. 3000 base pairs/second
3. 4000 base pairs/second
4. 1000 base pairs/second

Explanation: E. coli has 4.6 × 106 base pairs. Replication time is 18 minutes = 1080 seconds. Average rate = total base pairs / time = 4.6 × 106 / 1080 ≈ 4259 ≈ 4000 base pairs/second. Correct answer: Option 3.

1. Single Correct Answer MCQ:

Which enzyme is responsible for adding nucleotides during replication?
1. DNA ligase
2. DNA polymerase
3. Helicase
4. Topoisomerase

Explanation: DNA polymerase catalyzes the addition of nucleotides to the growing DNA strand during replication, ensuring accurate synthesis. Helicase unwinds DNA, ligase joins fragments, and topoisomerase relieves supercoiling. Correct answer: Option 2.

2. Single Correct Answer MCQ:

The replication of E. coli DNA is completed in approximately:
1. 18 seconds
2. 18 minutes
3. 18 hours
4. 180 minutes

Explanation: E. coli completes DNA replication in 18 minutes, demonstrating rapid polymerization in prokaryotes. This short time is due to circular DNA and multiple replication forks. Answer: Option 2.

3. Single Correct Answer MCQ:

What is the total number of base pairs in E. coli genome?
1. 4.6 × 103
2. 4.6 × 106
3. 4.6 × 109
4. 46

Explanation: The E. coli genome contains approximately 4.6 × 106 base pairs forming a single circular DNA molecule. This small genome allows rapid replication. Correct answer: Option 2.

4. Single Correct Answer MCQ:

The time taken to replicate 1 base pair at the average rate is approximately:
1. 0.25 milliseconds
2. 0.25 seconds
3. 2.5 microseconds
4. 1 second

Explanation: At 4000 base pairs/second, time per base pair = 1/4000 sec = 0.00025 sec = 0.25 milliseconds. This shows the high efficiency of E. coli replication machinery. Correct answer: Option 1.

5. Single Correct Answer MCQ:

Which structure allows faster replication in prokaryotes?
1. Linear DNA
2. Circular DNA
3. Mitochondrial DNA
4. Plasmid only

Explanation: Circular DNA in prokaryotes like E. coli allows bidirectional replication from a single origin, leading to faster replication compared to linear DNA. Plasmids replicate independently but are smaller. Answer: Option 2.

6. Single Correct Answer MCQ:

What is the main reason E. coli replicates DNA rapidly?
1. Multiple chromosomes
2. Multiple replication origins
3. Single circular chromosome
4. Lack of DNA polymerase

Explanation: E. coli has a single circular chromosome with a single origin of replication, but replication is bidirectional, allowing rapid completion in 18 minutes. This efficiency supports fast bacterial growth. Correct answer: Option 3.

7. Assertion-Reason MCQ:

Assertion (A): E. coli completes replication in 18 minutes.
Reason (R): Its genome is small and circular, allowing fast bidirectional replication.
1. Both A and R true, R correct explanation
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true

Explanation: E. coli replicates its 4.6 × 106 base pair circular genome in 18 minutes due to bidirectional replication from a single origin. Both assertion and reason are correct, and reason explains assertion accurately. Answer: Option 1.

8. Matching Type MCQ:

Column I    Column II
(a) DNA polymerase    (i) Unwinds DNA
(b) Helicase    (ii) Adds nucleotides
(c) Ligase    (iii) Joins Okazaki fragments
(d) Topoisomerase    (iv) Relieves supercoiling

1. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
3. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
4. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Explanation: DNA polymerase adds nucleotides, helicase unwinds DNA, ligase joins Okazaki fragments, and topoisomerase relieves supercoiling. Correct match: (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv). Answer: Option 1.

9. Fill in the blanks:

The bidirectional replication in E. coli begins at a single ______ of replication.
1. Origin
2. Terminus
3. Fork
4. Centromere

Explanation: E. coli DNA replication begins at a single origin of replication, proceeding bidirectionally around the circular chromosome until termination. This strategy allows rapid genome duplication. Answer: Option 1.

10. Choose the correct statements MCQ:

Select the correct statements about E. coli DNA replication:
(a) Replication is bidirectional
(b) Rate ≈ 4000 base pairs

Subtopic: Chromosomal Theory of Inheritance

Keyword Definitions:

• Chromosomal Theory of Inheritance: Theory stating that genes are located on chromosomes and segregation of chromosomes explains inheritance patterns.

• Sutton: Scientist who studied grasshopper chromosomes and proposed the role of chromosomes in heredity.

• Boveri: Scientist who independently concluded that chromosomes carry genetic material.

• Bateson: Geneticist who coined the term “genetics” and studied inheritance patterns.

• Punnett: Developed Punnett squares to predict genotype ratios.

• T. H. Morgan: Worked on Drosophila, showed sex-linked inheritance and role of chromosomes.

• Watson and Crick: Discovered DNA double helix structure, not chromosomal theory.

• Inheritance: Transmission of genetic information from parents to offspring.

Lead Question - 2020 (COVID Reexam)

Chromosomal theory of inheritance was proposed by:

1. Sutton and Boveri
2. Bateson and Punnet
3. T. H. Morgan
4. Watson and Crick

Explanation: The chromosomal theory of inheritance was proposed by Sutton and Boveri. They concluded that genes reside on chromosomes, and their segregation during meiosis explains inheritance patterns. Morgan contributed later with sex-linked inheritance. Watson and Crick discovered DNA structure. Correct answer: Option 1.

1. Single Correct Answer MCQ:

Who demonstrated sex-linked inheritance in Drosophila?
1. Sutton
2. Morgan
3. Bateson
4. Boveri

Explanation: T. H. Morgan studied fruit flies (Drosophila) and demonstrated sex-linked inheritance, confirming the role of chromosomes in transmitting specific traits. Sutton and Boveri proposed the chromosomal theory, Bateson studied genetics but not sex linkage. Answer: Option 2.

2. Single Correct Answer MCQ:

Which scientists independently proposed chromosome involvement in heredity?
1. Sutton and Boveri
2. Morgan and Punnett
3. Watson and Crick
4. Bateson and Mendel

Explanation: Sutton and Boveri independently concluded that chromosomes carry genetic material and explain Mendelian inheritance patterns. Other scientists contributed to genetics but did not propose chromosomal theory. Answer: Option 1.

3. Single Correct Answer MCQ:

Punnett is known for:
1. Chromosomal theory
2. Punnett square for predicting genotypes
3. DNA structure
4. Sex-linked inheritance

Explanation: Punnett developed the Punnett square to predict genotypic ratios of offspring. He did not propose chromosomal theory or study DNA. Sex-linked inheritance was demonstrated by Morgan. Answer: Option 2.

4. Single Correct Answer MCQ:

Watson and Crick are credited with:
1. Chromosomal theory
2. Drosophila genetics
3. DNA double helix structure
4. Punnett square

Explanation: Watson and Crick discovered the DNA double helix structure in 1953, explaining the molecular basis of inheritance. Chromosomal theory and Drosophila genetics were contributed by Sutton, Boveri, and Morgan. Answer: Option 3.

5. Single Correct Answer MCQ:

Chromosomal theory explains:
1. DNA replication only
2. Mendelian inheritance via chromosomes
3. Protein synthesis
4. Mutation mechanism

Explanation: Chromosomal theory links Mendelian inheritance to the behavior of chromosomes during meiosis, explaining segregation and independent assortment of genes. DNA replication, protein synthesis, and mutations are explained separately. Answer: Option 2.

6. Single Correct Answer MCQ:

Sutton’s observations were made on:
1. Drosophila
2. Grasshopper chromosomes
3. Pea plants
4. Human cells

Explanation: Sutton studied grasshopper chromosomes and observed their segregation, leading him to propose that chromosomes carry hereditary units. Morgan worked on Drosophila, Mendel on pea plants. Answer: Option 2.

7. Assertion-Reason MCQ:

Assertion (A): Chromosomal theory links chromosomes to inheritance.
Reason (R): Genes are located on chromosomes and segregate during meiosis.
1. Both A and R true, R correct explanation
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true

Explanation: Chromosomal theory states genes reside on chromosomes, and their segregation during meiosis explains inheritance. Both assertion and reason are true, and reason correctly explains the assertion. Answer: Option 1.

8. Matching Type MCQ:

Column I    Column II
(a) Sutton    (i) Sex-linked traits
(b) Morgan    (ii) Chromosomal theory
(c) Boveri    (iii) Chromosome study in sea urchin
(d) Punnett    (iv) Punnett square

1. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
3. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
4. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

Explanation: Sutton proposed chromosomal theory, Morgan studied sex-linked traits in Drosophila, Boveri studied sea urchin chromosomes, and Punnett developed Punnett square. Correct matching: (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv). Answer: Option 1.

9. Fill in the blanks:

The scientist who studied grasshopper chromosomes and proposed chromosomal inheritance is ________.
1. Morgan
2. Sutton
3. Boveri
4. Watson

Explanation: Sutton studied grasshopper chromosomes, observed segregation during meiosis, and proposed that chromosomes carry hereditary material. Morgan studied Drosophila, Boveri sea urchins, Watson DNA structure. Answer: Option 2.

10. Choose correct statements:

(a) Chromosomal theory was proposed by Sutton and Boveri
(b)

Topic: Enzymes in DNA Metabolism

Subtopic: DNA-modifying Enzymes

Keyword Definitions:

• Nucleases – Enzymes that cleave the phosphodiester bonds in nucleic acids, separating DNA strands.

• Exonucleases – Enzymes that remove nucleotides from the ends of DNA molecules.

• Endonucleases – Enzymes that make cuts at specific internal sites in DNA.

• Ligases – Enzymes that join DNA fragments by forming phosphodiester bonds.

• Polymerases – Enzymes that synthesize DNA or RNA from nucleotides.

• DNA fragments – Pieces of DNA resulting from enzymatic cleavage.

• DNA replication – Process of producing two identical copies of DNA from one original molecule.

• Recombinant DNA technology – Technique using enzymes to cut and join DNA fragments for genetic engineering.

• Restriction enzymes – Endonucleases that recognize specific DNA sequences and cleave them.

• Phosphodiester bond – Covalent bond connecting nucleotides in DNA or RNA strand.

• DNA metabolism – Collective processes of replication, repair, and recombination involving DNA.

Lead Question - 2020

Choose the correct pair from the following:
(1) Nucleases – Separate the two strands of DNA
(2) Exonucleases – Make cuts at specific positions within DNA
(3) Ligases – Join the two DNA molecules
(4) Polymerases – Break the DNA into fragments

Explanation: Ligases are enzymes that catalyze the joining of DNA fragments by forming phosphodiester bonds, which is essential in replication and repair processes. Correct answer is (3) Ligases – Join the two DNA molecules. Other options incorrectly describe functions of respective enzymes.

1. Single Correct Answer: Enzyme that synthesizes DNA from nucleotides is:
(1) Nuclease
(2) Polymerase
(3) Ligase
(4) Exonuclease

Explanation: DNA polymerase catalyzes the formation of DNA by adding nucleotides complementary to the template strand during replication. Correct answer is (2) Polymerase.

2. Single Correct Answer: Enzyme that cleaves DNA at internal sites is:
(1) Endonuclease
(2) Ligase
(3) Polymerase
(4) Exonuclease

Explanation: Endonucleases recognize specific DNA sequences and cut at internal positions to produce fragments. Correct answer is (1) Endonuclease.

3. Single Correct Answer: Exonucleases act on:
(1) DNA ends
(2) DNA internal sites
(3) RNA only
(4) Protein

Explanation: Exonucleases remove nucleotides sequentially from DNA ends, playing roles in DNA repair and degradation. Correct answer is (1) DNA ends.

4. Single Correct Answer: Function of ligase is:
(1) Break DNA into fragments
(2) Separate DNA strands
(3) Join DNA fragments
(4) Make cuts at specific sites

Explanation: Ligase catalyzes the formation of phosphodiester bonds to join DNA fragments during replication and repair. Correct answer is (3) Join DNA fragments.

5. Assertion-Reason:
Assertion (A): DNA polymerase is essential for replication.
Reason (R): It synthesizes DNA using nucleotides complementary to the template strand.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true