Molecular Basis of Inheritance (NCERT Practice Paper)
Q1. Which is true for 'VNTR'? (i) Belongs to mini-satellite DNA. (ii) Size varies from 0.1 to 20 kb. (iii) Copy number varies from chromosome to chromosome.
Correct Answer: (a)
VNTRs are minisatellites (i), their length ranges from 0.1 to 20 kb (ii), and the number of repeats (copy number) varies among individuals and chromosomes (iii). All are correct.
Q2. The enzyme used to join discontinuously synthesised fragments (Okazaki fragments) is:
Correct Answer: (c)
On the lagging strand, DNA is synthesised in short segments called Okazaki fragments. These fragments are later sealed together by DNA ligase, which forms phosphodiester bonds between the fragments, creating a continuous strand.
Q3. Correct statements about 'DNA Polymerase' accuracy: (i) Must be fast. (ii) Must be highly accurate. (iii) Mistakes result in mutations.
Correct Answer: (a)
DNA polymerase must be fast to replicate large genomes quickly (i), highly accurate to maintain genetic fidelity (ii), and errors can lead to mutations (iii). All are correct.
Q4. Which bacteria produced smooth shiny colonies (S) in Griffith's experiment?
Correct Answer: (b)
The smooth (S) strain produces colonies with a shiny appearance because each cell is surrounded by a polysaccharide capsule. This capsule protects the bacteria from the host immune system, making them virulent. The rough (R) strain lacks this coat.
Q5. Assertion: Adenine pairs with Thymine with three H-bonds.
Reason: Guanine pairs with Cytosine with two H-bonds.
Correct Answer: (a)
Adenine‑thymine base pairing involves two hydrogen bonds, while guanine‑cytosine pairing involves three hydrogen bonds. The assertion states A‑T has three (false) and the reason says G‑C has two (false). Both statements are incorrect, so the answer is both false.
Q6. Which of the following is correct? (i) George Gamow proposed triplet code. (ii) Har Gobind Khorana synthesised RNA homopolymers. (iii) Severo Ochoa enzyme is polynucleotide phosphorylase.
Correct Answer: (a)
Gamow proposed the triplet nature of the code (i). Khorana synthesised RNA polymers for codon assignment (ii). Ochoa discovered polynucleotide phosphorylase (iii). All statements are correct.
Q7. Regarding the 'Structure of Proteins' (Biologist's view): (i) Biologists describe structure at 4 levels. (ii) Primary is the sequence of amino acids. (iii) Tertiary is a hollow woolen ball-like fold.
Correct Answer: (a)
Protein structure is described at primary, secondary, tertiary, and quaternary levels (i). Primary is amino acid sequence (ii). Tertiary structure is the overall 3D fold, often globular; the 'hollow woolen ball' is a rough description (iii). All statements are correct.
Q8. Which are correct for 'Genetic Mapping' tools? (i) Information on polymorphism of restriction sites. (ii) microsatellites. (iii) sequencing of whole genomes.
Correct Answer: (a)
Genetic mapping uses RFLPs (polymorphism of restriction sites) (i), microsatellites (ii), and full genome sequencing (iii). All are tools used in mapping genomes.
Q9. Match the codon features:
Column-I
Column-II
A. Degenerate
(I) Same codon from bacteria to human
B. Universal
(II) More than one codon for one amino acid
C. Stop codons
(III) UAA, UAG, UGA
D. Initiator codon
(IV) AUG
Correct Answer: (a)
Degeneracy means multiple codons for one amino acid (A‑II). Universality means same codon codes same amino acid across species (B‑I). Stop codons are UAA, UAG, UGA (C‑III). Initiator codon is AUG (D‑IV). So A‑II, B‑I, C‑III, D‑IV.
Q10. Assertion: Human Genome Project (HGP) was called a mega project.
Reason: It required 13 years and approximately 9 billion US dollars.
Correct Answer: (a)
The HGP was a massive international effort that took about 13 years and cost roughly 9 billion dollars. Such scale, complexity, and funding justify its designation as a 'mega project'. The reason provides the quantitative evidence that supports the assertion.
Q11. Match the following for HGP facts:
Column-I
Column-II
A. Human genome size
(I) 3164.7 million bp
B. Genome coding for proteins
(II) Less than 2 per cent
C. Chromosome 1 genes
(III) 2968
D. Chromosome Y genes
(IV) 231
Correct Answer: (a)
Human genome size is ~3164.7 million bp (A‑I). Only <2% codes for proteins (B‑II). Chromosome 1 has 2968 genes (C‑III). Chromosome Y has 231 genes (D‑IV). Thus A‑I, B‑II, C‑III, D‑IV.
Q12. Identify the correct features of the DNA double helix: (i) Back-bone is made of sugar and phosphate. (ii) Two chains have anti-parallel polarity. (iii) The pitch is 3.4 nm. (iv) Coiled in a left-handed fashion.
Correct Answer: (a)
DNA has a sugar‑phosphate backbone, anti‑parallel strands, and a pitch of 3.4 nm per turn. However, the helix is right‑handed (B‑DNA), not left‑handed (Z‑DNA is left‑handed but not the typical form). Thus (iv) is false, leaving (i), (ii), (iii) correct.
Q13. Regarding 'HGP' goals: (i) identify all 20,000-25,000 genes. (ii) Determine sequence of 3 billion base pairs. (iii) Address ELSI (Ethical, Legal, Social Issues).
Correct Answer: (a)
The HGP aimed to identify all genes (estimated at 20,000‑25,000), sequence the entire genome (~3 billion bp), and address ELSI arising from the project. All three are explicit goals of the HGP.
Q14. Which of the following is correct regarding 'The DNA' as described in the text? (i) Length is usually defined as number of nucleotides. (ii) E. coli has 4.6 × 10^6 bp. (iii) Haploid human DNA is 3.3 × 10^9 bp. (iv) λ bacteriophage has 5386 nucleotides.
Correct Answer: (a)
Statements (i), (ii), and (iii) are correct. DNA length is often given in base pairs or nucleotides. E. coli has ~4.6×10^6 bp, and human haploid genome is ~3.3×10^9 bp. However, λ phage has 48502 bp, not 5386 nucleotides (that is φ×174). Thus (iv) is incorrect.
Q15. Assertion: The genetic code is nearly universal.
Reason: Exceptions have been found in mitochondrial codons.
Correct Answer: (b)
The genetic code is nearly universal, meaning the same codons specify the same amino acids in most organisms. However, minor variations exist, particularly in mitochondrial genomes (e.g., AGA and AGG as stop codons in human mitochondria). These exceptions justify the term 'nearly universal' and directly explain why it is not absolute.
Q16. In the lac operon, the i gene codes for:
Correct Answer: (b)
The i gene is the regulatory gene of the lac operon; it produces the lac repressor protein. This repressor binds to the operator region and prevents transcription of the structural genes (z, y, a) when lactose is absent. The inducer (allolactose) inactivates the repressor.
Q17. According to Chargaff's rule, for a double-stranded DNA, the ratio of Adenine to Thymine and Guanine to Cytosine is:
Correct Answer: (b)
Chargaff's rule states that in double‑stranded DNA, the amount of adenine equals thymine and guanine equals cytosine, so the ratios are constant and equal to one. This reflects complementary base pairing (A‑T, G‑C) and is crucial for the double‑helix structure.
Q18. The average rate of polymerisation by DNA polymerase in E. coli is approximately:
Correct Answer: (c)
DNA polymerase in E. coli adds about 2000 nucleotides per second (2000 bp/s) during replication. This high speed is necessary to replicate the entire genome (4.6×10^6 bp) in about 40 minutes, given that replication occurs bidirectionally from a single origin.
Q19. In eukaryotes, the monocistronic structural genes have interrupted coding sequences called:
Correct Answer: (b)
Eukaryotic genes contain exons (coding sequences) that are expressed and introns (non‑coding sequences) that are removed by splicing. Exons are the segments that carry the information for protein synthesis and are joined together after intron removal to form mature mRNA.
Q20. Regarding 'Satellite DNA' in fingerprinting: (i) Form small peaks in density centrifugation. (ii) Categorized into micro and mini satellites. (iii) High degree of polymorphism.
Correct Answer: (a)
Satellite DNA appears as small peaks separate from bulk DNA in density gradients (i). It includes micro‑ and minisatellites (ii). These sequences show high polymorphism (iii), making them useful for fingerprinting. All correct.
Q21. Assertion: Splicing is an essential step in eukaryotic transcription.
Reason: Primary transcripts in eukaryotes contain non-functional introns.
Correct Answer: (a)
Eukaryotic primary transcripts (hnRNA) contain both exons and introns. Introns are non‑coding sequences that must be removed to produce a functional mRNA. Splicing is essential to join the exons in the correct order; without it, the mRNA would not be translated properly. The reason correctly identifies the presence of introns as the need for splicing.
Q22. Regarding 'Transcription Unit', identify the correct statements: (i) Promoter is located towards 5' -end (upstream) of coding strand. (ii) Terminator is towards 3' -end (downstream) of coding strand. (iii) Structural gene is flanked by promoter and terminator.
Correct Answer: (a)
A transcription unit includes a promoter (upstream, 5' side of coding strand), a structural gene, and a terminator (downstream, 3' side). The promoter is the site for RNA polymerase binding, and the terminator signals the end of transcription. All three statements are correct.
Q23. Identify the correct statements regarding RNA world: (i) RNA was the first genetic material. (ii) RNA used to act as a catalyst. (iii) DNA evolved from RNA with chemical modifications.
Correct Answer: (a)
The RNA world hypothesis proposes that RNA was the first genetic material (i). RNA can also catalyse reactions (ribozymes) (ii), and DNA evolved later from RNA by gaining greater stability (iii). All statements are correct.
Q24. Assertion: DNA replication is semiconservative.
Reason: After replication, each DNA molecule has one parental and one newly synthesised strand.
Correct Answer: (a)
Semiconservative replication means that each daughter DNA double helix consists of one original (parental) strand and one newly synthesised strand. This was experimentally verified by Meselson‑Stahl. The reason precisely describes the outcome of semiconservative replication, thereby correctly explaining the assertion.
Q25. DNA was first identified as an acidic substance 'Nuclein' in 1869 by:
Correct Answer: (c)
Friedrich Meischer isolated a substance rich in phosphorus from cell nuclei and named it 'nuclein' in 1869. This was the first recognition of DNA as a distinct chemical entity. Watson, Crick, and Wilkins later worked on its structure, while Chargaff discovered base‑pairing rules.
Q26. Identify the correct phylum/group being described? (i) They are polymers of nucleotides. (ii) Deoxyribonucleic acid and Ribonucleic acid are two types. (iii) They are the genetic material of living systems.
Correct Answer: (c)
Nucleic acids (DNA and RNA) are polymers of nucleotides and serve as the genetic material. Proteins are polymers of amino acids, carbohydrates are sugars, and lipids are fats. The description matches nucleic acids.
Q27. Which are correct for 'tRNA'? (i) Called adapter molecule. (ii) Has an anticodon loop. (iii) Has an amino acid acceptor end. (iv) Looks like an inverted L in actual structure.
Correct Answer: (a)
tRNA is the adapter (i), contains an anticodon loop for codon recognition (ii), has a 3'‑CCA end for amino acid attachment (iii), and its three‑dimensional structure is an inverted L (iv). All statements are correct.
Q28. The length of DNA double helix in a typical mammalian cell is approximately:
Correct Answer: (b)
If stretched out, the DNA in a single human cell would be about 2.2 metres long. This is calculated from the haploid genome size (3.3×10^9 bp) multiplied by the distance between base pairs (0.34 nm). The cell must compact this enormous length into a tiny nucleus.
Q29. Match the DNA structure features:
Column-I
Column-II
A. Purine
(I) Adenine, Guanine
B. Pyrimidine
(II) Cytosine, Thymine, Uracil
C. DNA Back-bone
(III) Sugar and Phosphate
D. Stability
(IV) Stacking of base pairs
Correct Answer: (a)
Purines are adenine and guanine (A‑I). Pyrimidines are cytosine, thymine, and uracil (B‑II). The DNA backbone consists of alternating sugar and phosphate groups (C‑III). Stability is enhanced by base‑pair stacking (D‑IV). So A‑I, B‑II, C‑III, D‑IV.
Q30. The pitch of the DNA double helix is:
Correct Answer: (b)
The pitch of a DNA double helix is the distance for one complete turn, which is 3.4 nm (about 10 base pairs per turn). The vertical rise per base pair is 0.34 nm, and the diameter of the helix is about 2 nm. The total length of DNA in a cell is much larger.
Q31. Assertion: tRNA is called an adapter molecule.
Reason: It reads the code on mRNA and binds to specific amino acids.
Correct Answer: (a)
tRNA acts as an adapter because it has an anticodon that recognises the mRNA codon and a 3'‑end that carries a specific amino acid. This dual role bridges the nucleotide sequence of mRNA and the amino acid sequence of proteins, directly explaining why it is called an adapter.
Q32. Match the Following for RNA types:
Column-I
Column-II
A. mRNA
(I) Adapter molecule
B. tRNA
(II) Provides template
C. rRNA
(III) Catalyst in translation
Correct Answer: (a)
mRNA carries the genetic template from DNA (A‑II). tRNA acts as an adapter, bringing amino acids to the ribosome (B‑I). rRNA is the catalytic component of ribosomes, facilitating peptide bond formation (C‑III). Hence A‑II, B‑I, C‑III.
Q33. Which of the following is the first genetic material?
Correct Answer: (b)
The RNA world hypothesis suggests that RNA was the first genetic material because it can both store information (like DNA) and catalyse reactions (like proteins). Over time, DNA evolved as a more stable storage molecule, while proteins took over catalytic roles.
Q34. In the Meselson-Stahl experiment, the heavy isotope used was:
Correct Answer: (c)
They used ¹⁵N (heavy isotope of nitrogen) to label DNA. Nitrogen is a component of the nitrogenous bases, so DNA synthesized in ¹⁵N medium becomes heavier. When shifted to ¹⁴N medium, newly synthesized strands incorporate the lighter isotope, allowing separation by density.
Q35. Histones are rich in basic amino acid residues such as:
Correct Answer: (b)
Histones are positively charged proteins due to an abundance of basic amino acids lysine and arginine. This positive charge allows them to bind tightly to the negatively charged phosphate backbone of DNA, facilitating packaging into nucleosomes.
Q36. Regarding 'Translation', identify true statements: (i) Polymerisation of amino acids to form polypeptide. (ii) Ribosome acts as a catalyst. (iii) UTRs are required for efficient translation.
Correct Answer: (a)
Translation is the polymerisation of amino acids (i). Ribosomes catalyse peptide bond formation (peptidyl transferase activity) (ii). Untranslated regions (UTRs) at the 5' and 3' ends of mRNA are essential for translation efficiency and stability (iii). All are correct.
Q37. The codon AUG has dual functions; it codes for Methionine and acts as:
Correct Answer: (b)
AUG serves as the start codon, signalling the beginning of translation. It also codes for methionine (in eukaryotes, formylmethionine in prokaryotes). This dual role ensures that translation initiation and the first amino acid are coordinated.
Q38. Match the following for the lac operon:
Column-I (Gene)
Column-II (Product)
A. i gene
(I) Permease
B. z gene
(II) Repressor
C. y gene
(III) Transacetylase
D. a gene
(IV) Beta-galactosidase
Correct Answer: (a)
The lac operon has genes: i (repressor), z (β‑galactosidase), y (permease), a (transacetylase). The repressor binds operator; β‑galactosidase breaks lactose; permease transports lactose; transacetylase has a minor role. So A‑II, B‑IV, C‑I, D‑III.
Q39. Regarding 'HGP', which salient features are true? (i) Total number of genes is estimated at 30,000. (ii) Less than 2% of genome codes for proteins. (iii) Functions are unknown for over 50% of discovered genes. (iv) 99.9% nucleotide bases are same in all humans.
Correct Answer: (a)
The HGP estimated ~30,000 genes (i), protein‑coding regions <2% (ii), >50% of genes have unknown functions (iii), and all humans share 99.9% of their base sequences (iv). All these are salient features of the human genome.
Q40. Match the enzymes in Column-I with their functions in Column-II:
Column-I (Enzyme)
Column-II (Function)
A. DNA Polymerase
(I) Joining Okazaki fragments
B. DNA Ligase
(II) Synthesis of DNA strand
C. RNA Polymerase I
(III) Transcribes snRNAs
D. RNA Polymerase III
(IV) Transcribes rRNAs
Correct Answer: (a)
DNA polymerase synthesises new DNA strands (A‑II). DNA ligase seals Okazaki fragments (B‑I). RNA polymerase I transcribes ribosomal RNA (C‑IV), while RNA polymerase III transcribes tRNA, 5S rRNA, and small nuclear RNAs (D‑III). Thus A‑II, B‑I, C‑IV, D‑III.
Q41. Unequivocal proof that DNA is the genetic material came from the experiments of:
Correct Answer: (b)
Hershey and Chase used radioactive isotopes (³²P for DNA and ³⁵S for protein) in bacteriophage T2 infection of E. coli. Only the DNA entered the bacterial cells and directed the production of new phages, providing definitive evidence that DNA, not protein, is the genetic material.
Q42. Which statements about 'Lac Operon' are correct? (i) Elucidated by Jacob and Monod. (ii) Lactose is the inducer. (iii) Repressor binds to operator in absence of inducer. (iv) Regulation by repressor is positive regulation.
Correct Answer: (a)
Jacob and Monod proposed the lac operon model (i). Lactose (or its derivative allolactose) acts as the inducer (ii). In the absence of inducer, the repressor binds the operator, blocking transcription (iii). Regulation by repressor is negative (iv is false). So (i),(ii),(iii) are correct.
Q43. Assertion: 2'-OH group in RNA makes it easily degradable.
Reason: DNA lacks this hydroxyl group at 2' position of the sugar.
Correct Answer: (b)
The presence of the 2'‑hydroxyl group makes RNA susceptible to alkaline hydrolysis and enzymatic degradation, as it can participate in intramolecular cleavage. DNA lacks this group (having only H at 2'), which confers chemical stability. Thus the reason correctly explains why RNA is less stable than DNA.
Q44. The technique of DNA Fingerprinting was initially developed by:
Correct Answer: (a)
Alec Jeffreys developed DNA fingerprinting in 1984, using probes that detect hypervariable minisatellite regions. His technique revolutionised forensic science and paternity testing. Sanger developed DNA sequencing, Jacob and Monod studied the lac operon, and Nirenberg deciphered the genetic code.
Q45. The transcriptionally active, loosely packed chromatin is called:
Correct Answer: (c)
Euchromatin is loosely packed and stains lightly; it contains genes that are actively transcribed. In contrast, heterochromatin is densely packed, stains dark, and is generally transcriptionally silent. Nucleosomes are the structural units, not a form of chromatin.
Q46. Identify correct features of the 'Genetic Code': (i) It is nearly universal. (ii) It is read in a contiguous fashion. (iii) One codon codes for only one amino acid (unambiguous).
Correct Answer: (a)
The genetic code is nearly universal (i), is read without punctuation (contiguous) (ii), and is unambiguous (each codon specifies a single amino acid) (iii). All statements are correct.
Q47. Correct statements for 'Eukaryotic Transcription' include: (i) RNA Polymerase I transcribes rRNAs. (ii) RNA Polymerase II transcribes hnRNA. (iii) RNA Polymerase III transcribes tRNA and 5srRNA.
Correct Answer: (a)
Eukaryotes have three RNA polymerases: Pol I transcribes most rRNAs (i); Pol II transcribes mRNA precursors (hnRNA) (ii); Pol III transcribes tRNA, 5S rRNA, and snRNAs (iii). All statements are correct.
Q48. Match the DNA lengths in Column-I with the corresponding organisms in Column-II:
Column-I (DNA Length)
Column-II (Organism)
A. 5386 nucleotides
(I) Human (Haploid)
B. 48502 base pairs
(II) Bacteriophage φ × 174
C. 4.6 × 10^6 bp
(III) Bacteriophage Lambda
D. 3.3 × 10^9 bp
(IV) E. coli
Correct Answer: (a)
Bacteriophage φ × 174 has 5386 nucleotides (small, single‑stranded genome). Lambda phage has 48502 base pairs (double‑stranded). E. coli has 4.6×10^6 bp, and the haploid human genome is about 3.3×10^9 bp. The correct pairing is A‑II, B‑III, C‑IV, D‑I.
Q49. Identify true statements for 'Deoxyribonucleoside triphosphates': (i) Act as substrates for replication. (ii) Provide energy for polymerisation. (iii) Have high-energy phosphates.
Correct Answer: (a)
dNTPs are the building blocks for DNA synthesis (i). The cleavage of pyrophosphate provides the energy for polymerisation (ii). They contain high‑energy phosphoanhydride bonds (iii). All are correct.
Q50. Assertion: RNA was the first genetic material.
Reason: Essential life processes such as metabolism and splicing evolved around RNA.
Correct Answer: (a)
The RNA world hypothesis posits that RNA predates DNA because it can both store genetic information and catalyse reactions (ribozymes). Key cellular processes like splicing and translation are RNA‑based, supporting the idea that early life relied on RNA. The reason explains why RNA is considered the first genetic material.
Q51. Correct statements for 'DNA Packaging': (i) DNA is negatively charged. (ii) Histones are positively charged proteins. (iii) chromatin fibres are condensed at metaphase.
Correct Answer: (a)
DNA's phosphate groups give it a negative charge (i). Histones are rich in basic amino acids and are positively charged (ii). Chromatin condenses into visible chromosomes during metaphase (iii). All are correct.
Q52. Assertion: The genetic code is degenerate.
Reason: Some amino acids are coded by more than one codon.
Correct Answer: (a)
Degeneracy of the genetic code means that most amino acids are specified by multiple codons (synonymous codons). This redundancy provides a buffer against point mutations, as some nucleotide changes may not alter the amino acid sequence. The reason correctly defines degeneracy and directly supports the assertion.
Q53. Assertion: In eukaryotes, transcription and translation can be coupled.
Reason: Transcription and translation take place in the same compartment in eukaryotes.
Correct Answer: (b)
In eukaryotes, transcription occurs in the nucleus while translation occurs in the cytoplasm, so they are spatially separated. Coupling (simultaneous transcription and translation) is a feature of prokaryotes, where no nuclear envelope exists. Thus both the assertion (coupling in eukaryotes) and the reason (same compartment) are false.
Q54. Identify the correct phylum/group being described? (i) It is the flow from RNA to DNA. (ii) Occurs in some viruses. (iii) Reverse of the Central Dogma.
Correct Answer: (c)
Reverse transcription is the process of synthesising DNA from an RNA template, carried out by reverse transcriptase in retroviruses. It reverses the usual DNA→RNA flow of the Central Dogma.
Q55. Which of the following are salient features of the genetic code? (i) The codon is triplet. (ii) 61 codons code for amino acids. (iii) UAA, UAG, UGA are stop codons. (iv) Code is ambiguous and specific.
Correct Answer: (a)
The code is triplet (i), 61 codons specify amino acids (ii), and UAA, UAG, UGA are stop codons (iii). The code is unambiguous (one codon specifies one amino acid) and specific, not ambiguous (iv is false). Thus (i),(ii),(iii) are correct.
Q56. Which are correct for 'Human Genome' features? (i) 3164.7 million bp. (ii) 99.9% bases are same in all humans. (iii) Average gene has 3000 bases.
Correct Answer: (a)
Human genome size is ~3164.7 million bp (i). All humans share 99.9% of their DNA (ii). The average gene is about 3000 bases long (iii). All are correct.
Q57. Assertion: DNA is a better genetic material than RNA.
Reason: DNA is chemically less reactive and structurally more stable than RNA.
Correct Answer: (a)
DNA's stability arises from its deoxyribose sugar (no 2'‑OH) and the presence of thymine (which is more resistant to damage than uracil). These features reduce chemical reactivity and hydrolysis, making DNA more suitable as the long‑term repository of genetic information. RNA, while more versatile, is less stable and thus serves as a transient messenger.
Q58. In the Hershey-Chase experiment, viruses grown on radioactive Phosphorus contained:
Correct Answer: (b)
Phosphorus is a component of DNA (and RNA), but not of most proteins. When phages were grown in medium containing radioactive ³²P, their DNA became labelled. This allowed the experimenters to track DNA and show it entered the host cell, while protein (labelled with ³⁵S) remained outside.
Q59. The process of removing introns and joining exons in a defined order is:
Correct Answer: (c)
Splicing is the post‑transcriptional modification in which introns are excised and exons are ligated in the correct sequence. This process is essential for producing a functional mRNA in eukaryotes and is mediated by the spliceosome, a complex of snRNPs.
Q60. Match the following for DNA packaging:
Column-I
Column-II
A. Histone octamer
(I) 200 bp of DNA
B. Nucleosome
(II) 8 histone molecules
C. Euchromatin
(III) Densely packed, inactive
D. Heterochromatin
(IV) Loosely packed, active
Correct Answer: (a)
Histone octamer is composed of 8 histone proteins (A‑II). A nucleosome contains ~200 bp of DNA (B‑I). Euchromatin is loosely packed and transcriptionally active (C‑IV). Heterochromatin is densely packed and inactive (D‑III). So A‑II, B‑I, C‑IV, D‑III.
Q61. Assertion: DNA polymerase catalyse polymerisation only in one direction (5'->3').
Reason: This creates a replication fork at the origin.
Correct Answer: (a)
DNA polymerase indeed adds nucleotides only to the 3'‑OH end, synthesising DNA in the 5'→3' direction. However, the replication fork is formed by the unwinding of the helix, not by the directionality of polymerase. Thus the assertion is true, but the reason is false.
Q62. Match the scientists with their contributions:
Column-I
Column-II
A. Frederick Griffith
(I) Biochemical nature of transforming principle
B. Avery, MacLeod, McCarty
(II) Semiconservative replication proof
C. Meselson and Stahl
(III) Transforming Principle
D. Alec Jeffreys
(IV) DNA Fingerprinting
Correct Answer: (a)
Griffith discovered transformation (A‑III). Avery et al. proved DNA is the transforming substance (B‑I). Meselson‑Stahl confirmed semiconservative replication (C‑II). Jeffreys developed DNA fingerprinting (D‑IV). The matching is A‑III, B‑I, C‑II, D‑IV.
Q63. The nitrogenous base Uracil is found in RNA at the place of:
Correct Answer: (d)
In RNA, uracil replaces thymine as a complementary base to adenine. Thymine contains a methyl group at position 5, whereas uracil lacks this group. This substitution is a key chemical difference between DNA and RNA, affecting stability and function.
Q64. Which chromosome has the most genes (2968)?
Correct Answer: (b)
Chromosome 1 is the largest human autosome and contains the most genes, with approximately 2968 protein‑coding genes. In contrast, the Y chromosome has only about 231 genes, and chromosome 21 has fewer than 300. This gene density reflects the size and evolutionary history of each chromosome.
Q65. Assertion: Repetitive DNA sequences have no direct coding functions.
Reason: They are used as markers in DNA fingerprinting.
Correct Answer: (a)
Repetitive sequences, such as VNTRs and microsatellites, do not code for proteins but are highly polymorphic. Their variability makes them useful as genetic markers for identification. However, the fact that they are used as markers does not explain why they have no coding function; that is an independent property. Thus both statements are true, but the reason is not the explanation for the assertion.
Q66. Sickle cell anemia is caused by a point mutation in the beta globin chain resulting in the change of ______ to ______.
Correct Answer: (b)
A single nucleotide substitution in the β‑globin gene changes the sixth codon from GAG (glutamic acid) to GUG (valine). This replacement of a charged glutamic acid with a hydrophobic valine causes haemoglobin to polymerise under low oxygen conditions, distorting red blood cells into a sickle shape.
Q67. Which of the following is correct? (i) S. pneumoniae has S and R strains. (ii) S strain has a polysaccharide coat. (iii) Transformation was due to DNA transfer.
Correct Answer: (a)
S. pneumoniae has smooth (S) and rough (R) strains (i). The S strain has a polysaccharide capsule (ii). Griffith’s transformation was later shown to be due to DNA transfer (iii). All are correct.
Q68. A typical nucleosome contains how many base pairs of DNA helix?
Correct Answer: (b)
A nucleosome core particle consists of about 146 bp of DNA wrapped around a histone octamer, plus about 54 bp of linker DNA, totaling approximately 200 bp per nucleosome. This packaging unit is the fundamental repeating unit of chromatin.
Q69. Which are true for 'Euchromatin'? (i) Loosely packed. (ii) Stains light. (iii) Transcriptionally active.
Correct Answer: (a)
Euchromatin is less condensed (i), appears light under staining (ii), and contains actively transcribed genes (iii). These are the defining characteristics of euchromatin.
Q70. Assertion: Chromosome 1 was the last human chromosome to be sequenced.
Reason: Its sequence was completed only in May 2006.
Correct Answer: (a)
Chromosome 1, the largest human chromosome, was the last to be fully sequenced; its completion in May 2006 marked a milestone in the Human Genome Project. The reason provides the specific timeline, which explains why it was considered the last to be finished.
Q71. In a transcription unit, the strand with 3' -> 5' polarity is called:
Correct Answer: (b)
The template strand (also called antisense strand) has 3'→5' polarity and is used by RNA polymerase to synthesise RNA in the 5'→3' direction. The coding strand (sense strand) has 5'→3' polarity and its sequence is identical to the RNA transcript (with U instead of T).
Q72. Assertion: The structural gene in eukaryotes is monocistronic.
Reason: They have interrupted coding sequences (exons and introns).
Correct Answer: (a)
Eukaryotic structural genes are typically monocistronic, meaning each gene codes for a single polypeptide. The presence of exons and introns is a distinct characteristic of eukaryotic gene structure, but it does not explain monocistronicity. Prokaryotic genes are often polycistronic, but the presence of introns is not the reason for monocistronicity. Hence both are true, but the reason is not the explanation.
Q73. Regarding 'Replication Fork', identify correct facts: (i) Replication is continuous on 3' -> 5' template. (ii) Replication is discontinuous on 5' -> 3' template. (iii) DNA dependent DNA polymerase catalyse only in 5' -> 3' direction.
Correct Answer: (a)
On the 3'→5' template, synthesis is continuous (leading strand, i). On the 5'→3' template, synthesis is discontinuous (lagging strand, ii). DNA polymerase always adds nucleotides to the 3'‑OH, so it synthesises in the 5'→3' direction (iii). All are correct.
Q74. The length of DNA in a bacteriophage known as φ × 174 is:
Correct Answer: (b)
Bacteriophage φ × 174 has a relatively small genome of 5386 nucleotides, which was one of the first DNA sequences to be determined. This length is typical for a small single‑stranded DNA phage. The other options correspond to larger genomes: lambda phage (48502 bp), E. coli (4.6×10^6 bp), and human haploid genome (3.3×10^9 bp).
Q75. Regarding the Meselson-Stahl experiment, which statements are true? (i) 15N is a radioactive isotope. (ii) CsCl density gradient was used for separation. (iii) E. coli divides in 20 minutes. (iv) After 40 minutes, DNA was composed of equal amounts of hybrid and light DNA.
Correct Answer: (a)
15N is a stable heavy isotope, not radioactive (i is false). They used CsCl density gradient centrifugation (ii true). E. coli doubles every 20 minutes (iii true). After 40 minutes (two generations), half the DNA is hybrid and half is light (iv true). So (ii), (iii), (iv) are correct.
Q76. Assertion: Heterochromatin is transcriptionally inactive.
Reason: It is densely packed and stains dark.
Correct Answer: (a)
Heterochromatin is highly condensed, which restricts access of transcription factors and RNA polymerase, making it transcriptionally silent. Its dense packing also causes it to stain darkly with DNA‑binding dyes. The reason (dense packing) explains why it is inactive (transcription machinery cannot access DNA).
Q77. Assertion: Lac operon is under negative regulation.
Reason: The repressor protein binds to the operator to prevent transcription.
Correct Answer: (a)
Negative regulation means that a repressor protein inhibits transcription by binding to the operator site. In the lac operon, the repressor (produced by the i gene) blocks RNA polymerase from transcribing the structural genes in the absence of inducer. The reason describes this mechanism, directly explaining the negative regulation.
Q78. Regarding 'Transcription' vs 'Replication': (i) Total DNA duplicated in replication. (ii) Only a segment of DNA copied in transcription. (iii) Complementarity governs both.
Correct Answer: (a)
Replication copies the entire DNA (i). Transcription copies only specific genes (ii). Both processes rely on complementary base pairing (iii). All statements are correct.
Q79. Identify the correct steps of DNA Fingerprinting technique: (i) Isolation of DNA. (ii) Digestion by restriction endonucleases. (iii) Separation by electrophoresis. (iv) Blotting to synthetic membranes.
Correct Answer: (a)
The standard steps are: DNA isolation → restriction digestion → gel electrophoresis → Southern blotting → hybridisation with probe → detection. The listed steps (i‑iv) are correct and in the proper order, so all are correct in sequence.
Q80. Regarding 'Hershey and Chase' experiment: (i) Used bacteriophages. (ii) Radioactive 35S was used for protein. (iii) Radioactive 32P was used for DNA.
Correct Answer: (a)
They used T2 bacteriophages (i). ³⁵S labels protein (sulphur is in methionine/cysteine) (ii). ³²P labels DNA (phosphate) (iii). All statements are correct.
Q81. Assertion: Translation is energetically an expensive process.
Reason: Amino acids are activated in the presence of ATP before joining.
Correct Answer: (a)
Translation requires energy at several steps: aminoacyl‑tRNA synthesis (charging) consumes ATP (converted to AMP), and peptide bond formation and ribosome translocation use GTP. The activation of amino acids by ATP is a major energy‑consuming step. The reason explains why translation is energetically costly.
Q82. Which of the following is used as a probe in DNA Fingerprinting?
Correct Answer: (c)
Variable Number of Tandem Repeats (VNTR) are specific DNA sequences that vary in repeat number among individuals. They are used as probes in DNA fingerprinting because they produce unique banding patterns after restriction digestion and hybridisation, enabling individual identification.
Q83. Identify correct features of 'Avery, MacLeod and McCarty' work: (i) Worked on transforming principle. (ii) Used DNase to inhibit transformation. (iii) Proved DNA is the transforming substance.
Correct Answer: (a)
They extended Griffith's work (i). They used enzymes to destroy DNA, RNA, and protein; only DNase treatment prevented transformation (ii). Thus they proved DNA is the transforming substance (iii). All are correct.
Q84. Assertion: DNA polymorphism is the basis of DNA fingerprinting.
Reason: Polymorphism arises due to mutations which accumulate in non-coding DNA.
Correct Answer: (a)
DNA fingerprinting exploits differences in repetitive DNA sequences (VNTRs) that vary among individuals. These variations (polymorphisms) arise from mutations in non‑coding regions, which accumulate without selective pressure. The reason explains why these polymorphisms exist and why they are useful for identification, thereby supporting the assertion.
Q85. The largest known human gene is:
Correct Answer: (c)
The dystrophin gene, located on the X chromosome, is the largest known human gene, spanning about 2.4 million base pairs. It contains 79 exons and encodes a protein essential for muscle integrity; mutations cause Duchenne muscular dystrophy.
Q86. Which are correct for 'Mutations' as per the text? (i) Frameshift mutations insert or delete one/two bases. (ii) Sickle cell anemia is a point mutation. (iii) Chromosomal aberrations are common in cancer cells.
Correct Answer: (a)
Frameshift mutations (insertions/deletions not multiples of three) alter the reading frame (i). Sickle cell is a point mutation (GAG→GUG) (ii). Cancer cells often have chromosomal aberrations (iii). All statements are correct.
Q87. Which statements about 'DNA Fingerprinting' are correct? (i) Involves identifying repetitive DNA. (ii) Satellite DNA shows high degree of polymorphism. (iii) Alec Jeffreys developed the technique using VNTR. (iv) It is the basis of paternity testing.
Correct Answer: (a)
DNA fingerprinting relies on repetitive sequences that are highly polymorphic (i, ii). Jeffreys pioneered it using VNTR probes (iii). The technique is widely used in paternity testing (iv). All statements are correct.
Q88. In the Central Dogma, genetic information flows from:
Correct Answer: (b)
The Central Dogma, proposed by Francis Crick, describes the flow of genetic information as DNA → RNA → Protein. Transcription converts DNA to RNA, and translation converts RNA to protein. Some viruses reverse the flow (RNA → DNA), but the standard dogma is DNA → RNA → Protein.
Q89. Which linkage joins a nitrogenous base to the 1' C of a pentose sugar?
Correct Answer: (b)
The bond between the nitrogenous base and the sugar is an N‑glycosidic linkage, formed between the anomeric carbon (1') of the sugar and a nitrogen atom of the base. Phosphodiester linkages connect nucleotides in a strand, while hydrogen bonds hold the two strands together.
Q90. Match the following for DNA replication:
Column-I
Column-II
A. 3' -> 5' template
(I) Discontinuous synthesis
B. 5' -> 3' template
(II) Continuous synthesis
Correct Answer: (a)
The template with 3'→5' polarity allows continuous synthesis of the new strand in the 5'→3' direction (leading strand, A‑II). The template with 5'→3' polarity forces discontinuous synthesis (lagging strand, Okazaki fragments, B‑I). Thus A‑II, B‑I.
Q91. Correct statements for 'RNA Splicing' complexity: (i) Representative of ancient feature of genome. (ii) reminiscent of antiquity. (iii) Dominance of RNA-world.
Correct Answer: (a)
Splicing, especially self‑splicing introns, is considered a relic of the RNA world (i and ii). The existence of ribozymes and spliceosomes supports the idea that RNA was once dominant (iii). All are correct.
Q92. Which are correct for 'mRNA' translation? (i) read in combination of three (codons). (ii) Translation occurs in ribosomes. (iii) Site of protein synthesis.
Correct Answer: (a)
mRNA is read in triplets (codons) (i). Translation takes place on ribosomes (ii), which are the sites of protein synthesis (iii). All statements are correct.
Q93. Frederick Griffith conducted his 'Transforming Principle' experiments using:
Correct Answer: (b)
Griffith worked with Streptococcus pneumoniae, which has two strains: the virulent S strain (smooth colonies with a polysaccharide capsule) and the avirulent R strain (rough colonies). His experiments showed that heat‑killed S strain could transform R strain into virulent forms.
Q94. Identify the correct phylum/group being described? (i) They are small nuclear RNAs. (ii) Transcribed by RNA Polymerase III. (iii) Part of eukaryotic transcription complexity.
Correct Answer: (c)
Small nuclear RNAs (snRNAs) are transcribed by RNA polymerase III and are involved in splicing (as part of the spliceosome). They are a distinct class of RNA molecules, not mRNA, tRNA, or rRNA.
Q95. Which are correct regarding 'Capping and Tailing'? (i) Capping adds methyl guanosine triphosphate to 5'-end. (ii) Tailing adds adenylate residues to 3'-end. (iii) Occurs in hnRNA.
Correct Answer: (a)
Capping (5' cap) adds a modified guanine nucleotide to the 5' end (i). Tailing (polyadenylation) adds a poly‑A tail to the 3' end (ii). These modifications occur on hnRNA (primary transcripts) before splicing and export (iii). All are correct.
Q96. The experimental proof for semiconservative replication of DNA was first shown in:
Correct Answer: (c)
Meselson and Stahl used E. coli to demonstrate semiconservative replication by growing bacteria in a heavy nitrogen isotope (¹⁵N) and then shifting to light nitrogen (¹⁴N). Density gradient centrifugation showed hybrid DNA after one generation, consistent with semiconservative replication.
Q97. Regarding 'Polynucleotide chain': (i) 5'-end has a free phosphate group. (ii) 3'-end has a free OH group. (iii) Sequence of bases read from 5' to 3'.
Correct Answer: (a)
A polynucleotide has polarity: 5' end has a phosphate, 3' end has a hydroxyl. The sequence is conventionally written from 5' to 3'. All statements are correct.
Q98. Which of the following is being described? (i) It is the regulatory gene of lac operon. (ii) Expressed constitutively (all-the-time). (iii) Produces a protein that binds to operator.
Correct Answer: (c)
The i gene is the regulator gene; it is expressed constitutively to produce the repressor protein, which binds the operator. The z, y, and a genes are structural genes under the operator control.
Q99. Assertion: The distance between two polynucleotide chains in DNA remains almost constant.
Reason: A purine always comes opposite to a pyrimidine in base pairing.
Correct Answer: (a)
The constant distance between the two DNA strands (about 2 nm) is maintained because a purine (two‑ring) always pairs with a pyrimidine (one‑ring), making the combined width uniform. If purine‑purine or pyrimidine‑pyrimidine pairing occurred, the distance would vary. Thus the reason explains the constant distance.
Q100. Identify correct facts for 'SNPs': (i) Single nucleotide polymorphism. (ii) 1.4 million locations in human DNA. (iii) Used for tracing human history.
Correct Answer: (a)
SNPs are single nucleotide variations (i). The HGP identified about 1.4 million SNPs (ii). They are used as markers in population genetics and tracing human migrations (iii). All are correct.