Chapter 5: Molecular Basis of Inheritance Long Questions | biology Class 12 CBSE

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Chapter 5: Molecular Basis of Inheritance Long Questions | biology Class 12 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Describe the Double Helix model of DNA, including its structural features and the significance of this model for understanding genetic inheritance.

DNA is composed of two polynucleotide chains with antiparallel polarity (one 5'→3', the other 3'→5'); the backbone is formed by sugar-phosphate alternation, with nitrogenous bases projecting inward.
Bases pair specifically by hydrogen bonds: A with T (two H-bonds) and G with C (three H-bonds); this complementarity ensures each strand can act as a template for the other, and Chargaff's rules (A=T, G=C) are satisfied.
The chains coil right-handedly with a pitch of 3.4 nm, ~10 bp per turn, and 0.34 nm between consecutive base pairs; base stacking in addition to H-bonds stabilises the helix.
The complementarity between strands explains how genetic information can be faithfully duplicated: each strand serves as a template for a new daughter strand, underpinning the concept of semiconservative replication.
Watson and Crick's model was immediately linked by Francis Crick to the Central Dogma (DNA→RNA→Protein), making the Double Helix not just a structural model but the foundation of molecular genetics.

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Long Answer

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Hard

Question 2

Long Form

Compare the experimental contributions of Griffith, Avery-MacLeod-McCarty, and Hershey-Chase in establishing DNA as the genetic material.

Griffith (1928) demonstrated transformation: heat-killed virulent S-strain bacteria converted live avirulent R-strain into virulent S-strain in mice, indicating a transferable 'transforming principle,' but he could not identify its biochemical nature.
Avery, MacLeod, and McCarty (1933–44) biochemically purified the transforming principle and showed that DNA alone caused transformation; DNase abolished transformation while proteases and RNases did not, pointing to DNA as the genetic material.
Hershey and Chase (1952) used radioactive labelling (32P in phage DNA, 35S in phage protein) to show that only 32P-labelled DNA entered E. coli during infection, providing unequivocal proof that DNA — not protein — is the genetic material.
Each experiment progressively built on the previous: Griffith established the phenomenon, Avery et al. identified the molecule, and Hershey-Chase provided definitive physical evidence separating the genetic material from protein.
Together, these experiments shifted scientific consensus from 'protein is the genetic material' to 'DNA is the genetic material,' laying the foundation for the Watson-Crick structural model and modern molecular biology.

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Long Answer

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Hard

Question 3

Long Form

Explain the Meselson-Stahl experiment. How did it prove the semiconservative mode of DNA replication?

E. coli was grown for many generations in 15NH4Cl medium so that all DNA was 15N-labelled (heavy DNA); cells were then shifted to 14N medium and DNA samples were taken after each generation.
DNA was isolated and centrifuged in CsCl density gradients: heavy (15N), intermediate (hybrid 15N/14N), and light (14N) DNA bands at different positions based on density.
After one generation in 14N, all DNA appeared at an intermediate (hybrid) density — one 15N strand and one newly synthesised 14N strand — consistent with semiconservative replication.
After two generations, equal amounts of hybrid and light DNA appeared, with no heavy DNA remaining; this pattern is only possible if each original strand served as a template and was retained in one of the daughter molecules.
Conservative replication would have predicted heavy and light DNA bands (no hybrid) after one generation; dispersive replication would have given all-intermediate DNA in all generations — neither fitted the results, confirming semiconservative replication.

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Long Answer

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Question 4

Long Form

Describe the steps of transcription in bacteria, highlighting the role of sigma (σ) and rho (ρ) factors.

In bacteria, a single DNA-dependent RNA polymerase catalyses all three types of RNA (mRNA, tRNA, rRNA) using nucleoside triphosphates as substrates and synthesising in the 5'→3' direction.
Initiation: the sigma (σ) factor associates with RNA polymerase, enabling it to recognise and bind to the specific promoter sequence upstream of the structural gene, and initiates RNA synthesis at the correct start site.
Elongation: RNA polymerase unwinds the DNA, reads the template strand (3'→5'), and polymerises a complementary RNA chain; only a short stretch of RNA-DNA hybrid exists at any time as the polymerase moves forward.
Termination: when the polymerase reaches the terminator region, the rho (ρ) factor associates with the enzyme and facilitates dissociation of the RNA transcript and RNA polymerase from the DNA template.
Unlike eukaryotes, bacterial mRNA does not require post-transcriptional processing and can be directly translated; transcription and translation can even be coupled since both occur in the same cellular compartment.

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Long Answer

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Hard

Question 5

Long Form

Contrast the process of transcription in prokaryotes and eukaryotes with respect to RNA polymerases, post-transcriptional processing, and coupling with translation.

Prokaryotes have a single RNA polymerase (with σ and ρ factors for initiation and termination) that transcribes all RNA types; eukaryotes have three nuclear RNA polymerases: Pol I (rRNA: 28S, 18S, 5.8S), Pol II (pre-mRNA/hnRNA), and Pol III (tRNA, 5S rRNA, snRNAs).
Prokaryotic mRNA is polycistronic and functional immediately after transcription; eukaryotic primary transcripts (hnRNA) are monocistronic and require extensive processing: 5'-capping (methylguanosine triphosphate), 3'-poly-A tailing (~200–300 adenylate residues), and splicing (removal of introns, joining of exons).
In eukaryotes, genes are split (exons interrupted by introns); the introns in hnRNA are removed and exons are joined in a defined order during splicing, producing mature mRNA that exits the nucleus for translation.
In bacteria, the absence of a nuclear membrane allows transcription and translation to occur simultaneously in the cytoplasm (they are coupled); in eukaryotes, transcription occurs in the nucleus and mRNA must be exported before translation in the cytoplasm.
The processing steps in eukaryotes add complexity and an additional regulatory layer; splicing is important for generating protein diversity through alternative splicing, and the capping and tailing protect mRNA from degradation and assist ribosome binding.

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Long Answer

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Question 6

Long Form

Outline the process of translation from initiation to termination, highlighting the role of each component.

Initiation: the small ribosomal subunit binds to mRNA at the 5'-end and moves to the start codon (AUG); the initiator tRNA (carrying methionine) recognises AUG by anticodon–codon complementarity, and the large subunit joins to form the initiation complex.
Elongation: charged tRNAs enter the A-site (aminoacyl site) of the ribosome and pair with successive mRNA codons; the 23S rRNA (ribozyme) catalyses peptide bond formation between the incoming amino acid and the growing peptide chain on the P-site tRNA.
Translocation: the ribosome moves one codon at a time along the mRNA in the 5'→3' direction; the deacylated tRNA exits from the E-site (exit site), and the peptidyl-tRNA shifts from the A-site to the P-site.
Energy: aminoacylation of tRNA requires ATP; GTP is used for translocation and ribosome movements, making translation energetically costly.
Termination: when a stop codon (UAA, UAG, or UGA) enters the A-site, a protein release factor binds (no tRNA exists for stop codons), causing hydrolysis of the peptide-tRNA bond and release of the complete polypeptide, followed by dissociation of the ribosome from mRNA.

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Long Answer

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Hard

Question 7

Long Form

Analyse the regulation of the lac operon in E. coli under conditions of (i) absence of lactose and (ii) presence of lactose. Explain why this regulation is termed negative regulation.

The lac operon consists of a regulatory i gene (encoding the lac repressor) and three structural genes z (β-galactosidase), y (permease), and a (transacetylase), controlled by a common promoter and operator.
In the absence of lactose: the i gene constitutively produces active repressor protein, which binds with high affinity to the operator region adjacent to the promoter, physically blocking RNA polymerase from accessing the structural genes — the operon is OFF.
In the presence of lactose: transported lactose is converted to allolactose (by the small basal level of β-galactosidase), which acts as an inducer by binding to and inactivating the repressor; the conformationally altered repressor cannot bind the operator, allowing RNA polymerase to transcribe the structural genes — the operon is ON.
Once all lactose is metabolised, allolactose levels fall, the repressor regains activity and rebinds the operator, shutting off the operon — a form of product feedback regulation (the substrate induces the enzymes needed for its own metabolism).
This is termed negative regulation because the repressor (a negative regulator) normally keeps the operon OFF; expression requires relief from repression (de-repression) rather than active stimulation — in contrast to positive regulation where an activator is required to turn the operon ON.

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Long Answer

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Medium

Question 8

Long Form

Describe the goals, methodologies, and at least four major findings of the Human Genome Project.

Goals: to identify all ~20,000–25,000 human genes, determine the sequence of ~3 billion bp, store data in accessible databases, improve analytical tools and bioinformatics, transfer technologies to industry, and address ethical, legal, and social issues (ELSI).
Methodology: two approaches were used — Expressed Sequence Tags (ESTs) sequencing transcribed genes, and whole-genome shotgun sequencing (sequencing total fragmented DNA and using computational tools for alignment); BAC and YAC vectors amplified fragments; automated Sanger sequencing was employed, followed by bioinformatics-based assembly.
Key findings: the human genome contains ~3164.7 million bp; total genes ~30,000 (far fewer than earlier estimates); less than 2% of the genome codes for proteins; repeated sequences constitute a large portion; chromosome 1 has most genes (2968), Y chromosome fewest (231).
99.9% of base sequences are identical across all humans; ~1.4 million SNP locations were identified, valuable for disease-gene mapping and evolutionary studies; functions are unknown for over 50% of discovered genes.
The project gave birth to Bioinformatics as a field, enabled whole-genome approaches to biology, opened new avenues in medicine (diagnosis, treatment, and prevention of genetic disorders), and led to sequencing of non-human model organisms (bacteria, yeast, Drosophila, rice, Arabidopsis) for comparative genomics.

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Long Answer

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Hard

Question 9

Long Form

Explain the molecular basis of DNA fingerprinting. Discuss the steps involved in Southern blot hybridisation used in this technique.

DNA fingerprinting exploits the highly polymorphic Variable Number of Tandem Repeats (VNTR) — mini-satellite DNA sequences where a short motif is repeated in tandem; copy numbers differ significantly between individuals (except monozygotic twins) and are inherited from parents, making them ideal identity markers.
Step 1 — DNA isolation and digestion: DNA is extracted from any tissue (blood, hair follicle, saliva, sperm, bone), digested with restriction endonucleases to cut it into fragments of varying sizes depending on VNTR copy number.
Step 2 — Electrophoresis and blotting: DNA fragments are separated by size using gel electrophoresis, then transferred (blotted) onto a nitrocellulose or nylon membrane (Southern blot), preserving the pattern of separation.
Step 3 — Hybridisation: the membrane is hybridised with a radiolabelled VNTR probe complementary to the repeat sequence; probe binds specifically to fragments containing VNTR sequences.
Step 4 — Autoradiography: X-ray film is exposed to detect the radiolabelled hybridised bands; the resulting pattern of bands of different sizes constitutes the unique DNA fingerprint; PCR can be used to amplify trace DNA, making the technique extremely sensitive from a single cell.

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Long Answer

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Medium

Question 10

Long Form

Discuss the packaging of DNA in eukaryotic cells from the nucleosome level to the chromosome, explaining the proteins involved at each level.

The total DNA in a mammalian cell, if extended, would be ~2.2 metres long — far exceeding the nuclear diameter of ~10⁻⁶ m; this necessitates hierarchical compaction.
Level 1 — Nucleosome: negatively charged DNA wraps ~200 bp around a positively charged histone octamer (8 histone proteins rich in lysine and arginine); nucleosomes appear as 'beads-on-string' under electron microscopy and constitute the basic unit of chromatin.
Level 2 — Chromatin fibre: the beads-on-string are further coiled and compacted with the help of Non-Histone Chromosomal (NHC) proteins to form a 30 nm chromatin fibre, and further into looped domains held together by a protein scaffold.
Level 3 — Metaphase chromosome: during cell division, chromatin fibres are maximally condensed and organised into visible chromosomes; this requires additional NHC proteins and involves further coiling.
Functional significance: loosely packed euchromatin is transcriptionally active, while densely packed heterochromatin is inactive; the degree of packaging thus regulates gene expression, linking chromatin structure directly to cellular function.

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Chapter 5: Molecular Basis of Inheritance | Long Questions

Describe the Double Helix model of DNA, including its structural features and the significance of this model for understanding genetic inheritance.

Compare the experimental contributions of Griffith, Avery-MacLeod-McCarty, and Hershey-Chase in establishing DNA as the genetic material.

Explain the Meselson-Stahl experiment. How did it prove the semiconservative mode of DNA replication?

Describe the steps of transcription in bacteria, highlighting the role of sigma (σ) and rho (ρ) factors.

Contrast the process of transcription in prokaryotes and eukaryotes with respect to RNA polymerases, post-transcriptional processing, and coupling with translation.

Outline the process of translation from initiation to termination, highlighting the role of each component.

Analyse the regulation of the lac operon in E. coli under conditions of (i) absence of lactose and (ii) presence of lactose. Explain why this regulation is termed negative regulation.

Describe the goals, methodologies, and at least four major findings of the Human Genome Project.

Explain the molecular basis of DNA fingerprinting. Discuss the steps involved in Southern blot hybridisation used in this technique.

Discuss the packaging of DNA in eukaryotic cells from the nucleosome level to the chromosome, explaining the proteins involved at each level.

Unlock all Long Questions