A base is the nitrogenous ring (purine or pyrimidine) alone.
A nucleoside = base + sugar (no phosphate). Cytidine = cytosine + ribose.
A nucleotide = base + sugar + phosphate. Cytidine monophosphate (CMP).
Exercise 2 (NCERT)
If a double-stranded DNA has 20% cytosine, calculate the percent of each of the other bases.
Step 1 — Apply Chargaff's rule: G = C, so G = 20%. Step 2: A + T + G + C = 100%. Therefore A + T = 100 − (20+20) = 60%. Step 3: A = T (Chargaff), so A = T = 30% each.
Why is the Watson-Crick model double helix described as 'right-handed'? What is meant by the antiparallel nature of strands?
Right-handed: If you look along the helix axis from one end, the strands wind in a clockwise direction (like a typical screw or staircase). Twist your right thumb up — your fingers curl in the same direction the helix winds. This contrasts with left-handed Z-DNA, which is rare.
Antiparallel strands: The two DNA strands run in OPPOSITE chemical directions:
One strand: 5' → 3'
The other strand: 3' → 5' (alongside)
This means the 5' end of one strand pairs with the 3' end of the other. The 5' end has a free phosphate group; the 3' end has a free OH group. Antiparallel orientation is essential for proper base pairing geometry.
Exercise 4 (NCERT)
Briefly describe the experimental work of Hershey and Chase.
Hershey and Chase (1952) — used bacteriophage T2 to identify the genetic material:
Setup: Bacteriophage T2 has only DNA + protein. Phages were grown in two batches:
Batch 1: with radioactive sulfur (³⁵S) — labels protein only (DNA has no S).
Batch 2: with radioactive phosphorus (³²P) — labels DNA only (protein has no P).
Infection: Each batch infected E. coli separately.
Blender: A waring blender was used to shake the phage coats off the bacterial cells. Then centrifugation separated the bacteria (heavy, in pellet) from the phage coats (light, in supernatant).
Results:
³⁵S (protein) — found mostly in supernatant; bacteria not radioactive.
³²P (DNA) — found in the bacterial pellet; supernatant not radioactive.
Conclusion: Only DNA enters the bacterium during infection. Therefore DNA, not protein, is the genetic material.
This experiment finally convinced biologists (combined with Avery's enzyme work) that DNA carries genetic information.
Exercise 5 (NCERT)
Differentiate between heterochromatin and euchromatin. Which of these is transcriptionally active?
Feature
Euchromatin
Heterochromatin
Density of staining
Loose, light-staining
Dense, dark-staining
Compaction
Less condensed
Highly condensed
Transcriptional activity
Active — transcribed
Inactive — not transcribed
Location
Mostly inner nucleus
Nuclear periphery and around centromere
Replication timing
Early in S phase
Late in S phase
Transcriptionally active: Euchromatin — the loose form is accessible to RNA polymerase. Heterochromatin is too tightly packed for transcription machinery to access.
Exercise 6 (NCERT)
List the salient features of the genetic code.
Salient features of the genetic code:
Triplet code: Three nucleotides specify one amino acid. 4³ = 64 codons total.
Universal: Same code in nearly all organisms (bacteria, plants, animals, humans). Strong evidence for common ancestry.
Degenerate: Most amino acids have multiple codons. E.g., Leucine has 6, Serine has 6.
Unambiguous: Each codon codes for only one amino acid (no codon codes for two).
Non-overlapping: Codons are read consecutively without any base sharing.
Comma-less: No punctuation between codons; reading proceeds continuously from start.
Has start and stop signals: AUG = start (Met); UAA, UAG, UGA = stop.
Wobble at 3rd position: Some flexibility in pairing the third base — explains degeneracy.
Exercise 7 (NCERT)
Following are the features of genetic code. What does each one indicate? (a) Stop codon (b) Unambiguous codon (c) Degenerate codon (d) Universal codon.
(a) Stop codon: A codon that does not specify an amino acid but signals end of translation. The three stop codons are UAA, UAG, UGA. They cause release of the polypeptide from the ribosome.
(b) Unambiguous codon: Each codon codes for ONLY ONE specific amino acid. There is no codon that means two different amino acids — the code is precise.
(c) Degenerate codon: An amino acid has more than one codon. For example, Leucine is coded by UUA, UUG, CUU, CUC, CUA, CUG — six different codons. Provides protection against mutations.
(d) Universal codon: The genetic code is the same across nearly all organisms — bacteria, fungi, plants, animals all read AUG as Met. This is strong evidence that all life shares a common ancestor.
Exercise 8 (NCERT)
Write the function of (a) Methylated guanosine cap (b) Poly-A tail.
Both are post-transcriptional modifications added to eukaryotic mRNA.
(a) Methylated guanosine (5' cap):
Added to the 5' end of pre-mRNA.
Protects mRNA from degradation by exonucleases.
Helps the mRNA bind to the small ribosomal subunit during translation initiation.
Marks the mRNA for export from nucleus to cytoplasm.
(b) Poly-A tail (3' end):
~200 adenine residues added to the 3' end.
Stabilises mRNA — increases its half-life.
Promotes nuclear export.
Enhances translation efficiency by interacting with poly-A binding protein.
Both modifications work together to ensure efficient translation of eukaryotic mRNAs.
Exercise 9 (NCERT)
What is DNA fingerprinting? Mention its applications.
Definition: DNA fingerprinting is a technique to identify individuals based on unique patterns in their DNA, particularly in regions called VNTRs (Variable Number Tandem Repeats). Developed by Sir Alec Jeffreys in 1984.
Principle: The number of tandem repeats at various VNTR loci varies dramatically between individuals. Combining the patterns at multiple loci gives a profile that is essentially unique (except for identical twins).
Steps: DNA isolation → restriction digestion → gel electrophoresis → Southern blot → hybridisation with VNTR probe → autoradiography → band pattern analysis.
Applications:
Forensic identification — match crime scene DNA to suspects.
Paternity / maternity testing.
Identification of disaster victims and war casualties.
The HGP is considered one of the greatest scientific achievements of the 20th century.
Exercise 11 (NCERT)
Discuss the process of transcription in prokaryotes.
Transcription in prokaryotes (e.g., E. coli):
Single RNA polymerase synthesises all three RNA types (mRNA, tRNA, rRNA).
Initiation: The RNA polymerase holoenzyme has a sigma (σ) factor that recognises and binds the promoter (typically -10 and -35 regions). DNA unwinds locally.
Elongation: The sigma factor leaves; RNA polymerase moves along the template strand 3'→5', synthesising mRNA 5'→3'. Adds ribonucleotides complementary to template (A→U, T→A, G→C, C→G).
Termination: Two mechanisms:
Rho-independent: RNA forms a hairpin loop followed by a poly-U sequence — RNA polymerase falls off.
Rho-dependent: Rho protein binds RNA and pulls polymerase off.
Coupled translation: No nucleus, so ribosomes can begin translating the mRNA while it's still being transcribed.
Polycistronic mRNAs: Many bacterial mRNAs encode multiple proteins (operons like lac).
No introns: Bacterial genes typically don't have introns — no splicing needed.
This rapid, streamlined process allows bacteria to respond quickly to environmental changes.
Exercise 12 (NCERT)
Describe the lac operon model of E. coli.
Lac operon — discovered by Jacob and Monod (1961). It controls how E. coli digests lactose.
Components:
i gene — codes for the repressor protein (constitutively expressed).
p — promoter where RNA polymerase binds.
o — operator where repressor binds (between p and structural genes).
z gene — codes for β-galactosidase (splits lactose).
y gene — codes for permease (transports lactose into cell).
a gene — codes for transacetylase.
How it works:
Without lactose (OFF): Repressor is active → binds operator → blocks RNA polymerase → no transcription. The cell saves energy.
With lactose (ON): Some lactose is converted to allolactose (the inducer) → allolactose binds repressor → repressor changes shape and falls off → RNA polymerase transcribes z, y, a → enzymes are made → lactose is digested for energy.
This is an inducible operon — turned on by its substrate. The lac operon was the FIRST gene regulation system understood at molecular level.
📐 Activity 6.5 — Final Synthesis
Setup: A gene's coding strand reads 5'-ATGGCATTGAAATGA-3'.
Predict: (a) Write the template strand. (b) Write the mRNA. (c) Translate to protein. (d) What does the gene code for in size?
(b) mRNA (5'→3'): 5'-AUGGCAUUGAAAUGA-3' (same as coding with T→U).
(c) Translate codons:
AUG → Met (start)
GCA → Ala
UUG → Leu
AAA → Lys
UGA → STOP
Protein: Met–Ala–Leu–Lys (a 4-amino-acid peptide).
(d) Gene size: 15 bp = 5 codons (4 amino acids + 1 stop). One of the smallest possible functional genes.
🎯 Competency-Based Questions — Chapter Review
Q1. The number of hydrogen bonds between G and C is:L1 Remember
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3. G and C share 3 hydrogen bonds, while A and T share 2. The extra H-bond makes GC pairs more stable than AT pairs — DNA segments rich in GC require higher temperatures to denature.
Q2. Match the column: Connect each enzyme to its function. L2 Understand
Enzyme
Function
Helicase
Unwinds the double helix
Primase
Synthesizes RNA primers
DNA polymerase III
Adds DNA nucleotides 5'→3'
DNA ligase
Joins Okazaki fragments
RNA polymerase
Transcribes DNA into RNA
Q3. mRNA: 5'-AUGGCGUAA-3'. Translate it. L3 Apply
Codons: AUG-GCG-UAA AUG → Met (start); GCG → Ala; UAA → STOP. Protein: Met–Ala (a dipeptide).
Q4. Compare: List 4 differences between DNA replication and transcription. L4 Analyse
Feature
Replication
Transcription
Product
DNA
RNA
Enzyme
DNA polymerase
RNA polymerase
Template strands used
Both strands
Only one strand
Primer needed
Yes (RNA primer)
No
Product length
Entire chromosome
Single gene/operon
Frequency
Once per cell cycle
Many times
Q5. HOT (Create): A drug that selectively binds RNA polymerase prevents bacterial growth. Predict why this drug is selective for bacteria. L6 Create
Why selective for bacteria:
Structural difference: Bacterial RNA polymerase is structurally distinct from human RNA polymerase II. The drug-binding pocket exists in the bacterial form but not the human form.
Single vs multiple polymerases: Bacteria have ONE RNA polymerase. Humans have THREE (Pol I, II, III). Even if a drug affects one, the others still function.
Cellular localisation: Bacterial transcription happens in the cytoplasm. Human transcription is in the nucleus, with permeability barriers.
Real-world example: Rifampicin selectively inhibits bacterial RNA polymerase β-subunit. It is used to treat tuberculosis. Human RNA polymerase is not affected — the basis of selective toxicity.
This is the principle behind nearly all antibiotics: target essential bacterial processes that differ structurally from human equivalents.
🧠 Assertion–Reason Questions — Chapter Review
A: The genetic code is degenerate.
R: There are 64 codons but only 20 amino acids.
Answer: (A). Both true; R explains A. Sixty-four codons divided among 20 amino acids (and 3 stop signals) means most amino acids have multiple codons — this is degeneracy.
A: The lac operon is repressed when glucose is the only sugar available.
R: The repressor binds the operator in the absence of allolactose.
Answer: (A). Both true; R explains A. With glucose only and no lactose, no allolactose is made → repressor stays active and bound → operon OFF. (Catabolite repression involving CAP-cAMP also contributes, but the question is at the basic level.)
A: The Human Genome Project sequenced ~3.2 billion base pairs.
R: All 3.2 billion bases code for proteins.
Answer: (C). A is TRUE — the genome size is correct. R is FALSE — only ~1.5% codes for proteins; the rest is regulatory regions, introns, repetitive sequences, and other non-coding DNA. The HGP showed our genome is much more complex than a simple gene catalogue.
Frequently Asked Questions - NCERT Exercises and Solutions: Molecular Basis of Inheritance
What are the most-asked NCERT exercise questions in Chapter Molecular Basis of Inheritance?
NCERT Class 12 Biology Chapter on Molecular Basis of Inheritance exercises cover definitions, mechanisms, labelled diagrams, comparison tables, and application-based questions. The MyAiSchool solution set provides full step-by-step solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should master scientific terminology, mechanism flowcharts, and concept comparison to score full marks.
How should students approach diagram-based questions in Molecular Basis of Inheritance?
For diagram-based questions in NCERT Class 12 Biology Chapter on Molecular Basis of Inheritance: (1) draw clean, proportional, large diagrams with sharp pencil lines, (2) label all parts horizontally on the right side using a ruler, (3) use scientific terminology (Latin/Greek names where applicable), (4) write a 1-2 line description if asked. The MyAiSchool solutions provide editable reference diagrams aligned with NCERT textbook figures.
What types of CBSE board questions come from Molecular Basis of Inheritance?
CBSE Class 12 Biology board questions from Molecular Basis of Inheritance typically include: (1) 1-mark MCQs on definitions and processes, (2) 2-mark short-answer mechanism/comparison questions, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining mechanism + diagram + significance. The MyAiSchool exercise set tags each question by mark weight and Bloom level (L1-L6).
How do I compare two biological processes in 5-mark questions?
For 5-mark comparison questions in NCERT Class 12 Biology on Molecular Basis of Inheritance: (1) use a two-column table with feature headings down the left side, (2) compare on at least 5-6 features (definition, location, mechanism, control, outcome, significance), (3) include one labelled diagram if relevant, (4) end with biological significance. The MyAiSchool solutions follow this CBSE-aligned tabular format consistently for full marks.
What are common mistakes in Molecular Basis of Inheritance exercises?
Common mistakes in NCERT Class 12 Biology Molecular Basis of Inheritance include: (1) confusing similar terms (mitosis vs meiosis, syngamy vs triple fusion, etc.), (2) skipping intermediate steps in mechanisms, (3) missing examples or species names, (4) writing essays when tables are expected, (5) forgetting biological significance. The MyAiSchool solutions highlight these traps with red flags so students avoid losing marks.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Biology Molecular Basis of Inheritance solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with biological reasoning, fully labelled SVG diagrams, comparison tables, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes.
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