This MCQ module is based on: Dna Structure Griffith Experiment
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Dna Structure Griffith Experiment
🎓 Class 12
Biology
CBSE
Theory
Ch 5 – Molecular Basis of Inheritance
⏱ ~14 min
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Dna Structure Griffith Experiment
6.1 The Search for the Genetic Material
By the early 1900s, scientists knew that genes existed on chromosomes — but chromosomes contain BOTH protein and DNA. Which is the actual genetic material? For decades, most scientists believed proteins (with their 20 amino acids) were "complex enough" to carry hereditary information; DNA (with only 4 nucleotides) seemed too simple.
6.1.1 Griffith's Transforming Principle (1928)
Frederick Griffith worked with two strains of Streptococcus pneumoniae:
- S strain (Smooth) — has polysaccharide capsule, virulent (causes pneumonia, kills mice).
- R strain (Rough) — no capsule, non-virulent (mice live).
Griffith conducted four experiments:
| # | Inject mice with | Result |
|---|---|---|
| 1 | Live S strain | Mice die (S recovered from blood) |
| 2 | Live R strain | Mice live |
| 3 | Heat-killed S strain | Mice live |
| 4 | Heat-killed S + Live R (mixed) | Mice DIE — live S strain recovered! |
The shocking result: dead bacteria + harmless live bacteria → killer bacteria. Griffith concluded that some chemical substance in the dead S strain "transformed" the live R strain into virulent S strain. He called this the transforming principle. He didn't know what the chemical was.
6.1.2 Avery, MacLeod & McCarty (1933–44)
Sixteen years later, three scientists at Rockefeller University set out to identify Griffith's mysterious transforming principle. Their approach: take heat-killed S strain extract and selectively destroy each component (proteins, RNA, DNA) using specific enzymes:
- Add protease (digests proteins) → R still transformed to S. So protein is NOT the transforming principle.
- Add RNase (digests RNA) → R still transformed to S. So RNA is NOT it.
- Add DNase (digests DNA) → transformation STOPPED. No more S strain produced!
Conclusion: DNA was the transforming principle — the genetic material that converted R into S. Despite this elegant proof, many biologists still resisted, citing DNA's "simplicity."
6.1.3 Hershey-Chase Experiment (1952)
Alfred Hershey and Martha Chase used bacteriophage T2 — a virus made only of protein and DNA. Their goal: which one enters the bacterium during infection and directs the production of new viruses?
They used radioactive labelling:
- Batch A: Phages grown with radioactive sulfur (³⁵S). Only proteins (which contain S) become radioactive.
- Batch B: Phages grown with radioactive phosphorus (³²P). Only DNA (which contains P) becomes radioactive.
- They infected E. coli with each batch separately, then sheared off the phage coats with a blender and centrifuged.
| Phage label | Found inside bacterium | Found in supernatant |
|---|---|---|
| ³⁵S (protein) | No radioactivity | All ³⁵S — protein stayed outside |
| ³²P (DNA) | Radioactive — DNA entered the cell | No radioactivity |
Conclusion: Only DNA enters the bacterium during infection — DNA must be the genetic material. This experiment, combined with Avery's findings, finally convinced biologists that DNA is the genetic material.
6.2 Structure of DNA — The Watson-Crick Double Helix (1953)
In 1953, James Watson and Francis Crick published their model of DNA structure in the journal Nature. They built on:
- Erwin Chargaff's rules: In any DNA, A=T and G=C (and total purines = total pyrimidines).
- Rosalind Franklin's X-ray diffraction (Photo 51): showed DNA was helical with regular spacing.
6.2.1 Components of DNA
DNA is a polymer of nucleotides. Each nucleotide has three parts:
- Pentose sugar — deoxyribose (5-carbon sugar).
- Phosphate group.
- Nitrogenous base — one of four:
- Purines (double ring): Adenine (A) and Guanine (G).
- Pyrimidines (single ring): Cytosine (C) and Thymine (T).
6.2.2 Key Features of the Double Helix
- Two polynucleotide strands wound around each other in a right-handed double helix.
- The two strands are antiparallel — one runs 5'→3', the other 3'→5'.
- Sugar-phosphate backbone is outside; bases face inward.
- Complementary base pairing: A pairs with T (via 2 H-bonds), G pairs with C (via 3 H-bonds).
- Pitch (one full turn) = 3.4 nm; ~10 base pairs per turn; rise per base = 0.34 nm.
- Diameter = 2 nm (constant — purine + pyrimidine width is uniform).
6.2.3 Chargaff's Rules
Chargaff's rules were essential clues:
- Amount of A = Amount of T.
- Amount of G = Amount of C.
- Total purines (A + G) = Total pyrimidines (T + C).
- (A + T) / (G + C) varies between species but is fixed within a species.
🧬 Interactive: Complementary Strand Generator
Enter a DNA sequence (5'→3') and see its complementary strand:
📐 Activity 6.1 — Test Chargaff's Rule
Setup: A scientist analyses DNA from a new bacterium and finds 22% adenine.
Predict: Using Chargaff's rules, calculate the percentage of (a) thymine, (b) guanine, (c) cytosine.
(a) Thymine: Since A = T → T = 22%.
(b) Guanine + Cytosine: Total = 100% − 44% (A+T) = 56%. Since G = C → G = C = 28%. So G = 28%.
(c) Cytosine: C = G = 28%.
Verification: A(22) + T(22) + G(28) + C(28) = 100% ✓; Purines (A+G = 50%) = Pyrimidines (T+C = 50%) ✓.
Worked Examples
Worked Example 1: Chargaff Calculation
A double-stranded DNA molecule contains 1200 base pairs. If guanine constitutes 30% of the bases, find the number of each base.
Total bases: 1200 bp × 2 strands = 2400 bases.
G = 30% = 0.30 × 2400 = 720.
By Chargaff's rule, G = C, so C = 720.
A + T = 100% − (30% + 30%) = 40%; A = T = 20% each.
A = T = 0.20 × 2400 = 480 each.
Total: 720 + 720 + 480 + 480 = 2400 ✓.
G = 30% = 0.30 × 2400 = 720.
By Chargaff's rule, G = C, so C = 720.
A + T = 100% − (30% + 30%) = 40%; A = T = 20% each.
A = T = 0.20 × 2400 = 480 each.
Total: 720 + 720 + 480 + 480 = 2400 ✓.
Worked Example 2: Complementary Sequence
Write the complementary strand and indicate polarity for: 5'-ATGCATTCG-3'.
Original strand (5'→3'): A T G C A T T C G
Complementary base of each: T A C G T A A G C
BUT — the complementary strand is antiparallel (runs 3'→5' opposite the original).
Writing it in conventional 5'→3' orientation: 5'-CGAATGCAT-3' (reverse).
Pairing display:
5'-ATGCATTCG-3'
3'-TACGTAAGC-5'
GC content = 4/9 ≈ 44%.
Complementary base of each: T A C G T A A G C
BUT — the complementary strand is antiparallel (runs 3'→5' opposite the original).
Writing it in conventional 5'→3' orientation: 5'-CGAATGCAT-3' (reverse).
Pairing display:
5'-ATGCATTCG-3'
3'-TACGTAAGC-5'
GC content = 4/9 ≈ 44%.
🎯 Competency-Based Questions
Q1. The transforming principle in Griffith's experiment was identified as DNA by:L1 Remember
Answer: (c). Avery, MacLeod & McCarty (1944) used DNase, RNase, and protease to systematically eliminate components of heat-killed S strain and showed only DNase destruction prevents transformation — proving DNA is the genetic material.
Q2. Fill in the blank: The two strands of DNA are held together by _____ between complementary bases. L2 Understand
Answer: Hydrogen bonds. A pairs with T via 2 H-bonds; G pairs with C via 3 H-bonds. These weak bonds allow strand separation during replication and transcription, while still being strong enough collectively to maintain stability.
Q3. If a DNA strand has 30% adenine, calculate the percentage of guanine. L3 Apply
Answer: By Chargaff's rule, A = T → T = 30%. So A + T = 60%. Therefore G + C = 40%. Since G = C, G = C = 20% each.
Q4. Analyse: Why was the Hershey-Chase experiment a more decisive proof than Avery's? L4 Analyse
- Avery's experiment used enzymes which could be impure — critics argued residual protein contamination might be the real transforming agent.
- Hershey-Chase directly tracked which molecule entered the bacterium using radioactive isotopes — no enzymatic intermediate steps.
- Radioactive labeling provided direct physical evidence: ³²P (DNA) entered the host; ³⁵S (protein) stayed outside.
- The experiment used a virus, not bacteria — broader generality supporting "DNA is genetic material in all cellular life forms".
Q5. HOT (Create): Design an experiment to demonstrate that DNA, not RNA, is the primary genetic material in most cellular organisms. L6 Create
Sample experimental design:
- Hypothesis: DNA, not RNA, is the primary genetic material in cellular organisms.
- Method: Use a yeast cell that expresses a fluorescent reporter gene.
- Treatments:
- Group A: DNase added (degrades DNA only)
- Group B: RNase added (degrades RNA only)
- Group C: No treatment (control)
- Observation: Track gene expression over generations.
- Predicted outcome:
- Group A: Gene expression eventually lost — daughter cells lack DNA template, cannot maintain inheritance.
- Group B: Cells lose immediate function but recover (RNA is regenerated from DNA template).
- Group C: Normal expression continues.
- Conclusion: Permanent disruption only by DNase confirms DNA carries hereditary information; RNA loss is recoverable, indicating its role is intermediary not primary.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: In DNA, the diameter is constant at 2 nm regardless of the base pair sequence.
R: Each base pair always involves one purine and one pyrimidine.
Answer: (A). Both true; R explains A. A purine (2-ring, larger) + a pyrimidine (1-ring, smaller) always sum to the same width. Two purines would be too wide; two pyrimidines too narrow. The strict pairing rule keeps the helix uniform.
A: GC base pairs are more stable than AT base pairs.
R: GC pairs share three hydrogen bonds while AT pairs share two.
Answer: (A). Both true; R explains A. The extra hydrogen bond in G-C makes those pairs more thermally stable. DNA segments rich in GC require higher temperatures to denature.
A: The Hershey-Chase experiment proved that proteins are the genetic material.
R: Bacteriophages contain only DNA and protein.
Answer: (D). A is FALSE — Hershey-Chase proved that DNA (not protein) is the genetic material. R is TRUE — phages are made of only DNA and protein. The experiment showed only ³²P (DNA) entered the bacterial cell during infection.
Frequently Asked Questions - Dna Structure Griffith Experiment
What is the main concept covered in Dna Structure Griffith Experiment?
In NCERT Class 12 Biology Chapter on Molecular Basis of Inheritance, "Dna Structure Griffith Experiment" covers the core biological structures, processes, and pathways students need for board exam success. The MyAiSchool lesson explains the topic with definitions, labelled diagrams, comparison tables, and interactive simulations. Scientific terminology and physiological/genetic significance are highlighted throughout to build conceptual depth aligned with CBSE 2025-26 syllabus.
How is Dna Structure Griffith Experiment useful in real-life or applied biology?
Real-life applications of "Dna Structure Griffith Experiment" from NCERT Class 12 Biology Molecular Basis of Inheritance include medical diagnostics, agriculture, biotechnology, public health, evolutionary insights, and ecological monitoring. The MyAiSchool lesson links every biological concept to a tangible application so students see biology as a problem-solving framework for living systems and real-world challenges.
What are the key terms students should memorize for Dna Structure Griffith Experiment?
Key terms in "Dna Structure Griffith Experiment" (NCERT Class 12 Biology Molecular Basis of Inheritance) are tabulated in the MyAiSchool key-terms grid. Students should memorize each term with its precise definition, function, and example. Terminology is high-yield in CBSE board exams — 1-mark MCQs and 2-mark short answers test definitions directly. The Summary section provides a printable quick-reference card.
How does this part connect to other parts of the chapter?
NCERT Class 12 Biology Molecular Basis of Inheritance is structured so each part builds biological understanding sequentially. "Dna Structure Griffith Experiment" connects to neighbouring parts via shared mechanisms, structural hierarchies, and physiological processes. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected biological story rather than disconnected fragments.
What types of CBSE board questions come from Dna Structure Griffith Experiment?
CBSE board questions from "Dna Structure Griffith Experiment" typically include: (1) 1-mark MCQs on definitions and processes, (2) 2-mark short-answer differences/comparisons, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining mechanism + diagram + significance. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Dna Structure Griffith Experiment" lesson allows students to explore biological processes, classifications, or pathways using selectors and sliders, with live visual feedback. To use it effectively: (1) explore each option/state, (2) compare with textbook diagrams, (3) note the function/outcome changes, (4) try the integrated practice quiz. The simulation reinforces visual-spatial understanding that pure text-based study cannot.
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