BiologyISC Class 12 · Molecular Basis of Inheritance

Structure of DNA and Chargaff's Rules: ISC Class 12 Biology

Updated 15 October 2026 · Fact-checked

DNA is a double helix of two antiparallel polynucleotide chains with bases paired inside: A with T by two hydrogen bonds, G with C by three. Chargaff's rules state that in double-stranded DNA A = T and G = C, so purines equal pyrimidines. The helix has a 3.4 nm pitch, about 10 bp per turn.

Understand Structure of DNA and Chargaff's Rules

Building blocks. A nucleotide = nitrogenous base + pentose sugar + phosphate.

Purines: adenine (A), guanine (G). Pyrimidines: cytosine (C), thymine (T, in DNA), uracil (U, in RNA; thymine is 5-methyl uracil).
Sugar: deoxyribose in DNA, ribose in RNA (RNA has an extra 2'-OH).
Base + sugar (joined at the 1' C by an N-glycosidic linkage) = nucleoside (e.g. adenosine, deoxyadenosine). Phosphate joined to the 5'-OH of the sugar (phosphoester linkage) gives a nucleotide.
Nucleotides join by 3'-5' phosphodiester linkages. A chain has a free phosphate at the 5' end and a free 3'-OH at the 3' end. The backbone is sugar-phosphate; bases project from it.

History (ISC lists these contributors).

Friedrich Miescher (1869): first identified DNA as an acidic substance in the nucleus and named it 'nuclein'.
Erwin Chargaff: in double-stranded DNA the ratios A : T and G : C are constant and equal to one.
Maurice Wilkins and Rosalind Franklin: X-ray diffraction data of DNA.
James Watson and Francis Crick (1953): the double helix model, based on the X-ray data and Chargaff's observations.

Salient features of the double helix.
1. Two polynucleotide chains; sugar-phosphate backbone outside, bases inside.
2. Antiparallel polarity: one chain 5' → 3', the other 3' → 5'.
3. Bases pair by hydrogen bonds: A = T (two H-bonds), G ≡ C (three H-bonds). A purine always faces a pyrimidine, so the distance between the strands is approximately uniform.
4. Right-handed coil; pitch 3.4 nm; about 10 bp per turn; so about 0.34 nm between adjacent base pairs.
5. The planes of base pairs stack over one another; stacking plus H-bonds stabilise the helix.
6. The strands are complementary: knowing one sequence gives the other, and each can act as a template, which immediately suggested a copying mechanism (semi-conservative replication).

Chargaff's rules (double-stranded DNA).

A = T and G = C.
So A + G = T + C (purines = pyrimidines) and A + C = G + T.
The ratio (A + T)/(G + C) is not fixed; it varies from species to species, and is characteristic of a species.
Given any one base percentage you can find all four. Given G and T you can find A (= T) and C (= G).
Not applicable to RNA or to single-stranded DNA (such as bacteriophage φX174), because bases are not paired across two strands; RNA also has U instead of T.

Why GC-rich DNA is more heat-stable. Each G≡C pair has three H-bonds against two for A=T, so more energy (higher temperature) is needed to separate GC-rich strands. Organisms of very hot habitats tend to have GC-rich DNA, which resists denaturation.

Size and length. Length of a double helix = number of bp × 0.34 nm. E. coli has 4.6 × 10⁶ bp; human haploid DNA 3.3 × 10⁹ bp; diploid 6.6 × 10⁹ bp, about 2.2 m long, far longer than a nucleus (about 10⁻⁶ m), hence packaging into nucleosomes.

DNA versus RNA.

	DNA	RNA
Sugar	deoxyribose	ribose (2'-OH)
Pyrimidines	C, T	C, U
Strands	usually double	usually single
Stability	more stable	reactive, labile; can be catalytic
Role	stores genetic information	transmits and expresses it

Types of RNA (ISC). mRNA: carries the code (template for translation). tRNA: adapter; reads the codon with its anticodon and brings a specific amino acid. rRNA: structural and catalytic role in ribosomes (23S rRNA in bacteria acts as a ribozyme). hnRNA (heterogeneous nuclear RNA): the primary transcript in eukaryotes, containing exons and introns, processed into mRNA. snRNA (small nuclear RNA): helps in splicing.

Draw and label: part of a DNA double helix (ladder form). Checklist: two vertical backbones of alternating sugar (S) and phosphate (P); 5' at the top of the left strand and 3' at the top of the right (antiparallel arrows); rungs A=T (two lines) and G≡C (three lines); labels: phosphodiester bond, N-glycosidic bond, hydrogen bonds, purine, pyrimidine, 0.34 nm between base pairs, 3.4 nm pitch on a helix sketch.

Structure of DNA and Chargaff's Rules: key facts

Chargaff's rules: $A = T, G = C, A + G = T + C = 50%$; Double-stranded DNA only.
Finding all bases from one: $% A = % T = \frac{100 - 2 ( % G )}{2}$; Similarly %G = %C = (100 − 2 %A)/2.
Hydrogen bonds in a segment: $H-bonds = 2 \times (A-T pairs) + 3 \times (G-C pairs)$; Number of A-T pairs = number of A in the duplex.
Length and turns: $length = bp \times 0.34 nm; turns = \frac{bp}{10} = \frac{length}{3.4 nm}$; 1 nm = 10⁻⁹ m.
Species-specific ratio: $\frac{A + T}{G + C} \neq = constant across species$; GC fraction = 1/(1 + ratio).

How to answer Structure of DNA and Chargaff's Rules questions

ISC asks base-composition sums, features of the Watson-Crick model, reasons (GC stability, why Chargaff fails for RNA) and DNA-RNA differences.

1For base percentages, write A = T and G = C, then A + G = 50%; one value gives the rest.
2For a segment in bp, convert to nucleotides (× 2) before using a percentage of 'bases'.
3For H-bonds, count A-T pairs and G-C pairs separately; multiply by 2 and 3.
4For length, multiply bp by 0.34 nm and convert units carefully (nm → µm → mm).
5For reasoning about heat stability, link GC content to three hydrogen bonds per pair.
6For RNA, say Chargaff's rules do not apply because RNA is single-stranded (bases unpaired) and has uracil.

Structure of DNA and Chargaff's Rules in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

GC content and life in hot springs

Suppose a student compares DNA from a bacterium living in a hot spring at about 80 °C with DNA from a soil bacterium. Analysis shows the hot-spring DNA has 64% G + C and the soil DNA has 42% G + C.

Find the percentage of each base in the hot-spring DNA, and explain how its base composition suits its habitat.

Try it first, then show the solutionSolution

G + C = 64%, and G = C, so G = 32% and C = 32%.
A + T = 36%, and A = T, so A = 18% and T = 18%.
Each G-C pair is held by three hydrogen bonds and each A-T pair by two, so the hot-spring DNA has many more H-bonds per unit length.
More energy (higher temperature) is needed to separate its strands, so it stays double-stranded and functional at temperatures that would denature AT-rich DNA.

Answer: G = C = 32%, A = T = 18%; the high GC content gives more hydrogen bonds and resists heat denaturation.

Memory hook: Three bonds beat two in the heat.

Tricks to remember Structure of DNA and Chargaff's Rules

Short memory aids. Say each one aloud once, then use it on a question the same day.

Pure As Gold
PURines are A and G ('Pure As Gold'); the rest (C, T, U) are pyrimidines.
Use it when: Classifying bases.
AT 2, GC 3
A-T has two letters with straight lines: 2 bonds; G-C 'curls' three times: 3 bonds.
Use it when: Counting hydrogen bonds.
3.4, 10, 0.34
One turn is 3.4 nm and holds 10 bp, so each bp is 0.34 nm.
Use it when: Length and turn problems.

Structure of DNA and Chargaff's Rules: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: One base gives four · 2 marks · Short answerBoard level

A double-stranded DNA sample contains 22% guanine. Find the percentages of cytosine, adenine and thymine.

Try it first, then show the solutionSolution

G = C, so C = 22% (½ mark).
A + T = 100 − 44 = 56%; A = T, so A = 28% and T = 28% (1½ marks).

Answer: C 22%, A 28%, T 28%.

Example 2: Features of the Watson-Crick model · 2 marks · Short answerBoard level

State four salient features of the Watson and Crick model of DNA.

Try it first, then show the solutionSolution

Two polynucleotide chains with sugar-phosphate backbones outside and bases inside; antiparallel polarity (1 mark).
Complementary base pairing, A=T by two and G≡C by three hydrogen bonds; right-handed helix with 3.4 nm pitch and about 10 bp per turn (1 mark).

Answer: Two antiparallel chains, backbone outside, complementary base pairing, right-handed helix of pitch 3.4 nm with 10 bp per turn.

Example 3: Counting hydrogen bonds · 3 marks · Short answerHarder board level

A 500 bp segment of DNA contains 30% adenine. Calculate (a) the number of each base, (b) the number of hydrogen bonds holding the strands together, (c) the length of the segment and the number of turns.

Try it first, then show the solutionSolution

(a) 500 bp = 1000 nucleotides. A = 30% of 1000 = 300; T = 300; G + C = 400, so G = C = 200 (1 mark).
(b) A-T pairs = 300 (2 H-bonds each) and G-C pairs = 200 (3 each): 600 + 600 = 1200 H-bonds (1 mark).
(c) Length = 500 × 0.34 nm = 170 nm; turns = 500 ÷ 10 = 50 (1 mark).

Answer: (a) A 300, T 300, G 200, C 200. (b) 1200. (c) 170 nm, 50 turns.

Example 4: Why Chargaff fails for RNA · 2 marks · Short answerHarder board level

A double-stranded DNA sample contains 1800 thymine and 1200 guanine bases. How many adenine and cytosine bases does it contain, and how many nucleotides in all? Why could the same reasoning not be used to find A and C from the G and U content of an RNA sample? (A question of this kind was asked in 2025.)

Try it first, then show the solutionSolution

By Chargaff's rules, A = T and C = G in double-stranded DNA, so A = 1800 and C = 1200; total = 2 × (1800 + 1200) = 6000 nucleotides (1 mark).
RNA is usually single-stranded, so its bases are not paired one-to-one; it also has uracil in place of thymine. A need not equal U, nor G equal C (1 mark).

Answer: A = 1800, C = 1200 (6000 nucleotides); RNA is single-stranded (with U), so no such equality holds.

Example 5: Antiparallel complement · 3 marks · Short answerHarder board level

One strand of a DNA segment reads 5'-ATGCCGTAAC-3'. Write the complementary strand with its polarity, and state how many hydrogen bonds join the two strands.

Try it first, then show the solutionSolution

Pair each base: A-T, T-A, G-C, C-G; the partner strand runs the opposite way: 3'-TACGGCATTG-5' (1 mark).
Written 5' to 3' it is 5'-GTTACGGCAT-3' (1 mark).
The given strand has A 3, T 2, G 2, C 3: A-T pairs = 5, G-C pairs = 5; H-bonds = 5 × 2 + 5 × 3 = 25 (1 mark).

Answer: 3'-TACGGCATTG-5' (5'-GTTACGGCAT-3'); 25 hydrogen bonds.

Example 6: Which DNA melts first? · 3 marks · Short answerHOTS / competency

Suppose DNA sample X has (A + T)/(G + C) = 1.5 and sample Y has (A + T)/(G + C) = 0.67. (a) Find the GC percentage of each. (b) Which sample needs a higher temperature to separate its strands, and why? (c) Can these ratios show whether X and Y come from the same species? Explain.

Try it first, then show the solutionSolution

(a) GC% = 100/(1 + ratio): X = 100/2.5 = 40%; Y = 100/1.67 ≈ 60% (1 mark).
(b) Y: more G-C pairs, each with three hydrogen bonds, so more energy is needed to separate the strands (1 mark).
(c) The ratio (A + T)/(G + C) is characteristic of a species; such different values suggest different species (though equal values alone would not prove the same species) (1 mark).

Answer: (a) X 40%, Y about 60%. (b) Y, more three-bond G-C pairs. (c) Different ratios indicate different species.

Where students lose marks in Structure of DNA and Chargaff's Rules

The slips examiners see most often, and what to write instead.

Mistake: Crediting heat-resistant proteins for thermostable DNA
In 2025 examiners noted that candidates repeated the question or credited heat-resistant proteins or temperature gradients, rather than the thermal stability of GC pairs.
Do this: Write: a G-C pair has three hydrogen bonds (A-T two), so GC-rich DNA needs more heat to denature.
Usually costs: 1 to 2 marks.
Mistake: Applying Chargaff's rules to RNA
Examiners (2025) found candidates overlooking that RNA is single-stranded and has uracil.
Do this: Chargaff's equalities hold only for double-stranded DNA.
Usually costs: 1 mark.
Mistake: Forgetting to double bp when a percentage of bases is given
Students apply 30% to 500 instead of to 1000 nucleotides.
Wrong working
A = 30% of 500 = 150.
Right working
500 bp = 1000 nucleotides; A = 30% of 1000 = 300.
Do this: Convert bp to nucleotides first.
Usually costs: Every later step.
Mistake: Writing the complementary strand with the same polarity
Students forget the strands are antiparallel.
Do this: Mark 5' and 3' ends on both strands; they run in opposite directions.
Usually costs: ½ to 1 mark.

Exam tips

Learn the five contributors and their years: Miescher 1869; Watson and Crick 1953; Wilkins and Franklin (X-ray data); Chargaff (base ratios).
Quote the numbers: 3.4 nm pitch, 10 bp per turn, 0.34 nm per bp, 2 and 3 H-bonds.
Show working for base sums: rule, substitution, answer.
Name both bonds correctly: N-glycosidic (base-sugar), phosphodiester (between nucleotides).
In diagrams, mark antiparallel polarity with arrows and 5'/3' labels.

Practice questions on Structure of DNA and Chargaff's Rules

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of hydrogen bonds between guanine and cytosine in DNA is:
(a) 1 (b) 2 (c) 3 (d) 4
Hint
More than between A and T.
Answer
(c) 3.
Q2 · 1 mark · Very short answerBoard level
Name the scientist who first isolated DNA and what he called it.
Hint
1869.
Answer
Friedrich Miescher (1869); he called it 'nuclein'.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The distance between the two strands of DNA is almost uniform.
Reason (R): A purine always pairs with a pyrimidine.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Large base + small base.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
The genome of bacteriophage lambda has 48 502 bp. Calculate the length of its DNA in micrometres.
Hint
0.34 nm per bp.
Answer
48 502 × 0.34 nm = 16 490.68 nm ≈ 16.5 µm.
Q5 · 2 marks · Short answerHarder board level
Give four differences between DNA and RNA.
Hint
Sugar, base, strands, stability.
Answer
DNA has deoxyribose, RNA ribose (with 2'-OH). DNA has thymine, RNA uracil. DNA is usually double-stranded, RNA single-stranded. DNA is more stable and stores genetic information; RNA is reactive, can be catalytic, and transmits and expresses information.
Q6 · 2 marks · Short answerHarder board level
Name the RNA that (a) acts as an adapter during translation, (b) is the primary transcript in eukaryotes, (c) helps in splicing, (d) has a catalytic role in the ribosome.
Hint
Five types are in the ISC list.
Answer
(a) tRNA (b) hnRNA (c) snRNA (d) rRNA.
Q7 · 3 marks · Short answerHOTS / competency
Suppose analysis of a viral nucleic acid gives A 25%, T 33%, G 24%, C 18%. (a) Is it DNA or RNA? (b) Is it single- or double-stranded? (c) Justify both answers.
Hint
Look for T or U, then for A = T.
Answer
(a) DNA: it contains thymine, not uracil. (b) Single-stranded. (c) A ≠ T and G ≠ C, so the bases are not paired as Chargaff's rules require for double-stranded DNA (like bacteriophage φX174).
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a DNA segment of 34 nm is found to have 40% G + C.
(a) How many base pairs and how many helical turns does it have? (1)
(b) How many A-T and G-C pairs does it contain? (1)
(c) How many hydrogen bonds hold its strands together? (1)
Hint
0.34 nm per bp; 3.4 nm per turn.
Answer
(a) 34 ÷ 0.34 = 100 bp; 34 ÷ 3.4 = 10 turns. (b) G-C pairs = 40; A-T pairs = 60. (c) 60 × 2 + 40 × 3 = 120 + 120 = 240 hydrogen bonds.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the pitch of the DNA double helix?Show answer
3.4 nm.
2.Which bond joins a base to the sugar?Show answer
N-glycosidic linkage.
3.Who produced the X-ray diffraction data of DNA?Show answer
Maurice Wilkins and Rosalind Franklin.
4.What is the other chemical name of thymine?Show answer
5-methyl uracil.
5.Which ratio varies between species?Show answer
(A + T)/(G + C).
6.How long is the DNA of a human diploid cell?Show answer
About 2.2 metres (6.6 × 10⁹ bp × 0.34 nm).

Structure of DNA and Chargaff's Rules: frequently asked questions

What are Chargaff's rules?

In double-stranded DNA, A = T and G = C, so the total purines equal the total pyrimidines; the (A + T)/(G + C) ratio varies between species.

Why is DNA with more GC pairs more stable?

Each GC pair is held by three hydrogen bonds compared with two for AT, so more energy is needed to separate the strands.

What does antiparallel mean in DNA?

The two strands run in opposite directions: one 5' to 3', the other 3' to 5'.

Why do Chargaff's rules not apply to RNA?

RNA is usually single-stranded, so its bases are not paired one-to-one, and it has uracil instead of thymine.

What are the main types of RNA?

mRNA (carries the code), tRNA (adapter), rRNA (ribosomes, catalytic), hnRNA (primary transcript) and snRNA (splicing).