Nucleic acid structure — DNA double helix, base pairing, Chargaff's rules

Question

Describe the structure of DNA. What are Chargaff’s rules, and how does base pairing work in the Watson-Crick model?

(NEET and CBSE 12 — DNA structure is one of the most asked topics across biology and chemistry)

Solution — Step by Step

Each nucleotide has three components:

Phosphate group
Deoxyribose sugar (pentose sugar lacking -OH at 2’ position)
Nitrogenous base: Purines (Adenine, Guanine — double ring) or Pyrimidines (Cytosine, Thymine — single ring)

The backbone is a sugar-phosphate chain — phosphodiester bonds link the 3’ carbon of one sugar to the 5’ carbon of the next sugar through a phosphate group. The bases extend inward from this backbone.

Erwin Chargaff analysed DNA from many organisms and found:

[A] = [T] \quad \text{and} \quad [G] = [C]

Therefore:

$\dfrac{A + G}{T + C} = 1$ (purines = pyrimidines)
$\dfrac{A + T}{G + C}$ varies between species (this ratio is characteristic of each organism)

These rules provided the key clue for Watson and Crick — adenine pairs with thymine, and guanine pairs with cytosine.

Two antiparallel polynucleotide chains wound around each other in a right-handed double helix
One strand runs 5’ to 3’, the other runs 3’ to 5’
Base pairing: A=T (2 hydrogen bonds), G≡C (3 hydrogen bonds)
The helix has a major groove and a minor groove (where proteins interact)
One complete turn = 3.4 nm (34 angstroms), containing 10 base pairs
Distance between adjacent base pairs = 0.34 nm (3.4 angstroms)
Diameter of the helix = 2 nm (20 angstroms)

graph TD
    A["DNA Structure"] --> B["Backbone: Sugar-Phosphate"]
    A --> C["Base Pairing"]
    A --> D["Double Helix"]
    C --> E["A = T: 2 H-bonds"]
    C --> F["G ≡ C: 3 H-bonds"]
    D --> G["Right-handed helix"]
    D --> H["10 bp per turn"]
    D --> I["3.4 nm per turn"]
    D --> J["2 nm diameter"]
    K["Chargaff's Rules"] --> L["A=T, G=C"]
    K --> M["Purines = Pyrimidines"]

Why This Works

The Watson-Crick model elegantly explains DNA replication. Because A always pairs with T and G with C, each strand serves as a template for the other. During replication, the strands separate, and each strand directs the synthesis of a complementary strand — producing two identical DNA molecules from one.

The G-C base pair has 3 hydrogen bonds (stronger), while A-T has 2 hydrogen bonds (weaker). DNA with a higher G-C content has a higher melting temperature ( $T_m$ ) — it takes more energy to separate the strands. This is a favourite JEE and NEET factoid.

Alternative Method

For NEET numerical problems on Chargaff’s rules: if given one base percentage, you can find all others.

Example: If A = 30%, then T = 30% (Chargaff), so G + C = 100% - 60% = 40%, meaning G = C = 20%.

Also, total H-bonds = $2 \times [A\text{-}T \text{ pairs}] + 3 \times [G\text{-}C \text{ pairs}]$ . If given total base pairs and % composition, you can calculate total H-bonds.

Common Mistake

The most common error: applying Chargaff’s rules to single-stranded DNA or RNA. Chargaff’s rules ( $A = T$ , $G = C$ ) apply only to double-stranded DNA. In single-stranded DNA or RNA, there is no complementary strand, so $A \neq T$ and $G \neq C$ .

Also, students confuse the number of hydrogen bonds between base pairs. A-T has 2 H-bonds, G-C has 3 H-bonds. A common JEE trick: “which base pair is easier to break?” — A-T, because it has fewer hydrogen bonds. DNA denaturation begins at A-T rich regions.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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