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MHT-CET Chemistry · Biomolecules

Nucleic Acids: Nucleotides, Bases and the Double Helix

A nucleotide is a pentose sugar (ribose in RNA, 2-deoxyribose in DNA) carrying a nitrogen base at C-1 prime and a phosphate at C-5 prime; the bases are purines (adenine, guanine — nitrogens at 1, 3, 7, 9, joined through N-9) and pyrimidines (cytosine, thymine, uracil — nitrogens at 1 and 3, joined through N-1), and nucleotides link by 3 prime to 5 prime phosphodiester bonds into strands that pair A with T and G with C in the right-handed Watson–Crick double helix.

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Why this matters

16 PYQs, none HARD. Fourteen are the nucleotide and its bases — which base is a purine, which has the most nitrogens or none of the amino group, the atom counts of cytosine, thymine and uracil, the ring positions of N in purine and pyrimidine, which N bonds the sugar, which sugar carbon lacks oxygen in DNA, how many OH a ribonucleoside has, where the phosphate sits in AMP; two are the Watson–Crick model and the phosphodiester backbone. Two cards.

Concept 1 of 2

Nucleotide Anatomy: Sugar, Base and Phosphate by Position

Intuition

Draw the sugar as a five-membered furanose with its carbons primed, 1 prime to 5 prime. The base goes on C-1 prime; the phosphate on the C-5 prime CH₂OH; C-2 prime has OH in ribose and only H in 2-deoxyribose. Purines are the two-ring bases (adenine, guanine), pyrimidines the one-ring bases (cytosine, thymine in DNA, uracil in RNA). Count nitrogens from the ring positions plus any NH₂: the purine ring has N at 1, 3, 7, 9 — four — so adenine and guanine have five N each; the pyrimidine ring has N at 1 and 3, so cytosine (with NH₂) has three, thymine and uracil two.

Definition

  • Sugar: RNA = D-ribose (5 C; OH at 2′, 3′, 5′), DNA = 2-deoxyribose (C-2′ lacks the OH). A ribonucleoside has 3 OH (2′, 3′, 5′); a deoxyribonucleoside 2.
  • Nucleoside = sugar + base (N-glycosidic bond at C-1′): purine N-9, pyrimidine N-1. Nucleotide = nucleoside + phosphate at C-5′ (AMP = adenosine 5′-monophosphate).
  • Purines (two rings, N at 1, 3, 7, 9): adenine C5H5N5\text{C}_5\text{H}_5\text{N}_5, guanine C5H5N5O\text{C}_5\text{H}_5\text{N}_5\text{O} — five N each, guanine has the most N of the four common bases.
  • Pyrimidines (one ring, N at 1 and 3): cytosine C4H5N3O\text{C}_4\text{H}_5\text{N}_3\text{O} (3 N, 1 O); thymine C5H6N2O2\text{C}_5\text{H}_6\text{N}_2\text{O}_2 (2 N, 2 O); uracil C4H4N2O2\text{C}_4\text{H}_4\text{N}_2\text{O}_2 (2 N, 2 O).
  • Amino group on the ring: adenine, guanine, cytosine have –NH₂; thymine and uracil do not (two C=O instead; thymine adds a CH₃).
  • DNA bases A, G, C, T; RNA bases A, G, C, U.

Ring nitrogens and the glycosidic N

purine: N-1, 3, 7, 9 (sugar at N-9);pyrimidine: N-1, 3 (sugar at N-1);phosphate at C-5′\text{purine: N-1, 3, 7, 9 (sugar at N-9)};\quad \text{pyrimidine: N-1, 3 (sugar at N-1)};\quad \text{phosphate at C-5}'

Worked example

How many moles of N atoms and of O atoms are in one mole of guanine, and how many OH groups does deoxyguanosine carry?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1BiomoleculesMODERATE
What are the positions of ' NN ' atoms present in purine ring of nucleic acids?

[Q68 · 26 April Shift I · 2025]

Purine N at 1, 3, 5

There is no N-5 in a purine — the six-membered ring's nitrogens are 1 and 3, the five-membered ring's 7 and 9. Position 5 is the shared carbon.
Drill 13 more on nucleotide anatomy: sugar, base and phosphate by position

Concept 2 of 2

The Phosphodiester Backbone and the Watson–Crick Double Helix

Intuition

Nucleotides join when the phosphate on C-5 prime of one esterifies the OH on C-3 prime of the next: a 3 prime to 5 prime phosphodiester link, so every strand has a free phosphate at its 5 prime end and a free OH at its 3 prime end, and the backbone is sugar–phosphate–sugar (C–O–P–O–C), not C–O–C. Two such strands wind into a right-handed double helix with the backbone OUTSIDE and the bases INSIDE, held by hydrogen bonds between complementary pairs: A with T by two bonds, G with C by three. RNA is one strand with U in place of T.

Definition

  • Backbone: sugar–phosphate, linked by phosphodiester bonds (C–O–P–O–C, not C–O–C). 5′ end = free phosphate; 3′ end = free OH.
  • Watson–Crick: two antiparallel polynucleotide strands, right-handed double helix; backbone outside, bases inside; stabilised by hydrogen bonds between complementary pairs A=T (2 H-bonds) and G≡C (3 H-bonds).
  • Consequences: equal A and T, equal G and C (Chargaff); one strand fixes the other's sequence — the basis of replication.
  • RNA: single strand, ribose, uracil for thymine; messenger, ribosomal, transfer RNA.
  • The false statement to expect: 'backbone inside, bases outside' or 'backbone of –C–O–C– linkage'.

Base pairing

A=T (2 H-bonds),G≡C (3 H-bonds);5′-phosphate⋯3′-OH\text{A}=\text{T}\ (2\ \text{H-bonds}),\quad \text{G}\equiv\text{C}\ (3\ \text{H-bonds});\quad 5'\text{-phosphate} \cdots 3'\text{-OH}

Worked example

A DNA sample is 22% adenine. Give the percentage of each of the other three bases.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2BiomoleculesMODERATE
Which from following is NOT a salient feature of Watson and Crick model for DNA?

[Q58 · 10th May Shift 1 · 2023]

–C–O–C– as the backbone

An ether link is what a glycosidic bond is; the nucleic-acid backbone runs through PHOSPHORUS — sugar 3′-O–P–O-5′ sugar. The statement 'backbone formed of –C–O–C– linkage' is the false one.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Nucleotide Anatomy: Sugar, Base and Phosphate by Position

    Ring nitrogens and the glycosidic N

    purine: N-1, 3, 7, 9 (sugar at N-9);pyrimidine: N-1, 3 (sugar at N-1);phosphate at C-5′\text{purine: N-1, 3, 7, 9 (sugar at N-9)};\quad \text{pyrimidine: N-1, 3 (sugar at N-1)};\quad \text{phosphate at C-5}'
  • The Phosphodiester Backbone and the Watson–Crick Double Helix

    Base pairing

    A=T (2 H-bonds),G≡C (3 H-bonds);5′-phosphate⋯3′-OH\text{A}=\text{T}\ (2\ \text{H-bonds}),\quad \text{G}\equiv\text{C}\ (3\ \text{H-bonds});\quad 5'\text{-phosphate} \cdots 3'\text{-OH}

Watch out for (2)

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