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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper I

Q8(a). T and R forms of haemoglobin

  • 10 marks
  • Answered
Question

Differentiate between T and R conformation of hemoglobin.

Answer in brief

T (tense) is the low-affinity form of deoxyhaemoglobin: high-spin Fe²⁺ lies out of the porphyrin plane and salt bridges lock the four subunits. R (relaxed) is the high-affinity form of oxyhaemoglobin: low-spin iron sits in the plane, the salt bridges are broken and the subunits have turned.

Key points

  • In T the two e_g electrons of high-spin Fe²⁺ point at the ring nitrogens. The ion is too large for the hole of the porphyrin and sits below the plane.
  • O₂, a strong-field π-acceptor, binds at the sixth site. Iron becomes low spin, the e_g orbitals empty, the ion shrinks and moves into the plane.
  • The iron pulls His F8 and helix F with it. The shift reaches the contacts between the subunits and breaks the salt bridges.
  • Once one or two haems have bound O₂, the whole tetramer turns to R, and the remaining haems bind O₂ much more readily. Binding is cooperative and the curve is a sigmoid (Hill coefficient about 2.8).

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

FeatureT (tense)R (relaxed)
Found indeoxyhaemoglobinoxyhaemoglobin
Fe(II)high spin, t_(2g)⁴e_g², paramagnetic; largerlow spin, t_(2g)⁶, diamagnetic; smaller
Coordinationfive: four ring N and His F8 (square pyramid)six: O₂ added end-on, bent (octahedral)
Position of Feabout 60 pm out of the ring plane, towards His F8; ring domedin the ring plane; ring flat
Subunit contactssalt bridges between the subunits intact; 2,3-BPG held between the β chainssalt bridges broken; one αβ pair turned by about 15° against the other; BPG released
Affinity for O₂lowhigh
Favoured byH⁺, CO₂, BPG, low p(O₂): tissueshigh p(O₂), higher pH: lungs

How binding of O₂ switches T to R

  1. In T the two e_g electrons of high-spin Fe²⁺ point at the ring nitrogens. The ion is too large for the hole of the porphyrin and sits below the plane.
  2. O₂, a strong-field π-acceptor, binds at the sixth site. Iron becomes low spin, the e_g orbitals empty, the ion shrinks and moves into the plane.
  3. The iron pulls His F8 and helix F with it. The shift reaches the contacts between the subunits and breaks the salt bridges.
  4. Once one or two haems have bound O₂, the whole tetramer turns to R, and the remaining haems bind O₂ much more readily. Binding is cooperative and the curve is a sigmoid (Hill coefficient about 2.8).
  5. Myoglobin, one haem with no T/R switch, gives a hyperbola instead.

What the switch achieves: near 13 kPa (lungs) haemoglobin is nearly saturated; near 5 kPa (working muscle) it drops back towards T and gives up a large share of its O₂. A single-state carrier would either load poorly or unload poorly.

Application. H⁺ and CO₂ strengthen the salt bridges of T, so acidic, working tissue receives more oxygen (Bohr effect); CO also binds the sixth site and keeps the protein in R, which blocks unloading in CO poisoning.

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Model answer to Q8(a), 2026 Paper I: solution page 49
Solution page 49 of 55
Read the whole 2026 solution2026 question paper

All parts of Question 8