Question

The correct statement(s) regarding myoglobin (Mb) and haemoglobin (Hb) is(are)

• A. At low partial pressure of O\(_2\) (e.g., 5 kPa), the O\(_2\) affinity of Hb lowers upon lowering the pH
• B. Binding of the first O\(_2\) molecule to Hb results in lower affinity for the binding of second O\(_2\) molecule
• C. Metal center in deoxy-Mb is low-spin whereas it is high-spin in the case of oxy-Mb
• D. One end of O\(_2\) binds to the metal center in oxy-Mb and the other end of the bound O\(_2\) is H-bonded with imidazole-NH of a distal histidine

Hint:

Bohr effect: \(\downarrow\)pH or \(\uparrow\)CO\(_2\) \(\Rightarrow\) \(\downarrow\) Hb–O\(_2\) affinity.
Hb: positive cooperativity; Mb: hyperbolic, non-cooperative binding.
Deoxy Fe(II) is high-spin; O\(_2\) binding leads to low-spin and movement of Fe into the porphyrin plane.

Answer 1

The Correct Option is A, D

Step 1: Bohr effect. Lowering the pH decreases Hb’s affinity for O\(_2\) (right shift of the O\(_2\)-binding curve), especially relevant at low \(p\)O\(_2\). Hence (A) is true.
Step 2: Cooperativity of Hb. Hb displays positive cooperativity: binding of the first O\(_2\) increases affinity for subsequent O\(_2\) molecules. Thus (B) is false.
Step 3: Spin state changes. In deoxy-Mb, Fe(II) is high-spin; upon O\(_2\) binding (oxy-Mb), iron becomes low-spin due to strong-field interaction. Statement (C) reverses this and is false.
Step 4: Distal His H-bond. In oxy-Mb/Hb, bound O\(_2\) is hydrogen-bonded to the distal histidine (His E7) N–H, stabilizing the adduct. Therefore (D) is true.