Question

The side chain of Asp 94 on the b chain of hemoglobin is in close proximity to the imidazole ring of His 146 on the b chain in the deoxy form of hemoglobin but not the oxy form.

1. What kind of interaction occurs between Asp 94 and His 146 in the deoxy form of hemoglobin?

2. Is the imidazole ring of His protonated or unprotonated? Explain.

3. The proximity of Asp 94 alters the pK value of the imidazole ring of His. In what way?

Answer 1

1) Haemoglobin exists in the T (tense) and R (relaxed) state. The T form is the quaternary structure of deoxyhaemoglobin while the R state is the structure of oxyhaemoglobin. The side chain of the C-terminal His 146 residue of each beta subunit interacts with Asp 94 residue of the same chain to form ionic salt bridges. These amino acids form an intrachain ion pair. Asparate has a carboxylic acid side chain and histidine has an imidazole side chain. The positively charged imidazole group forms an ionic bond with the negatively charged carboxylic acid group. Ionic bonds between these amino acids are broken when Hb transitions from T to R state.

2) Imidazole ring in His 146 is in the protonated form in T state. The affinity of haemoglobin (Hb) for oxygen is decreased when pH of solution decreases. Hence, Hb releases more oxygen when pH is decreased. In deoxyHb, the pH is decreased in the tissue and is much lower than the pH of lungs (due to metabolic activity). At low pH, histidine is protonated and participates in salt bridge formation with Asp94. Protonated form of imidazole side chain in Histidine has a pKa of 6. At a pH below 6, this imidazole is protonated as the pH is higher than the pKa.

3) Asp94 increases the pK of His146. The Asp 94 is an acidic amino acid with a negatively charged carboxylic acid side chain. When Asp94 comes in close proximity to His146, it will reduce the pK of the positively charged imidazole side chain and pKa becomes more than the pH.