The Bohr effect describes the rightward shift of the oxyhemoglobin dissociation curve with increasing CO2/H+. At tissue level, CO2 enters RBCs and is hydrated to carbonic acid by carbonic anhydrase. The protons stabilize the T-state (deoxy) conformation through specific salt bridges. Which histidine residue at the alpha-beta1 subunit interface is primarily responsible for the alkaline Bohr effect?
- A His 63 alpha (E7 proximal histidine)
- B His 146 beta (C-terminal histidine of beta chains) ✓
- C His 97 beta
- D His 58 beta (E7 distal histidine)
Explanation
The alkaline Bohr effect (CO2/H+-mediated right shift of the O2 dissociation curve) primarily involves His 146 beta (the penultimate C-terminal histidine of the beta chain). In the T-state (deoxy form), His 146 beta forms a salt bridge with Asp 94 beta after being protonated at its imidazole ring. The pKa of His 146 beta is approximately 0.5 units higher in the T-state than in the R-state, meaning protons (H+) preferentially bind to His 146 beta when Hb transitions to the T-state, stabilizing the deoxy conformation and promoting O2 release at tissues. His 143 beta also contributes. The proximal His (F8, position 87 in alpha and 92 in beta) coordinates iron directly and is not involved in the Bohr effect. This structural understanding is fundamental to interpreting hemoglobin function.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
Written and medically reviewed by the StethoPrep medical team.