Question

Oxygen-Proton Exchange in Hemoglobin 11.0 10.0 As shown in the graph, hemoglobin exchanges oxygen and protons due to an inverse relationship between oxygen binding and proton binding known as the Bohr effect. The binding of oxygen to hemoglobin causes a conformational change that disrupts salt bridges, making oxyhemoglobin more acidic than deoxyhemoglobin. For purposes of calculation, hemoglobin can be modeled as a simple monoprotic buffer, dissociating one proton per subunit as illustrated in the graph and equilibrium equations. Tissues - H+ + HCO3 Hb H₂O <co, HHD Blood pH (7.4) A pH 7.0 pk, -7.8 HHb H+Hb HHBO, HBO H20 H+ + HCO3 CO. Lungs HHOH TTT PR, -6.7 + HbO, You will now calculate the quantity (in millimoles) of protons that will be released when 2.65 mmol of oxygen binds to deoxyhemoglobin at pH 7.4 and the pH then returns to 7.4 (i.e., going from point A to point B on the curve). 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Fraction Unprotonated

Answer 1

(i) According to the Henderson-Hasselbulch equation:

pH = pKa + Log[base/acid]

i.e. 7.4 = 7.8 + Log[Hb/HHb]

i.e. Log[Hb/HHb] = -0.4

i.e. [Hb/HHb] = 0.398

i.e. [HHb/Hb] = 2.512

(ii) (2.65-HHb)/HHb = 0.398, i.e. n_HHb = 2.65/(1+0.398) = 1.895 mmol

(iii) 7.4 = 6.7 + Log[HbO2/HHbO2]

i.e. Log[HbO2/HHbO2] = 0.7

i.e. [HbO2/HHbO2] = 10^0.7 = 5.012

i.e. [HHbO2/HbO2] = 0.200

(iv) (2.65-HHbO2)/HHbO2 = 5.012, i.e. n_HHbO2 = 2.65/(1+5.012) = 0.441 mmol