Map The figure at the right illustrates the inverse relationship oxygen-Proton E

Question

Map The figure at the right illustrates the inverse relationship oxygen-Proton Exchange in Hemoglobin between oxygen and proton binding to hemoglobin, as Bohr effect. Oxyhemoglobin more because of salt bridges broken when oxygen binds 0.0 For purposes of calculation, hemoglobin can be modeled as a monoprotic buffer dissociating one proton per subunit 9.0 as illustrated in the figure and in the following two equilibria: pKa 7.8 HHb Hb pH 7.0 pKa 6.7 HHbO H Hbo Calculate the quantity (in mmol) of protons that will be released when 2.15 mmol of oxygen bind to deoxyhemoglobin at pH 7.4 and the pH then returns to 7.4, i.e. in 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 STRATEGY: Step 1: Calculate the [HHbVIHbl ratio at pH 7,4 Step 2: Calculate the mmoi of HHb in 2.15 mmol of deoxyhemoglobin at pH 7.4 Step 3: Calculate the [HHbo2l/Hbool ratio at pH 7.4. Step 4: Calculate the mmol of HHbo2 in 2.15 mmol of oxyhemoglobin at pH 7.4 Step 5: Calculate the protons released as the difference between nHHb and nHHbo2. Step 1: Calculate the IHHblVIHb ratio at pH 7.4. HA Expressing the Henderson-Hasselbalch equation in the form: HHb 7.8-7.4 2.5 10' Hb Step 2: Calculate the mmol of HHb in 2.15 mmol of yhe globin at pH 7.4. Number HHB" 66 mmol

Explanation / Answer

pKa 7.8 = -log Ka

==> Ka = 10-7.8 = 1.58 * 10-8

HHb ----------> H+ + Hb

2.15 - - I

-x +x +x C

(2.15-x) +x +x E

Ka = 1.58 * 10-8  = x2

   (2.15-x)

Igonoring x in the denominator, after solving x = 1.846 * 10-4 mM

so at equilibrium HHb is (2.15 - x) = (2.15 - 1.58 * 10-8 ) = 2.1498 (answer)

  

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