Calculate the quantity (in mmol) of protons that will be released when 1.70 mmol

Question

Calculate the quantity (in mmol) of protons that will be released when 1.70 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.

The figure at the right illustrates the inverse relationship between oxygen and proton binding to hemoglobin, known as the Bohr effect. Oxyhemoglobin is more acidic because of salt bridges broken when oxygen binds Oxygen-Proton Exchange in Hemoglobin 11.0 '· Tissues 10.0 H2O For purposes of calculation, hemoglobin can be modeled as a Hb simple monoprotic buffer dissociating one proton per subnt 9.0 as illustrated in the figure and in the following two equilibria HHb Hb 8.0 pH 7.0 6.0 5.0 pKa = 7.8 Blood pH (74) A HbO. pKa = 6.7 Lungs Calculate the quantity (in mmol) of protons that will be released when 1.70 mmol of oxygen bind to deoxyhemoglobin at pH 7.4 4.0 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 Number protons mmol

Explanation / Answer

Calculate the quantity (in mmol) of protons that will be released when 1.70 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.

[HA]/[A-]=10(pK-pH)

[HHb]/[Hb]=10(7.8-7.4)=2.51

fraction of HHb= 2.51/1+2.51=.715

nHHb= .715x 1.70 = 1.2155 mm

2) [HHbO2]/[HbO2]=10(6.7-7.4)=.200

fraction HHbO2= .200/1+.002=.374mm

nHHb= .374 x 1.70 = 0.6358 mm

1.2155 - 0.6358 = 0.5797 mmol

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