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. pK₂=7.8 HHb — H+ + Hb pK₂=6.7 HHbO₂ H+ + HbO₂ You will now calculate the quantity (in millimoles) of protons that will be released when 1.55 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).
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. pK₂=7.8 HHb — H+ + Hb pK₂=6.7 HHbO₂ H+ + HbO₂ You will now calculate the quantity (in millimoles) of protons that will be released when 1.55 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).
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Transcribed Image Text: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.
pK₁=7.8
HHb
H+ + Hb
pK₁=6.7
HHbO₂
H+ + HbO₂
You will now calculate the quantity (in millimoles) of protons
that will be released when 1.55 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).
11.0
10.0
9.0
8.0
pH 7.0
6.0
5.0
4.0
Oxygen-Proton Exchange in Hemoglobin
Blood pH (7.4) A
HHb
HHbO₂
Tissues
H+ + HCO3
Hb
HbO₂
H* + HCO3
Lungs
H₂O
CO₂
H₂O
B
CO₂
Hb
HbO₂
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Fraction Unprotonated
Transcribed Image Text:Step 1: Calculate the ratio of [HHb] to [Hb] at pH 7.4.
Step 2: Calculate the millimoles of HHb in 1.55 mmol
deoxyhemoglobin at pH 7.4.
Step 3: Calculate the ratio of [HHbO₂] to [HbO₂] at pH 7.4.
Step 4: Calculate the millimoles of HHbO₂ in 1.55 mmol of
oxyhemoglobin at pH 7.4.
Step 5: Calculate the millimoles of protons that will be
released when 1.55 mmol of oxygen binds to
deoxyhemoglobin at pH 7.4.
[HHb]
[Hb]
иннь =
[HHbO₂]
[HbO₂]
иHHbO₂ =
nprotons =
2.51
2.0002
0.2
0.4648
1.54
mmol
mmol
mmol
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![Step 1: Calculate the ratio of [HHb] to [Hb] at pH 7.4.
Step 2: Calculate the millimoles of HHb in 1.55 mmol
deoxyhemoglobin at pH 7.4.
Step 3: Calculate the ratio of [HHbO₂] to [HbO₂] at pH 7.4.
Step 4: Calculate the millimoles of HHbO₂ in 1.55 mmol of
oxyhemoglobin at pH 7.4.
Step 5: Calculate the millimoles of protons that will be
released when 1.55 mmol of oxygen binds to
deoxyhemoglobin at pH 7.4.
[HHb]
[Hb]
иннь =
[HHbO₂]
[HbO₂]
иHHbO₂ =
nprotons =
2.51
2.0002
0.2
0.4648
1.54
mmol
mmol
mmol](https://content.bartleby.com/qna-images/question/25d5c273-ce26-424b-bd36-a8a6b930ab8d/22374f50-40f8-4aa1-8221-4c97f43e0bb1/o7tprgu_processed.png)
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