The binding of oxygen by hemoglobin (Hb), giving oxy-hemoglobin (HbO 2 ), is partially regulated by the concentration of H 3 O+ and dissolved CO 2 in the blood. Although the equilibrium is complicated, it can be summarized as HbO 2 ( a q ) + H 3 O + ( a q ) + CO 2 ( g ) ⇌ CO 2 − Hb − H + + O 2 ( g ) + H 2 O ( l ) (a) Write the equilibrium constant expression for this reaction. (b) Explain why the production of lactic acid and CO 2 in a muscle during exertion stimulates release of O 2 from the oxy-hemoglobin in the blood passing through the muscle.
The binding of oxygen by hemoglobin (Hb), giving oxy-hemoglobin (HbO 2 ), is partially regulated by the concentration of H 3 O+ and dissolved CO 2 in the blood. Although the equilibrium is complicated, it can be summarized as HbO 2 ( a q ) + H 3 O + ( a q ) + CO 2 ( g ) ⇌ CO 2 − Hb − H + + O 2 ( g ) + H 2 O ( l ) (a) Write the equilibrium constant expression for this reaction. (b) Explain why the production of lactic acid and CO 2 in a muscle during exertion stimulates release of O 2 from the oxy-hemoglobin in the blood passing through the muscle.
The binding of oxygen by hemoglobin (Hb), giving oxy-hemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as
HbO
2
(
a
q
)
+
H
3
O
+
(
a
q
)
+
CO
2
(
g
)
⇌
CO
2
−
Hb
−
H
+
+
O
2
(
g
)
+
H
2
O
(
l
)
(a) Write the equilibrium constant expression for this reaction.
(b) Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxy-hemoglobin in the blood passing through the muscle.
need help please and thanks dont understand only need help with C-F
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal…
need help please and thanks dont understand only need help with C-F
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal…
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