|
PHA365 |
||||
| Home
Membership
|
Bioenergetics_problemPosted by Terese Wignot, 3/27/03 at 12:40:07 PM. The phosphorylation of glucose to glucose-6-phosphate is the first step in the catabolism of glucose. The reaction for the direct phosphorylation of glucose by inorganic phosphate is described by the equation:glucose + Pi ---------> glucose-6-phosphate + H2O DeltaGo/ = +3.3kcal/mol a)Calculate the equilibrium constant (Keq) for the above reaction. b)In a rat liver cell, the physiological concentrations of glucose and Pi are maintained at 4.8 x 10-3 M. What is the equilibrium concentration of gluose-6-phosphate? c)Is the direct phosphorylation of glucose by Pi a reasonable route for the catabolism of glucose? Explain.
2.As biochemists, we know that the phosphorylation of glucose does not occur by direct phosphorylation but is coupled to the hydrolysis of ATP:
glucose + Pi ---------> glucose-6-phosphate + H2O DeltaGo/ = +3.3kcal/mol
ATP + H2O -------> ADP + Pi DeltaGo/ = -7.3kcal/mol
a) Calculate DelatGo/ and Keq for the net reaction. b) When the ATP dependent phosphorylation of glucose is carried out, what concentration of glucose is needed to achieve a 2.5 x 10-6 M intracellular concentration of glucose-6-phosphate when the concentrations of ATP and ADP are 3.38 x 10-3M. and 1.32 x 10-3 M, respectively?
|
|||
|
Last update: Thursday, November 6, 2003 at 9:33:57 AM. |
||||
This site is using the Original Default theme.