The enzyme hexokinase catalyzes the first reaction in glycolysis. In this reacti

Question

The enzyme hexokinase catalyzes the first reaction in glycolysis. In this reaction, glucose is phosphorylated to form glucose-6-phosphate (G6P), with ATP as the phosphate donor and energy source:

Glucose + ATP----Hexokinase---> Glucose-6-phosphate +ATP

Change in G'= -4kcal/mol

In a typical bacterial cell the following concentrations are observed:

[ATP]= 2.0 mM

[ADP]= 0.15 mM

[G6P]= 0.05 mM

a) What is the Keq for this reaction at 25C?

b) Assuming that the concentration of glucose in the bacterial cell is 5.0 mM what is the change in G for this reaction at 25C? Is the reaction spontaneous at the concentrations?

c) The above reaction could be considered a coupled reaction. Rewrite it so that it's broken into two separate reactions that were coupled (hydrolysis of ATP by water, and phosphorylation of glucose by inorganic phosphate Pi). Knowing that the change in G' for ATP hydrolysis is -7.3kcal/mol, calculate the change in G' for the direct phosphorylation of glucose by Pi. Is this reaction spontanteous? Would you expect it to ever take place in a cell

Explanation / Answer

Glucose + ATP----Hexokinase---> Glucose-6-phosphate +ATP
Change in G'= -4kcal/mol

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