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Solving Chemical Equilibrium Problems

Solving Chemical Equilibrium Problems

1.) The catabolism (break-down) of glucose is an important source of energy for all cells. It begins with the following transformation which is the first step of the glycolytic pathway:

Glucose --> Glucose-6-Phosphate

Theoretically, the cell could phosphorylate glucose directly with inorganic phosphate (Pi) like so:

a) Calculate the equilibrium constant for this reaction (include correct units).

b) Is it favorable under standard conditions? Why or why not?

No. This is not a favorable reaction under standard conditions because , meaning that energy must be expended to accomplish the reaction.

c) In a typical cell, glucose and phosphate are maintained at 4.8 mM each. (1 mM = 10-3 M) What would be the equilibrium concentration of Glucose-6-Phosphate if the cells used the reaction as written above to make it?

d) Does this direct phosphorylation of glucose represent a reasonable route for the catabolism of glucose? Explain briefly.

No. This concentration is vanishingly small. Even if the reaction were forced to proceed by draining the product with subsequent reactions, the cell could not make practical use of such an unfavorable reaction.

e) The cell actually accomplishes the phosphorylation of glucose by coupling it to the hydrolysis of ATP in a reaction catalyzed by the enzyme hexokinase:

Write a balanced equation for the coupled reactions showing the for the net reaction.

f) Calculate the Keq for the net reaction.

g) The concentration of Glucose-6-Phosphate typically found in cells is 250 micromolar. If the [ATP] were 3.38 mM and the [ADP] were 1.32 mM, what concentration of glucose would be necessary to yield the observed Glucose-6-Phosphate concentration if the reactions are coupled?

h) In addition to being phosphorylated by hexokinase, glucose may also be phosphorylated by an enzyme called glucokinase. Glucokinase has a much higher KM for glucose than does hexokinase. Given that the KM's for glucose are 10.0 mM (glucokinase) and 0.10 mM (hexokinase) and given that glucokinase and hexokinase are present in equal concentrations, what would be the molar ratio of glucose-hexokinase complexes to glucose-glucokinase complexes in a cell when the concentration of glucose is 4.8 mM?

G = Glucose

G-GK = Glucose-glucokinase complex

G-HK = Glucose-hexokinase complex

According to the Michaelis-Menten model of enzyme kinetics, the fraction of enzyme involved in ES complexes is:

The problem asks for the ratio of [G-HK] : [G-GK] which is :

Glucose-hexokinase complexes outnumber glucose-glucokinase complexes by 3 : 1 because hexokinase binds more tightly to glucose than does glucokinase.

This is the end of the section on chemical equilibria.

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