The first reaction in glycolysis is the phosphorylation of glucose: Pi+glucosegl

Question

The first reaction in glycolysis is the phosphorylation of glucose:

Pi+glucoseglucose6phosphate+H2O

This is a thermodynamically unfavorable process, with G=+13.8kJ/mol.

Part A

In a liver cell at 37 C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would be the equilibrium concentration of glucose-6-phosphate, according to the above data?

Express your answer to two significant figures and include the appropriate units.

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Part B

This very low concentration of the desired product would be unfavorable for glycolysis. In fact, the reaction is coupled to ATP hydrolysis to give the overall reaction

ATP+glucoseglucose6phosphate+ADP+H+

What is G for the coupled reaction?

Express your answer to three significant figures and include the appropriate units.

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Part C

If, in addition to the constraints on glucose concentration listed previously, we have in the liver cell ATPconcentration = 3 mM and ADP concentration = 1 mM, what is the theoretical concentration of glucose-6-phosphate at equilibrium at pH=7.4 and 37 C?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

1. The formula to calculate the equilibrium concentration G = G0 + RTln (Q)

G (Gibb’s energy) = 0 at equilibrium

G0 (free energy change) = 13.8 kJ/ mol

R (Universal Gas constant)= 8.31 J/mol K

T (Temperature in K) = 310 K

Q = [products]/[reactants] = [Glu-6-phosphate]/[glucose][Pi]

Substitute the above values in the equation,

0 = 13.8 + (8.31 J/mol K)(310 K)ln([Glu-6-phos]/[0.005M* 0.005M])

-5.35 = ln([Glu-6-phos]/[0.005M* 0.005M]) (Since, e-5.35 = 0.00474)

0.00474 = [Glu-6-phos]/[0.000025]

Equilibrium concentration of [Glu-6-phos] = 1.185 * 10-7 M

1.

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