The enzyme hexokinase catalyzes the phosphorylation of 6 carbon surgars, and is

Question

The enzyme hexokinase catalyzes the phosphorylation of 6 carbon surgars, and is one of the first steps in glycolysis. In this reaction fructose is phosphorylated to form fructose-6-phosphate (F6P), with ATP as the phosphate donor and energy source:

Fructose+ATP-----hexokinase-------> Fructose-6-phosphate+ADP

G^0=-3.5kcal/mol

In a typical bacterial cell the following concentrations are observed:

[ATP]=2.0 mM

[ADP]=0.15 mM

[F6P]=0.03 mM

(a). What is Keq for this reaction at 25 degrees celsius?

(b). Assuming that the concentration of fructose in the bacterial cell is 3.5 mM what is the change in G for this reaction at 25 degrees celsius? Is the reaction spontaneous at these 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 the fructose by inorganic phosphate Pi). Knowing that the change in G intitial for ATP hydrolysis is -7.3 kcal/mol, calculate the change in G initial for the direct phosphorylation of fructose by Pi. Is this reaction spontaneous? Would you expect it to ever take place in a cell?

Explanation / Answer

a) Keq is the equilibrium constant and is given by the ratio of concentration of products and reactants.
Go= -RT ln Keq.

considering that the value given for Go is -3.5kcal/mol…

1kcal/mol=4.184kJ/mol

Go = -3.5kcal/mol=-(3.5 x 4.184)kJ/mol=-14.644kJ/mol= -RT ln Keq;

R is the gas constant; R=8.315 J/mol K
T is the temperature; T=25degree celsius= 25+273.15K=298.15K

therefore, using the formula for Go

Go= -RT ln K’eq

-14.644kJ/mol = -(8.315 J/mol K)(298.15K)(ln Keq)

ln Keq=(14.644 x 1000kJ/mol)/((8.315 J/mol K)(298.15K))=5.9069

Keq=367.565

b) presence of catalysis does not affect the value of Keq.

given that :

[ATP]=2.0 mM

[ADP]=0.15 mM

[F6P]=0.03 mM

[F]=3.5 mM
Keq=([F6P][ADP])/([F][ATP])=(0.03x0.15)/(3.5x2.0)=0.0045/7=6.43x10^-4

Go= -RT ln K’eq
= -(8.315 J/mol K)(298.15K)(ln 0.000643)
= -2479.11725 J/mol x (-7.35)
= 18221.51J/mol=18kJ/mol

Since Go is positive, it is not a spontaneous reaction.

c) ATP + water ----(ATPase)------->ADP + Pi + H+   Go=-7.3 kcal/mol
Fructose + Pi -----(Hexokinase)----> Fructose-6-phosphate+H2O

Go from answer a) is -3.5kcal/mol

hence Go for phosphorylation of fructose = -3.5+7.3kcal/mol = 3.8kcal/mol

Since the Go for phosphorylation of fructose is positive, it is not a spontaneous reaction.
Non spontaneous reactions are catalyzed and coupled with spontaneous reactions. An unfavorable reaction can be coupled together with a favorable one to make the overall reaction favorable.
Conversion of glucose to glucose-6-phosphate is a typical example. Also the reaction given in this question follows the same principle.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.