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

Question

The enzyme hexokinase catalyzes the phosphorylation of 6 carbon sugars, 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 Fructose-6-phosphate + ADP Delta G degree = -3.5kcal/mol In a typical bacterial cell the following concentrations are observed: [ATP] = 2.0 mM [ADPJ = 0.15 mM [F6P] = 0.03 mM What is K_eq for this reaction at 25 degree C? Show all calculations Assuming that the concentration of fructose in the bacterial cell is 3.5 mM what is Delta G for this reaction at 25 degree C? Show all calculations. Is the reaction spontaneous at these concentrations? 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 fructose by inorganic phosphate P_i). Knowing that Delta G degree for ATP hydrolysis is -7.3 kcal/mol, calculate the Delta G degree for the direct phosphorylation of fructose by P_i. Is this reaction spontaneous? Would you expect it to ever take place in a cell?

Explanation / Answer

a)

Keq can be calcualted via

dG° = -RT*lnKeq

Keq = exp(-dG°/(RT))

dG° = 3.5 kcal/mol --> = -3.5*4.184 = -14.644kJ/mol = -14644 J/mol

Keq = exp(--dG°/(RT))

Keq = exp(14644 /(8.314*298))

Keq = 368.9

b)

[Fructose] = 3.5*10^-3 M

find dG for this reaction...

Q = product s/ reactants

Q = [Fructose][ATP]/([Fructo-6-phosph][ADP]

Q = (3.5)(2)/((0.03)(0.15)) = 1555.5555

Apply

dG = dG° + RT¨ln(Q)

so

dG = - 14644 + 8.314*298 * ln(1555.5555)

dG = 3565.13 J/mol = 3.56 kJ/mol

this must NOT be spontaneous, since dG > 1, meaning this will favour reactants

c)

ATP + H2O --> ADP + Pi

Pi + fructose --> Fructose-6-PO4 + H2O

overall:

ATP + H2O + Pi + fructose  --> ADP + Pi+ Fructose-6-PO4 + H2O

cancel common terms

ATP +fructose    --> ADP+Fructose-6-PO4

dG for hydrolysis = -7.3 kcal/mol --> -7.3 *4.184 = -30.5432 kJ/molo

dG for phosphorilation

dG = dG - dG1 = 3.56 -30.5432 = -26.9832kJ/mol

since this iis negative, this should be spontanoeus

In a cell, yopu could expect this to happen, since it is an organized set of pahtways that favour these reactions

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