1.) The first reaction in glycolysis is the phosphorylation of glucose Pi+glucos

Question

1.) The first reaction in glycolysis is the phosphorylation of glucose

Pi+glucose <>glucose-6-phosphate delta G knot prime=+14 kj/mol

A.) In a liver cell at 37 degrees C the concentration of both phosphate and glucose are normally maintained at about 5 mM each. What would the equilibrium concentration of glucose-6-p?

B.) this very low concentration of product would bevery unfavorable for glycolysis.in fact the reaction is coupled to ATP hydrolysis.

ATP>ADP + Pi DELTA G KNOT prime = -30.5 KJ/MOL

to give the overall reaction:

glucose + ATP > g-6-p+ ADP

assuming the free energy of the two reactions are additive, what is the delta G prime of the reaction now?

C.) in addition to the constraintsof glucose concentration, liver cell ATP =3mM and ADP =1mM. Taking this into account what is the equilibrium concentration of g-6-p

Explanation / Answer

1A) K = [G-6-P] /[P] [Glu]  

we have dGo = -RT ln K

14x1000 = -8.314 x 310 x ln K

K = 0.0043745 = [G-6-P] / ( 5x10^-3) ( 5x10^-3)

[G-6-P] = 1.09 x 10^-7 M

B) overall reaction dGo = -30.5 + 14 = -16.5 KJ

C) dGo = - RT ln K

-16.5 x 1000 = -8.314 x 310ln K

K = 603 = [G-6-P] ( 3x10^-3) / ( 1x10^-3) x ( 5x 10^-3)

[G-6-P] = 1 M

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