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A highly regulated reaction in glycolysis involves the addition of a second phosphate to fructose-6-phosphate, which...

A highly regulated reaction in glycolysis involves the addition of a second phosphate to fructose-6-phosphate, which could be be written as:
Fructose-6-phostphate + Pi à Fructose-1,6-bisphosphate + H2O
The ∆Go’ for this reaction is +16.8 kJ/mol   
In a muscle cell at 37 oC, assume the concentration of Fructose-6phosphate is 0.014 mM and the concentration of phosphate is 1 mM.
a. What would be the equilibrium concentration of fructose-1,6bisphosphate under these conditions?
b. What is Keq for this reaction?

Show all work for numerical calculations and units where appropriate. 1. A highly regulated reaction in glycolysis involves t

Show all work for numerical calculations and units where appropriate. 1. A highly regulated reaction in glycolysis involves the addition of a second phosphate to fructose-6-phosphate, which could be be written as: Fructose-6-phostphate + P; → Fructose-1,6-bisphosphate + H2O 0 6014 Imm The AG for this reaction is +16.8 kJ/mol In a muscle cell at 37 °C, assume the concentration of Fructose-6- phosphate is 0.014 mM and the concentration of phosphate is 1 mM. Ma. What would be the equilibrium concentration of fructose-1,6- bisphosphate under these conditions? s. What is Key for this reaction?
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Answer #1

Solution :

First of all, we can calculate equilibrium constant (Keq) :

Part B) For the given reaction, equilibrium constant (keq) is calculated as,

ΔG° = - R T ln Keq

Given,

ΔG° = + 16.8 kJ/mol = + 16800 J/mol

T = 37 + 273 = 310 K

R = 8.314 J K-1 mol-1

Thus,

ln Keq = - ΔG°/ RT

ln Keq = (-16800 J/mol)/ 8.314 J K-1mol-1 x 310 K

ln Keq = - 6.518

2.303 log Keq = - 6.518

log keq = - 2.83

keq = antilog (-2.83) = 1.48 x 10^-3

Part A)

The equilibrium constant is related with the concentrations as,

Keq = [Products] / [Reactants ]

Keq = [Fructose-1,6-bisphosphate] / [Fructose-6-phosphate] [Phosphate]

1.48 x10^-3 = [Fructose-1,6-bisphosphate] / (0.014 mM x 1 mM)

[Fructose-1,6-bisphosphate] = 2.07 x 10^-5 mM

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