One of the key steps in glycolytic pathway in mitochondria in cells is the conve

Question

One of the key steps in glycolytic pathway in mitochondria in cells is the conversion (isomerization) of glucose-6-phosphate to fructose-6-phosphate Glucose-6-phosphate Fructose-6-phosphate At 298 K (25 degree C) the change in Gibbs Free Energy for the reaction is delta G degree = 1.7 kJ/mol Calculate the equilibrium constant, K_eq for this reaction Which of the compounds in the reaction will be present in a higher concentration? The next step in glycolysis is the conversion of fructose-6-phosphate to 1.6-fructose-bisphosphate. By itself, this is not a spontaneous reaction, but in the cell it occurs with a change in free energy of delta G degree = -14.2 kJ/mol. What is the value of equilibrium constant for this 2^nd reaction? What chemical reaction is responsible for driving this non-spontaneous reaction?

Explanation / Answer

G6P= Glucose 6 phosphate F6P= Fructose 6 phosphate and 1,6F = 1,6 fructose

For the reaction G6P-<-->F6P

deltaGO= -1.7*1000 j/mol

deltaGo=-RT lnK= -1.7*1000

where R= 8.314 J/mole.K and K= equilibrium constant

T=298.15K

lnK= 1.7*1000/(8.314*298.15)

K= 1.98 = [F6P]/ [G6P]

Since K>1, G6P concentration is less than f6P at equilibrium

2

For the reaction F6P<-----> 1,6F

deltaGo = -14.2*1000 j/mole

lnK= 14.2*1000/(8.314*298.15)

K=307. 52

K= [1,6F]/ [F6P] >1

Hence 1,6F is more at equilibrium than F6P

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