amus. (10 pts) Last step of the glycolytic pathway is transfer of a phosphate gr

Question

amus. (10 pts) Last step of the glycolytic pathway is transfer of a phosphate group from phosphoenolpyruvate to ADP to give pyruvate. Phosphoenolpyruvate ADP pyruvate+ATP (a) Use the information below to calculate AG o for the last step of glycolysis above: Phosphoenolpyruvate + H20 - pyruvate + P ATP H20ADP P &G;?,--2.3 kcal/mol Is the last step of the glycolytic pathway spontaneous or non-spontaneous? (b) Calculate the equilibrium constant for the reaction at 25 °C, Assume that 2.303 RT 1.36 kcal/mole.

Explanation / Answer

delta Go is -7.5 kcal/mol which is negative so the reaction is spontaneous.

Keq = 3.266 x 105

Phosphoenolpyruvate + H2O ---> Pyruvate + Pi , delta Go = -14.8 kcal/mol

ATP + H2O ---> ADP + Pi, delta Go = -7.3 kcal/mol

Reaction

Phosphoenolpyruvate + ADP = Pyruvate + ATP

Inversing above equation and adding to first equation

ADP + Pi ---> ATP + H2O, delta Go = 7.3 kcal/mol

Phosphoenolpyruvate + H2O ---> Pyruvate + Pi , delta Go = -14.8 kcal/mol

Phosphoenolpyruvate + H2O + ADP + Pi   ---> Pyruvate + Pi + ATP + H2O, delta Go = -14.8+7.3 = -7.5 kcal/mol

Rearranging,

Phosphoenolpyruvate + ADP   ---> Pyruvate + ATP, delta Go = -7.5 kcal/mol,

delta Go is -7.5 kcal/mol which is negative so the reaction is spontanious.

delta Go = –RT In K(eq), delta Go = –2.303 RT log K(eq)

Keq = 10-deltaGo/(2.303RT)

Since 2.303 RT = 1.36 kcal/mole at 25 oC

Keq = 10-(-7.5)/(1.36) = 10(5.514) = 3.266 x 105

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