Question

Consider the standard free energies shown below(T =37 C, R=8.3145 J/mole K) :

UDP-Mannose + H2O → Mannose 1-phosphate + UMP + 2H+ delta G' = -34.9 kJ/mol

UTP + H2O → UMP + PPi + 2H+ delta G' = -32.5 kJ/mol

PPi + H2O → 2Pi delta G' = -33.6 kJ/mol

(A) ( 2 pts) Calculate delta G' & Keq’ for the following reaction.

Mannose 1-phosphate + UTP + H2O → UDP-mannose + 2Pi

(B) ( 2 pts) Which of the following statements best describe this reaction?

(A)The reaction is irreversible under standard conditions.

(B) The reaction is reversible under standard conditions.

(C) The reaction is at equilibrium under intracellular conditions.

(D)The reaction is reversible under intracellular conditions.

(E) K’eq = 0

(C) (2 pts) This reaction is catalyzed by phosphomannose pyrophosphorylase. How would this enzyme effect delta G' and Keq’?

(D) ( 3 pts) The physiological concentration of these three metabolites in a mammalian cell at physiological temperature ( 37 C) are: [UDP-mannose] = 0.130 mM; [UMP] = 0.140 mM , [H2O ] = 55.6 M, [H+ ] = 10^-7 M and [mannose 1-phosphate] = 2.33 mM [UTP] = 4.35 mM; [Pi] = 8.90 mM [PPi] = 8.14 microM

What would be the actual free energy change (delta G) for the reaction under these cellular conditions?

Mannose 1-phosphate + UTP + H2O → UDP-mannose + 2Pi

Answer 1

UDP-Mannose + H2O → Mannose 1-phosphate + UMP + 2H+ delta G' = -34.9 kJ/mol (1)

Reversing the reaction gives

Mannose 1-phosphate + UMP + 2H+ -------àUDP- Mannose + H2O, deltaG= 34.9 Kj/mole (1A)

UTP + H2O → UMP + PPi + 2H+ delta G' = -32.5 kJ/mol (2)

PPi + H2O → 2Pi delta G' = -33.6 kJ/mol (1)

Eq.1A+Eq.2+Eq.3 gives

Mannose 1-phosphate + UMP + 2H+UTP+H2O+PPi+H2O-----à UDP- Mannose + H2O+UMP+PPi+2H++2Pi, deltaG= 34.9-32.5-33.6 KJ/mole =-31.2 KJ/mole

Mannose 1-phosphate + UTP+H2O-----à UDP- Mannose +2Pi, deltaG= 34.9-32.5-33.6 KJ/mole =-31.2 KJ/mole

deltaGo= -RTlnK, given T= 37deg.c= 37+273= 310K,

lnK= -deltaG/RT= 31.2*1000/(8.314*310)=12.11

K= 181680

K= [UDP-Mannose] [PI]²/ [UTP] [Mannose-1-phosphate] (4)

Since K is very high, the concentrations of reactants is very less and the reaction is irreversible under standard conditions. ( A is correct)

The presence of catalyst does not have any effect on the deltaG and Keq. The reaction proceeds through an alternative path and reduces the activation energy.

K from 4= 0.130*10^-3*(8.9*10^-3)²/ (4.35*10^-3*2.33)=1.016*10^-6

deltaG=-RTlnK= -8.314*(37+273)*ln(1.016*10^-6)=35566 j/mole =35.566 Kj/mole