Consider the following reaction catalyzed by phosphoribose mutase at 37 oC. Ribo
ID: 714799 • Letter: C
Question
Consider the following reaction catalyzed by phosphoribose mutase at 37 oC. Ribose 1-Phosphate Ribose 5-Phosphate (a) ( 2 pts) At equilibrium, the ratio of Ribose 1-phosphate / Ribose 5-phosphate = 0.16667. Calculate K’eq and G o ’for the above reaction as written above. (b) ( 1 pt) Would the reaction proceed as written under standard conditions?_______________ (c) (3 pts) If you started with 1 mM of Ribose 1-Phosphate and 1 mM of Ribose 5-Phosphate, what would be the equilibrium concentrations of Ribose 1-Phosphate and Ribose 5-Phosphate? (d) ( 2 pts) The physiological concentration of Ribose 1-Phosphate is 3.419 mM and the physiological concentration of Ribose 5-Phosphate is 0.440 mM. Calculate G. (e) ( 1 pt)Which direction would the reaction proceed as written under physiological conditions?
Explanation / Answer
(a) Ribose -1 - phosphate = Ribose-5-phosphate
K'eq = [Ribose-5-phosphate ]/ [ Ribose -1 - phosphate]
Given that: Ribose 1-phosphate / Ribose 5-phosphate = 0.16667
So, [Ribose-5-phosphate ]/ [ Ribose -1 - phosphate] = 1/0.16667 = 6
K'eq = 6
delG0 = -RT* lnK'eq= - 8.31 Joule mole-1K-1 * 310 K * ln6 = -4616 J /mole = - 4.616 kJ/ mole
(b) Would the reaction proceed as written under standard conditions?
Yes, the reaction proceeds spontaneously as the change in stabdard gibbs freee energy is negative.
(c) Suppose x mM of Ribose -1- phosphate converts to x mM Ribose-5-phosphate
So, at equilibrium: [ Ribose-5-phosphate] = (1+x) mM and [ Ribose -1- phosphate] = (1-x) mM
So, K'eq = [Ribose-5-phosphate ]/ [ Ribose -1 - phosphate] = ( 1+x)/(1-x) = 6
Or, 1+x = 6 - 6x
Or, 7x = 5
Or, x = 5/7 = 0.7142
Equilibrium concentrations of Ribose 1-Phosphate is: (1-0.7142) mM = 0.2858 mM
Equilibrium concentrations of Ribose-5-phosphate is: ( 1+ 0.7142) mM = 1.714 mM
(d) K'eq = [Ribose-5-phosphate ]/ [ Ribose -1 - phosphate] = 0.440/ 3.419 = 0.129
delG0 = -RT* lnK'eq= - 8.31 Joule mole-1K-1 * 310 K * ln(0.129) = 5282 J/mole = 5.282 kJ/ mole
(e) As delG0 is positive at physiological conditions, the reaction is not spontaneous for the conversion from Ribose -1- phosphate to Ribose -5- phosphate.
Rather the reverse reaction i.e. the conversion from Ribose-5-phosphate to Ribose-1-phosphate is spontaneous.