7. The enzyme mannose 6-P isomerase catalyzes the reversible reaction: Mannose 6

Question

7. The enzyme mannose 6-P isomerase catalyzes the reversible reaction: Mannose 6 P (M6P) Fructose-6-P (FER ). The fructose-6-P (F6P) formed can then enter glycolysis. The K 3.1. A) Calculate G, for the reaction above at 25°C (Use 2 significant figures), what does your answer tell you regarding spontaneity? (3pts) B) Assume you incubate 30 ml of a solution containing 0.075 M M6P overnight at 25'C with an excess of the enzyme mannose-6-P isomerase. How many millimoles of F6P will be in the 30 ml solution the next morning? (5 pts) C) Experimental values for the actual steady-state concentrations of these compounds in the cell are [M6P) 3 uM and [F6P]-14 Calculate G at 25 C. (Use 2 significant figures in which direction will the reaction proceed to reach equilibrium? Explain S pts)

Explanation / Answer

Ans. #A. Using the equation dG0’ = - RT lnKeq             - equation 1

Where, dG0’ = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq = equilibrium constant under given condition

R = (0.001987 kcal mol-1K-1 or 0.008315 kJ mol-1 K-1)

Putting the values in equation 1-

            dG0’ = - (0.008315 kJ mol-1 K-1) x 298.15 K x ln 3.1

            Or, dG0’ = - (0.008315 kJ mol-1K-1) x 298.15 K x 1.1314021115

            Hence, dG0’ = -2.805 kJ/mol

#B. The reaction goes on to establish the equilibrium.

Create an ICE table as shown in figure-

            Keq = [F6P] / [M6P]             - all [conc.] at equilibrium, in terms of molarity

            Or, 3.1 = X / (0.075 - X)

            Or, 3.1 (0.075 - X) = X

            Or, 3.1X – 0.2325 = X

            Or, 3.1X – X = 0.2325

            Or, X = 0.2325 / 2.1

            Hence, X = 0.1107

Therefore, [F6P] at equilibrium = 0.1107 M

# Now, moles of F6P at equilibrium = Molarity x Volume of solution in liters

                                                            = 0.1107 M x 0.030 L

                                                            = (0.1107 mol / L) x 0.030 L

                                                            = 0.003321 mol

                                                            = 3.321 millimoles

#C. Experimental equilibrium constant, Keq = [F6P] / [M6P]

            Or, Keq = 14 x 10-6 M / 3 x x103 M = 4.67

Using the equation dG’ = dG0’ + RT lnKeq                    - equation 2            

Where, dG’ = calculated/ experimental free energy change

dG0’ = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq = equilibrium constant under given condition

Putting the values in equation 2-

            dG’= (-2.805 kJ mol-1) + (0.008315 kJ mol-1 K-1) x 298.15K x ln 4.67

            Or, dG’ = -2.805 kJ mol-1 + 3.821 kJ mol-1

            Hence, dG’ = +1.02 kJ mol-1

# Reaction quotient, Q = [F6P] / [M6P] = 4 x 10-6 M / 3 x x103 M = 4.67

Since, Q (4.67) > Keq (3.1), - the reaction goes to the left to attain equilibrium.

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