Provide step-wise calculations and the answer for the following problem; provide

Question

Provide step-wise calculations and the answer for the following problem; provide the best explanation/answer for the different conditions and catalyzation processes

1. The conversion of fructose-1,6-bisphosphate (FbP) to glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP) has a Keq of 6.8 x 10-5. a. Calculate AGo' for this reaction. b. Caclulate AG for this reaction if [GAP] 19 uM, [FbP] -2 [GAP], and [GAP] - [DHAP]. Under the conditions in (b), is it thermodynamically possible to couple this reaction to synthesis of ATP? This reaction is catalyzed by aldolase. In the presence of aldolase, is ?G of the reaction higher, lower, or the same? c. d.

Explanation / Answer

Ans. #a. Using the equation dG0’ = - RT ln Keq                         - equation 1

Where, dG0’ = standard/ theoretical free energy change

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

Keq = equilibrium constant under standard condition

R = (0.001987 kcal mol-1K-1 or 0.008314 kJ mol-1 K-1)

Note: The question does not mentions temperature. Let it be 25.00C, standard temperature for a cell.

Putting the values in equation 1-

            dG0’ = - (0.008314 kJ mol-1K-) x 298.0 K x ln (6.8 x 10-5)

            Or, dG0’ = -2.477572 kJ mol-1 x 2.303 log (6.8 x 10-5)

            Or, dG0’ = -2.477572 kJ mol-1 x (-9.5960)

            Hence, dG0’ = +23.775 kJ mol-1

# B:     F-6-P ------> G-3-P + DHAP           OR       F-6-P ------> 2 G-3-P

Now,

            Keq = [G-3-P]2 / [F-6-P]                  - all [conc.] in terms of M

            Or, Keq = (19 x 10-6 M)2 / (2 x 10-6 M) = 1.805 x 10-4

# Using, dG = dG0 + RT (ln Keq)

Where, dG = free energy change of the reaction

dG0’ = standard free energy change of the reaction

R = universal gas constant = 0.008314 kJ mol-1K-1

T = temperature in kelvin

Keq = equilibrium constant

Putting the values in above equation –

dG = 23.775 kJ mol-1 + (0.008314 kJ mol-1K-1 x 298 K) x ln 1.805 x 10-4

Or, dG = 23.775 kJ mol-1 + (-21.356 kJ mol-1)

            Or, dG = +2.419 kJ mol

# C. No.

dG0’ for ATP synthesis is -30.5 kJ mol-1. To catalyze synthesis of ATP, the given reaction must have its dG0’ value greater than that of -30.5 kJ mol-1.

Since dG of the given reaction is +2.419 kJ mol-1, it can’t be coupled to ATP synthesis from ADP and Pi.

# D. The same.

Note the enzymes act as biocatalyst, they increase the rate of a reaction but does not affect the equilibrium.

Note that the equation for dG calculation depends on Keq but not on [E].

So, because Keq remains unaffected, the dG for also remains unaffected.

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