Question

If the intracellular concentrations of a metabolite (M-COO-) and its phosphorylated form (M-COOPO32-) were 2.9 mM and 0.09 mM, respectively, and if the intracellular concentrations of ATP and ADP were 3.8 mM and 0.16 mM, respectively, what would be the numerical value of \DeltaΔG (in kcal per mol to the nearest hundredth) for the following reaction: M-COO- + ATP <--> M-COOPO32- + ADP? Assume a temperature of 37 °C and a pH of 7.4. To solve this problem, you will need to know the standard free energies of hydrolysis of the phosphorylated metabolite and of ATP. These values are –10.3 kcal/mol and –7.3 kcal/mol, respectively.

Answer 1

Ans. Given:

M-COO-P + H₂O --------> M-COO^- + Pi   ; dG⁰ = -10.3 Kcal/mol - reaction 1

ATP + H₂O --------> ADP + Pi                     ; dG⁰ = -7.3 Kcal/mol - reaction 2

Note: M-COO-P = M-COOPO₃^2-

Now,

# Step 1: Reverse reaction 1-

M-COO^- + Pi --------> M-COO-P + H₂O ; dG⁰ = +10.0 Kcal/mol - reaction 1’

# Step 2: Add reaction 1’ and 2 to get the desired reaction-

M-COO^- + Pi --------> M-COO-P + H₂O             ; dG⁰ = +10.0 Kcal/mol

     (+) ATP + H₂O --------> ADP + Pi                              ; dG⁰ = -7.3 Kcal/mol

Net Reaction : M-COO^- + ATP --------> M-COO-P + ADP ; dG⁰_net = + 2.7 Kcal/mol

# Step 3: Convert all concentrations in terms of molarity-

            [M-COO-P] = 0.09 mM = 0.00009 M

            [M-COO^-] = 2.9 mM = 0.0029 M

            [ATP] = 3.8 mM = 0.0038 M

            [ADP] = 0.17 mM = 0.00016 M

Now,

Equilibrium constant, Keq = [M-COO-P] [ADP] / ([M-COO^-] [ATP])

            Or, Keq = (0.00009 x 0.00016) / (0.0029 x 0.0038)

            Hence, Keq = 0.0000000144 / 0.00001102

            Or, Keq = 0.0013067

# Step 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 = 0.001987 kcal mol^-1K^-1

T = temperature in kelvin = 310.15 K

Keq = equilibrium constant

Or, dG = +2.7 kcal mol^-1 + (0.001987 kcal mol^-1K^-1) x 310.15 K x ln (0.0013067)

Or, dG = 2.7 Kcal mol^-1 + 0.61626805 kcal mol^-1 x (-6.640250404)

Or, dG = 2.7 kcal mol^-1 – 4.09 kcal mol^-1

Or, dG = - 1.39 kcal mol^-1

Hence, dG = - 1.39 kcal/ mol