Question

If the intracellular concentrations of a metabolite (M-OH) and its phosphorylated form (M-OPO32-) were 2.9 mM and 0.09 mM, respectively, and if the intracellular concentrations of ATP and ADP were 3.8 mM and 0.17 mM, respectively, what would be the numerical value of \DeltaΔG (in kcal per mol to the nearest hundredth) for the following reaction: M-OH + ATP <--> M-OPO32- + ADP + H+? Assume a temperature of 37 °C and a pH of 7.3. 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 –3.3 kcal/mol and –7.3 kcal/mol, respectively. The correct answer is -8.48

Answer 1

Ans. Given:

M-O-P + H₂O --------> M-OH + Pi   dG0 = -3.3 Kcal/mol - reaction 1

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

Note: M-O-P = M-OPO₃^2-

Now,

# Step 1: Reverse reaction 1-

M-OH + Pi --------> M-O-P + H₂O , dG0 = +3.3 Kcal/mol - reaction 1’

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

M-OH + Pi --------> M-O-P + H₂O , dG0 = +3.3 Kcal/mol - reaction 1’

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

Net Reaction : M-OH + ATP --------> M-O-P + ADP ; dG⁰_net = - 4.0 Kcal/mol

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

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

            [M-OH] = 2.9 mM = 0.0029 M

            [ATP] = 3.8 mM = 0.0038 M

            [ADP] = 0.17 mM = 0.00017 M

Given, pH = 7.3

So, [H⁺] = antilog (-7.3) = 5.012 x 10^-8 M - on the product side as [H+] is shown on product side

On the reactant side, [H⁺] = 1.07 x 10^-7 - the [H+] in neutral water

Now,

Equilibrium constant, Keq = [M-O-P] [ADP] [H⁺_,product] / ([M-OH] [ATP] [H⁺,_reactant])

            Or, Keq = (0.00009 x 0.00017 x 5.012 x 10^-8) / (0.0029 x 0.0038 x 1.0 x 10^-7)

            Hence, Keq = 0.000696

# 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 = - 4.0 kcal mol^-1 + (0.001987 kcal mol^-1K^-1) x 310.15 K x ln (0.000696)

Or, dG = - 4.0 Kcal mol^-1 + 0.61626805 kcal mol^-1 x (-7.2701608976)

Or, dG = - 4.0 kcal mol^-1 – 4.480 kcal mol^-1

Or, dG = - 8.48 kcal mol^-1

Hence, dG = - 8.84 kcal/ mol