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Gibbs Free Energy: G

D G = D H –T/ D S. D G = amount of energy available to do work. D G = G products - G reactants. A + B C + D. D G = (G C + ) - (G A + G B ). G D. Gibbs Free Energy: G. (See pages 13-15; 309- 318 Horton). G = H –TS H = heat of system

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Gibbs Free Energy: G

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  1. DG = DH –T/DS DG = amount of energy available to do work DG = Gproducts - Greactants A + B C + D DG = (GC + ) - (GA + GB) GD Gibbs Free Energy: G (See pages 13-15; 309- 318 Horton) G = H –TS H = heat of system S = entropy of system Determine the change in free energy of a reaction DG DG < 0 exergonic, rx tends to be spontaneous DG > 0 endergonic, rs requires input of energy DG = 0 at equilibrium Units: Joules or kiloJoules Joule = amount of energy required to apply 1 newton of force over 1m

  2. A B Reaction may be: exergonic and A B DGo’ < 0 Keq = [B] / [A] Reaction may be: endergonic and B A DGo’ > 0 Reaction may be at equilibrium DGo’ = 0 Standard free energy change: DGo’ Reactants and products present at Concentration of 1M and pH = 7.0

  3. A B C or A B E F G C Both pathways have the same DGo’ • DGo’ is independent of pathway

  4. 1. DGo’ = -RTlnKeq R = 8.315 J/mol/oK T = 298oK (25oC) = 2.48lnKeq (kJ/mol) • A B DGo’ =+16.7 KJ/mol Calculation ofDG: change in Gibbs free energy What is Keq for this reaction? DGo’ = -RTlnKeq 16.7 = -2.48lnKeq lnKeq = -(16.7/2.48) = -6.73 Keq = 1.19 x 10-3 = [B] / [A]

  5. DGo’produce large changes • in Keq • small changes in If DGo’ =+22.4 kJ/mol 22.4 = -2.48lnKeq lnKeq = -(22.4/2.48) = -9.03 Keq = 1.19 x 10-4 Increase in DGo’ from 16.7 to 22.4, a 35% increase, results in a 10-fold change in Keq.

  6. e.g. A B DGo’ = +15kJ/mol B C DGo’ = -20kJ/mol Net DGo’ = -5kJ/mol Net rx: A C • An unfavorable reaction may be made to • proceed by coupling it to a favorable reaction

  7. A B DGo’ =+16.7 kJ/mol • ATP ADP + Pi DGo’ =-30kJ/mol A + ATP B + ADP + Pi DGo’ =-13.8kJ/mol Keq = X [B] [B] [ADP][Pi] [ATP] [ATP] [A] [A] [ATP] Assume that = 500 [ADP][Pi] [ADP][Pi] and: = Keq x Coupling of unfavorable reaction to a favorable one. DGo’ = -RTlnKeq -13.8 = -2.48lnKeq Keq = 2.6 x 102 = 2.6 x 102 x 500 = 1.32 x 105

  8. Keq in the absence of ATP hydrolysis: = 1.19 x 10-3 Keq in the presence of ATP hydrolysis: = 1.32 x 105 An increase of 108-fold

  9. [B] DG = DGo’ + RTln 2. [A] Keq = 0.0475 e.g. A B from DGo’ = -RTlnKeq DGo’ = 7.55kJ/mol standard conditions [B] = Q, the mass action ratio [A] [B] Then DG = DGo’ + RTln [A] 3 x 10-6 M = 7.55 + RTln 2 x 10-4 M = -2.86kJ/mol actual conditions Actual free energy change BUT If [A] = 2 x 10-4 M , and [B] = 3 x 10-6 M

  10. Control of metabolic flux • Reactions that operate near equilibrium are • readily reversible - rate and direction of • reaction effectively controlled by concentrations • of substrate and products • Reactions that operate far from equilibrium • are metabolically irreversible – rate can only • be altered by changing enzyme activity

  11. F-1,6-bisP + ADP F-6-P + ATP This a potential control point e.g. Phosphofructokinase Keq = 300 But under intracellular conditions Q = 0.03 Insuffcient enzyme activity to equilibrate reaction and enzyme operates near Vmax at all times Can only increase rate of product formation by increasing enzyme activity

  12. A B C D E 3 4 2 1 In metabolic pathways intermediates are not allowed to “pile up” All reactions in a sequence proceed at the same rate and concentration of intermediates is constant – the steady state condition This is achieved by having several points of control First enzyme of a pathway does not feed substrate into the pathway at a rate that is faster than the slowest enzyme downstream

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