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

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

 G o ’ is independent of pathway A B C or A B E F G C Both pathways have the same  G o ’

Calculation of  G: change in Gibbs free energy  G o’ = -RTlnKeq R = J/mol/ o K T = 298 o K (25 o C) = 2.48lnKeq (kJ/mol) 1. i.A B  G o’ = KJ/mol  G o’ = -RTlnKeq 16.7 = -2.48lnKeq lnKeq = -(16.7/2.48) = Keq = 1.19 x = [B] / [A] What is Keq for this reaction?

If  G o’ = kJ/mol 22.4 = -2.48lnKeq lnKeq = -(22.4/2.48) = Keq = 1.19 x small changes in  G o’ produce large changes  in Keq Increase in  G o’ from 16.7 to 22.4, a 35% increase, results in a 10-fold change in Keq.

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

Coupling of unfavorable reaction to a favorable one. i.A B  G o’ = kJ/mol ii.ATP ADP + Pi  G o’ = -30kJ/mol A + ATP B + ADP + Pi  G o’ = -13.8kJ/mol  G o’ = -RTlnKeq -13.8= -2.48lnKeq Keq= 2.6 x 10 2 Keq = [A] [B] [ADP][Pi] [ATP] X Assume that [ATP] [ADP][Pi] = 500 and: [A] [B] = Keq x [ATP] [ADP][Pi] = 2.6 x 10 2 x 500 = 1.32 x 10 5

Keq in the presence of ATP hydrolysis: = 1.32 x 10 5 Keq in the absence of ATP hydrolysis: = 1.19 x An increase of fold

 G =  G o’ + RTln [B] [A] 2. [B] [A] = Q, the mass action ratio Actual free energy change If [A] = 2 x M, and [B] = 3 x M Then  G =  G o’ + RTln [B] [A] =  + RTln 3 x M 2 x M = -2.86kJ/mol actual conditions BUT e.g. A B from  G o’ = -RTlnKeq  G o’ = 7.55kJ/mol standard conditions Keq =

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

e.g. Phosphofructokinase F-6-P + ATP F-1,6-bisP + ADP 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 This a potential control point

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 A B C D E