.Chemical.Examples.EnzymeKinetics

Information

Be carefull, the assumption for Michaelis-Menton are very strong:

The substrate must be in sufficiently high concentration and the product must be in very low concentration to reach almost all enzyme in enzyme-substrate complex all time. ([S] >> Km) && ([P] << K2)

To recalculate the enzyme kinetics from Michaelis-Menton parameters Km, tE a k_cat is selected the same half-rate of the reaction defined as:

E = ES = tE/2 .. the amount of free enzyme is the same as the amount of enzyme-substrate complexes

S = Km .. the amount of substrate is Km

r = Vmax/2 = tE*k_cat / 2 .. the rate of reaction is the half of maximal rate

Conversions of molar concentration to mole fraction (MM is molar mass of the solvent in solution -> 55.508 kg/mol for water):

x(Km) = Km/MM

x(tE) = tE/MM

xS = S/MM = Km/MM

The new kinetics of the system defined as:

uS° = DfG(S) = 0

uE° = DfG(E) = 0

uES° = DfG(ES) = DfG(S) + DfG(E) - R*T*ln(2/x(Km))

from dissociation coeficient of the frist reaction 2/x(Km) = xSE/(xS*xE) = exp((uE° + uS° - uES°)/(RT))

uP° = DfG(P)

r = Vmax/2

r = -kC1 * (uES° - uE° - uS° + R*T*ln(xES/(xE*xS) ) = -kC1 * (-R*T*ln(2/x(Km)) + R*T*ln(xS) ) = kC1 * R * T * ln(2)

because xES=xE this time

r = -kC2 * (uP° + uE° - uES° + R*T*ln(xP*xE/xES) ) = -kC2 * (DfG(P) - uES° + R*T*ln(xP) ) = kC2 * (-DfG(P) - R * T * ln(2))

kC1 = (Vmax/2) / (R * T * ln(2))

kC2 = (Vmax/2) / ( -DfG(P) - R * T * ln(2) )

For example in case of C=AmountOfSolution/(Tau*ActivationPotential) we can rewrite C to ActivationPotential (Be carefull: this energy is not the same as in Arrhenius equation or in Transition State Theory):

ActivationPotential1 = AmountOfSolution/(Tau*(Vmax/2)) * R * T * ln(2)

ActivationPotential2 = AmountOfSolution/(Tau*(Vmax/2)) * ( -DfG(P) - R * T * ln(2) )

where

AmountOfSolution = MM = 55.508 (for water)

Tau = 1 s (just to be physical unit correct)

DfG(P) = -R*T*50 is Gibbs energy of formation of product (setting negative enough makes second reaction almost irreversible)

The maximum of the new enzyme kinetics

The enzymatic rate must have a maximum near of Vmax.

The new maximum is a litle higher: Vmax * (1 + 1/( -uP°/(R*T*ln(2)) - 1) ), for example if -uP°/RT = 50, the new maximum is around 1.014*Vmax, where Vmax is the maximum of Michaelis Menten.

The proof:

We want to sutisfied the following inequality:

-kC2 * (uP° + uE° - uES° + R*T*ln(xP*xE/xES) ) ?=<? Vmax * (1 + 1/( -uP°/(R*T*ln(2)) - 1) )

(Vmax/2) * (uP° + uE° - uES° + R*T*ln(xP*xE/xES) ) / ( - uP° - R * T * ln(2) ) ?=<? Vmax*(1 + R*T*ln(2) / ( -uP° - R*T*ln(2)) )

(uP° + R*T*ln(2/x(Km)) + R*T*ln(xP*xE/xES) ) ?=<? 2*( - uP° - R * T * ln(2) ) + 2*R*T*ln(2)

R*T*ln(xP*xE/xES) ?=<? - uP° - R*T*ln(2/x(Km))

xP*xE/xES ?=<? exp((- uP° - R*T*ln(2/x(Km))/(R*T))

The equality is the equation of the equilibrium: xP*xE/xES = exp((- uP° - uE° + uES° )/(R*T)) = exp((- uP° - R*T*ln(2/x(Km))/(R*T))

If the equilibrium of the reaction is reached only by forward rate then xP*xE/xES must be less than the dissociation constant.

Revisions

2015-2018

Marek Matejak, Charles University, Prague, Czech Republic