CHAPTER 6 - TRANSPORT AND KINETICS
B: Kinetics of Simple and
BIOCHEMISTRY - DR. JAKUBOWSKI
Learning Goals/Objectives for Chapter 6B: After class and
this reading, students will be able to
- write appropriate chemical and differential equations for
the rate of disappearance of reactants or appearance of products
for 1st order, pseudo first order, second order, reversible
first order reactions
- draw and interpret graphs for integrated rate equations
(showing reactant or product concentrations as a function of
time) and initial rate equations (showing the initial velocity
vo as a function of reactant ;
- derive kinetic rate constants from data and graphs of
integrated rate and initial rate equations;
- write appropriate chemical, differential equations, and
initial rate equations for the rate of disappearance of
reactants or appearance of products for simple enzyme catalyzed
- differentiate between rapid and steady state assumptions;
- simplify the initial rate equation containing rate constants
for an enzyme catalyzed reactions to one replacing the rate
constants with kcat and KM, and give operational and
mathematical definitions of those constants;
B5. Analysis of the General Michaelis-Menten Equation
This equation can be simplified and studied
under different conditions. First notice that (k2 + k3)/k1 is a constant
which is a function of relevant rate constants. This term is usually
replaced by Km which is called the Michaelis constant (which was used in the
Mathematica graph above). Likewise, when S approaches infinity (i.e. S
>> Km, equation 5 becomes v = k3(Eo) which is also a constant, called Vm for
maximal velocity. Substituting Vm and Km into equation 5 gives the
Equation 10) v = Vm(S)/(Km+ S)
It is extremely important to note that Km in the general equation does
not equal the Ks, the dissociation constant used in the rapid equilibrium
assumption! Km and Ks have the same units of molarity,
however. A closer examination of Km shows that under the limiting case when
k2 >> k3 (the rapid equilibrium assumption) then,
Equation 11) Km = (k2 + k3)/k1 = k2/k1 =
Kd = Ks.
If we examine Equations 9 and 10 under several different scenarios, we
can better understand the equation and the kinetic parameters:
- when S = 0, v = 0.
- when S >> Km, v = Vm = k3Eo. (i.e. v is
zero order with respect to S and first order in E. Remember, k3 has
units of s-1since it is a first order rate constant. k3 is often called
the turnover number, because it describes how many molecules of S "turn
over" to product per second.
- v = Vm/2, when S = Km.
- when S << Km, v = VmS/Km = k3EoS/Km (i.e.
the reaction is bimolecular, dependent on both on S and E. k3/Km has
units of M-1s-1, the same as a second order rate constant.
Notice that equations 9 and 10 are exactly analysis to the previous
equations we derived:
- ML = MoL/(Kd + L) for binding of L to M
- Jo = JmA/(Kd + A) for
rapid equilibrium binding and facilitated transport of A
- vo = VmS/(Ks + S) for
rapid equilbirum binding and catalytic conversion of A to P.
- vo = Vm(S)/(Km+ S) for
steady state binding and catalytic conversion of A to P.
Please notice that all these equations give
hyperbolic dependencies of the y dependent variable (ML, Jo, and vo) on the
ligand, solute, or substrate concentration, respectively.
Return to B:
Kinetics of Simple and Enzyme-Catalyzed Reactions Sections
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Kinetics of Simple and Enzyme-Catalyzed Reactions
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