380 Chapter 10. Enzymes and molecular machines[[Student version, January 17, 2003]]
Now suppose we throw a single enzyme molecule into a vat initially containing substrate at
concentrationcS,iand a negligible amount of product.^9 This system is far from equilibrium, but it
soon comes to a quasi-steady state: The concentration of substrate remains nearly constant, and
that of product nearly zero, since substrate molecules enormously outnumber the one enzyme. The
enzyme spends a certain fractionPEof its time unoccupied, and the rest,PES=1−PEbound to
substrate, and these numbers too are nearly constant in time. Thus the enzyme converts substrate
at a constant rate, which we’d like to find.
Let us summarize the discussion so far in a reaction formula:
E+Sk 1
k- 1
ES
k 2
⇀E+P. (10.16)It’s a cyclic process: The starting and ending states in this formula are different, but in each the
enzyme itself is in the same state. The notation associates rate constants to each process (see
Section 8.2.3 on page 269). We are considering only one molecule of E; thus the rate of conversion
for E+S⇀ES isk 1 cS,notk 1 cScE.
In a short time interval dt,the probabilityPEto be in the state E can change in one of three
ways:
1.If the enzyme is initially in the unbound state E, it has probability per unit timek 1 cSof
binding substrate and hence leaving the state E.
2.If the enzyme is initially in the enzyme-substrate complex state ES, it has probability per
unit timek 2 of processing and releasing product, hence reentering the unbound state E.
3.The enzyme-substrate complex also has probability per unit timek- 1 of losing its bound
substrate, reentering the state E.Expressing the above argument in a formula (see Idea 6.29 on page 196),
d
dt
PE=−k 1 cS×(1−PES)+(k- 1 +k 2 )×PES. (10.17)Make sure you understand the units on both sides of this formula.
The quasi-steady state is the one for which Equation 10.17 equals zero. Solving gives the
probability to be in the state ES as
PES=
k 1 cS
k- 1 +k 2 +k 1 cS. (10.18)
According to Equation 10.16, the rate at which a single enzyme molecule creates product is Equa-
tion 10.18 timesk 2 .Multiplying by the concentrationcEof enzyme then gives the reaction velocity
v,defined in Section 10.1.2.
The preceding paragraph outlined how to get the initial reaction velocity as a function of the ini-
tial concentrations of enzyme and substrate, for a reaction with an irreversible step (Reaction 10.16).
Wecan tidy up the formula by defining theMichaelis constantKMand maximum velocityvmax
of the reaction to be
KM≡(k- 1 +k 2 )/k 1 and vmax≡k 2 cE. (10.19)
(^9) Even if there are many enzyme molecules, we can expect the same calculations to hold as long as their concen-
tration is much smaller than that of substrate.