6.3: Kinetics with Enzymes
 Page ID
 25227
An enzyme alters the pathways for converting a reactant to a product by binding to the reactant and facilitating the intramolecular conversion of bound substrate to bound product before it releases the product. Enzymes do not affect the thermodynamics of reactions. For reversible reactions (as an example), the equilibrium constant, K_{eq,} is unchanged. What is charged is the rate at which equilibrium is achieved. Enzymes lower the activation energy for bound transition states and change the reaction mechanism.
Figure \(\PageIndex{1}\) shows the simplest chemical reaction that can be written to show how an enzyme catalyzes a reaction.
Rapid Equilibrium EnzymeCatalyzed Reactions
We have previously derived equations for the reversible binding of a ligand to a macromolecule. Next, we derived equations for the receptormediated facilitated transport of a molecule through a semipermeable membrane. This latter case extended the former case by adding a physical transport step. Now, in what hopefully will seem like deja vu, we will derive almost identical equations for the chemical transformation of a ligand, commonly referred to as a substrate, into a product by an enzyme. We will study two scenarios based on two different assumptions, each enabling a straightforward mathematical derivation of kinetic equations:
 Rapid Equilibrium Assumption  enzyme E (macromolecule) and substrate S (ligand) concentrations can be determined using the dissociation constant since E, S, and ES are in rapid equilibrium, as we previously used in our derivation of the equations for facilitated transport. Sorry about the switch from A to S in the designation of the substrate. Biochemists use S to represent the substrate (ligand) and A, B, P, and Q to represent reactants and products in the case of multisubstrate and multiproduct reactions.
 Steady State Assumption (more general)  enzyme and substrate concentrations are not those determined using the dissociation constant.
Enzyme kinetics experiments, as we will see in the following chapters, must be used to determine the detailed mechanism of the catalyzed reaction. Using kinetic analysis, you can determine the order of binding/dissociation of substrates and products, the rate constants for individual steps, and clues to the mechanism used by the enzyme in catalysis.
Consider the following reaction mechanism for the enzymecatalyzed conversion of substrate S into product P. (We will assume that the catalyzed rate is much greater than the noncatalyzed rate.)
\[\ce{E +S <=>[K_s] ES >[k_3] E + P} \nonumber \]
As we did for the derivation of the equations for the facilitated transport reactions under rapid equilibrium conditions, this derivation is based on the assumption that the relative concentrations of S, E, and ES can be determined by the dissociation constant, K_{S}, for the interactions and the concentrations of each species during the early part of the reaction (i.e. under initial rate conditions). Assume also the S >> E_{0}. Remember that under these conditions, S does not change much with time. Is this a valid assumption? Examine the mechanism shown above. S binds to E with a secondorder rate constant k_{1}. ES has two fates. It can dissociate with a firstorder rate constant k_{2} to S + E, or it can be converted to product with a firstorder rate constant of k_{3} to give P + E. If we assume that k_{2} >> k_{3} (i.e. that the complex falls apart much more quickly than S is converted to P), then the relative ratios of S, E, and ES can be described by Ks. Alternatively, you can think about it this way. If S binds to E, most of S will dissociate, and a small amount will be converted to P. If it does, then E is now free, and will quickly bind S and reequilibrate since the most likely fate of bound S is to dissociate, not be converted to P (since \(k_3 \ll k_2\)). This also makes sense if you consider that the physical step, characterized by k_{2}, is likely to be quicker than the chemical step, characterized by k_{3}. Hence the following assumptions have been used:
 \(S \gg E_0\)
 \(P_0 = 0\)
 \(k_3\) is rate limiting (i.e. the slow step)
We will derive equations showing the initial velocity v as a function of the initial substrate concentration, S_{0}, assuming that P is negligible over the time period used to measure the initial velocity. Also assume that \(v_{catalyzed} \gg v_{noncatalyzed}\). In contrast to the firstorder reaction of S to P in the absence of E, v is not proportional to S_{0} but rather to S_{bound}. Therefore, \(v \propto [ES]\), or
\[v = {const} [ES] = k_3 [ES] \label{EQ10} \]
where \(v\) is the velocity (i.e., reaction rate).
Now, let's get \(ES\) from the dissociation constant K_{S} (assuming rapid equilibrium of E, S and ES) and mass balance for E (E_{0 }= E + ES , so E = E_{0}  ES). We will use mass balance for S when we derive the equation of the steady state. In many cases, S is approximately equal to S_{0} or the total amount of substrate. This makes sense if you consider that the enzyme is a catalyst acts repeatedly to produce the product, so you don't need significant amounts of the enzyme. The resulting equation of [ES] is shown below.
\begin{equation}
(E S)=\frac{\left(E_0\right)(S)}{K_S+(S)}
\end{equation}
Here it is!
 Derivation

\begin{equation}
\begin{gathered}
K_S=\frac{[E][S]}{[E S]}=\left(\frac{\left.\left(\left[E_0\right][E S]\right)[S]\right)}{[E S]}\right. \\
(E S) K_S=\left(E_0\right)(S)(E S)(S) \\
(E S) K_S+(E S)(S)=\left(E_0\right)(S) \\
(E S)\left(K_S+(S)\right)=\left(E_0\right)(S)
\end{gathered}
\end{equation}
This derivation assumes that we know S (which is equal to S_{0}) and E_{tot} (which is E_{0}).
Let us assume that S is much greater than E, as is the likely biological case. We can calculate ES using the following equations and the same procedure we used for the derivation of the binding equation, which gives the equation below:
\begin{equation}
[E S]=\frac{\left[E_0\right][S]}{K_S+S}
\end{equation}
which is analogous to
\begin{equation}
[M L]=\frac{\left[M_0\right][S]}{K_D+L}
\end{equation}
This leads to
\begin{equation}
v_0=\frac{\left(k_3\right)\left(E_0\right)(S)}{K_S+(S)}=\frac{\left(V_M\right)(S)}{K_S+(S)}
\end{equation}
where
\begin{equation}
V_M=k_3 E_0
\end{equation}
This is the worldfamous HenriMichaelisMenten Equation. It is a hyperbola just like the graph for binding of a ligand to a macromolecule with a given dissociation constant, K_{D. }
Move the sliders in the interactive graph below to see how changing Km and Vm alters the graph. Note that this graph is identical to the graph for M + L ↔ ML.
Just as in the case with noncatalyzed firstorder decay, it is easiest to measure the initial velocity of the reaction when [S] does not change much with time and the velocity is constant (i.e. the slope of the dP/dt curve is constant). A plot of [P] vs t (called a progress curve) is made for each different substrate concentration studied. From these curves, the initial rates at each [S] is determined.
Alternatively, one reaction time that gives a linear rise in [P] with time is determined for all the different substrate concentrations. At that specified time, the reaction can be stopped (quenched) with a reagent that does not cause any change in S or P. Then initial rates can be easily calculated for each [S] from a single data point.
Under these conditions:
 a plot of v vs S is hyperbolic
 v = 0 when S = 0
 v is a linear function of S when S<<Ks.
 v = V_{max}_{ }(or V_{M}) when S is much greater than Ks
 S = K_{S} when v = V_{M}/2.
These are the same conditions we detailed for our understanding of the binding equation
\begin{equation}
(M L)=\frac{\left(M_0\right) L}{K_D+L}
\end{equation}
Note that when S is not >> K_{S}, the graph does not reach saturation and does not look hyperbolic. It should be apparent from the graph that only if S >> KS (or when S is approximately 100 x > K_{s}) will saturation be achieved.
The K_{S} constant is usually called the Michaelis constant, K_{M}. We will see in a bit that the K_{M} for most enzymecatalyzed reactions is not equal to the dissociation constant for ES, which we called K_{S}.
Very often, these graphs are transformed into double reciprocal or LineweaverBurk plots as shown below.
\begin{equation}
\frac{1}{v_0}=\frac{K_M+S}{V_M S}=\left(\frac{K_M}{V_M}\right) \frac{1}{S}+\frac{1}{V_M}
\end{equation}
These plots are used to estimate V_{M} from the 1/v intercept (1/V_{M}) and K_{M} from the 1/S axis (1/K_{M}). These values should be used as "seed" values for a nonlinear fit to the hyperbola that models the actual v vs S curve. An interactive graph of 1/v_{0} vs 1/[S] is shown below. Change the sliders for K_{M} and V_{M} and note the change in the slope and intercepts of the plot
As we saw in the graph of A or P vs t for a noncatalyzed, firstorder reaction, the velocity of the reaction, given as the slope of those curves, is always changing. Which velocity should we use in Equation 5? The answer invariably is the initial velocity, v_{0}, measured in the early part of the reaction when little substrate is depleted. Hence v vs S curves for enzymecatalyzed reactions invariably are really v_{0} vs [S] curves.
Steady State EnzymeCatalyzed Reactions
In this derivation, we will consider the following equations and all the rate constants, and will not arbitrarily assume that k_{2} >> k_{3}. We will still assume that S >> E_{0} and that P_{0} = 0. An added assumption, however, is that d[ES]/dt is approximately 0. Look at this assumption this way. When an excess of S is added to E, ES is formed. In the rapid equilibrium assumption, we assumed that it would fall back to E + S (a physical step) faster than it would go onto product (a chemical step). In the steady state case, we will assume that ES might go on to product either less or more quickly than it will fall back to E + S. In either case, a steady state concentration of ES arises within a few milliseconds, and its concentration does not change significantly during the initial part of the reaction under which the initial rates are measured. Therefore, d[ES]/dt is about 0. For the rapid equilibrium derivation, v = k_{3}[ES]. We then solved for ES using K_{S} and mass balance of E. In the steady state assumption, the equation v= k_{3}[ES] still holds, but now we will solve for [ES] using the steady state assumption that d[ES]/dt =0.
\begin{equation}
\frac{d[E S]}{\mathrm{dt}}=\mathrm{k}_1[E][S]\mathrm{k}_2[E S]\mathrm{k}_3[E S]=0
\end{equation}
We can solve this and obtain the MichaelisMenten equation for reaction.
\begin{equation}
\mathrm{v}=\mathrm{k}_3[\mathrm{ES}]=\frac{\mathrm{k}_3\left[\mathrm{E}_0\right][\mathrm{S}]}{\frac{\mathrm{k}_2+\mathrm{k}_3}{\mathrm{k}_1}+\mathrm{S}}=\frac{\mathrm{V}_{\mathrm{M}}[\mathrm{S}]}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}}
\end{equation}
To see how to derive this, click below.
 Answer

Applying mass balance for E (i.e E = E_{0}ES) in appropriate term below gives
\begin{array}{l}{\mathrm{k}_{1}[E][S]=\left(\mathrm{k}_{2}+\mathrm{k}_{3}\right)[E S]} \\ {\mathrm{k}_{1}\left[\mathrm{E}_{0}\mathrm{ES}\right][S]=\left(\mathrm{k}_{2}+\mathrm{k}_{3}\right)[E S]} \\ {\mathrm{k}_{1}\left[\mathrm{E}_{0}\right][S]\mathrm{k}_{1}[\mathrm{ES}][S]=\left(\mathrm{k}_{2}+\mathrm{k}_{3}\right)[E S]} \\ {\mathrm{k}_{1}\left[\mathrm{E}_{0}\right][S]=\left(\mathrm{k}_{2}+\mathrm{k}_{3}\right)[E S]+\mathrm{k}_{1}[\mathrm{ES}][S]} \\ {\mathrm{k}_{1}\left[\mathrm{E}_{0}\right][S]=[E S]\left(\mathrm{k}_{2}+\mathrm{k}_{3}+\mathrm{k}_{1} \mathrm{S}\right)}\end{array}
Solving for [ES] gives
\begin{equation}
[\mathrm{ES}]=\left(\frac{\left[\mathrm{E}_0\right][\mathrm{S}]}{\frac{\mathrm{k}_2+\mathrm{k}_3}{\mathrm{k}_1}+\mathrm{S}}\right)
\end{equation}Substituting into the Henri Michaelis Menten equation gives
\begin{equation}
\mathrm{v}=\mathrm{k}_3[\mathrm{ES}]=\frac{\mathrm{k}_3\left[\mathrm{E}_0\right][\mathrm{S}]}{\frac{\mathrm{k}_2+\mathrm{k}_3}{\mathrm{k}_1}+\mathrm{S}}=\frac{\mathrm{V}_{\mathrm{M}}[\mathrm{S}]}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}}
\end{equation}
Note that
\begin{equation}
V_M=k_3 E_0
\end{equation}
and
\begin{equation}
K_M=\frac{k_2+k_3}{k_1}
\end{equation}
Now let's look at a progress curve simulation that compares the MichaelisMenten equation derived from rapid equilibrium assumptions (when k_{2}>>k_{3}) and in which K_{M} is the actual dissociation constant to the steady state approximation, when k_{2} is not >>k_{3}.
The initial conditions in the graph are set so the graphs of the rapid equilibrium and steady state are identical.
Recommendations:
 Scale the yaxis to 50
 change the slider k_{2_r2} for the steady state graph to other values. Watch the curves separate. Although these plots are only for 1 substrate concentration (50 uM), the effects of changing k_{2} for the steady state plot are very dramatic. This should convince you that in general, unless k_{3} << k_{2}, the calculated value of K_{M }is not equal to the thermodynamic dissociation constant, K_{D}.
Analysis of the General MichaelisMenten Equation
This equation can be simplified and studied under different conditions. First, notice that (k_{2} + k_{3})/k_{1} is a constant which is a function of relevant rate constants. This term is usually replaced by K_{M} which is called the Michaelis constant. Likewise, when S approaches infinity (i.e. S >> K_{M}, equation 5 becomes v = k_{3}(E_{0}) which is also a constant, called V_{M} for maximal velocity. Substituting V_{M} and K_{M} into equation 5 gives the simplified equation:
\begin{equation}
\mathrm{v}=\frac{\mathrm{V}_{\mathrm{M}}[\mathrm{S}]}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}}
\end{equation}
It is extremely important to note that K_{M} in the general equation does not equal the K_{S}, the dissociation constant used in the rapid equilibrium assumption! K_{M} and K_{S} have the same units of molarity, however. A closer examination of K_{M} shows that under the limiting case when k_{2} >> k_{3} (the rapid equilibrium assumption) then,
\begin{equation}
\mathrm{K}_{\mathrm{M}}=\frac{\mathrm{k}_2+\mathrm{k}_3}{\mathrm{k}_1}=\frac{\mathrm{k}_2}{\mathrm{k}_1}=\mathrm{K}_{\mathrm{D}}=\mathrm{K}_{\mathrm{S}}
\end{equation}
If we examine these equations under several different scenarios, we can better understand the equation and the kinetic parameters:
 when S = 0, v = 0.
 when S >> K_{M}, v = V_{M} = k_{3}E_{0}. (i.e. v is zero order with respect to S and first order in E. Remember, k_{3} has units of s^{}^{1> }since it is a firstorder rate constant. k_{3} is often called the turnover number, because it describes how many molecules of S "turn over" to product per second.
 v = V_{M}2, when S = K_{M}.
 when S << K_{M}, v = V_{M}S/K_{M} = k_{3}E_{0}S/K_{M} (i.e. the reaction is bimolecular, dependent on both on S and E. k3/K_{M} has units of M^{1}s^{1}, the same as a second order rate constant.
More Complicated Enzymecatalyzed Reactions
A reversiblycatalyzed reaction
You have learned previously that enzymes don't change the equilibrium constant for a reaction, but rather lowers the activation energy barrier to move from reactants to products. This implies that the activation energy to move in the reverse direction, from products to reactants, is also lowered. Hence the enzyme speeds up both the forward and reverse reaction. We haven't accounted for that yet in our kinetic equations. Many reactions in metabolic reactions that do not have large negative values of ΔG (ie. they are not significantly favored) are reversible, allowing the enzyme to be used in the reverse direction. Take for example the pathway to break down glucose to pyruvate (glycolysis). It has 9 different steps, of which 5 are reversible, allowing them to be used in the reverse pathway to take pyruvate to glucose (gluconeogenesis).
Let's set up the equations for the reversible reaction of substrate S to product P catalyzed by enzyme E. Assume that the K_{M} for the forward reaction is K_{MS} (or K_{S}) and for the reverse reaction is K_{MP} (or K_{P}) as shown in the reaction scheme in Figure \(\PageIndex{2}\). The rate constant k_{2} is the k_{cat }(forward rate constant) for conversion of ES to EP and k_{2} is the k_{cat} (reverse rate constant) for conversion of EP to ES
The following simple MichaelisMenten equations can be written for just the forward reaction and for the reverse reaction:
\begin{equation}
\begin{aligned}
&v_f=\frac{V_f S}{K_{M S}+S}=\frac{\frac{V_f S}{K_{M S}}}{1+\frac{S}{K_{M S}}} \\
&v_r=\frac{V_r P}{K_{M P}+P}=\frac{\frac{V_r P}{K_{M P}}}{1+\frac{P}{K_{M P}}}
\end{aligned}
\end{equation}
Now you might think that simply substracting the two would give the net velocity in the forward direction, but that is NOT the case.
\begin{equation}
v \neq\left[\frac{\frac{V_f S}{K_{M S}}}{1+\frac{S}{K_{M S}}}\frac{\frac{V_r P}{K_{M P}}}{1+\frac{P}{K_{M P}}}\right]
\end{equation}
The reason is that in the derivation, the equations for both the forward and reverse rates must have terms for the reverse and forward reactions, respectively.
A simple derivation shows that this is the equation for the reversible conversion of substrate to product.
\begin{equation}
v=k_2[E S]k_{2}[E P]=\frac{V_f \frac{[S]}{K_S}}{\left[1+\frac{[S]}{K_S}+\frac{[P]}{K_p}\right]}\frac{V_r \frac{[P]}{K_P}}{\left[1+\frac{[S]}{K_S}+\frac{[P]}{K_P}\right]}=\frac{V_f \frac{[S]}{K_S}V_r \frac{[P]}{K_P}}{\left[1+\frac{[S]}{K_S}+\frac{[P]}{K_P}\right]}
\end{equation}
This reversible form of the MichaelisMenten equation and other equations, written in the format shown in equation 6.24, are commonly used in programs such as VCell and Copasi to model the kinetics of whole pathways of biological interactions and reactions. The figures below show a reaction diagram, graphical results showing S and P vs time for the selected K_{M} and V_{M} values shown, and animations for the reaction. The chemical (S and P) are shown as green spheres connected by a line. The red dot again represents the enzyme (shown as a node of connection between S and P). Equation 6.24 was used to model the reversible reaction (even though the arrows shown between S and P are unidirectional.
If you reflect on it, this reaction is very similar to the reversible reaction of A ↔ P in the absence of an enzyme, which we explored in Chapter section 6.2. Just for comparison, the graph for that reaction is shown below. Change the sliders to produce a curve similar to the enzymecatalyzed reaction shown in the figure above.
Reaction with intermediates
Not all reactions can be characterized so simply as a simple substrate interacting with an enzyme to form an ES complex, which then turns over to form product. Sometimes, intermediates form. For example, a substrate S might interact with E to form a complex, which then is cleaved to products P and Q. Q is released from the enzyme, but P might stay covalently attached. This happens often in the hydrolytic cleavage of a peptide bond by a protease, when an activated nucleophile like Ser reacts with the sessile peptide bond in a nucleophilic substitution reaction, releasing the amine end of the former peptide bond as the leaving group while the carboxy end of the peptide bond remains bonded to the Ser as an Seracyl intermediate. Water then enters and cleaves the acyl intermediate, freeing the carboxyl end of the original peptide bond. This is shown in the written reaction in Figure \(\PageIndex{3}\):
Even for this seemingly complicated reaction, you get the standard MichaelisMenten equation.
To simplify the derivation of the kinetic equation, let's assume that E, S, and ES are in rapid equilibrium defined by the dissociation constant, Ks. Assume Q has a visible absorbance, so it is easy to monitor. Assume from the steady state assumption that:
\begin{equation}
\frac{\mathrm{d}[\mathrm{E}\mathrm{P}]}{\mathrm{dt}}=\mathrm{k}_2[\mathrm{ES}]\mathrm{k}_3[\mathrm{E}\mathrm{P}]=0
\end{equation}
assuming that k_{3} is a pseudo firstorder rate constant and that [H_{2}O] doesn't change.
The velocity depends on which step is ratelimiting. If k_{3}<<k_{2}, then the k_{3} step is ratelimiting. Then
\begin{equation}
\mathrm{v}=\mathrm{k}_3[\mathrm{E}\mathrm{P}]
\end{equation}
If k_{2}<<k_{3}, then the k_{2} step is ratelimiting. Then
\\begin{equation}
\mathrm{v}=\mathrm{k}_2[\mathrm{ES}]
\end{equation}
The following kinetic equation for this reaction can be derived, assuming v = k_{2}[ES].
\begin{equation}
\mathrm{v}=\frac{\frac{\mathrm{k}_2 \mathrm{k}_3}{\mathrm{k}_2+\mathrm{k}_3}\left[\mathrm{E}_0\right][\mathrm{S}]}{\left(\frac{\mathrm{k}_3}{\mathrm{k}_2+\mathrm{k}_3}\right) \mathrm{K}_{\mathrm{S}}+\mathrm{S}}=\frac{\mathrm{k}_{\mathrm{cat}}\left[\mathrm{E}_0\right][\mathrm{S}]}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}}=\frac{\mathrm{V}_{\mathrm{M}} \mathrm{S}}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}}
\end{equation}
You can verify that you get the same equation if you assume that v = k_{3}[EP].he Derivation:
This equation looks quite complicated, especially if you substitute for Ks, k_{1}/k_{1}. All the kinetic constants can be expressed as functions of the individual rate constants. However, this equation can be simplified by realizing the following:
 When \(\mathrm{S}>>\frac{\mathrm{K}_{\mathrm{S}} \mathrm{k}_{3}}{\mathrm{k}_{2}+\mathrm{k}_{3}}, \mathrm{v}=\left(\frac{\mathrm{k}_{2} \mathrm{k}_{3}}{\mathrm{k}_{2}+\mathrm{k}_{3}}\right) \mathrm{E}_{0}=\mathrm{V}_{\mathrm{M}}\)
 \(\frac{\mathrm{K}_{\mathrm{S}} \mathrm{k}_{3}}{\mathrm{k}_{2}+\mathrm{k}_{3}}=\mathrm{constant}=\mathrm{K}_{\mathrm{M}}\)
Substituting these into equation 7 gives:
\[\mathrm{v}=\frac{\mathrm{V}_{\mathrm{M}} \mathrm{S}}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}} \nonumber \]
This again is the general form of the MichaelisMenten equation
The expression for V_{M} in the first bulleted expression above is more complicated than our earlier definition of V_{M} = k_{3}E_{0}. They are similar in that the term E_{0} is multiplied by a constant which is itself a function of rate constant(s). The rate constants are generally lumped together into a generic constant called k_{cat}.
 For the simple reaction k_{cat} = k3
 For the more complicated reaction above with a covalent intermediate, \(\mathrm{k}_{\mathrm{cat}}=\frac{\mathrm{k}_{2} \mathrm{k}_{3}}{\mathrm{k}_{2}+\mathrm{k}_{3}}\)
 For all reactions, V_{M} = k_{cat}E_{0}.
Figure \(\PageIndex{4}\) compares the MichaelisMenten kinetic equations for the rapid equilibrium, steady state assumptions, and covalent intermediate cases .
Meaning of Kinetic Constants
It is important to get a "gutlevel" understanding of the significance of the rate constants. Here they are:
 K_{M}: The Michaelis constant with units of molarity (M), is operationally defined as the substrate concentration at which the initial velocity is half of V_{M}. It is equal to the dissociation constant of E and S only in if E, S and ES are in rapid equilibrium. It can be thought of as an "effective" (but not actual) K_{D }in other cases.
 k_{cat}: The catalytic rate constants, with units of s^{1} is often called the turnover number. It is a measure of how many bound substrate molecules "turnover" or form product in 1 second. This is evident from equation v_{0} = k_{cat}_{[}ES]
 k_{cat}/K_{M}: Under condition when [S] << K_{M}, the MichaelisMenten equation becomes v_{0} = (k_{cat}/K_{M})[E_{0}][S]. This really describes a biomolecular rate constant (k_{cat}/K_{M}), with units of M^{1}s^{1}, for conversion of free substrate to product. Some enzymes have k_{cat}/Km values around 10^{8}, indicating that they are diffusion controlled. That implies that the reaction is essentially done as soon as the enzyme and substrate collide. The constant k_{cat}/Km is also referred to as the specificity constant in that it describes how well an enzyme can differentiate between two different competing substrates. (We will show this mathematically in the next chapter.)
Table \(\PageIndex{1}\) below shows K_{M} and k_{cat} values for various enzymes
K_{M} values  
enzyme  substrate  Km (mM) 
catalase  H_{2}O_{2}  25 
hexokinase (brain)  ATP  0.4 
DGlucose  0.05  
DFructose  1.5  
carbonic anhydrase  HCO_{3}^{}  9 
chymotrypsin  glycyltyrosinylglycine  108 
Nbenzoyltyrosinamide  2.5  
bgalactosidase  Dlactose  4.0 
threonine dehydratase  LThr  5.0 
k_{cat} values  
enzyme  substrate  k_{cat} (s^{1}) 
catalase  H_{2}O_{2}  40,000,000 
carbonic anhydrase  HCO_{3}^{}  400,000 
acetylcholinesterase  acetylcholine  140,000 
blactamase  benzylpenicillin  2,000 
fumarase  fumarate  800 
RecA protein (ATPase)  ATP  0.4 
Table \(\PageIndex{1}\): K_{M} and k_{cat} values for various enzymes
Table \(\PageIndex{2}\) below show k_{cat}, K_{M} and k_{cat}/K_{M} values for diffusioncontrolled enzymes
Enzymes with k_{cat}/K_{M} values close to diffusion controlled (10^{8}  10^{9} M^{1}s^{1})  
enzyme  substrate  kcat (s^{1})  Km (M)  kcat/Km (M^{1}s^{1})  
acetylcholinesterase  acetylcholine  1.4 x 10^{4}  9 x 10^{5}  1.6 x 10^{8}  
carbonic anhydrase  CO_{2}  1 x 10^{6}  1.2 x 10^{2}  8.3 x 10^{7}  
HCO_{3}^{}  4 x 10^{5}  2.6 x 102  1.5 x 10^{7}  
catalase  H_{2}O_{2}  4 x 10^{7}  1.1  4 x 10^{7}  
crotonase  crotonylCoA  5.7 x 10^{3}  2 x 10^{5}  2.8 x 10^{8}  
fumarase  fumarate  8 x 10^{2}  5 x 10^{6}  1.6 x 10^{8}  
malate  9 x 10^{2}  2.5 x 10^{5}  3.6 x 10^{7}  
triose phosphate isomerase  glyceraldehyde3P  4.3 x 10^{3}  4.7 x 10^{4}  2.4 x 10^{8}  
blactamase  benzylpenicillin  2.0 x 10^{3}  2 x 10^{4}  1 x 10^{8} 
Table \(\PageIndex{2}\) below show k_{cat}, K_{M} and k_{cat}/K_{M} values for diffusion controlled enzymes
Experimental Determination of V_{M} and K_{M}
How can V_{M} and K_{M} be determined from experimental data?
From initial rate data
The most common way to determine V_{M} and K_{M} is through initial rates, v_{0}, obtained from P or S vs time curves. Hyperbolic graphs of v_{0} vs [S] can be fitted or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine K_{D}. These included:
 MichaelisMenten plot: nonlinear hyperbolic fit
 LineweaverBurk double reciprocal plot
 Scatchard plot
 EadieHofstee plot
We discussed all of these plots except for the EadieHofstee plot, in the chapter on binding. The EadieHofstee plot is another linearized version of MichaelisMenten equation
Here is a derivation of that equation, which starts with each side of the doublereciprocal plot being multiplied by v_{0}V_{M}.
\begin{equation}
\begin{aligned}
\left(v_0 V_M\right) \frac{1}{v_0} &=\left(v_0 V_M\right)\left(\frac{K_M}{V_M}\right) \frac{1}{S}+\left(v_0 V_M\right) \frac{1}{V_M} \\
V_M &=\left(v_0\right) K_M \frac{1}{S}+\left(v_0\right) \\
v_0 &=K_M\left(\frac{v_0}{S}\right)+V_M
\end{aligned}
\end{equation}
Note that a graph of v_{0} vs v_{0}/S is linear so slopes and intercepts can be used to obtain values for V_{M} and K_{M}.
The doublereciprocal plot is commonly used to analyze initial velocity vs substrate concentration data. When used for such purposes, the graphs are referred to as LineweaverBurk plots, where plots of 1/v vs 1/S are straight lines with slope m = K_{M}/V_{M}, and yintercept b = 1/V_{M}. Figure \(\PageIndex{5}\) common graphs used to display initial rate enzyme kinetic data.
The straightline plots shown above should not be analyzed using linear regression, since simple linear regression assumes constant error in v_{0} values. A weighted linear regression or even better, a nonlinear fit to a hyperbolic equation should be used. (Common Error in Biochemistry Textbooks: The Shape of the Hyperbola). A rearrangement of the corresponding Scatchard equations in the EadieHofstee plot is also commonly used, especially to visualize enzyme inhibition data as we will see in the next chapter.
An Extension:
K_{M} and V_{M} could be theoretically extracted from progress curves of A or P as a function of t at one single A concentration by deriving an integrated rate equation for A or P as a function of t, as we did in equation 2 (the integrated rate equation for the conversion of A → P in the absence of enzyme). In principle, this method would be better than the initial rates methods. Why? It is not easy to be certain you are measuring the initial rate for each and every [S] which should vary over a wide range. It's also time intensive. In addition, think how much data is discarded if you take an entire progress curve at each substrate concentration, especially if you quench the reaction at a given time point, which effectively limits the data to one time point per substrate.
In practice, the mathematics is complicated and it is not possible to get a simple explicit function of [P] or [S] as a function of time. A slight variant of a progress curve can be derived. Let us consider the simple case of a single substrate S (or A) being converted to product P in an enzymecatalyzed reaction. The analogous equations for firstorder, noncatalyzed rates were A=A_{0}e^{k1t} or P = A_{0}(1e^{k1t}).
We can derive the equation for the enzymecatalyzed reaction shown below.
Here it is!
\begin{equation}
\frac{\mathrm{P}}{\mathrm{t}}=\frac{\mathrm{K}_{\mathrm{M}} \ln \left(\frac{\mathrm{S}_0\mathrm{P}}{\mathrm{S}_0}\right)}{\mathrm{t}}+\mathrm{V}_{\mathrm{M}}
\end{equation}
Click below to see the derivation
 Derivation

\begin{equation}
\begin{array}{r}
\mathrm{v}=\frac{\mathrm{dS}}{\mathrm{dt}}=+\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{V}_{\mathrm{M}} \mathrm{S}}{\mathrm{K}_{\mathrm{M}}+\mathrm{S}} \\
\int_{\mathrm{S}_0}^{\mathrm{S}} \frac{\mathrm{K}_{\mathrm{M}}+\mathrm{S}}{\mathrm{V}_{\mathrm{M}} \mathrm{S}} \mathrm{d} S=\int_0^{\mathrm{t}} \mathrm{t}
\end{array}
\end{equation}\begin{equation}
\mathrm{t}=\frac{\mathrm{S}+\mathrm{K}_{\mathrm{M}} \ln \mathrm{S}\mathrm{S}_0\mathrm{K}_{\mathrm{M}} \ln S_0}{\mathrm{~V}_{\mathrm{M}}}
\end{equation}On rearrangement, this gives:
\begin{equation}
\mathrm{S}_0\mathrm{S}+\mathrm{K}_{\mathrm{M}} \ln \frac{\mathrm{S}_0}{\mathrm{~S}}=\mathrm{V}_{\mathrm{M}} \mathrm{t}
\end{equation}This equation is an implicit equation, not an explicit one, as it does NOT give S(t) explicitly as a function of t.
Equation yy can be written with respect to product P as follows:
\begin{equation}
\begin{aligned}
&P=\mathrm{S}_0S{ }^{\prime \prime} \text { or }{ }^{\prime \prime} S=\mathrm{S}_0P \\
&\mathrm{~S}_0\left(\mathrm{S}_0\mathrm{P}\right)+\mathrm{K}_{\mathrm{M}} \ln \frac{\mathrm{S}_0}{\mathrm{~S}_0\mathrm{P}}=\mathrm{V}_{\mathrm{M}} \mathrm{t} \\
&\mathrm{P}\mathrm{K}_{\mathrm{M}} \ln \left(\frac{\mathrm{S}_0\mathrm{P}}{\mathrm{S}_0}\right)=\mathrm{V}_{\mathrm{M}} \mathrm{t}
\end{aligned}
\end{equation}Rearranging this gives
\begin{equation}
\frac{\mathrm{P}}{\mathrm{t}}=\frac{\mathrm{K}_{\mathrm{M}} \ln \left(\frac{\mathrm{S}_0\mathrm{P}}{\mathrm{S}_0}\right)}{\mathrm{t}}+\mathrm{V}_{\mathrm{M}}
\end{equation}
S(t) explicitly as a function of t.
This equation does not give P(t) explicitly as a function of time. Rather one can get a graph of P/t vs [ln (1P/S_{0})]/t (shown below) from the derived equation, which does give a straight line with a slope of Km and a yintercept of Vm. Note that the calculated values of V_{M} and K_{M} are derived from only one substrate concentration, and the values may be affected by product inhibition.
Figure \(\PageIndex{6}\) shows the comparison of a firstorder noncatalyzed conversion of A → P to the enzymecatalyzed rate. The V_{M} for the enzymecatalyzed reaction was chosen to be small to make the two graph comparable.
Note that the curves are similar but not identical. If you didn't know an enzyme was present, you could fit the data to a firstorder rise in [P] with time, but it would not be the optimal fit. The progress curves are a lot more complicated to analyze if the product, which shares structural similarities with the substrate, binds the enzyme tightly and inhibits it (called product inhibition).
Comparison of progress Curves of uncatalyzed and catalyzed reactions.
Let's explore progress curves for enzymecatalyzed reactions a bit further. Students usually see v_{0} vs [S] MichaelisMenten plots in textbooks. These plots are in some ways less intuitive than seeing P vs t curves, which are more in line with how we might contemplate how a reaction proceeds. Hence, it would be illuminating to compare progress curve graphs of A → P (irreversible) for the uncatalyzed and S → P for the enzymecatalyzed reactions. What might you expect? We saw one example in Figure \(\PageIndex{6}\).
In each case, P should increase with time. In the uncatalyzed reaction, S exponentially decreases to 0 and P concomitantly rises to a value of P = S_{0}. You also get a rise in P vs t for the enzymecatalyzed rate, but you would think it would be a faster rise with time since the enzyme catalyze the reaction. Figure \(\PageIndex{7}\) shows a comparison of the progress curves for the uncatalyzed firstorder reaction of A → P_{1} (red) and S → P_{2} (blue) for enzym catalyzed reaction (blue, right) for these conditions: uncatalyzed reaction A → P_{1}, k_{1} = 0.1; Catalyzed reaction: S → P_{2}, V_{M}=10, K_{M}=5. Note that the rate at which bound S (i.e ES) goes to P for the catalyzed rate is 100x faster than the rate constant for the catalyzed rate.
Note that the curves are somewhat similar in shape but also clearly different in comparison to the one shown in Figure \(\PageIndex{6}\).
Now let's use Vcell to compare the reactions for different values for the kinetic constants for the uncatalyzed and the enzymecatalyzed reaction. Change the constants and try to find a set of conditions so that the catalyzed and catalyzed rates for conversion of reactions to products are superimposable. How can that be?
Animations
Now let's look at an animation of the same irreversible reactions in which enzymecatalyzed reaction is no faster than the noncatalyzed rate (a worthless enzyme!). Here are the reactions:
A→P, k_{1} =0.1; S → P, V_{M}=?, K_{M} = ?.
The animations show just the accumulation of product. Animations are by Shraddha Nakak and Hui Liu.
Use the Vcell model above to find a set of values for K_{M} and V_{M} that would make the two graphs superimposable  i.e. when the graphs in the absence and presence of E are identical. (Hint: that would be a really bad enzyme if it didn't increase the reaction over the uncatalyzed rate!)
 Answer

K_{M} = 96, V_{M} = 10