6.3 Enzyme Kinetics as an Approach to Understanding Mechanism

Biochemists commonly use several approaches to study the mechanism of action of purified enzymes. The three-dimensional structure of the protein provides important information, which is enhanced by traditional protein chemistry and modern methods of site-directed mutagenesis (changing the amino acid sequence of a protein by genetic engineering; see Fig. 9-10). These technologies permit enzymologists to examine the role of individual amino acids in enzyme structure and action. However, the oldest approach to understanding enzyme mechanisms, and one that remains very important, is to determine the rate of a reaction and how it changes in response to changes in experimental parameters, a discipline known as enzyme kinetics. We provide here a basic introduction to the kinetics of enzyme-catalyzed reactions.

Substrate Concentration Affects the Rate of Enzyme-Catalyzed Reactions

At any given instant in an enzyme-catalyzed reaction, the enzyme exists in two forms: the free or uncombined form E and the substrate-combined form ES. When the enzyme is first mixed with a large excess of substrate, there is an initial transient period, the pre–steady state, during which the concentration of ES builds up. For most enzymatic reactions, this period is very brief. The pre-steady state is frequently too short to be observed easily, lasting only the time (often microseconds) required to convert one molecule of substrate to product (one enzymatic turnover). The reaction quickly achieves a steady state in which [ES] (and the concentrations of any other intermediates) remains approximately constant for much of the remainder of the reaction (Fig. 6-10). As most of the reaction reflects the steady state, the traditional analysis of reaction rates is referred to as steady-state kinetics.

A graph plots concentration against time to show the course of an enzyme-catalyzed reaction.

FIGURE 6-10 The course of an enzyme-catalyzed reaction. In a typical reaction, product will increase as substrate declines. The concentration of free enzyme, E, declines rapidly, as the concentration of the ES complex increases and reaches a steady state. The steady-state concentration of ES remains nearly constant for much of the remainder of the reaction.

A key factor affecting the rate of a reaction catalyzed by an enzyme is the concentration of substrate, [S]. However, studying the effects of substrate concentration is complicated by the fact that [S] changes during the course of an in vitro reaction as substrate is converted to product. One simplifying approach is to measure the initial rate (or initial velocity), designated V0bold-italic upper V bold 0 (Fig. 6-11). In a typical reaction, the enzyme may be present in nanomolar quantities, whereas [S] may be five or six orders of magnitude higher. If only the beginning of the reaction is monitored, over a period in which only a small percentage of the available substrate (<2%−3%) is converted to product, [S] can be regarded as constant, to a reasonable approximation. V0upper V 0 can then be explored as a function of [S], which is adjusted by the investigator. Note that even the initial rate reflects a steady state.

A graph shows initial velocities of enzyme-catalyzed reactions by plotting product concentration against time.

FIGURE 6-11 Initial velocities of enzyme-catalyzed reactions. A theoretical enzyme that catalyzes the reaction S  Pupper S right harpoon over left harpoon upper P is present at a concentration of S sufficient to catalyze the reaction at a maximum velocity, defined as Vmaxupper V Subscript max, of 1 μm/min. The rate observed at a given concentration of S depends on the Michaelis constant, Kmupper K Subscript m, explained in more detail in the next section. Here, the Kmupper K Subscript m is 0.5 μm. Progress curves are shown for substrate concentrations below, at, and above the Kmupper K Subscript m. The rate of an enzyme-catalyzed reaction declines as substrate is converted to product. A tangent to each curve taken at time = 0 (dashed line) defines the initial velocity, V0upper V 0, of each reaction.

The effect on V0upper V 0 of varying [S] when the enzyme concentration is held constant is shown in Figure 6-12. At relatively low concentrations of substrate, V0upper V 0 increases almost linearly with an increase in [S]. At higher substrate concentrations, V0upper V 0 increases by smaller and smaller amounts in response to increases in [S]. Finally, a point is reached beyond which increases in V0upper V 0 are vanishingly small as [S] increases. This plateau-like V0upper V 0 region is close to the maximum velocity, Vmaxbold-italic upper V Subscript bold max.

A graph plots substrate concentration, open square bracket S close square bracket, in millimolar on the horizontal axis against initial velocity, V subscript 0, in micromolar per minute on the vertical axis, to show the effect of substrate concentration on initial velocity of an enzymatic reaction.

FIGURE 6-12 Effect of substrate concentration on the initial velocity of an enzyme-catalyzed reaction. The maximum velocity, Vmaxupper V Subscript max, is extrapolated from the plot, because V0upper V 0 approaches but never quite reaches Vmaxupper V Subscript max. The substrate concentration at which V0upper V 0 is half-maximal is Kmupper K Subscript m, the Michaelis constant. The concentration of enzyme in an experiment such as this is generally so low that [S]≫[E] even when [S] is described as low or relatively low. The units shown are typical for enzyme-catalyzed reactions and are given only to help illustrate the meaning of V0upper V 0 and [S]. (Note that the curve describes part of a rectangular hyperbola, with one asymptote at Vmaxupper V Subscript max. If the curve were continued below [S] = 0, it would approach a vertical asymptote at [S]=Kmleft-bracket upper S right-bracket equals minus upper K Subscript m.)

The ES complex is the key to understanding this kinetic behavior, just as it was a starting point for our discussion of catalysis. The kinetic pattern in Figure 6-12 led Victor Henri, following a proposal by Adolphe Wurtz a few decades earlier, to propose in 1903 that the combination of an enzyme with its substrate molecule to form an ES complex is a necessary step in enzymatic catalysis. This idea was expanded into a general theory of enzyme action, particularly by Leonor Michaelis and Maud Menten in 1913. They postulated that the enzyme first combines reversibly with its substrate to form an enzyme-substrate complex in a relatively fast reversible step:

E+S k1k1 ESupper E plus upper S right harpoon over left harpoon Underscript k Subscript negative 1 Baseline Overscript k 1 Endscripts ES

(6-7)

The ES complex then breaks down in a slower second step to yield the free enzyme and the reaction product P:

ES k2k2 E+PES right harpoon over left harpoon Underscript k Subscript negative 2 Baseline Overscript k 2 Endscripts upper E plus upper P

(6-8)

Because the slower second reaction (Eqn 6-8) must limit the rate of the overall reaction, the overall rate must be proportional to the concentration of the species that reacts in the second step — that is, ES.

Two photos. The first photo shows Leonor Michaelis, 1875-1949 and the second photo shows Maud Menten, 1879-1960.

Left: Leonor Michaelis, 1875–1949; Right: Maud Menten, 1879–1960

At low [S], most of the enzyme is in the uncombined form E. Here, the rate is proportional to [S] because the equilibrium of Equation 6-7 is pushed toward formation of more ES as [S] increases. The maximum initial rate of the catalyzed reaction (Vmax)left-parenthesis upper V Subscript max Baseline right-parenthesis is observed when virtually all the enzyme is present as the ES complex and [E] is vanishingly small. Under these conditions, the enzyme is “saturated” with its substrate, so that further increases in [S] have no effect on rate. This condition exists when [S] is sufficiently high that essentially all the free enzyme has been converted to the ES form. After the ES complex breaks down to yield the product P, the enzyme is free to catalyze the reaction of another molecule of substrate (and will do so rapidly under saturating conditions). The saturation effect is a distinguishing characteristic of enzymatic catalysts and is responsible for the plateau observed in Figure 6-12, and the pattern seen in the figure is sometimes referred to as saturation kinetics.

The Relationship between Substrate Concentration and Reaction Rate Can Be Expressed with the Michaelis-Menten Equation

The curve expressing the relationship between [S] and V0upper V 0 (Fig. 6-12) has the same general shape for most enzymes (it approaches a rectangular hyperbola), which can be expressed algebraically by the Michaelis-Menten equation. Michaelis and Menten derived this equation starting from their basic hypothesis that the rate-limiting step in enzymatic reactions is the breakdown of the ES complex to product and free enzyme. The equation is

V0=Vmax [S]Km+[S]upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-9)

This is the Michaelis-Menten equation, the rate equation for a one-substrate enzyme-catalyzed reaction. It is a statement of the quantitative relationship between the initial velocity V0upper V 0, the maximum velocity Vmaxupper V Subscript max, and the initial substrate concentration [S], all related through a constant, Kmupper K Subscript m, called the Michaelis constant. All these terms — [S], V0upper V 0, Vmaxupper V Subscript max, and Kmupper K Subscript m — are readily measured experimentally.

Here we develop the basic logic and the algebraic steps in a modern derivation of the Michaelis-Menten equation, which includes the steady-state assumption, a concept introduced by G. E. Briggs and J. B. S. Haldane in 1925. The derivation starts with the two basic steps of the formation and breakdown of ES (Eqns 6-7 and 6-8). Early in the reaction, the concentration of the product, [P], is negligible, and we make the simplifying assumption that the reverse reaction, PSupper P right-arrow upper S (described by k2k Subscript negative 2), can be ignored. This assumption is not critical but it simplifies our task. The overall reaction then reduces to

E+S k1k1 ES k2 E+Pupper E plus upper S right harpoon over left harpoon Underscript k Subscript negative 1 Baseline Overscript k 1 Endscripts ES right-arrow Overscript k 2 Endscripts upper E plus upper P

(6-10)

V0upper V 0 is determined by the breakdown of ES to form product, which is determined by [ES]:

V0 =k2 [ES]upper V 0 equals k 2 left-bracket ES right-bracket

(6-11)

Because [ES] in Equation 6-11 is not easily measured experimentally, we must begin by finding an alternative expression for this term. First, we introduce the term [Et]left-bracket upper E Subscript t Baseline right-bracket, representing the total enzyme concentration (the sum of free and substrate-bound enzyme). Free or unbound enzyme [E] can then be represented by [Et][ES]left-bracket upper E Subscript t Baseline right-bracket minus left-bracket ES right-bracket. Also, because [S] is ordinarily far greater than [Et]left-bracket upper E Subscript t Baseline right-bracket, the amount of substrate bound by the enzyme at any given time is negligible compared with the total [S]. With these conditions in mind, the following steps lead us to an expression for V0upper V 0 in terms of easily measurable parameters.

Step 1 The rates of formation and breakdown of ES are determined by the steps governed by the rate constants k1k 1 (formation) and k1+k2k Subscript negative 1 Baseline plus k 2 (breakdown to reactants and products, respectively), according to the expressions

Rate of ES formation=k1([Et][ES])[S]Rate of ES formation equals k 1 left-parenthesis left-bracket upper E Subscript t Baseline right-bracket en-dash left-bracket ES right-bracket right-parenthesis left-bracket upper S right-bracket

(6-12)

Rate of ES breakdown=k1[ES]+k2[ES]Rate of ES breakdown equals k Subscript negative 1 Baseline left-bracket ES right-bracket plus k 2 left-bracket ES right-bracket

(6-13)

Step 2 We now make an important assumption: that the initial rate of reaction reflects a steady state in which [ES] is constant — that is, the rate of formation of ES is equal to the rate of its breakdown. This is called the steady-state assumption. The expressions in Equations 6-12 and 6-13 can be equated for the steady state, giving

k1([Et][ES])[S]=k1[ES]+k2[ES]k 1 left-parenthesis left-bracket upper E Subscript t Baseline right-bracket minus left-bracket ES right-bracket right-parenthesis left-bracket upper S right-bracket equals k Subscript negative 1 Baseline left-bracket ES right-bracket plus k 2 left-bracket ES right-bracket

(6-14)

Step 3 In a series of algebraic steps, we now solve Equation 6-14 for [ES]. First, the left side is multiplied out and the right side simplified to give

k1[Et] [S]k1[ES] [S]=(k1+k2) [ES]k 1 left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket minus k 1 left-bracket ES right-bracket left-bracket upper S right-bracket equals left-parenthesis k Subscript negative 1 Baseline plus k 2 right-parenthesis left-bracket ES right-bracket

(6-15)

Adding the term k1[ES][S]k 1 left-bracket ES right-bracket left-bracket upper S right-bracket to both sides of the equation and simplifying gives

k1[Et] [S]=(k1[S]+k1+k2) [ES]k 1 left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket equals left-parenthesis k 1 left-bracket upper S right-bracket plus k Subscript negative 1 Baseline plus k 2 right-parenthesis left-bracket ES right-bracket

(6-16)

We then solve this equation for [ES]:

[ES]=k1[Et] [S]k1[S]+k1+k2left-bracket ES right-bracket equals StartFraction k 1 left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket Over k 1 left-bracket upper S right-bracket plus k Subscript negative 1 Baseline plus k 2 EndFraction

(6-17)

This can now be simplified further, combining the rate constants into one expression:

[ES]=[Et] [S][S]+(k1+k2)/k1left-bracket ES right-bracket equals StartFraction left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket Over left-bracket upper S right-bracket plus left-parenthesis k Subscript negative 1 Baseline plus k 2 right-parenthesis slash k 1 EndFraction

(6-18)

The term (k1+k2)/k1left-parenthesis k Subscript negative 1 Baseline plus k 2 right-parenthesis slash k 1 is defined as the Michaelis constant, Kmbold-italic upper K Subscript bold m. Substituting this into Equation 6-18 simplifies the expression to

[ES]=[Et] [S]Km+[S]left-bracket ES right-bracket equals StartFraction left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-19)

Step 4 We can now express V0upper V 0 in terms of [ES]. Substituting the right side of Equation 6-19 for [ES] in Equation 6-11 gives

V0 =k2[Et] [S]Km+[S]upper V 0 equals StartFraction k 2 left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-20)

This equation can be simplified further. Because the maximum velocity occurs when the enzyme is saturated (that is, when [ES]=[Et]left-bracket ES right-bracket equals left-bracket upper E Subscript t Baseline right-bracket), Vmaxupper V Subscript max can be defined as k2[Et]k 2 left-bracket upper E Subscript t Baseline right-bracket. Substituting this in Equation 6-20 gives Equation 6-9:

V0 =Vmax[S]Km+[S]upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

Note that Kmupper K Subscript m has units of molar concentration. Does the equation fit experimental observations? Yes; we can confirm this by considering the limiting situations where [S] is very low or very high, as shown in Figure 6-13.

A graph plots substrate concentration, open square bracket S close square bracket, in millimolar on the horizontal against initial velocity, V subscript 0, in micromolar per minute on the vertical axis to show the relationship between initial velocity and substrate concentration.

FIGURE 6-13 Dependence of initial velocity on substrate concentration. This graph shows the kinetic parameters that define the limits of the curve at low [S] and high [S]. At low [S], Km[S]upper K Subscript m Baseline much-greater-than left-bracket upper S right-bracket, and the [S] term in the denominator of the Michaelis-Menten equation (Eqn 6-9) becomes insignificant. The equation simplifies to V0=Vmax[S]/Kmupper V 0 equals upper V Subscript max Baseline left-bracket upper S right-bracket slash upper K Subscript m Baseline, and V0upper V 0 exhibits a linear dependence on [S], as observed here. At high [S], where [S]Kmleft-bracket upper S right-bracket much-greater-than upper K Subscript m, the Kmupper K Subscript m term in the denominator of the Michaelis-Menten equation becomes insignificant and the equation simplifies to V0=Vmaxupper V 0 equals upper V Subscript max; this is consistent with the plateau observed at high [S]. The Michaelis-Menten equation is therefore consistent with the observed dependence of V0upper V 0 on [S], and the shape of the curve is defined by the terms Vmax/Kmupper V Subscript max Baseline slash upper K Subscript m Baseline at low [S] and Vmaxupper V Subscript max at high [S].

An important numerical relationship emerges from the Michaelis-Menten equation in the special case when V0upper V 0 is exactly one-half Vmaxupper V Subscript max (Fig. 6-13). Then

Vmax2=Vmax[S]Km+[S]StartFraction upper V Subscript max Baseline Over 2 EndFraction equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-21)

On dividing by Vmaxupper V Subscript max, we obtain

12=[S]Km+[S]one-half equals StartFraction left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-22)

Solving for Kmupper K Subscript m, we get Km+[S]=2[S]upper K Subscript m Baseline plus left-bracket upper S right-bracket equals 2 left-bracket upper S right-bracket, or

Km=[S],upper K Subscript m Baseline equals left-bracket upper S right-bracket comma when V0=½Vmaxupper V 0 equals one half upper V Subscript max

(6-23)

This is a very useful, practical definition of Km:Kmupper K Subscript m Baseline colon upper K Subscript m Baseline is equivalent to the substrate concentration at which V0upper V 0 is one-half Vmaxupper V Subscript max.

Michaelis-Menten Kinetics Can Be Analyzed Quantitatively

There are several ways to determine Vmaxupper V Subscript max and Kmupper K Subscript m from a plot relating V0upper V 0 to [S]. A traditional approach is to algebraically transform the Michaelis-Menten equation into equations that convert the hyperbolic curve in the V0upper V 0 versus [S] plot into a linear form from which values of Vmaxupper V Subscript max and Kmupper K Subscript m may be obtained by extrapolation. The most common of these approaches takes the reciprocal of both sides of the Michaelis-Menten equation (Eqn 6-9):

V0 =Vmax[S]Km+  [S]1V0=Km+[S]Vmax[S]StartLayout 1st Row upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction 2nd Row StartFraction 1 Over upper V 0 EndFraction equals StartFraction upper K Subscript m Baseline plus left-bracket upper S right-bracket Over upper V Subscript max Baseline left-bracket upper S right-bracket EndFraction EndLayout

(6-24)

Separating the components of the numerator on the right side of the equation gives

1V0=KmVmax[S]+[S]Vmax[S]StartFraction 1 Over upper V 0 EndFraction equals StartFraction upper K Subscript m Baseline Over upper V Subscript max Baseline left-bracket upper S right-bracket EndFraction plus StartFraction left-bracket upper S right-bracket Over upper V Subscript max Baseline left-bracket upper S right-bracket EndFraction

(6-25)

which simplifies to

1V0=KmVmax[S]+1VmaxStartFraction 1 Over upper V 0 EndFraction equals StartFraction upper K Subscript m Baseline Over upper V Subscript max Baseline left-bracket upper S right-bracket EndFraction plus StartFraction 1 Over upper V Subscript max Baseline EndFraction

(6-26)

This form of the Michaelis-Menten equation is called the Lineweaver-Burk equation. For enzymes obeying the Michaelis-Menten relationship, a plot of 1/V01 slash upper V 0 versus 1/[S] (the “double reciprocal” of the V0upper V 0 versus [S] plot we have been using up to this point) yields a straight line (Fig. 6-14). This line has a slope of Km/Vmaxupper K Subscript m Baseline slash upper V Subscript max Baseline, an intercept of 1/Vmax1 slash upper V Subscript max Baseline on the 1/V01 slash upper V 0 axis, and an intercept of 1/Kmnegative 1 slash upper K Subscript m Baseline on the 1/[S] axis. The Lineweaver-Burk plot is a useful way to display data and can provide some mechanistic information, as we will see. However, the double-reciprocal transformation tends to give undue weight to data obtained at low substrate concentration, and can distort errors in the extrapolated values of Vmaxupper V Subscript max and Kmupper K Subscript m.

A Lineweaver-Burk graph plots the reciprocal of the substrate against the reciprocal of the initial velocity.

FIGURE 6-14 A double-reciprocal, or Lineweaver-Burk, plot. Plotting 1/V01 slash upper V 0 versus 1/[S] puts the data into a linear form. Intercepts on the 1/V01 slash upper V 0 and 1/[S] axes are 1/Vmax1 slash upper V Subscript max Baseline and 1/Kmnegative 1 slash upper K Subscript m Baseline, respectively.

More commonly, the parameters Vmaxupper V Subscript max and Kmupper K Subscript m are derived directly from the V0upper V 0 versus [S] plot via nonlinear regression, using one of a multitude of curve-fitting programs available online. These are generally easy to use and offer accuracy superior to Lineweaver-Burk or related approaches.

Kinetic Parameters Are Used to Compare Enzyme Activities

It is important to distinguish between the Michaelis-Menten equation and the specific kinetic mechanism on which the equation was originally based. The Michaelis-Menten equation does not depend on the relatively simple two-step reaction mechanism proposed by Michaelis and Menten (Eqn 6-10). All enzymes that exhibit a hyperbolic dependence of V0upper V 0 on [S] are said to follow Michaelis-Menten kinetics. Many enzymes that follow Michaelis-Menten kinetics have quite different reaction mechanisms, and enzymes that catalyze reactions with six or eight identifiable steps often exhibit the same steady-state kinetic behavior. The practical rule that Km =[S]upper K Subscript m Baseline equals left-bracket upper S right-bracket when V0=½Vmaxupper V 0 equals one half upper V Subscript max (Eqn 6-23) holds for all enzymes that follow Michaelis-Menten kinetics. The most important exceptions to Michaelis-Menten kinetics are the regulatory enzymes, discussed in Section 6.5.

The parameters Kmupper K Subscript m and Vmaxupper V Subscript max can be obtained experimentally for any given enzyme, and their measurement is often a key first step in enzyme characterization. However, the mechanistic insight they offer is limited. Obtaining information about the number, rates, or chemical nature of discrete steps in the reaction usually requires additional complementary approaches. Steady-state kinetics nevertheless is the standard language through which biochemists compare and characterize the catalytic efficiencies of enzymes.

Interpreting Kmsans-serif-bold-italic upper K Subscript bold-sans-serif m and Vmaxsans-serif-bold-italic upper V Subscript bold-sans-serif max

Both the magnitude and the meaning of Kmupper K Subscript m and Vmaxupper V Subscript max can vary greatly from enzyme to enzyme.

Kmupper K Subscript m can vary even for different substrates of the same enzyme (Table 6-6). The term is sometimes used (often inappropriately) as an indicator of the affinity of an enzyme for its substrate. The actual meaning of Kmupper K Subscript m depends on specific aspects of the reaction mechanism, such as the number and relative rates of the individual steps. For reactions with two steps,

Km =k2+k1k1upper K Subscript m Baseline equals StartFraction k 2 plus k Subscript negative 1 Baseline Over k 1 EndFraction

(6-27)

TABLE 6-6 Kmbold-italic upper K Subscript bold m for Some Enzymes and Substrates

Enzyme Substrate Km (mM)bold-italic upper K Subscript bold m Baseline bold left-parenthesis bold m bold upper M bold right-parenthesis
Hexokinase (brain)
  • ATP
  • d-Glucose
  • d-Fructose
  • 0.4
  • 0.05
  • 1.5
Carbonic anhydrase HCO3HCO Subscript 3 Superscript minus 26
Chymotrypsin
  • Glycyltyrosinylglycine
  • N-Benzoyltyrosinamide
  • 108
  • 2.5
β-Galactosidase d-Lactose 4.0
Threonine dehydratase l-Threonine 5.0

When k2k 2 is rate-limiting, k2k1k 2 much-less-than k Subscript negative 1, and Kmupper K Subscript m reduces to k1/k1k Subscript negative 1 Baseline slash k 1, which is defined as the dissociation constant, Kdbold-italic upper K Subscript bold d, of the ES complex. Where these conditions hold, Kmupper K Subscript m does represent a measure of the affinity of the enzyme for its substrate in the ES complex. However, this scenario does not apply for most enzymes. Sometimes k2k1k 2 much-greater-than k Subscript negative 1, and then Km=k2/k1upper K Subscript m Baseline equals k 2 slash k 1. In other cases, k2k 2 and k1k Subscript negative 1 are comparable, and Kmupper K Subscript m remains a more complex function of all three rate constants (Eqn 6-27). The Michaelis-Menten equation and the characteristic saturation behavior of the enzyme still apply, but Kmupper K Subscript m cannot be considered a simple measure of substrate affinity. Even more common are cases in which the reaction goes through several steps after formation of ES; Kmupper K Subscript m can then become a very complex function of many rate constants.

The quantity Vmaxupper V Subscript max depends upon the rate-limiting step of the enzyme-catalyzed reaction. If an enzyme reacts by the two-step Michaelis-Menten mechanism, Vmax =k2[Et]upper V Subscript max Baseline equals k 2 left-bracket upper E Subscript t Baseline right-bracket, where k2k 2 is rate-limiting. However, the number of reaction steps and the identity of the rate-limiting step(s) can vary from enzyme to enzyme. For example, consider the common situation where product release, EPE+PEP right-arrow upper E plus upper P, is rate-limiting. Early in the reaction (when [P] is low), the overall reaction can be adequately described by the scheme

E+S k1k1 ES k2k2 EP k3 E+Pupper E plus upper S right harpoon over left harpoon Underscript k Subscript negative 1 Baseline Overscript k 1 Endscripts ES right harpoon over left harpoon Underscript k Subscript negative 2 Baseline Overscript k 2 Endscripts EP right-arrow Overscript k 3 Endscripts upper E plus upper P

(6-28)

In this case, most of the enzyme is in the EP form at saturation, and Vmax =k3[Et]upper V Subscript max Baseline equals k 3 left-bracket upper E Subscript t Baseline right-bracket.

It is useful to define a more general rate constant, kcatbold-italic k Subscript bold cat, to describe the limiting rate of any enzyme-catalyzed reaction at saturation. If the reaction has several steps, one of which is clearly rate-limiting, kcatk Subscript cat is equivalent to the rate constant for that limiting step. For the simple reaction of Equation 6-10, kcat =k2k Subscript cat Baseline equals k 2. For the reaction of Equation 6-28, when product release is clearly rate-limiting, kcat =k3k Subscript cat Baseline equals k 3. When several steps are partially rate-limiting, kcatk Subscript cat can become a complex function of several of the rate constants that define each individual reaction step. In the Michaelis-Menten equation, kcat =Vmax/[Et]k Subscript cat Baseline equals upper V Subscript max Baseline slash left-bracket upper E Subscript t Baseline right-bracket. Rearranged, Vmax =kcat[Et],upper V Subscript max Baseline equals k Subscript cat Baseline left-bracket upper E Subscript t Baseline right-bracket comma and Equation 6-9 becomes

V0=kcat[Et] [S]Km+[S]upper V 0 equals StartFraction k Subscript cat Baseline left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-29)

The constant kcatk Subscript cat is a first-order rate constant and hence has units of reciprocal time. It is also called the turnover number. It is equivalent to the number of substrate molecules converted to product in a given unit of time on a single enzyme molecule when the enzyme is saturated with substrate. The turnover numbers of several enzymes are given in Table 6-7.

TABLE 6-7 Turnover Number, kcatbold-italic k Subscript bold cat, of Some Enzymes

Enzyme Substrate kcat (s1)bold-italic k Subscript bold cat Baseline bold left-parenthesis bold s Superscript negative bold 1 Baseline right-parenthesis
Catalase H2O2upper H Subscript 2 Baseline upper O Subscript 2 40,000,000
Carbonic anhydrase HCO3HCO Subscript 3 Superscript minus 400,000
Acetylcholinesterase Acetylcholine 14,000
β-Lactamase Benzylpenicillin 2,000
Fumarase Fumarate 800
RecA protein (an ATPase) ATP 0.5
Comparing Catalytic Mechanisms and Efficiencies

The kinetic parameters kcatk Subscript cat and Kmupper K Subscript m are useful for the study and comparison of different enzymes, whether their reaction mechanisms are simple or complex. Each enzyme has values of kcatk Subscript cat and Kmupper K Subscript m that reflect the cellular environment, the concentration of substrate normally encountered in vivo by the enzyme, and the chemistry of the reaction being catalyzed.

The parameters kcatk Subscript cat and Kmupper K Subscript m also allow us to evaluate the kinetic efficiency of enzymes, but either parameter alone is insufficient for this task. Two enzymes catalyzing different reactions may have the same kcatk Subscript cat (turnover number), yet the rates of the uncatalyzed reactions may be different and thus the rate enhancements brought about by the enzymes may differ greatly. Experimentally, the Kmupper K Subscript m for an enzyme tends to be similar to the cellular concentration of its substrate. An enzyme that acts on a substrate present at a very low concentration in the cell usually has a lower Kmupper K Subscript m than an enzyme that acts on a substrate that is more abundant.

The best way to compare the catalytic efficiencies of different enzymes or the turnover of different substrates by the same enzyme is to compare the ratio kcat/Kmk Subscript cat Baseline slash upper K Subscript m Baseline for the two reactions. This parameter, sometimes called the specificity constant, is the rate constant for the conversion of E + S to E + P. When [S]Kmleft-bracket upper S right-bracket much-less-than upper K Subscript m, Equation 6-29 reduces to the form

V0 =kcatKm [Et] [S]upper V 0 equals StartFraction k Subscript cat Baseline Over upper K Subscript m Baseline EndFraction left-bracket upper E Subscript t Baseline right-bracket left-bracket upper S right-bracket

(6-30)

V0upper V 0 in this case depends on the concentration of two reactants, [Et]left-bracket upper E Subscript t Baseline right-bracket and [S], so this is a second-order rate equation, and the constant kcat/Kmk Subscript cat Baseline slash upper K Subscript m Baseline is a second-order rate constant with units of M1s1upper M Superscript negative 1 Baseline s Superscript negative 1. There is an upper limit to kcat/Kmk Subscript cat Baseline slash upper K Subscript m Baseline, imposed by the rate at which E and S can diffuse together in an aqueous solution. This diffusion-controlled limit is 10810 Superscript 8 to 109 M1s110 Superscript 9 Baseline upper M Superscript negative 1 Baseline s Superscript negative 1, and many enzymes have a kcat/Kmk Subscript cat Baseline slash upper K Subscript m Baseline near this range (Table 6-8). Such enzymes are said to have achieved catalytic perfection. Note that different values of kcatk Subscript cat and Kmupper K Subscript m can produce the maximum ratio.

TABLE 6-8 Enzymes for Which kcat/Kmbold-italic k Subscript bold cat Baseline slash bold-italic upper K Subscript bold m Baseline Is Close to the Diffusion-Controlled Limit (108bold 10 Superscript bold 8 to 109 M1 S1bold 10 Superscript bold 9 Baseline bold upper M Superscript negative 1 Baseline bold upper S Superscript negative 1)

Enzyme Substrate kcat(S1)bold-italic k Subscript bold cat Baseline bold left-parenthesis bold upper S Superscript negative 1 Baseline bold right-parenthesis

Km(M)bold-italic upper K Subscript bold m Baseline left-parenthesis bold upper M bold right-parenthesis

kcat/Km(M1 S1)bold-italic k Subscript bold cat Baseline bold slash bold-italic upper K Subscript bold m Baseline bold left-parenthesis bold upper M Superscript negative 1 Baseline bold upper S Superscript negative 1 Baseline bold right-parenthesis
Acetylcholinesterase Acetylcholine 1.4×1041.4 times 10 Superscript 4 9×1059 times 10 Superscript negative 5 1.6×1081.6 times 10 Superscript 8
Carbonic anhydrase
  • CO2CO Subscript 2
  • HCO3HCO Subscript 3 Superscript minus
  • 1×1061 times 10 Superscript 6
  • 4×1054 times 10 Superscript 5
  • 1.2×1021.2 times 10 Superscript negative 2
  • 2.6×1022.6 times 10 Superscript negative 2
  • 8.3×1078.3 times 10 Superscript 7
  • 1.5×1071.5 times 10 Superscript 7
Catalase H2O2upper H Subscript 2 Baseline upper O Subscript 2 4×1074 times 10 Superscript 7 1.1×1001.1 times 10 Superscript 0 4×1074 times 10 Superscript 7
Crotonase Crotonyl-CoA 5.7×1035.7 times 10 cubed 2×1052 times 10 Superscript negative 5 2.8×1082.8 times 10 Superscript 8
Fumarase
  • Fumarate
  • Malate
  • 8×1028 times 10 squared
  • 9×1029 times 10 squared
  • 5×1065 times 10 Superscript negative 6
  • 2.5×1052.5 times 10 Superscript negative 5
  • 1.6×1081.6 times 10 Superscript 8
  • 3.6×1073.6 times 10 Superscript 7
β-Lactamase Benzylpenicillin 2.0×1032.0 times 10 cubed 2×1052 times 10 Superscript negative 5 1×1081 times 10 Superscript 8

Information from A. Fersht, Structure and Mechanism in Protein Science, p. 166, W. H. Freeman and Company, 1999.

WORKED EXAMPLE 6-1 Determination of Kmbold-italic upper K Subscript bold m

An enzyme is discovered that catalyzes the chemical reaction

SAD  HAPPYSAD right harpoon over left harpoon HAPPY

A team of motivated researchers sets out to study the enzyme, which they call happyase. They find that the kcatk Subscript cat for happyase is 600 s1600 s Superscript negative 1 and they carry out several additional experiments.

When [Et]=20 nMleft-bracket upper E Subscript t Baseline right-bracket equals 20 n upper M and [SAD] = 40 μm, the reaction velocity, V0upper V 0, is 9.6 μM s19.6 mu upper M s Superscript negative 1. Calculate Kmupper K Subscript m for the substrate SAD.

SOLUTION:

We know kcatk Subscript cat, [Et]left-bracket upper E Subscript t Baseline right-bracket, [S], and V0upper V 0. We want to solve for Kmupper K Subscript m. We will first derive an expression for Kmupper K Subscript m, beginning with the Michaelis-Menten equation (Eqn 6-9). We will calculate Vmaxupper V Subscript max by equating it to kcat[Et]k Subscript cat Baseline left-bracket upper E Subscript t Baseline right-bracket. We can then substitute the known values to calculate Kmupper K Subscript m.

V0 =Vmax[S](Km+[S])upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over left-parenthesis upper K Subscript m Baseline plus left-bracket upper S right-bracket right-parenthesis EndFraction

First, multiply both sides by Km+[S]upper K Subscript m Baseline plus left-bracket upper S right-bracket:

   V0(Km+[S])=Vmax[S]V0Km+V0[S]=Vmax[S]StartLayout 1st Row upper V 0 left-parenthesis upper K Subscript m Baseline plus left-bracket upper S right-bracket right-parenthesis equals upper V Subscript max Baseline left-bracket upper S right-bracket 2nd Row upper V 0 upper K Subscript m Baseline plus upper V 0 left-bracket upper S right-bracket equals upper V Subscript max Baseline left-bracket upper S right-bracket EndLayout

Subtract V0[S]upper V 0 left-bracket upper S right-bracket from both sides to give

V0Km =Vmax[S]V0[S]upper V 0 upper K Subscript m Baseline equals upper V Subscript max Baseline left-bracket upper S right-bracket minus upper V 0 left-bracket upper S right-bracket

Factor out [S] and divide both sides by V0upper V 0.

Km=(VmaxV0)[S]V0upper K Subscript m Baseline equals StartFraction left-parenthesis upper V Subscript max Baseline minus upper V 0 right-parenthesis left-bracket upper S right-bracket Over upper V 0 EndFraction

Now, equate Vmaxupper V Subscript max to kcat[Et]k Subscript cat Baseline left-bracket upper E Subscript t Baseline right-bracket, and substitute the given values to obtain Kmupper K Subscript m.

Vmax=kcat[Et]=(600 s1) 0.02 μM=12 μM s1Km=(12 μM s19.6 μM s1) 40 μM /9.6 μM s1=96 μM2 s1/9.6 μM s1 =10 μMStartLayout 1st Row 1st Column upper V Subscript max 2nd Column equals 3rd Column k Subscript cat Baseline left-bracket upper E Subscript t Baseline right-bracket equals left-parenthesis 600 s Superscript negative 1 Baseline right-parenthesis 0.02 mu upper M equals 12 mu upper M s Superscript negative 1 Baseline 2nd Row 1st Column upper K Subscript m 2nd Column equals 3rd Column left-parenthesis 12 mu upper M s Superscript negative 1 Baseline minus 9.6 mu upper M s Superscript negative 1 Baseline right-parenthesis 40 mu upper M slash 9.6 mu upper M s Superscript negative 1 Baseline 3rd Row 1st Column Blank 2nd Column equals 3rd Column 96 mu upper M squared s Superscript negative 1 Baseline slash 9.6 mu upper M s Superscript negative 1 Baseline equals 10 mu upper M EndLayout

Once you have worked with this equation, you will recognize shortcuts to solve problems like this. For example, rearranging Equation 6-9 by simply dividing both sides by Vmaxupper V Subscript max gives

V0Vmax=[S]Km+[S]StartFraction upper V 0 Over upper V Subscript max Baseline EndFraction equals StartFraction left-bracket upper S right-bracket Over upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

Thus, the ratio V0/Vmax =9.6 μM s1/12 μM s1=[S]/(Km+[S])upper V 0 slash upper V Subscript max Baseline equals 9.6 mu upper M s Superscript negative 1 Baseline slash 12 mu upper M s Superscript negative 1 Baseline equals left-bracket upper S right-bracket slash left-parenthesis upper K Subscript m Baseline plus left-bracket upper S right-bracket right-parenthesis. This sometimes simplifies the process of solving for Kmupper K Subscript m, in this case, giving 0.25[S], or 10 μm.

WORKED EXAMPLE 6-2 Determination of [S]

In a separate happyase experiment using [Et]=10 mMleft-bracket upper E Subscript t Baseline right-bracket equals 10 m upper M, the reaction velocity, V0upper V 0, is measured as 3 μM s13 mu upper M s Superscript negative 1. What is the [S] used in this experiment?

SOLUTION:

Using the same logic as in Worked Example 6-1 — equating Vmaxupper V Subscript max to kcat[Et]k Subscript cat Baseline left-bracket upper E Subscript t Baseline right-bracket — we see that Vmaxupper V Subscript max for this enzyme concentration is 6 μM s16 mu upper M s Superscript negative 1. Note that V0upper V 0 is exactly half of Vmaxupper V Subscript max. Recall that Kmupper K Subscript m is by definition equal to the [S] at which V0 =½ Vmaxupper V 0 equals one half upper V Subscript max Baseline. Thus, in this example, the [S] must be the same as Kmupper K Subscript m, or 10 μm. If V0upper V 0 were anything other than ½Vmaxone half upper V Subscript max, it would be simplest to use the expression V0/Vmax =[S]/(Km+[S])upper V 0 slash upper V Subscript max Baseline equals left-bracket upper S right-bracket slash left-parenthesis upper K Subscript m Baseline plus left-bracket upper S right-bracket right-parenthesis to solve for [S].

Many Enzymes Catalyze Reactions with Two or More Substrates

We have seen how [S] affects the rate of a simple enzymatic reaction with only one substrate molecule (SP)left-parenthesis upper S right-arrow upper P right-parenthesis. In most enzymatic reactions, however, two (and sometimes more) different substrate molecules bind to the enzyme and participate in the reaction. Nearly two-thirds of all enzymatic reactions have two substrates and two products. These are generally reactions in which a group is transferred from one substrate to the other, or one substrate is oxidized while the other is reduced. For example, in the reaction catalyzed by hexokinase, ATP and glucose are the substrate molecules, and ADP and glucose 6-phosphate are the products:

ATP+glucoseADP+glucose 6-phosphateATP plus glucose right-arrow Overscript Endscripts ADP plus glucose 6 hyphen phosphate

A phosphoryl group is transferred from ATP to glucose. The rates of such bisubstrate reactions can also be analyzed by the Michaelis-Menten approach. Hexokinase has a characteristic Kmupper K Subscript m for each of its substrates (Table 6-6).

Enzymatic reactions with two substrates proceed by one of several different types of pathways. In some cases, both substrates are bound to the enzyme concurrently at some point in the course of the reaction, forming a noncovalent ternary complex (Fig. 6-15a); the substrates bind in a random sequence or in a specific order. Ordered binding occurs when binding of the first substrate creates a condition, often a conformation change, required for the second substrate to bind. In other cases, the first substrate is converted to product and dissociates before the second substrate binds, so no ternary complex is formed. An example of this is the Ping-Pong, or double-displacement, mechanism (Fig. 6-15b).

A four-part figure, a, b, c, and d, shows four common mechanisms for enzyme-catalyzed bisubstrate reactions. Part a shows an enzyme reaction involving a ternary complex, part b shows an enzyme reaction in which no ternary complex is formed, part c shows ordered bi bi and random bi bi reactions using Cleland nomenclature, and part d shows a ping-pong reaction using Cleland nomenclature.

FIGURE 6-15 Common mechanisms for enzyme-catalyzed bisubstrate reactions. (a) The enzyme and both substrates come together to form a ternary complex. In ordered binding, substrate 1 must bind before substrate 2 can bind productively. In random binding, the substrates can bind in either order. Product dissociation can also be ordered or random. (b) An enzyme-substrate complex forms, a product leaves the complex, the altered enzyme forms a second complex with another substrate molecule, and the second product leaves, regenerating the enzyme. Substrate 1 may transfer a functional group to the enzyme (to form the covalently modified Eupper E prime), which is subsequently transferred to substrate 2. This is called a Ping-Pong or double-displacement mechanism. (c) Ternary complex formation depicted using Cleland nomenclature. In the ordered bi bi and random bi bi reactions shown here, the release of product follows the same pattern as the binding of substrate — both ordered or both random. (d) The Ping-Pong or double-displacement reaction described with Cleland nomenclature.

A shorthand notation developed by W. W. Cleland can be helpful in describing reactions with multiple substrates and products. In this system, referred to as Cleland nomenclature, substrates are denoted A, B, C, and D, in the order in which they bind to the enzyme, and products are denoted P, Q, S, and T, in the order in which they dissociate. Enzymatic reactions with one, two, three, or four substrates are referred to as uni, bi, ter, and quad, respectively. The enzyme is, as usual, denoted E, but if it is modified in the course of the reaction, successive forms are denoted F, G, and so on. The progress of the reaction is indicated with a horizontal line, with successive chemical species indicated below it. If there is an alternative in the reaction path, the horizontal line is bifurcated. Steps involving binding and dissociating substrates and products are indicated with vertical lines.

Common reactions with two substrates and two products (bi bi) are described with the shorthand forms illustrated in Figure 6-15c for an ordered bi bi reaction and a random bi bi reaction. In the latter example, the release of product is also random, as indicated by the two sets of bifurcations. Rarely, the binding of substrates is ordered and the release of products is random, or vice versa, eliminating the bifurcation at one end or the other of the progress line. In a Ping-Pong reaction, lacking a ternary complex, the pathway has a transient second form of the enzyme, F (Fig. 6-15d). This is the form in which a group has been transferred from the first substrate, A, to create a transient covalent attachment to the enzyme. As noted above, such reactions are often called double-displacement reactions, as a group is transferred first from substrate A to the enzyme and then from the enzyme to substrate B. Substrates A and B do not encounter each other on the enzyme.

Michaelis-Menten steady-state kinetics can provide only limited information about the number of steps and intermediates in an enzymatic reaction, but the approach can be used to distinguish between pathways that have a ternary intermediate and pathways — including Ping-Pong pathways — that do not (Fig. 6-16). As we will see when we consider enzyme inhibition, steady-state kinetics can also distinguish between ordered and random binding of substrates and products in reactions with ternary intermediates.

A two-part figure, a and b, shows Lineweaver-Burk plots that illustrate a steady-state kinetic analysis of bisubstrate reactions. Part a shows a reaction involving a ternary complex and part b shows a reaction involving a ping-pong pathway.

FIGURE 6-16 Steady-state kinetic analysis of bisubstrate reactions. In these double-reciprocal plots, the concentration of substrate 1 is varied while the concentration of substrate 2 is held constant. This is repeated for several values of [S2]left-bracket upper S Subscript 2 Baseline right-bracket, generating several separate lines. (a) Intersecting lines indicate that a ternary complex is formed in the reaction; (b) parallel lines indicate a Ping-Pong (double-displacement) pathway.

Enzyme Activity Depends on pH

In general, steady-state kinetics provides information required to characterize an enzyme and assess its catalytic efficiency. Additional information can be gained by examination of how the key experimental parameters kcatk Subscript cat and kcat/Kmk Subscript cat Baseline slash upper K Subscript m Baseline change when reaction conditions change, particularly pH. Enzymes have an optimum pH (or pH range) at which their activity is maximal (Fig. 6-17); at higher or lower pH, activity decreases. This is not surprising. Amino acid side chains in the active site may act as weak acids and bases only if they maintain a certain state of ionization. Elsewhere in the protein, removing a proton from a His residue, for example, might eliminate an ionic interaction that is essential for stabilizing the active conformation of the enzyme. A less common cause of pH sensitivity is titration of a group on the substrate.

A graph plots p H on the horizontal axis against log V subscript 0 on the vertical axis for two enzymes, pepsin and glucose 6-phosphate.

FIGURE 6-17 The pH-activity profiles of two enzymes. These curves are constructed from measurements of initial velocities when the reaction is carried out in buffers of different pH. Because pH is a logarithmic scale reflecting 10-fold changes in [H+]left-bracket upper H Superscript plus Baseline right-bracket, the changes in V0upper V 0 are also plotted on a logarithmic scale. The pH optimum for the activity of an enzyme is generally close to the pH of the environment in which the enzyme is normally found. Pepsin, a peptidase found in the stomach, has a pH optimum of about 1.6. The pH of gastric juice is between 1 and 2. Glucose 6-phosphatase of hepatocytes (liver cells), with a pH optimum of about 7.8, is responsible for releasing glucose into the blood. The normal pH of the cytosol of hepatocytes is about 7.2.

The pH range over which an enzyme undergoes changes in activity can provide a clue to the type of amino acid residue involved (see Table 3-1). A change in activity near pH 7.0, for example, often reflects titration of a His residue. The effects of pH must be interpreted with some caution, however. In the closely packed environment of a protein, the pKap upper K Subscript a Baseline of amino acid side chains can be significantly altered. For example, a nearby positive charge can lower the pKap upper K Subscript a Baseline of a Lys residue, and a nearby negative charge can increase it. Such effects sometimes result in a pKap upper K Subscript a Baseline that is shifted by several pH units from its value in the free amino acid. In the enzyme acetoacetate decarboxylase, for example, one Lys residue has a pKap upper K Subscript a Baseline of 6.6 (compared with 10.5 in free lysine) due to electrostatic effects of nearby positive charges.

Pre–Steady State Kinetics Can Provide Evidence for Specific Reaction Steps

The mechanistic insight provided by steady-state kinetics can be augmented, sometimes dramatically, by an examination of the pre–steady state. Consider an enzyme with a reaction mechanism that conforms to the scheme in Equation 6-28, featuring three steps:

E+S k1k1 ES k2k2 EP k3 E+Pupper E plus upper S right harpoon over left harpoon Underscript k Subscript negative 1 Baseline Overscript k 1 Endscripts ES right harpoon over left harpoon Underscript k Subscript negative 2 Baseline Overscript k 2 Endscripts EP right-arrow Overscript k 3 Endscripts upper E plus upper P

Overall catalytic efficiency for this reaction can be assessed with steady-state kinetics, but the rates of the individual steps cannot be determined in this way, and the slow (rate-limiting) step can rarely be identified. For the rate constants of individual steps to be measured, the reaction must be studied during its pre–steady state. The first turnover of an enzyme-catalyzed reaction often occurs in seconds or milliseconds, so researchers use special equipment that allows mixing and sampling on this time scale (Fig. 6-18a). Reactions are stopped and protein-bound products are quantified, after the timed addition and rapid mixing of an acid that denatures the protein and releases all bound molecules. A detailed description of pre–steady state kinetics is beyond the scope of this text, but we can illustrate the power of this approach by a simple example of an enzyme that uses the pathway shown in Equation 6-28. This example also involves an enzyme that catalyzes a relatively slow reaction, so the pre–steady state is more conveniently observed.

A three-part figure, a, b, and c, shows how pre-steady state kinetics are studied. Part a shows a rapid-mixing device for combining the enzyme and substrate, part b shows a graph of experimental data plotting concentration against time, and part c is a plot of burst amplitude against concentration of E.

FIGURE 6-18 Pre–steady state kinetics. The transient phase that constitutes the pre–steady state often exists for mere seconds or milliseconds, requiring specialized equipment to monitor it. (a) A simple schematic for a rapid-mixing device, called a stopped-flow device. Enzyme (E) and substrate (S) are mixed with the aid of mechanically operated syringes. The reaction is quenched at a programmed time by adding a denaturing acid through another syringe, and the amount of product formed is measured, in this case with a spectrophotometer. (b) Experimental data for an enzyme reaction show the pre–steady state occurring in the first 5 to 10 seconds. This is a relatively slow reaction and is used as an example because the steady state can be conveniently monitored. The slope of the lines after 15 seconds reflects the steady state. Extrapolating this slope back to zero time (dashed lines) gives the amplitude of the burst phase. The progress of the reaction during the pre–steady state primarily reflects the chemical steps in the reaction (details of which are not shown). The presence of a burst implies that a step following the chemical step that produces P is rate-limiting — in this case, the product-release step. Notice that the extrapolated intercept at time = 0 increases as [E] increases. (c) A plot of burst amplitude (the intercepts from (b)) versus [E] shows that one molecule of P is formed in each active site during the burst (pre–steady state) phase. This provides evidence that product release is the rate-limiting step, because it is the only step following product formation in this simple enzymatic reaction. The enzyme used in this experiment was RNase P, one of the catalytic RNAs described in Chapter 26. [(b, c) Data from J. Hsieh et al., RNA 15:224, 2009.]

For many enzymes, dissociation of product is rate-limiting. In this example (Fig. 6-18b, c), the rate of dissociation of the product (k3)left-parenthesis k 3 right-parenthesis is slower than the rate of its formation (k2)left-parenthesis k 2 right-parenthesis. Product dissociation therefore dictates the rates observed in the steady state. How do we know that k3k 3 is rate-limiting? A slow k3k 3 gives rise to a burst of product formation in the pre–steady state, because the preceding steps are relatively fast. The burst reflects the rapid conversion of one molecule of substrate to one molecule of product at each enzyme active site. The observed rate of product formation soon slows to the steady-state rate as the bound product is slowly released. Each enzymatic turnover after the first one must proceed through the slow product-release step. However, the rapid generation of product in that first turnover provides much information. The amplitude of the burst — when one molecule of product is generated per molecule of enzyme present (Fig. 6-18c), measured by extrapolating the steady-state progress line back to zero time — is the highest amplitude possible. This provides one piece of evidence that product release is, indeed, rate-limiting. The rate constant for the chemical reaction step, k2k 2, can be derived from the observed rate of the burst phase.

Of course, enzymes do not always conform to the simple reaction scheme of Equation 6-28. Formally, the observation of a burst indicates that a rate-limiting step (typically, product release, or an enzyme conformational change, or another chemical step) occurs after formation of the product being monitored. Additional experiments and analysis can often define the rates of each step in a multistep enzymatic reaction. Some examples of the application of pre–steady state kinetics are included in the descriptions of specific enzymes in Section 6.4.

Enzymes Are Subject to Reversible or Irreversible Inhibition

Enzyme inhibitors are molecules that interfere with catalysis, slowing or halting enzymatic reactions. Enzymes catalyze virtually all cellular processes, so it should not be surprising that enzyme inhibitors are among the most important pharmaceutical agents known. For example, aspirin (acetylsalicylate) inhibits the enzyme that catalyzes the first step in the synthesis of prostaglandins, compounds involved in many processes, including some that produce pain. The study of enzyme inhibitors also has provided valuable information about enzyme mechanisms and has helped define some metabolic pathways. There are two broad classes of enzyme inhibitors: reversible and irreversible.

Reversible Inhibition

One common type of reversible inhibition is called competitive (Fig. 6-19a). A competitive inhibitor competes with the substrate for the active site of an enzyme. While the inhibitor (I) occupies the active site, the substrate is excluded, and vice versa. Many competitive inhibitors are structurally similar to the substrate and combine with the enzyme to form an unreactive EI complex. Even fleeting combinations of this type will reduce the efficiency of an enzyme. Competitive inhibition can be analyzed quantitatively by steady-state kinetics. In the presence of a competitive inhibitor, the Michaelis-Menten equation (Eqn 6-9) becomes

V0 =Vmax[S]αKm +[S]upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over alpha upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction

(6-31)

where

α=1+[I]KI    and    KI=[E] [I][EI]alpha equals 1 plus StartFraction left-bracket upper I right-bracket Over upper K Subscript upper I Baseline EndFraction and upper K Subscript upper I Baseline equals StartFraction left-bracket upper E right-bracket left-bracket upper I right-bracket Over left-bracket EI right-bracket EndFraction

Equation 6-31 describes the important features of competitive inhibition. The experimentally determined variable αKmalpha upper K Subscript m, the Kmupper K Subscript m observed in the presence of the inhibitor, is often called the “apparent” Kmupper K Subscript m.

A three-part figure, a, b, and c, describes competitive inhibition. Part a shows how competitive inhibitors interact with enzymes, part b plots v subscript 0 against concentration of substrate, and part c is a Lineweaver Burk plot.

FIGURE 6-19 Competitive inhibition. (a) Competitive inhibitors bind to the enzyme’s active site; KIupper K Subscript upper I is the equilibrium dissociation constant for inhibitor binding to E. (b) Competitive inhibitors affect the observed Kmupper K Subscript m, but not the Vmaxupper V Subscript max. This is readily evident in the plot of V0upper V 0 versus [S]. (c) In a Lineweaver-Burk plot, the lines generated + and − inhibitor intersect on the y axis, which reflects 1/Vmax1 slash upper V Subscript max Baseline. As the observed Kmupper K Subscript m increases in the presence of an inhibitor, the intercept on the x axis (1/Km)left-parenthesis negative 1 slash upper K Subscript m Baseline right-parenthesis moves to the right.

Bound inhibitor does not inactivate the enzyme. When the inhibitor dissociates, substrate can bind and react. Because the inhibitor binds reversibly to the enzyme, the competition can be biased to favor the substrate simply by adding more substrate. When [S] far exceeds [I], the probability that an inhibitor molecule will bind to the enzyme is minimized and the reaction exhibits normal Vmaxupper V Subscript max. However, in the presence of inhibitor, the [S] at which V0 =½Vmaxupper V 0 equals one half upper V Subscript max Baseline, the apparent Kmupper K Subscript m, increases in the presence of inhibitor by the factor α (Fig. 6-19b). This effect on apparent Kmupper K Subscript m, combined with the absence of an effect on Vmaxupper V Subscript max, is diagnostic of competitive inhibition and is readily revealed in a double-reciprocal plot (Fig. 6-19c). The equilibrium constant for inhibitor binding, KIupper K Subscript upper I, can be obtained from the same plot.

A medical therapy based on competition at the active site is used to treat patients who have ingested methanol, a solvent found in gas-line antifreeze. The liver enzyme alcohol dehydrogenase converts methanol to formaldehyde, which is damaging to many tissues. Blindness is a common result of methanol ingestion, because the eyes are particularly sensitive to formaldehyde. Ethanol competes effectively with methanol as an alternative substrate for alcohol dehydrogenase. The effect of ethanol is much like that of a competitive inhibitor, with the distinction that ethanol is also a substrate for alcohol dehydrogenase and its concentration will decrease over time as the enzyme converts it to acetaldehyde. The therapy for methanol poisoning is slow intravenous infusion of ethanol, at a rate that maintains a controlled concentration in the blood for several hours. This slows the formation of formaldehyde, lessening the danger while the kidneys filter out the methanol to be excreted harmlessly in the urine.

Two other types of reversible inhibition, uncompetitive and mixed, can be defined in terms of one-substrate enzymes, but in practice are observed only with enzymes having two or more substrates. An uncompetitive inhibitor (Fig. 6-20a) binds at a site distinct from the substrate active site and, unlike a competitive inhibitor, binds only to the ES complex. In the presence of an uncompetitive inhibitor, the Michaelis-Menten equation is altered to

V0=Vmax[S]Km +α[S]upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over upper K Subscript m Baseline plus alpha Superscript prime Baseline left-bracket upper S right-bracket EndFraction

(6-32)

where

α=1+[I]KI    and    KI=[ES] [I][ESI]alpha prime equals 1 plus StartFraction left-bracket upper I right-bracket Over upper K prime Subscript upper I EndFraction and upper K prime Subscript upper I Baseline equals StartFraction left-bracket ES right-bracket left-bracket upper I right-bracket Over left-bracket ESI right-bracket EndFraction

As described by Equation 6-32, at high concentrations of substrate, V0upper V 0 approaches Vmax/αupper V Subscript max Baseline slash alpha prime. Thus, an uncompetitive inhibitor lowers the measured Vmaxupper V Subscript max. Apparent Kmupper K Subscript m also decreases, because the [S] required to reach one-half Vmaxupper V Subscript max decreases by the factor αalpha prime (Fig. 6-20b, c). This behavior can be explained as follows. Because the enzyme is inactive when the uncompetitive inhibitor is bound, but the inhibitor is not competing with substrate for binding, the inhibitor effectively removes some fraction of the enzyme molecules from the reaction. Given that Vmaxupper V Subscript max depends on [E], the observed Vmaxupper V Subscript max decreases. Given that the inhibitor binds only to the ES complex, only ES (not free enzyme) is deleted from the reaction, so the [S] needed to reach ½Vmaxone half upper V Subscript max — that is, Kmupper K Subscript m — declines by the same amount.

A three-part figure, a, b, and c, describes uncompetitive inhibition. Part a shows how uncompetitive inhibitors interact with enzymes, part b plots v subscript 0 against concentration of substrate, and part c is a Lineweaver Burk plot.

FIGURE 6-20 Uncompetitive inhibition. (a) Uncompetitive inhibitors bind at a separate site, but bind only to the ES complex; KIupper K prime Subscript upper I is the equilibrium constant for an inhibitor binding to ES. (b) In the presence of an uncompetitive inhibitor, both the Kmupper K Subscript m and the Vmaxupper V Subscript max decline, and by equivalent factors. In the V0upper V 0 versus [S] plot, the decline in Vmaxupper V Subscript max is somewhat easier to discern than the decline in Kmupper K Subscript m. (c) The Lineweaver-Burk plot for an uncompetitive inhibitor is quite diagnostic, as the lines generated in the presence and absence of an inhibitor are parallel. Note that the lines seen in the presence of an inhibitor are always above the line generated in the absence of an inhibitor, reflecting the decline in both Kmupper K Subscript m and Vmaxupper V Subscript max brought about by the inhibitor (i.e., the intercepts move up on the y axis and to the left on the x axis).

A mixed inhibitor (Fig. 6-21a) also binds at a site distinct from the substrate active site, but it binds to either E or ES. The rate equation describing mixed inhibition is

V0=Vmax[S]αKm+α[S]upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over alpha upper K Subscript m Baseline plus alpha Superscript prime Baseline left-bracket upper S right-bracket EndFraction

(6-33)

where α and αalpha prime are defined as above. A mixed inhibitor usually affects both Kmupper K Subscript m and Vmaxupper V Subscript max (Fig. 6-21b, c). Vmaxupper V Subscript max is affected because the inhibitor renders some fraction of the available enzyme molecules inactive, lowering the effective [E] on which Vmaxupper V Subscript max depends. The Kmupper K Subscript m may increase or decrease, depending on which enzyme form, E or ES, the inhibitor binds to most strongly. The special case of α=αalpha equals alpha prime, rarely encountered in experiments, historically has been defined as noncompetitive inhibition. Examine Equation 6-33 to see why a noncompetitive inhibitor would affect the Vmaxupper V Subscript max but not the Kmupper K Subscript m.

A three-part figure, a, b, and c, describes mixed inhibition. Part a shows how mixed inhibitors interact with enzymes, part b plots v subscript 0 against concentration of substrate, and part c is a Lineweaver Burk plot.

FIGURE 6-21 Mixed inhibition. (a) Mixed inhibitors bind at a separate site, but they may bind to either E or ES. (b) The kinetic patterns brought about by a mixed inhibitor are complex. The Vmaxupper V Subscript max always declines. The Kmupper K Subscript m may increase or decrease, depending upon the relative values of α and αalpha prime. (c) In a Lineweaver-Burk plot, the lines generated + and − inhibitor always intersect, but not on an axis. The intercepts on the y axis always move up, as the Vmaxupper V Subscript max declines. The lines may intersect either above or below the x axis. When they intersect above, as shown here, α>αalpha greater-than alpha prime and the observed Kmupper K Subscript m is increasing. When the lines intersect below the x axis (not shown), α<αalpha less-than alpha prime and the observed Kmupper K Subscript m is decreasing.

Equation 6-33 is a general expression for the effects of reversible inhibitors, simplifying to the expressions for competitive inhibition and uncompetitive inhibition when α=1.0alpha prime equals 1.0 or α = 1.0, respectively. From this expression we can summarize the effects of inhibitors on individual kinetic parameters. For all reversible inhibitors, the apparent Vmax=Vmax/αupper V Subscript max Baseline equals upper V Subscript max Baseline slash alpha prime, because the right side of Equation 6-33 always simplifies to Vmax/αupper V Subscript max Baseline slash alpha prime at sufficiently high substrate concentrations. For competitive inhibitors, α=1.0alpha prime equals 1.0 and can thus be ignored. Taking this expression for apparent Vmaxupper V Subscript max, we can also derive a general expression for apparent Kmupper K Subscript m to show how this parameter changes in the presence of reversible inhibitors. Apparent Kmupper K Subscript m, as always, equals the [S] at which V0upper V 0 is one-half apparent Vmaxupper V Subscript max or, more generally, when V0=Vmax/2αupper V 0 equals upper V Subscript max Baseline slash 2 alpha prime. This condition is met when [S]=αKm/αleft-bracket upper S right-bracket equals alpha upper K Subscript m Baseline slash alpha prime. Thus, apparent Km =αKm/αupper K Subscript m Baseline equals alpha upper K Subscript m Baseline slash alpha prime. The terms α and αalpha prime reflect the binding of inhibitor to E and ES, respectively. Thus, the term αKm/αalpha upper K Subscript m Baseline slash alpha prime is a mathematical expression of the relative affinity of inhibitor for the two enzyme forms. This expression is simpler when either α or αalpha prime is 1.0 (for uncompetitive or competitive inhibitors), as summarized in Table 6-9.

TABLE 6-9 Effects of Reversible Inhibitors on Apparent Vmaxbold-italic upper V Subscript bold max and Apparent Kmbold-italic upper K Subscript bold m

Inhibitor type Apparent Vmaxbold-italic upper V Subscript bold max Apparent Kmbold-italic upper K Subscript bold m
None Vmaxupper V Subscript max Kmupper K Subscript m
Competitive Vmaxupper V Subscript max αKmalpha upper K Subscript m
Uncompetitive Vmax/αupper V Subscript max Baseline slash alpha prime Km/αupper K Subscript m Baseline slash alpha prime
Mixed Vmax/αupper V Subscript max Baseline slash alpha prime αKm/αalpha upper K Subscript m Baseline slash alpha prime

In practice, uncompetitive inhibition and mixed inhibition are observed only for enzymes with two or more substrates — say, S1upper S Subscript 1 and S2upper S Subscript 2 — and are very important in the experimental analysis of such enzymes. If an inhibitor binds to the site normally occupied by S1upper S Subscript 1, it may act as a competitive inhibitor in experiments in which [S1]left-bracket upper S Subscript 1 Baseline right-bracket is varied. If an inhibitor binds to the site normally occupied by S2upper S Subscript 2, it may act as a mixed or uncompetitive inhibitor of S1upper S Subscript 1. The actual inhibition patterns observed depend on whether the S1upper S Subscript 1- and S2upper S Subscript 2-binding events are ordered or random, and thus the order in which substrates bind and products leave the active site can be determined. Product inhibition experiments in which one of the reaction products is provided as an inhibitor are often particularly informative. If only one of two reaction products is present, no reverse reaction can take place. However, a product generally binds to some part of the active site and can thus serve as an effective inhibitor. Enzymologists can combine steady-state kinetic studies involving different combinations and amounts of products and inhibitors with pre–steady state analysis to develop a detailed picture of the mechanism of a bisubstrate reaction.

WORKED EXAMPLE 6-3 Effect of Inhibitor on Kmbold-italic upper K Subscript bold m

The researchers working on happyase (see Worked Examples 6-1 and 6-2) discover that the compound STRESS is a potent competitive inhibitor of happyase. Addition of 1 nm STRESS increases the measured Kmupper K Subscript m for SAD by a factor of 2. What are the values for α and αalpha prime under these conditions?

SOLUTION:

Recall that the apparent Kmupper K Subscript m, the Kmupper K Subscript m measured in the presence of a competitive inhibitor, is defined as αKmalpha upper K Subscript m. Because Kmupper K Subscript m for SAD increases by a factor of 2 in the presence of 1 nmSTRESS, the value of α must be 2. The value of αalpha prime for a competitive inhibitor is 1, by definition.

Irreversible Inhibition

The irreversible inhibitors bind covalently with or destroy a functional group on an enzyme that is essential for the enzyme’s activity, or they form a highly stable noncovalent association. Formation of a covalent link between an irreversible inhibitor and an enzyme is a particularly effective way to inactivate an enzyme. Irreversible inhibitors are another useful tool for studying reaction mechanisms. Amino acids with key catalytic functions in the active site can sometimes be identified by determining which residue is covalently linked to an inhibitor after the enzyme is inactivated. An example is shown in Figure 6-22.

A figure shows changes in the enzyme chymotrypsin associated with irreversible inhibition.

FIGURE 6-22 Irreversible inhibition. Reaction of chymotrypsin with diisopropylfluorophosphate (DIFP) modifies Ser195Ser Superscript 195 and irreversibly inhibits the enzyme. This has led to the conclusion that Ser195Ser Superscript 195 is the key active-site Ser residue in chymotrypsin.

A special class of irreversible inhibitors is the suicide inactivators. These compounds are relatively unreactive until they bind to the active site of a specific enzyme. A suicide inactivator undergoes the first few chemical steps of the normal enzymatic reaction, but instead of being transformed into the normal product, the inactivator is converted to a very reactive compound that combines irreversibly with the enzyme. These compounds are also called mechanism-based inactivators, because they hijack the normal enzyme reaction mechanism to inactivate the enzyme. Suicide inactivators play a significant role in rational drug design, an approach to obtaining new pharmaceutical agents in which chemists synthesize novel substrates based on knowledge of substrates and reaction mechanisms. A well-designed suicide inactivator is specific for a single enzyme and is unreactive until it is within that enzyme’s active site, so drugs based on this approach can offer the important advantage of few side effects (Box 6-1).

An irreversible inhibitor need not bind covalently to the enzyme. Noncovalent binding is enough, if that binding is so tight that the inhibitor dissociates only rarely. How does a chemist develop a tight-binding inhibitor? Recall that enzymes evolve to bind most tightly to the transition states of the reactions that they catalyze. In principle, if one can design a molecule that looks like that reaction transition state, it should bind tightly to the enzyme. Even though transition states cannot be observed directly, chemists can often predict the approximate structure of a transition state based on accumulated knowledge about reaction mechanisms. Although the transition state is by definition transient and thus unstable, in some cases stable molecules can be designed that resemble transition states. These are called transition-state analogs. They bind to an enzyme more tightly than does the substrate in the ES complex, because they fit into the active site better (that is, they form a greater number of weak interactions) than the substrate itself.

The idea of transition-state analogs was suggested by Linus Pauling in the 1940s, and it has been explored using a variety of enzymes. For example, transition-state analogs designed to inhibit the glycolytic enzyme aldolase bind to that enzyme more than four orders of magnitude more tightly than do its substrates (Fig. 6-23). A transition-state analog cannot perfectly mimic a transition state. Some analogs, however, bind to a target enzyme 10210 squared to 10810 Superscript 8 times more tightly than does the normal substrate, providing good evidence that enzyme active sites are indeed complementary to transition states. The concept of transition-state analogs is important to the design of new pharmaceutical agents. As we shall see in Section 6.4, the powerful anti-HIV drugs called protease inhibitors were designed in part as tight-binding transition-state analogs.

A figure shows a transition state analog that resembles the proposed enediolate transition state produced during glycolysis.

FIGURE 6-23 A transition-state analog. In glycolysis, a class II aldolase (found in bacteria and fungi) catalyzes the cleavage of fructose 1,6-bisphosphate to form glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. The reaction proceeds via a reverse aldol-like mechanism. The compound phosphoglycolohydroxamate, which resembles the proposed enediolate transition state, binds to the enzyme nearly 10,000 times better than does the dihydroxyacetone phosphate product.

SUMMARY 6.3 Enzyme Kinetics as an Approach to Understanding Mechanism