In the Classroom
Enzyme Catalysis and the Gibbs Energy Addison Ault Department of Chemistry, Cornell College, Mount Vernon, IA 52314;
[email protected] I read with interest Brian Bozlee’s recent article “Reformulation of the Michaelis–Menten Equation: How EnzymeCatalyzed Reactions Depend on Gibbs Energy” (1). In this article he considers the paradoxical result of a change in the energy of the enzyme–substrate complex, ES. “Is it possible that raising G3 could both speed and slow the rate of product formation?” (1, p 106). We can distinguish two rates for an enzyme-catalyzed reaction. One is the rate at low substrate concentration (extrapolated to zero substrate concentration), and the second is the rate at high substrate concentration (extrapolated to infinite substrate concentration). When we make this distinction we can resolve the paradox presented in Bozlee’s article. Reaction of Interest We can represent the reaction of interest (1) as
E+S
k1
k−1
ES
k2
E+P
where E is the enzyme, S is the substrate, and P is the product. The unusual, but characteristic, feature of the typical enzymecatalyzed reaction is that the initial rate increases with increased substrate concentration, but at high substrate concentrations the initial rate approaches a maximum that is independent of further increases in the concentration of the substrate. This behavior is illustrated by the example in Figure 2 of Bozlee’s article (1, p 107). Gibbs-Energy Profile The connections between rates and Gibbs energies are often summarized as a plot of Gibbs energy versus “progress of reaction”. While such a graph, or “Gibbs-energy profile”, is a vast simplification of what actually happens, it does represent in a useful way the average behavior of the reacting species. Bozlee shows such a graph in Figure 1 of his article (1, p 106). Kinetics of a Simple Enzyme-Catalyzed Reaction The kinetic behavior of the reaction1 of interest can be represented by the following equation in which [E0] is sum of the concentrations of all forms of the enzyme, and [S] is the concentration of unbound substrate (2–4): 1 rate = k2 [E 0 ] k−1 + k2 (1) 1 + k1 [S ] When the constants in the second term of the denominator are combined and designated as KM, k + k2 KM = −1 (2) k1 the rate equation can be rewritten as
rate = k2 [E 0 ]
1 K 1 + M [S ]
(3)
Rate at High [S] We can see that when [S] becomes high the term in the parentheses, which is always fractional, approaches 1. Under these conditions [E0] is equal to [ES] because [E] has been driven down by the large excess of S, which converts E to ES. The equation can then be rewritten as
rate = k2 [E 0 ]
(4)
This corresponds to Bozlee’s eq 2, in which [E]T corresponds to [E0] . However, where Bozlee says that “[E]T is the total concentration of bound and unbound enzyme” we must remember that here [E]T is actually [ES] alone since, at high [S], “all” the enzyme is present as ES and “none” is present as E. This increase and then leveling-off of the initial rate with increasing [S] is sometimes called a “saturation” effect; E becomes “saturated” with S. Rate at Low [S] When [S] is low the second term in the denominator of eq 1 becomes large and eq 1 can then be rewritten as
rate =
k2 k1 [E 0 ][S ] k−1 + k2
(5)
Alternatively, eq 3 can be rewritten as
rate =
k2 [E 0 ][S ] KM
(6)
These equations indicate that when [S] is low the initial rate of the enzyme-catalyzed reaction will be proportional to [S]. Under these conditions, when [S] is low, [E0] is equal to [E] because “all” the enzyme is present as E and “none” as ES. Gibbs Energy at High [S] In Figure 1 the vertical distance k2 is the Gibbs energy of activation for the rate at high [S], conditions under which the experimental first-order rate constant equals k2. The vertical distance k2 is the Gibbs energetic distance from the “higher valley” to the “high pass” on the path to product. The reaction appears to start at the “higher valley” because the great “pressure” of S when [S] is high forces S onto E to give ES; all E is ES. Gibbs Energy at Low [S] In Figure 1 the vertical distance k2k1/k‒1 is the Gibbs energy of activation for the rate at low [S], conditions under which the experimental second-order rate constant is approximately k2k1/k‒1. When Is This Approximation Valid? The experimental second-order rate constant, from eq 5, is exactly k2k1/(k‒1 + k2), or, from eq 6, it is exactly k2/KM. The approximation of the experimental second-order rate constant as k2k1/k‒1 is valid when we can assume that the steady-state con-
© Division of Chemical Education • www.JCE.DivCHED.org • Vol. 86 No. 9 September 2009 • Journal of Chemical Education
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In the Classroom
centration of ES is the same as what would be the equilibrium concentration of ES if no ES could go on over the “high pass”. Remembering that the Gibbs energy scale is a log scale, the apparent 5-fold difference in vertical distance from the “higher valley” to the transition state, ts2, (and on to product) and the vertical distance from the “higher valley” to ts1 (and back to starting materials) corresponds to a 100,000-fold difference in rate constant. If we take k‒1 as relative 1, k2 would be relative 0.00001. Setting k2k1∙(k‒1 + k2) equal to k2k1/k‒1 is then an excellent approximation. Why Do We Need a “Higher Valley”? If there were no “higher valley”, and only the “high pass”, the rate of the reaction would be proportional to [S] at all concentrations of S. We would not see the “saturation” effect that characterizes enzymatic reactions. How Important Is the Gibbs Energy of the “Higher Valley”? The “higher valley”, indicated by G3 in Bozlee’s Figure 1 (1, p 106), simply divides the total Gibbs activation energy under conditions of low [S] into two parts. The first part is the Gibbs energy difference (k1/k‒1 in Figure 1) that corresponds to the pre-equilibrium, actually steady-state, in which ES is formed. The second part is the Gibbs activation energy (k2 in Figure 1) that corresponds to the further reaction of ES, the product of the pre-equilibrium. It really does not matter where the division is made. The sum of the two parts will be the same; k2k1/k‒1 would be unchanged because a change in k2 would be exactly balanced by a compensating change in k1/k‒1. How Important Is the Gibbs Energy of the Transition State? The Gibbs energy of ts1, indicated as G2 in Bozlee’s Figure 1 (1, p 106), does not enter the calculation. The Gibbs energy of ts1 should be higher than the Gibbs energy of the “higher valley” (otherwise there would be no valley), but not so high that it approaches the energy of ts2. If the Gibbs energy of ts1 were to approach the Gibbs energy of ts2 the rate of equilibration of
ts2
Gibbs Energy
G4
k2
k2 k1 k∙1 G3
k1 k∙1
G1
E∙S
k1 k∙1
ES
k2
E∙P
Progress Figure 1. Plot of Gibbs energy for the progress of a standard reaction.
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Conclusions These conclusions apply only to the very first part of the reaction; they apply only to the initial rate, the first moments of the reaction. The initial rate is typically calculated by extrapolation back to zero time.
1. The rate-limiting process under all conditions is passage through the transition state of highest Gibbs energy, the “high pass”.
2. At high [S] the Gibbs energy of activation is the Gibbs energy difference between the “higher valley” and the “high pass”. At high [S] all E starts as ES from the “higher valley” to go to the “high pass”. The “higher valley” acts as a “base camp” that gives E a “head start” toward the “high pass”. All E gets pushed up into the high valley as ES by the huge excess of S.
3. At low [S] the Gibbs energy of activation is the Gibbs energy difference between the “lower valley” and the “high pass”. This Gibbs energy difference equals the sum of the both the Gibbs energy difference between the “lower valley” and the “higher valley” and the Gibbs energy difference between the “higher valley” and the “high pass”. At low [S] all E starts from the “lower valley” to go to the “high pass.” At low [S] the concentration of ES at the “base camp” is negligible.
4. When [S] = KM the initial rate, as we can see from eq 3, will be half of the maximum possible initial rate, the rate at high [S]. We account for this by saying that when [S] = KM only half of the enzyme is present as ES. The implication is that the half of the enzyme present as E makes a negligible contribution to the rate under these conditions.
Summary We can summarize in this way:
• One, above, is always true: the highest energy point on the path to product is the “high pass”.
• Two, above, is the special case when [S] is so high that in the steady state “all” the enzyme is present as [ES]. Here, the trip over the highest pass starts from the “high valley”.
• Three, above, is the special case when [S] is so low that in the steady state “all” the enzyme is present as [E]. This time the trip over the highest pass starts from the “low valley”.
We see, then, that there are two “special” rates for an enzymecatalyzed reaction, a low (small [S]) rate and a high (large [S]) rate. We need to think of them separately.
ts1
G2
E with ES would decrease, and if the Gibbs energy of ts1 were to exceed the Gibbs energy of ts2 the reaction of E with S to give ES would become the rate-limiting step of the reaction.
Resolution of the Paradox Realizing now that there are two extreme rates for an enzyme-catalyzed reaction, a low (small [S]) rate, and a high (large [S]) rate, we are ready to answer Bozlee’s question: “Is it possible that raising G3 could both speed and slow the rate of product formation?” (1, p 106). Effect on Vmax As Bozlee says (1), the fast (high [S]) rate will increase Vmax because this change will move the “higher valley”, or G3, closer to the “high pass”, or G4. The difference between G3 and G4 will be less; k2 will be greater.
Journal of Chemical Education • Vol. 86 No. 9 September 2009 • www.JCE.DivCHED.org • © Division of Chemical Education
In the Classroom
Effect on the Initial Rate There will, however, be no effect on the rate of the slow (low [S]) rate because a change in G3 does not change the energetic distance between the “low valley” and the “high pass”, the difference between G1 and G4. The second-order rate constant will still be k2k1/k‒1, which equals k2/KM, as indicated in eq 4. Effect on KM A slight upward adjustment of the “high valley”, a slight increase in G3, will not only increase k2, but it will also increase k‒1. Since k1 will not be changed by this adjustment, this adjustment will also increase KM since k KM ≈ −1 (7) k1 We can see this also by looking at the Gibbs-energy profile. If the “high valley” is given a slight upward adjustment the ES/E ratio in the steady state will become less favorable, and a greater concentration of substrate will be required to bring the rate up to half of the new, higher Vmax. Bozlee’s Figure 2 illustrates this nicely (1, p 107). You Cannot Get Something for Nothing Although a slight upward adjustment of the “high valley” will increase Vmax, which can be perceived to be a “benefit”, there will be, however, an associated “cost”. That cost is the need for a greater concentration of substrate S to raise the rate of the reaction to any particular fraction of Vmax. Thus to achieve a certain rate, determined by the height of the highest pass, there can be different combinations of concentration and rate constant, and the price of a more favorable rate constant is the requirement for a higher concentration. Comparison with Acid Catalysis We see this effect in many reactions, and the acid-catalyzed hydrolysis of an ester is a familiar example. As the reaction mixture is made more acidic, the reaction speeds up because in a more acidic solution there is a higher concentration of a more reactive species, the “ester–proton complex”. The price paid for this benefit is a higher concentration of hydronium ion, a price the chemist is quite willing to pay. However, since esters are only weakly basic, we do not observe “saturation”. Pedagogy Three basic ideas must be kept in mind while thinking about connections between the rates of reactions and the Gibbs energies of intermediates and transition states.
• The rate and the rate constant are not the same thing.
The rate is one or more concentrations times the rate constant. That is,
rate = concentration
rate constant
As in many kinetic descriptions, it is important to make a clear distinction between a rate constant and the absolute rate for a particular step; the latter quantity is a function of the rate constant for that step, the rate and equilibrium constants for preceding steps, and the concentrations of the reactants. This point may seem obvious, but uncertainty regarding the meaning of “rate-determining step” has repeatedly led to confusion in description of reaction mechanism.
• The rate-determining step is passage over the highest pass on the path to product.
We quote again from Jencks (5): it is the highest point on the overall Gibbs-energy profile, not the highest energy barrier for an individual step, that determines which step is rate-determining.
• In the next sentence Jencks (5) tells us why it is k2/k‒1 and not k1/k‒1.
Thus, under steady-state conditions in which the intermediate does not accumulate, the critical factor in determining which step of a reaction is rate-determining is the relative rate of breakdown of the unstable intermediate in the forward and reverse direction, that is, the ratio k2/k‒1 rather than the relative rates of formation and breakdown of the intermediate, k1/k‒1.
Note 1. In this simple model we assume the absence of any inhibitor, and so we do not include [I], [EI], or [EIS] in our calculations (2–4). When we consider the forms of the enzyme, E, we can then say that the enzyme will be present only as the free enzyme, E, and as the enzyme substrate complex, ES. When this is true the total concentration of all forms of the enzyme, [E0], equals [E] + [ES]. We now consider the forms of the substrate, S. The substrate can be present as free substrate, S, bound substrate, ES, and, eventually, as product, P. Thus, in general and in the absence of an inhibitor, the concentration of all forms of the substrate, [S0], equals [S] + [ES] + [P]. Even though some substrate will be present as ES the concentration of the substrate, [S], is typically orders of magnitude greater than the concentration of the enzyme. Therefore, even under saturating conditions, [ES] will be orders of magnitude less than [S]. It is for this reason that the concentration of free substrate, [S] is typically taken to be equal to the total concentration of substrate, [S0] (2). Restricting ourselves, however, to “initial rates” we can assume that [P] = 0.
Literature Cited Bozlee, B. J. J. Chem. Educ. 2007, 84, 106–107. Shaw, W. H. R. J. Chem. Educ. 1957, 34, 22–25. Ault, A. J. Chem. Educ. 1974, 51, 381–386. Ault, A. J. Chem. Educ. 2008, 85, 1432–1434; online supplement pp 15–16. 5. Jencks, W. P. Catalysis in Chemistry and Enzymology; McGrawHill Book Company: New York, 1969; p 475.
1. 2. 3. 4.
Supporting JCE Online Material
http://www.jce.divched.org/Journal/Issues/2009/Sep/abs1069.html
or
Jencks (5) puts it this way:
rate =
how much
how well
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© Division of Chemical Education • www.JCE.DivCHED.org • Vol. 86 No. 9 September 2009 • Journal of Chemical Education
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