An Introduction to Enzyme Kinetics, Part Deux - ACS Publications

ARTICLE pubs.acs.org/jchemeduc

An Introduction to Enzyme Kinetics, Part Deux Addison Ault* Department of Chemistry, Cornell College, Mount Vernon, Iowa 52314, United States ABSTRACT: Rate equations for enzyme-catalyzed reactions are derived and presented in a way that makes it easier for the nonspecialist to see how the rate of an enzyme-catalyzed reaction depends upon macroscopic kinetic constants and concentrations. Distribution equations are used to show how the rate of the reaction depends upon the relative quantities of the various enzyme species in the reaction mixture. This approach is introduced and reviewed as it applies to single-substrate reactions without inhibition, with competitive inhibition, with uncompetitive inhibition, and with noncompetitive inhibition. Then, this approach is extended to the ordered two-substrate reaction, the two-substrate random rapid equilibrium reaction, and the two-substrate ping-pong reaction. KEYWORDS: Second-Year Undergraduate, Upper-Division Undergraduate, Biochemistry, Analytical Chemistry, Textbooks/Reference Books, Bioanalytical Chemistry, Catalysis, Enzymes, Kinetics, Mechanisms of Reactions ome years ago this Journal published “An Introduction to Enzyme Kinetics”,1 in which a simple, intuitive, yet accurate approach to the some of the fundamental concepts of enzyme kinetics was described. In that article, the four most common types of single-substrate enzyme-catalyzed reaction were considered: the single-substrate no-inhibition reaction, the single-substrate with competitive inhibition reaction, the single-substrate with uncompetitive inhibition reaction, and the single-substrate with noncompetitive inhibition reaction. In this second article, the same approach can be extended to the three most common types of two-substrate enzyme-catalyzed reactions: the ordered two-substrate reaction, the two-substrate rapid random equilibrium reaction, and the twosubstrate ping-pong reaction.

S

The total concentration of the enzyme, [Et], is the sum of the concentrations of all the forms in which the enzyme can be present, ES and E in this example ½Et ¼ ½ES þ ½E ð4Þ ½ES ½ES þ ½E ½ES rate ¼ Vmax ½ES þ ½E ðfractionES Þ ¼

and, finally,

B B rate ¼ Vmax B @

’ THE FOUR TYPES OF SINGLE-SUBSTRATE REACTION Before extending this approach to two-substrate reactions, its application to the four single-substrate reactions will be reviewed. This reaction involves only one substrate and no inhibition, k1

k2

B rate ¼ Vmax B @

ð1Þ

k1

where E is the enzyme, S is the substrate, ES is the enzyme substrate complex, and P is the product. The rate of product formation can be written ð2Þ rate ¼ k2 ½ES and rewritten as

1 1 þ

C C C ½E A

ð7Þ

C C KM A 1 þ ½S 1

ð8Þ

The Macroscopic Kinetic Constants Vmax and KM

The experimental double-reciprocal plots provide values for the macroscopic kinetic constants Vmax and KM. If the reaction is adequately represented by eq 1, Vmax will equal k2[Et] and KM will equal (k1 + k2)/k1. If the total enzyme concentration, [Et], is known, the microscopic rate constant, k2 , can be calculated from Vmax. KM, however, will not equal k1/k1, the dissociation constant for ES; k2 will be larger than zero and so KM will be larger than k1/k1 .

where [ES] is represented by the total concentration of enzyme, [Et], times the fraction of the enzyme that is present as ES. When “all” the enzyme is present as ES (that is, when the fraction present as ES is 1), the rate will be maximal, and this maximum rate, k2[Et], is Vmax. The rate can be represented as ð3Þ

This indicates that the steady-state rate is the maximum rate times the fraction of the enzyme that is present in the form that gives the product of the reaction. In this example, that form is ES. Copyright r 2010 American Chemical Society and Division of Chemical Education, Inc.

1

This expression shows that when [S] is low the rate will be proportional to [S], the proportionality constant being Vmax/KM, that when [S] = KM the rate will be half-maximal, and that as [S] is further increased the rate will approach Vmax.1

rate ¼ k2 ½Et ðfractionES Þ

rate ¼ Vmax ðfractionES Þ

ð6Þ

½ES As shown in the first article,1 [E]/[ES] = KM/[S], where KM = (k1 + k2)/k1, and so eq 7 can be rewritten as 0 1

The Single-Substrate No-Inhibition Reaction

E þ S h ES f E þ P

0

ð5Þ

Published: November 01, 2010 63

dx.doi.org/10.1021/ed1004432 | J. Chem. Educ. 2011, 88, 63–66

Journal of Chemical Education

ARTICLE

and substituting as shown in the first article1 leads to eq 22. 0 1

The Single-Substrate with Competitive-Inhibition Reaction

The second example also involves a single substrate but adds competitive inhibition k1

k2



ð9Þ

k1 k3

ð10Þ

E þ I h EI k3

ð11Þ

ðfractionES Þ ¼

½ES ½ES þ ½E þ ½EI

rate ¼ Vmax

½ES ½ES þ ½E þ ½EI

C C C ½I A

ð22Þ

KM þ 1 þ K4 ½S

This time a high [S] cannot overcome the effect of I, the enzyme cannot be entirely converted to ES when I is present, and the new Vmax will be less than the original Vmax when I is present.

where I is the inhibitor and EI is the enzymeinhibitor complex. This time there are three enzyme forms, ES, E, and EI, and so ½Et ¼ ½ES þ ½E þ ½EI

1

The Single-Substrate with Noncompetitive Inhibition Reaction

This is the last of the four single-substrate reactions

ð12Þ

k1

k2


ð23Þ

k1

ð13Þ

k3

E þ I h EI

ð24Þ

k3

and

0


1 C 1 C C ½E ½EI A þ 1 þ ½ES ½ES

k4

I þ ES h IES This time the inhibitor can bind to both E and ES and so ½Et ¼ ½ES þ ½E þ ½EI þ ½IES

ð14Þ

Substituting as shown in the first article1 leads to 0 1 B B rate ¼ Vmax B @

C C C KM KM ½IA þ 1 þ ½SK3 ½S 1

with

k2

½ES ½ES þ ½E þ ½EI þ ½IES

0

1


C C 1 C ½E ½IE ½IESA þ þ 1 þ ½ES ½ES ½ES

k4

I þ ES h IES


Again, there are three enzyme forms, but this time the inhibitor is bound to ES rather than to E, and so ½Et ¼ ½ES þ ½E þ ½IES

ð18Þ

½ES ðfractionES Þ ¼ ½ES þ ½E þ ½IES

ð19Þ

½ES ½ES þ ½E þ ½IES

1


C 1 C C ½E ½IESA þ 1 þ ½ES ½ES

ð28Þ

C C 1 C KM KM ½I ½I A þ þ 1 þ ½SK3 K4 ½S

ð29Þ

’ THE THREE TYPES OF TWO-SUBSTRATE REACTIONS This approach is now applied to the three most common types of two-substrate enzymatic reactions. These are the ordered twosubstrate reaction, the two-substrate rapid equilibrium reaction, and the ping-pong reaction.

0

ð27Þ

Again the original Vmax cannot be attained in the presence of the inhibitor because some of the enzyme will then be present as IES, which does not form product. Also, because the inhibitor competes with S for E, it will take a higher concentration of S to attain half of the new, lower, Vmax; the new KM will be higher.

ð17Þ

k4

ð26Þ

Finally, substituting as shown in the first article1 0 1

ð16Þ

k1

Uncompetitive-Inhibition


rate ¼ Vmax

½ES ½ES þ ½E þ ½EI þ ½IES

and

The third example also involves a single substrate but adds competitive inhibition. k1

ðfractionES Þ ¼ rate ¼ Vmax

ð15Þ

Equation 15 indicates that a high [S] can completely overcome the effect of I, which is the characteristic of competitive inhibition. The Single-Substrate Reaction

ð25Þ

k4

ð20Þ

The Ordered Two-Substrate Reaction

The two-substrate reaction is examined in which one substrate, A, must bind before the second substrate, B, can bind. Thus, EA must form before B can bind, and there is no EB.

ð21Þ

k1

k2

k1

k2

k3

E þ A h EA þ B h EAB f E þ P 64

ð30Þ

dx.doi.org/10.1021/ed1004432 |J. Chem. Educ. 2011, 88, 63–66


ARTICLE

The rate of product formation can be written rate ¼ k3 ½Et ðfractionEAB Þ

and proceeding as before the rate can be re-expressed as ð31Þ

rate ¼ Vmax ðfractionEAB Þ

and rewritten

There are four enzyme forms and so

rate ¼ Vmax ðfractionEAB Þ

ð32Þ

½Et ¼ ½EAB þ ½EA þ ½EB þ ½E

The three forms in which the enzyme can be present are EAB, E, and EA, and so ½Et ¼ ½EAB þ ½E þ ½EA

ðfractionEAB Þ ¼

ð33Þ

½EAB ðfractionEAB Þ ¼ ½EAB þ ½E þ ½EA and the rate can be represented as ½EAB rate ¼ Vmax ½EAB þ ½E þ ½EA and

ð41Þ

0

1


C C C ½E ½EA A þ 1 þ ½EAB ½EAB 1

C 1 C C KA KB KIA A þ 1 þ 1 þ ½A ½B ½A

ð34Þ

and ð35Þ

ð36Þ

k1

k2

k3

k4

k3

k4

k5

k5

E þ B h EB þ A h EAB f E þ P

0

1


C 1 C C ½EA ½EB ½E A þ þ 1 þ ½EAB ½EAB ½EAB

ð45Þ

KA

KB

k5

ð46Þ

KB

KA

k5

ð47Þ

and the rate as

0

1

B rate ¼ Vmax B @

C 1 C KB KA KA KB A þ þ 1 þ ½B ½A ½A½B

ð37Þ

ð48Þ

Following the argument by Segel,3 that the binding of A will change the binding constant of B by a factor of α, and vice versa, the reaction can be written KA

αKB

k5

ð49Þ

KB

αKA

k5

ð50Þ

E þ A h EA þ B h EAB f E þ P E þ B h EB þ A h EAB f E þ P and the rate as

0

1

B rate ¼ Vmax B @

C 1 C αKB αKA αKA KB A þ þ 1 þ ½B ½A ½A½B

ð51Þ

The second term in the denominator indicates that [EA] can be driven down by saturation with B, and the third term in the denominator indicates that [EB] can be driven down by saturation with A. The fourth term, however, indicates that the rate will not approach Vmax until both A and B are saturating so that [E] will be driven down as well.

ð38Þ

The Third Two-Substrate Mechanism

ð39Þ

Reaction:

The

Ping-Pong

According to the ping-pong mechanism, the enzyme, E, reacts with the first substrate, A, to give EA (ping), which then releases the first product, P, leaving the enzyme in a different form, F:

The assumption is that all forms prior to EAB are in equilibrium. The rate of product formation is rate ¼ k5 ½Et ðfractionEAB Þ

ð44Þ

E þ B h EB þ A h EAB f E þ P

In this type of reaction, the two-substrate rapid equilibrium reaction, the two substrates, A and B, can bind in either order to give the ternary complex, EAB, which then goes on to products in the rate-determining step. The reactions are k2

E þ A h EA þ B h EAB f E þ P

The Random Rapid Equilibrium Two-Substrate Reaction

k1

½EAB ½EAB þ ½EA þ ½EB þ ½E

ð43Þ

Assuming that the binding of A does not affect the binding of B, and vice versa, the reaction can be represented as

Thus, a high concentration of A can minimize the concentration of the free enzyme, but those high concentrations of both A and B are required to convert “all” the enzyme to EAB in the steady state and thereby cause the rate to approach Vmax. When KB/[B] is very small (when [B] is “saturating”), the rate will be half maximal when [A] = KA. Thus, KA is the Michaelis constant for A. When KA[A] is very small (when [A] is saturating), the rate will be half maximal when [B] = KB, provided that KIA/[A] is also small. Thus, KB is the Michaelis constant for B, but its apparent value depends upon KIA[A]. KIA is the dissociation constant for EA, and a high value for KIA (a low affinity of the enzyme for A) will lead to a larger apparent KB.

E þ A h EA þ B h EAB f E þ P

½EAB ½EAB þ ½EA þ ½EB þ ½E

rate ¼ Vmax

At this point an article by Paul Sims2 is cited in which his eq 34 is divided top and bottom by [A][B] and the [KB]/[B] term in the denominator is factored out. Substituting the corresponding terms gives 1 0 B B rate ¼ Vmax B @

ð42Þ

k1

k2

ping : E þ A h EA f F þ P

ð40Þ

k1

65

ð52Þ



ARTICLE

equals Vmax. The relative concentrations of EA and FB will always be determined by

This different form of the enzyme, F, then reacts with the second substrate, B, to give FB (pong), which then releases the second product, Q, returning the enzyme to its original form, E. pong :

k3

k4

F þ B h FB f E þ Q

rate ¼ k2 ½EA ¼ k4 ½FB

ð53Þ

k3

’ SUMMARY The rate equations for enzyme-catalyzed reactions have been derived and presented in a form that makes it easier for the nonspecialist to see how the rate of an enzyme-catalyzed reaction depends upon macroscopic experimental kinetic constants and concentrations. This was accomplished with distribution equations that show how the rate of the reaction depends upon the relative quantities of the various enzyme species in the reaction mixture. There is a pair of equations for each mechanism. Both equations represent the steady-state rate of the reaction as the maximum rate of the reaction, Vmax, times a fraction. The fraction represents the distribution of the enzyme species. The first equation of each pair represents the fraction in terms of ratios of species. The second equation of each pair represents the fraction in terms of macroscopic kinetic constants and concentrations. The correspondence in terms between the two equations reveals the way in which the ratio of species and hence the rate of the reaction depends upon macroscopic kinetic constants and concentrations. Although there must be a connection between macroscopic observed kinetic constants and microscopic constants, it is generally not possible to determine all of the microscopic rate and equilibrium constants from kinetic data alone. For example, as shown previously, KM does not equal k1/k1, the dissociation constant for ES.

In the steady state, the rates of formation of P and Q must be equal. If this were not true, there would not be a steady state, and the slower reaction would speed up, and the faster reaction would slow down until the rates became equal. There are, therefore, two equivalent ways to express the steady-state rate of product formation. rate ¼ rate of formation of P ¼ rate of formation of Q

ð54Þ

rate ¼ k2 ½EA ¼ k4 ½FB

ð55Þ

rate ¼ k2 ½Et ðfractionEA Þ ¼ k4 ½Et ðfractionFB Þ

ð56Þ

Because k2 and k4 will not be equal, the two fractions will not be equal, and the larger fraction will be associated with the smaller rate constant. k2 ðfractionEA Þ ¼ k4 ðfractionFB Þ

ð57Þ

Although neither fraction can be 1, their sum will be 1 when the rate equals Vmax. “All” the enzyme will be present as EA and FB when the rate equals Vmax. The rate can be represented by rate ¼ Vmax ðfractionEAþFB Þ

ð58Þ

Because there are four enzyme forms, EA, FB, E, and F ½Et ¼ ½EA þ ½FB þ ½E þ ½F and

ðfractionEAþFB Þ ¼

½EA þ ½FB ½EA þ ½FB þ ½E þ ½F

ð59Þ

Corresponding Author

*E-mail: [email protected]. ð61Þ

0

1


C C C A ½E ½F þ 1 þ ½EA þ ½FB ½EA þ ½FB 1

’ AUTHOR INFORMATION

ð60Þ

Then, the rate can be expressed as ½EA þ ½FB rate ¼ Vmax ½EA þ ½FB þ ½E þ ½F and

’ REFERENCES (1) Ault, A. An Introduction to Enzyme Kinetics. J. Chem. Educ. 1974, 51, 381–386. (2) Sims, P. A. An “Aufbau” Approach to Understanding How the King-Altman Method of Deriving Rate Equations for Enzyme-Catalyzed Reactions Works. J. Chem. Educ. 2009, 86, 385–389. (3) Segel, I. H. Enzyme Kinetics; Wiley-Interscience: New York, 1975; p 274.

ð62Þ

Because [E] is minimized by saturation with A, [F] is minimized by saturation with B, and both [EA] and [FB] are maximized by saturation with A and B, eq 62 can be rewritten as eq 63: 0 1 B rate ¼ Vmax B @

C 1 C KA KB A þ 1 þ ½A ½B

ð64Þ

’ NOTE ADDED AFTER ISSUE PUBLICATION The originally submitted version of this article was published in the January 2011 issue, instead of the accepted version. The Online version was corrected on December 16, 2011. An Addition and Correction also appears in the December 16, 2011 (Vol. 89, Issue 1), print issue.

ð63Þ

Again, although neither [EA] nor [FB] alone will equal [Et] when the rate equals Vmax, their sum will equal [Et] when the rate 66


An Introduction to Enzyme Kinetics, Part Deux - ACS Publications

Recommend Documents