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Home , Enzyme kinetics, Uncompetitive inhibitor

Recommended problems from the end of chapter 13: 1,2,3,4,5,9,10,11,12,13,16,17

Enzyme kinetics: what do actually do? •Enzymes lower the activation energy barrier, thus make the reaction more Likely. This is observed as as an increased .

•Enzymes DO NOT change ∆G0’, i.e., whether or not the reaction is favorable.

•Hydrolysis of ATP is a favorable reaction But it is very slow in the absence of a catalyst (usually a ).

•This means that the hydrolysis reaction is unlikely in the absence of a catalyst.

∆G0’ •This is due to the high activation energy barrier that must be overcome to achieve ATP hydrolysis. Statistically, very few manage to overcome this high activation barrier. In the absence of a catalyst, therefore, this situation results in a low rate of hydrolysis.

1 Characteristics of enzymatic reactions

1. Speed. Enzymes speed up reaction rates, usually by factors of 106 to 1010 but the reactions can go up to 1014 faster.

2. Mild reaction conditions. Protein and RNA enzymes typically do not work at high temperatures because the enzyme structure is denatured at high temp. Biochemical must work at physiological salt, pH, pressure and temperature.

3. Reaction specificity. Biological enzymes exhibit geometrical and chiral specificity. Substrates that mimic a reactant can be bound by the enzyme but often cannot serve as substrates and sometimes inhibit the enzyme.

4. Capacity for regulation. Enzymes can be regulated on the transcriptional, translational, and post-translational levels. Post translational regulation include allosteric control, covalent modifications, and non-covalent modifications.

Substrate specificity

• specificity is determined by the geometries and non-covalent interactions around the substrate . This site is referred to as the “” of the enzyme. Substrate binding within the active site is followed by a that changes the substrate to . Thus a kinetic framework must take into account at least two steps, the binding step and the chemical step.

•Remember, though, that the active site may NOT be a stationary, rigid binding pocket in all cases. The so-called “lock and key” idea implies a rigid substrate binding site, but such a binding site is not ubiquitous.

•Active sites can be dynamic structures, meaning they can undergo structural changes during the course of a reaction, typically before the chemical step takes place. This notion has been termed “induced fit” because the presence of the substrate results in one or more structural changes that are followed by the chemical step. This notion is not consistent with the “lock and key” idea.

2 Enzyme nomenclature and classification

Classification Reactions catalyzed 1. Oxidation-reduction 2. Transfer of functional groups 3. Hydrolysis reactions 4. Group elimination from double bonds 5. isomerization 6. Bond formation coupled with high energy compound (ATP, GTP) hydrolysis

For further classification details see textbook.

Cofactors and coenzymes In some cases the enzyme alone cannot catalyze a reaction without using another /. The co-factor can be bound covalently to the enzyme, as well as via van der waal interactions. Cofactors can be metals ions such as Cu2+, Fe3+, or Zn2+.

On the right is an illustration of a horse heart cytochrome C protein, with the iron atom (bright white) within the heme group at the catalytic center of the protein.

Heavy metals, such as Pb2+, Hg2+ and Cd2+ can be toxic. What could be the molecular basis for such toxicity?

3 Cofactors and coenzymes

Some cofactors are small organic molecules, which are known as co-enzymes. These often are vitamins, which, by definition, are coenzymes that an organism cannot synthesize and must obtain through the diet.

Niacin forms the redox-active component of NAD+ (nicotinamide adenine dinucleotide) and NADP+ (nicotinamide adenine dinucleotide phosphate) .

Chemical kinetics, reaction order

For the general reaction

aA + bB ↔ cC + dD

The overall rate law that describes this reaction does NOT ALWAYS have terms for all the reactants (although usually it does). Therefore, one cannot just look at the molar coefficients (a, b) of the reactants and decide the order of the reaction. To derive the reaction order, one has to obtain the rate law experimentally.

For example, for this reaction:

A + 3B ↔ 2C + 2D

If Rate = k [A][B]3 , then the reaction is first order with respect to A and 3rd order with respect to B. Overall, this is a 4th order reaction. If Rate = k [A]2[B] , then the reaction is second order with respect to A and first order with respect to B. Overall, this is a 3th order reaction. If Rate = k [A]0[B]1/2 , then the reaction is 0th order with respect to A and half order with respect to B. see reaction order worksheet, part 1.

4 - first order reactions

Elemental reactions are those which involve a simple rearrangement of (including bonding rearrangements).

Consider the irreversible, elemental reaction, A → P.

The instantaneous rate of appearance of the product, P, is equal to the rate of disappearance of the reactant, A. This is called the velocity of the reaction.

dP dA V = = − = k[A] 0 dt dt

This means that the reaction velocity is proportional to the concentration of A. This is an example of€ a first order reaction. The order of the reaction is determined by the number of molecules that must collide to give a product. This also is called the molecularity of the reaction. A first order reaction is a unimolecular reaction, i.e., a reaction involving a simple rearrangement within a reactant.

Chemical kinetics - first order reactions

d[A] From the previous slide you can see that = −k × dt [A]

d[A]/[A]≡ d ln [A]. € If A0 is the initial concentration of A, and At is the concentration at time t, then integration from A0 to At gives

A t ∫ t dln[A] = ∫ −k × dt A0 0

Which results in ln[A]t − ln[A]0 = −kt €

or ln[A]t = ln[A]0 − kt €

€

5 Chemical kinetics - first order reactions

ln[A]t = ln[A]0 − kt

Plotting ln[A]t as a function of time should give the curve on the right for a first order € reaction.

Taking the antilog of both sides yields

−kt [A]t = [A]0e

which is another convenient way of plotting a first order reaction. This equation describes an . €

Half life of first order reactions

The half life of a reaction is the time at which half the original amount of reactant has been converted to product. Substituting A0/2 for A and t0.5 for t, we get

[A] ln( 0 ) = ln[A] − kt 2 0 0.5

Rearranging in order to solve for t0.5 gives €

[A] [A]0 ln[A] ln( 0 ) ln( ) 0 − [A] /2 ln2 t = 2 = 0 = 0.5 k k k

So the half life in a first order reaction is independent of substrate concentration. €

6 Reversible first order reactions and the Consider the reversible elemental reaction A ↔ P. A is consumed by the reaction to the right, with a rate constant k1. A also is forms by the reaction to the left, with a rate constant k-1. This reaction also can be written as: k1 A ↔ P k-1 The is given by d[P] −d[A] v = = = k [A]− k [P] dt dt 1 −1

By definition, at equilibrium€ the overall rate of the reaction is zero, so k1[A] = k-1 [P]

This can be rearranged€ to give k1 [P]eq Keq = = k−1 [A]eq The equilibrium constant for a first order reaction is always the ratio of the forward and reverse rate constants. € See reaction order worksheet, part 2.

Chemical kinetics - second order reactions

In second order reactions, two molecules must collide productively to undergo a reaction. This can be written as 2A → P (class I 2nd order reaction), or A + B → P if the reactants are different (class II 2nd order reaction.)

The instantaneous velocity of a class I second-order reactions can be described by:

d[A] d[A] v = k[A]2 = − or − = k × dt dt [A]2

The reaction velocity is proportional to the square of the concentration of A, and this is a bimolecular reaction. The reaction has a molecularity of 2. € €

This can be integrated over time t, and the solution gives

1 1 1 1 − = kt or = + kt [A] [A]0 [A]t [A]0

€ €

7 Second order reaction half life (Class I) 1 1 = + kt [A]t [A]0

At the substrate half-life, [A]t = [A]0 / 2 and t = t0.5.

€1 1 2 1 1 = + kt0.5 and t0.5 = ( − ) × [A]0 /2 [A]0 [A]0 [A]0 k

1 t € so €0.5 = k ×[A]0

In a bimolecular reaction the substrate€ half time exhibits concentration dependence.

This allows for the distinction between a first order and second order reactions.

Chemical kinetics – class II second order reactions (1)

A class II second order reaction involves two reactants forming a product, where the reactant concentrations are not necessarily the same.

A + B ↔ P

The overall rate depends on both A and B:

dP v = = k[A][B] dt

This is not as simple as the class I case, but if we recognize that for every product P formed one A and one B reactants are consumed, we can write

dP v = = k([A] −[P] )×([B] −[P] ) dt 0 t 0 t

8 Chemical kinetics – class II second order reactions (2)

Rearranging to include like terms on the same side gives

dP = k × dt ([A]0 −[P]t )×([B]0 −[P]t )

Integrating this equation from t=0 to t gives

1 [A] [B] × ln( t 0 ) = k × t [A]0 −[B]0 [A]0[B]t

And some rearranging yields

[A]t [B]0 ln( ) = k([A]0 −[B]0 )× t − ln( ) [B]t [A]0

Chemical kinetics – class II second order reactions (3)

[A]t [B]0 ln( ) = k([A]0 −[B]0 )× t − ln( ) [B]t [A]0

1. A plot of ln ([A]t/[B]t) as a function of time yields a straight line for a second order reaction.

2. If the initial concentrations of the reactants are the same, this equation fails. It is then appropriate to treat the reaction as a class I second order reaction.

3. A plot of 1/[A] as a function of time gives a straight line for a class I second order reaction (see rate equation of 2A ↔ P).

9 Simulation of first and second order reactions

1st and 2nd order reactions

1.00 0.90 0.80 0.70 0.60 [A] / [A] 0, first order [A]0 0.50 / [A] / [A] 0, second order 0.40 [A] 0.30 0.20 0.10 0.00 0 2 4 6 8 10 12 14 16 18 20 time (sec)

For both reactions k = 0.1 and the Y axis represents the fraction of substrate remaining at time t.

−kt For a first order reaction [A]t = [A]0e

[A]0 For a second order reaction [A]t = 1+ [A]0 kt €

€

Simulation of first and second order reactions, same first t0.5

-1 For both reactions k = 0.1 s , [A]0 = 1.443

For a first order reaction −kt [A]t = [A]0e [A] For a second order reaction 0 [A]t = 1+ [A]0 kt First and second€ order reactions, same 1st half life (6.9 sec)

1.00 € 0.90 0.80 0.70 [A] / [A] 0, first order 0.60

[A]0 0.50 / 0.40 [A] / [A] 0,

[A] second 0.30 order 0.20 0.10 0.00 0 2 4 6 8 10 12 14 16 18 20 time (sec)

10 Chemical kinetics: Eyring theory •Consider the bimolecular reaction of a atom with diatomic hydrogen. •According to transition state theory, during the reaction a high energy unstable intermediate is formed. The HA–HB is in the process of breaking, and the HB – HC is in the process of forming.

•Resolution of the high-energy intermediate results in the formation of the HB – HC bond and breakage of the HA–HB bond. •∆ G‡ is the free energy of activation. The energy of activation is the difference in free energy between the transition state and the reactants.

Chemical kinetics: Eyring transition state theory The reactants and products are rarely the same. This can be generalized by

A + B ↔ X‡ ↔ P + Q

The reaction coordinate is no longer symmetrical because the free energy of the reactants is different from the free energy of the products.

The high energy transition state lifetime is very short, so its concentration at any time also is very small. Consequently, the rates of X‡ formation are postulated to be the slowest, or limiting, rates in the reaction.

11 Chemical kinetics: Eyring transition state theory

k’ A + B ↔ X‡ ↔ P + Q Transition state theory postulates that X‡ is in rapid equilibrium with the reactants A and B. The association equilibrium constant for this reaction, K‡, is

[X‡] K‡ = [A][B] Therefore,

‡ ‡ ‡ or ‡ −ΔG / RT € ΔG = −RT lnK K = e where ∆G‡ is the transition state free energy.

Because k’ is the overall rate constant for the second part of the reaction, then the rate of product formation, d[P]/dt, is given by € € d[P] = k'[X‡ ] dt

€ Chemical kinetics: Eyring transition state theory

Because [X‡] K‡ = [A][B] then [X‡ ] = K‡[A][B] € and the overall rate of the reaction is € d[P] ‡ v = = k'[X ‡ ] = k'K ‡[A][B] = k'e−ΔG / RT [A][B] dt This equation demonstrates two important points:

1. The€ overall rate of the reaction increases with the concentrations of A and B. 2. The rate of the reaction increases as ∆G‡ becomes smaller. Stated differently, as the transition state become more stable, the reaction rate becomes faster. As the the transition state becomes less stable, the reaction rate becomes slower. Enzymes stabilize the transition state, thus make the reaction faster. Enzymes do not change the ∆G of a reaction. Enzymes alter only ∆G‡.

12 Two step reactions and the rate-liming step A ↔ I ↔ P

A‡

I‡ G

A I P Reaction coordinate

Many chemical reactions have more than one step. The rate-limiting step is the step of the reaction where ∆G‡ is greatest. This is where most of the reactants accumulate Which is the rate limiting step in the reaction described by the blue curve? Which is the rate limiting step in the reaction described by the red curve?

Enzyme kinetics In 1902 Adrian Brown reported studies on the kinetics of yeast (β-fructofuranosidase)

sucrose + H2O ↔ glucose + fructose

•Brown observed that when the sucrose concentration was much higher than the enzyme concentration, the rate of the reaction (v) became independent of the sucrose concentration. The rate could not be made faster even if the concentration of sucrose was raised. The reaction rate is said to be zero order with respect to sucrose under these conditions. •Brown proposed that the overall reaction is composed of two elemental reactions. •The substrate forms a complex with the enzyme, and, •the substrate-enzyme complex then decomposes to give free enzyme and product.

E +S ↔ ES → E +P

•This explained Brown’s observations because at high substrate concentrations the enzyme was fully bound by substrate, i.e., there was no free enzyme left.

•Under these conditions k2 becomes rate-limiting. The overall reaction rate is not sensitive to any further increase€ in substrate concentration.

13 Michaelis-Menten equilibrium assumption

k1 k2 E +S ↔ ES → E +P k-1 The rate of product formation is the overall reaction rate, which is d[P] v = = k [ES] dt 2 This is not easy€ to measure directly because [ES] and [S] usually cannot be distinguished. Expressing d[ES]/dt gives: d[ES] € = k [E][S]− k [ES]− k [ES] dt 1 −1 2 This also is not easy to measure (because ES and E are hard to distinguish), and not easy to integrate, except when two simplifying assumptions are made: 1. Assumption of€ equilibrium. In 1913 and Maude Menten proposed that ES reaches an equilibrium with E + S if k2 << k-1. If k2<

[E][S] k−1 Ks = = [ES] k1

where Ks is the equilibrium for the enzyme-substrate complex. Although this assumption is not always correct, ES is known as the Michaelis complex.

€

Briggs-Haldane steady-state assumption 2. In 1925 George Briggs and James Haldane presented their assumption of , which builds on the equilibrium assumption. Steady state is reached when [S] > [E]. [ES] remains approximately constant throughout the reaction, except at the very early stages of the reaction (the pre-steady state conditions). This is reasonable for most biological reactions where substrate is in great excess over enzyme. This assumption is expressed by d[ES]/dt = 0

14 Michaelis-Menten equation

If d[ES]/dt = 0, then k1[E][S] = [ES] (k-1 + k2).

[E]T = [E] + [ES], so [E] = [E]T - [ES]

Substituting the expression for [E] into the above equation gives

k1([E]T - [ES])[S] = [ES] (k-1 + k2)

k1[E]T[S] - k1[ES][S] = [ES] (k-1 + k2)

k1[E]T[S] = [ES] (k-1 + k2) + [ES]k1[S] = [ES] (k-1 + k2 + k1[S])

k [E] [S] [E] [S] [E] [S] [ES] = 1 T = T = T (k k k [S]) k-1 + k2 K [S] -1 + 2 + 1 +[S] M + k1

k−1 + k 2 where KM = € k1

€

Michaelis-Menten equation

The initial velocity of the reaction can be expressed in terms of [S] and [E]T:

k2[E]T [S] V0 = k2[ES] = KM + [S]

The initial velocity, v0, is defined as the velocity of the reaction before 90% of the substrate has been converted to product. The use of initial velocity, rather than overall velocity, avoids possible complications€ such as inhibition of activity by product, progressive inhibition of the enzyme, and reverse reactions.

The maximal velocity, Vmax, is defined as the reaction velocity when the enzyme is saturated, i.e., when the enzyme is entirely in the ES form. For such conditions [E]T = [ES], and V0 = Vmax = k2[E]T

Vmax [S] v0 = KM + [S]

This is the Michaelis-Menten equation, the basic equation of enzyme kinetics under steady state conditions. €

15 Michaelis-Menten equation

Vmax [S] v0 = KM + [S]

€

A plot of initial velocity, v0, against substrate concentration, [S], yields the classic steady-state enzyme activity curve.

From this curve you can see that when the substrate concentration is equal to KM, the initial velocity is half the maximal velocity.

See Reaction kinetics worksheet, part 3.

Analysis of kinetic data - Lineweaver-Burk plot

It is sometimes difficult to determine Vmax from the vo vs. [S] plots because even at very high [S] Vmax may not be reached and cannot be calculated directly (without fitting the curve). To overcome this problem, Hans Lineweaver and Dean Burk used the reciprocal of the Michaelis-Menten equation to obtain the following expression that allows a linear regression to calculate Vmax and KM:

The Lineweaver-Burk plot is useful because plotting

1/v0 as a function of 1/[S] yields the curve on the right.

The slope is KM/Vmax. The X intercept is -1/KM. The Y intercept is 1/Vmax.

16 Analysis of kinetic data, kcat / KM (1)

Why is the expression kcat / KM considered an indicator of the catalytic efficiency of the enzyme?

Vmax kcat is the catalytic constant of the enzyme and is defined as: k cat = [E]T

kcat also is called the turnover number because by the above definition it is the number of catalytic processes catalyzed by each active site per unit time. For the simple Michaelis-Menten model k = k . For enzymes with more complicated cat 2 € mechanisms kcat may be a combination of several rate constants.

Under conditions where [S] << KM, the concentration of ES is small, and [E]T ≈ [E] the initial rate equation then becomes

Vmax [S] kcat [E]T [S] kcat v0 = ≈ ≈ [E][S] KM + [S] KM KM

This is a second order rate equation, with a rate constant of kcat / KM.

€

Analysis of kinetic data, limit of kcat / KM (2)

k k k k cat = 2 = 2 1 k + k KM −1 2 k−1 + k2

k1

This expression reaches maximum when k2>>k-1. This means that the rate of ES decomposition to E + S is very small in comparison to the rate of product formation from ES. Under these conditions, (kcat / KM) ≈€ k 1. k1 is the second order rate constant in the formation of ES. For second order reactions in liquid solvent, the formation of ES is limited by diffusion, so k1 cannot exceed the value of the diffusion-limited reaction constant, which is ~109 M-1 • sec-1.

Some enzymes have reached such values of kcat / KM. These enzymes must catalyze a reaction essentially every time a Michaelis complex (ES) forms. These enzymes are said to have reached a state of catalytic “perfection” or “complete efficiency.” Such enzymes include , , and possibly .

17 Inhibition of catalysis •A substance that can bind an enzyme reversibly and adversely affect binding of a substrate or reduce the turnover number is an inhibitor. •A competitive inhibitor directly competes with binding of substrate. Normally, this inhibitor resembles the substrate such that it can bind within the active site but cannot be modified by the enzyme. For example, malonate is a competitive inhibitor of succinate .

- - malonate COO — CH2 — COO

- - succinate COO — CH2 —CH2 — COO The general model for , where EI, the enzyme-inhibitor complex, is catalytically inactive, is given below. KI is the dissociation equilibrium constant of the inhibitor from the enzyme-inhibitor complex.

E + S ↔ ES → E + P + [E][I] I where K = I [EI] KI ! EI + S → no reaction €

Competitive inhibition of catalysis

The fact that EI is inactive means that the competitive inhibitor reduces the concentration of enzyme available for catalysis.

The Michaelis-Menten equation is derived as before, except that there is a term for EI, such that

[E]T = [E] + [ES] + [EI]. The modified Michaelis-Menten equation looks like this:

Vmax [S] [I] v0 = where α =1+ αKM +[S] KI

€ €

18 By measuring the Competitive inhibition of catalysis values of α at different [I] one can obtain KI. By comparing KI for different inhibitors with their structures one can obtain some structural insights into the 3- dimensional arrangement of the active site, in the absence of an x-ray crystal or NMR structural information. Inhibitors can be designed to bind at the active site and inhibit the enzyme, e.g., HIV inhibitors.

Uncompetitive inhibition

The does not bind the free enzyme, rather, it binds the enzyme-substrate complex.

E + S ↔ ES → E + P + [ES][I] I where K'I = [ESI] ’ € K I  ESI → no reaction € The uncompetitive inhibitor need not resemble the substrate. Instead, it should have a binding affinity to the filled€ active site. Presumably, the binding of this type of inhibitor can “distort” the active site and prevent the reaction from proceeding further. The Michaelis-Menten equation for an uncompetitive inhibitor becomes:

Vmax[S] [I] v0 = where α'=1+ KM +α'[S] K'I

€ €

19 Uncompetitive inhibition

Notice that the ratio of KM / Vmax does not change as the concentration of inhibitor is increased. The apparent KM and Vmax are decreased by the same factor because k2 (which is kcat) is a part of KM. The effects on Vmax cannot be changed by increasing the substrate concentration.

Noncompetitive inhibition A mixed inhibitor may bind an enzyme site that participate in substrate binding AND catalysis. A mixed noncompetitive inhibitor is characterized by different values of KI and K’I.

Vmax [S] v0 = αKM + α'[S]

The lines intersect to the left of the

€ 1/v0 axis. Vmax is decreased. KM may decrease or increase.

If of KI and K’I are equal, the inhibition is characterized as pure noncompetitive. This type of inhibition is rare.

20