BCH 5045 Graduate Survey of Biochemistryhort.ifas.ufl.edu/faculty/guy/bch5045/Lecture Files/Lecture 14.pdfFIGURE 6-7 Rate enhancement by entropy reduction. Shown here are reactions

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BCH 5045

Graduate Survey of Biochemistry

Instructor: Charles GuyProducer: Ron ThomasDirector: Glen Graham

Lecture 14Slide sets available at:

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• LEHNINGER• PRINCIPLES OF BIOCHEMISTRY

• Fifth Edition

David L. Nelson and Michael M. Cox

© 2008 W. H. Freeman and Company

CHAPTER 6

Enzymes





The shapes of a substrate and its binding site on dihydrofolatereductase are complimentary with substrate NADP+ (red), unbound (right) and bound (left) and tetrahydrofolate (yellow). NADP+ binds to the active site that is complementary to its shape and ionic properties and illustrates the "lock and key" hypothesis of enzyme action

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FIGURE 6-4 Complementary shapes of a substrate and its binding site on an enzyme. The enzyme dihydrofolate reductase with its substrate NADP+ (red), unbound (top) and bound (bottom); another bound substrate, tetrahydrofolate (yellow), is also visible (PDB ID 1RA2). In this model, the NADP+ binds to a pocket that is complementary to it in shape and ionic properties, an illustration of Emil Fischer's "lock and key" hypothesis of enzyme action. In reality, the complementarity between protein and ligand (in this case substrate) is rarely perfect, as we saw in Chapter 5.



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FIGURE 6-5 An imaginary enzyme (stickase) designed to catalyze breakage of a metal stick. (a) Before the stick is broken, it must first be bent (the transition state). In both stickase examples, magnetic interactions take the place of weak bonding interactions between enzyme and substrate. (b) A stickase with a magnet-lined pocket complementary in structure to the stick (the substrate) stabilizes the substrate. Bending is impeded by the magnetic attraction between stick and stickase. (c) An enzyme with a pocket complementary to the reaction transition state helps to destabilize the stick, contributing to catalysis of the reaction. The binding energy of the magnetic interactions compensates for the increase in free energy required to bend the stick. Reaction coordinate diagrams (right) show the energy consequences of complementarity to substrate versus complementarity to transition state (EP complexes are omitted). ΔGM, the difference between the transition-state energies of the uncatalyzed and catalyzed reactions, is contributed by the magnetic interactions between the stick and stickase. When the enzyme is complementary to the substrate (b), the ES complex is more stable and has less free energy in the ground state than substrate alone. The result is an increase in the activation energy.



Effect of a catalyst on the free energy of activation compared to the noncatalyzedreaction. In the catalyzed reaction it takes a lot less energy input for the reaction to proceed to product than the noncatalyzed reaction. Note that the initial energy state and the final energy state is the same for both catalyzed and noncatalyzed reaction.

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FIGURE 6-6 Role of binding energy in catalysis. To lower the activation energy for a reaction, the system must acquire an amount of energy equivalent to the amount by which ΔG‡ is lowered. Much of this energy comes from binding energy (ΔGB) contributed by formation of weak noncovalent interactions between substrate and enzyme in the transition state. The role of ΔGB is analogous to that of ΔGM in Figure 6-5.



This is a simple reaction, actually a tautomerization that readily can go in either direction. Would you expect the activation energy of the catalyzed reaction to be large? Enzymes exhibit specificity for substrates derived from forming many weak bonds with the substrate and thus changes shape upon interacting with the substrate, a process known as induced fit.



Enzymes (catalysts) mediate rate enhancement of reactions, while at the same time, the catalyst is not consumed by the reaction.

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FIGURE 6-7 Rate enhancement by entropy reduction. Shown here are reactions of an ester with a carboxylate group to form an anhydride. The R group is the same in each case. (a) For this bimolecular reaction, the rate constant k is second order, with units of M–1 s–1. (b) When the two reacting groups are in a single molecule, and thus have less freedom of motion, the reaction is much faster. For this unimolecular reaction, k has units of s–1. Dividing the rate constant for (b) by the rate constant for (a) gives a rate enhancement of about 105 M. (The enhancement has units of molarity because we are comparing a unimolecular and a bimolecular reaction.) Put another way, if the reactant in (b) were present at a concentration of 1 M, the reacting groups would behave as though they were present at a concentration of 105 M. Note that the reactant in (b) has freedom of rotation about three bonds (shown with curved arrows), but this still represents a substantial reduction of entropy over (a). If the bonds that rotate in (b) are constrained as in (c), the entropy is reduced further and the reaction exhibits a rate enhancement of 108 M relative to (a).



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FIGURE 6-8 (part 1) How a catalyst circumvents unfavorable charge development during cleavage of an amide. The hydrolysis of an amide bond, shown here, is the same reaction as that catalyzed by chymotrypsin and other proteases. Charge development is unfavorable and can be circumvented by donation of a proton by H3O+ (specific acid catalysis) or HA (general acid catalysis), where HA represents any acid. Similarly, charge can be neutralized by proton abstraction by OH– (specific base catalysis) or B: (general base catalysis), where B: represents any base.



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FIGURE 6-8 (part 2) How a catalyst circumvents unfavorable charge development during cleavage of an amide. The hydrolysis of an amide bond, shown here, is the same reaction as that catalyzed by chymotrypsin and other proteases. Charge development is unfavorable and can be circumvented by donation of a proton by H3O+ (specific acid catalysis) or HA (general acid catalysis), where HA represents any acid. Similarly, charge can be neutralized by proton abstraction by OH– (specific base catalysis) or B: (general base catalysis), where B: represents any base.



Amino acids that function in acid/base catalysis reactions as proton donors (general acids) or proton acceptors (general bases).

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FIGURE 6-9 Amino acids in general acid-base catalysis. Many organic reactions are promoted by proton donors (general acids) or proton acceptors (general bases). The active sites of some enzymes contain amino acid functional groups, such as those shown here, that can participate in the catalytic process as proton donors or proton acceptors.





Enzyme KineticsEnzymes, we may know them well. We know they are proteins; composed of amino acids and that they act as catalysts for most of the reactions that occur in cells.

But did you know that some RNAs (no protein) can be catalysts and act like enzymes? RNAs with enzyme-like activities are known as ribozymes.



Catalyst, what is a catalyst? A catalyst is a substance or agent (enzyme), which alters the velocity of a reaction, but is not itself changed in the process. For example the enzyme catalase accelerates the conversion of hydrogen peroxide to water and oxygen.

1. 2H2O2 → 2H2O + O2

Only a little more than 70 years ago James Sumner purified and crystallized the first enzyme jack bean urease, and Northrup soon afterward purified digestive enzymes and showed they too were proteins. Before this, it was known that cell extracts contained catalysts, but their exact nature was something for biochemists to argue about.



So imagine studying a reaction, finding a catalyst of the reaction and having no idea of what that catalyst might be, or how it catalyzes the transformation of reactant to product. However, even not knowing the nature of the catalyst, it would still be possible for you to learn something about what the catalyst is able to do in the acceleration of a given reaction.

For example the enzyme catalase accelerates the conversion of hydrogen peroxide to water and oxygen.

1. 2H2O2 → 2H2O + O2



In solution the rate of conversion of H2O2 to water and oxygen is slow. In contrast, one molecule of the catalase protein can stimulate the conversion of up to 40 X 106 H2O2 molecules per second, which makes it one of the fastest enzymes in cells.

Similarly, carbonic anhydrase accelerates the hydration of CO2 to H2CO3 over what would occur in the absence of the enzyme. Carbonic anhydrase is slower than catalase being only able to convert 106

molecules of substrate to product second per enzyme molecule.



Chemical Kinetics

The basic principles of kinetics apply to uncatalyzed reactions as well as catalyzed reactions.

2. Reactant → Product



If we plot the [R] or [P] vs time we find that the velocity (v) at any given time is the slope of the plot at the given time and is described by equation 3:

3. ν = - d[R] or ν = d[P] dt dt

If we follow reactant concentration, then to get a positive ν value we take the negative of d[R]/dt because the [R] is declining while if we follow [P] which is increasing we don’t need to take the negative of d[P]/dt because the value will be positive.



For this simple first-order reaction, ν is proportional to [R] by the relationship:

4. ν = k[R]

where k is the rate constant and has units of (s−1) seconds. The rate constant is fixed and independent of concentration, but sensitive to temperature, pH and other factors.

If we substitute k[R] for ν and change the sign we get:

5. -k[R] = d[R] dt



What happens if the reaction is reversible?

k1

6. Reactant ↔ Product k−1

then equation 5 takes the form of equation 7

7. –k1[R] + k−1[P] = d[R] dt

At equilibrium, d[R]/dt must equal zero, therefore k1[R]eq = k−1[P]eq. Stated another way

8. [P]eq = k1 = Keq

[R]eq k−1



What happens if the reaction is reversible?

k1

6. Reactant ↔ Product k−1

then equation 5 takes the form of equation 7

7. –k1[R] + k−1[P] = d[R] dt

At equilibrium, d[R]/dt must equal zero, therefore k1[R]eq = k−1[P]eq. Stated another way

8. [P]eq = k1 = Keq

[R]eq k−1



This is a typical hyperbolic curve, we need not know whether the catalyst is an enzyme or not, we can still characterize the kinetics of the reaction.

However, by plotting the progress of the reaction with respect to substrate concentration, it remains difficult to precisely know the Vmax.

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FIGURE 6-11 Effect of substrate concentration on the initial velocity of an enzyme-catalyzed reaction. The maximum velocity, Vmax, is extrapolated from the plot, because V0 approaches but never quite reaches Vmax. The substrate concentration at which V0 is half maximal is Km, 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 V0 and [S]. (Note that the curve describes part of a rectangular hyperbola, with one asymptote at Vmax. If the curve were continued below [S] = 0, it would approach a vertical asymptote at [S] = –Km.)





Henri-Michaelis-Menten Derivation

For a simple reaction that follows a hyperbolic curve, when the substrate concentration is plotted against the reaction velocity, Michaelis and Menten made an assumption that the catalyst that accelerated the reaction would form a complex with the substrate and the complex would be in equilibrium with the free catalyst and free substrate.

They further postulated that the only way to product is to proceed from ES.



Steady-State Kinetics

An alternative way to approach kinetics that avoids the requirement that the ES be in equilibrium with E and S, invokes a steady-statesituation that was first formulated by Briggs and Haldane in the 1920s. In this derivation, immediately upon mixing E and S, a steady state is established and the reaction velocity is constant for a brief period of time. During this time, ES remains constant, but S and P are changing.

The rate of ES formation can be written as:

1. vf = k1[E][S]

Defining a new term Km, the Michaelis Constant, Km is the substrate concentration where

2. υ = Vmax/2



Km, we can substitute for ESand this permits the formulation of the Briggs-Haldane equation or what is most often known as the Michaelis-Menten equation:

3. υ = Vmax[S] or v = [S] Km + [S] Vmax [S] + Km

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FIGURE 6-12 Dependence of initial velocity on substrate concentration. This graph shows the kinetic parameters that define the limits of the curve at high and low [S]. At low [S], Km >> [S] and the [S] term in the denominator of the Michaelis-Menten equation (Eqn 6-9) becomes insignificant. The equation simplifies to V0 = Vmax[S]/Km and V0 exhibits a linear dependence on [S], as observed here. At high [S], where [S] >> Km, the Km term in the denominator of the Michaelis-Menten equation becomes insignificant and the equation simplifies to V0 = Vmax; this is consistent with the plateau observed at high [S]. The Michaelis-Menten equation is therefore consistent with the observed dependence of V0 on [S], and the shape of the curve is defined by the terms Vmax/Km at low [S] and Vmax at high [S].



Documents

BCH 5045 Graduate Survey of Biochemistryhort.ifas.ufl.edu/faculty/guy/bch5045/Lecture Files/Lecture 14.pdfFIGURE 6-7 Rate enhancement by entropy reduction. Shown here are reactions