Enzyme Kinetics

GLUCOSE BINDING TO A GLUCOSE SENSOR

ANTHONY CARRUTHERS, BLOCK 1

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The chemistries that drive biological processes are reversible, occur on a time-scale of 10-12 to 109 sec and are most frequently catalyzed by enzymes. The study of reversible biological reactions, their time-dependence and the mechanisms of enzyme-mediated catalysis is called enzyme kinetics.Enzyme-kinetics is central to every biological process that ever has or will be studied and is the basis for a great many assays that are routinely undertaken in every research laboratory. Understanding enzyme kinetics is, therefore, important if we are to understand biological processes and the limitations of the assays we undertake in order to study these processes.

Graduate students often characterize kinetics as uninteresting. I have observed several explanations for this: The subject matter may be dicult to the

mathematically-challenged biology student. Students may think that they will never undertake a

kinetic analysis and thus the study of kinetics is unnecessary.

The student may find the subject matter accessible but nevetheless uninteresting.

My goals in developing this guide to kinetics are: To show that kinetic analysis is simply one more

very powerful tool that biologists use to study biological problems.

To show that the use of the tools is straightforward when the underlying principles and assumptions are appreciated.

To emphasize that the student does NOT have to memorize equations and derivations - they are included in this guide for your reference.

To provide examples of analyses. To emphasize key points that student should

understand. To provide formative, self-evaluation questions with

keys in order that the student can evaluate their understanding of the concepts.

i

Foreword

CHAPTER 1

Equilibria

This chapter considers reversible chemical reactions. We ask:

1. What is a reversible chemical reaction and what are its characteristics?

2. When can a chemical reaction occur spontaneously (by itself)?

3. How do enzymes accelerate reactions?

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Loss () and uptake () of sugar by human red cells (lower graph) and human red cell ghosts (upper graph) at 4C.

LEARNING OBJECTIVES

1. Reversible reactions never come to a halt - they achieve an equilibrium in which the forward and reverse reactions are quantitatively balanced.

2. For any reversible reaction, there is a fixed relationship between the conentration of products formed and substrates remaining at equilibrium.

3. This relationship is unique to each reaction and reaction conditions.

SECTION 1

Reversible Reactions Law of Mass ActionA reversible reaction is one in which a product can be formed from starting material (substrate) and substrate can be formed from the ending material (product). This reaction never stops but reaches an equilibrium in which the rate of product formation from substrate is identical to the rate of substrate formation from product.

In a reversible reaction, there is a fixed relationship between the concentrations of reactants (substrate, S) and products (P) at a given temperature in the equilibrium mixture.

3

For the reaction

where [ ]eq denote equilibrium concentrations, k1 and k-1 are constants describing the rates of forward and reverse reactions respectively and Keq is the equilibrium constant.

For example, the reaction between hydrogen and iodine

H2 + I2 2HI;

or, in general

pA + qB + rC xD + yE + zF;

The equilibrium constant for any specific reaction (at a given temperature) is rather like a fingerprint - it is unique.

4

Keq =

[Glucose 6 phosphate]e[ADP]e[Glucose]e[ATP]e

=k1k1

k1

k-1Glucose + ATP Glucose-6-phosphate + ADP

Keq =

[HI]2

[H2 ][I2 ]

Keq =

[D]x[E]y[F]z

[A]p[B]q[C]r

Displacement of the position of equilibrium

If the concentration of any one of the substances is altered in an equilibrium mixture, the concentrations of the other substances must change so as to keep Keq constant.

For example, if [ADP] is increased in an equilibrium mixture by adding exogenous ADP, [glucose-6-phosphate] will fall by combination of glucose-6-phosphate and ADP to increase [Glucose] and [ATP]. The net eect, of course, is that Keq is unchanged.

Formative self-evaluation questions

In the reaction

1. Define Keq in terms of k1, k-1

2. Define Keq in terms of [S]eq, [P]eq

3. What would happen if the reaction were at equilibrium and more S were added?

4. What would happen if the reaction were at equilibrium and more P were added?

5

Keq =

[Glucose 6 phosphate]e[ADP]e[Glucose]e[ATP]e

S Pkk

1

1

-

LEARNING OBJECTIVES

1. The chemical potential of the reactants determines the net direction of the reaction.

2. The Gibbs free energy change for a reaction, G, is the chemical potential of the product(s) minus the chemical potential of the reactant(s)

3. The reaction proceeds in the direction of high chemical potential to low chemical potential (G < 0)

SECTION 2

In which direction will a reversible reaction proceed?

The chemical reaction S P can be described by the following energy diagram:

We return to the chemical intermediate X and the term GA later.

This reaction (Example 1) will proceed in the net direction left to right. Why? Because the chemical potential (energy) of S > P. (Note G is independent of the path of the reaction).

Now consider Examples 2 and 3 below

Reaction 2 will proceed, net right to left (chemical potential of P > S). Reaction 3 above will proceed equally in both directions (no net reaction) because the chemical potential of S = P.

6

Example 1

Progress of Reaction

Ener

gy

S

X

GA

GP

Example 2

Progress of ReactionEn

ergy

Ener

gy

S

X

GA

GP

Example 3

Progress of Reaction

S

X

GA

G = 0P

Chemical potential and GThe chemical potential of a molecule is a measure of the ability of the molecule to perform work.

G is the change in Gibbs free energy for the reaction.

Consider the reaction:S P

The chemical potential or partial molar free energy of S, for example is given by:

S = S + RT ln[S]We will discuss the meaning of s very shortly, but you can see that the chemical potential of S is directly proportional to its concentration.The free energy change (G) for the reaction S P is given by:

G = P - SAt equilibrium, the rate of P formation is matched exactly by the rate of S formation. Thus the abilities of P to form S (to do work) and vice versa are identical. Hence, at equilibrium,

P = SThus,

G = 0

Forms of EnergyFree Energy (G) - performs work at constant temperature and pressure.Heat Energy (enthalpy, H) - performs work only through a change in temperature. Entropy (S) is the unavailable energy. Changes in the free energy (G) and enthalpy (H) of a system (the reactants) and changes in the entropies (S) of the system plus the surroundings (universe) are related in the following manner:

G = H - TS

Thus: G T When TS > H, G < 0 When TS < H, G > 0What is the importance of the entropy change?The irreversible increase in Entropy gives direction to the reaction!Imagine we have two glass flasks connected by a valve. One flask has an internal volume of 10 mL and the second has an internal volume of 100 mL. The 10 mL flask is filled with an ideal gas.

7

When the valve is opened between the two flasks, the gas immediately expands to occupy both flasks. We have all observed this type of behavior in one form or another and know this to be a spontaneous reaction ( G < 0).

Since with our ideal gas there are no interactions between gas molecules (H = 0). Thus:

G = -TSThe fall in free energy is due to increased S. The molecules redistribute to maximize system entropy.Summary1. A reaction occurs spontaneously only if G is negative (P < S).2. A system is at equilibrium (forward and reverse reactions are balanced) and no net change occurs when G = 0 (P = S)3. The forward reaction cannot occur spontaneously when G is positive. An input of free energy is required to drive the reaction.4. G depends upon the free energy of the products (final state) minus that of the reactants (initial state). i.e. G is independent of the reaction mechanism.5. The irreversible increase in entropy provides a

directional driving force for the reaction.

Formative self-evaluation questions

In the reversible reaction S P

1. Define G and the relationship between S and P when the reaction is at equilibrium.

2. Define G and the relationship between S and P when the reaction proceeds in a net direction from left to right.

3. Define the G and the relationship between S and P when the reaction proceeds in a net direction from right to left.

4. At equilibrium, what is the relationship between H and TS?

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LEARNING OBJECTIVES

1. Understanding starting, standard and equilibrium conditions.

2. The relationships between G, G and Keq.

3. Understanding that standard free energy changes are additive and how this is exploited in nature.

SECTION 3

G and equilibriaTo proceed with the following discussion, we must first understand 3 conditions. The starting condition refers to the concentration of reactants at the beginning of the reaction. The equilibrium condition refers to the concentration of reactants the reaction is at equilibrium. The standard condition refers to the concentration of reactants under standard conditions. In the reaction:

A + B C + D

G G RTln [A] [B][C] [D]o= +

G is the standard free energy changeR = gas constant (1.98 cal/mol/degree)1 T = absolute temperature[A], [B], [C] and [D] are the molar activities of the reactants under starting conditions.

1one calorie (cal) is that amount of heat required to raise the temperature of 1 gram of water from 14.5 C to 15.5 C. One joule (J) is that energy needed to apply a 1 newton force over a distance of 1 meter. 1 kcal (1000 cal) = 4.184 kJ (4184 J)

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G is the free energy change under standard conditions

i.e. [A] = [B] = [C] = [D] = 1 MP = 1 AtmospherepH = 7.0T = 298K = 25C

This condition does not, however, describe [A], [B], [C] and [D] (and therefore G) under the conditions of the reaction but is used only to describe G.

At equilibrium, G = 0 and the ratio remains constant

(i.e. the forward reaction is balanced by the reverse reaction)Thus the free energy equation becomes

0 G RTln [A] [B][C] [D]o= +

and

G RTln [A] [B][C] [D]o =-

Defining Keq as

K [A][B][C] [D]

eq =

G = -RT ln Keq= -2.303 RT log10 Keq

Rearranging gives usKeq = e-G/RT

=10-G/(2.303 RT)

Substututing R and T (25C) we obtainKeq =10-G/1.36

when G has units of kcal/mol.Thus an equilibrium constant of 10 corresponds to a G of -1.36 kcal/mol (see Table below)

Note: H-bond energies range from 3 to 7 kcal/mol. van der Waals bond energies are approximately 1 kcal/mol.

10

[A] [B][C] [D]

GoGoKeq per M kcal/mol kJ/mol

1,000,000 -8.2 -34.210,000 -5.5 -22.8100 -2.7 -11.410 -1.4 -5.71 0.0 0.00.1 1.4 5.70.01 2.7 11.40.0001 5.5 22.80.0000001 9.5 39.9

Standard Free Energy Changes are AdditiveConsider the following reactions: A B G1

A C Gtotal B C G2The G of sequential reactions are additive, thus

Gtotal = G1 + G2This principle of bioenergetics explains how an endergonic reaction (Keq < 1) can be improved (more product formed) by coupling it to a highly exergonic reaction (Keq >>1) through a common intermediate.Consider the synthesis of glucose6phosphate a reaction that occurs in all cells:

G = 13.8 kJ/mol

G = -30.5 kJ/mol

Glucose + Pi Glucose6phosphate + H2O

ATP + H2O ADP + Pi

These reactions share common intermediates (Pi and H20) and may be expressed as the sequential reactions:

Glucose + Pi Glucose6phosphate + H2O

ATP + H2O ADP + Pi

1)

2)

sum: ATP + glucose ADP + glucose6phosphate

Thus Gtotal = +13.8 kJ/mol +(-30.5 kJ/mol) = -16.7 kJ/mol.

The overall reaction is therefore exergonic.We can also compute Keq for each reaction.

K [glucose] [P][G6P] 3.9 x10 Meq1

i

3 1= = - -

(note H20 is not included)

K [ATP][ADP][P] 3.9 x10 Meq2 i 5= =

K [glucose] [P] [ATP][G6P][ADP][P]

eqtotali

i

=Keq1 * Keq2 = 7.8 x 102

Thus by coupling ATP hydrolysis to G6P synthesis, the Keq for G6P formation is raised by a factor of 2 x 105.This strategy of coupling one reaction with a low Keq (or G > 0) to a second with a high Keq (or G < 0) is used by all living cells in the synthesis of metabolic intermediates.

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Summary

1. At equilibrium, G = 0

2. Exergonic reactions are characterized by Keq > 1 and G < 0

3. Endergonic reactions are characterized by Keq < 1 and G > 0

4. An endergonic reaction can be made more favorable (i.e. Keq increases) by coupling it to a second exergonic reaction via common chemical intermediates. Here the G of the reactions are summative.

5. At 25C, and equlibrium constant of 10 corresponds to a G of -1.36 kcal/mol

Formative self assessmentsWe will solve an exam-type question to illustrate relationships between starting and equilibrium levels of reactants and reaction spontaneity (direction of net flux).We use the isomerization of dihydroxyacetone phosphate (DHAP) to glyceraldehyde 3-phosphate (G3P) as our example.Question:Keq for the reaction DHAP G3P is 0.1 (log10 Keq = -1.). Defining: G = G+ {1.36 x log10[G3P]/[DHAP]} kcal/mol. When the starting concentrations of DHAP and G3P are 0.2 mM and 2 M respectively (log10[G3P]/[DHAP] = -2).F. G = 1.36 kcal/mol and G = -2.72 kcal/molG. Under the starting conditions stated above, the

reaction cannot occur spontaneously.H. G = 2.72 kcal/mol and G = -1.36 kcal/molI. Under the starting conditions stated above, the

reaction can occur spontaneously.J. The net reaction will proceed from right to left.SolutionA. At equilibrium, [G3P]/[DHAP] = 0.1 thus Keq = 0.1

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B. Thus G = - 1.36 x log10 0.1 = -1.36 x -1 = +1.36 kcal/mol.

C. When the initial concentrations of DHAP and G3P are 2 x 10-4 M and 2 x 10-6 M respectively we can substitute these concentrations and G into the equation to obtain

D. G = 1.36 kcal/mol + (1.36 x -2) = -1.36 kcal/mol

E. The answer is, therefore, D.The negative value for G indicates that the reaction can occur spontaneously when the species are present at the concentration stated above. Note however that no calculations were necessary. Since Keq > starting [G3P]/[DHAP], the forward reaction must occur spontaneously (more G3P must be formed) since all reactions must proceed to equilibrium. If starting [G3P]/[DHAP] had been > Keq the reaction would have proceeded spontaneously from right to left.

Answer the following:1. In biochemical reactions:A. A reaction can occur spontaneously only when

the sum of the entropies of the system and its surroundings < 0.

B. The most important criterion that determines whether a reaction can occur spontaneously is Go.

C. The change in energy of a system is independent of the path of a reaction.

D. At chemical equilibrium where no net change in [products] or [reactants] can occur, G > 0.A reaction can occur spontaneously only if the standard free energy change of the reaction is < 0.

2. When a reversible reaction between substrate and product has achieved equilibrium:

A. The standard free energy change of the reaction is always zero.

B. Subsequent addition of product will drive the reaction to the right.

C. The chemical potentials of substrate and product are equal.

D. The forward and reverse reactions halt.E. Subsequent addition of an enzyme that

accelerates the reaction will drive the reaction to the right.

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3. Keq for the reaction A is 0.01 (log10 Keq = -2). Defining: G = Go+ {1.36 x log10 [B]/[A]} kcal/mol. When the starting concentrations of A and B are 1 mM and 0.1 mM respectively (log10[B]/[A] = -1). A. G = 1.36 kcal/mol and Go = -2.72 kcal/molB. Under the starting conditions stated above, the

reaction can occur spontaneously.C. Go = 2.72 kcal/mol and G = -1.36 kcal/molD. The reaction will proceed from left to right.E. The reaction will proceed from right to left.

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LEARNING OBJECTIVES

1. Enzymes accelerate a chemical reaction but in doing so are neither chemically transformed at the completion of the reaction nor do they alter the equilibrium of the reaction.

2. Enzymes introduce althernative reaction pathways with lower energy barriers to catalysis.

SECTION 4

Enzymes are biological catalysts

Explanations of Catalytic Action

Enzymes lower the free energy of activation necessary for a reaction to occur.How can an enzyme accelerate a reaction without shifting its equilibrium? To understand this we return to the "Transition state theory" (Eyring, 1935). Reactant molecules must overcome an energy barrier and pass through an activated complex before proceeding on to the product of the reaction.The isomerization reaction

S Pis best represented by

S X Pwhere X is the activated complex or transition state.In terms of an energy diagram

Example 1

Progress of Reaction

Ener

gy

S

X

GA

GP

Reactant molecules that achieve only a fraction of the activation energy (GA) fall back to the ground state. Those that achieve the transition state energy are committed to form product.

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The fraction of S that achieves the transition state X at any temperature T is given by

S eF RTGA

= -

If T = 0C (273K) and GA = 10,000 cal/molSF = 9.9 x 10-9

If GA is in some way reduced to 1,000 cal/mol SF = 0.158

A 10-fold decrease in GA results in a 16,000,000-fold increase in the fraction of S that can achieve the transition state! In principle, the reaction would be accelerated by 16,000,000-fold.Enzyme thus provide alternative routes of reaction which have lower energy barriers.An enzyme could act in the following manner:

Progress of Reaction

Ener

gy

S

X

XX

X

P

Progress of Reaction

Ener

gy

S

X

P

Either way, less energy is needed to form transition state species but the ground states of substrate S and product P are unchanged (G is unchanged). Thus the reaction is accelerated and the equilibrium is unchanged.

How can an enzyme introduce alternative reaction pathways?A number of mechanisms are observed:1. Covalent CatalysisA nucleophile (electron-rich group with a strong tendency to donate electrons to an electron-deficient nucleus) on the enzyme displaces a leaving group on the substrate. The enzyme-substrate bond is then hydrolyzed to form product and free enzyme.2. Acid-base Catalysise.g. Lysozyme cleaves the glycosidic bond between C1 of N-acetylmuramic acid and C4 of N-acteylglucosamine of bacterial cell wall polysaccharides. Glu35 of lysozyme donates a proton to the oxygen of the polysaccharide glycosidic bond thereby hydrolyzing the bond.3. ProximityAn enzyme may bind two reactants and in doing so increase their proximity. Reaction rate is related to the number of collisions of correct orientation. When an enzyme binds its substrates it insures that their orientation is precisely that required for reactivity.4. Molecular DistortionThe enzyme active site undergoes a conformational change upon binding substrate distorting the substrate into a conformation resembling the transition state species.

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Summary

1. Reactant molecules must overcome an energy barrier and pass through an activated complex before proceeding on to the product of the reaction.

2. Enzymes accelerate reactions by lowering this energy barrier by introducing alternative reaction pathways

3. Enzymes do not aect the ground state of reactants and products therefore do not aect the equilibrium of a reaction.

4. Enzymes therefore accelerate forward and reverse reactions equally.

5. Enzyme introduce alternative reaction pathways through covalent catalysis, acid-base catalysis, proximity eects or by molecular distortion or combinations thereof.

Formative self-evaluation questions

1. What is the Transition state theory?

2. By how much could a 20-fold reduction in GA from 10 kcal/mol to 0.5 kcal/mol accelerate a reaction at 0C?

3. Name 4 mechanisms by which enzymes lower energy barriers for catalysis.

4. Do enzymes aect Keq?

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SECTION 5

Key to formative evaluations

Section 1

1. Keq = k1/k-1

2. Keq = [P]eq/[S]eq

3. [S] would fall and [P] would increase so that Keq remained unchanged.

4. [S] would rise and [P] would fall so that Keq remained unchanged.

Section 2

1. At equilibrium, G =0 and S = P

2. When the reaction proceeds left to right, G < 0 and S > P

3. When the reaction proceeds right to left, G > 0 and S < P

4. At equlibrium, H = T S

Section 3

1. C (S increases; G not G; G=0 at equilibrium)

2. C (G=0, not G; more P produces more S; the reaction continues; enzyme leaves Keq unaltered)

3. E. Keq = 0.01; starting [B]/[A] = 0.1 > Keq thus the reaction must proceed right to left.

Section 4

1. Transition state theory states that reactant molecules must overcome an energy barrier and pass through an activated complex before proceeding to product formation.

2. At 0 C & GA = 10kcal/mol, SF = e-10,000/(1.987*273) = 9.8 x 10-9. When GA falls to 0.5 kcal/mol, SF = 0.398. Thus SF increases 40.6 x106-fold.

3. Covalent catalysis, acid-base catalysis, proximity eects or by molecular distortion.

4. No! Keq is unchanged.18

CHAPTER 2

Analysis of time dependent processesLorem ipsum dolor sit amet, suspendisse nulla pretium, rhoncus tempor placerat fermentum, enim integer ad vestibulum volutpat. Nisl rhoncus turpis est, vel elit, congue wisi enim nunc ultricies dolor sit, magna tincidunt. Maecenas aliquam est maecenas ligula nostra.

This chapter considers time-dependent processes. We seek to:

1. provide a set of tools to analyze time-dependent processes

2. understand the underlying mechanisms that govern reaction rates

kobs

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DNPA + NaOH DNP + acetate

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Some constants are given multiple names in this and subsequent chapters. This reflects: 1) my poor editing; 2) the reality that multiple names for the same constant are also found in the literature:

KS KD; KM Km KM(app); sometimes KM KM(app) can be equal to KS KD (I will explain when this is true)

kf kon; kr ko

Vmax Vm Vm(app)

kp kcat (I will explain when this is true)

Bmax Bm; sometimes Bmax Bm = [Et] (I will explain when this is true)

Your challenge will be to understand when dierent names mean the same thing or dierent terms.

In general you will encounter 2 types of kinetic constants in this and subsequent chapters.

Constants shown in lowercase (e.g. k1, k-1, kon, ko etc) are typically rate constants with units of per unit time (e.g. s-1) or per concentration per unit time (e.g. M-1.s-1).

Constants shown in uppercase (e.g. KM, KD, KS, Ki, Vmax etc) can have more complex meanings, have units of concentration (M) or are rates (mol/sec) and are related in some predictable way to multiple rate constants

e.g. KM = (k-1+kp)/k1

You will also encounter a great many equations. You do not need to memorize these (although you may do so involuntarily particularly if you use them frequently). My goal is for this pamphlet to serve as a useful resource for you.

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Constants, equations and glossary

Glossary

catalysis - the process by which an enzyme or catalyst accelerates a reaction.

catalyst - an agent that accelerates a chemical reaction but which is unchanged in amount or chemistry at the end of the reaction.

chemical equilibrium - a reaction in which forward and reverse reactions continue to proceed but are quantitatively balanced

enzyme - a biological catalyst

equilibrium - a state in which opposing forces are balanced.

products - molecules produced by the action of enzymes on substrates

reaction order - the dependence of the reaction rate on [substrates]n when n is the number of substrates which must interact to form a single molecule of product.

substrates - molecules that are acted upon by enzymes

thermodynamic equilibrium - Keq = [Product]e/[Substrate]e i.e. the ratio of product formed : substrate remaining at equilibrium

thermodynamics of reaction rates - enzymes introduce alternative reaction pathways in which the Gibbs free energy of activation is reduced

velocity or rates - amount of material (substrate, product, # cells etc) consumed or produced or number of events occurring per unit time

xxi

LEARNING OBJECTIVES

1. Biological reactions occur over intervals ranging from psec (10-12 s) to days (105 s) or even longer

2. The tools used to analyze these reactions, however, are invariant and fall under the general umbrella of kinetic analysis.

SECTION 1

ContextSome examples

22

time for heat to propagate from the base to theends of the chains, as a function of the length hof the alkane molecules.

An 800-nm, 500-fs-duration laser pulse froman amplified titanium-dopedsapphire laser (5)incident on the Au/glass interface (the backside) of the 50-nm-thick Au layer generated hotelectrons within a skin depth of ~15 nm (6).Because the hot electrons have a large diffusioncoefficient, the electron temperatures at the frontand back of the Au layer equalized even beforeelectron-phonon coupling brought the hot elec-trons into equilibrium with the lattice (6). Within~1 ps, the Au layer was in thermal equilibri-um and uniformly heated throughout (6). Toimprove the adhesion of Au to glass it wasnecessary to add a Cr layer beneath the Au.Unfortunately, heat transfer from a Cr layer toAu is relatively slow; to minimize this effect, wemade the Cr layer just 0.8 nm thick. An ultrafastthermoreflectance apparatus (2, 7) was used tocharacterize the temperature rise of the Au layer.As shown in Fig. 1C, there is a fast increase ofthe Au surface temperature to 80% of the finaltemperature within 1 ps. There is also a slower(1.5 ps time constant) rise to the final temper-ature due to the Cr layer. The same transientresponse was observed with either front-side orback-side flash-heating and with or without aSAM. The Au layer remained at an approxi-mately constant high temperature for severalnanoseconds, subsequently cooling by heatdiffusion into the glass. In the SFG experiments,the intensity of the heating pulse was varied tolocate the threshold for melting the Au, and thenthe pulse was attenuated by 20%. Because themelting temperature of Au Tm = 1064C, thisprocedure resulted in flash-heating of the Aulayer to ~800C.

SAMs have been studied extensively bySFG since 1991 (8), but ultrafast probing of aflash-heated SAM requires some elaboration. Inthe SFG technique we used, a femtosecond in-frared (IR) pulse at 3.3 mm with a bandwidth of150 cm1 is incident on the SAM, coherentlyexciting all the alkane CH-stretch transitions inthe 2850 to 3000 cm1 range, along withelectrons in the Au skin layer, producing anoscillating polarization in both the Au and theSAM layers. At the same time, a picosecond-duration 800-nm pulse (visible) with a band-width of 7 cm1 is incident on the sample. Thevisible pulse interacts with this oscillating polar-ization through coherent Raman scattering tocreate a coherent output pulse at the IR + visiblefrequency. This combined IR-Raman interactionis forbidden (in the dipole approximation) incentrosymmetric media because the second-order susceptibility c(2) vanishes in such media.Because the methylene CH2- groups of thealkane SAM form a nearly centrosymmetricsolid, the SFG signal that we observed originatedpredominantly from the Au surface and theterminal methyl CH3 groups. The well-knownSFG spectrum obtained in ppp polarization (4),

from a SAM with n = 17 (i.e., an 18-carbon orC18 SAM), is shown in Fig. 1D. Molecularvibrational transitions appear as dips against abroad nonresonant background from Au. Thesemethyl transitions have a spectral width Dn = 15cm1, corresponding to a coherence decay timeconstant T2 = 0.7 ps, which indicates that SFGsignals are emitted during an ~1 ps time win-dow. Thus the time resolution of these SFGmeasurements is ~1 ps.

Three intense vibrational transitions wereobserved, originating from the symmetric nsCH3and antisymmetric naCH3 methyl stretching vi-brations and from the dCH3 bending overtonetransition, which draws intensity from a 2:1Fermi resonance with the CH stretches (4, 8).All methylene transitions are weak, which isindicative of a high degree of order (4). Figure1D shows the spectrum of a C18 SAM ~400 psafter flash-heating, where the SAM is in equi-librium with Au at ~800C. All three methyltransitions have lost intensity as a result ofthermal disordering of the methyl groups. The2dCH3 band evidences a red shift. The red shiftis caused by thermal excitation of the ~1500 cm1

v = 1 state, which introduces an additionalcontribution from the anharmonically red-shiftedv = 1 v = 3 transition. It is notable thatmethylene transitions remain weak at hightemperature and that the transient intensity lossis reversible once the SAM returns to ambienttemperature. This indicates that chains remainupright and remain bonded to their originalsites. Under ordinary circumstances, alkaneSAMs on Au desorb to form the disulfideCH3-(CH2)n-S-S-(CH2)n-CH3 at 175 to 225C(9, 10), which displays enhanced methyleneSFG transitions, so the unexpected stability ofthese SAMs at 800C must be attributed to thebrief duration of the temperature increase.

We performed molecular simulations of aC16 SAM on Au (27 molecules with periodicboundary conditions) to better understand ther-mal disordering of the terminal methyl groups.When the SAM was equilibrated at 300 K,the well-known (11) all-trans structure with achain tilt of ~35 and a zenith angle (anglebetween surface normal and final C-C bond)of ~25 was obtained. The nsCH3 transitionhas an IR transition dipole moment of mag-

Fig. 2. Results of molecular simu-lations of alkanethiol SAMs. (A)Structure of alkanethiol SAM (n =15). Simulations were performedon a unit cell of 27 alkanes withperiodic boundary conditions. WhenT is increased to a high tempera-ture, the methyl head groups be-come orientationally disordered.(B) The SFG intensity for the nsCH3transition is approximately propor-tional to the square of the nor-malized ensembleaverage IR dipolemoment (m/mIR)

2, which is temper-ature dependent. (C) With aninstantaneous temperature jump to1100 K, the methyl head groupsbecome orientationally disordered in less than 2 ps.

Fig. 3. (A) SFG spectraof C8 (n = 7) and C18(n = 17) SAMs withoutheating pulses (blue) andwith flash-heating to800C (red). (B) VRF fora C8 monolayer. (C) VRFfor a C18 monolayer.

10 AUGUST 2007 VOL 317 SCIENCE www.sciencemag.org788

REPORTS

Vibrational Response Functions (VRFs) of Self-Assembled Monolayers. B VRF for a C8 monolayer. C VRF for a C18 monolayer

From: Wang et al., SCIENCE VOL 317, pp 787-790, 2007

W.Stuhmer et al.

ARCK1 ==

1 SOpA

RCK3 1

J6pA

RCK4

4 0 0 p A~~~~~~4Op

RCK5 1

,,,J@20pA

20ms

BG/Gm

1.2 T +

-80 -40

Fig. 5. Conductance-voltage relations of RCK channels. (A) Familiesof outward currents in response to depolarizing voltage steps. Fromtop to bottom RCKI, RCK3, RCK4, RCK5. The traces are responsesto 50 ms voltage steps from -50 to 40 mV in 10 mV intervals.Ensemble currents recorded from macro-patches. Sampling at 10 kHz,filtering at 3 kHz low pass. (B) Plots of normalized conductance(G/Gm) versus test potential for different RCK channels (RCK1: opencircles; RCK3: crosses; RCK4: diamonds; RCK5: filled circles). Toobtain the conductance values the current at a particular test potentialwas divided by the driving potential assuming a reversal potential of- 100 mV. The lines showed the results of a non-linear least-squaresfit of a Boltzmann isotherm (see Materials and methods) to theconductance values. The maximal conductance (Gm) obtained by the fitwas used to normalize the data. The half-activation voltages in thisplot are -24 mV (RCK1), -37 mV (RCK3), -30 mV (RCK4) and-40 mV (RCK5).

to voltage steps from -60 to 0 mV are shown in Figure 7.The step size of elementary currents at 0 mV varied between0.46 pA (RCK4) and 1.02 pA (RCK5). The single channelcurrent - voltage relations were measured in cell attachedpatches with normal frog Ringer's solution on the extra-cellular side. For all channels, the current-voltage relationis linear in the voltage range -20 to 20 mV. However, sincethis is a rather narrow range for conductance estimation, wemeasured the average amplitudes at 0 mV membranepotential. While the RCK1, RCK3 and RCK5 channels haverather similar single-channel current amplitudes, that of theRCK4 channel is considerably lower (Table I).

Pharmacology of RCK channelsA profile of the pharmacological sensitivity of the differentRCK channels to the K+ channel blockers 4-aminopyridine(4-AP) and tetraethylammonium (TEA) and several basicpeptide toxins was determined. The concentration

RCK1

5LOOpA

RCK3

lOpA

RCK4

I 200pA

RCK5

300pA

1 s

Fig. 6. Inactivation time course of currents mediated by different RCKchannels. Ensemble currents from macro-patches recorded at 0 mV testpotential from oocytes expressing RCKI, RCK3, RCK4 and RCK5channels at 0 mV test potential. Duration of the test pulse was 3.2 s.Holding potential was -80 mV. Note difference in the degree ofinactivation at the end of the 3.2 s pulse. Sampling at 62.5 Hz andlow pass filtering at 120 Hz.

dependence of the block of outward currents by a particularsubstance was determined in whole-cell current recordingsat 20 mV test potential and the results are summarized inTable II. A striking difference in the inhibition of K+currents by TEA is observed between channels formed byRCK1 and RCK5 proteins. The RCK4 channels have a lowersensitivity to 4-AP than the other RCK channels. Both slowlyinactivating channels, RCK1 and RCK5, are much moresensitive to DTX than the inactivating channels RCK3 andRCK4. A different profile is observed for CTX, whichblocks RCK 1, RCK3 and RCK5 well, but is much lesseffective on RCK4 channels.

Discussion

Comparison between K+ channels in neurones andRCK channels expressed in Xenopus oocytesAn important question resulting from the molecular andfunctional diversity of RCK proteins is their relation to K+channels in native membranes. To establish the molecularidentity of a K+ channel in its native cell membrane anda particular RCK channel expressed in Xenopus oocytes,properties such as the voltage and time dependence, thesingle-channel amplitude and the susceptibility to blockersshould be compared. This comparison assumes that the K+channels expressed in oocytes accurately reflect thefunctional properties of K+ channels in the nativemembrane. This has not yet been shown and onlypreliminary conclusions can be drawn on the molecularstructure of K+ channels in native membranes.Delayed K + outward currents which inactivate only

slowly, on a time-scale of hundreds of milliseconds, arefound in neurones of different origins (Hille, 1984). Non-inactivating outward currents, e.g. in PC12 cells or frogspinal cord, are mediated by channels with a low (5-15 pS)channel conductance (Harris et al., 1988; Hoshi and Aldrich,1988a,b). Low conductance non-inactivating K+ channelswhich are DTX sensitive were found in rat sensory neurones(Feltz and Stansfeld, 1988). Non-inactivating K+ channelswhich are DTX sensitive participate in regulating transmitter

3240

Conductance-voltage relations of RCK channels. (A) Outward currents in response to depolarizing voltage steps. From top to bottom RCKI, RCK3, RCK4, RCK5. The traces are responses to 50 ms voltage steps from -50 to 40 mV in 10 mV intervals.

From: Sthmer et al. EMBO Journal vol.8, pp.3235 - 3244, 1989.

Chemical control of metabolically-labeled Shaker channels with CTXBiotin. (A) CTXBiotin (10 nM) inhibits Shaker-IR K+ currents in CHO-K1 cells treated with either 50 M thiol sugar 1 (h) or 0.5% ethanol as a vehicle (s). Only metabolically-labeled (+thiol sugar) channels were irreversibly blocked by CTXBiotin after a simple washout; inhibition was completely reversed by an application of 1 mM TCEP. Reaction profiles were monitored with a 200 ms, 40 mV pulse every 15s.

From: Hua Z, Lvov A, Morin TJ, Kobertz WR. Chemical control of metabolically-engineered voltage-gated K(+) channels. Bioorg Med Chem Lett 2011.

needed to be chemically reversible. Although disulfide bond forma-tion between CTX and thiol-containing sialic acids on the cell sur-face is an obvious chemoselective and cell friendly reaction, wechose to label CTX with a bismaleimide that had an internal disul-fide bond because maleimides are inherently more stable in waterthan MTS reagents. Moreover, this subtle difference would allowfor delivery of a small molecule probe to the modified K+ channelsubunit after cleavage with reductant, which would be useful insubsequent biochemical, biophysical or imaging experiments. Tosimplify the synthesis of the bismaleimide, derivatization of CTX,and ensure delivery of a molecular probe to a K+ channel subunit,we set out to synthesize a symmetrical bismaleimide (Scheme 1)from cystamine dihydrochloride 2 that would allow for the facileincorporation of a molecular probe in the final step of the synthe-sis. The amino groups of cystamine 2were capped with 2 equiv of adoubly amino-protected, activated ester of L-lysine 3. Selectivedeprotection of the Fmoc protecting groups gave the symmetricdiamine 4. Addition of 2 equiv of the NHS-ester of 3-(maleimi-do)propionic acid and N-Boc deprotection afforded bismaleimide5, which was subsequently biotinylated with 2 equiv of NHSBio-tin to give biotin bismaleimide 6. CTX was then derivatized bylabeling a cysteine mutant of CTX (R19C)32 with 100-fold molar ex-cess of biotin bismaleimide 6 to yield CTXBiotin, which was puri-fied by reverse phase HPLC as we have previously described.17,19

With CTXBiotin in hand, we decided to change both the chan-nel (ShakerIR)33 and the expression system (CHO-K1 cells) todemonstrate that our approach was versatile and could be usedto label glycosylated ion-conducting subunits. Inactivation-re-moved Shaker is similar to Q1 in that it is an archetypical volt-age-gated K+ channel with a pore-forming domain flanked by fourvoltage-sensing domains.1 However, each Shaker voltage sensor isdoubly N-glycosylated on the extracellular loop connecting the firsttwo transmembrane segments (S1 and S2 loop).2 Previous studieshave shown that the N-glycans on Shaker contain sialic acid resi-dues.34 Therefore, each tetrameric Shaker channel will have eightN-glycans and potentially up to 2432 sialic acids if every antennaon each glycoconjugate is capped. Figure 3 shows the effects of10 nM CTXBiotin on CHO cells expressing Shaker K+ channelsincubated with either 50 lM thiol sugar 1 or vehicle in perfo-rated-patch whole cell recordings.35 CTXBiotin inhibition ofShaker in vehicle treated cells (circles) was completely reversibleafter 3 min. In striking contrast to control and Q1 channels( Fig. 2C), metabolically-labeled Shaker channels (squares) wereirreversibly inhibited after identical treatment with CTXBiotin.Irreversible inhibition of the Shaker K+ currents was reversed with

tris(2-carboxyethyl)phosphine (TCEP) (1 mM for 2 min), whichreduced the disulfide bond, freeing the toxin moiety and leaving abiotin moiety on a Shaker K+ channel subunit. Rewardingly, both

H2NS

SNH2 BocHN COOPfp

NHFmoc

2 HCl

NHBiotin

NH2

NHBoc

TFA

NH

HN

S

R3

ON

OO

O

NH

HN

S

R2

ON

OO

O

H2N

HN

S

R1

O 2

2

R2=

R3=

4

5

6

a, b

c, d

2

R1=

2 3

+

e

Scheme 1. Reagents and conditions: (a) MeOH, Et3N, rt; (b) piperidine, CH2Cl2, rt; (c) 3-(maleimido)propionic acid N-hydroxysuccinimide ester, DMF, rt, 22% for three steps(ac); (d) TFA, CHCl3, H2O, rt; (e) NHSbiotin, Et3N, rt, 32% for two steps (d and e).

Time (s)0 200 400 600 800 1000 1200

Nor

mal

ized

cur

rent

0.0

0.5

1.0

TCEP

CTX-Biotin

Washout

A

BInitial

Washout

After TCEP

1 nA

50 ms

+ Thiol Sugar

- Thiol Sugar +Thiol Sugar

1 nA

50 ms

- Thiol Sugar

InitialWashout

C

V, mV

I/Imax

AfterTCEP

Initial

1.0

0.5

+ Thiol Sugar

V, mV-80

I/Imax1.0

0.5

- Thiol Sugar

InitialWashout

-60 -40 -20 0 20 40 60 -80 -60 -40 -20 0 20 40 60

Figure 3. Chemical control of metabolically-labeled Shaker channels with CTXBiotin. (A) CTXBiotin (10 nM) inhibits Shaker-IR K+ currents in CHO-K1 cellstreated with either 50 lM thiol sugar 1 (h) or 0.5% ethanol as a vehicle (s). Onlymetabolically-labeled (+thiol sugar) channels were irreversibly blocked by CTXBiotin after a simple washout; inhibition was completely reversed by an applicationof 1 mM TCEP. Reaction profiles were monitored with a 200 ms, 40 mV pulse every15 s. (B) Raw current traces before (black) and after (light gray) CTXBiotintreatment. Traces of the recovered current after TCEP treatment are shown in thedark gray. (C) IV relationship obtained from the same cell before inhibition withCTXBiotin and after the current recovery by either washout (!thiol sugar) orreduction with TCEP (+thiol sugar).

Z. Hua et al. / Bioorg. Med. Chem. Lett. xxx (2011) xxxxxx 3

Please cite this article in press as: Hua, Z.; et al. Bioorg. Med. Chem. Lett. (2011), doi:10.1016/j.bmcl.2011.04.099

Time course of 3-0-methylglucose uptake in isolated muscle cells of Balanus nubilis. Ordinate: ratio of intracellular activity to extracellular activity of 3-0-methylglucose per equivalent volume of bulk external solution. Abscissa: time in hours. External sugar concentration, 1 mM. Uptake was measured using conventional (filled squares) and scintillator probe (open circles) methods. The water content of isolated fibres (70%) is shown by the continuous line above the points. The time at half-equilibration is shown by the dashed line. Mean fiber diameter, 1352 m ; 21 C.

From: Carruthers, A. J. Physiol. VOL 336, pp 377-396, 1983

where Jt is uptake at time t , Jm is uptake a t complete equilibration and k is the rate constant. The experiment'al results of Fig. 2 are wnsisknt with a value of k of 0.195 x s-l. Fig. 1 also shows that uptake is essentially linear during the initial

45 min of exposure to sugar. 3feasurements of sugar uptake in subsequent experiments

were made using an incubation period of 30 min in order to obtain accurate

measurements of t2he initial rate of sugar uptake.

Time (h)

Fig. I . Time course of 3-0-methylglucose uptake in isolated muscle fibres. Ordinate: ratio of i n t r a ~ l l u l a r activity t o extracellula~ activity of 3-0-methy1glucose per equixra1ent volume of bulk external solution, .4bscissa: time in hours. External sugar concentration, 1 m M Uptake was measured using rc)nventional and scintillator probe (0) methods. Kumber of points per conventional determination, five or more. The water content of isolated fibres (70%) is shown by the continuous line above the points. The time a t half-equilibration is shown by thedmhed line. Mean fibre diame*r, 1352 p m ; k m p r a t u r e . 21 OC.

The calculated rate of sugar uptake a t 30 min is 2 pmol . cm-*. s-I. Assuming uptake is not saturated, the permeability of the barnacle muscle fibre, 2' (em. s-I), to 3-0-methy~g~ucose is related to the sugar flux, J (mol. ernv2. s-I), by

where So is the external sugar concentration (in01 . emp3). This corresponds to a value of P of 2 x cm. s-I which is some 3 4 orders of magnitude larger than the

permeability of artificial lipid bilayers to sugars (Jung & Snell, 1968 ; Lidgard & Jones,

19'75).

At equilibrium, the 3-0-methylglucose space of the fibre is 70 yo, which is in close sgreement with estimates of the xra%er contents of barnacle muscle (71 & 1 7; ; R = 5 ) .

Assuming this water is not bound, these results shour that 3-0-methylglucose is not

accumulated by barnacle muscle and that the transfer of the sugar across the

sarco1emma is mediated by a passive, facilitated process. EJat ofphloretia on mqzr uptuke. Re18tively low concentrations of phloretin inhibit

First-order decay analysis of Drosophila embryonic total RNA during normal and slow larval growth conditions. The percentage of embryonic RNA remaining at various times during a chase under normal growth conditions (solid circles) or during slow growth conditions (open circles) is plotted. The regression lines indicate that the stability of embryonic RNA increases from 48 h 115 h if the larval growth rate is reduced.

From: Winkles et al., J. Biol. Chem., Vol 269, pp 7716-7720, 1985

7718 Ribosomal R N A Stability and Growth Conditions

Fraction

FIG. 1. Stability of embryonic total RNA during normal larval growth conditions. Adult females were fed a density-labeled Chlorella paste containing [3H]uridine for 84 h. Embryos were col- lected during the final 14 h of labeling, and some were transferred into a yeast medium containing ['*C]uridine for subsequent develop- ment. Larval chase times were estimated by assuming that the average age of collected embryos was the length of the collection period divided by two, that the duration of embryonic development is 23 h (Poulson, 1950), and that larvae begin feeding immediately after hatching. RNA was prepared from embryos and larvae and centri- fu.ged in CsHC02 equilibrium gradients. The arrow in panel A denotes the position of a ["CIRNA light buoyant density marker which was added to the embryonic RNA sample. A , embryonic RNA; B , 64-h larval RNA; C, 84-h larval RNA.

The autoradiogram is shown in Fig. 3C. The signals evident in the total RNA samples reflect hybridization to rRNA; hybridization to the tRNA carrier is negligible. The signal from sample 2, reflecting the relative amount of rRNA in the stable RNA population whose half-life we have measured above, is approximately 18% higher than the signal from sample 1, representing embryonic RNA. Another experiment using dense RNA from a 50-h chase point gives a similar result. The apparent increase in signal in the RNAs from the chase points is probably not significant, however, given slight variations seen in this kind of hybridization experiment. The results indicate that the fraction of RNA which is rRNA is

Length of chose, h C""

FIG. 2. First-order decay analysis of embryonic total RNA during normal and slow larval growth conditions. The per- centage of embryonic RNA remaining a t various times during a chase with light yeast (normal growth conditions, solid circles) or dense algae (slow growth conditions, open circles) has been plotted. The values in solid circles are derived from the experiment shown in Fig. 1 as well as four other independent experiments (not shown). The values in open circles are derived from the experiment shown in Fig. 4 as well as one other independent experiment (not shown). The regression lines indicate that the stability of embryonic RNA in- creases from 48 h ( r2 = 0.915) to 115 h ( r2 = 0.892) if the larval growth rate is reduced.

not depleted during the chase. The dense RNA sample may be slightly enriched in rRNA, as would be expected if non- rRNA species decay faster than rRNA. Therefore, the half- life of total RNA is a good estimate of rRNA half-life.

To investigate whether the stability of embryonic rRNA varied under different larval growth conditions, we then mea- sured the half-life of this RNA using the L + D protocol as described under "Materials and Methods." A representative experiment using a dense chase is shown in Fig. 4. Adult females were fed a yeast paste containing [3H]uridine for 46 b. Some of the embryos laid during the last 12 h of this radioactive labeling period were transferred into a medium containing 50% 13C,15N-labeled Chlorella cells, 50% 13C,2H,'5N-labeled Chlorella cells, and [14C]uridine. First-in- star larvae were collected after 31 and 67 h of development. Total RNA was prepared and centrifuged to equilibrium in KI/NaI gradients. Since larvae develop and grow more slowly in algal medium, we might expect that the incorporation of dense isotopic precursors into the originally light embryonic RNA pools would be a gradual process. In the extreme case, this would result in an ineffective chase and an inaccurate estimate of stable RNA. As seen in Fig. 4B, after a chase of 31 h the newly synthesized larval [14C]RNA had a wide band width, indicating an RNA population of heterogeneous buoy- ant density. However, at this early chase time, the chase was still effective enough so that the buoyant density of the new larval RNA was greater than that of the inherited embryonic RNA and the amount of stable RNA could be calculated. During the next 36 h of larval development, continued incor- poration of stable isotope produced larval [14C]RNA with a more homogeneous buoyant density distribution and a greater density shift (Fig. 4C).

The half-life of embryonic rRNA under slow larval devel- opment conditions was calculated, as previously discussed, by quantitating the percentage of embryonic RNA remaining a t various times of larval development. A decay curve was gen- erated assuming first-order decay kinetics; experimental val- ues from two independent experiments were used in this analysis (Fig. 2). This analysis indicates that the half-life of embryonic rRNA inherited by slowly developing larvae is

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These time courses may be typical of the measurements you may undertake in your own research or observe when reading the work of other researchers.

The analysis of these time courses has 2 elements:

1. The time courses may be predictable from first principles (e.g. an assumed reaction mechanism) and thus an analysis yields important information about mechanism.

2. The data can be analyzed using theory but the method of analysis may involve:

a. linearization of the data followed by linear regression to obtain the constants related to the underlying mechanism

b. nonlinear fitting to obtain the constants related to the underlying mechanism

c. quality analysis to determine whether the analysis is appropriate.

Goals

1. This chapter will review the classification of reaction orders and the limitations of this classification.

2. The tools available to analyze reactions and the quality or appropriateness of the underlying analysis.

23

LEARNING OBJECTIVES

1. Reaction orders

1. Zero-order

2. First order

3. Second order

1. Class 1

2. Class 2 (pseudo first-order)

SECTION 2

Reaction orderRate of a reaction

The rate or velocity, v, of a reaction or process describes how fast it occurs. The velocity is expressed as a change in concentration (C) per unit time (t),

v dtdC=

but may also express the change in a population of cells with time, the increase or decrease in the pressure of gas with time or the change in absorption of light by a colored solution with time.The order of a reactiondescribes how the velocity of the reaction depends upon the concentration of reactants.In the (irreversible) isomerization reaction

A Bk1the theory of chemical kinetics tells us that

dtd[B] k [A] k [A]1 m 1= =

Since m = 1, this reaction is first order with respect to A and since A is the only independent concentration variable in the rate equation, the reaction is overall first-order.

24

The units for this first order reaction are derived from moles of product formed per second per mole of reactant or,

Mper sec k M1=

MMper sec k per sec1= =

In a reaction of the type

E S E Sk1+ $The rate of the reaction is proportional to [E].[S].

dtd[ES] k [E] [S] k [E] [S]1 1 1 1= =

Because m = 1 for both species the reaction is first-order with respect to E or S but is second order overall as one single step is involved in the reaction of two species.The units of this second order reaction are derived from moles of product formed per second per mole2 of reactants or

molarity per sec = k1 (molarity)2

MM per sec k perM per sec2 1= =

The Order of a Reaction must be determined experimentallyUnderstanding the stoichiometry of a reaction is not sucient to predict the rate law of the reaction. This is illustrated in the table below.

If the concentration of a reactant remains unchanged by the reaction, it is frequently omitted in the rate-law expression. For example, with the first reaction, a more complete rate law is:

v = k[sucrose][H+][H2O](the reaction is 3rd order overall). However, H+ is a catalyst and its [ ] is constant during the run; [H2O] (solvent) is little changed because of its vast excess (55.5 M). Thus the terms [H+] and [H2O] are omitted in the rate law. If the reaction were carried out at varying [H+] or in an inert solvent, a first-order dependence of the reaction on [H+] and on [H2O] is seen.

25

The first column in the table indicates only stoichiometry, NOT reaction mechanism or order.Reaction 3 Dinitrogen pentoxide decomposition

N2O5 NO2 + NO3

NO3 NO2 + O

2O O2

Hence, the first step is first order. 2N2O5 was indicated to balance the equation. It could just as easily have been written as: N2O5 2NO2 + O2

Reaction 4 Nitrogen dioxide decomposition to nitric oxide and oxygen.This involves formation of an intermediate thought to be the one shown in the reaction below

2NO2 ONOONO 2NO + O2

The first step is therefore second order.

Examples of reaction orders in natureTo understand how we can distinguish reaction orders experimentally, we will examine the following reactions1

Zero-order kinetics First order kinetics True Second order kinetics Second order kinetics characterized by pseudo-

first order behavior.

1The illustrated reactions are available in the file: CoreKinetics.pzf This file is a GraphPad Prism file that contains the data for each type of plot shown and the types of analysis made.If you wish to plot these data yourself, you should download the file from the Core Curriculum website:http://inside.umassmed.edu/uploadedFiles/gsbs/courses/2012-2013_Core_Course_Files/CoreKinetics.pzf%20-%20for%20students%20only.zipand download the GraphPad Prism version 6 software fromPrism 6 Winhttp://cdn.graphpad.com/downloads/prism/6/InstallPrism6.exePrism 6 Windows serial number: GPW6-200512-LEM5-16772Prism 6 Machttp://cdn.graphpad.com/downloads/prism/6/InstallPrism6.dmgPrism 6 Mac serial number: GPM6-200513-LEM5-F3EF2

26

A zero-order reaction

Zero-order reaction

0 2 4 6 8 100

2

4

6

8

10

TIME

[Sub

stra

te] o

r [Pr

oduc

t]

SubstrateProduct

Note that [substrate] decreases linearly with time and [product] increases linearly with time. An example of such a reaction is ethanol conversion to acetaldehyde by the liver enzyme, alcohol dehydrogenase (ALDH). The oxidizing agent is nicotinamide adenine dinucleotide (NAD+) and the reaction can be written:

CH3CH2OH + NAD+ CH3CHO + NADH + H+

At saturating [alcohol] (about 2 beers) and with NAD+ buered via metabolic reactions that restore it rapidly, the rate of this reaction in the liver is zero-order over

most of its course (ALDH is saturated by ethanol & NAD+).

zero-order kinetics

0 100 2000

20

40

60

80

100

[S] M

v (

d[P

]/dt)

v dtd[ethanol]

dtd[acetaldehyde]

k0=- = =

The negative sign is used with reactant - ethanol - because its concentration decreases with time. The concentration of its product, acetaldehyde, increases with time.

t00

C0

C

CH3CH2OH

CH3CHO

27

ALDH

Theory of Zero-order ReactionsA zero-order reaction corresponds to the dierential rate law

dtdC k0=

The units of k0 are molarity per sec. This is a zero-order reaction because there is no concentration term in the right hand of the equation.Defining C0 as the concentration at zero time and C as the concentration at any other time, the integrated rate law is:

C C k t0 0= +

y = y-intercept + slope * xThis is the equation for a linear relation between the independent (time) and dependent (concentration) variables. We can therefore subject the raw data to linear regression analysis to obtain C0 (y-intercept) and k0 (the slope).

Zero-order reaction

0 2 4 6 8 100

2

4

6

8

10

TIME

[Sub

stra

te] o

r [Pr

oduc

t]

SubstrateProduct

y = x-intercept + slope * x

Best-t values" " " " Substrate" " Product" " Units Slope" " " " " -1 0 " " 1 0 mols/sec Y-intercept when X=0.0" " 10 0" " 0 0"" " mols X-intercept when Y=0.0" " 10.00"" " 0" " " sec Goodness of Fit" " R squared" " " " 1.000"" " 1.000

General rules for zero-order reactions1. Plot of St or Pt vs time produces a straight line with

slope = -k (for St) or k (for Pt)2. k has units of mols produced or consumed per unit

time3. Zero-order, enzyme catalyzed kinetics are typically

observed at saturating [S]

28

A first-order reaction

1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B] [Substrate]

[Product]

Note here that [substrate] decreases in a curvilinear fashion with time and [product] increases in a curvilinear manner with time. This observation indicates that the reaction is NOT zero-order. How can we analyze this further?

Theory of First-order ReactionsA first-order reaction corresponds to the dierential rate-law:

dtdC k C1=

The units of k1 are time-1 (e.g. s-1). There are no concentration units in k1 so we do not need to know absolute concentrations - only relative concentrations are needed.The reaction

A Bk1

has the rate law

v dtd[A]

dtd[B] k A][1=- = =

where k1 is the rate constant for this reaction. The velocity may be expressed in terms of either the rate of disappearance of reactant (-d[A]/dt) or the rate of appearance of product (d[P]/dt).

29

First Order reactions - loss of substrateTheory

d[A] k [A] t01- =Defining [A]0 as [A] at time = 0 and integrating between A at time zero and time t gives

ln [A] k t ln[A]1 0=- +

y = slope x + intercept

First Order reactions - loss of substrateIntegrated rate law

[A] [A] e0 k t1= -

Half-lifeDefining [A] at t1/2 as [A]0/2

t kln2

k0.693

1/21 1

= =

and because = 1/k1, t1/2 = 0.693

30

First Order reactions - product formationTheoryDefining [B] as [B] at equilibrium and assuming that all A is converted to B, [A] at any time t can be calculated from [B] at time t as [A]t = [B] - [B]t

Thus we obtain

ln([B] [B] ) k t ln[B]1 t - =- +

y = slope x + intercept

First Order reactions - product formationIntegrated rate law

[B] [B] e{ }1 k t1= - -

Half-lifeDefining [B] at t1/2 as [B]/2

t kln2

k0.693

1/21 1

= =

and because = 1/k1, t1/2 = 0.693

31

Returning to our example of a first order reaction,1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B] [Substrate]

[Product]

The raw data suggest that [substrate] falls from 5 mM to an equilibrium value of 0 mM.If we plot the log [substrate] vs time (or show the y-axis data on a log scale), we obtain

1stOrder

0 2 4 6 8 100.01

0.1

1

10

TIME

[A]

[Substrate]

This produces a linear plot which is consistent with 1st order kinetics!

A second clue comes from the measurement of half-times. As [Substrate] declines from 5 - 2.5 mM, from 2.5 - 1.25 mM and from 1.25 to 0.625 mM, the time required for each 50% reduction is unchanged at 1.4 sec.This is characteristic of first-order decay as observed with radioactive decay.

1st Order

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

TIME

[Sub

stra

te]

1.4 sec

1.4 sec

1.4 sec

Constant decay times and the linear relationship between log {[A]t - [A]} vs time indicate a first order process. Let us now check this by applying a first-order analysis to the data.

32

Non-linear regression analysisTo do this we subject the data to nonlinear regression (the plot is nonlinear) using an appropriate equation for first-order reactions.The integrated rate law for first-order substrate loss is

[A] [A] e0 k t1= -

Nonlinear regression finds the values of those parameters of the equation (k1 and [A]0) that generate a curve that comes closest to the data. The result is the best possible estimate of the values of those parameters.To use nonlinear regression, therefore, you must choose a model or enter one. GraphPad Prism oers a model for first-order reactions called One-Phase DecayThe equation is:

Y=(Y0 - Plateau)*exp(-k*X) + PlateauIn which the parameters are defined as:1. Y0 is the Y value when X (time) is zero or [A]0 in this case.2. Plateau is the Y value at infinite time (0 for our data set).3. k is the rate constant k1 (per unit time).4. Span is the dierence between Y0 and Plateau and is [A]0 in this

case because Y0-plateau = [A]0

Every nonlinear regression method follows these steps:1. Start with initial estimated values for each parameter in the

equation.2. Generate the curve defined by the initial values. Calculate the

sum-of-squares - the sum of the squares of the vertical distances of the points from the curve.

3. Adjust the parameters to make the curve come closer to the data points - to reduce the sum-of-squares. There are several algorithms for adjusting the parameters - Prism uses the Marquardt algorithm.

4. Adjust the parameters again so that the curve comes even closer to the points. Repeat.

5. Stop the calculations when the adjustments make virtually no dierence in the sum-of-squares.

6. Report the best-fit results. The precise values you obtain will depend in part on the initial values chosen in step 1 and the stopping criteria of step 5. This means that repeat analyses of the same data will not always give exactly the same results.

33

1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[Sub

stra

te] o

r [Pr

oduc

t][Substrate][Product]

Y=(Y0 - Plateau)*exp(-k*t) + PlateauOne phase decay"Perfect t" " [Substrate]" [Product]" " UnitsBest-t values" " Y0"" " " " " 5.000"" " 0 mM Plateau" " " " " 0" " " 5.000 mM k" " " " " " 0.5000" " 0.5000 per sec Half Life"" " " " 1.386 " " 1.386 sec Tau = 1/k" " " " 2.000"" " 2.000 sec Goodness of Fit" " Degrees of Freedom" " 48" " " 48 R square"" " " " 1.000"" " 1.000

General rules for 1st order reactions1. First-order enzyme catalyzed kinetics are typically

observed at subsaturating [S] 2. Plot of log (St-S) vs time produces a straight line

with slope = -k3. The half-time (t1/2) and k are invariant of the

starting value of St chosen.4. Plot of log (P-Pt) vs time produces a straight line

with slope = -k5. t1/2 = 0.693/k

6. k has units of time-1 (e.g. s-1). There are no concentration units in k so we need not know absolute concentrations - only relative concentrations are needed.

7. k may be obtained by direct curve fitting procedures using nonlinear regression

8. The full equation for loss of substrate is [S]t = {[S]0 - [S]} e-(k.t) + [S]

9. The full equation for product formation is [P]t = [P] (1 - e-(k.t))

10. When a first order reaction is reversible (as most are), e.g.

A Bkk

2

1

The equations are unchanged but now k = k1 + k2

34

Second-order reactionsFall into 2 categories in which the rate law depends upon: 1. the second power of a single reactant species, or 2. the product of the concentrations of two dierent

reagents.Class 1 reactions (A+A P)

The dierential rate law is

v=k[A]2

Although one or more reactants may be involved, the rate law for many reactions depends only on the second power of a single component. e.g.

2 proflavin [proflavin]2v k[proflavin] 2=

2 AAGCUUAAGCUU

UUCGAA

v k[A GCU ]2 2 2=

2nd Order class 1

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B]

[Substrate]

[Product]

[substrate] decreases and [product] increases in a curvilinear fashion with time. This indicates that the reaction is NOT zero-order. How can we analyze this further?The curves drawn through the points were computed by nonlinear regression assuming first order kinetics (one-phase decay equation). Note the systematic deviations from the fit. This strongly suggests that this reaction does not follow first order kinetics.We can investigate this further by using GraphPad Prism to plot the residuals of the fit (how each point deviates from the calculated fit) vs time.A non-random scatter of residuals around the origin (perfect fit) would confirm a poor fit and that we should consider either an error in data sampling or another model for the data.

35

Nonlin fit of 2ndOrderIrrev:Residuals

0 1 2 3 4 5 6 7 8 9 10-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

TIME

[Substrate][Product]

This plot therefore shows that this is not a 1st order reaction

Theory of Class 1 Second-order ReactionsDefining [A] at zero-time = [A]0, it can be shown that the integrated rate law is

[A]1

[A]1 k t

0- =

adding 1/[A]0 to both sides gives

[A]1 k t [A]

10

= +multiplying both sides by [A]0 gives

[A][A] [A] k t 10 0= +

Thus one expects a linear relation between the reciprocal of the reactant concentration and time.

Class 1, 2nd order Transform of data

0 2 4 6 8 100

2

4

6

8

TIME

[A] 0

/[A]

1st order data2nd order data

Here we re-plot the data from the previous page (open circles, a second order reaction) as well as data from a true first order reaction (closed circles) as suggested by the 2nd-order linearized equation. As you can see, transformation of the 2nd order data produces a straight line with slope [A]0 k and y-intercept = 1.The slope [A]0 k indicates that the rate of loss of [A] will increase linearly with [A]0This is infact observed

36

How starting [A] affects rate of 2nd order reaction

0 2 4 6 8 100

5

10

15

TIME

[A] 0

/[A]

12345678910

Increasing[A]0

A0k (slope) vs A0 second order

0 2 4 6 8 100.0

0.5

1.0

1.5

[A]o

[A] 0

k p

er s

ec

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

[A]0 k per sec

0.1320 2.842e-009-3.974e-009 1.763e-0083.010e-0087.576

General rules for 2nd order reactions (Class 1)1. Standard 1st order analysis does not work2. Plotting [A]0/[A] vs time produces a straight line

with slope [A]0 k3. Plotting slope vs [A]0 produces a straight line with

slope k and y-intercept 0.4. The units of k are concentration-1.time-1.

Although we do not show it here, this analysis breaks down when the reaction is reversible. Thus in the reaction

A A Pkk

r

f+the kinetics more closely resemble 1st-order kinetics when when kr kf/10In fact, this general analysis of 2nd-order Class 1 kinetics derives from classical irreversible chemical kinetics which have only limited application in biology.2nd-order reactions - Class 2 (A+BP) AKA pseudo-first order A reaction that is 2nd order overall is 1st order with respect to each of the two reactants.For example, in the reaction

E S E Skk

2

1+ $

If the enzyme E were maintained at a constant low [ ] (e.g. [E] < [S]/100) and the substrate were varied, the reaction dierential rate law is:

v dtd[ES] [E]k [S] k [ES]1 2= = -

37

Let us review this by examining ligand (L) binding to a receptor (R).

R L R Lkk

r

f+ $(note the forward and reverse rate constants have now been called kf and kr but this name change is purely arbitrary - they could have been called kon and ko or k1 and k2)

Upon rapid mixing of R and L, the receptor may undergo a fluorescence change allowing measurement of ligand binding. Alternatively, it may be possible to measure ligand binding by use of radiolabeled ligand and filter-bound receptor. Either way, the time course of ligand binding may be examined to determine whether it displays first or second order kinetics.At zero-time, various concentrations of L (M) were mixed with 1 nM R. The time course of LR formation was monitored at each [L].

Pseudo 1st Order

0 5 100.0000

0.0002

0.0004

0.0006

0.0008

0.0010

TIME

[LR

] M

.167

.278

.1

.464

.774

1.292

2.154

3.594

5.995

10

[L]

This can also be plotted with the x-axis (time) shown as a log scale - this allows us to observe the data at short time intervals more closely

Pseudo 1st Order

0.1 1 100.0000

0.0002

0.0004

0.0006

0.0008

0.0010

TIME

[LR

] M

.167

.278

.1

.464

.774

1.292

2.154

3.594

5.995

10

[L]

The data were fitted by nonlinear regression using GraphPad Prism and the one-phase association equation. The fit is excellent in each case (the residuals < [LR]/100)You can also see that the reaction becomes faster at higher [L] i.e. k increases and t1/2 falls with increasing [L].Each curve fit produces a value of k (typically called kobs because it is an experimentally observed k) for each starting [L].We can analyze this further by plotting kobs vs [L]

38

kobs vs L

0 2 4 6 8 100

5

10

15

20

25

k obs

per

sec

[L] M

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

kobs per sec

1.999 0.00018610.5012 0.0007352-0.25070.5002

We will show below that:1. The slope is kf2. The y-intercept is kr3. The x-intercept is -kr/kf

Theory for pseudo first-order reactionsFor our reaction

R L R Lkk

r

f+ $

The rate of LR formation is given by:

v dtd[LR] [R]k [L] k [LR]f r= = -

If [R]0 is the amount of receptor at t = 0, it can be shown that the integrated rate law is:

[LR] [R] constant(1 e )0 t(k k [L])r f= - - +

1. The time dependent component of this expression is e-t (kr + kf [L]) = e-t kobs.

2. Thus kobs = (kr+kf[L])3. In a plot of kobs versus [L], kobs increases linearly

with [L] (slope = kf) and the y-intercept = kr.4. The x-intercept (when kobs = 0) = -kr/kf. Why?

0=kr + kf[L]; thus -kf[L] = kr; thus -[L] = kr/kf5. Analysis of the time course of L binding to R at

varying [L] permits computation of kf, kr and kf/kr = Keq for the reaction.

6. This is ONLY true when [L] >> [R]. Here, first-order kinetics are observed because [L] does not change significantly. If [L] [R] the system will behave like a class 1 second order reaction.

39

SECTION 3

Recap of Key points

1. What is the dierence between a first-order reaction and a second-order reaction that behaves like a first order reaction?1. A true first-order reaction is characterized by a

rate-constant, k, that is independent of [substrate] or [product]. t1/2 is independent of [substrate].

2. A second-order reaction that behaves like a first order reaction (e.g. see this plot) is called a pseudo-first-order reaction. Its rate constant, kobs, increases linearly with [S] (i.e. kobs = kr+kf[S]). t1/2 falls with increasing [substrate].

2. What is the dierence between a class 1 second order reaction and a class 2 second-order (pseudo-first-order) reaction?1. A class 1 2nd order reaction is not described

accurately by first order equations but when 1/[S] is plotted vs time, the plot is linear.

2. kobs for a class 1 2nd order reaction is k[S]0 and when [S]0 is 0, kobs = 0

3. A class 2 2nd order reaction is described accurately by first order equations.

4. kobs for a class 2 2nd order reaction is kf[S]0 + kr and when [S]0 is 0, kobs = kr.

Summary for Reaction order and kinetics1. The reaction order describes how the velocity of

the reaction depends upon the concentration of reactants.1. In the reaction, E+S ES, the reaction is first

order with respect to [E] at fixed [S], first order with respect to [S] at fixed [E] but second-order overall.

2. In the reaction, S+S P, the reaction is second-order with respect to [S].

2. Zero-order reactions occur at a constant rate even as substrate levels fall. 1. A plot of [S] vs time for a zero-order reaction is

linear with slope = -k2. k has units of mol consumed or produced per

unit time.3. First order reactions are non linear with time

1. A plot of St vs time is described by[S]t = {[S]0 - [S]} e-(k.t) + [S]

2. Plot of Pt vs time is described by[P]t = [P] (1 - e-(k.t))

40

3. t1/2 = 0.693/k

4. k has units of time-1 (e.g. s-1).5. The half-time (t1/2) and k are invariant of the

starting value of [S].4. There are two classes of second order reactions.

1. In reactions where two molecules of a substrate combine to form a product (Class 1 reactions), the reaction is non linear with time1. Plots of [S]0/[S] vs time are linear with slope

= [S]0 k. This slope has units of per sec2. Plots of [S]0 k vs [S]0 are linear with y-

intercept = 0 and slope = true k.3. These rules break down for reversible

reactions.4. k has units of concentration-1 time-1 (e.g.

M-1.s-1).2. In reactions where two dierent molecular

species combine to form a product (Class 2 reactions), the reaction is non linear with time1. If one species (e.g. an enzyme or receptor)

is held at a fixed and very low [ ] relative to its substrate or ligand, the reaction is pseudo-first order with respect to [substrate] or [ligand].

2. If [S] is varied and [ES] is plotted as a function of time, each curve is described by first-order kinetics but now:1. kobs increases and t1/2 decreases with [S]2. kobs = kr + kf [S]3. kf has units of M-1.s-1 and kr units of s-14. Keq can be obtained as kf / kr

41

SECTION 4

Formative self-evaluation questionsTest your understanding of time-dependent processes by answering the following questions1. What are the units of zero-, first- and second-

order rate constants?2. Does the stoichiometry of a reaction always

predict reaction order and mechanism?3. What is the defining characteristic of a zero-order

reaction?4. How do you compute the value of a zero-order

rate constant?5. Do first-order reactions show a linear dependence

on time?6. What is the t1/2 of a first-order reaction?7. How does t1/2 of a true first-order reaction vary

with [S]?8. How is t1/2 related to the rate constant k of a first-

order reaction?9. In the 1st order reaction S P, why is a log plot of

St vs time not always the best approach to confirm a 1st order reaction?

10. In a 1st order reaction, why is a log plot of Pt vs time never the best approach to confirm a 1st order reaction? At fixed [enzyme], what concentration of [S] produce zero-order kinetics?

11. At fixed [enzyme], what concentration of [S] produce zero-order kinetics?

12. At fixed [enzyme], what concentration of [S] produce first-order kinetics?

13. What is the dierence between Class 1 and Class 2 second-order kinetics?

14. Why is it that a Class 2 second order reaction can behave like a first-order reaction?

15. How may we compute the rate constant k for a Class 1 second-order reaction?

16. How does the rate constant k for a Class 1 second-order reaction vary with [S]?

17. When does this type of analysis break down?18. How does kobs vary with [S] for a Class 2 second-

order reaction?19. How can we use this relationship to compute Kd, kf

and kr for a Class 2 second-order reaction?20. What are the major dierences between first-order

kinetics and Class 2 second-order kinetics?21. Why, then are Class 2 second-order kinetics called

pseudo-first order kinetics?

42

SECTION 5

Key to formative evaluations

Section 2 - Reaction order

1. The units are:

1. zero-order = mol/s (amount per unit time)

2. first-order = per sec (per unit time)

3. second-order = per M per sec (per amount per unit time)

2. No. The order of a reaction and mechanism must be determined experimentally. Stoichiometry simply shows a balanced reaction.

3. A zero-order reaction proceeds ([S] falls or [P] increases) linearly with time.

4. Plotting [S] vs time yields a straight line with slope = -k. Plotting [P] vs time yields a straight line with slope = k.

5. No. Plots of [S] or [P] vs time are curvilinear.

6. t1/2 or the half-time of a reaction is the time required for [S] to decrease by 50%.

7. t1/2 for a true first-order reaction is independent of the starting [S].

8. t1/2 = 0.693/k

9. A log plot of [S]t vs time will only produce a straight line if all of S is converted to P. What will work in all cases is to plot log([S]t - [S]) vs time where [S] is that concentration of S that remains when the reaction achieves equilibrium.

10. A log plot of [P]t vs time can never produce a straight line because [P] increases with time. You have to invert the [P]t vs time data so that it now resembles [S]t vs time data. This can be achieved by measuring [P] (that concentration of P produced when the reactions attains equilibrium) then calculating [P]-[P]t and plotting that result vs time. Thus a plot of log([P]-[P]t) vs time will

43

produce a straight line. This analysis assumes that [P]0 (that [P] present at zero-time) is 0. If [P]0 > 0 then [P]0 must be subtracted from [P] and [P]t for this analysis to work.

11. Zero-order kinetics are observed when [S] >>> Km.

12. First-order kinetics are observed when [S]

CHAPTER 3

Steady-state kinetics of enzyme-catalyzed Lorem ipsum dolor sit amet, suspendisse nulla pretium, rhoncus tempor placerat fermentum, enim integer ad vestibulum volutpat. Nisl rhoncus turpis est, vel elit, congue wisi enim nunc ultricies dolor sit, magna tincidunt. Maecenas aliquam est maecenas ligula nostra.

This chapter considers receptor-ligand equilibria and enzyme catalyzed reactions. We seek to:

1. rationalize simple Michaelis-menten behavior

2. provide a set of tools for routine analysis of enzyme-catalyzed reactions

E

E + P

E + I EI

ES + I ESIE

I

I

I

I

S S

S S

EC C

C C

kp

E + P

kp

+S

+S

Ki

Ks

A key step in enzyme function is the formation of the enzyme (E)/substrate (S) complex, ES.A number of observations point to the existence of ES prior to the release of product (P) and regeneration of free E. These are:1. ES complexes have been directly visualized by EM

and X-ray crystallography.Jack Grith developed techniques that let scientists see the finer details of DNA. In 1971 he and Arthur Kornberg published this photo, the first electron microscope image of DNA bound to a known protein - DNA polymerase.

Several observation tell us that an enzyme E and its substrate S combine to form a complex.1. The physical properties of an enzyme can change

upon binding S.2. The spectroscopic characteristics of E and S can

change upon ES formation.3. High specificity for ES formation is observed.4. The ES complex may be isolated in pure form. 5. At constant [E], increasing [S] results in increased

product formation to a point where product formation no longer increases. This saturation is presumed to reflect the fact that all E is now in the form ES. This is illustrated below.

Vm

0.5 Vm

Km

0 10 20 30 40 500

20

40

60

80

100

120

[S] mM

v (d

[P]/d

t), r

ate

of r

eact

ion

xlvi

Context

LEARNING OBJECTIVES

1. Enzymes are biological catalysts

2. Enzymes combine reversibly with substrates to form products

3. Enzymes show saturability

4. The behavior of simple enzymes can be rationalized in terms of the Michaelis Menten equation

0.01 0.1 1 10 100 1000

0

25

50

75

100

[S] M

v (

d[P

]/dt)

[S] mM

Vm is a theoretical, maximum value for v.Km is that concentration of [S] producing a v of Vm/2.

The reaction velocity curve is a section of a single, rectangular hyperbola which in this instance takes the generic form

v K [S]V [S]m

m= +

This equation is called the Michaelis-Menten equation.Our challenge in this chapter is to understand why the phenomenon of enzyme-mediated catalysis is well approximated by this relationship.

xlvii

LEARNING OBJECTIVES

1. Breaking down an enzyme-catalyzed reaction into its various parts

2. Understanding enzyme-substrate interactions

1. Analyzing enzyme-substrate interactions

2. Introducing the catalytic step

3. Analyzing enzyme-catalyzed reactions

3. Inhibition of enzyme-catayzed reactions

1. Competitive

2. Noncompetitive

3. Uncompetitive

4. What does Km represent?

5. What does Vm represent?

6. Kinetic perfection

SECTION 1

Michaelis-Menten KineticsThe steps in enzyme-mediated catalysisLet us examine an enzyme catalyzed reaction.

E S ES EP E Pkk

k

k

k

k

r

f

2

2

p

p+ +- -

Here enzyme E reacts with substrate S to form the complex ES. ES is then converted to EP which dissociates to E and product (P). The rate constants kf, kr, k2, k-2, kp and k-p describe the various steps involved in the reaction:kr, k2, k-2, and kp are first-order rate constantskf and k-p are second-order rate constantsThis may be represented as a reaction coordinate - an abstract one-dimensional coordinate representing progress along a reaction pathway.

E+S

ES

EP

E+P

(E..S)

(ES..EP)(E...P)

1

2

34 5

6

7

0

+

-Energy

Reaction coordinate

48

Deriving the equations that describe this reaction scheme is a very significant undertaking. Moreover, the equation describing the rate of reaction in terms of substrate and product levels and rate constants is quite complex. However, we can make a number of simplifying assumptions in order to more readily obtain a solution to this scheme. How do we do this?

Assume that the reverse reaction (P S) is negligible. While enzymes accelerate both forward and reverse reactions to the same extent, we (as the biochemists working with this enzyme) can establish experimental conditions that preclude or minimize the reverse reaction.

e.g. adding an additional enzyme which converts P into another species Q which cannot react with our enzyme. Or, we can measure the rate of reaction at very early time points where the reverse reaction is insignificant.

Assume only a single central complex (ES) exists. i.e. ES breaks down directly to E + P.

Make certain that [S] >> [E]. Thus the instantaneous interaction of S and E to form ES does not significantly aect free [S] (although [S] will slowly fall due to its conversion to P).

Typically, ([S]-[ES])/[S] 99.9%

With these assumptions in mind, the reaction scheme now becomes:

E S ES E Pkk k

r

f p+ +There are two parts to this reaction:1) Formation of ES2) ES breakdown to product P and free enzyme E

The formation of ES is a second order process and the breakdown of ES to E + S or to E + P are first order processes. The units are:1. kr = kp = per sec. 2. kf = per M per sec. You will see later why these

units are important.

49

Enzyme/substrate interactions Receptor/ligand interactionsThe first step in enzyme function is the formation of the enzyme (E)/substrate (S) complex, ES.Consider the reaction:

E S ESkk

r

f+

The rate of ES formation is given by:

dtd[ES] k [E] [S]f=

The rate of ES breakdown is given by:

dtd[ES] k [ES]r- =

Thus at equilibrium

dtd[ES]

dtd[ES] k [E] [S] k [ES]f r=- =`

Hence

[ES] kk [E] [S]r

f=

If we seek to understand how much substrate is bound (i.e. [ES]) at any given [S] and [E], we can express [ES] as a fraction of total enzyme [E]t as:

[E][ES]

[E] [ES][ES]

t= +

Substituting for [ES] we obtain

[E][ES]

[E] kk [E] [S]

kk [E] [S]

t

r

f

r

f

=+

Cancelling [E] we find

[E][ES]

1 kk [S]

kk [S]

t

r

f

r

f

=+

and rearranging we obtain the fraction of enzyme complexed with S at any given [S], [E]t as

[E][ES]

kk [S]

[S