Enzyme Kinetics and Ligand Binding

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catalyst - an agent that accelerates a chemical reaction but which is unchanged in amount or chemistry at the end of the reaction. enzyme - a biological catalyst.
ANTHONY CARRUTHERS, BLOCK 1

Enzyme Kinetics

Fluorescence (%)

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0 mM 1 mM 2 mM 3 mM 4 mM 5 mM 12.5 mM

8 6 4 2 0 0.001

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GLUCOSE BINDING TO A GLUCOSE SENSOR

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Foreword

Graduate students often characterize “kinetics” as uninteresting. I have observed several explanations for this: • The subject matter may be difficult 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:

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 timedependence and the mechanisms of enzymemediated 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.

• 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

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Equilibria

Fractional equilibration

C HAPTER 1

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3. How do enzymes accelerate reactions?

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Fractional Equilibration

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

10.0 Time (min)

This chapter considers reversible chemical reactions. We ask: 1. What is a reversible chemical reaction and what are its characteristics?

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

S ECTION 1

Reversible Reactions

L EARNING 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.

Law of Mass Action A 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. This relationship is unique to each reaction and reaction conditions. 3

For example, the reaction between hydrogen and iodine

For the reaction

H2 + I2 Glucose + ATP

k1

Glucose-6-phosphate + ADP

[HI]2 K eq = [H 2 ][I 2 ]

k-1

K eq =

2HI;

[Glucose − 6 − phosphate]e [ADP]e k1 = [Glucose]e [ATP]e k−1

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.

or, in general pA + qB + rC

xD + yE + zF;

[D]x [E] y [F]z K eq = [ A] p [B]q [C]r The equilibrium constant for any specific reaction (at a given temperature) is rather like a fingerprint - it is unique.

4

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.

Formative self-evaluation questions In the reaction

S

k1 k -1

P

1. Define Keq in terms of k1, k-1 2. Define Keq in terms of [S]eq, [P]eq

[Glucose − 6 − phosphate]e [ADP]e K eq = [Glucose]e [ATP]e

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

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

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

5

S ECTION 2

The chemical reaction S ⇌ P

In which direction will a reversible reaction proceed?

can be described by the following energy diagram:

X

Energy

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

Example 1

∆GA S ∆G

P Progress of Reaction

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 Example 3

Example 2

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

Energy

L EARNING O BJECTIVES

X

∆GA P S

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)

∆GA ∆G

S

Energy

X

P ∆G = 0

Progress of Reaction

Progress of Reaction

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

Chemical potential and ∆G

Forms of Energy

The chemical potential of a molecule is a measure of the ability of the molecule to perform work.

Free Energy (G) - performs work at constant temperature and pressure.

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

Heat Energy (enthalpy, H) - performs work only through a change in temperature.

Consider the reaction: S ⇌ P The “chemical potential” or “partial molar free energy” of S, for example is given by:

µS = µºS + RT ln[S]

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:

We will discuss the meaning of µºs 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 = µP - µS At 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,

µP = µS Thus,

∆G = 0

∆G = ∆H - T∆S Thus:



∆G α T







When T∆S > ∆H, ∆G < 0







When T∆S < ∆H, ∆G > 0

What 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:

Formative self-evaluation questions In the reversible reaction S⇌P 1.

Define ∆G and the relationship between µS and µP when the reaction is at equilibrium.

2.

Define ∆G and the relationship between µS and µP when the reaction proceeds in a net direction from left to right.

3.

Define the ∆G and the relationship between µS and µP when the reaction proceeds in a net direction from right to left.

4.

At equilibrium, what is the relationship between ∆H and T∆S?

∆G = -T∆S The fall in free energy is due to increased ∆S. The molecules redistribute to maximize system entropy. Summary 1. A reaction occurs spontaneously only if ∆G is negative (µP < µS). 2. A system is at equilibrium (forward and reverse reactions are balanced) and no net change occurs when ∆G = 0 (µP = µS) 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.

8

S ECTION 3

∆G and equilibria

To 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

[C] [D] G = G o + RTln [A] [B] ∆Gº is the standard free energy change R = gas constant (1.98 cal/mol/degree)1 T = absolute temperature LEARNING

O BJECTIVES

[A], [B], [C] and [D] are the molar activities of the reactants under starting conditions.

1. Understanding starting, standard and equilibrium conditions. 2. The relationships between ∆G, ∆Gº and Keq. 3. Understanding that standard free energy changes are additive and how this is exploited in nature.

1one

calorie (cal) is that amount of heat required to raise the temperature of 1 gram of water from 14.5 ºC to 15.5 ºC. 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)

9

∆Gº is the free energy change under standard conditions i.e.

Defining Keq as

[C] [D] K eq = [A] [B] ∆Gº = -RT ln Keq

[A] = [B] = [C] = [D] = 1 M

= -2.303 RT log10 Keq

P = 1 Atmosphere pH = 7.0

Rearranging gives us

T = 298ºK = 25ºC

Keq = e-∆Gº/RT

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 ∆Gº.

At equilibrium, ∆G = 0 and the ratio [C] [D] [A] [B] remains constant (i.e. the forward reaction is balanced by the reverse reaction) Thus the free energy equation becomes

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

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

=10-∆Gº/(2.303 RT) Substututing R and T (25ºC) we obtain Keq =10-∆Gº/1.36 when ∆Gº has units of kcal/mol. Thus an equilibrium constant of 10 corresponds to a ∆Gº of -1.36 kcal/mol (see Table below) ∆Go Keq per M 1,000,000 10,000 100 10 1 0.1 0.01 0.0001 0.0000001

kcal/mol -8.2 -5.5 -2.7 -1.4 0.0 1.4 2.7 5.5 9.5

kJ/mol -34.2 -22.8 -11.4 -5.7 0.0 5.7 11.4 22.8 39.9

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

10

Standard Free Energy Changes are Additive

Thus

Consider the following reactions:

∆G˚total = +13.8 kJ/mol +(-30.5 kJ/mol) = -16.7 kJ/mol.



The overall reaction is therefore exergonic.

A ⇌ B ∆Gº1





We can also compute Keq for each reaction.

A ⇌ C ∆Gºtotal

[G6P] K eq1 = [glucose] [P] = 3.9 x 10 -3 M -1 i

B ⇌ C ∆Gº2



The ∆G˚ of sequential reactions are additive, thus ∆Gºtotal = ∆Gº1 + ∆Gº2 This 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 glucose–6–phosphate – a reaction that occurs in all cells: Glucose + Pi

Glucose–6–phosphate + H2O

∆G˚ = 13.8 kJ/mol

ATP + H2O

ADP + Pi

∆G˚ = -30.5 kJ/mol

These reactions share common intermediates (Pi and H20) and may be expressed as the sequential reactions: 1)

Glucose + Pi

Glucose–6–phosphate + H2O

2)

ATP + H2O

ADP + Pi

sum:

ATP + glucose

(note H20 is not included)

K eq2 =

[ADP] [P]i 5 = 3.9 x 10 M [ATP]

[G6P] [ADP] [P]i K eqtotal [glucose] [P] [ATP] i =Keq1 * Keq2 = 7.8 x 102 Thus by coupling ATP hydrolysis to G–6–P synthesis, the Keq for G–6–P formation is raised by a factor of 2 x 105. This strategy of coupling one reaction with a low Keq (or ∆Gº > 0) to a second with a high Keq (or ∆Gº < 0) is used by all living cells in the synthesis of metabolic intermediates.

ADP + glucose–6–phosphate 11

Summary 1.

At equilibrium, ∆G = 0

2.

Exergonic reactions are characterized by Keq > 1 and ∆Gº < 0

3.

4.

5.

Endergonic reactions are characterized by Keq < 1 and ∆Gº > 0 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 ∆Gº of the reactions are summative. At 25ºC, and equlibrium constant of 10 corresponds to a ∆Gº of -1.36 kcal/mol

Formative self assessments We 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 = ∆Gº+ {1.36 x log10[G3P]/ [DHAP]} kcal/mol. When the starting concentrations of DHAP and G3P are 0.2 mM and 2 µM respectively (log10[G3P]/[DHAP] = -2). F.

∆G = 1.36 kcal/mol and ∆Gº = -2.72 kcal/mol

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

Under the starting conditions stated above, the reaction can occur spontaneously.

J.

The net reaction will proceed from right to left.

Solution A.

At equilibrium, [G3P]/[DHAP] = 0.1 thus Keq = 0.1

12

B. Thus ∆Gº = - 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 ∆Gº 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. 13

3. Keq for the reaction A ⇌ Β 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/mol B. Under the starting conditions stated above, the reaction can occur spontaneously. C. ∆Go = 2.72 kcal/mol and ∆G = -1.36 kcal/mol D. The reaction will proceed from left to right. E. The reaction will proceed from right to left.

14

S ECTION 4

Explanations of Catalytic Action

Enzymes are biological catalysts

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⇌P is best represented by S⇌X⇌P where X is the activated complex or transition state. In terms of an energy diagram Example 1

L EARNING O BJECTIVES Energy

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.

X

∆GA S P

∆G

Progress of Reaction

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

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. 15

The fraction of S that achieves the transition state X at any temperature T is given by GA S F = e - RT

How can an enzyme introduce alternative reaction pathways? A number of mechanisms are observed:



If T = 0ºC (273ºK) and ∆GA = 10,000 cal/mol SF = 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: X

X

1. Covalent Catalysis A 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 Catalysis e.g. Lysozyme cleaves the glycosidic bond between C1 of N-acetylmuramic acid and C4 of Nacteylglucosamine of bacterial cell wall polysaccharides. Glu35 of lysozyme donates a proton to the oxygen of the polysaccharide glycosidic bond thereby hydrolyzing the bond.

Energy

Energy

3. Proximity X S

X X

S P

Progress of Reaction

P Progress of Reaction

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.

An 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 Distortion The enzyme active site undergoes a conformational change upon binding substrate distorting the substrate into a conformation resembling the transition state species. 16

Summary

Formative self-evaluation questions

1.

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 0ºC?

3.

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

4.

Do enzymes affect Keq?

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 affect the ground state of reactants and products therefore do not affect 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 effects or by molecular distortion or combinations thereof.

17

S ECTION 5

Key to formative evaluations

3. When the reaction proceeds right to left, ∆G > 0 and µS < µP 4. At equlibrium, ∆H = T ∆S Section 3 1. C (S increases; ∆G not ∆Gº; ∆G=0 at equilibrium) 2. C (∆G=0, not ∆Gº; more P produces more S; the reaction continues; enzyme leaves Keq unaltered)

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 µS = µP 2. When the reaction proceeds left to right, ∆G < 0 and µS > µP

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 ºC & ∆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 effects or by molecular distortion. 4. No! Keq is unchanged. 18

C HAPTER 2

relative fluorescence (Fr)

Analysis of time dependent processes

39 38 37 kobs

36

F

35 34 33 32 0

0.2 0.4 0.6 0.8

1

time in seconds

DNPA + NaOH → DNP + acetate

tempor placerat fermentum, enim integer 1. provide a set of tools to analyze ad vestibulum volutpat. Nisl rhoncus time-dependent processes turpis est, vel elit, congue wisi enim 2. understand the underlying nunc ultricies dolor sit, magna tincidunt. mechanisms that reaction Maecenas aliquam estgovern maecenas ligula rates nostra.

1

Fraction reacted

This chapter time-dependent Lorem ipsumconsiders dolor sit amet, processes. We seek to: suspendisse nulla pretium, rhoncus

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0.8 0.6 0.4 0.2 0 10

100

1000

time, msec

10

4

Constants, equations and glossary

Bmax ≡ Bm; sometimes Bmax ≡ Bm = [Et] (I will explain when this is true) Your challenge will be to understand when different names mean the same thing or different 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, koff 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).

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 ≡ koff Vmax ≡ Vm ≡ Vm(app)

• 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.

kp ≡ kcat (I will explain when this is true) xx

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

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

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 xxi

S ECTION 1

Context

SAMs have been studied extensively by SFG since 1991 (8), but ultrafast probing of a flash-heated SAM requires some elaboration. In the SFG technique we used, a femtosecond infrared (IR) pulse at 3.3 mm with a bandwidth of 150 cm−1 is incident on the SAM, coherently exciting all the alkane CH-stretch transitions in the 2850 to 3000 cm−1 range, along with electrons in the Au skin layer, producing an oscillating polarization in both the Au and the SAM layers. At the same time, a picosecondduration 800-nm pulse (“visible”) with a bandwidth of 7 cm−1 is incident on the sample. The visible pulse interacts with this oscillating polarization through coherent Raman scattering to create a coherent output pulse at the IR + visible frequency. This combined IR-Raman interaction is forbidden (in the dipole approximation) in centrosymmetric media because the secondorder susceptibility c(2) vanishes in such media. Because the methylene –CH2- groups of the alkane SAM form a nearly centrosymmetric solid, the SFG signal that we observed originated predominantly from the Au surface and the terminal methyl –CH3 groups. The well-known SFG spectrum obtained in ppp polarization (4),

malized ensemble–average IR dipole moment (〈m〉/mIR)2, which is temperature dependent. (C) With an instantaneous temperature jump to 1100 K, the methyl head groups become orientationally disordered in less than 2 ps.

Some examples

W.Stuhmer et al. A

Fig. 3. (A) SFG spectra of C8 (n = 7) and C18 (n = 17) SAMs without heating pulses (blue) and with flash-heating to 800°C (red). (B) VRF for a C8 monolayer. (C) VRF for a C18 monolayer.

==

RCK1

1 SOpA

RCK3

1

J6pA

RCK4 4 0

0

p

RCK5

A~~~~~~4Op

1

,,,J@20pA

Fig.

20ms

chann

B

poten

G/Gm

788

10 AUGUST 2007

VOL 317

SCIENCE

www.sciencemag.org

1.2 T

Vibrational Response Functions (VRFs) of SelfAssembled Monolayers. B VRF for a C8 monolayer. C VRF for a C18 monolayer From: Wang et al., SCIENCE VOL 317, pp 787-790, 2007

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

+

Normalized current

CTX-Biotin 1.0

0.5

0.0

Washout

0

200

-40

From: Stühmer et al. EMBO Journal vol.8, pp.3235 - 3244, Fig. 5. Conductance-voltage relations1989. of RCK channels. (A) Families of outward currents in response to depolarizing voltage steps. From 3 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. - Thiol Sugar Ensemble currents recorded +Thiol Sugar 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: open circles; RCK3: crosses; RCK4: diamonds; RCK5: filled circles). To obtain the conductance values the current at a particular test potential was divided by the driving potential assuming a reversal potential of - 100 mV. The lines showed the results of a non-linear least-squares TCEP fit of a Boltzmann isotherm (see Materials and methods) to the conductance values. The maximal conductance (Gm) obtained by the fit was used to normalize the data. The half-activation voltages in this 400 600 are -24 800mV (RCK1), 1000 1200 plot -37 mV (RCK3), -30 mV (RCK4) and -40 (s) mV (RCK5). Time

Sugar - Thiol Sugar of metabolically-labeled control Shaker to voltage+ Thiol from -60 to 0 mV are shown in Figure 7. B Chemical steps channels with CTX–Biotin. Washout(A) CTX–Biotin (10 nM) After TCEP The of size step currents at 0 mV varied between elementary Initial Initial inhibits Shaker-IR K+ currents in CHO-K1 cells treated 0.46 and 1.02 pA (RCK4) pA (RCK5). The single channel with either 50 µM thiol sugar 1 (h) or 0.5% ethanol as a current - voltage relations were measured in cell attached vehicle (s). Only metabolically-labeled (+thiol sugar) patchesbywith normal after frog Washout channels were irreversibly blocked CTX–Biotin aRinger's solution on the extracellular side. For all channels, simple washout; inhibition was completely reversed by an the current-voltage relation 50 ms 50 ms in is linear application of 1 mM TCEP. Reaction profiles were range -20 to 20 mV. However, since the voltage monitored with a 200 ms, 40 this mV is pulse every 15s. range for conductance estimation, we a rather narrow measured the average amplitudes at 0 mV membrane - Thiol Sugar + Thiol Sugar CFrom: Hua Z, Lvov A, Morinpotential. TJ, Kobertz WR. While theChemical RCK1, RCK3 and RCK5 channels have I/I I/I control of metabolically-engineered voltage-gated K(+) rather similar 1.0single-channel current amplitudes, that of the 1.0 channels. Bioorg Med Chem RCK4 Lett 2011. channel is considerably lower (Table I).

1 nA

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

A

1 nA

L EARNING OBJECTIVES

needed to be chemically reversible. Although disulfide bond formation between CTX and thiol-containing sialic acids on the cell surface is an obvious chemoselective and cell friendly reaction, we chose to label CTX with a bismaleimide that had an internal disulfide bond because maleimides are inherently more stable in water than MTS reagents. Moreover, this subtle difference would allow for delivery of a small molecule probe to the modified K+ channel subunit after cleavage with reductant, which would be useful in subsequent biochemical, biophysical or imaging experiments. To simplify the synthesis of the bismaleimide, derivatization of CTX, and ensure delivery of a molecular probe to a K+ channel subunit, we set out to 5 synthesize a symmetrical bismaleimide (Scheme 1) from cystamine dihydrochloride 2 that would allow for the facile incorporation of a molecular probe in the final step of the synthesis. The amino groups of cystamine 2 were capped with 2 equiv of a doubly amino-protected, activated ester of L-lysine 3. Selective deprotection of the Fmoc protecting groups gave the symmetric diamine 4. Addition of 2 equiv of the NHS-ester of 3-(maleimido)propionic acid and N-Boc deprotection afforded bismaleimide 5, which was subsequently biotinylated with 2 equiv of NHS–Biotin to give biotin bismaleimide 6. CTX was then derivatized by labeling a cysteine mutant of CTX (R19C)32 with 100-fold molar excess of biotin bismaleimide 6 to yield CTX–Biotin, which was purified by reverse phase HPLC as we have previously described.17,19 With CTX–Biotin in hand, we decided to change both the channel (Shaker–IR)33 and the expression system (CHO-K1 cells) to demonstrate that our approach was versatile and could be used to label glycosylated ion-conducting subunits. Inactivation-removed Shaker is similar to Q1 in that it is an archetypical volt-

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. -80

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

+

max

Initial Washout

0.5

max

Initial After TCEP

Pharmacology 0.5 of RCK channels 22 A profile of the pharmacological sensitivity of the different RCK channels to the K+ channel blockers 4-aminopyridine

chann

Hold

inact low

depe

subs

at 2

Tabl

curr

RCK

sensi inac

sensi RCK bloc effec

Dis

Comp

RCK An funct chan

ident pa

a

prope sing shou chan func memb prel

struc

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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. 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 ractivity t o extracellula~activity of 3-0-methy1glucose per equixra1ent volume of bulk external solution, .4bscissa: time in hours. External sugar concentration, Time course of 3-0-methylglucose uptake in isolated muscle of Balanus 1 m M Uptake was measured using rc)nventional and scintillator probe (cells 0 )methods. Kumber of points per conventional determination, five or more. The water content of nubilis. Ordinate: ratio of intracellular activity to extracellular activity of 3-0isolated fibres (70%) is shown by the continuous line above the points. The time a t diame*r, 1352 p m ;k m pAbscissa: rature. half-equilibration is shown by thedmhed methylglucose per equivalent volumeline. ofMean bulkfibre external solution. 21 OC.

time in hours. External sugar concentration, 1 mM. Uptake was measured using conventional (filled squares) and scintillator probe (open circles) The calculated rate of sugar uptake a t 30 min is 2 pmol .cm-*. s-I. Assuming methods. The water content of isolated fibres (70%) is shown by the uptake is not saturated, the permeability of the barnacle muscle fibre, '2 (em. s-I), continuous line above the points. timeflux, at half-equilibration to 3-0-methy~g~ucose is related t o The the sugar J (mol. ernv2.s-I), byis shown by the dashed line. Mean fiber diameter, 1352 µm ; 21 ºC. where So is the external sugar concentration (in01 . emp3).This corresponds to a value From: Carruthers, cm. A. J.s-IPhysiol. VOL 336, pp 377-396, 1983 which is some 3 4 orders of magnitude larger than the of P of 2 x

of artificial lipid bilayers to sugars (Jung & Snell, 1968; Lidgard & Jones, Ribosomal R N A Stability permeability and Growth Conditions 19'75). At equilibrium, the 3-0-methylglucose space of the fibre is 70 yo,which is in close contentsof barnacle muscle (71 & 1 7; ; R = 5 ) . sgreement with estimates of the xra%er 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

C""

Length of chose, h

Downloaded from www.jbc.org at Univeristy of Massachusetts Medical Center/The La

tion

45 min of exposure t o sugar. 3feasurements of sugar uptake in subsequent experiments were made using an incubation period of 30 min in order to obtain accurate measurements of t2heinitial rate of sugar uptake.

FIG. 2. First-order decay analysis of embryonic total RNA during normal and slow larval growth conditions. The perFirst-order decay analysis of Drosophila embryonic total RNA during centage of embryonic RNA remaining a t various times duringa chase normal and slow larval conditions. The percentage of embryonic with light yeast (normalgrowth growth conditions, solid circles) or dense RNAalgae remaining at various times open during a chase normalThe growth (slow growth conditions, circles) has under been plotted. values in(solid solid circles from thegrowth experiment shown in (open Fig. circles) conditions circles)areorderived during slow conditions 1 as well as regression four other independent experiments (not shown). is plotted. The lines indicate that the stability ofThe embryonic values in open circles are derived from the experiment shown in Fig. RNA4 increases from 48 h 115 h if the larval growth rate as well as one other independent experiment (not shown). The is reduced. regression lines indicate that the stability of embryonic RNA increases from48h ( r 2 = 0.915) to 115 h ( r 2 = 0.892) if the larval From: Winkles al., J. Biol. Chem.,Vol 269, pp 7716-7720, 1985 growth rate iset reduced.

not depleted during the chase. The dense RNA samplemay be slightly enriched in rRNA, as would be expected if nonrRNA species decay faster than rRNA. Therefore, the halflife of total RNA isa good estimate of rRNA half-life. To investigate whether the stability of embryonic rRNA varied under differentlarval growth conditions, we then measured 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]uridinefor 46 b.Some of the embryoslaid 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-labeledChlorella cells, and [14C]uridine. First-in-

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.

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. 23

S ECTION 2

Reaction order

Rate 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),

dC v = dt 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 reaction describes how the velocity of the reaction depends upon the concentration of reactants. In the (irreversible) isomerization reaction L EARNING O BJECTIVES 1. Reaction orders 1. Zero-order 2. First order 3. Second order 1. Class 1 2. Class 2 (pseudo first-order)

A

k1

B

the theory of chemical kinetics tells us that

d [B] m = = k 1 [A] k 1 [A] dt 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,

M per sec = k 1 M

M per sec = k 1 = per sec M

The Order of a Reaction must be determined experimentally Understanding the stoichiometry of a reaction is not sufficient to predict the rate law of the reaction. This is illustrated in the table below. ⇌

In a reaction of the type

E+S



k1

E$S

⇌ ⇌

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

d [ES] 1 1 = k 1 [E] [S] = k 1 [E] [S] dt Because m = 1 for both species the reaction is firstorder 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

M per sec = k 1 = per M per sec M2

⇌ ⇌ ⇌ ⇌

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 + ½O2 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

Examples of reaction orders in nature To 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 pseudofirst 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:

2NO2

ONOONO

The first step is therefore second order.

2NO + O2

http://inside.umassmed.edu/uploadedFiles/gsbs/courses/ 2012-2013_Core_Course_Files/CoreKinetics.pzf%20-%20for %20students%20only.zip and download the “GraphPad Prism” version 6 software from  Prism 6 Win http://cdn.graphpad.com/downloads/prism/6/ InstallPrism6.exe Prism 6 Windows serial number: GPW6-200512-LEM5-16772 Prism 6 Mac http://cdn.graphpad.com/downloads/prism/6/ InstallPrism6.dmg Prism 6 Mac serial number: GPM6-200513-LEM5-F3EF2 26

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

A zero-order reaction

zero-order kinetics

Zero-order reaction Substrate Product

8

80

v (d[P]/dt)

[Substrate] or [Product]

10

100

60 40 20

6

0

0

100

4

200

[S] µM

d [ethanol] d [acetaldehyde] v == = k0 dt dt

2 0 0

2

4

6

8

10

TIME

Note that [substrate] decreases linearly with time and [product] increases linearly with time.

The negative sign is used with reactant - ethanol because its concentration decreases with time. The concentration of its product, acetaldehyde, increases with time. C0

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:

C

ALDH +

CH3CH2OH + NAD ⇌ CH3CHO + NADH + H+

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

CH3CHO



0

CH3CH2OH 0

t

27

Theory of Zero-order Reactions

dC dt = k 0 The units of k0 are molarity per sec. This is a “zeroorder 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 = C0 + k0 t

y = y-intercept + slope * x This 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).

10 [Substrate] or [Product]

A zero-order reaction corresponds to the differential rate law

Zero-order reaction Substrate Product

8 6 4 2 0 0

2

4

6

8

10

TIME

y = x-intercept + slope * x Best-­‐fit  values" " "          Slope" " " "          Y-­‐intercept  when  X=0.0"          X-­‐intercept  when  Y=0.0"       Goodness  of  Fit" "          R  squared" " "

" " " "

Substrate" " -­‐1  ±  0  " " 10  ±  0" " 10.00"" "

Product" 1  ±  0     0 ± 0"" 0" "

"

1.000""

1.000

"

"   " "

Units mols/sec mols sec

General rules for zero-order reactions 1. 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 time 3. Zero-order, enzyme catalyzed kinetics are typically observed at saturating [S] 28

A first-order reaction

Theory of First-order Reactions A first-order reaction corresponds to the differential rate-law:

1stOrder

dC dt = k 1 C

5

[A] or [B]

4 3

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.

[Substrate] [Product]

2

The reaction

A

1 0 0

k1

B

has the rate law 2

4

6

8

10

TIME

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?

d [A] d [B] v =- dt = dt = k 1 [A] 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 substrate

First Order reactions - loss of substrate

Theory

Integrated rate law

- d [A] = k 1 [A] 0 t

[A] = [A] 0 e -k t 1

Defining [A]0 as [A] at time = 0 and integrating between A at time zero and time t gives

ln [A] =- k 1 t + ln [A] 0 Half-life Defining [A] at t1/2 as [A]0/2





y

=

slope x + intercept

ln2 0.693 t 1/2 = k = k 1 1 and because τ = 1/k1, t1/2 = 0.693 τ

30

First Order reactions - product formation

First Order reactions - product formation

Theory

Integrated rate law

Defining [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







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

[A]t = [B]∞ - [B]t

Thus we obtain

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







y

=



slope x + intercept

Half-life Defining [B] at t1/2 as [B]∞/2

ln2 0.693 t 1/2 = k = k 1 1 and because τ = 1/k1, t1/2 = 0.693 τ

31

Returning to our example of a first order reaction, 1stOrder 5

[A] or [B]

4 3

[Substrate]

This is characteristic of “first-order decay” as observed with radioactive decay.

[Product]

2

A second clue comes from the measurement of halftimes. 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.

1

1st Order 5

0 0

2

4

6

8

10

TIME

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 yaxis data on a log scale), we obtain 1stOrder

3 2 1.4 sec

1 0 0

10

1.4 sec 1.4 sec

1

2

3

4

5

6

7

8

9

10

TIME

[Substrate] 1

[A]

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 firstorder analysis to the data.

0.1

0.01 0

[Substrate]

4

2

4

6

8

10

TIME

This produces a linear plot which is consistent with 1st order kinetics! 32

Non-linear regression analysis To 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] 0 e -k t 1

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 offers a model for first-order reactions called “One-Phase Decay” The equation is: Y=(Y0 - Plateau)*exp(-k*X) + Plateau In 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 difference 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 difference 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

General rules for 1st order reactions

1stOrder

[Substrate] or [Product]

5 4

1. First-order enzyme catalyzed kinetics are typically observed at subsaturating [S]

[Substrate] [Product]

2. Plot of log (St-S∞) vs time produces a straight line with slope = -k

3

3. The half-time (t1/2) and k are invariant of the starting value of St chosen.

2 1 0 0

4. Plot of log (P∞-Pt) vs time produces a straight line with slope = -k 2

4

6

8

10

TIME

Y=(Y0 - Plateau)*exp(-k*t) + Plateau One  phase  decay"Perfect  fit" Best-­‐fit  values" "          Y0"" " " "          Plateau" " " "          k" " " " "          Half  Life"" " "          Tau  =  1/k" " "          Goodness  of  Fit" "          Degrees  of  Freedom"          R  square"" " "

"

[Substrate]"

[Product]" "

Units

" " " " "

5.000"" 0" " 0.5000" 1.386  " 2.000""

" " " " "

0     5.000     0.5000   1.386     2.000    

mM mM per  sec sec sec

" "

48" " 1.000""

" "

48 1.000

         

5. 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

k1 k2

B

The equations are unchanged but now k = k1 + k2 34

Second-order reactions Fall into 2 categories in which the rate law depends upon: 2. the product of the concentrations of two different reagents. Class 1 reactions (A+A ⇌ P)

[Substrate] [Product]

3 2 1

The differential rate law is

0 0

2

4

6

8

10

TIME

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]2 v = k [proflavin] 2 A–A–G–C–U–U 2 A–A–G–C–U–U

4

[A] or [B]

1. the second power of a single reactant species, or

2nd Order class 1

5

U–U–C–G–A–A

v = k [A 2 GCU 2] 2

[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 0.4

8

0.3

[Substrate]

6

[Product]

[A]0/[A]

0.2

Class 1, 2nd order Transform of data

0.1

4

1st order data 2nd order data

0.0

2

-0.1 -0.2

0

1

2

3

4

5 TIME

6

7

8

9

10

This plot therefore shows that this is not a 1st order reaction Theory of Class 1 Second-order Reactions Defining [A] at zero-time = [A]0, it can be shown that the integrated rate law is

1 1 [A] [A] 0 = k t adding 1/[A]0 to both sides gives

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

multiplying both sides by [A]0 gives

0 0

2

4

6

8

10

TIME

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 yintercept = 1. The slope [A]0 k indicates that the rate of loss of [A] will increase linearly with [A]0 This is infact observed

[A] 0 [A] = [A] 0 k t + 1 Thus one expects a linear relation between the reciprocal of the reactant concentration and time. 36

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

How starting [A] affects rate of 2nd order reaction 15

1 2 3 4 5 6 7 8 9 10

[A]0/[A]

10

Increasing [A]0

5

0 0

2

4

6

8

10

TIME

[A]0 k per sec

1.5

1.0

A0k (slope) vs A0 second order Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 1/slope

kr

P

the kinetics more closely resemble 1st-order kinetics when when kr ≥ kf/10 In 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+B⇌P)

[A]0 k per sec 0.1320 ± 2.842e-009 -3.974e-009 ± 1.763e-008 3.010e-008 7.576

AKA pseudo-first order A reaction that is 2nd order overall is 1st order with respect to each of the two reactants.

0.5

0.0 0

kf

A+A

For example, in the reaction 2

4

6

8

E+S

10

[A]o

General rules for 2nd order reactions (Class 1) 1. Standard 1st order analysis does not work 2. Plotting [A]0/[A] vs time produces a straight line with slope [A]0 k 3. 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.

k1 k2

E$S

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

v=

d [ES] dt = [E] k 1 [S] - k 2 [ES]

37

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

R+L

kf kr

R$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

(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 koff or k1 and k2)

At zero-time, various concentrations of L (µM) were mixed with 1 nM R. The time course of LR formation was monitored at each [L]. Pseudo 1st Order

[L]

0.0010

10 5.995

0.0008

3.594

[LR] µM

2.154

0.0006

1.292 .774

0.0004

.464 .278

0.0002

.167 .1

0.0000

0

5 TIME

10

0.0010

10 5.995

0.0008

3.594 2.154

[LR] µM

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.

[L]

0.0006

1.292 .774

0.0004

.464 .278

0.0002

.167 .1

0.0000

0.1

1

10

TIME

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

Theory for pseudo first-order reactions

kobs vs L

25

For our reaction

kobs per sec

kobs per sec

20 15

Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 1/slope

R+L

1.999 ± 0.0001861 0.5012 ± 0.0007352 -0.2507 0.5002

kr

R$L

The rate of LR formation is given by:

d [LR] v = dt = [R] k f [L] - k r [LR]

10 5 0 0

kf

2

4

6 [L] µM

We will show below that: 1. The slope is kf 2. The y-intercept is kr 3. The x-intercept is -kr/kf

8

10

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

[LR] = [R] 0 constant (1 - e -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/kf 5. 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

S ECTION 3

Recap of Key points

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 kinetics

1. What is the difference 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 difference 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

1. 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 secondorder 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 = -k 2. 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 time

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-1 4. Keq can be obtained as kf / kr

1. Plots of [S]0/[S] vs time are linear with slope = [S]0 k. This slope has units of per sec 2. Plots of [S]0 k vs [S]0 are linear with yintercept = 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 different molecular species combine to form a product (Class 2 reactions), the reaction is non linear with time 1. 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].

41

S ECTION 4

Formative self-evaluation questions

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?

Test your understanding of time-dependent processes by answering the following questions

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

1.

What are the units of zero-, first- and secondorder rate constants?

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

2.

Does the stoichiometry of a reaction always predict reaction order and mechanism?

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

3.

What is the defining characteristic of a zero-order reaction?

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

4.

How do you compute the value of a zero-order rate constant?

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

5.

Do first-order reactions show a linear dependence on time?

17. When does this type of analysis break down?

6.

What is the t1/2 of a first-order reaction?

18. How does kobs vary with [S] for a Class 2 secondorder reaction?

7.

How does t1/2 of a true first-order reaction vary with [S]?

19. How can we use this relationship to compute Kd, kf and kr for a Class 2 second-order reaction?

8.

How is t1/2 related to the rate constant k of a firstorder reaction?

20. What are the major differences between first-order kinetics and Class 2 second-order kinetics?

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?

21. Why, then are Class 2 second-order kinetics called pseudo-first order kinetics? 42

S ECTION 5

Key to formative evaluations

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].

Section 2 - Reaction order

8. t1/2 = 0.693/k

1. The units are:

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.

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.

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]