Enzymes/ Kinetics

Study of enzyme catalyzed reactionsRate of reactionsEnzyme specificityMechanism of catalysis

There are several good reasons for adding kinetic modeling. Mechanism and dynamics are intimately related, and kinetic modeling reveal new and useful information about biological control systems.

Observation during steady state is not informative about the process

Catalyst

Catalyst=

Enzyme= biocatalyst

e.g.Fermentation of sugar to ethanol by yeast enzymes

Conversion of fatty acids to polyketide antibiotics by filamentous fungi

Conversion of milk to cheese by microorganisms

Conversion of sugar to CO2 by bakers yeast to make leavened bread

BioCatalystMOST Enzymes are Proteins

Enzyme advantages over chemical catalysis:

Enzyme concentration is very small and enzyme is almost always limiting

Enzyme function

S <--------->P

1

2

3

Go =standard free energy change for chemical reactionsG’o =standard free energy change for biological reactions298o k (RT); 1 atm, 1M substrate, pH7.0

Active site- pocket in enzyme that binds substrates- most complimentary in structure to transition state

Rationale for enzyme kinetics

Conformation of proteins and positions of side chains are important for enzyme-substrate interactions and catalysis.

Forces involved in protein folding and structure are also involved in catalysis- enzyme-substrate specificity

Trypsin hydrolyses proteins by cleaving peptide bond adjacent to Lys/Arg.Aspartate residue in trypsin active site mediates ionic interaction with Lys /Arg and this arranges protein residue at which hydrolysis occurs.

To use enzymes in biotechnology, pharmaceutics or drugs that inhibit enzymes in medicine, you NEED TO KNOW KINETIC PARAMETERS OF THE ENZYME REACTION.

We may want enzymes that WORK FAST- convert more substrate in a fixed unit of time. To do this optimization we have to perform and analyze the enzyme catalyzed reaction.

You can adjust pH, temperature and add co-factors to optimize enzyme activity. You cannot adjust substrate selectivity.

Just like chemical reactions, enzyme catalyzed reactions have kinetics and ratesReaction kinetics is Michaelis-Menten kinetics.

Enzyme-substrate cycle

Ground state= starting point (energetically speaking)

-

-

-

-

-

Free e

nerg

y

Activation energy

Free e

nerg

y

[S]

[P]

-

-

-

-

-

-

Energy Barriers

Sucrose to CO2 and H2O

C12H22O11 + 12O2 <-----> 12CO2 + 11 H2O

Reaction has large negative G’o Therefore Product prevails at equilibrium

In your pantry you can store sucrose in the presence of oxygen. It does not spontaneously convert into CO2 and water!ACTIVATION energy of this reaction is HUGE

In a cell sucrose is rapidly converted to CO2 and H2OENZYMES!!!

Activation energy barriers are required for ordered life!!!Without barriers molecules would spontaneously interconvert- there would be no regulation

/=G

Reaction coordinate

Overall std free energy change(S goes to P)

Catalyzed Vs uncatalyzed reaction

Rate enhancements

Enzymatic rate enhancements are 103 to 1017 x

How do ENZYMES carry out catalysis?

•

•

•

~

~

~

-

-

Whats the Bill?5.7 kJ/mol is needed to achieve a 10x increase in rate of a reactionTypical weak interactions are 4-30 kJ/molTypical binding event yields 60-100 kJ/molMORE THAN ENOUGH ENERGY!!!

A to B to C

A------------> B------------>C

The rate constant for B to C is smaller. B to C is slow (rate determining)A is rapidly converted to B but B accumulates because its conversion to C is slowReaction is in only one direction because G of B is lower than A and C is lower than BB to C is rate limiting

conc

time

G

Rxn coordinate

A to B to C

A------------> B------------>C

The rate constant for A to B is smaller. A to B is slow (rate determining)A is not rapidly converted to B because its conversion to B is slow. B never accumulates because it is rapidly converted to CReaction is in only one direction because G of B is lower than A and C is lower than BA to B is rate limiting

conc

time

G

Rxn coordinate

A to B to C

A------------> B------------>C <-----------

The rate constant for A to B is small. The rate constant for B to A is large. A to B is slow but B to A is very fast and B to C is kinda fast.A is not rapidly converted to B. B never accumulates because it is very rapidly reconverted to A and C accumulation is not very rapid because most of B is reconverted back to A rather than to CReaction is in two directions because G of B is higher than A and C is lower than B and A

conc

time

G

Rxn coordinate

A to B to C

A------------> B------------>C <-----------

The rate constant for A to B is small. The rate constant for B to A is large. A to B is slow but B to A is kinda fast and B to C is very fast.A is not rapidly converted to B. B never accumulates because it is very rapidly converted to C. or reconverted to A. Accumulation of C is rapid because most of B is converted to C rather than reconverted to AReaction is primarily in one directions because while G of B is higher than A, C is lower than B and AA to B is rate limiting

conc

time

G

Rxn coordinate

Enzyme catalyzed Steady State

Enz + Sub <-------> Enz-Sub <-------> Enz + ProdC

onc

Time

[P][S]

[E]

[ES]

Presteadystate

Steady state

During steady state formation of ES equals its breakdownk1[E][S]=k-1[ES]+k2[ES]

Km= k-1+k2/k1

What is the rate of a chemical reactionS---->P

-

-

-

-

-

A+B----.>P is a bimolecular reaction (there are two reactants) and this is a second order reaction-

Sometimes second order reactions appear as first order. E.g. If conc of B is very large and does not change much then reaction rate is entirely dependent on conc of A only

Rate of a Reaction

Rate of catalysis

Vo=

For a fixed amount of enzyme,

The rate of catalysis

E+S ESk1

k-1

All measurements are done at very high substrate conc and very low product conc so the reverse reaction is rare. We can simplify the above reaction scheme as

Rate constants

k1 k2A -----> B ------> C <----- k-1If k-1 is greater than k2 (B reforms A faster than it forms C) then there will be a rate-determining pre-equilibrium and overall rate of formation of C will depend on ALL THREE RATE CONSTANTS((((d[C]/dt= [k1][k2]/[k-1][A]))))

If k2 is vastly greater than k-1 (B forms C faster than it reforms A) then we can effectively ignore the back reaction (B to A) and the only question then is whether k1 or k2 is rate limiting.

E+S ES E+Pk1

k-1

k2

k1 is rate constant for formation of ESk-1 is rate constant for conversion of ES to E+Sk2 is rate constant for formation of product = kcat

1) Binding of substrate to enzyme is reversible

2) Product release from enzyme is instantaneous

[ES]

Rate of formation = k1[E][S]

Rate of breakdown = k-1[ES] + k2[ES]

Assume steady state

–As ES is produced, it reacts

–[ES] remains constant

–Rate formation = rate breakdown

So, k1[E][S] = k-1[ES] + k2[ES]

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

Rearrange

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

Where (k-1 + k2) / k1 = KM (Michaelis constant)

Define total enzyme concentration [E]T = [E] + [ES]

Substitute for [E]

([E]T – [ES])[S] / [ES] = KM

Solve for [ES]

[ES] = [E]T[S] / KM + [S]

V0

= initial velocity/rate

[ES] = [E][ES] = [E]TT[S] / K[S] / KMM + [S] + [S]

Since VSince V00 = k = k22[ES][ES]

VV00 = k = k22[E][E]TT[S] / K[S] / KMM + [S] + [S]

Define maximum velocity VDefine maximum velocity Vmaxmax

Occurs at high [S]Occurs at high [S]

Enzyme is all in ES formEnzyme is all in ES form

[ES] = [E][ES] = [E]TT

VVmaxmax = k = k22[E][E]TT

Therefore Therefore

VV00 = V = Vmaxmax[S] / K[S] / KMM + [S] + [S]

Rate of reaction increase with [S]

Rate levels off as approach Vmax

–More S than active sites in E

–Adding S has no effect

At V0 = ½ Vmax

–> [S] = KM

Michaelis-Menten The classic Michaelis-Menten rate equation

The first order rate constant kcat is k2

Vmax is product of kcat and enzyme conc

Km is mixture of rate constants that describe formation and dissociation of enzyme-substrate complexes

Km gives a sense of the affinity of enzyme for substrate

When [substrate] is >>> Km, the reaction kinetics are equal to max rate (Vmax).When [substrate] is <<< Km, the reaction rate is kcat/Km

All these analyzes are done at saturation kinetics

Kcat and Km are PRIMARY INDICATORS of how well an enzyme will react with a particular substrate. High kcat= fast reactions. low kcat= slow reactionsHigh Km = low affinity of enzyme for substrate. Low Km = high affinity of enzyme for substrate

Michaelis-Menton Equation

-

-

–V0 =

-

What is the [ES]?

What do we measure?-

--

-

----

-----

[S]1

[S]2

[S]3

[S]4

Prod

uct (

mol

/lit

)

Time

Vo(4)Vo(3)

Vo(2)

Vo(1)

A+B<--->AB

At low [S], Km >> [S]At high [S], [S] >> Km

Initial Velocity

Velocity measured at very beginning of reaction when very little product is madeInitial velocity is measured at saturation kinetics- at high [S], enzyme is saturated with respect to substrate

[Pro

duct

]

Time [Substrate]

Vo

Vmax

Vmax and Km

V0 = Vmax [S] [S]+Km

k-1 + k2

k1Km=

Km is unique to each Enzyme and Substrate. It describes properties of enzyme-substrate interactionsIndependent of enzyme conc. Dependent on temp, pH etc.Vmax is maximal velocity POSSIBLE. It is directly dependent on enzyme conc. It is attained when all of the enzyme binds the substrate. (Since these are equilibrium reactions enzymes tend towards Vmax at high substrate conc but Vmax is never achieved. So it is difficult to measure).When an enzyme is operating at Vmax, all enzyme is bound to substrate and adding more substrate will not change rate of reaction (enzyme is saturated). (adding more enzyme will change the reaction).

When V0=Vmax, Km= [S]

Measuring Km and Vmax

[Substrate]

vo

Vmax

You can use a curve fitting algorithm to determine Km and Vmax from a V vs [S] plot (need a computer)

Reaction rates are initial rates determined when the substrate is in vast excess and isn’t changing much.

Alternatively you can convert the curve to a straight line via a double reciprocal plot (1/Vmax and 1/[S])

1/[S]

1/vo

1/Vmax

-1/Km

ReciprocalReciprocal

It is not easy to extrapolate a hyperbola to its limiting value It is not easy to extrapolate a hyperbola to its limiting value (computers can do this)(computers can do this)

The Michaelis-Menten equation can be recast into a linear formThe Michaelis-Menten equation can be recast into a linear form

The y-intercept gives the Vmax value and the slope gives Km/Vmax

To obtain parameters of interestReciprocal form of equation

1 = Km 1 + 1V Vmax S Vmax

Y= m x + b

Vmax is determined by the point where the line crosses the 1/Vi = 0 axis (so the [S] is infinite).Km equals Vmax times the slope of line. This is easily determined from the intercept on the X axis.

1/Vo

It is difficult to accurately measure Vmax

V0 = Vmax [S]

[S]+Km

Reciprocal

1

V0

Km

= Vmax

1S + 1

Vma

Km values of enzymes range from 10-1M to 10-7M for their substrates. It varies depending on substrate, pH, temp, ionic strength etc. Kcat is turnover number for the enzyme-number of substrate molecules converted into product per unit time by that enzyme

Vmax = k2[Et]k2 is also called Kcat

Et is conc of active sites in enzyme

Km and Vmax

Km is [S] at 1/2 VmaxIt is a constant for a given enzyme at a particular temp and pressureIt is an estimate of equilibrium constant for substrate binding to enzymeSmall Km= tight binding, large Km=weak bindingIt is a measure of substrate concentration required for effective catalysis

Vmax is THEORETICAL MAXIMAL VELOCITYVmax is constant for a given enzymeTo reach Vmax, ALL enzyme molecules have to be bound by substrate

Kcat is a measure of catalytic activity- direct measure of production of product under saturating conditions.Kcat is turnover number- number of substrate molecules converted to product per enzyme molecule per unit time

Catalytic efficiency = kcat/kmAllows comparison of effectiveness of an enzyme for different substrates

Enzyme Km examples

Hexokinase prefers glucose as a substrate over ATP

Kcat

Catalase is very efficient-it generates 40 million molecules of product per second.Fumarase is not efficient-it generates only 800 molecules/per second

kkcatcat = V = Vmaxmax / [E] / [E]TTTurnover numberTurnover number

Number of reaction processes each active site catalyzes per unit timeNumber of reaction processes each active site catalyzes per unit timeMeasure of how quickly an enzyme can catalyze a specific reactionMeasure of how quickly an enzyme can catalyze a specific reaction

For M-M systems kFor M-M systems kcatcat = k = k22

Kcat/Km

Rate constant of rxn E + S Rate constant of rxn E + S --->---> E + P E + PSpecificity constantSpecificity constantGauge of catalytic efficiencyGauge of catalytic efficiencyCatalytic perfection ~ 10Catalytic perfection ~ 1088 --10109 9 MM-1 -1 ss-1 -1 (close to diffusion)(close to diffusion)

Enzyme cofactors

Coenzymes

How do ENZYMES carry out catalysis?

•Entorpy reduction- holds substrates in proper positionBringing two reactants in close proximity (reduce entropy & increase effective reactant concentrat)•Substrate is desatbilized when bound to enzyme favoring reaction-(change of solvent, charge-charge interactions strain on chemical bonds).•Desolvation of substrate- H bonds with water are replaced by H bonds with active site

Enzymes form a covalent bond with substrate which stabilizes ES complex (Transition state is stabilized)Enzyme also interacts non-covalently via MANY weak interactionsBond formation also provides selectivity and specificity (H bonds- substrates that lack appropriate groups cannot form H bonds and will be poor substrates) (Multiple weak interactions between enzyme and substrate)

Free energy released by forming bonds is used to activate substrate (decrease energy barrier/lower activation energy of reaction)

Induced fit-binding contributes to conformation change in enzyme

Whats the Bill?5.7 kJ/mol is needed to achieve a 10x increase in rate of a reactionTypical weak interactions are 4-30 kJ/molTypical binding event yields 60-100 kJ/molMORE THAN ENOUGH ENERGY!!!

A Hypothetical reaction

Breaking a stick

Imagine you have to break a stick. You hold the two ends of the stick together and apply force. The stick bends and finally breaks.You are the catalyst. The force you are applying helps overcome the barrier.

A stickase with a pocket complementary in structure to the stick (the substrate) stabilizes the substrate. Bending is impeded by the attraction between stick and stickase.

An enzyme with a pocket complementary to the reaction transition state helps to destabilize the stick, contributing to catalysis of the reaction. The binding energy of the interactions between stickase and stick compensates for the energy required to bend the stick.

Role of binding energy in catalysis. The system must acquire an amount of energy equivalent to the amount by which G‡ is lowered. Much of this energy comes from binding energy (GB) contributed by formation of weak noncovalent interactions between substrate and enzyme in the transition state.

Lock/Key or Induced Fit

Lock/Key- Complementary shape

The enzyme dihydrofolate reductase with its substrate NADP+

NADP+ binds to a pocket that is complementary to it in shape and ionic properties, an illustration of "lock and key" hypothesis of enzyme action. In reality, the complementarity between protein and ligand (in this case substrate) is rarely perfect,

Induced Fit

Hexokinase has a U-shaped structure (PDB ID 2YHX). The ends pinch toward each other in a conformational change induced by binding of D-glucose (red).

Substrate specificity

The specific attachment of a prochiral center (C) to an enzyme binding site permits enzyme to differentiate between prochiral grps

Enzyme-substrate

•Acid-Base Catalysis- donate or accept protons/electrons from and to substrate•Covalent Catalysis-transient covalent link between substrate and enzyme side chain•Metal-Ion Catalysis-Metal in active site donate or accept protons with substrate

•Proximity & Orientation Effects (reduction in entropy-two mol brought together and oriented in specific manner)•Transition State Preferential Binding

Catalysis

R GroupsThe active sites of enzymes contain amino acid R groups.Active site is lined with hydrophobic residues

Polar amino acid residues in active site are ionizable and participate in the reaction. Anion/cation of some amino acids are involved in catalysis

Lysozyme: Cleaves glycosidic bonds in carbohydrates

Covalent Catalysis

All or part of a substrate is transiently covalently bound to the enzyme to form a reactive intermediateGroup X can be transferred from A-X to B in two steps via the covalent ES complex -EX

A-X+ E <-----> X-E + A

X-E + B <-----> B-X+ E

Without a catalyst the intermediate converts back to the reactants and does not proceed forward (high barrier).Donation of a proton by water or an acid helps the process move forward.

The active sites of enzymes contain amino acid R groups, that participate in the catalytic process as proton donors or proton acceptors.

Catalysts

Proton donor/acceptor (Nucleophile/electrophile)

Asp and Glu are negatively charged at pH7.0 and their side chains are acidic.These side chains ACCEPT protons which neutralize the charge.Lys, Arg, His are positively charged at pH 7.0 and their side chains are basic.These side chains DONATE protons to neutralize their charge.

Asp/Glu COO- + H+ <-----------> COOH Lys/Arg NH3+ <----------> NH2 + H+

NucleophilesR-OH <---> R-O: + H+ (hydroxyl)R-SH <---> R-S: + H+ (sulphydryl)R-NH3 <---> R-NH2: + H+ (amino)

ElectrophilesH+ ProtonM+ Metal ion

+C O

R

R’

Carbonyl

Nucleophiles-groups rich in and capable of donating electron (attracted to nucleus)Electrophile- group deficient in electron (attracted to electron)Reactions are promoted by proton donors (general acids) or proton acceptors (general bases). The active sites of some enzymes contain side groups, that can participate in the catalytic process as proton donors or proton acceptors.

•A general acid (BH+) can donate a proton. A covalent bond may break more easily if one of its atoms is protonated.

Reaction acceleration achieved by catalytic transfer of protonA general base (B:) acts as proton acceptor to remove proton from OH, NH, CH (XH)This produces a stronger nucleophilic reactant (X:)A general base(B:) removes a proton from water thereby generating the equivalent of OH-in neutral solution.

X H

:B

:X- HB+

Acid-base catalysis

Acid base catalysis

RNaseA cleavage of RNA

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Enzyme Inhibition

SS

EE

ES

PP

EE

EP

PP

EE

E + P

EE

SS

E + S

EE

Enzyme Inhibition

Many molecules inhibit enzymes

Competitive Inhibitor

Most commonMost commonInhibitor competes with natural substrate for binding to active siteInhibitor competes with natural substrate for binding to active siteInhibitor similar in structure to natural substrate and binds active site of enzyme Inhibitor similar in structure to natural substrate and binds active site of enzyme (reducing effective enzyme conc)(reducing effective enzyme conc)

Binds more stronglyBinds more stronglyMay or may not reactMay or may not reactIf reacts, does so very slowlyIf reacts, does so very slowlyGives info about active site through comparison of structuresGives info about active site through comparison of structures

Competitive Inhibitor

Drug targets

Gleevec

Gleevec: How it works

HIV protease structure

Protease Inhibitors

Protease + Inhibitor

Reversible Inhibition (competitive)

[Substrate]

vo

Vmax

-Inh+inh

1/2 Vmax

Km Km(app)

+Inh

-Inh

-1/Km -1/Km (app)

1/Vmax

1/v

1/[S]

Inhibitor competes with substrates for binding to active siteInhibitor is similar in structure to substrate

binds more stronglyreacts more slowly

Increasing [I] increases [EI] and reduces [E] that is available for substrate bindingNeed to constantly keep [I] high for effective inhibition (cannot be metabolized away in body)

Slope is larger (multiplied by Slope is larger (multiplied by ))

Intercept does not change (VIntercept does not change (Vmaxmax is the same) is the same)

KKMM is larger (multiplied by is larger (multiplied by ))

Uncompetitive Inhibitor

Binds only to ES complex but not free enzyme Binds at location other than active site

Does not look like substrate. Binding of inhibitor distorts active site thus preventing substrate binding and catalysis

Cannot be competed away by increasing conc of substrate (Vmax is affected by [I])

Increasing [I] lowers Vmax and lowers Km.

Increasing [I]Increasing [I]Lowers VLowers Vmaxmax (y-intercept (y-intercept

increases)increases)Lowers KLowers KMM (x-intercept (x-intercept

decreases)decreases)Ratio of KRatio of KMM/V/Vmaxmax is the same is the same

(slope)(slope)

Mixed

Inhibitor binds E or ESIncreasing [I]Increasing [I]

Lowers VLowers Vmaxmax (y-intercept (y-intercept

increases)increases)Raises KRaises KMM (x-intercept (x-intercept

increases)increases)Ratio of KRatio of KMM/V/Vmaxmax is not the same is not the same

(slope changes)(slope changes)

Non-competitive inhibition

Inhibitor binds ES or EIt is a special case of mixed inhibition where Vmax is lowered when [I] increases but Km does not change

Reversible Inhibition (non-competitive)

A inhibitor binds the enzyme but not in its active site. It affects the Kcat because substrate can still bind the active site.Rate of catalysis is affected

[Substrate]

vo

Vmax-Inh

+inh

1/2 Vmax

Km Km(app)

1/2 Vmax(app)

Vmax(app)_

+Inh

-Inh

-1/Km

1/Vmax

1/v

1/[S]

1/Vmax(app)

Vmax is decreased proportional to inhibitor conc

Example

When a slice of apple is cut, it turns brown- enzyme o-diphenol oxidase oxidizes phenols in the appleLets determine max rate at which enzyme functions (Vmax), and Km1 When it acts alone (we will use catechol as substrate. Enz converts this to o-quinone which is dark

and can be measured via absorbance at 540 nm 2 when it acts in presence of competitive inhibitor para hydroxy benzoic acid which bind active site

but is not acted upon3 when it acts in the presence of a non-competitive inhibitor- phenylthiourea which binds copper in

the enzyme which is necessary for enzyme activity

Make a supernatant of the apple-enzyme. Measure color produced (product)Set up 4 tubes with different conc of cathecol and a fixed amount of enzyme (apple pulp).Measure change in absorbance at 1 min intervals for several minutes and record average change in

absorbance. Absorbance is directly proportional to product, we can measure rate of reaction (velocity)

TubeA TubeB TubeC TubeD

[S] mM mM mM mM

1/[S]

Vi (OD)

1/Vi

1/Vmax=10Vmax=0.1-1/Km=-0.8Km=1.25mM

Example

TubeA TubeB TubeC TubeD

[S] mM mM mM mM

1/[S]

Vi (OD)

1/Vi

Each tube also has a fixed amount of PHBA (competitive inhibitor)

1/Vmax=10Vmax=0.1-1/Km=-0.4Km=2.5 mM

TubeA TubeB TubeC TubeD

[S] mM mM mM mM

1/[S]

Vi (OD)

1/Vi

Each tube has a fixed amount of phenylthiourea (non competitive inhibitor)

1/Vmax=20Vmax=0.05-1/Km=-0.8Km=1.25 mM

Irreversible

Inhibitor Inhibitor Binds covalently, orBinds covalently, orDestroys functional Destroys functional group necessary for group necessary for enzymatic activity, orenzymatic activity, orVery stable Very stable noncovalent bindingnoncovalent binding

Suicide InactivatorsSuicide InactivatorsStarts steps of Starts steps of chemical reactionchemical reactionDoes not make Does not make productproductCombines irreversibly Combines irreversibly with enzymewith enzyme

Regulation of enzymes

Catalytic activity is increased or decreased1) Enzyme synthesis or degradation2) Covalent modification3) Non-covalent binding and conformational change (allosteric)

Usually located early in multi-enzyme reaction pathwayUsually located early in multi-enzyme reaction pathwayKinetics differ for allosteric enzymes- sigmoidal curve and K1/2 instead of Km

Usually large; multiple subunitsUsually large; multiple subunitsCompare to HbCompare to Hb

Site for allosteric modulator (Site for allosteric modulator (R = regulatoryR = regulatory) generally different from active site () generally different from active site (C = catalyticC = catalytic))Can be positive or negativeCan be positive or negative

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