lec11_enzkin_08-ppt

BIOC 460, Spring 2008

LEC 11, Enzymes - Kinetics 1

Lecture 11Enzymes: Kinetics

Reading: Berg, Tymoczko & Stryer, 6th ed., Chapter 8,pp. 216-225

Key Concepts• Kinetics is the study of reaction rates (velocities).• Study of enzyme kinetics is useful for measuring

– concentration of an enzyme in a mixture (by its catalytic activity),– its purity (specific activity),– its catalytic efficiency and/or specificity for different substrates– comparison of different forms of the same enzyme in different tissues or

organisms,– effects of inhibitors (which can give information about catalytic

mechanism, structure of active site, potential therapeutic agents...)• Dependence of velocity on [substrate] is described for many enzymes by the

Michaelis-Menten equation:• kinetic parameters:

– Km (the Michaelis constant)– kcat (the turnover number, which relates Vmax, the maximum velocity,

to [Et], the total active site concentration)– kcat/Km (the catalytic efficiency of the enzyme)– can't be greater than limit imposed by diffusion control, ~108-109 M–1sec–1

• Kinetic parameters can be determined graphically by measuring velocity ofenzyme-catalyzed reaction at different concentrations of substrate(Vo vs. [substrate]).



Learning Objectives• Terminology: active site, enzyme-substrate complex, induced fit, initial

velocity, steady state, Vmax , Km , kcat , turnover number, KES , enzymeefficiency.

• Write out a simple Michaelis-Menten kinetic mechanism for an enzyme-catalyzed reaction.

• Recognize the Michaelis-Menten equation, and sketch a graph of Vo vs. [S]for an enzyme-catalyzed reaction that illustrates Vmax and Km.

• Define Km in terms of the rate constants in the Michaelis-Menten kineticmechanism; give the operational definition of Km that holds no matterwhat the actual kinetic mechanism is for a particular enzyme.

• Explain the relationship of kcat to Vmax, and the relationship of Km to KES.• State the units of Km, kcat, and Vmax.• Express the ratio of occupied active sites to total enzyme active sites

([ES]/[ET]) in terms of Vo and Vmax. What is the maximum possible value ofthat ratio?

• Given a plot of Vo/Vmax vs. [S], find the value of Km from the plot.• What two things is the parameter kcat/ Km used to indicate?

– What sets the upper limit for the value of kcat/Km for an enzyme?– What is the approximate range of numerical values for that upper limit of

kcat/Km, with units?

REVIEW: How do enzymes reduce activation energy (ΔG‡)?1. by lowering the free energy of the transition state (‡), e.g., by

binding the transition state tightly2. by changing the reaction pathway by which reactants react to

form products; e.g., taking a 1-step uncatalyzed reaction andaccomplishing the same result by a different route, with severalintermediate reactions en route.

• Each reaction step has its own transition state with its own activation energy (ΔG‡).

• If all of the individual steps' ΔG‡s are lower than the activation energy of the uncatalyzed reaction, the overall rate of product formation will be greater in the presence of the catalyst.

• The overall rate of the catalyzed reaction is dictated by the slowest step in a multistep reaction. Given a free energy diagram like the one in Nelson & Cox, Lehninger Principles of Biochemistry, 4th ed. (2004) Fig. 6-3 (previous lecture notes), how do you identify the rate-limiting (slowest) step on the reaction coordinate?



BINDING = the essence of enzyme action!• binding of SUBSTRATE to form an ES COMPLEX• binding of TRANSITION STATE more tightly than the substrate• Binding occurs at ACTIVE SITE of enzyme.• Subsequent chemical events can then occur.

• Active site:–relatively small part of whole enzyme structure–3-dimensional cleft with participating components from different parts of primary structure–water often excluded so substrates and intermediates are in non-aqueous environment (unless H2O is a reactant)

• Binding uses multiple weak interactions:1. hydrogen bonds2. salt links3. van der Waals interactions4. hydrophobic effect

Lysozyme (residues from different parts of AA sequencecome together in active site)

Berg et al., Fig. 8-7



Specificity of binding• depends on active site crevice being sterically and chemically

complementary to groups it is binding (best complementarity may bepresent in ES complex but NOT in free enzyme -- induced fit)

• Enzymes flexible -- conformational changes can occur when substratebinds during the reaction, to get maximal complementarity to thetransition state.

• induced fit: conformational changes giving tighter binding in a newconformation

• For many (probably most) enzymes, the active site assumes shape complementary to S only when S is bound.

Berg et al., Fig. 8-10

Why study enzyme kinetics (reaction rates)?• measurement of velocity = reaction rate• compare enzymes under different conditions, or from different

tissues or organisms– understand how differences relate to physiology/function of organism– e.g., physiological reason for different Km values for hexokinase vs.

glucokinase (discussed later in course)• compare activity of same enzyme with different substrates (understand

specificity)• measure amount or concentration of one enzyme in a mixture by its

activity• measure enzyme purity (specific activity = amount of activity/amount of

protein)• study/distinguish different types of inhibitors

– info about enzyme active sites and reaction mechanism– development of specific drugs (enzyme inhibitors)



Simple Enzyme-Catalyzed Reaction:

• Measurement of velocity:– V = rate of appearance of product = change in [P] per unit time– units of velocity (V)? _______________

Plot of [P] vs. time• Experimentally, forward velocity VF = slope of plot of [P] vs. time because VF = kF[S] = Δ[P]/Δt

Berg et al., Fig. 8.11(slightly modified)

Determining initial velocity, Vo• Why we measure initial velocity, Vo, the slope of [P] vs. [time] at

very early time after mixing enzyme with substrate• Problem:

– As S is converted to P, concentration of S decreases, so forwardvelocity gets slower and slower.

– Furthermore, as [P] increases, rate of reverse reaction (P --> S)becomes significant, and eventually, when VF = VR, reaction is atequilibrium: d[P]/dt = 0.

• Solution: –Measure V at very early times in reaction, before [S] decreases signficantly (so [P] = ~0).–Velocity measured immediately after mixing E + S, at beginning of reaction (initial velocity), is called Vo.



Vo vs. [S] plots• Simple uncatalyzed S → P reaction

shows linear dependence of Vo on [S]:

• Enzyme-catalyzed reactions show a hyperbolic dependence of Vo on [S]: 1. Rate is saturable: there's a

maximum rate (velocity = Vmax) for any single concentration of enzyme.

2. Rate (velocity) is proportional to [Etotal]: at any single concentration of [S], Vo = c[E] (double [E] → double Vo).

3. Half-maximal velocity occurs at a specific substrate concentration, independent of [E].Km = substrate concentration that gives Vo = 1/2 Vmax.

What’s the slope of plot of VF vs. [S]?

Dependence of Vo on [S] in enzyme-catalyzed reaction:Plot of Vo vs. [S]) for an enzyme that follows Michaelis-Menten kinetics

• plots as a rectangular hyperbola• approaches Vmax (= maximum velocity) at high [S]

• 3 parts of [S] concentration range

1. At very low [S]: Vo is proportional to [S]; doubling [S] → double Vo.

2. In mid-range of [S], Vo is increasing less as [S] increases (where Vo is around 1/2 Vmax).Km = [S] that gives Vo = 1/2 Vmax.

3. At very high [S], Vo is independent of [S]: Vo = Vmax. 1

2

3

Berg et al., Fig. 8.12



Michaelis-Menten model to explain hyperbolicdependence of Vo on [S] in enzyme catalyzed reactions

• Before the chemistry occurs, enzyme binds substrate tomake a noncovalent ES complex.

• Turnover number (def.): number of substrate molecules converted into product by one molecule of enzyme active site per unit time, when enzyme is fully saturated with substrate. = k2 in M-M scheme above = kcat (general term) • kcat = turnover number (general term used, independent of specific kinetic mechanism)• NOTE: kcat (the turnover number) = k2 in the specific Michaelis-Menten kinetic mechanism above.

The Michaelis-Menten Equation describes a rectangular hyperbola.

• The Michaelis-Menten Equation

• where Vmax = k2[ET] (so Vmax is indeed proportional to [ET])• Km: an "aggregate" constant (sum of rate constants for breakdown of ES divided by rate constant for formation of ES):

• Michaelis-Menten equation explains hyperbolic Vo vs. [S] curve:1. At very low [S] ([S] << Km), Vo approaches (Vmax/Km)[S].

Vmax and Km are constants, so linear relationship between Vo and [S]at low [S].

2. When [S] = Km, Vo = 1/2 Vmax3. At very high [S], ([S] >> Km), Vo approaches Vmax (velocity

independent of [S])



Meaning of Km• By definition,

Km = substrate concentration at which velocity (V) is exactly 1/2 of Vmax(operational definition that holds for ANY kinetic mechanism).

• Km is a SUBSTRATE CONCENTRATION.

Compare:

Km = KES (dissoc. constant for ES complex) only if k2 << k–1.

• What is the equilibrium dissociation constant for ES complex (KES)?

Kinetic parameters used to characterize enzyme activityKinetic parameters useful for many comparisons, e.g., effects of enzyme

inhibitors on Km and Vmax = basis of “diagnosis” of type of inhibition.– effects immediately apparent on a double reciprocal plot (see

Enzyme Inhibition notes)1. Km (units: concentration units, e.g., M or mM or µM)

IF k2 << k–1 (not always true), thenKm = k-1/k1 = KES andKm can be taken as a measure of the dissociation constant for the EScomplex (an INVERSE measure of the binding affinity of enzyme for thatsubstrate).

2. kcat (directly related to Vmax) (units: inverse time, e.g., s–1)calculate kcat from Vmax and concentration of enzyme active sites [ET]:Vmax = k2[ET] = kcat[ET] so

3. kcat/ Km (the criterion of catalytic efficiency and "kinetic perfection")(units: inverse conc.•inverse time, e.g., M–1•s–1)



• Wide range of catalytic rate constants, kcat (turnover numbers) amongenzymes:Lysozyme: kcat = 0.5 s–1

Catalase: kcat = 4 x 107 s–1

• In vivo, most enzymes operate below saturation (V < Vmax)(important for regulation)Activity of some metabolic enzymes is regulated: V can be "tuned" up ordown depending on cell's metabolic needs.

• Remember:Vo = kcat[ES] = rate of appearance of P, andVmax = kcat[ET] so(divide 1st equation by the second):

• Ratio of actual velocity at a given substrate concentration (Vo) to Vmax indicates ratio of occupied active sites to total active sites at that [S].

= fractional saturation!

Michaelis-Menten Equation:

Divide both sides by Vmax:

What is the maximum possible value of [ES]/[ET]?

What would be the shape of a plot of Vo/Vmax vs. [S], and how would you find Km on the plot?

Can calculate fractional saturation {ratio of (occupied sites / total active sites)} as



kcat/Km

• kcat/Km is the criterion of substrate specificity, catalytic efficiency and"kinetic perfection”.– units of kcat/Km = conc–1time–1.– k1 sets an upper limit on value of kcat/Km.

• kcat/Km can never be greater than k1 (rate constant for associationof E + S) for a given enzyme-substrate system. (You're notresponsible for the explanation of this.)

– How fast 2 molecules can react (or bind in the case of E + S) islimited by how fast the molecules can diffuse to "bump into eachother" in solution so they can react.

– For molecules the size of an enzyme and a typical substrate, themaximum (diffusion-limited) k1 = ~ 108–109 M–1sec–1.

– Thus max. possible kcat/Km for an enzyme = ~ 108–109 M–1sec–1.• Any enzyme with kcat/Km value in that range approaches the limit of

diffusion control, and thus has achieved something very close to "kineticperfection": the rate at which that enzyme's active site can convertsubstrate into product is limited only by the rate at which it encountersthe substrate in solution.

Uses of kcat/Kmkcat/Km used as a measure of 2 things:1. enzyme's substrate preference2. enzyme's catalytic efficiency

1. enzyme's preference for different substrates (substrate specificity)– The higher the kcat/Km, the better the enzyme works on that substrate.– e.g., chymotrypsin: protease that clearly "prefers" to cleave after bulky hydrophobic and aromatic side chains.

– chymotrypsin specificity: active site best accommodates substrates with a bulky hydrophobic or aromatic residue contributing carbonyl group to peptide bond to be hydrolyzed.

Berg et al., 5th ed., Fig. 9.1



Substrate preferences of chymotrypsin• Chymotrypsin also catalyzes hydrolysis of ESTER bonds whose

carboxylic acid component has a bulky, hydrophobic and/or aromatic Rgroup.

• Amino acids listed in Berg et al. Table 8-6 (below) are the groups thatcontribute the carbonyl group to the ester bond being hydrolyzed by theenzyme.

Which amino acid is most preferred by chymotrypsin for contribution ofcarbonyl group in ester bond to be cleaved, among substrates above?

Catalytic efficiency of enzymes• the higher the kcat/Km, the more "efficient" the enzyme• maximum possible kcat/Km dictated by the diffusion limit

NOTE: kcat/KM for an enzyme can have a value close to the limit of diffusion control either because its kcat is very high, or because its Km is very low, or some combination.

If KM = KES, which enzyme above binds its substrate the most tightly?

Which enzyme has the most rapid catalytic turnover when the enzyme is saturated with substrate?

Which enzymes have the highest catalytic efficiency? Are they near the limit of diffusion control?



Appendix:How was the Michaelis-Menten Equation Derived?

To derive Michaelis-Menten Equation (describe hyperbolic Vo vs. [S] plot):Start with the following 3 equations/assumptions:1. Enzyme active site concentration: [ETotal] = [Efree] + [ES]

[total active sites] = [empty sites] + [occupied sites]2. Total concentration of [S] is high enough that [Sfree] = [Stotal]

i.e., an insignificant fraction of total [S] is bound even when all theenzyme active sites are occupied.

3. In the STEADY STATE, concentration of [ES] is constant(the steady state assumption):rate of formation of ES = rate of breakdown of ES(breakdown rate = sum of rates of breakdown forward + backward)

(See last 2 slides, Appendix, if you’re interested in where to start the derivation, but you’re not responsible for it.)

M-M Equation Derivation, Steady State Assumption• Changes in conc. of enzyme-catalyzed reaction participants with time:• "Steady state" is the part of the reaction period that concentration of ES is not changing: rate of formation of ES = rate of breakdown of ES

k1[E][S] = k–1[ES] + k2[ES]

Berg et al., 5th ed.Fig. 8-13a

lec11_enzkin_08-ppt

Documents

reaction step

particular enzyme

reaction pathway

multistep reaction

study of enzyme kinetics

enzymesubstrate complex

total enzyme active

enzymes kinetics