Lecture 10 Enzyme Kinetics

Rate constant (k) measures how rapidly a reaction occurs

A B + Ck1

k-1

Rate (v, velocity) = (rate constant) (concentration of reactants)

v= k1 [A] 1st order reaction (rate dependent on concentration of 1 reactant)

v= k-1[B][C]2nd order reaction (rate dependent on concentration of 2 reactants)

Zero order reaction (rate is independent of reactant concentration)

Rate constants and reaction order

Sample questions

The rate of a second order reaction depends on the concentration of _________.

• (a) one substrate • (b) two substrates • (c) three substrates • (d) none of the above

E + S ES E + Pk1

k-1

k2

k-2

E S+ E S E + P

Initial Velocities

[S] = 1 mM

d[P]/dT = Vo1 mM

[P]

time

Hold [E] constant[E]<<<<<[S]

Initial Velocities

[S] = 1 mM

[S] = 5 mM

[S] = 10 mMd[P]/dT = Vo10 mM

d[P]/dT = Vo5 mM

d[P]/dT = Vo1 mM

[P]

time

Plot Vo vs. [S]

Vo 1 mM

Vo 5 mM

Vo 10 mM

1) Measurements made to measure initial velocity (vo). At vo very little product formed. Therefore, the rate at which E + P react to form ES is negligible and k-2 is 0. Therefore

Initial Velocity Assumption

E + S ES E + Pk1

k-1

k2

E S+ E S

k-2

E + P

Steady State Assumption

E + S ES E + Pk1

k-1

k2

Steady state Assumption = [ES] is constant. The rate of ES formation equals the rate of ES breakdown

E S+ E S E + P

Data from a single experiment performed with at a single [S].(single point on Vo vs. [S] plot)

E + S ESk1

E S+ E S

Rate of ES formation

Rate = k1 [E] [S]

ES E + Pk2

E S E + P

ES E + Sk-1

E S+E S

Rate of ES breakdown

Rate = (k2 [ES]) + (k-1[ES])

Rate = [ES](k2 + k-1)

If the rate of ES formation equals the rate of ES breakdown

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

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

3) (k-1+ k2) / k1 = Km (Michaelis constant)

Michaelis-Menton Derivation

1. The overall rate of product formation: v = k2 [ES]

2. Rate of formation of [ES]: vf = k1[E][S]

3. Rate of decomposition of [ES]:

vd = k-1[ES] + k2 [ES]

4. Rate of ES formation = Rate of ES decomposition (steady state)

5. So: k1[E][S] = k-1[ES] + k2 [ES]

E + S ES E + P k1 k2

k-1

Not required to know

Michaelis-Menton Derivation

6. In solving for [ES], use the enzyme balance to eliminate [E]. ET = [E] + [ES]

7. k1 (ET - [ES])[S] = k-1[ES] + k2 [ES]

k1 ET[S] - k1[ES][S] = k-1[ES] + k2 [ES]

8. Rearrange and combine [ES] terms:

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

k1 ET[S] 9. Solve for [ES] = -----------------------

(k-1 + k2 + k1 [S])

Not required to know

Michaelis-Menton Derivation

ET[S]10. Divide through by k1: [ES] = -----------------------

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

11. Defined Michaelis constant: KM = (k-1 + k2) / k1

12. Substitute KM into the equation in step 10.

13. Then substitute [ES] into v = k2 [ES] from step1 and replace Vmax with k2 ET to give:

Vmax[S] vo = ----------- KM + [S]

Not required to know

Km = [S] at ½ Vmax

(units moles/L=M)(1/2 of enzyme bound to S)

Vmax = velocity where all of the enzyme is bound to substrate(enzyme is saturated with S)

Understanding Vmax

The theoretical maximal velocity

• Vmax is a constant

• Vmax is the theoretical maximal rate of the reaction - but it is NEVER achieved in reality

• To reach Vmax would require that ALL enzyme molecules are tightly bound with substrate

• Vmax is asymptotically approached as substrate is increased

What does Km mean?

1. Km = [S] at ½ Vmax

2. Km is a constant; Km is a combination of rate constants describing the formation and breakdown of the ES complex

3. Km is usually a little higher than the physiological [S]

What does Km mean?

4. Km represents the amount of substrate required to bind ½ of the available enzyme (binding constant of the enzyme for substrate)

5. Km can be used to evaluate the specificity of an enzyme for a substrate

6. Small Km means tight binding; high Km means weak binding

Glucose Km = 8 X 10-6

Allose Km = 8 X 10-3

Mannose Km = 5 X 10-6

Hexose KinaseGlucose + ATP <-> Glucose-6-P + ADP

Sample questions

• How does the Michaelis-Menten equation explain why the rate of an enzyme-catalyzed reaction reaches a maximum value at high substrate?

• At high So, Km <<<< So (numerically), so the term Km + So in the M-M equation becomes equal to So. Vo = (Vmax So)/So, and So cancels. Therefore at high So then, Vo = Vmax.

The turnover number

A measure of catalytic activity

• kcat, the turnover number, is the number of substrate molecules converted to product per enzyme molecule per unit of time, when E is saturated with substrate.

• If the Michaelis-Menten model fits, k2 = kca

t = Vmax/Et

What does kcat mean?1. kcat is the 1st order rate constant describing

ES E+P

2. Also known as the turnover number because it describes the number of reactions that a molecule of enzyme can catalyze per second under optimal condition.

3. Most enzyme have kcat values between 102 and 103 s-1

4. For simple reactions k2 = kcat , for multistep reactions kcat = rate limiting step

E + S ES E + Pk1

k-1

kcat

What does kcat/Km mean?

• It measures how the enzyme performs when S is low

• kcat/Km describes an enzymes preference for different substrates = specificity constant

• The upper limit for kcat/Km is the diffusion limit - substrate diffuse into the active site, or product diffuse out

• Catalytic perfection when kcat/Km = diffusion rate

The catalytic efficiency

kcat/KM

kcat/KM is taken to be a measure of the efficiency of an enzyme.

Rewriting kcat/KM in terms of the kinetic constants gives:

kcat k1k2 ---- = ----------- KM k-1 + k2

So, where k2 is small, the denominator becomes k-

1 and kcat/KM is small.

Short summary

• Km substrate specificity; substrate binding

• kcat, the turnover number

• kcat/Km the catalytic efficiency

Sample questions

Which of the following kinetic parameters best describes how well suited a specific compound functions as a substrate for a particular enzyme?

• (a) Km • (b) Vmax • (c) kcat

• (d) kcat/Km

Sample questions

The rate-determining step of Michaelis Menten kinetics is• A.the complex formation step• B.the complex dissociation step to produce product• C.the product formation step• D.Both (a)and(c)

Limitations of Michaelis-Menten model

1. Some enzyme catalyzed reactions show more complex behaviorE + S<->ES<->EZ<->EP<-> E + P

Michaelis-Menten can look only at rate limiting step

2. Often more than one substrate E+S1<->ES1+S2<->ES1S2<->EP1P2<-> EP2+P1<-> E+P2 Must optimize one substrate then calculate kinetic parameters for the other

3. Assumes k-2 = 0

4. Assume steady state conditions

The dual nature of the Michaelis-Menten equation

Combination of 0-order and 1st-order kinetics

• When S is low, the equation for rate is 1st order in S

• When S is high, the equation for rate is 0-order in S

• The Michaelis-Menten equation describes a hyperbolic dependence of v on S

How do you get values for Vmax, Km and kcat?

• Can determine Km and Vmax experimentally

• Km can be determined without an absolutely pure enzyme

• Kcat values can be determined if Vmax is known and the absolute concentration of enzyme is known (Vma

x = kcat[Etotal]

BB

B

BB B

B

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6 7 8 9 10

Vo

[S]

[S] Vo0.5 0.0750.75 0.092 0.1524 0.1966 0.218 0.21410 0.23

V max

KmKm ~ 1.3 mM

Vmax ~ 0.25

Lineweaver-Burke Plots(double reciprocal plots)

•Plot 1/[S] vs 1/Vo

•L-B equation for straight line

•X-intercept = -1/Km

•Y-intercept = 1/Vmax

•Easier to extrapolate values w/ straight line vs hyperbolic curve

Sample questions• For an enzyme (5 μM) , the following in

itial velocities have been reported depending on the substrate concentration:

• (a) Draw a Michaelis-Menten plot for this enzyme.

• (b) Draw a Lineweaver-Burke plot for this enzyme.

• (c) Determine Km and Vmax for this enzyme

• (d) Indicate in both graphs (a & b) where Vmax and Km can be recognized.

• (e) Calculate the turnover number and the catalytic efficiency for this enzyme.

[Substrate], mM

v0, mM/s

0.02 10.83

0.04 18.57

0.07 26.76

0.1 32.50

0.15 39.00

0.2 43.33

0.3 48.75

0.5 54.17

0.7 56.88

kcat = Vmax / [E]total catalytic efficiency: kcat/Km

Answer

• (a)

• (b)

• (c) Km and Vmax can be determined from the intercepts in the Lineweaver-Burke plot: 1/Vmax = 0.015 s/mM Vmax = 66 mM/s -1/Km = -10 mM Km = 0.1 mM

• (e) kcat = Vmax / [E]total = 65 mM/s ÷ 5 μM = 65 mM/s ÷ 0.005 mM = 13000/s catalytic efficiency: kcat/Km = 13000/s ÷ 0.1 mM = 13000/s ÷ 0.0001 M = 1.3×108 M/s

Enzyme Inhibition

• Inhibitor – substance that binds to an enzyme and interferes with its activity

• Can prevent formation of ES complex or prevent ES breakdown to E + P.

Reversible versus Irreversible

• Reversible inhibitors interact with an enzyme via noncovalent associations

• Irreversible inhibitors interact with an enzyme via covalent associations

Reversible Inhibitors

E + S <-> ES -> E + PE + I <-> EI

Ki = [E][I]/[EI]• Competitive• Uncompetitive• Non-competitive

Classes of Reversible Inhibition

Two real, one hypothetical • Competitive inhibition - inhibitor (I) competes with the

substrate for the active site of the enzyme

• Non-competitive inhibition - inhibitor (I) binds to an enzyme somewhere other than the active site. It can binds either ES or E. A non-competitive inhibitor reacts with the enzyme-substrate complex, and slows the rate of reaction to form the enzyme-product complex.

• Uncompetitive inhibition - inhibitor (I) binds only to ES, not to E. This is a hypothetical case that has never been documented for a real enzyme, but which makes a useful contrast to competitive inhibition

• Competitive inhibitor: Vmax stays the same, but Km increases• Non-competitive inhibitor decreases the turnover number of the enz

yme rather than preventing substrate binding- Vmax decreases but Km stays the same. This cannot be overcome with an increase in substrate concentration.

Enzyme Inhibition

Noncovalent binding:

Competitive (I binds only to E)

Uncompetitive (I binds only to ES)

Noncompetitive (I binds to E or ES)

Covalent binding – irreversible

Group Specific

Substrate Analogs

(bound to the active site and prevent further reactions)

Competitive Inhibitor (CI)

•CI binds free enzyme

•Competes with substrate for enzyme binding.

•Raises Km without effecting Vmax

•Can relieve inhibition with more S

Competitive Inhibitors look like substrate

NH2C

O

HO NH2S

O

H2N

O

PABASulfanilamide

Non-competitive Inhibitor (NI)

•NI can bind free E or ES complex

•Lowers Vmax, but Km remains the same

•NI’s don’t bind to S binding site therefore don’t effect Km

•Alters conformation of enzyme to effect catalysis but not substrate binding

Uncompetitive Inhibitor (UI)

•UI binds ES complex •Prevents ES from proceeding to E + P or back to E + S.•Lowers Km & Vmax, but ratio of Km/Vmax remains the same•Occurs with multisubstrate enzymes

Sample questions

Which of the following binds to an enzyme at its active site? • A) irreversible inhibitor • B) reversible competitive inhibitor • C) reversible noncompetitive inhibitor • D) more than one correct response • E) no correct response

An uncompetitive inhibitor binds to _____. • (a) E • (b) ES • (c) P • (d) a and b • (e) a and c

Sample questions

A reversible inhibitor that can bind to either E alone or the ES complex is referred to as a _____.

• (a) competitive inhibitor. • (b) non-competitive inhibitor. • (c) uncompetitive inhibitor. • (d) suicide inhibitor. • (e) irreversible inhibitor.

Sample questions

A competitive inhibitor of an enzyme is usually• A.a highly reactive compound• B.a metal ion such as Hg2+ or Pb2+

• C.structurally similar to the substrate.• D.water insoluble

The enzyme inhibition can occur by• A.reversible inhibitors• B.irreversible inhibitors• C.Both (a) and (b)• D.None of these

Sample questions

In a Lineweaver-Burk Plot, competitive inhibitor shows which of the following effect?

• A.It moves the entire curve to right• B.It moves the entire curve to left• C.It changes the x-intercept• D.It has no effect on the slope

Sample questions

Non-competitive inhibitor of an enzyme catalyzed reaction

• A.decreases Vmax

• B.binds to ES• C.both (a) and (b)• D.can actually increase reaction velocity in rare cases

Sample questions

A classical uncompetitive inhibitor is a compound that binds• A.reversibly to the enzyme substrate complex yielding an inactiv

e ESI complex• B.irreversibly to the enzyme substrate complex yielding an inacti

ve ESI complex• C.reversibly to the enzyme substrate complex yielding an active

ESI complex• D.irreversibly to the enzyme substrate complex yielding an activ

e ESI complex

Kinetics of Multisubstrate Reactions

E + A + B <-> E + P + Q

• Sequential Reactions

a) ordered

b) random

• Ping-pong Reactions

Sequential Reactions

E EA (EAB) (EPQ) EQ E

A B P Q

AB PQ

A B

E

EA

EB

(EAB)(EPQ)

P Q

EQ

EP

E

Ordered

Random

Ordered Sequential

Ordered Random

Ping-Pong Reactions

E (EA)(FP) (F) (FB)(EQ) E

A BP Q

•In Ping-Pong reactions first product released before second substrate binds

•When E binds A, E changes to F

•When F binds B, F changes back to E

Lineweaver-Burke Plot of Multisubstrate Reactions

Increasing [B]

Increasing [B]

Sequential Ping-Pong

Vmax doesn’t changeKm changes

Both Vmax & Km change

1/Vo

1/[S]

1/Vo

1/[S]