Bio 98 - Lecture 9

Enzymes II: Enzyme Kinetics

Amino acid side chains with titratable groups

Appr. pKa

4

10-12

8

6

13 (lecture 8!)

10

Carboxylate

Amine/guanidinium

Sulfhydryl

Imidazole

Hydroxyl

Hydroxyl

1. Enzymes do not alter the equilibrium or G.2. They accelerate reactions by decreasing G‡.3. They accomplish this by stabilizing the transition state.

Enzyme catalyzed reaction

G (

free

ene

rgy)

Reaction coordinate

E+P

ES‡

E+S G‡

G

S‡

S PE

binding step catalytic step - rapid - slower - reversible - irreversible (often)

I. Enzyme reactions have at least two steps

k-1

k1 k2E + S ES E + P

ES = “enzyme-substrate complex” ≠ transition state (ES‡)

(1) What is the physical meaning of the constants? What do they tell us about effectiveness of binding & catalysis?

(2) How can we determine experimentally the value of these constants for a given enzyme?

II. Enzyme kinetics: Michaelis-Menten equation

d[P] k2 [E]t [S] = ––– = vo = ––––––––– dt Km + [S]

[E]t = concentration of total enzyme

[S] = concentration of free substrate

k-1 + k2Km = –––––– k1

Information obtained from the study of vo vs [S]

k2: catalytic power of the enzyme (turnover rate), aka kcat; unit: 1/s

Km: effectiveness (affinity) with which enzyme E binds S; unit: M

k-1

k1 k2E + S ES E + P

Rate of breakdown of ES

Rate of formation of ES

Michaelis-Menten equationInitial reaction rate

urea(mM)

5 10 20 50 etc

velocity/rate     (M CO2/min)

30 50 80 100 etc

Raw data

III. How do we measure k2 and Km values?

urease (0.1 M) (urea) + H2O CO2 + 2 NH3

A. Typical experiment

vo

[urea]

Vmax

50

100

III. Why is there a Vmax?

urease (0.1 M) (urea) + H2O CO2 + 2 NH3

vo

[urea]

Vmax

50

100

vo

[S]

Vmax

B. How do we get k2 and Km from this graph?

k2 [E]t [S]vo = ––––––––– Km + [S]

Km

Consider three special cases

1. [S] = 0 vo = 0

2. [S] ≈ ∞ vo ≈ k2 [E]t = Vmax, so k2 = Vmax / [E]t

3. [S] = Km when vo = ½ Vmax

Vmax/2

Remember a finite number (Km) becomes negligible in the face of infinity

k-1

k1 k2E + S ES E + P

Assumptions for steady-state kinetics

The Michaelis-Menten equation assumes that the chemical reaction has reached steady state:

• [ES] remains constant over time• presteady state (the build up of the ES complex) happens in microseconds• Usually nM [enzyme] but mM [substrate] in reaction, so [S] >> [E]

k2 [E]t [S]vo = ––––––––– Km + [S]

with k2 [E]t = Vmax, then Vmax [S]vo = ––––––––– Km + [S]

Case 2 from previous slide!

Vmax 200 µM/min k2 = –––––– = –––––––––––– = 20,000 min-1

[E]t 0.01 µM

IV. What is the physical meaning of k2?

so 20,000 moles of P produced per min per mole of E

k2 = kcat = “catalytic constant” or “turnover number”, expressed in catalysis events per time.

k2 is the # of reactions a single enzyme molecule can catalyze per unit time

Suppose [E]t = 0.01 µM, Vmax = 200 µM/min

V. What is the physical meaning of Km?

k-1 + k2 k-1 Km = –––––– ≈ ––– = Kdiss provided (k2 << k-1) k1 k1

1. Km is a measure of how tightly an enzyme binds its substrate.

2. It is the value of [S] at which half of the enzyme molecules have their active sites occupied with S, generating ES.

3. For a given enzyme each substrate has its own Km.

4. Lower Km values mean more effective binding. Consider Km =  10-3 vs. 10-6 M

(high affinity vs low affinity, compare to P50s for T and R states of hemoglobin, lecture 7)

k-1

k1 k2E + S ES E + P

Remember: k2 is rate-limiting thus rate is slower than k-1 and thus k2 numerically much smaller than k-1

Rate of breakdown of ES

Rate of formation of ES

VI. A better way to plot vo vs [S] data.

[S]

vo

Vmax

vo vs [S] plot

?

Km 1/[S]

1/Vmax

-1/Km

1/vo

Lineweaver-Burk plot

Vmax [S]vo = ––––––––– Km + [S]

1 Km 1 1–– = –––– ––– + ––––– vo Vmax [S] Vmax

Lineweaver-Burk eliminates uncertainty in estimating Vmax.The estimates of Vmax and Km are thus greatly improved.

Vmax [S]vo = ––––––––– Km + [S]

Take reciprocal of both sides of equation

Expand

= ––––––––vo Vmax [S]

Km + [S]1

1 Km 1 1–– = –––– ––– + ––––– vo Vmax [S] Vmax

Thus

y = ax + b

Lineweaver-Burk

= ––––––––vo Vmax [S]

Km1 +[S]

Vmax[S]

Vmax [S]vo = ––––––––– Km + [S]

1 Km 1 1–– = –––– ––– + ––––– vo Vmax [S] Vmax

y = a x + b

Solve for y at x=1/[S]=0: y =

Solve for x at y=1/v0=0: x =

1/[S]

1/Vmax

-1/Km

1/vo

Lineweaver-Burk plot

1 1–– = –––– = b vo Vmax

1 1–– = - –––– [S] Vmax

Vmax b –– = - –– Km a

VII. Enzyme efficiency

Efficiency = kcat / Km (specificity constant)

Combines an enzyme’s catalytic potential with its ability to bind substrate at low concentration.

Example – which enzyme is more efficient?

Enzyme Km kcat kcat/Km

Chymotrypsin 0.015 M 0.14 s-1 9.3Ac-Phe-Gly Ac-Phe + Gly

Pepsin 0.0003 M 0.50 s-1 1,700Phe-Gly Phe +Gly

VIII. Enzyme inhibition - what to know

1. Reversible vs. irreversible inhibition• What is the difference?

2. Competitive inhibition • Know how to recognize or draw the model.• Know how vo vs [S], and Lineweaver-Burk plots

are affected by competitive inhibition.• What are and Ki?

3. Irreversible inhibition• What is it; how does it work; what is its use?       • What are suicide inhibitors, how do they work?• Know one example.

Classical competitive inhibitionwhere I is the inhibitor

K1

How do you measure competitive inhibition?

Vmax [S]vo = ––––––––– Km + [S]

[E][I]KI = –––––– [EI]

where [I]

= 1 + ––– KI

-1/Km

K1

Vmax remains unchanged, but apparent Km increases with increasing [I]

Inactivation of chymotrypsin by diisopropylfluorophosphate, an irreversible or suicide inhibitor

R2

Chymotrypsin is a serine protease that cleaves a peptide at Phe/Tyr/Trp (C) -

leaving a COO- on Phe/Tyr/Trp

Inhibitor (diisopropylfluorophosphate)

Aspirin acts as an acetylating agent where an acetyl group is covalently attached to a serine residue in the active site of the cyclooxygenase enzyme, rendering it inactive.