1. Enzymes II: Enzyme Kinetics
Bio 98 - Lecture 9
Enzymes II: Enzyme Kinetics

2. Amino acid side chains with titratable groups
Appr. pKa 4
10-12 8 6
13 (lecture 8!) 10
Carboxylate
Amine/guanidi nium
Sulfhydryl
Imidazole
Hydroxyl

3. Enzyme catalyzed reactionS P
G (free energy)
Reaction coordinate
E+P
ES‡
E+S
G‡ G S‡
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.

4. I. Enzyme reactions have at least two steps
k-1 k1 k2
E + S ES E + P
binding step catalytic step
- rapid slower
- reversible irreversible (often)
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?

5. II. Enzyme kinetics: Michaelis-Menten equation
E + S ES E + P
d[P] k2 [E]t [S]
= ––– = vo = –––––––––
dt Km + [S]
Michaelis-Menten
equation
Initial reaction rate
[E]t = concentration of total enzyme
[S] = concentration of free substrate
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
Rate of breakdown of ES
k-1 + k2
Km = –––––– k1
Rate of formation of ES

6. III. How do we measure k2 and Km values?
urease (0.1 M)
(urea) + H2O CO NH3
A. Typical experiment
urea
(mM) 5 10 20 50
etc
velocity/rate    
(M CO2/min) 30 80
100
Raw data
Vmax
100 vo 50
[urea]

7. III. Why is there a Vmax? urease (0.1 M) (urea) + H2O CO2 + 2 NH3
100 vo 50
[urea]

8. B. How do we get k2 and Km from this graph?
Vmax vo
k2 [E]t [S]
vo = –––––––––
Km + [S]
Vmax/2 Km
[S]
Consider three special cases
1. [S] = vo = 0
[S] ≈ ∞ vo ≈ k2 [E]t = Vmax, so k2 = Vmax / [E]t
3. [S] = Km when vo = ½ Vmax
Remember a finite number (Km) becomes negligible in the face of infinity

9. Assumptions for steady-state kinetics
E + S ES E + P
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]
Vmax [S]
vo = –––––––––
Km + [S]
with k2 [E]t = Vmax, then
Case 2 from previous slide!

10. IV. What is the physical meaning of k2?
Suppose [E]t = 0.01 µM, Vmax = 200 µM/min
Vmax µM/min
k2 = –––––– = –––––––––––– = 20,000 min-1
[E]t µM
so 20,000 moles of P produced per min per mole of E
k2 is the # of reactions a single enzyme molecule can catalyze per unit time
k2 = kcat = “catalytic constant” or “turnover number”, expressed in catalysis events per time.

11. V. What is the physical meaning of Km?
E + S ES E + P
k-1 + k k-1
Km = –––––– ≈ ––– = Kdiss provided (k2 << k-1)
k k1
Rate of breakdown of ES
Remember: k2 is rate-limiting thus rate is slower than k-1 and thus k2 numerically much smaller than k-1
Rate of formation of ES
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 M (high affinity vs low affinity, compare to P50s for T and R states of hemoglobin, lecture 7)

12. VI. A better way to plot vo vs [S] data.
Vmax
vo vs [S] plot ? Km
Lineweaver-Burk plot
1/vo
1/Vmax
-1/Km
1/[S]
Vmax [S]
vo = –––––––––
Km + [S] Km
–– = –––– ––– –––––
vo Vmax [S] Vmax
Lineweaver-Burk eliminates uncertainty in estimating Vmax.
The estimates of Vmax and Km are thus greatly improved.

13. Vmax [S] vo = ––––––––– Km + [S] 1 Km + [S] = –––––––– vo Vmax [S]
Take reciprocal of both sides of equation 1
Km + [S]
= ––––––––
vo Vmax [S]
Expand
= ––––––––
vo Vmax [S] Km 1
[S] +
Vmax[S]
Thus Km
–– = –––– ––– –––––
vo Vmax [S] Vmax
y = ax + b
Lineweaver-Burk

14. Solve for y at x=1/[S]=0: y =
Lineweaver-Burk plot
Vmax [S]
vo = –––––––––
Km + [S]
1/vo Km
–– = –––– ––– –––––
vo Vmax [S] Vmax
1/Vmax
-1/Km
1/[S]
y = a x b
Solve for y at x=1/[S]=0: y =
Solve for x at y=1/v0=0: x =
–– = –––– = b
vo Vmax
–– = - ––––
[S] Vmax
Vmax b
–– = - ––
Km a

15. Efficiency = kcat / Km (specificity constant)
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 M s
Ac-Phe-Gly Ac-Phe + Gly
Pepsin M s ,700
Phe-Gly Phe +Gly

16. VIII. Enzyme inhibition - what to know
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?
Irreversible inhibition
• What is it; how does it work; what is its use?       
• What are suicide inhibitors, how do they work?
• Know one example.

17. Classical competitive inhibition
where I is the inhibitor K1

18. How do you measure competitive inhibition?K1
-1/Km
Vmax [S]
vo = –––––––––
aKm + [S]
[I]
a = 1 + ––– KI
[E][I]
KI = ––––––
[EI]
where
Vmax remains unchanged, but apparent Km increases with increasing [I]

19. Inactivation of chymotrypsin by diisopropylfluorophosphate,
an irreversible or suicide inhibitor

20. Inhibitor (diisopropylfluorophosphate)
Chymotrypsin is a serine protease that cleaves a peptide at Phe/Tyr/Trp (C) - leaving a COO- on Phe/Tyr/Trp R2
Inhibitor (diisopropylfluorophosphate)

21. Structure of COX-2 inactivated by Aspirin
Structure of COX-2 inactivated by Aspirin. In the active site of each of the two enzymes, Serine 516 has been acetylated. Also visible is the salicylic acid which has transferred the acyl group, and the heme cofactor.
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.