Enzymes II: Enzyme Kinetics 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/guanidi nium Sulfhydryl Imidazole Hydroxyl

Enzyme catalyzed reaction S 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.

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?

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

III. How do we measure k2 and Km values? urease (0.1 M) (urea) + H2O CO2 + 2 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]

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

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] = 0 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

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!

IV. What is the physical meaning of k2? Suppose [E]t = 0.01 µM, Vmax = 200 µM/min Vmax 200 µM/min k2 = –––––– = –––––––––––– = 20,000 min-1 [E]t 0.01 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.

V. What is the physical meaning of Km? E + S ES E + P k-1 + k2 k-1 Km = –––––– ≈ ––– = Kdiss provided (k2 << k-1) k1 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. 10-6 M (high affinity vs low affinity, compare to P50s for T and R states of hemoglobin, lecture 7)

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] 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] 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 1 Km 1 1 –– = –––– ––– + ––––– vo Vmax [S] Vmax y = ax + b Lineweaver-Burk

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

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 0.015 M 0.14 s-1 9.3 Ac-Phe-Gly Ac-Phe + Gly Pepsin 0.0003 M 0.50 s-1 1,700 Phe-Gly Phe +Gly

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.

Classical competitive inhibition where I is the inhibitor K1

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]

Inactivation of chymotrypsin by diisopropylfluorophosphate, an irreversible or suicide inhibitor

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)

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.