Recitation week 8 (Chapter 6) PDF

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enzyme kinetics...

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Recitation week 8 Chapter 6: Enzyme kinetics A. Key points to remember: •

Rate of the reaction: A + B ↔ C

rate = −

∆[𝐴]

∆[𝐵]

∆𝑡

∆𝑡

=−

=

∆[𝐶] ∆𝑡

In other words, the rate of consumption of A or B (negative because concentration of A/B is decreasing) is equal to the rate of formation of C (positive) • • •

•

Enzyme kinetics: study of the rate of the reaction Why study enzyme kinetics? Enzyme kinetics provide useful information of mechanism, regulation, the order of binding and the potential inhibitors of enzymes. Importance of initial velocity (Vo): o Rate of formation of product BEFORE equilibrium is reached. (Think about what will happen when you measure the rate at equilibrium? [P] / [S] will not change at equilibrium, which means the velocity under these conditions is zero => no information of the reaction) o Ignore the dead time between mixing and start of measurement (figure below): during this time, [S] is low and enzyme would wait for enough substrate to bind to initiate a reaction => measure V during this time will yield inaccurate results  1st Assumption (to develop Michaelis-Menten equation): Saturation of enzyme with substrate will give a maximal velocity of the reaction or [S]>>>[E}

Michaelis-Menten equation:

o Etot = ES + E 𝑑[𝑃] o Vo = =k2[ES] 𝑑𝑡

o 3 important assumptions: ▪ 1st assumption (to develop Michaelis-Menten equation) => Saturation of enzyme with substrate will give a maximal velocity of the reaction or [S]>>>[E} and [S] ≈ [Stot] (free ligand approximation) ▪ 2nd assumption (Steady-state theory) => during this stage, [ES] remains constant while [S] is decreasing linearly and [P] is increasing linearly. ▪ 3rd assumption: Back reaction (E+P → ES) is ignored o Start from the rate law (rate of consumption of ES = rate of formation of ES) 𝑑[𝐸𝑆] = 𝑘−1 [𝐸𝑆] + 𝑘2 [𝐸𝑆] = 𝑘1 [𝐸 ][𝑆] 𝑑𝑡 Try to solve for [ES] in terms of measurable variables [E], [S] to get: 𝑉max [𝑆] 𝑉𝑜 = 𝐾𝑀 + [𝑆] ▪ [S] = KM => Vo = Vmax/2 ▪ [S]>>KM => Vo = Vmax ▪

•

𝑉max [𝑆] 𝐾𝑀

Turnover number (kcat or k2): number of substrate molecules converted to product per enzyme molecule per unit of time, when [Etot] = [ES] => 𝑉𝑜 = 𝑘2 [𝐸𝑆] = 𝑘𝑐𝑎𝑡[𝐸𝑆] => 𝑤ℎ𝑒𝑛 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑: [𝐸𝑆] = [𝐸𝑡𝑜𝑡 ] 𝑎𝑛𝑑 𝑉𝑜 = 𝑉𝑚𝑎𝑥 => 𝑘𝑐𝑎𝑡 =

•

[S] = [𝑆] 𝑀

𝑉𝑜

𝑉max

𝑉max

(x,y), you should be able to fit them on a plot and determine a and b)

=>

•

Enzyme inhibitors: o Normal reaction:

o Competitive inhibition:

• • • •

Where to bind? The inhibitor binds to the active site Properties? It has the similar structure and chemical properties compared to substrate Action? Binds to E and compete with substrate for the binding at the active site Effect? Reduce binding affinity of S to E => KM increases => can be overcome by increasing [S]

o Noncompetitive (rarely found in reality) and mixed inhibition • • • •

Mixed inhibition KM increased Vmax reduced

Where to bind? The inhibitor binds to the allosteric sites (other than active site) Properties? Similar or different structure to regulators Action? Binds to E or ES, and causes conformational change of E => cannot bind to S Effect? o Non-competitive: Binding of I to S or ES does not affect the binding affinity of S to E => not directly compete with S; reduce Vmax o Mixed: Binding of I to S or ES reduces the binding affinity of S to E => KM increases, Vmax decreases

o

• • • •

Uncompetitive inhibition

Where to bind? The inhibitor binds to site other than the active site Properties? Does not share the similar structure to substrate Action? Binds to ES and does not allow the reaction of S to P to occur even though S can still binds to E Effect? Causes E to be inactive => Vmax decreases; a portion of ES is deleted => [S] required to achieve ½ Vmax declined => KM decreases

B. Problems 1. Derive Michaelis-Menten equation for uncompetitive inhibition. 𝐾𝐼′ = 𝑉𝑜 = 1 𝑉𝑜

=

[𝐸𝑆][𝐼] [𝐸𝑆𝐼] 𝑉max [𝑆] 𝐾𝑀 +𝛼′[𝑆] 𝛼′ 𝑉max

𝛼′ = 1 +

𝐾𝑀

+𝑉

1

max [𝑆]

[𝐼] 𝐾′𝐼

2. What is the chemical basis of enzyme catalysis? Or which steps are involved in enzyme catalysis?

3. a) You measure the kinetics of an enzyme E as a function of substrate concentration first without any inhibitor (see Table) and plot the data using the double-reciprocal (Lineweaver-Burk) plot (Figure below). The enzyme concentration is maintained constant at a level of 1 µM (=10 -6 M)

From these data, determine Vmax, KM, kcat, and turnover number for the enzyme. b) Now you study enzyme inhibition by measuring enzyme kinetics in the presence of 10 mM of inhibitor A or inhibitor B (separately). The Lineweaver-Burk plots in the presence of these inhibitors are indicated by “+A” or “+B” in the Figure below: Determine the type of inhibition for A and B. Explain. c) Which of the two inhibitors is more efficient at high substrate concentrations? At low substrate concentrations? Show you reasoning. d) Using the data you have, determine the binding constant (KI) for each of the inhibitors. Show the calculations. (hint: For competitive inhibition: Vo = Vmax[S]/(αKm+[S]) For noncompetitive inhibition: Vo = (Vmax/α)[S]/(Km+[S] α = 1 + [I] /KI

4. Meat tenderizer contains an enzyme that interacts with meat. If meat is coated with tenderizer and then placed in a refrigerator for a short time, how would the enzyme be affected? A. It would be broken down. B. Its activity would slow down. C. Its shape would change. D. It would no longer act as an enzyme 5. The effect of pH on a certain enzyme is shown in the graph. At what pH would the enzyme be most effective? A. above 10 B. between 8 and 10 C. between 5 and 7 D. below 5...