Question

1. MICHAELIS-MENTON-(REQUIRED) a. Draw a simple graph, showing the classic Michaelis-Menton plot of enzyme activity as a function of substrate concentration; label both axes. Write the associated Michaelis-Menton equation and show the location of Km and Vmax on your graph. b. Draw a second graph showing the classic Lineweaver-Burk plot; label both axes. Show the location of Km and Vmax on your graph. Discuss which plot is the most useful to determine Vmax. Draw a second line on each graph representing the results when a non-competitive inhibitor is present. Label which line corresponds to the inhibitor. d. Using the Michaelis-Menton equation, calculate the velocity of an uninhibited enzyme-catalyzed reaction and the enzyme's turnover number given the following data: (Show your calculations and the units in your answer.) Km= 4x108 M [S]= 1 km Vmax= 6.0 nmole/L.sec Et = 1.0x10-10M e. Draw a graph of enzyme velocity as a function of total enzyme concentration (EU) in the presence of infinitely high substrate concentration. Label the axes and the slope of the line.

Answer 1

AE MICHAELIS MENTEN EQUATION (a) Ets K Ets I ES K ES P + E P + E K2 The rate of formation of product: v=aff - Omax [5] [5] + km dut v= d[P ] - K₃ [E]. [5] [5] + km dt Then, concentration of [5] is very large, [5].7 km all = K₂ [E]. ; Voniax = K3 [E] - Low Concentration of substrate [5] v [f], 15). & smak o Zero order Pate of reaction -- First onder region Concentration of Substrate

At beginning, the reaction to [5] and its slope is is first onder with max . At very high respect concentration km Ploti and its slope is zero i (6) Line weaver and Burk v =- - K₂ [E]. [5] [3] + km Taking inverse of both sides. km te = km [E]o: K₃ [E] [5]. Km t o reman + remar (5) slope = Umar intercept = Umax Thus, Line weaver - Burk determine Vmax. plot is most useful to inhibiton Umar, 12 Una km

When a non- Competitive inhibitor is added the Umax is changed, while the km remains unchanged. (6) we know, we Umax [5] [5] tkm ver, km = 4x108 M Umax = 6.0 nmole/L. See = 6.0x109 mol/l see Et = 10x10 toer . 10 lme xx km - Umax. Km - Umar km + 28m km + km 2 60x10 9 = 3 - 2x159 mol/Lisee. • Turnover number = Umat Et 6.0 x 10 mol/Lisee 1x 10-10 M - 60 See! So, the velocity of an uninhibited enzyme. Catalyzed reaction is 2x10 - mollisee and the enzyme's turnover number is 60 see.