Question

a. what are the values of Vmax and Km in the abscence if the inhibitor what are the values of Vmax and Km in the presence of the inhibitor?

b. what type of inhibition is it?

c. what is the dissociation constant (Ki) of the inhibition?

***d. graph a linear scatter plot including equation.

Homework (CHE 407) The initial velocity for an enzyme-catalyzed reaction is measured at various initial substrate concentration [S]o, in the absence and in the presence of a reversible inhibitor. Data are presented below. 4-mxtc c=1/Ump m-km/Umap 15 INITIAL VELOCITIES (M/Miny Sl. (M) v (no inilibitor) (2.0 mit inhibitor) 4-0.28% +0.021 4.10 0.244 6.40 0.15 9-0.47y +0.021 3.00 0.553 5.00 0.2 10.00 0-1 30.00 0.033 90.00 0.011 10.40 0.01 14.50 DOUB 22.50 0. Out 33.80 D-021 40.50 0.024 11.30 0.080 40.50 -027 33.80 0.029 11 0.0474 (a) What are the values of V and Ky in the absence of the inhibitor? What are the values of V and in the presence of the inhibitor? (b) What type of inhibition is it? (c) What is the dissociation constant (K) of this inhibitor? Wanawanan Willy MAMA 47. micromol mes 0.021 Kam 0000 0.28 oylove = 10.95 - 10.95 x 10- 1-X 10-S Voy @ presence of india ter Vmay = = 47.6 pocomellom 70.670 47.4 = 3489 - 31.09% 106 = 3.1 x 10-5

Answer 1

(A) from Lime wearer eanation we get V = k tos & Jamax ... (1) where, vo- Initia velocity of the reaction Vmax= maximum velocity of the reaction (S = Substrate concentration. K = Michaelis constant an presence of inhibitor the value of max & kr may change as. are & & all respectively. 3o, in foresence of inhibitor the line- -weaner canation is expressed as M solito Anax VMAX (i) & (e) both are linear cauation. Now plotting on y axis & 75 on u axis are will get following canation: 2.0mm inhibitor No inhibitor y =0 2240 +0.022 ... (i compairing canation 02 ül Intercopt = 0.022 Vmax= = 45.45 Vmax = 45.45 Malmin MX mir = 0.6600+ 0.022 compairing Cultiv) Time : 022 : Verde = 45.45 mm kan = 0.666. kapp map x 0 666 - 45.458 0:666 - 30.27 MM KM Vorax - 0.224 SA Kes-Vmax xorry - 45.458 0.224 = 10-18 MM

[S]uM WA Volum/min) 1/Vo(min/uM) No Inhib 2 mM Inhib 1/[S]uM No Inhib 2mM inhib 4.1 0.333333333 0.096153846 0.243902439 0.2 0.068965517 0 .15625 0.1 0.044444444 0.088495575 22.6 0.033333333 0.029585799 0.044247788 33.8 0.011111111 0.024691358 0.029585799 1112 Linweaver Burk plot 0.25 y= 0.666x+0.022 0. 015 = 0.224x+0.022 0.1 1/Vo(min/UM) 005 no inhib 22.0mm Inhib 122 -01 0:15 1/[] (17uM) TV

AD Cost here, &max= no & kape yks This is the feature of competitive in hibitor competitive inhibitor in . Inhibitor & inhibitor # For competitive where s h ) = inhibitor = 2.0mm =) 36-27 sylim = 1048 2010 ( 14 ) Pornst 7 lt L - 3902 2.97 =) LED 52.97 -1 = 197 - *; - teen 20mm i. K;= d.o15 mm Ki