Question

b) (10 points) Acetylcholinesterase is a hydrolase that hydrolyzes the neurotransmitter acetylcholine. Caffeine inhibits the action of Acetylcholinesterase and clinical studies have indicated that it can be involved in the slowing of Alzheimer disease pathology. P11149 is a galanthamine analogue that has also been studied as a potential drug for treating Alzheimer disease.

The initial reaction rate for acetylcholine hydrolysis was measured at an enzyme concentration of 5×10-8 M with no inhibitor present, and in the presence of caffeine or P11149 (separately). Both inhibitors were evaluated at the same concentration (2×10-4 M). The Lineweaver-Burk plot below summarizes the experimental observations.

b1. Determine vmax and km for the hydrolysis of acetylcholine by acetylcholinesterase.
b2. Explain the main type of enzyme inhibition and from experimental results, determine the type of inhibition for: Caffeine: P11149:
b3. Which of the two inhibitors is more efficient at high substrate concentrations? At low substrate concentrations? Show your reasoning.
b4. Using the data you have, determine the inhibition constant (k I) for each inhibitor. Show your calculations.

0.35 No Inhibitor y = 2970x + 33333 0.30 Caffeine P11149 0.25 0.20 y = 1786x + 71429 1/v (Ls/mol) x 106 0.15 0.10 y = 833x + 33333 0.05 0.00 -0.05 -60 -10 -20 0 40 60 80 100 120 20 1/S [L/mol]

Answer 1

vo Vmax Vmax bl. t = km x + tmax for cauty1 Choline, y = 833x +3333 so se cone. is so, then y = (833x20 + 8333) = 0.05x106 So, x 2 / 20, then = (t = 0.05x106 similesly, when /= 40, then yo = (833x 40t 3333) = 666134 So, putting, &ls and to value to equation no. for, I = 40, /o=66653 66653 = Km x 40 & ton ③ mar For Y=20, /vo = $9993 49993 - Km x 20 Usax Bogy So, substalting eq & with eqo 49993-66683=120km - Yokm 116660 = f 20km, km = 16.660 Nmax = 833 may So, km = 833kmax - pert this egno. 4 in ego 49993= 833 Vmaxx 20 +t more imaxe = (49993-833470) 9993–16660)= 33,333 I may hoon I har man Visart Vmax = (1 3333) = 3x10-5 - purring this value ey 9 Kan = 833x Knay = (833x3X107) = 0.025 so, for allly scholine, Borax= 3x10-5mlins= 30mm/ utsee km2 o.org (mol/wt) = 28mmut so, to kom? Dieedilm b im Shle P

Vom For competitive Inhibitor, vor Knts Vor Vmax [57 to = 4 km, t + 1. a kmts o komory s max mxtc so interupt at ; ax axis is an So, for P-11149, km 20 km = (to )= 0.05 0 Similarly for Acetylcholine, 1 t = 40, km = 6 ) = 0.025 - So, from eq 0 [email protected] 0.06 12 000) = 2 So, q=2 for competitive inlibitur, lo 0 - 1 + 2 1 + 2x1024. Is inbibitor com. 2x10rum KF 2x10-4) So, KI for competitive inhibitor = 2410-4 M. Sinolarly for mixed intisitor, s takim &. & 9 & Kont It's a So, intercept at ax axis will be at which is -40 so, & -40 kosa 0.025, 222, &-? & ko - 21 – 40 - 2 = (40 x 0.028+2) = 2 2X/04 KI KIJ kr vo Vnaris], 1 S horts Vmax km - 2 = Now, &le 1+12 7 TK 17' = (KIJ = 2410-4 1 + 2-410-4 TKI m/mixed type inbibitor will be so, ks] for competitive & same 2x10mm.