Question

[Alcohol] mMAD Activity AD Activity AD Activity mM/min Agent A Agent B mM/minmM/min 15.0 2.5 5.0 7.5 45.0 20.0 0.5 63.0 10.0 32.0 71.0 2.0 11.7 48.0 4.0 79.0 13.4 66.0 8.0 14.3 83.0 84.0 32.0 90.0 15.5 90.0 1. Using the data provided draw a Michaelis-Menten and Lineweaver-Burk Plot for each of the enzyme activities? When you produce the 2 graphs make sure you have plotted 3 curves for the Michaelis-Menten plot on one graph and the three lines for the Lineweaver-Burk plot on one graph. The point is to compare the data to one another to complete the analvsis 2. Determine K and V.. for each of the three experiments condition'? 3. Agent A and B are inhibitors of the enzyme. Classify each type of inhibitor based on your analysis of the data set. Remember this is representative of real data so support your justification based on the best (maybe not exact) fit to the definitions of what each inhibitor should demonstrate.

Answer 1

Michaelis-Menten Plot 100 90 80 70 AD activity mM/mi Activity, mM/min 501 -AD activity +Agent AD activity +Agent 40 30 20 10 0 30 35 20 25 10 5 0

for, michaela-menten pit- to O. .oS 0.06(+ 0, y 0.13y 0:023 o, o 31ar O,ols80.1 0.5 0.2 0.01265 0046, 015TS O180 0.35. o, o 639 o, o l 19 o, 125 ㅡ 0.0.3 tr o.on, au o.oosい( + u.o " f

Lineweaver-Burk Plot 0.45 y- 0.0334x + 0.0663 0.4 0.35 0.3 AD activity mM/min 0.25 --AD activity +Agent mM/mirn 1/Activity y- 0.0189x +0.011 -AD activity +Agent B mM/min 0.2 Linear (AD activity +Agent mM/min) Linear (AD activity +Agent B mM/min) 0.15 Linear (AD activity mM/min) 0.1 y 0.0056x+0.011 0.05 0 10 12 1/[Alcohal]

0,033 Vett erent 031892 8-0, 013 gt to,o1IJ inhibitr. em.