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Sample V_max and K_m calculations
Protease inhibitors
Sample lab data & graphs

Sample V_max and K_m calculations | Go to Lockey Program in Flash | in HTML

A Run the experiment at different concentrations of acetylcholine (Ach). Here are some sample data (these are not actual data from the experiment):

Time (min)	Amount of HAc produced
	at 0.5 mM Ach	at 1.0 mM Ach	at 2.0 mM Ach	at 3.0 mM Ach
0.0	0.0	0.0	0.0	0.0
0.1	0.2	0.2	0.3	0.3
0.2	0.4	0.5	0.5	0.6
0.3	0.5	0.7	0.8	0.8
0.4	0.5	0.8	1.1	1.1
0.5	0.5	0.9	1.3	1.4
0.6		1.0	1.5	1.6
0.7		1.0	1.7	1.9
0.8			1.9	2.2
0.9			2.0	2.4
1.0			2.0	2.6
1.1			2.0	2.8
1.2				2.9
1.3				3.0
1.4				3.0
1.5				3.0

B. For each concentration, plot the acetic acid (HAc) produced versus time.
C. Draw the tangent (thick red line) to the fastest part of the reaction, out to 1 minute. To get the amount of HAc produced per minute (thin magenta line).
	At 0.5 mM Ach, the rate of HAc production was 2.25mM/min.
D. Continue steps A and B for each concentration of Ach for which you have data.
	How much HAc is produced per minute? _____  (The answer is in E, below.)
	mM HAc/min = _____ (The answer is in E, below.)
	HAc/min = _____ (The answer is in E, below.)
E. You have now collected the following data:
	[Ach], mM mM HAc/min 0.0  0.00 0.5 2.25 1.0 2.40 2.0 2.70 3.0 3.00
F. Plot these data to determine the Km and Vmax for this enzyme.
	The Vmax is 3.0 mM HAc/min. The Km is 0.3 mM Ach.
G. Assume you ran this reaction with different inhibitors and got the following data:
	Inhibitor Km Vmax #1 0.56 1.5 #2 3.33 3.0
	A competitive inhibitor shows the max. rate at a higher concentration (Km) than normal because it binds the enzyme causing the enzyme "wastes" time bound to the inhibitor.
	A noncompetitive inhibitor gives a slower rate (Vmax) because it binds the enzyme changing the active site of the enzyme.
	Which inhibitor (#1 or #2) is a competitive inhibitor? A noncompetitive inhibitor?