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Name: ___________________________
CHEMISTRY 333 HOMEWORK # 2
DUE FRIDAY OCTOBER 23, 2009 (NO LATER THAN 5 PM!) SHOW ALL OF YOUR WORK AND HAND IN ALL GRAPHS
(35 points) 1. (20 points) The kinetics of an enzyme were analyzed in both the absence and presence of Inhibitor A and Inhibitor B. Given the following data, calculate or construct the following for A and B on separate graphs.
a) Plot the data as a Michaelis-Menten saturation curve. b) Estimate the Km and Vmax from these curves both in the presence and
absence of inhibitors. c) Plot the data in the Lineweaver-Burk format
*Make sure to label the both the inhibitor line and the no inhibitor line d) Mathematically determine the Km and Vmax. e) What types of inhibitors are A and B? How can you tell? f) On each graph draw a line that would indicate an increase in the
concentration of inhibitor. *Make sure to turn in all your plots, corresponding equations and calculations.
V (mmol/min) [S] (mM) No Inhibitor B (0.1mM)
.2 5.0 2.0
.4 7.5 3.0
.8 10.0 4.0 1.0 10.7 4.3 2.0 12.5 5.0 4.0 13.6 5.5
2. (15 points) Draw the following disaccharides in Haworth projections. Label the anomeric carbon with a * and circle any reducing ends.
a) galactose β(1,4) mannose b) glucose β,α(1,1) galactose
c) fructose α(2,5) ribose
V (mmol/min) [S] (mM) No Inhibitor A (5mM)
.2 5.0 3.0
.4 7.5 5.0
.8 10.0 7.5 1.0 10.7 8.3 2.0 12.5 10.7 4.0 13.6 12.5
V (mmoVmin
t ' ,
Vii
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16
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V (mmoUmin)
lsl (mM) No Inhibitor A (5mM) 1/ ts l LIV (no I) Llv (+A)0.2 5 a
J 5 o.2 0.3333333330.4 7.5 5 2.5 0.133333333 0.20.8 l0 7.5 L.25 0.1 0.1333333331 t0.7 8.3 1 0.0934579440. 12048 L9282 12.5 10.7 0.5 0.08 0.0934579444 13.6 12.s 0.25 o.o735294L2 0.08
I
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i
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V (mmoUmin)
lSl (mM) No Inhibitor B (0.1mM)
0 0 00.2 5 20.4 7.5 J
0.8 l0 4I r0.7 4.32 12.5 54 13.6 5.5
M \ kt**'QL -N'-l--*I^
1_.522.s33.544.5
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V (mmoVmin)
tSl (mM) No Inhibitor B (0.1mM) 1/ rs l LIV (no I) Llv (+B)0.2 5 2 5 0.2 0.50.4 7.5 a
J 2.5 0.1333333330.3333333330.8 l0 4 t .25 0.1 0.25t r0.7 4.3 1 0.093457944 0.232558142 12.5 5 0.5 0.08 0.24 13.6 5.5 0.25 0.0735294L2 0.181818182
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