1.The kinetics of an enzyme are measured as a function of substrate concentration in the presence or absence of an inhibitor. The following results are obtained: (nm/mn) Unitsi Km > mm) Substrate Concentration units: Vmas > Velocity (nanomoles/min) No inhibitor in millimoles (mM) With inhibitor 55.8 x 10 54.9 x 10 51 x 10 47 x 10 40 x 10' 28 x 10 18.7 x 10 11.2 x 10 2.7 x 10 55.7 x 10 53.3 x 103 45 x 10 37.3 x 10 28 x 103 16 x 10 9.3 x 10 100 10 1 0.5 0.2 0.1 0.05 Not tested 0.01 Not tested o.2 05 S. 2. A. What are the Vmax and Km values for the enzyme-catalyzed reaction without inhibitor? Please give units and explain your reasoning. (See note.) Vmons > Km B. (i). What are the apparent Vmax and Km Values plus inhibitor? (ii). Is this a case of competitive or noncompetitive inhibition? How did you know?

Biochemistry
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ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
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Chapter21: Photosynthesis
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Please complete Boothe parts of part B thoroughly with detailed explanations since only part A was done already. Thank you
ENZYME KINETICS; PATHWAYS & FEEDBACK INHIBITION
3-1. The kinetics of an enzyme are measured as a function of substrate concentration in the presence or absence of
an inhibitor. The following results are obtained:
UnitsiKm> (mm)
units: Vmon > (nm/mn)
Velocity (nanomoles/min)
No inhibitor
55.8 x 103
54.9 x 10
51 x 10
47 x 10
40 x 103
28 x 10
18.7 x 10
11.2 x 103
2.7 x 10°
Substrate Concentration
in millimoles (mM)
With inhibitor
55.7 x 10
53.3 x 10
45 x 10
37.3 x 10
28 x 103
16 x 10
9.3 x 10
100
10
1
0.5
0.2
0.1
0.05
Not tested
0.01
Not tested
0.2 0°5
2.
A. What are the Vmax and Km Values for the enzyme-catalyzed reaction without inhibitor? Please give
units and explain your reasoning. (See note.)
Vmons >
Km →
B. (i). What are the apparent Vmax and Km values plus inhibitor?
(ii). Is this a case of competitive or noncompetitive inhibition? How did you know?
Note: You do not need a calculator or graph paper to figure any of this out. However you may want to plot the results on
graph paper to get a better feel for what is going on. If you do it that way, try to do it again without the graph, using the
numbers shown.
Transcribed Image Text:ENZYME KINETICS; PATHWAYS & FEEDBACK INHIBITION 3-1. The kinetics of an enzyme are measured as a function of substrate concentration in the presence or absence of an inhibitor. The following results are obtained: UnitsiKm> (mm) units: Vmon > (nm/mn) Velocity (nanomoles/min) No inhibitor 55.8 x 103 54.9 x 10 51 x 10 47 x 10 40 x 103 28 x 10 18.7 x 10 11.2 x 103 2.7 x 10° Substrate Concentration in millimoles (mM) With inhibitor 55.7 x 10 53.3 x 10 45 x 10 37.3 x 10 28 x 103 16 x 10 9.3 x 10 100 10 1 0.5 0.2 0.1 0.05 Not tested 0.01 Not tested 0.2 0°5 2. A. What are the Vmax and Km Values for the enzyme-catalyzed reaction without inhibitor? Please give units and explain your reasoning. (See note.) Vmons > Km → B. (i). What are the apparent Vmax and Km values plus inhibitor? (ii). Is this a case of competitive or noncompetitive inhibition? How did you know? Note: You do not need a calculator or graph paper to figure any of this out. However you may want to plot the results on graph paper to get a better feel for what is going on. If you do it that way, try to do it again without the graph, using the numbers shown.
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