Question
Asked Nov 27, 2019
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The following table indicates the rates at which a substrate reacts as catalyzed by an enzyme that follows the Michaelis-Menten mechanism: (1) in the absence of inhibitor; (2) and (3) in the presence of 10 mM concentration, respectively, of either of two inhibitors. Assume [E]T is the same for all reactions.

(a) Determine KM and Vmax for the enzyme. For each inhibitor determine the type of inhibition and KI and/or K’I. What additional information would be required to calculate the turnover number of the enzyme? (b) For [S] = 5 mM, what fraction of the enzyme molecules have a bound substrate in the absence of inhibitor, in the presence of 10 mM inhibitor of type (2), and in the presence of 10 mM inhibitor of type (3)?

(1) Vo
(uM.S
(2) Vo
(3) Vo
(mM)
0.77
1
2.5
1.17
2
4.0
2.10
1.25
5
6.3
4.00
2.00
2.50
10
7.6
5.7
20
9.0
7.2
2.86
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(1) Vo (uM.S (2) Vo (3) Vo (mM) 0.77 1 2.5 1.17 2 4.0 2.10 1.25 5 6.3 4.00 2.00 2.50 10 7.6 5.7 20 9.0 7.2 2.86

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Step 1

In order to obtain a linear graph between th

In order to obtain a linear graph between the rate and substrate concentration, a Lineweaver -Burk graph is made in which the inverse of rate is plotted against inverse of substrate concentration. The Lineweaver-Burk relation is shown below. In the given relation, V is the reaction rate, VM is the maximum reaction velocity and KM is the Michaelis-Menten constant.  

e rate and substrate concentration, a Lineweaver -Burk graph is made in which the inverse of rate is plotted against inverse of substrate concentration. The Lineweaver-Burk relation is shown below. In the given relation, V is the reaction rate, VM is the maximum reaction velocity and KM is the Michaelis-Menten constant.  

Км
1
1
и и [5]
V
Мас
м
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Км 1 1 и и [5] V Мас м

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Step 2

The graphs between substrate concentration and rate of reaction with and without inhibitors are shown below:

1/IS
[S](mM)
(HM)
Rate 2
Rate 1
Rate 3
1/Rate 3
(uM/s)
1/Rate 1
(uM/s)
1.17 0.854701
1/Rate 2
(uM/s)
2.5
1
0.001
0.4
0.77
1.298701
2.1
0.8
2
0.0005
4
0.25
0.47619
1.25
6.3 0.158730159
0.25
5
0.0002
4
2.
0.5
10
0.0001
7.6 0.131578947
5.7
0.175439
2.5
0.4
9 0.111111111
20
0.00005
7.2
0.138889
2.86
0.34965
1/Rate 1
1/Rate 2
0.45
0.9
0.4
y 301.67x +0.0987
y 754.13x +0.1
0.8
0.35
0.7
0.3
0.6
0.25
0.5
0.2
0.4
0.3
0.15
0.2
0.1
0.1
0.05
0
0
0
0.0006
0.0002
0.0004
0.0008
0.001
0.0012
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
1/Rate 3
1.4
y= 998.86x +0.3001
1.2
0.8
0.6
0.4
0.2
0
0.0002
0.0008
0
0.0004
0.0006
0001
0.0012
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1/IS [S](mM) (HM) Rate 2 Rate 1 Rate 3 1/Rate 3 (uM/s) 1/Rate 1 (uM/s) 1.17 0.854701 1/Rate 2 (uM/s) 2.5 1 0.001 0.4 0.77 1.298701 2.1 0.8 2 0.0005 4 0.25 0.47619 1.25 6.3 0.158730159 0.25 5 0.0002 4 2. 0.5 10 0.0001 7.6 0.131578947 5.7 0.175439 2.5 0.4 9 0.111111111 20 0.00005 7.2 0.138889 2.86 0.34965 1/Rate 1 1/Rate 2 0.45 0.9 0.4 y 301.67x +0.0987 y 754.13x +0.1 0.8 0.35 0.7 0.3 0.6 0.25 0.5 0.2 0.4 0.3 0.15 0.2 0.1 0.1 0.05 0 0 0 0.0006 0.0002 0.0004 0.0008 0.001 0.0012 0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 1/Rate 3 1.4 y= 998.86x +0.3001 1.2 0.8 0.6 0.4 0.2 0 0.0002 0.0008 0 0.0004 0.0006 0001 0.0012

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Step 3

According to the Lineweaver-Burk plot, the intercept is equal to 1/Vmax and slope is KM/Vmax. Therefore, from the equation on graph, the value of K...

1
0.1
V
max
=10 M/s
max
KM
= 754.13
max
K
754.13 x10 M/s
= 7541.3 LM
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1 0.1 V max =10 M/s max KM = 754.13 max K 754.13 x10 M/s = 7541.3 LM

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