In general, Please explan how to get > Condition No inhibitor max Thus, the y-intercept is equal to 1/Vmax and the x-intercept is equal to -1/Km. (a) 5 mM inhibitor (2) 5 mM inhibitor (3) 1 YOL ט K 1 1 10 X+ V [S] V max Vmax (umol/mL-sec) 51.2 Vmax,app 49.9 22.8 Km (MM) 3.2 Km,app 10.6 3.1 mst
Catalysis and Enzymatic Reactions
Catalysis is the kind of chemical reaction in which the rate (speed) of a reaction is enhanced by the catalyst which is not consumed during the process of reaction and afterward it is removed when the catalyst is not used to make up the impurity in the product. The enzymatic reaction is the reaction that is catalyzed via enzymes.
Lock And Key Model
The lock-and-key model is used to describe the catalytic enzyme activity, based on the interaction between enzyme and substrate. This model considers the lock as an enzyme and the key as a substrate to explain this model. The concept of how a unique distinct key only can have the access to open a particular lock resembles how the specific substrate can only fit into the particular active site of the enzyme. This is significant in understanding the intermolecular interaction between proteins and plays a vital role in drug interaction.
![4. Graphing the results from kinetics experiments with enzyme inhibitors
The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the
presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [Er] is the same in
each experiment.
[S]
(mM)
1
2
4
8
12
1
2
4
8
12
[S]
(mm)
a. Determine Vmax and Km for the enzyme.
b. Determine the type of inhibition and the Kıfor each inhibitor.
1/[S]
M-1
Plot these data as double-reciprocal Lineweaver-Burk plots and use your graph to answers a.
and b.
1000
500
250
125
83.3
Answer: The data may be analyzed using double-reciprocal variables. For each [S] and corresponding u, we
will calculate 1/[S] and 1/v.
sed to get mo
Helmien
(1)
v (μmol/mL sec
12
20
29
35
40
1/v (1)
v (1)
μmol/mL se mL-sec/umol
C
12
20
29
35
40
Please explain
how to get
K
1/v
20 × 104
10 × 104
-400
8.33x104
5.00x104
3.45x104
2.86x104
2.50x104
Condition
No inhibitor
0
5 mM inhibitor (2)
5 mM inhibitor (3)
Plots of 1/v vs. 1/[S] indicate straight lines given by
(1) 1/v = 63.2(1/[S] ) + 1.95 x 104
(2) 1/v = 211.8(1/[S]) +2.00 x 104
(3) 1/ 137.2(1/[S)) + 4.38 x 104
In general,
v (2)
μmol/mL sec
400
(2)
v (µmol/mL sec
4.3
8
14
1/[S]
1- K
4.3
8
14
21
26
800
26
m x
1 1
+
V [S] V
max
max
Thus, the y-intercept is equal to 1/Vmax and the x-intercept is equal to -1/Km.
(a)
Vmax,app
49.9
22.8
1/v (2)
mL sec/µmol
Vmax (umol/mL sec)
51.2
2.33x105
1.25x105
7.14x104
4.76x104
3.85x104
1200
(3)
v (µmol/mL sec
5.5
9
13
16
18
v (3)
1/v (3)
μmol/mL se mL-sec/μmol
Km (mm)
3.2
Km,app
10.6
3.1
inhibitor 3
inhibitor 2
no inhibitor
с
5.5
9
13
16
18
1.82×105
1.11×105
7.69x104
6.25x104
5.56x104
(b) Inhibitor (2) increases the apparent Km of the enzyme without affecting Vmax. This is characteristic of a
competitive inhibitor. In this case Ki is calculated as follows](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeadb183-1175-48ee-be1d-be4cbd36e1e9%2Fc8dd7a13-bb09-4c99-8327-4ddfd8db2b53%2Ff9zhrku_processed.jpeg&w=3840&q=75)
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