Enzyme Inhibition Kinetic 1. What is the advantage of using a substrate concentr
ID: 209179 • Letter: E
Question
Enzyme Inhibition Kinetic
1. What is the advantage of using a substrate concentration near the Km in I50 determinations?
2. Estimate the I50 from the data below. How did you come to this conclusion?
[I] (mM) Activity (A/minute)
0 0.53
3.75 0.38
7.5 0.29
15 0.19
30 0.11
60 0.05
120 0.02
3. The data in the table below were obtained by performing Km determination experiments in the absence of inhibitor and in the presence of two different inhibitors.
Velocity (µmol/min)
Substrate (mM)
No Inhibitor
Inhibitor 1
(1 mM)
Inhibitor 2
(1 mM)
0.25
0.06
0.02
0.02
0.5
0.10
0.05
0.04
1
0.17
0.08
0.07
2
0.25
0.14
0.10
4
0.33
0.22
0.13
8
0.40
0.31
0.16
a. What type of inhibitor is Inhibitor 1? Why?
b. What type of inhibitor is Inhibitor 2? Why?
c. What are the Ki values for Inhibitor 1 and Inhibitor 2?
Velocity (µmol/min)
Substrate (mM)
No Inhibitor
Inhibitor 1
(1 mM)
Inhibitor 2
(1 mM)
0.25
0.06
0.02
0.02
0.5
0.10
0.05
0.04
1
0.17
0.08
0.07
2
0.25
0.14
0.10
4
0.33
0.22
0.13
8
0.40
0.31
0.16
Explanation / Answer
Answer
Enzymes are the biomolecule which increase the rate of conversion of substrate into product the process studying under enzyme kinetics. In this process enzyme bindes to the substrate and formed enzyme-substrate complex then after convert into product on their active site and then release the product.
According to Michaelis-menton equation it forms rectangular hyperbole :
Vo = Vmax [S]/Km + [S]
Km = is the value when half of the enzyme active site is bound with the substrate. At this point the rate of reaction is half the Vmax value i.e. Vmax/2
Vmax is the max rate of reaction.
I50 is the concentration of the inhibitors which binds to the 50% of the enzyme under set assay condition.
Then if we take substrate concentration near Km then we can find inhibitors competition to half of the active site and we can calculate the appropriate inhibitors constant using I50.
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