you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?

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Asked Oct 11, 2019
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you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?

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Expert Answer

Step 1

The value for Ki′ cannot be 0/2 ×10-5, so we are assuming it as 0.2×10-5­M. We are also assuming that we have to calculate the apparent Km in the presence of 1.0x10-6 M inhibitor.

Step 2

Non-competitive inhibitors are those molecules that bind to a site other than the active site on the enzyme. The binding results in the change in the conformation of the active site of the enzyme, which prevents the enzyme to react to the substrate.

Step 3

The apparent Km value for a mixed non-competitive inhibitor can be calcula...

Кm(1+1ук)
(1+TVK;)
where
app
кар.
I-Inhi bitor
Kapp-Apparent km
app_1x10-(1+1.0x10-/0.4x10-6)
(1+1.0x10/0.2x10-5)
1+1.0x10- 0.2x 10-5
m
0.4
1+1.0x10-6
0.2x10-5
0.4
=0.5x10M
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Кm(1+1ук) (1+TVK;) where app кар. I-Inhi bitor Kapp-Apparent km app_1x10-(1+1.0x10-/0.4x10-6) (1+1.0x10/0.2x10-5) 1+1.0x10- 0.2x 10-5 m 0.4 1+1.0x10-6 0.2x10-5 0.4 =0.5x10M

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