Atorvastatin is a cholesterol lowering drug whose mechanism of action is the inh
ID: 64807 • Letter: A
Question
Atorvastatin is a cholesterol lowering drug whose mechanism of action is the inhibition of HMG-CoA reductase. If a patient takes a 10.0 mg dose of atorvastatin, what is the concentration of atorvastatin in the blood? Assume the patient has 5.0 L of blood. The molecular weight of atorvastatin is 559
Atorvastatin is a competitive inhibitor of HMG-CoA reductase with a Ki of 7.5 × 10-3 ?M. What fraction of the enzyme will have a bound substrate under the conditions described above? Assume the concentration of HMG-CoA is 15 ?M, and the KM for HMG-CoA reductase is 1.4 ?M.
After a period of time, all but 0.10% of the atorvastatin in the blood is metabolized. What fraction of the enzyme will have bound substrate under these conditions?
Explanation / Answer
Answer
= 10 * 10-3g
Its concentration can be calculated by using the number of moles.
Therefore, number of moles of the inhibitor = mass / molecular weight
= 10 * 10-3 / 559
= 0.01788 *10-3 moles.
Given volume of solution (blood) = 5 liters.
Therefore, concentration of the inhibitor = number of moles / volume of solution
= 0.01788 *10-3 / 5
= 0.0035778 moles/lit
2. Given, inhibitor is Atorvastatin, I
Substrate is HMG-CoA, S
Enzyme is HMG-CoA reductase, E
The given concentration of substrate is 15
The calculated concentration of inhibitor is 0.0035778 moles/lit
Given Km for the enzyme is 1.4
for the enzyme is 7.5*10-3 microM
There will be an apparent Km value for an enzyme in presence of a competitive inhibitor.
This can be calculated as follows:
= 1.4 (1+ 0.0035778 / 7.5*10-3)
= 1.4 *103 microM
From this apparent Km, the fraction of enzyme bound substrate can be calculated using the formula,
Fes = [s] / [s] + Km
= 15 / (15 + 1.4) *103
= 0.914*10-3 = 0.000914%
Therefore, 0.914*10-3 th portion of the available enzyme is bound to the substrate.
3. After a period of time, all the inhibitor is metabolized except 0.10% . That means only 0.1% inhibitor is available. This alters app. Km and fraction of the enzyme bound substrate.
Therefore, [I] = 0.0035778 * 0.1 /100
= 0.0035778 * 10-3
Km app = 0.0035778 * 10-3 / 7.5*10-3)
= 1.4006 microM
= 0.9146 %
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