Quinidine sulphate (F= 0.8, k=0.1/hr, and Vd = 180L) is a class IAantiarrhythmic drug. Barbiturates are enzyme inducers that canenhance the elimination of quinidine by 50% (k=0.15/hr). What will bethe average steady state concentration in a patient who is taking 300mgq 8hr quinidine alone and in a different patient who is taking the samedose of quinidine with barbiturates?
Question
Quinidine sulphate (F= 0.8, k=0.1/hr, and Vd = 180L) is a class IAantiarrhythmic drug. Barbiturates are enzyme inducers that canenhance the elimination of quinidine by 50% (k=0.15/hr). What will bethe average steady state concentration in a patient who is taking 300mgq 8hr quinidine alone and in a different patient who is taking the samedose of quinidine with barbiturates?
Solution
To solve this problem, we need to use the formula for the average steady state concentration (Css) of a drug, which is:
Css = (F * Dose) / (Cl * Tau)
Where:
- F is the bioavailability of the drug
- Dose is the amount of drug administered
- Cl is the clearance of the drug, which is calculated as Vd * k
- Tau is the dosing interval
First, let's calculate the Css for a patient taking quinidine alone:
- Calculate the clearance (Cl) of the drug: Cl = Vd * k = 180L * 0.1/hr = 18 L/hr
- Calculate the Css: Css = (0.8 * 300mg) / (18 L/hr * 8hr) = 1.67 mg/L
Next, let's calculate the Css for a patient taking quinidine with barbiturates:
- Calculate the new clearance (Cl) of the drug: Cl = Vd * k = 180L * 0.15/hr = 27 L/hr
- Calculate the Css: Css = (0.8 * 300mg) / (27 L/hr * 8hr) = 1.11 mg/L
So, the average steady state concentration in a patient who is taking 300mg q 8hr quinidine alone is 1.67 mg/L, and in a different patient who is taking the same dose of quinidine with barbiturates is 1.11 mg/L.
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