T15: K from Dose, AUC and V
50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC)
50.1.3.1) AUC basic concepts Normal 1
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Determine the elimination rate constant of a drug from the provided information. The volume of distribution is 42.12 L after administering 63.6 mg of the drug. The area under the plasma drug concentration over time (AUC) was found to be 55.93 mg × h/L.
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lb equals 0.03 h-1 kg
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We know that, `AUC=(dose)/(K × V)`.
`⇒ K=(dose)/(V × AUC)`
Plugging the values in this equation we get:
` K = (63.6 \quad mg)/(42.12 \quad L × 55.93 (mg × h)/L)`
` ∴ K=0.027 \quad h^{-1}`. Ans.