T16: Mixed types from T12-T18
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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193.11 mg of a drug (elimination rate constant of 0.223 h-1) is administered as an IV bolus to a patient. The weight of the patient is 84 Kg. If the volume of distribution of the drug is 0.79 L/Kg, then calculate the AUC.
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lb equals 13.049 mg × h/L kg
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We know that:
`AUC=(dose)/(cl earance)=(dose)/(K × V)`.
However, the total volume of distribution (V) will be equal to the volume of distribution per Kg body weight times the weight of the patient in Kg. Therefore, `V=0.79 \quad L/(Kg) × 84 \quad Kg = 66.36 \quad L`.
`∴ AUC=(193.11 \quad mg)/(0.223 h^{-1} × 66.36 \quad L)`.
`∴ AUC=13.05 \quad (mg × h)/L`. Ans.