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Pharmacokinetics lecture 5 Contents ...

Pharmacokinetics lecture 5 Contents . Single i.v. bolus into a one compartment system Area Under the Curve. Single i.v. bolus dose into one compartment. D. Dose = 400 mg V = 100 Litres K = 0.3 h -1 Initial conc (C 0 ) = 4 mg/L. V. K. D = 400 mg V = 100 L K = 0.3 h -1.

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Pharmacokinetics lecture 5 Contents ...

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  1. Pharmacokinetics lecture 5Contents ... • Single i.v. bolus into a one • compartment system • Area Under the Curve

  2. Single i.v. bolus dose into one compartment D Dose = 400 mg V = 100 Litres K = 0.3 h-1 Initial conc (C0) = 4 mg/L V K

  3. D = 400 mg V = 100 L K = 0.3 h-1 Mass = 4 mg/L x 100 L = 400 mg Rate of elim = 400 mg x 0.3 h-1 = 120 mg/h 4 3 2 1 0 Mass = 2 mg/L x 100 L = 200 mg Rate of elim = 200 mg x 0.3 h-1 = 60 mg/h Conc (mg/L) Mass = 1 mg/L x 100 L = 100 mg Rate of elim = 100 mg x 0.3 h-1 = 30 mg/h Exponential curve 0 2 4 6 8 10 Time (h)

  4. Half-life and K 4 3 2 1 0 C0 = 4 mg/L Inefficientlyeliminated Efficiently eliminated Conc (mg/L) Low K / Long t ½ t ½ = 2.3 h t ½ = 5.4 h High K / Short t ½ 0 2 4 6 8 10 Time (h)

  5. Predicting concentration at a given time point C0 4 3 2 1 0 Ct = C0.e-Kt Conc (mg/L) Ct t 0 2 4 6 8 10 Time (h)

  6. Predicting concentration at a given time point D = 10 mg V = 50 L K = 0.05 h-1 What conc. after 12 hours? C0 = D/V = 10mg/50L = 0.2mg/L = 200µg/L Ct = C0 .e-Kt = 200µg/L . e-0.05h-1 x 12h = 200µg/L . e-0.6 = 200µg/L . 0.55 = 110 µg/L

  7. Predicting the time to reach a given concentration C0 = 5 mg/L K = 0.02h-1 How long until C = 1 mg/L? Ct = C0 .e-Kt Ct / C0 = e-Kt 1/5 = e-Kt 0.2 = e-Kt (Take natural logs of both sides) Ln(0.2) = -Kt -1.609 = -Kt (Drop minus from both sides) 1.609 = 0.02h-1 x t t = 1.609 / 0.02h-1 = 80.5 hours

  8. Area Under the Curve (AUC) 3 2 1 0 • Area under the concentration • versus time curve • ‘Area Under the Curve’ • AUC Conc (mg/L) 0 2 4 6 8 10 Time (h)

  9. Units of AUC 3 2 1 0 Area is height x width Units are mg/L x h mg.L-1.h Usually rearranged as: mg.h .L-1 In general: Mass.Time.Volume-1 Conc (mg/L) 0 2 4 6 8 10 Time (h)

  10. AUC between time limits 3 2 1 0 ‘AUC’ assumed to mean AUC 0-¥ (The complete area under the whole curve) AUC2-4h Conc (mg/L) 0 2 4 6 8 10 Time (h)

  11. Relationship between AUC, Dose and Clearance Total amount of drug entering the body. (More drug = greater AUC) AUC = F.D Cl Efficiency of removal. (Greater clearance = lower AUC)

  12. Calculation of AUC following single i.v. bolus AUC = F.D Cl e.g. D = 5 mg & Cl = 4 L/h AUC = 1.0 x 5mg 4 L/h = 1.25 mg.h.L-1 Conc (mg/L) 0 2 4 6 8 10 Time (h)

  13. Example calculation 1mg of drug is administered (i.v) at time zero. Vol Dis = 50 L. Elim rate constant = 0.015h-1 What will blood concentration be 2 days later? Tip: Calculate C0 first, then Ct.

  14. Answer C0 = D/V = 1mg / 50L = 0.02 mg/L = 20 µg/L Ct = C0.e-Kt = 20µg/L x e-0.015h-1 x 2days = 20µg/L x e-0.015h-1 x 48h = 20µg/L x e-0.72 = 20µg/L x 0.487 = 9.7µg/L Mixed units £2 calculator is no good. Need a good one!

  15. Terms with which you should be familiar ... Area Under the Curve (AUC)

  16. What you should beable to do • For a single i.v. bolus injection into a one compartment system ... • Describe why the graph of concentration versus time adopts its distinctive shape • Calculate remaining concentration from initial concentration, elimination rate constant and time elapsed (etc) • Apply appropriate units to an AUC • Calculate AUC following single i.v. injection

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