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Solutions to homework problems related to pharmacokinetics, including calculating cmax, cmin, accumulation factors, loading doses, maintenance doses, and steady state reach times.
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Homework 5
A. The Cmax B. The accumulation factor C. Elimination
A=Dose/Vd B=1/(1-e^(-ketau)) C=e^(-ket)
A. Calculate the accumulation factor at steady state (1pt).
ke= 0.693/T1/2=0.693/4= 0.173/hr Rss=1/(1-e^(-ke12)=1/(1-e^(-0.17312))=1.
B. Calculate the average concentration for a dose of 500 mg (1pt).
Cpss=D/(CLτ)=D/(keVd*τ)=500mg/(0.173hr -1^ 18L12hr)=13.36mg/L
C. Calculate the maximum and minimum plasma concentrations (Cmax , C min ) in the body at steady state if dose of 500mg (2pts).
Cmax=D/VdRss=500/181.14=24.37 mg/L Cmin=Cmaxe^(-keτ)=24.37mg/Lexp(-0.17312)=3.06 mg/L
A. Calculate the loading dose (1pt).
Loading dose=ConcentrationVd=15mg/L30L=450mg
B. Calculate the maintance dose (infusion rate) (1pt).
R0=ConcentrationCL=15mg/L7L/hr=105mg/hr
C. How long until steady state is reached? (0.5pt)
Steady state=5 half-lives=5(0.693/(Cl/Vd))=5(0.693/(7L/hr/30L))=14.85hr
D. The patient remains on the constant infusion for 5 days and it is then stopped. Predict the concentration 5 hours after the infusion is stopped (1pt).
Once the infusion has stopped, the concentration at any given time can be calculated using the iv bolus equation.
Css=R0/CL=105mg/hr/7L/hr=15mg/L C=C0e^(-ket)=15mg/Le(-7L/hr/30L5hr)=4.67mg/L
E. It is decided that the infusion should be restarted when the concentration equals 2mg/L. How long after the stop of the infusion should it be restarted (1pt)?
C=C0e^(-ket)= Ln(C/C0)=-ke*t Ln(C/C0)/-ke=t Ln(2mg/L/15mg/L)/-(7L/hr/30L)=8.63hr