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Pharmacokinetics Case Study: Calculating Clearance, Volume of Distribution, and Half-Life, Assignments of Health sciences

University of Florida (UF)Health sciences

A case study for calculating the clearance, volume of distribution, and half-life of two different drugs using given patient information and drug concentrations. The study involves the use of creatinine clearance, elimination rate constant, and the concept of steady state. It also includes assumptions about the drug distribution and elimination.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

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koofers-user-fxo 🇺🇸

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Case Study 5
1. A 500 mg IV dose of getamicin is given as a short-term IV infusion to an 80kg,
52 year old, 5’5” male patient. He has a serum creatinine of 0.8mg/dL. The
clearance of this drug is equivalent to creatinine clearance.
A. Calculate the clearance.
IBW=50+2.3*5=61.5
Is he obese? TBW/IBW*100=80kg/61.5kg*100%~130%. Yes, he is clinically
obese.
Therefore use ABW in CLcr calculation.
ABW=IBW+0.4*(TBW-IBW)=68.9kg
Clcr=(140-52)*68.9/(72*0.8)=105mL/min
B. Assuming this drug is only distibutes into extracellular fluid (assume the
volume of an average person). What is the half-life?
Volume of distribution(Vd)=18L
Half-life=0.693/(Cl/Vd)=0.693/(105mL/min*60/100)/18L=1.98hours~2hours
2. Drug X has a narrow therapeutic window (~5-40mg/L). Two plasma samples are
drawn for monitoring; one at 2 hours after the first iv bolus administration and
one at 12 hours. The concentrations are 33mg/L and 8mg/L, respectively. This
drug displays a one compartment body model and 1g was administered.
A. Calculate the following: Vd, Cl, and half-life.
Ke=Ln(C1/C2)/(t2-t1)=Ln(33mg/L/8mg/L)/(12hr-2hr)=0.1417hr-1
After Ke is know we can back calculate the concentration to find Cmax and
calculate the Vd.
C=C0*e^(-ke*t) C0=C/e^(-ke*t) C0=33mg/L/e^(-0.14*2)=43.66mg/L
Vd=Dose/C0=1000mg/43.66mg/L=22.9~23L
Cl=Ke*Vd=0.1417*22.9L=3.24L/hr
Half-life=0.693/ke=0.693/0.1417hr-1=4.89hr
B. Assumuing linear kinetics what would the effect be on clearance, volume of
distribution, half-life, AUC, and Cmax if the dose were decreased by 25%.
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Case Study 5

  1. A 500 mg IV dose of getamicin is given as a short-term IV infusion to an 80kg, 52 year old, 5’5” male patient. He has a serum creatinine of 0.8mg/dL. The clearance of this drug is equivalent to creatinine clearance.

A. Calculate the clearance.

IBW=50+2.35=61. Is he obese? TBW/IBW100=80kg/61.5kg*100%~130%. Yes, he is clinically obese. Therefore use ABW in CLcr calculation.

ABW=IBW+0.4*(TBW-IBW)=68.9kg

Clcr=(140-52)68.9/(720.8)=105mL/min

B. Assuming this drug is only distibutes into extracellular fluid (assume the volume of an average person). What is the half-life?

Volume of distribution(Vd)=18L Half-life=0.693/(Cl/Vd)=0.693/(105mL/min*60/100)/18L=1.98hours~2hours

  1. Drug X has a narrow therapeutic window (~5-40mg/L). Two plasma samples are drawn for monitoring; one at 2 hours after the first iv bolus administration and one at 12 hours. The concentrations are 33mg/L and 8mg/L, respectively. This drug displays a one compartment body model and 1g was administered.

A. Calculate the following: Vd, Cl, and half-life.

Ke=Ln(C1/C2)/(t2-t1)=Ln(33mg/L/8mg/L)/(12hr-2hr)=0.1417hr-

After Ke is know we can back calculate the concentration to find Cmax and calculate the Vd.

C=C0e^(-ket) C0=C/e^(-ket) C0=33mg/L/e^(-0.142)=43.66mg/L

Vd=Dose/C0=1000mg/43.66mg/L=22.9~23L

Cl=KeVd=0.141722.9L=3.24L/hr

Half-life=0.693/ke=0.693/0.1417hr -1=4.89hr

B. Assumuing linear kinetics what would the effect be on clearance, volume of distribution, half-life, AUC, and Cmax if the dose were decreased by 25%.

Cl↔ Vd↔ Half-life↔ AUC↓by 25% Cmax↓by 25%

  1. Patient AB and Patient CD are on a multiple IV bolus regimen of the same drug and steady state has been reached. This drug is cleared only by glomerular filtration and is not bound to plasma proteins. Both patients have a CLcr of 130mL/min.

A. Patient AB has a longer half-life. Explain.

Because both patients have the same clearance value patient AB must have a larger volume of distribution.

B. Would you expect the Cmax to be higher for patient AB or CD.

CD would have a higher Cmax because at steady state

and this patient has a smaller Vd (represented by V in this equation).

C. Which patient would have more fluctuation in plasma concentration between each dose?

Patient CD because this patient has a shorter half-life.

D. Which patient has a higher average steady state concentration.

The average concentration is the same for both patients at steady state. Based on the formula Css=Dose/(CL*Tau), where tau is the dosing interval.

True or False State if the following are True or False

  1. If the volume of distribution increases the clearance can remain the same. (T)
  2. At steady state (equal dose, equal interval multiple IV) accumulation stops because the amount of drug eliminated during the dosing interval is more than the dose given at each dose time. (F)