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Q1:
Drug A is a small and hydrophilic compound that distributes to
extracellular fluids only. It has a volume of distribution of 5.6 L
in a healthy 70-kg person . The free fraction of the drug in
plasma is 0.1. What is the free fraction of the drug in the
tissues?
fu = 0.1
VT = 13 L
0.1
V  3  13 x
 5.6 L
fuT
fuT = 0.5
1
Drug A (100 mg) is intravenously injected to a patient. The AUC
of plasma drug concentration vs. time curve is 20 mg/ml·h.
Drug A is eliminated via hepatic metabolism and renal excretion
only, and the fraction of the unchanged drug excreted in urine is
0.3. What is the total body clearance (CLT), hepatic clearance
(CLH) and renal clearance (CLR) of drug A?
CLT = Dose/AUC = 100 X 103 mg / (20 mg/ml·h) = 5 L/h
CLR = fe · CL = 0.3 X 5 L/h = 1.5 L/h
CLH = CLT - CLR = 3.5 L/h
EH= CLH/QH = (3.5 L/h)/(1.35 L/min) = (3.5 L/h)/(81 L/h)
= 0.043
In an average 70-kg adult, Drug B has a hepatic blood clearance
of 1.2 L/min and is 95% bound to plasma protein. What is the
new hepatic blood clearance of drug B, (a) if the plasma protein
binding of the drug is decreased to 90%? (b) if the hepatic
blood flow is decreased to 1.2 L/min?
a. CLH = 1.2 L/min, fu = 0.05, Q = 1.35 L/min
EH= CLH/QH = (1.2 L/min)/(1.35 L/min) = 0.89
Drug B is a high extraction ratio drug.
The hepatic clearance of drug A is insensitive to changes
in plasma protein binding.
fu’=0.1
CLH’≈ 1.2 L/min
b. CLH’ = QH’  EH = 1.2 L/min x 0.89 = 1.07 L/min
In an average 70-kg adult, Drug C has a hepatic blood clearance
of 10 ml/min and is 95% bound to the plasma protein. What is
the new hepatic blood clearance of drug C (a) if the plasma
protein binding of the drug is decreased to 90% ? (b) if the
hepatic blood flow is decreased to 1.2 L/min?
a. CL = 10 ml/min, f = 0.05, Q = 1.35 L/min
H
u
EH= CLH/Q = (10 ml/min)/(1.35 L/min) = 0.007
Drug C is a low extraction ratio drug.
CLint ≈ CLH/fu = (10 ml/min)/0.05 = 200 ml/min
fu’ = 0.1
CLH’ ≈ fu’ CLint = 0.1 X 200 ml/min = 20 ml/min
b. CLH’ = QH’  EH = 1.2 L/min x 0.007 = 8.4 ml/min
In an average 70-kg adult, Drug D is completely absorbed
into the intestinal epithelium following oral administration and
does not undergo intestinal metabolism. The oral
bioavailability of Drug D is 25%. If the protein binding of the
drug is decreased from 99% to 98%, what is the new oral
bioavailability?
F = 0.25, fu = 0.01
EH = 1-F = 0.75
QH
F
fu CL int
Drug D is a high extraction ratio drug.
fu’=0.02, fu’/fu = 2, i.e. the free fraction increases by 2 fold
F’ = F/2 = 12.5%
i.e. the new bioavailability decreases by a half.
In an average 70-kg adult, Drug E is completely absorbed
into the intestinal epithelium following oral administration and
does not undergo intestinal metabolism. The oral
bioavailability of Drug E is 75%. If the protein binding of the
drug is decreased from 99% to 98%, what is the new oral
bioavailability?
F = 0.75, fu = 0.01
EH = 1-F = 0.25
Drug E is a low extraction ratio drug
Even though fu’=0.02, fu’/fu = 2, i.e. the free fraction increases
by 2 fold
F’ = F = 75%
i.e. the bioavailability is insensitive to changes in fu
Drug F is administered via intravenous infusion to a patient at a
rate of 100 mg/h for 10 hours. The steady-state plasma drug
concentration is 20 mg/L, a total of 300 mg of the drug is
excreted unchanged in the urine. Drug F is only eliminated via
renal excretion and hepatic metabolism. What is the total
clearance (CLT), renal clearance (CLR) and hepatic clearance
(CLH)?
CLT = Ro/Css = (100 mg/h)/(20mg/L) =5 L/h
fe = excreted unchanged in the urine/IV dose
= 300 mg / (100 mg/h X 10 h) = 0.3
CLR = fe · CLT = 0.3 X 5 L/h = 1.5 L/h
CLH = CLT - CLR = 3.5 L/h
In an average 70-kg adult, intravenous Drug G is eliminated by renal
excretion only. When Drug G is given as i.v. infusion at 1 mg/min, an
steady-state plasma concentration of 10 mg/L
(fu =0.1) is
achieved.
a) What is the renal clearance of Drug G?
CLR= CLT = Ro/Css = (1mg/min)/(10 mg/L) = 0.1 L/min
= 100 ml/min
CLR = fu GFR + (CLtubular secretion – CLtubular reabsorption)
fu GFR = 0.1 X 120 ml/min =12 ml/min
CLR > fu GFR, there is a net tubular secretion of 88 ml/min
b) If the plasma protein binding of Drug G is decreased to 80%,
what is the new renal clearance of Drug G? Assume no
saturation in tubular secretion or reabsorption.
CLR = fu GFR + (CLtubular secretion – CLtubular reabsorption)
= 0.2 X 120 ml/min + 88 ml/min
= 112 ml/min