PHA 5127 Answers Homework 4 Jeff Stark

1. What is the maximum extraction ratio for renal if neither reabsorption nor active secretion take place? Explain.

The drug enters the kidney at the rate of renal blood flow (RBF), 1200 ml/min. The rate of filtration for a drug depends on the glomerular filtration rate (GFR) and the unbound fraction of drug in the plasma (fu). The extraction ratio for the kidney when filtration is the only renal process occurring is:

GFR × fu E = Re n RBF

The maximum extraction ratio is observed when there is no protein binding, fu=1. In this situation,

125ml / min×1.0 E = = 0.10 Re n 1200ml / min

Thus, about 10% of the drug entering the kidneys is filtered through the glomerular capillaries.

2. The renal clearances and the fractions unbound to plasma proteins of three drugs in a 70kg 40 year old man are listed as follows: Renal (ml/min) Fraction Unbound Drug 1 62.5 0.5 Drug2 30 0.01 Drug3 0.45 0.3

Explain the likely contribution of filtration, secretion, and reabsorption to the renal handling of each of these drugs. Assume a GFR of 125 ml/min and a flow of 1.5 ml/min.

When approaching problems such as this, ask yourself what the renal clearance would be if you have only filtration taking place. This is a good starting point and will give you some idea of the relative magnitude of secretion and reabsorption as well.

Drug 1.

If only glomerular filtration occurs, Clren = GFR·fu = 125 ml/min·0.5 = 62.5 ml/min

D:\Homeworks\Homework4\Fall-99\ans-hw-4-99.doc 1 Since the renal clearance of this drug is stated as 62.5 ml/min, we may conclude (I) that glomerular filtration is the only renal process involved or (ii) that active secretion and tubular reabsorption occur at the same rate.

Recall: Clren ~ Clfiltration + Clsecretion + Clurine flow - reabsorption

Drug 2

Clfiltration = 125 ml/min · 0.01 = 1.25 ml/min

The clearance due to glomerular filtration of this drug is very small since it is highly protein- bound (only free drug is filtered). The renal clearance is given to be 30 ml/min, much larger than clearance due only to filtration. We may conclude that active secretion occurs in addition to filtration. If tubular reabsorption occurs, the amount absorbed is outweighed by the process of active secretion.

Drug 3

Clfiltration = 125 ml/min · 0.3 = 37.5 ml/min

For drug 3, renal clearance is 0.45 ml/min. Since this value is so small compared to the clearance due to filtration, we may conclude that there is a large degree of tubular reabsorption. For some drugs, tubular reabsorption occurs to such a high degree that the only drug eliminated in the kidneys is that lost in normal urine flow,

Clurine flow = urine flow · fu = 1.5 ml/min · 0.3 = 0.45 ml/min

Since this value is equal to the renal clearance given, we may conclude that "complete" reabsorption takes place and drug lost in the urine flow is the only effective process of renal clearance.

3. A 5'10" 40 year old 90kg man has a serum level of 1.5 mg/100 ml and is being treated with aminoglycosides. Estimate the clearance for the aminoglycoside and the half-life for this patient. If the desired peak plasma level is 6 mg/L and the volume of distribution is 0.24 L/kg (IBW), what dose should be given (just consider one dose). How long will it take to reach a plasma level of 1 mg/L.

D:\Homeworks\Homework4\Fall-99\ans-hw-4-99.doc 2 Given Information · 5'10" tall male · 90 kg · Scr = 1.5 mg/100ml

Estimate clearance of aminoglycosides and t1/2 for this patient.

For aminoglycosides, ClCreatinine » ClAminoglycosides since these compounds are eliminated only by filtration and protein binding is minimal. The empirical equation for creatinine clearance is

(140 - age) × weight ClCreat (male) = [units : ml/min ] 72 × CpCreat

Weight in this equation is ideal body weight (IBW). For males IBW is calculated by

IBW = 50 kg + 2.3 kg for each inch over 5 ft in height = 50 kg + 2.3 kg · 10 = 73 kg

Clcreat may now be calculated:

(140 - 40) × (73) Cl = Creat (72) × (1.5) = 67.6 ml/min

Since Clcreat is related to the glomerular filtration rate (GFR), we know that there is some renal failure in the patient (normal GFR = 125 ml/min).

The aminoglycoside t1/2 may be calculated easily after ke is determined. For aminoglycosides,

-1 k e = 0.00293× ClCreat + 0.014 [units : hr ] = (0.00293)·(67.6) + 0.014 = 0.212 hr-1

The half-life is then ln 2 0.693 t1/ 2 = = -1 = 3.3hr ke 0.212

To obtain a peak plasma concentration of 6 mg/L, what dose should be given?

Assuming the infusion time is sufficiently short and IV bolus equations are adequate approximations,

D:\Homeworks\Homework4\Fall-99\ans-hw-4-99.doc 3 D Cp0 = Vd Solving for dose D gives

D = Cp0 ×Vd = (6 mg/L)·(0.24 L/kg)·(90kg) = 130 mg

How long will it take to reach a plasma concentration of 1 mg/L? For and IV bolus injection,

-ke ×t Cp(t) = Cp0 × e

In this equation, Cp0 is the initial plasma concentration (6mg/L), Cp(t) is the concentration at any later time point, and t is the time after injection.

Solving for t gives:

Cp(t) = e -ket Cp0

æ Cp(t) ö lnç ÷ = -k t ç ÷ e è Cp0 ø 1 æ Cp(t) ö t = - lnç ÷ ke è Cp0 ø 1 æ 1mg / L ö = - lnç ÷ -1 ç ÷ 0.212hr è 6mg / L ø = 8.5 hr

4. What are the assumptions made by the one-compartment model?

· Instantaneous distribution · Linear kinetics (AUC µ Dose)

D:\Homeworks\Homework4\Fall-99\ans-hw-4-99.doc 4 5. Mark the following statements true (T) or false (F):

T F The maximum value which renal clearance can approach is 1200 ml/min. T F Large drug molecules and low degree of protein binding could predispose a drug to glomerular filtration. T F Tubular secretion may be a competitive process. T F Tubular reabsorption can only be a passive transport process. T F Highly ionized drugs tend to have more tubular reabsorption. T F Molecular size, protein binding, and lipid solubility significantly influence glomerular filtration. T F Urinary excretion of amphetamine can be decreased by raising pH. T F With first-order elimination, the fraction of a drug in the body eliminated over a given time depends on the drug concentration.

Use the Excel Spreadsheets available to draw the Cp vs time curves for the following inputs. Use the IV-S.xls for the first two (one compartment, i.v. bolus, single dose) and Oral.xls for the third (one compartment, oral, single). Calculate and record the missing parameters and explain the results.

1. Input Results D[mg] 120 120 Peak(mg/L) 1.20 1.20 Vd(L) 100 100 Ke(1/h) 0.12 0.24 CL(L/h) 12 24 t1/2(h) 2.8 2.9 AUC(mg/L·h) 10.00 5.00

1.4

1.2

1

0.8 0.6

0.4

0.2

0 0 10 20 30

D C = e -ket Vd

D:\Homeworks\Homework4\Fall-99\ans-hw-4-99.doc 5 2. Input Results

D[mg] 180 60 Peak (mg/L) 1.80 0.60 Vd(L) 100 100 Ke(1/h) 0.24 0.24 CL(L/h) 24 24 t1/2 (h) 2.9 2.9 AUC(mg/L*h) 7.50 2.50

2

1.5

1

0.5

0 0 10 20 30

D C = e -ket Vd 3. Input Results D[mg] 60 60 tmax (h) 1.53 2.01 Bioavail. 1 1 Peak(mg/L) 0.65 0.40 Ka(1/h) 1 1 Ke(1/h) 0.4 0.2 Vd(L) 50 100 t1/2 (h) 1.73 3.47 CL(L/h) 20 20 AUC(mg/L*h) 3.00 3.00

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 6 12 18 24 30

f × D × k Cp = a (e -ket - e -kat ) Vd (ka - ke )

D:\Homeworks\Homework4\Fall-99\ans-hw-4-99.doc 6