# Chapter 16

# Routes of Excretion

return to the Course index

previous | next

## Cp_{average} Calculations

For patients with poor renal function taking drugs with high fe values, dosage regimen adjustment is essential. Drugs that are eliminated via the kidneys will have reduced elimination in patients with impaired renal function. Steady state and average drug concentration will rise dangerously unless the dosage regimen is adjusted. One way to make this adjustment is to adjust the dose or dosing interval to maintain a required average drug concentration.
For example consider the drug kanamycin. A patient of 70 kg with normal kidney
function may receive 250 mg IM every six hours (about 3 half-lives;
t_{1/2} = 2.3 hours). If F = 1.0 and V = 13.3 liter, kel = 0.693/2.3 = 0.30 hr^{-1}.
Then

**Equation 16.8.1 Average Drug Concentration**

If we assume that ka >> kel then

where R = e^{-0.3 * 6} = 0.165

Thus, Cp_{min} = 3.7 mg/L

These are the results you should expect in a patient with a normal creatinine
clearance value. However in a patient with a creatinine clearance of only 10
ml/min the elimination rate constant will be quite different and if the same
dosage regimen were used quite different plasma concentrations would be
achieved (see Figure 16.8.1). The elimination rate constant for this patient would
be 0.034 hr^{-1} (t_{1/2} = 20 hr). Using the same dosing regimen:

This average plasma concentration is well above the maximum recommended value of 35 mg/L which should be avoided (in the PDR 89 p740). Clearly some dosage adjustment should be made to the dosage regimen.

**Figure 16.8.1 Linear Plot of Cp ***versus* Time

We could consider

a) changing the dose

b) changing the dosing interval

or c) changing both the dose and the dosing interval.

We can make these alterations easily using the equation.

**Equation 16.8.2 Average Drug Concentration Equation**

From this we can see that decreasing the dose or increasing the dosing interval
will have the desired response.

### Altered dose

Assuming that a Cp of 10.4 mg/L (the value obtained in the normal patient on a
normal dosage regimen) is satisfactory we can calculate a dose to achieve this
value as:

Assuming ka >> kel, R = 0.815 and Cpmin = 9.3 mg/L

Thus this new dosing regimen of 28 mg every 6 hours should work

### Altered dose interval

Therefore giving 250 mg every 53 hours should achieve a satisfactory plasma
concentration profile.

R = 0.165; Cpmin = 3.7 mg/L

We would expect greater fluctuations with this method and dosing every 53 hours
is not all that convenient. Every 6 hours is not all that great either if a
longer dosing interval would work. We might consider dosing every 24 hours.

### Altered dose and interval

Using τ = 24 hours

R = 0.442; Cpmin = 6.7 mg/L.

**Figure 16.8.2 Plot of Cp ***versus* Time

The lines in Figure 16.8.2 were calculated to achieve a of 10.4 mg/L using 28 mg q6h, 250 mg q54h, or 113 mg q24h.

**Calculator 16.8.1 Calculations Using the Cp**_{average} Equation 16.8.2

return to the Course index

previous | next

This page (http://www.boomer.org/c/p4/c16/c1608.html) was last modified:
Saturday 28 Nov 2015 at 09:49 AM

Material on this website should be used for Educational or Self-Study Purposes Only

Copyright © 2001-2015 David W. A. Bourne (david@boomer.org)