     ### Fraction excreted or metabolized, fe or fm

#### Renal function

We can now use this information to start to understand dosage adjustments for patients with poor kidney function. These patients will have reduced ability to excrete some drugs. That is the ke value for a drug will be lower in these patients than in normal patients. Depending on the value of fe this may have a large effect on kel or it may be insignificant. Fortunately there are a number of clinical tests for renal function which can be used. One common one is creatinine clearance. Creatinine is formed in the body and is excreted almost entirely by filtration in the kidney. The normal creatinine clearance value is similar to the glomerular filtration rate of 120 to 130 ml/min. We can measure the CLCr prior to drug treatment and adjust the dosage accordingly.

For example. Vancomycin fe = 0.95 (Dose 500 mg; V = 33L; Cp0 = 15 mg/L)

The normal kel = 0.116 hr-1 (t1/2 = 6 hr) Equation V-16. Fraction Excreted into Urine

thus

ke = fe * kel = 0.95 * 0.116 = 0.110 hr-1

km = fm * kel = 0.05 * 0.116 = 0.006 hr-1

Now consider a patient with a creatinine clearance of 12 to 13 ml/min. About 1/10 th of the normal kidney function. ke should then be about 1/10 of normal in this patient.

therefore

kepatient = 0.011 hr-1

Now

kelpatient = kepatient + km

probably unchanged

= 0.011 + 0.006 hr-1 = 0.017 hr-1 (t1/2 = 41 hour)

Thus the half-life changes from 6 hours to 41 hours in a patient with impaired renal function. Therefore it takes seven times longer for the body to get rid of half the dose. If repeated doses were given based on a normal half-life the levels in this patient would rapidly reach toxic concentrations. Figure V-7. Plot of Cp versus Time after Multiple Doses in Normal Patient Figure V-8. Plot of Cp versus Time after Multiple Doses

Using a JAVA aware browser you can create your own version of Figures V-7 and V-8.

Plasma Concentration Plots

For another drug such as erythromycin, fe = 0.15 (Dose = 250 mg) the normal kel = 0.58 hr-1 (t1/2 = 1.2 hr)

kenormal = 0.58 x 0.15 = 0.087 hr-1 km = 0.58 x 0.85 = 0.493 hr-1

if the patient ke is reduced by tenth.

kepatient = 0.009 hr-1

kelpatient = 0.009 + 0.493 = 0.502 hr-1 (t1/2 = 1.4 hr)

Thus the half-life changes from 1.2 hour to 1.4 hours.

References:
Bennett et al 1977 Annuals of Int. Medicine 186, 754
Bennett et al 1974 J.A.M.A. 230 1544     