Chapter 5 Analysis of Urine Data

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Fraction Excreted or Metabolized, fe or fm

The equations provided on the two previous pages provide two new parameters, the fraction excreted unchanged in urine, fe, and the fraction metabolized, fm

Equation 5.4.1 Fraction Excreted into Urine as Unchanged Drug

and

Equation 5.4.2 Fraction Excreted into Urine as a Metabolite

Equation 5.4.2 could be repeated for any number of metabolites. Note:

fe + fm₁ + fm₂ + ... = 1

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 CL_Cr 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 (t_1/2 = 6 hr)

Equation 5.4.1, the equation for fe, leads to:

ke = kel^normal • fe = 0.116 x 0.95 = 0.110 hr^-1

km = kel^normal • fm = kel^normal • (1 - fe) = 0.116 x 0.05 = 0.006 hr^-1

If we now consider a patient with a creatinine clearance of 12 to 13 ml/min. That is, about 1/10 th of the normal kidney function, ke should then be about 1/10 of normal in this patient.

therefore

ke^patient = 0.011 hr^-1

Now assuming km is unchanged

kel^patient = ke^patient + km

= 0.011 + 0.006 hr^-1

= 0.017 hr^-1 (t_1/2 = 41 hour)

Thus the half-life changes from 6 hours to 41 hours in this patient with impaired renal function. Therefore it takes seven times longer for the body to eliminate half the dose. If repeated doses were given based on a normal half-life the levels in this patient would rapidly reach toxic concentrations.

With Dose = 100 mg, kel = 0.116 hr^-1, V = 33 L, Tau = 6 hr

Plot of Cp after three doses - normal renal function

Figure 5.4.1 Plot of Cp versus Time after Multiple Doses in Patient with Normal Renal Function

Compare Figure 5.4.1 with Figure 5.4.2

Note: The Change of Scale

Plot of Cp after three doses - poor renal function

Figure 5.4.2 Plot of Cp versus Time after Multiple Doses

Click on the figure to view the interactive graph

Another example

For another drug, erythromycin, fe = 0.15 (Dose = 250 mg) the normal kel = 0.58 hr^-1 (t_1/2 = 1.2 hr)

ke^normal = kel^normal • fe = 0.58 x 0.15 = 0.087 hr^-1

km = kel^normal • fm = kel^normal • (1 - fe) = 0.58 x 0.85 = 0.493 hr^-1

if the patient ke is reduced by a tenth

the ke^patient = 0.009 hr^-1

and again assuming that km is unchanged

kel^patient = ke^patient + km

= 0.009 + 0.493 hr^-1

= 0.502 hr^-1 (t_1/2 = 1.4 hour)

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

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Dose (mg)
U^∞ (mg)

fe (no units) is:
fm (no units) is:

Normal Elimination Rate Constant (hr^-1)
Normal Fraction excreted unchanged, fe
Percent Renal Function in Patient of Interest

Normal half-life t_1/2 (hr) is:		t_1/2 = ln(2)/kel
Normal ke value (hr^-1) is:		ke = fe • kel
Normal km value (hr^-1) is:		km = kel - ke
Patient ke value (hr^-1) is:		ke(patient) = ke • percent renal function
Patient km value (hr^-1) is:		km(patient) = km
Patient kel value (hr^-1) is:		kel(patient) = ke(patient) + km
Patient t_1/2 value (hr) is:		t_1/2(patient) = ln(2)/kel(patient)

kel^patient	= ke^patient + km
	= 0.011 + 0.006 hr^-1
	= 0.017 hr^-1 (t_1/2 = 41 hour)

kel^patient	= ke^patient + km
	= 0.009 + 0.493 hr^-1
	= 0.502 hr^-1 (t_1/2 = 1.4 hour)