Chapter 5

Analysis of Urine Data

return to the Course index
previous | next

Metabolism and Excretion - Parallel Pathways

Although a few drugs are eliminated as unchanged drug into urine or alternately may be completely metabolized, most drugs are eliminated by excretion AND metabolism. There is often more than one excretion or metabolism pathway.

Scheme or diagram

Schematically this can be represented as:-

Figure 5.2.1 Diagram Illustrating Multiple Elimination Pathways with a One Compartment Model

In Figure 5.2.1 ke is the excretion rate constant and km is the metabolism rate constant. Here we have two parallel pathways for elimination (with others as a shadow). We can write the differential equations for the four components shown in this diagram (X, U, M, Mu). There could be more pathways. It may be necessary to specify excretion by exhalation, in sweat, or as is commonly the case, more than one metabolic pathway.

The Equations

Drug in the Body, X

For X = V • Cp, amount of drug in the body

V*dCp/dt =

Equation 5.2.1 Rate of Change of the Amount of Drug in the Body

Equation 5.2.1 includes terms for excretion and metabolism. The number (and type) of these elimination processes can be changed to accommodate a variety of possible routes of excretion or metabolism. Some of these processes may not be first order, however many can be represented by first order parameters.

The elimination rate constant, kel, represents the sum of all the ('first-order') rate constants so we can substitute kel for (ke + km) in Equation 5.2.1 to give Equation 5.2.2.

Equation 5.2.2 Rate of Change of the Amount of Drug in the Body

Dividing by V gives the Equation 5.2.3, which is the same as Equation 4.4.1 in Chapter 4.

Equation 5.2.3 Rate of Change of Drug Concentration

dX/dt in Clearance terms

Equation 5.2.4 Rate of Elimination using Clearance Parameters

In Equation 5.2.4 the rate of elimination of drug from the central or plasma compartment is expressed in clearance terms, here, renal (CLR) and metabolic (CLM) clearance. The total body clearance (CL) is equal to the sum of the clearance terms in the model. In Equation 5.2.4 total body clearance, CL = CLR + CLM.

With more elimination pathways we sum all these process parameters to arrive at the elimination rate constant, kel, or total body clearance, CL. Thus the equation for rate of elimination, either with rate constant parameters or clearance parameters, is the same as before in Chapter 4. The integrated equation is given in Equation 5.2.5 or 5.2.6

Cp <i>versus</i> t with kel

Equation 5.2.5 Cp versus time with kel as the elimination rate parameter

Cp <i>versus</i> t with CL

Equation 5.2.6 Cp versus time with CL as the elimination rate parameter

Drug Excreted into Urine, U

The rate of excretion, dU/dt, can be derived from the model, Figure 5.2.1 in terms of ke or CLR

dU/dt = ke x X

dU/dt = ke x V x Cp

dU/dt = CLR x Cp

Equation 5.2.7 Rate of Change of Cumulative Amount Excreted into Urine

Then substituting for Cp (= (Dose/V) • e-kel • t or = (Dose/V) • e-CL • t / V) we get

Rate of excretion of drug into urine

Equation 5.2.8 Rate of Excretion of Unchanged Drug into Urine

after integrating using Laplace transforms we get:

Cumulative amount excreted into urine <i>versus</i> time

Equation 5.2.9 Cumulative Amount Excreted as Unchanged Drug versus Time

Note: ke or CLR are in the numerator of Equation 5.2.9. As time approaches infinity the exponential term, e-k • t, approaches zero. Setting the exponential terms in Equation 5.2.9 to zero gives Equation 5.2.10 for the total amount of unchanged drug excreted in urine, U

Total amount excreted as Unchanged Drug

Equation 5.2.10 Total Amount Excreted as Unchanged Drug into Urine

Amount of Drug Metabolized in the body, M, and excreted into urine, Mu

For M, the amount of drug which has been metabolized the differential equations are:-

Equation for dM/dt

Equation 5.2.11 Rate of Change of Amount of Metabolite in the Central Compartment

AND

Equation for dMu/dt

Equation 5.2.12 Rate of Excretion of Metabolite into Urine

After integrating Equation 5.2.11 and 12 using Laplace transforms we get Equation 5.2.13

Equation for Mu <i>versus</i> time

Equation 5.2.13 Cumulative Amount Excreted as Metabolite versus Time

Setting each exponential term, e-k • t, in Equation 5.2.13 to zero gives Equation 5.2.14 for the total amount excreted into urine as the metabolite.

Total amount excreted as metabolite

Equation 5.2.14 Total Amount Excreted as Metabolite into Urine

Adding Equations 5.2.10 and 5.2.14 gives Equation 5.2.15.

U(inf) + M(inf) = Dose

Equation 5.2.15 Mass Balance - Total Amount Eliminated equals Dose

Notice the total amount of the drug excreted and metabolized adds up to the Dose. This is based on the assumption that we have information from all the pathways of elimination.
Drug and metabolite versus Time

Figure 5.2.2 Plot of X, U, M and Mu versus Time

Click on the figure to view the interactive graph

In Figure 5.2.2 we have simulated amounts of drug in the body (X), unchanged drug in urine (U), metabolite in the body (M) and metabolite excreted in urine (Mu). The parameters used to create this figure were Dose = 500 mg, kel = 0.2 hr-1, fe = 0.25, fm = 0.75, and kmu = 0.5 hr-1. Note U and Mu, 125 and 375 mg respectively, total 500 mg, the dose given.


return to the Course index


This page was last modified: Wednesday, 20th Dec 2017 at 7:11 pm


Privacy Statement - 25 May 2018

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

iBook and pdf versions of this material and other PK material is available

Copyright © 2001-2019 David W. A. Bourne (david@boomer.org)