## Pharmacokinetics Of Drug Eliminated By Simultaneous Metabolism And Excretion

Although some drugs are excreted unchanged in urine, most are partially eliminated by metabolism. Usually both the urinary excretion of unchanged drug and the metabolism are first-order processes, with the rate of excretion and metabolism dependent on the amount of unchanged drug in the body. This results in a "branch" in the kinetic chain, representing exit of drug in the body as depicted in the accompanying pharmacokinetic model (Scheme 2).

ke drug in body-> unchanged drug in urine

km metabolite (Dm)

Scheme 2

In this scheme, the rate of loss of drug from the body is determined by both ke and km, and this can be written in differential form as follows:

Thus, the overall elimination rate constant (kel) here is the sum of the urinary excretion rate constant (ke) and the metabolism rate constant (km):

For drugs that are both metabolized and excreted unchanged, semilogarithmic plots of plasma concentrations versus time will provide values of kel.

Urine data are required to determine the individual values of ke and km. The required equations are derived next.

Derivation. From Scheme 2, the differential equation describing overall rate of disappearance of drug from the body may be written:

—— = -kelDB dt and the following integrated equation can be written as (see also equation 10):

Taking antilogs yields

It should be noted that equation (25) is another form of an integrated rate equation. This form makes use of an exp (—x) term and may be referred to as an exponential rate expression. These expressions are useful for visualizing the characteristics of a first-order process. For example, when t = 0, exp (—kelt) = 1, and DB = DB. When t = t1/2, exp (-kelt1/2) = 0.5, and DB = 0.5 x DB. When t = 1, exp (-kelt) = 0, and DB = 0. Thus, the value of exp (—kelt) varies from 1 to 0 as time varies from 0 to 1. At any time between 0 and i, the fraction of the dose remaining in the body is equal to exp (—kelt).

Exponential rate expressions are also useful in deriving kinetic equations because they can be substituted into differential equations that can then be integrated. For example, from Scheme 2 the differential equation describing the rate of appearance of unchanged drug in urine may be written as:

Substituting equation (25) into equation (25a) gives:

Integration yields:

^ ■ = +ke[DB exp(-fcelf)] dDU = +fce[DB exp(-kelt)]dt

-el at t = 0, Du = 0, and exp (—kelt) = 1; therefore, the constant equals (ke/kel) DB, and

At t = 1 after elimination is complete, the total amount of drug excreted unchanged in urine (Du1) can be calculated using equation (26) as follows:

ke kel'

D1 k

Equation (27) shows that the fraction of the dose appearing as unchanged drug in urine (fe) is equal to the fraction of kel attributable to ke. (An equation analogous to equation 27 for D1 and km could be derived in much the same way.)

Table 4 Drug Recovered in Urine Vs. Time | |||

Time interval |
Amount excreted |
Cumulative amount excreted |

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