Returning to the equation for Cp as a function of time

Equation IX-1 from earlier

we can calculate ka and kel given Cp versus time data. From the method of residuals, the intercept can be determined as

Since we know the DOSE and have calculated ka and kel, it is possible to calculate . However, with only data from a single administration available that's all we can determine, we cannot separate V and F. Of course if we have IV data for kel and V, we could use this to determine F.

Thus F must be determined by comparison with another dose administration. If
the other dosage form is an intravenous form then the F value is termed the
** absolute** bioavailability. In the case where the reference dosage
form is another oral product, the value for F is termed the

When a bioavailability study is conducted at least two dosage forms are administered to each subject. One dosage form is the product to be tested, while the other dosage form is a standard or reference dosage form. This may be an IV dose, oral solution or most commonly the original manufacturer's product. The doses are given with sufficient time between administrations for the drug to "washout" or be completely eliminated. We usually assume that each subject eliminates each dosage form at the same rate.

During the derivation of the Wagner-Nelson equations we calculated Amax, the maximum amount absorbed as:-

Amax = V * kel *

= V * kel * AUC

= F * DOSE

therefore

F =

Now by giving two dosage forms A and B, and calculating AUC values for each.

and if DOSE^{A} = DOSE^{B} and if we can assume that
kel^{A} = kel^{B} and V^{A} = V^{B} then:-

Thus a relative bioavailability, F, can be calculated. If the B dosage form is
given IV then F^{B} = 1 and F = F^{A} and thus F^{A}
can be called the absolute bioavailability.

**EXAMPLE:**

AUC^{A} = 12.4 mg.hr/L [Dose = 250 mg] and AUC^{B} = 14.1
mg.hr/L [Dose = 200 mg

We can do the same thing using urine data alone.

Since

fe =

Equation IX-2 Fraction Excreted Unchanged

therefore

F =

and for two dosage forms

again if DOSE^{A} = DOSE^{B} and fe^{A} =
fe^{B} then

F =

**EXAMPLE:**

250 mg dose; UA = 175 mg; UB = 183 mg; F = = 0.96

This page was last modified: 13 June 2002

Copyright 2001 David W.A. Bourne