The data obtained following the iv bolus dosing of KL25A as depicted in Fig. 2 are summarized in Table I.

Table I Pharmacokinetic Information Describing iv Bolus (Fig. 2) and Oral (Fig. 4) Dosing of KL25A to a Healthy SubjectDetermine the CLP,total, CLP,R and CLP,H

Determine the plasma (P) clearance parameters. From Eq. 1 where for an iv dose F=1

$$_=\frac_}=\frac\; mg}+\frac)\frac}=24.2\frac$$

$$_=\frac_}_}=\frac\; mg}+\frac)\frac}=13.9\frac$$

$$_=_-_=24.2-13.9=10.3 \frac$$

However, to relate clearance measures to organ blood flow, plasma clearance values must be converted to blood clearance values using Eq. 22 and the blood to plasma concentration ratio, B/P, which for KL25A is 0.9 as given in Table I

$$}}}_}}=\frac}}}_}}}}/}}$$

(22)

Determine Elimination Characteristics of the Kidney

\(_=\frac=15. 4\frac\) From Eq. 5 assuming kidney blood flow of 70 L/h.

\(\frac_}=\frac_}+\frac_\bullet GFR+(_-_)}\); \(\frac=\frac+\frac_-_)}\) Thus,

$$\left(_-_\right)=19.0\frac\text_=0.72 \frac$$

Secretory clearance is the major renal route of elimination.

Determine Elimination Characteristics of the Liver

\(_=\frac=11. 4\frac\) It is unlikely that KL25A is a substrate for acid transporting hepatobiliary membrane OATPs with a B/P of 0.9. Acids generally do not distribute into blood cells and typically exhibit a B/P ≈ 0.55.

From Eq. 7, assuming liver blood flow of 90 L/hr\(\frac_}=\frac_}+\frac_\bullet _}\);\(\frac=\frac+\frac_}\)

Thus, \(_=13.1\frac\), and the primary route of hepatic elimination is biliary excretion, since only minor metabolites are measured.

Overall, renal clearance by transporters (19.0 L/h) is about 50% greater that biliary clearance (13.1 L/h), yet both the kidney and liver are relevant for KL25A elimination.

Determine the Mean Residence Time of the Drug

For a drug following linear kinetics the mean residence time is calculated via Eq. 24, where AUMC (area under the moment curve, i.e., area under the product of C·t vs t) is the sum of the coefficients divided by the exponents squared (6) (Eq. 23) while Eq. 21 gives AUC.

$$\text=\frac_}_}=\frac_}_^} + \frac_}_^}}_}_} + \frac_}_}}=\frac^} + \frac^}} + \frac}=\frac=5.26\text$$

(24)

versus the terminal elimination half-life of 0.693/0.173 = 4.01 h

Determine the Volume of Distribution of the Drug

A single volume of distribution term is relevant (Eq. 25), designated previously as the volume of distribution steady-state (Vss), a volume term independent of elimination characteristic. It is an unfortunate accident that this volume term is designated as the volume of distribution steady-state, since it was first determined under steady-state conditions. In fact, Vss has no relation with steady-state. It is the total volume in Fig. 3 for the measurable and unmeasurable compartments and represents this volume under all conditions. Vss is the only relevant volume in clinical pharmacokinetics and it reflects the space in which the drug may distribute following an iv bolus dose.

$$_=CL\bullet MRT=\frac_}_}^}=\frac\bullet 5.26=142\; L$$

(25)

Multiplying measured systemic concentrations by this value gives the total amount of drug in the body. Traditional teaching of pharmacokinetics often includes extensive discussion of various volume parameters and the potential for protein binding within unmeasureable hypothetical compartments. However, we find limited clinical relevance in these analyses and instead view Vss as simply providing an important measure of the theoretical space available for drug distribution, while also allowing for the examination of changes in this measure as a result of drug-drug interactions. It is recognized that when drug-drug interactions only involve changes in measures of metabolic elimination, Vss is unchanged (8). However, when drug interactions involve transporters, then changes in Vss and CL can both occur (9). For example, OATP1B inhibition by single-dose rifampin significantly decreased both clearance and volume of distribution for fluvastatin, rosuvastatin, atorvastatin, and glyburide (9, 10). In some cases, the decrease in volume was greater than the decrease in clearance, leading to the paradoxical result that decreased clearance is accompanied by a shorter half-life (10).