Relationship Between AUC and Volume of Distribution

A few days ago on the pharmacokinetics listserve PharmPK, the following question appeared:

I’m having trouble wrapping my head around this question from an online pharmacology quiz.  The question is “The larger the volume of distribution, the smaller the AUC of a given drug.”  The answer is given as “False.”
As I look at the equation AUC = D/KV, it would seem that if V is larger, then AUC would be smaller.
Can anyone explain how the question is false instead of true?

This question is a common one that trips up many students and professionals in the pharmaceutical sciences. Once you understand the difference between primary and secondary PK parameters, you will always get the answer correct … and the tricks that professors play won’t work anymore.

Let’s start by restating the question:

(True or False) The larger the volume of distribution, the smaller the AUC of a given drug.

To evaluate this statement, we need to find the relationship between AUC and volume of distribution (V). The key is that we use only primary PK parameters. AUC is a secondary PK parameter, and V is a primary PK parameter. This means that V is determined by the physiology of the body (see previous post). It also means that AUC can be calculated from a collection of primary PK parameters. The idea of primary and secondary PK parameters is explained in an introduction to PK video tutorial on YouTube. The proper equation is:

 AUC=\frac{Dose}{CL}       Equation 1

The derivation for this equation can be found in my video tutorial on AUC on YouTube. As you can see, AUC is independent of Volume of distribution. The Volume of distribution can go up, down, or sideways … it will not affect AUC. Only Dose and Clearance will affect AUC. Thus, the answer to the question is False.

Now, there is a relationship that defines an elimination rate constant k, which is:

 k = \frac{CL}{V}       Equation 2

and if you do some mathematical substitution, you can get:

 AUC=\frac{Dose}{k*V}       Equation 3

This is the equation that the reader initially quoted, and mathematically, it is accurate. Unfortunately, the right-hand side of the equation contains primary PK parameters (Dose and V) and secondary PK parameters (k). Since we know that we can only rely on primary PK parameters, let’s substitute Equation 2 into Equation 3 for k, to get:

 AUC = \frac{Dose}{\frac{CL}{V}*V} = \frac{Dose}{CL}       Equation 4

When you define AUC by only using primary PK parameters, you always put yourself in the position of making a correct interpretation. When you rely on secondary PK parameters, you will often make mistakes.

The moral of the story is to know your primary PK parameters (Dose, V, CL, ka) and always think in terms of these parameters.

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