Medicinal Chemistry Applet

Cp vs time - iv bolus with metabolite formation

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Introduction

Special thanks for this exercise go to my colleague, Durwin Striplin, a physical chemist who derived Equation 4 from the diagram in Scheme 2.

Elimination is a term that refers to the removal of a drug (D) from circulation.  In simple treatments, elimination can be summarized in the umbrella term of kel, the elimination rate constant (Scheme 1).  This term is the sum of all the elimination processes that affect the drug.  A slightly more complicated discussion breaks down the elimination processes into two categories: excretion by the kidneys and metabolism by the liver.  Each metabolite (M) can then be excreted by the kidneys (Scheme 2).


Scheme 1 - simple drug elimination model


Scheme 2 - elimination model with excretion and metabolism

Drug metabolism is normally classified as either Phase I or Phase II.  Phase I includes hydrolyses, oxidations, and reductions.  Phase II includes conjugation of the drug with other compounds such as glutathione or glucuronic acid.  Generally, the effect of either Phase I or Phase II metabolism is to increase the polarity, or water-solubility, of the drug.  With increased polarity, the drug metabolite is more readily excreted through the kidneys and into the urine.  Unfortunately, not all metabolites are simply excreted.  Some metabolites are biologically active with effects that can be similar to or different from the original drug.  Identifying and tracing drug metabolites and their pharamcokinetics and pharmacodynamics is a key aspect in the FDA approval process.

Based on Scheme 2, the rate of change in concentration of the drug and metabolite can be expressed through Equations 1 and 2.  Equation 1 simplifies to the standard one-compartment, first-order expression for drug elimination.  Integration of Equations 1 and 2 affords Equations 3 and 4.  Once the required rate constants are known, the Cp of the metabolite can be determined at any time following an iv bolus with an drug concentration of [D]o.

     (1)

     (2)

     (3)

     (4)

Applet

This applet plots data for the Cp of a drug and a metabolite.  Common units for each parameter are shown in the form below.

Parameter (units) Value Description
Do (mg) initial dose
Vd (L) volume of distribution
kex (h-1) drug elimination rate constant
km (h-1) drug metabolism rate constant
kmex (h-1) metabolite elimination rate constant
calculation may be slow

Problem information

Morphine (1) is an analgesic is widely used to manage severe pain in hospitals.  The major metabolite for morphine is morphine-6-glucuronide (2) (Figure 1).  While the metabolism and elimination of morphine is not trivial, a few assumptions can simplify the analysis.


Figure 1 - Pharmacokinetic parameters for morphine (1) and morphine-6-glucuronide (2)

Morphine has a t1/2 of 1.9 h, corresponding to a kel of 0.36 h-1.  This kel is the sum of the kex and km.  Since only approximately 5% of morphine is excreted unchanged by the kidneys, we can assume that kex is 0.02 h-1 (5% of kel).  We will further assume that the remainder of kel is for metabolism of 1 to 2, so km is 0.34 h-1.  Metabolite 2 has a kex of about 0.17 h-1.

Problems

  1. Assuming that morphine (1) has a Vd of 230 L for a 70-kg patient and a typical iv bolus dose is 10 mg, generate the Cp vs time graph for these parameters.
  2. Morphine-6-glucuronide (2) is twice as potent as an analgesic as morphine itself.  At approximately what time point is the analgesic effect in the patient primarily from the activity of the metabolite and not morphine?
  3. With patients suffering from renal failure, the rate of elimination of metabolite 2 is much lower and kmex drops to about 0.014 h-1.  Generate the Cp vs time graph and note the change in the x-axis.  Which patient (healthy kidney or failed kidney) requires more morphine for effective pain control?
  4. Equation 4 has at least one problem.  It quietly assumes that the Vd of the metabolite is the same as the original drug.  The problem is in the [D]o term.  Replace the [D]o term with its equivalent, dose (drug)/Vd (drug).  Now, derive a version of Equation 4 that will use the proper Vd of the metabolite.  Of course, to use this new equation, one needs to know the Vd of the metabolite.

Reference

Goodman and Gilman's The Pharmacological Basis of Therapeutics, 10th ed.; Hardman, J. G., Limbird, L. E., Eds.; McGraw-Hill: New York, 2001.

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