- Ke, the elimination rate constant can be defined as the fraction of drug in an animal that is eliminated per unit of time, e.g., fraction/h.
- Elimination half-life is the time required for the amount of drug (or concentration) in the body to decrease by half.
- Although CL can be easily related to the function of a specific organ, it is more difficult to get a "minds-eye-view" of how fast a drug is removed from the whole animal from CL. It is also not easy to conceptualize what a Ke of 0.016 /h means. The elimination "half-life" is better suited for this because after 5 half-lives, approximately 97% of the drug is gone.Therefore, it is useful to be able to interconvert CL, Ke, and half-life.
- CL, Ke, Vd, and Half-life are all inter-related as
follows:
Ke = CL / Vd ---- [frcn/h = (L/h) / L]

Half-life = 0.693/Ke ---- [h/(half) = 0.693 / (frcn/h)]

- 0.693 is the natural logarithm of 2. Thus, half-life is an arbitrary measure of drug elimination that is useful for humans, but not easy to use in complex formulae. Therefore, many tables contain both, but most formulae require one to use Ke. You should be able to do this as well as to inter-convert CL and Ke when given Vd.
- Discovering Ke and Half-life from a graph

- Find the graph of "miraclemycin" that you made earlier.
- What is the "half-life" of the drug in this case?
- Label the slope of the line as being "-Ke"
- From the half-life, calculate the Ke
- Now, calculate the CL
- Answers

- Half-life = 2 h
- Ke = 0.693 / 2 h = 0.3465/h
- Remember that you calculated a Vd in this case of 140 L
in the 70 Kg patient
CL = Ke * Vd

*0.3465/h * 140 L = 48.5 L/h* - One can also express CL relative to weight:
CL/Kg = Ke * Vd(L/Kg)

*0.693 L/(h * Kg) = 0.3465/h * 2 L/Kg*

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Last modified: 03 Sep 1996 22:54 glc