Pharmacology for Anaesthesia and Intensive Care

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Section IBasic principles

resulting expression must be related to the original function by a constant of propor-
tionality. Experimentally we know that the rate of decline of plasma concentration
with time in a wash-out curve is dependent on the plasma concentration. The con-
stant of proportionality is k, defined as the rate constant for elimination:
dC/dt∝CordC/dt=−kC.
As we commented above, the only function that differentiates to itself is the expo-
nential function, y=ex, that is to say dy/dx=ex=y. From the differential equation
given above for the wash-out relationship we therefore know that the relationship
between concentration and time will be exponential and take the form: C=Be−bt.
The rate constant, b, will be k (the rate constant for elimination). We define the con-
centration at time t=0tobeC 0 so substituting these values in the equation the
constant B must take the value C 0 (because e^0 =1) giving the familiar equation
C=C 0 e−kt.

The hyperbolic function
The simple rectangular hyperbolic function is:
y= 1 /x.
It has asymptotes at x=0 and y=0, being symmetrical about the line y=x(Figure
6.3). We are usually interested in the positive range of the function, that is, for x>0. In
pharmacokinetics, we need to know that integration of this function gives the natural
logarithmic function (see below). The hyperbolic function is important in pharma-
codynamics, where the relationship between dose and response is hyperbolic. The
function relating fractional response (R/Rmax) and dose (D) is:
R/Rmax=D/(D+KD).
Here KDis the dissociation constant for the interaction of drug with receptor. This
is a hyperbolic relationship because the fractional response is proportional to 1/
(D+KD). Although the hyperbolic function is not of direct importance to pharma-
cokinetic modelling, many people confuse the shape of the hyperbolic and expo-
nential functions, as both are curves with asymptotes (Figure6.4).

Logarithms and the logarithmic function
Most people are familiar with the idea that any number can be expressed as a power
of 10; 1000 can be written 10^3 , 0.01 as 10−^2 and5as100.699.Inthese cases theexponent
(or power) is defined as the logarithm to the base 10 (simply written as log) of each
number. Thus the log of 1000 is 3, which can be written log(1000)=3, with log(0.01)
=−2 and log(5)=0.699. Note that log(10)=1 because 10^1 =10 and that log(1)=
0 because 10^0 =1. Any number can be written in this way so that for any positive
value of x
x= 10 log(x).
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