6 Extravascular routes of drug administration

114 Basic Pharmacokinetics
Mass of drug (mg)
120
100
80
60
40
20
0
Drug remaining
to be absorbed
Cumulative drug
absorbed
0 2 4 6
Time (h)
Drug in the body
Cumulative drug
eliminated
Cumulative drug
excreted unchanged
8 10 12
Figure 6.10 Rectilinear (RL) plot of disposition of extravascularly administered drug, showing the fraction of drug unabsorbed
versus time.
Mass of drug (mg)
100
10
1
0.1
Cumulative drug absorbed
Drug remaining
to be absorbed
Time (h)
Cumulative drug
eliminated
Cumulative drug
excreted unchanged
Drug in the body
0 2 4 6 8 10 12
Figure 6.11 Semilogarithmic (SL) plot of disposition of extravascularly administered drug, showing the fraction of drug
unabsorbed versus time.
Integration of Eq. (6.21) from t = 0 to t = ∞ yields
(XA)t = 1
dXu
+
Ku dt
K
(Xu)t. (6.22)
Ku
From Eq. (6.18), we see that the first term to the
right of the equal sign in Eq. (6.22) is equal to (X)t,
the amount of drug in the body at time t. In the second
term, (Xu)t is the cumulative amount of drug excreted
by the kidney at time t. We may view the ratio K/Ku
as a factor that will convert (Xu)t to the cumulative
amount of drug eliminated by all pathways, including
both renal excretion and metabolism. Thus Eq. (6.22)
shows that cumulative absorbed drug is the sum of
drug currently in the body plus the cumulative mass
of drug that has already been eliminated at time t.
At t = ∞, the rate of urinary excretion of drug
is zero; so, the first term of Eq. (6.22) drops out,
resulting in
(XA)∞ = K
(Xu)∞
Ku
Sample chapter for Basic Pharmacokinetics 2nd edition
(6.23)
where (Xu)∞ is the cumulative amount of excreted
drug at time infinity.