Since drugs can bind with proteins their distribution volume in the body can be
affected
Drug (Cu) + Protein (Cp) ↔ Drug-Protein (Cb)
Total drug concentration C = Cu + Cb
At equilibrium then K = Cb / (Cu Cp), but Cp is basically constant
So we let K’ = K Cp = Cb/Cu
The fraction unbound is defined as fu = Cu / C = Cu / (Cu + Cb) and we can
Use K’ to eliminate Cb and hence we obtain that
fu = 1 / (1+ K’) and then Cb = Cu (1 – fu) / fu and C = Cu / fu
important to remember that only unbound drug is available to move around in
the body
Acidic drugs
Basic drugs
Neutral drugs
testosterone
The volume into which a drug distributes within the body is called the
(apparent) distribution volume given by V or Vapp
Since usually it is the plasma concentration that we know, ie. C, then
the volume of distribution is really just a “volume” when multiplied by
C that gives the total amount of drug in the body at a particular time
Volume of Distributi on 
Amount of drug in body
Plasma drug concentrat ion
Apparent distribution volume linked to fluid volumes as well
as the effect of drug binding to proteins, so a drug that binds
strongly with plasma proteins will approach a distribution volume
of 3 liters, on the other hand a drug that binds tightly with extravascular
proteins will result in a very low plasma concentration and to account
for the mass of drug that means the value of V may greatly exceed the physical
volumes of fluid actually in the body, in the extreme could be 7000 L !
V = apparent distribution
volume
Vp = plasma volume, 3 L/70kg
VE = extracellular fluid volume
less plasma volume, 12 L/70kg
VR = remainder of the fluid volume
in the body
fU = fraction unbound in plasma, Cu/C
fUT = fraction unbound in R
RE/I = ratio of total drug binding sites
in the fluids outside of plasma to
those within the plasma, ~ 1.4
Oie – Tozer Equation
V  V P 1  R E / I   f U
 VE
V P 
 R E/I
 VP

VR f U
 

f UT

Using typical values for Vp and VE and R/EI we get this equation
V  7  8 f U  VR
fU
f UT
Some special cases:
1. If drug only goes to extracellular spaces and cannot enter the cells,
then VR = 0 and V = 7 + 8 fu and V has its smallest value and depends
only on the fraction unbound in the plasma, so if drug is
totally unbound (fu = 1) then V = 15 L, the extracellular fluid volume;
if it is completely bound (fu = 0) then the distribution volume cannot
be less than 7 L regardless of how tightly bound to albumin
2. Intracellular water (VR) is about 25-27 L so if the drug enters the cells but
is not significantly bound (fUT =1 and fU = 1), then the volume of
distribution is that of the total body water, ie. V = 40-42L, example
would be alcohol
3. Note that as fUT  0, then V gets really big
ADMET – absorption, distribution, metabolism, elimination, toxicity
Common routes for drug metabolism include oxidation, reduction, hydrolysis,
and conjugation. Can have many simultaneous pathways
Primary site for drug metabolism is the liver and sometimes this is the only
place metabolism occurs; other sites include the kidneys, lungs, blood,
and GI wall
Metabolism is good since it limits the time of drug action, may produce the
active form of the drug, metabolites may also be active drugs as well, can
even be a bioactivation making them more active or toxic than the parent
compound (aka prodrug)
Metabolism is the result of enzymatic reactions that occur within the cells
and the rate of metabolism can be described by Michaelis-Menten kinetics
rmetabolic 
V max C
Km C
 k metabolic C
Usually C << Km so the kinetics are just 1st order in drug concentration
Basic functional unit of the kidney is the
Nephron
About 1 million per kidney
Blood flow to kidneys is about 1100ml/min,
and the glomerulus produces (GFR) about
120ml/min of plasma filtrate, fortunately
most of the water is reabsorbed producing
a urine flow of about 1-2ml/min
Unbound drug will be filtered from the blood
flowing thru the glomerulus
GFR is the rate of filtration of plasma
across the fenestrated capillaries
of the glomerulus due to hydraulic
and osmotic pressure differences, so
at the glomerulus unbound drug is
removed at the rate given by:
Drug removal
Rate
= GFR x Cu = fu x GFR x C
Renal Clearance (CLrenal) = Drug removal rate
C
Since clearance by definition is that flowrate of
the fluid that is totally cleared of the solute
CLrenal = fu x GFR
So if the drug is only filtered out at the glomerulus, and all of the filtered
drug is excreted into the urine (not secreted or reabsorbed in the tubules),
then the rate of drug excretion (refers to urine) is the same as the
drug removal rate at the glomerulus, or:
CLrenal = fu x GFR, and if the drug is unbound (fu = 1), then the renal
clearance for the drug is the same as the GFR; for example, inulin is a
sugar-like substance with a molecular weight of about 6000 that is used
to determine GFR; in addition creatinine is also not secreted or
reabsorbed by the tubules and is also unbound and is a product of
endogenous protein degradation, its production rate (M) in the body is about
120 mg/min, so at steady state what is produced in the body has to be
removed by the kidneys, so we can write that
M = GFR x C = U x V or that GFR = U V / C, where U is the concentration
in the urine and V is the urine flowrate, also we have that
GFR = M/C, so everything else being equal, ie. M constant, then plasma
creatinine is a direct measure of kidney function via the GFR as shown
on the graph on the next slide
Normal
GFR is about
125 ml/min
Failure need dialysis
5 – 11 mg/100ml severe
Most drugs are secreted and reabsorbed so things are a bit more
complicated in terms of the renal clearance
Secretion may be inferred when the rate of excretion (CLrenal x C)
exceeds the rate of drug filtration (fu x GFR x C), or stated in a
different manner CLrenal > fu x GFR
Tubular reabsorption would be apparent whenever the rate of drug
excretion (CLrenal x C) is less than the rate of drug filtration (fu x GFR x C),
or CLrenal < fu x GFR
V apparent
dC
dt
  C L renal C
Which can be integrated from the IC of
t=0, C=C0 to give:
C(t)  C 0 e
 ( C L renal / V apparent ) t
 C0 e
 k renal t
C
krenal is the rate constant CLrenal/Vapparent
The drug removed by the kidneys then shows up the urine and we can
write that at any time:
C L renal C  Q urine C urine or C L renal 
Q urine C urine
C
Plasma clearance (CLplasma) relates to all possible drug elimination
pathways and would include the following:
Metabolism
CL i
Kidneys
Elimination rate constants
ki 
Sweating
V apparent
Bile
i = renal, metabolic, sweat, bile, respiration, feces
Respiration
Feces
C L plasma
l
k te   k i 
 CL i  V
V
i
i
apparent
apparent
The total elimination rate constant
C(t)  C 0 e
 k te t
Biological half life is the time needed for the plasma drug
concentration to decrease by ½. For first order processes, this
time is related to the first order elimination rate constant by
simply letting C/C0 equal 0.5. When solved for
t1/2, the following result is obtained.
t 1/ 2 
0 .693
k te

0 .693 V apparent
C L plasma
Intravenous Injection of a Drug
D
C0 
V apparent
AUC
AUC
dM
urine
dt
M urine 
k te
0 
0 




0
C0
k te

 D
k renal
k te
 AUC
0 
C L renal
 k te t
Measure of drug
exposure
C ( t) dt
D
V apparent k te
 k renal C 0 V apparent 
k t
M urine ( t )  
  1  e te 
k te


 k renal V apparent C
k renal C 0 V apparent
C(t)  C 0 e

D
CL plasma
V apparent
dC
dt
 I 0  C L plasma C  I 0  V apparent k te C


Io
k t
C(t)  
  1  e te 
 k te V apparent 
C ss 
I0
k te V apparent

I0
CL plasma
CSS = 873 cpm/ml
Example 7.1, inulin infusion study in a laboratory rat
i  0  8
time
i
number of data points

C 
time, minutes
inulin concentration, cpm/ml
i
30
849
40
845
60
903
70
888
75
873
80
882
90
565
100
412
110
271
C SS 
calculate the steady state inulin concentration
 1 
 
6
5

C
i
i  0
C SS  873.333
cpm/ml
perform linear regression on the data after the infusion stopped at 80 minutes assuming
the starting concentration was equal to that of Css
C  C SS
sets the 80 minute value to the average steady state value
j  0  3
only have 4 data points after infusion stops including the Css value
5
t  time
j
( 5 j)
y  ln C
j

 80
the time since the infusion stopped in minutes
( 5  j )

transform the concentration data
k renal   slope ( t y )
k renal  0.038
C 0  exp ( intercept ( t y ) )
r  corr ( t y )
1/min, the renal elimination rate constant
C 0  860.085
r   0.998
cpm/ml, the value of the starting inulin
concentration after the infusion stops,
compares quite well with the value of
Css determined above from the data
indicates that we have an excellent fit of the
single compartment model that describes the
elimination of inulin
time  0  30
running time, minutes
C pred ( time )  C 0  exp   k renal  time
Inulin Elimination
inulin concentr ation, cpm
1  10
exp
yj
C pred ( time )
3
800
600
400
200
0
10
20
t j time
time, minutes
30

the predicted
inulin levels
after the
infusion is
stopped