1. Since drugs can bind with proteins their distribution volume in the body can be
affected by their interaction with protein
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

2. Unbound
Bound

3. The volume into which a drug distributes within the body is called the
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

4. 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 !

5. Oie – Tozer Equation 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

6. Using typical values for Vp and VE and R/EI we get this equation
V is in liters
Some special cases:
If drug only goes to extracellular spaces and cannot enter the cells,
then VR = 0 and V = 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 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

7. 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
Usually C << Km so the kinetics are just 1st order in drug concentration

8. 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

9. GFR is the rate of filtration of plasma
across the fenestrated capillaries
of the glomerulus due to hydraulic
and osmotic pressure differences see
Eqn 3.4, 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
At the Glomerulus only
CLrenal = fu x GFR

10. 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, i.e. M constant, then plasma
creatinine concentration is a direct measure of kidney function via
the GFR as shown on the graph on the next slide. With creatinine about
1 mg/dL this gives a GFR = 120/ml/min

11. Creatinine Normal GFR is about 125 ml/min Failure need dialysis Urea
5 – 11 mg/100ml severe
(NH2)2CO
The liver produces urea in the urea cycle as a waste product of the digestion of protein. Normal human adult blood should contain
between 6 to 20 mg of urea nitrogen per 100 ml (6–20 mg/dL) of blood.
Creatinine (Greek: κρέας, "flesh") is a breakdown product of creatine phosphate in muscle, and is usually produced at a
fairly constant rate by the body (depending on muscle mass).

12. What is BUN?
Blood urea nitrogen (BUN) is an indication of renal (kidney) health.
The normal range is  mmol/L or 6–20 mg/dL.
The main causes of an increase in BUN are: high protein diet, decrease
in glomerular filtration rate (GFR) (suggestive of renal failure) and in
blood volume (hypovolemia), congestive heart failure, gastrointestinal
hemorrhage, fever and increased catabolism.
To convert from mg/dL of blood urea nitrogen to mmol/L of urea, multiply by
0.357 (each molecule of urea having 2 nitrogens, each of molar mass 14g/mol)
(BUN is the mass of nitrogen within urea/volume, not the mass of urea)
Urea [mmol/L] = BUN [mg/dL of nitrogen] x 10 [dL/L] / 14x2
[mg N/mmol urea] (the mass of nitrogen within urea is used)
convert BUN to urea in mg/dL by using following formula:
Urea [mg/dL]= BUN [mg/dL] * 2.14
(conversion factor derived by: MW of urea = 60,
MW of urea nitrogen = 14x2 => 60/28 = 2.14)

13. Most drugs are in the kidney are secreted and reabsorbed so
things are a bit more complicated in terms of the renal clearance in these cases
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
Renal Clearance (CLrenal) = Actual Drug removal rate C

14. Which can be integrated from the IC of t=0, C=C0 to give:
krenal is the rate constant CLrenal/Vapparent
Figure 8.4
Figure
The drug removed by the kidneys then shows up the urine and we can
write that at any time:
See pages 410 – 412 of 4th edition

15. Plasma clearance (CLplasma) relates to all possible drug elimination
pathways and would include the following:
Metabolism
Kidneys
Sweating
Bile
Respiration
Feces
See pages
Elimination rate constants
i = renal, metabolic, sweat, bile, respiration, feces
The total elimination rate constant
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.

17. Intravenous Injection of a Drug
See
Measure of drug
exposure
as time → ∞
Note that if Clrenal = Clplasma then Murine = D as time → ∞

18. Figure 8.5
See pages

19. Example 8.3 page
CSS = 873 cpm/ml
Find kte from these data

20. Figure 8.6 Measured and Predicted Inulin Concentrations
kte= /min
Css= cpm/ml
r-squared=