Since drugs can bind with proteins their distribution volume in the body can beAffected 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 canuse 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 inthe body

Bound

Unbound

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, thenthe volume of distribution is really just a “volume” when multiplied byC that gives the total amount of drug in the body at a particular time

Acidic drugsBasic drugs

Neutral drugs

testosterone

ionconcentratdrugPlasma

bodyindrugofAmountonDistributiofVolume

Apparent distribution volume linked to fluid volumes as wellas the effect of drug binding to proteins, so a drug that bindsstrongly with plasma proteins will approach a distribution volumeof 3 liters, on the other hand a drug that binds tightly with extravascularproteins will result in a very low plasma concentration and to accountfor the mass of drug that means the value of V may greatly exceed the physicalvolumes of fluid actually in the body, in the extreme could be 7000 L !

UT

URI/E

P

EPUI/EP f

fVR

V

VVfR1VV

V = apparent distribution volumeVp = plasma volume, 3 L/70kgVE = extracellular fluid volume less plasma volume, 12 L/70kgVR = remainder of the fluid volume in the bodyfU = fraction unbound in plasma, Cu/CfUT = fraction unbound in RRE/I = ratio of total drug binding sites in the fluids outside of plasma to those within the plasma, ~ 1.4

Oie – Tozer Equation

UT

URU f

fVf87V

Using typical values for Vp and VE and R/EI we get this equation

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 albumin2. 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 alcohol3. Note that as fUT 0, then V gets really big

V is in liters

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 onlyplace 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 theactive form of the drug, metabolites may also be active drugs as well, caneven be a bioactivation making them more active or toxic than the parentcompound (aka prodrug)

Metabolism is the result of enzymatic reactions that occur within the cellsand the rate of metabolism can be described by Michaelis-Menten kinetics

CkCK

CVr metabolic

mmetabolic

max

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

Basic functional unit of the kidney is theNephron

About 1 million per kidney

Blood flow to kidneys is about 1100ml/min,and the glomerulus produces (GFR) about 120ml/min of plasma filtrate, fortunatelymost of the water is reabsorbed producinga urine flow of about 1-2ml/min

Unbound drug will be filtered from the bloodflowing thru the glomerulus

GFR is the rate of filtration of plasmaacross the fenestrated capillariesof the glomerulus due to hydraulicand osmotic pressure differences, soat the glomerulus unbound drug is removed at the rate given by:

Drug removalRate = GFR x Cu = fu x GFR x C

Renal Clearance (CLrenal) = Drug removal rate CSince clearance by definition is that flowrate ofthe fluid that is totally cleared of the solute

CLrenal = fu x GFR

At the Glomerulus

So if the drug is only filtered out at the glomerulus, and all of the filtereddrug 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 renalclearance 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 orreabsorbed by the tubules and is also unbound and is a product of endogenous protein degradation, its production rate (M) in the body is about120 mg/min, so at steady state what is produced in the body has to beremoved 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 concentrationin 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 plasmacreatinine concentration is a direct measure of kidney function via the GFR as shown on the graph on the next slide

5 – 11 mg/100ml severe

Failure need dialysisNormalGFR is about125 ml/min

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).

(NH2)2CO

Urea [mg/dL]= BUN [mg/dL] * 2.14

Most drugs are secreted and reabsorbed so things are a bit morecomplicated 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 adifferent manner CLrenal > fu x GFR

Tubular reabsorption would be apparent whenever the rate of drugexcretion (CLrenal x C) is less than the rate of drug filtration (fu x GFR x C),or CLrenal < fu x GFR

Renal Clearance (CLrenal) = Drug removal rate C

C

VdC

dtCL Capparent renal

Which can be integrated from the IC oft=0, C=C0 to give:

C t C e C eCL V t k trenal apparent renal( ) ( / ) 0 0

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:

CL C Q C or CLQ C

Crenal urine urine renalurine urine

Plasma clearance (CLplasma) relates to all possible drug eliminationpathways and would include the following: Metabolism Kidneys Sweating Bile Respiration Feces

kCL

Vii

apparent

i = renal, metabolic, sweat, bile, respiration, feces

Elimination rate constants

k kl

VCL

CL

Vte ii apparent

ii

plasma

apparent

The total elimination rate constant

C t C e k tte( ) 0

Biological half life is the time needed for the plasma drugconcentration 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.

tk

V

CLte

apparent

plasma1 2

0 693 0 693/

. .

tk

V

CLte

apparent

plasma1 2

0 693 0 693/

. .

CD

Vapparent0

AUC C t dt0

0

( )

C t C e k tte( ) 0

AUCC

k

D

V k

D

CLte apparent te plasma

0 0

Measure of drugexposure

dM

dtk V Curine

renal apparent M tk C V

keurine

renal apparent

te

k tte( )

0

1

Mk C V

kD

k

kAUC CLurine

renal apparent

te

renal

terenal 0 0

Intravenous Injection of a Drug

Note that if Clrenal = Clplasma then Murine = D as time → ∞

as time → ∞

CCldt

dCV plasmaapp

VdC

dtI CL C I V k Capparent plasma apparent te 0 0

C tI

k Veo

te apparent

k tte( )

1

CI

k V

I

CLsste apparent plasma

0 0

CSS = 873 cpm/ml

Example 7.1, inulin infusion study in a laboratory rat

i 0 8 number of data points

time, minutes inulin concentration, cpm/mltimei

30

40

60

70

75

80

90

100

110

Ci

849

845

903

888

873

882

565

412

271

calculate the steady state inulin concentration CSS1

6

0

5

i

Ci

CSS 873.333 cpm/ml

perform linear regression on the data after the infusion stopped at 80 minutes assumingthe starting concentration was equal to that of Css

C5

CSS 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

tj

time5 j( )

80 the time since the infusion stopped in minutes

yj

ln C5 j( ) transform the concentration data

krenal slope t y( ) krenal 0.038 1/min, the renal elimination rate constant

C0 exp intercept t y( )( ) C0 860.085 cpm/ml, the value of the starting inulinconcentration after the infusion stops, compares quite well with the value ofCss determined above from the data

r corr t y( ) r 0.998 indicates that we have an excellent fit of thesingle compartment model that describes theelimination of inulin

time 0 30 running time, minutes

Cpred time( ) C0 exp krenal time the predictedinulin levelsafter theinfusion isstopped

0 10 20 30200

400

600

800

1 103

Inulin Elimination

time, minutes

inul

in c

once

ntra

tion,

cpm

exp y j Cpred time( )

tj time