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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 - PowerPoint PPT Presentation
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Since drugs can bind with proteins their distribution volume in the body can beaffected
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
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
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
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, ie. M constant, then plasmacreatinine is a direct measure of kidney function via the GFR as shownon the graph on the next slide
5 – 11 mg/100ml severe
Failure need dialysis
NormalGFR is about125 ml/min
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
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/
. .
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
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
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