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SUJITHA MARY
M PHARM
ST JOSEPH COLLEGE OF PHARMACY
ONE COMPARTMENT
OPEN MODEL
KUHS QUESTONS
1)Explain flip flop model. What are various
method for estimation of absorption rate
constant. Explain the method of residuals
2)Discuss assumptions of one compartment model
3) What is meant by one compartment model
4)Explain the method of calculating volume of
distribution Vd and elimination rate constant KE
of a drugfollowing one compartment model
5)Explain the feathering technique
Simplest model
Assumptions:-
1. Body as single kinetically homogenous unit.
Drugs move dynamically in an out of this
compartment
2. This unit has no barrier against movement
of drug
3.Final distribution equilibrium between drug
in plasma and other body fluid is attained
intravenously and maintained at all times.
INTRODUCTION
This model applies only to those drugs that
distributes rapidly throughout the body
Anatomical reference=Plasma
Change in plasma drug concentration
Change in concentration through out the body
Open ==>>Input (availability) and output
(elimination) are unidirectional
•Representation
Drug Ka
Input
(Absorption)
Blood & other
body tissues
KE
Output
(Elimination)
Metab
olism
Excre
tion
Based on Plasma levels of drugs taken by
administering a single dose of a drug
• Depending on rate of input, several one
compartment open models are :
• One compartment open model, i.v. bolus
administration
• One compartment open model , continuous i.v.
infusion .
• One compartment open model, e.v.
administration, zero order absorption.
• One compartment open model, e.v.
administration, first order absorption
INTRAVENOUS BOLUS ADMINISTRATION
When drugisgivenintheformofrapidi.v.injection(i.v.
bolus)ittakesonly onetothreeminutesforcomplete
circulation andthereforetherateofabsorptionis
neglected
KE
KE
Rateofdrugpresentationtothebody
dX = Rate In – Rate out
dX = -Rate Out
dX = -KEX
KE =first order
elimination rate
constant
X= amt of drug in
body at any time t
remaining to be
eliminaed
-ve sign
dt
dt
dt
Decline in plasma drug concentration is only
due to Elimination phase
Characterized by 3 parameters:
1)Elimination rate constant
2)Elimination half life/Biological half life
It is defined as time taken for the amount of
drug in the body as well as plasma
concentration to decline by ½ or 50 % its
initial value.
It is expressed in hrs or mins
3)Clearance
Estimationofpharmacokineticparameters
Elimination rate constant & Half life
dX = - KEX
dt
Integrating above equation yields
ln X = ln X0 – KEt
Where X0 = amount of drug at time t=0
Equation can be written in exponential form
as
X= X0 e-KE t
Transforming equation into logarithm form we
get,
Log X = Log X0 - KEt / 2.303
X is directly proportional to C
X = VdC
Vd=Apparentvolume of distribution
It permit the use of C in place of X
Log C = log C0-KEt/2.303
C0=Plsama drug concentration immediately
after i.v. injection
12
Half Life
t1/2=
Elimination
Rate Constant
KE = -Slope X
2.303
Unit min-1
0.693
KE
Unit min
/hour
Half Life
t1/2=
KE
Half life is secondary parameter that depends
upon the primary parameters clearance and
volume of distribution according to following
equation,
t1/2 = 0.693 Vd
ClT
Since they are closely related to the
physiological mechanism in the body, They are
called as primary parameter
ApparentVolumeof distribution
Vd=
Measure of extent of distribution of drug and
is expressed in Liters.
Amount of drug in the body
Plasma drug Concentration
=
X
C
Estimation:
Administer drug by rapid i.v. injection & use
the following equation
Vd =
Vd can only be estimated when distribution equilibrium
is achieved between the drug in plasma and that in
tissues
Thus they can only be used for drugs that obey one
compartment kinetics
C
0
i.v. bolus doseX0
C
0
=
Non compartmental method :
The Vd obtained is almost similar to compartmental
method
For drugs that are given i.v. bolus
Vd (area)= X0/KE.AUC
For drugs that are administered extra-vascularly
Vd(area)=FX0/KE.AUC
X0 =Dose administered
F=Fraction of drug absorbed into systemic
circulation
Clearance
Clearance is defined as the theoritical volume of
body fluid containing drug from which the drug is
completely removed in a given period of time.
Clearance = Rate of elimination
Plasma drug concentration
Cl = dX / dt
Cl=
Cl=KEVd
C
KEX
C
dx/dt = KE.X
X/C = Vd
Clearance is defined as the theoritical volume of
body fluid containing drug from which the drug is
completely removed in a given period of time.
Clearance = Rate of elimination
Plasma drug concentration
Cl = dX / dt
Cl=
Cl=KEVd
Total body clearance /Total systemic clearance:
Additive property of individual organ clearance
Renal clearance ClR =
Rate of elimination by kidney
C
Hepatic clearance ClH
Other organ clearance Clothers
Total body clearance ClT = ClR +ClH + Clothers
KEVd = KeVd +KmVd + KothersVd others
Relation between clearance and half life:
ClT=KEVd
ClT= 0.693Vd/t1/2
ClR= 0.693Vd/t1/2
ClH= 0.693Vd/t1/2
Urinary excretion half life of
unchanged drug
Metabolism half life
KE= 0.693/t1/2
Organ Clearance
Rate of Elimination=Rate of Presentation –Rate of
by an organ to organ (input) exit from
organ
(output)
Input =Organ blood flow X Entering concentration
=Q.Cin
Output=Organ blood flow X Exiting concentration
=Q.Cout
Rate of elimination
/Rate of extraction = Q.Cin-Q.Cout
=Q (Cin-Cout)
(1)
Divide the equation (1) by Cin:-
Rate of extraction
Cin
= Clorgan
= Q(Cin-Cout)
Cin
= Q.ER
ER is extraction ratio : ( Cin – Cout)
Cin
ER (Extraction ratio)
Index of how efficiently an eliminating organ
clears the blood flowing through it off drug
Eg. ER= 0.6 => 60% of the blood flowing
through the organ
has no unit
Range: - 0(NO ELIMINATION) to
1(COMPLETE ELIMINATION)
Based on ER values drugs can be classified as:
Drugs with high ER = above 0.7
Drugs with intermediate ER = between 0.7-
0.3
Drugs with low ER = below 0.3
High Intermediate Low
Hepatic
Extraction
Propanolol
Lidocaine
Nitroglycerin
Morphine
Aspirin
Codeine
Nortriptyline
Quinidine
Diazepam
Phenobarbital
Phenytoin
Procainamide
Theophilline
Renal
Extraction
Penicillins
Sulfates
Glucoronides
Procainamide
Cimetidine
Digoxin
Furosemide
Atenolol
Tetracycline
Clorgan= Q.ER
Systemic availability
Fraction of drug that escapes removal by organ
F=Systemic availability
when the eliminating organ is Liver
F=1-ER
INTRAVENOUS INFUSION
Constant rate (Zero order) administration
Duration of infusion > half life of the drug
Advantage:
No peak and valley plasma level (
especially for narrow therapeutic index
drugs)
Simultaneous administration by the same
infusion line in critically ill patients.
Control of rate according to individual
patient needs
Maintenance of stable drug concentration
in the body
• Flow chart
Drug
Blood and
other body
tissues
Elimination
R0
Zero
order
infusion
rate
KE
R0
Zero
order
infusion
rate
R0 Blood and
other body
tissuesZero
order
infusion
rate
R0 Blood and
other body
tissuesZero
order
infusion
rate
R0 Blood and
other body
tissuesZero
order
infusion
rate
R0 KEBlood and
other body
tissuesZero
order
infusion
rate
R0
Elimination
KEBlood and
other body
tissuesZero
order
infusion
rate
R0
Rate of change of amount of drug in the body:-
dX
dt
=
R0 -KEX
Zero order infusion rate-
First order elimination rate
=
By integrating the equation,
R0X = (1-eKE t)
KE
X =
By integrating the equation,
X =
Since, X=Vd C the above equation can be
transformed in to concentration terms ,
C =
R0
(1-e-KE t)
KEVd
= R0
ClT
(1-e-KE t)
At steady state/Plateau/Infusion equilibrium:-
Rate of infusion = Rate of elimination
dX
Zero = R0-KEXss
KEXss = R0
dt = 0
Transforming to concentration terms and
rearranging the equation
Css = R0 = R0
KEvd ClT
Where , XSS and Css are the amount of the drug
in body and concentration of the drug in
plasma at steady state respectively.
At the start of constant rate
infusion , the amt. of drug
in the body zero and hence
, there is no elimination. As
time passes , the amt. of
the drug in the body rises
gradually until a point after
which the rate of the
elimination equals the rate
of infusion i.e. the
concentration of drug in
plasma approaches a
constant value called as
steady-state.
For therapeutic purpose more than 90% of the
steady-state concentration in the blood is
desired which is reached in 3.3 half-lives
It take 6.6 half lives for the concentration to
reach 99% of the steady state
Thus the shorter the half life (eg. Penicillin G,
30 min) sooner is the steady state reached
Semi log Plot obtained by taking data after
stopping the infusion
t
Slope=
-KE/2.303
Semi log Plot obtained by taking data during the
infusion
t
Slope=
-KE/2.303
Css
Infusion plus loading dose
It takes very long time for the drugs having
longer half lives to reach the steady state
concentration.
An i.v. loading dose is given to yield the
desired steady-state immediately upon
injection prior to starting the infusion.
It should then be followed immediately by i.v.
infusion at a rate enough to maintain this
concentration.
The equation for the plasma concentration
time profile following i.v. loading dose and
constant rate i.v. infusion,
C =X0,L e-kEt + R0 (1- e-Ket)
Vd KEVd
Amountofdrug
inthebody
Half life
Start of infusion
Exponential decline
of loading dose
Asymptotic rise
of infused drug
Loading dose
Constant plasma level
Apparent Volume of distribution
Vd=
Clearance=KEVd
R0
KE Css
EXTRAVASCULAR ADMINISTRATION
Absorption is prerequisite for its therapeutic
activity. The rate of absorption may be described
mathematically as zero-order or first-order
process.
After e.v. administration (Oral,i.m.,rectal,etc) the
rate of change in the amount of drug in the body
is given by
dx = Rate of absorption – Rate of elimination
dt
dX = dXev - dXE
dt dt dt
During absorption phase, the rate of
absorption is greater than elimination phase.
dXev > dXE
dt dt
At peak plasma concentration,
dXev = dXE
dt dt
During post absorption phase,
dXev < dXE
dt dt
Absorpton
phase
Post absorption phase
Elimination phase
t
Plasmadrugconc.
Cmax
ZERO-ORDER ABSORPTION MODEL
Similar to that for constant rate infusion
All equations that explain the plasma
concentration time profile are also applicable to
this model
Elimination
KEBlood and
other body
tissuesZero order
absorption
R0
Drug at
e.v. Site
FIRST-ORDER ABSORPTION
MODEL
Equation. dX = dXev - dXE
dt dt dt
Differentiating above equation We get,
dX = Ka Xa – KEX, Ka= absorption rate const.
dt Xa= amt of drug remaining
to be absorbed.
Drug at
e.v. site
Blood & other
body tissue Elimination
Ka
First order
absorption
KE
Integrating above
equation:
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
X =
Transferring into
concentration terms:
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =
X=
Vd
F=Fraction of drug absorbed systematically
after e.v. administration
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =  tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =  tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =  tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =C =
Vd
Assessment of Pharmacokinetic
parameters-Extra vascular
administration
Elimination Rate constant
When the absorption is complete the change
in plasma concentration depends only on
elimination rate
In log form:
KaFX0
C=
Vd (Ka-KE)
e -KEt
C=
KaFX0
Vd (Ka-KE)
KEt
2.303
-
ABSORPTION RATE CONSTANT
This can be calculated by METHOD OF
RESIDUALS.
Method is also known as Feathering, stripping
and peeling.
Drug that follows one- compartment kinetics and
administered e.v. , the concentration of drug in
plasma is expressed by biexponential equation:
If KaFX0/Ka-KE=A
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
 tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C =
Vd
A  tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C = A  tKTK
Ea
oa aE
ee
KK
FXK 

 )(
C = A
C = A e-KEt – A e-Kat
During the elimination phase, when absorption
is most over, Ka >>KE
C = A e-KEt
In log form above equation is
Log C = Log A -
Where, C = back extrapolated plasma conc. values
2.303
KEt
Subtracting true plasma conc. From
extrapolated one:
C – C = Cr = A e-Kat
log Cr = log A –
Cr=Residual conc.
Kat
2.303
50
Absorption Rate
Ka = - Slope of residual curve X 2.303
Elimination Rate
KE = - Slope of terminal line X 2.303
This method works best when difference between Ka &
KE is large (Ka/KE >3)
In some cases KE obtained after i.v. bolus of the same
drug is very large much larger than the Ka obtained by
the method f residuals (eg. Isoprenaline) and If KE/Ka >
3 , the terminal slope estimates Ka and not KE whereas
the slope of residuals line gives KE and not Ka.
This is called as flip-flop phenomenon since the slopes
of the two lines have exchanged their meanings.
Flip-Flop Phenomenon
Time Lag
Ideally the extrapolated and residual lines:
Intersect at time=0 No lag in absorption
Intersect at time not =0Time lag(t0)
Time difference between the drug
administration and start of absorption
Not onset time
Curve Fitting Method
Above method for estimation of Ka is a crve
fitting method
Suited for:
Drugs which are rapidly and completely absorbed.
Follow on compartment kinetics even when given
i.v.
Not suited for:
If absorption of drug is affected by GI motility or
enzymatic degradation
Shows multicompartment characteristic after i.v.
administration (true for virtually all drugs)
REFERENCE
1.Biopharmaceutics and Pharmacokinetics-
D.M. Brahmankar;
Page No:230-250

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one compartment open model

  • 1. Presented by: SUJITHA MARY M PHARM ST JOSEPH COLLEGE OF PHARMACY ONE COMPARTMENT OPEN MODEL
  • 2. KUHS QUESTONS 1)Explain flip flop model. What are various method for estimation of absorption rate constant. Explain the method of residuals 2)Discuss assumptions of one compartment model 3) What is meant by one compartment model 4)Explain the method of calculating volume of distribution Vd and elimination rate constant KE of a drugfollowing one compartment model 5)Explain the feathering technique
  • 3. Simplest model Assumptions:- 1. Body as single kinetically homogenous unit. Drugs move dynamically in an out of this compartment 2. This unit has no barrier against movement of drug 3.Final distribution equilibrium between drug in plasma and other body fluid is attained intravenously and maintained at all times. INTRODUCTION
  • 4. This model applies only to those drugs that distributes rapidly throughout the body Anatomical reference=Plasma Change in plasma drug concentration Change in concentration through out the body Open ==>>Input (availability) and output (elimination) are unidirectional
  • 5. •Representation Drug Ka Input (Absorption) Blood & other body tissues KE Output (Elimination) Metab olism Excre tion Based on Plasma levels of drugs taken by administering a single dose of a drug
  • 6. • Depending on rate of input, several one compartment open models are : • One compartment open model, i.v. bolus administration • One compartment open model , continuous i.v. infusion . • One compartment open model, e.v. administration, zero order absorption. • One compartment open model, e.v. administration, first order absorption
  • 7. INTRAVENOUS BOLUS ADMINISTRATION When drugisgivenintheformofrapidi.v.injection(i.v. bolus)ittakesonly onetothreeminutesforcomplete circulation andthereforetherateofabsorptionis neglected KE KE
  • 8. Rateofdrugpresentationtothebody dX = Rate In – Rate out dX = -Rate Out dX = -KEX KE =first order elimination rate constant X= amt of drug in body at any time t remaining to be eliminaed -ve sign dt dt dt
  • 9. Decline in plasma drug concentration is only due to Elimination phase Characterized by 3 parameters: 1)Elimination rate constant 2)Elimination half life/Biological half life It is defined as time taken for the amount of drug in the body as well as plasma concentration to decline by ½ or 50 % its initial value. It is expressed in hrs or mins 3)Clearance
  • 10. Estimationofpharmacokineticparameters Elimination rate constant & Half life dX = - KEX dt Integrating above equation yields ln X = ln X0 – KEt Where X0 = amount of drug at time t=0 Equation can be written in exponential form as X= X0 e-KE t
  • 11. Transforming equation into logarithm form we get, Log X = Log X0 - KEt / 2.303 X is directly proportional to C X = VdC Vd=Apparentvolume of distribution It permit the use of C in place of X Log C = log C0-KEt/2.303 C0=Plsama drug concentration immediately after i.v. injection
  • 12. 12 Half Life t1/2= Elimination Rate Constant KE = -Slope X 2.303 Unit min-1 0.693 KE Unit min /hour Half Life t1/2= KE
  • 13. Half life is secondary parameter that depends upon the primary parameters clearance and volume of distribution according to following equation, t1/2 = 0.693 Vd ClT Since they are closely related to the physiological mechanism in the body, They are called as primary parameter
  • 14. ApparentVolumeof distribution Vd= Measure of extent of distribution of drug and is expressed in Liters. Amount of drug in the body Plasma drug Concentration = X C
  • 15. Estimation: Administer drug by rapid i.v. injection & use the following equation Vd = Vd can only be estimated when distribution equilibrium is achieved between the drug in plasma and that in tissues Thus they can only be used for drugs that obey one compartment kinetics C 0 i.v. bolus doseX0 C 0 =
  • 16. Non compartmental method : The Vd obtained is almost similar to compartmental method For drugs that are given i.v. bolus Vd (area)= X0/KE.AUC For drugs that are administered extra-vascularly Vd(area)=FX0/KE.AUC X0 =Dose administered F=Fraction of drug absorbed into systemic circulation
  • 17. Clearance Clearance is defined as the theoritical volume of body fluid containing drug from which the drug is completely removed in a given period of time. Clearance = Rate of elimination Plasma drug concentration Cl = dX / dt Cl= Cl=KEVd C KEX C dx/dt = KE.X X/C = Vd Clearance is defined as the theoritical volume of body fluid containing drug from which the drug is completely removed in a given period of time. Clearance = Rate of elimination Plasma drug concentration Cl = dX / dt Cl= Cl=KEVd
  • 18. Total body clearance /Total systemic clearance: Additive property of individual organ clearance Renal clearance ClR = Rate of elimination by kidney C Hepatic clearance ClH Other organ clearance Clothers Total body clearance ClT = ClR +ClH + Clothers KEVd = KeVd +KmVd + KothersVd others
  • 19. Relation between clearance and half life: ClT=KEVd ClT= 0.693Vd/t1/2 ClR= 0.693Vd/t1/2 ClH= 0.693Vd/t1/2 Urinary excretion half life of unchanged drug Metabolism half life KE= 0.693/t1/2
  • 20. Organ Clearance Rate of Elimination=Rate of Presentation –Rate of by an organ to organ (input) exit from organ (output) Input =Organ blood flow X Entering concentration =Q.Cin Output=Organ blood flow X Exiting concentration =Q.Cout Rate of elimination /Rate of extraction = Q.Cin-Q.Cout =Q (Cin-Cout) (1)
  • 21. Divide the equation (1) by Cin:- Rate of extraction Cin = Clorgan = Q(Cin-Cout) Cin = Q.ER ER is extraction ratio : ( Cin – Cout) Cin
  • 22. ER (Extraction ratio) Index of how efficiently an eliminating organ clears the blood flowing through it off drug Eg. ER= 0.6 => 60% of the blood flowing through the organ has no unit Range: - 0(NO ELIMINATION) to 1(COMPLETE ELIMINATION) Based on ER values drugs can be classified as: Drugs with high ER = above 0.7 Drugs with intermediate ER = between 0.7- 0.3 Drugs with low ER = below 0.3
  • 24. Systemic availability Fraction of drug that escapes removal by organ F=Systemic availability when the eliminating organ is Liver F=1-ER
  • 25. INTRAVENOUS INFUSION Constant rate (Zero order) administration Duration of infusion > half life of the drug Advantage: No peak and valley plasma level ( especially for narrow therapeutic index drugs) Simultaneous administration by the same infusion line in critically ill patients. Control of rate according to individual patient needs Maintenance of stable drug concentration in the body
  • 26. • Flow chart Drug Blood and other body tissues Elimination R0 Zero order infusion rate KE R0 Zero order infusion rate R0 Blood and other body tissuesZero order infusion rate R0 Blood and other body tissuesZero order infusion rate R0 Blood and other body tissuesZero order infusion rate R0 KEBlood and other body tissuesZero order infusion rate R0 Elimination KEBlood and other body tissuesZero order infusion rate R0
  • 27. Rate of change of amount of drug in the body:- dX dt = R0 -KEX Zero order infusion rate- First order elimination rate = By integrating the equation, R0X = (1-eKE t) KE X = By integrating the equation, X =
  • 28. Since, X=Vd C the above equation can be transformed in to concentration terms , C = R0 (1-e-KE t) KEVd = R0 ClT (1-e-KE t)
  • 29. At steady state/Plateau/Infusion equilibrium:- Rate of infusion = Rate of elimination dX Zero = R0-KEXss KEXss = R0 dt = 0
  • 30. Transforming to concentration terms and rearranging the equation Css = R0 = R0 KEvd ClT Where , XSS and Css are the amount of the drug in body and concentration of the drug in plasma at steady state respectively.
  • 31. At the start of constant rate infusion , the amt. of drug in the body zero and hence , there is no elimination. As time passes , the amt. of the drug in the body rises gradually until a point after which the rate of the elimination equals the rate of infusion i.e. the concentration of drug in plasma approaches a constant value called as steady-state.
  • 32. For therapeutic purpose more than 90% of the steady-state concentration in the blood is desired which is reached in 3.3 half-lives It take 6.6 half lives for the concentration to reach 99% of the steady state Thus the shorter the half life (eg. Penicillin G, 30 min) sooner is the steady state reached
  • 33. Semi log Plot obtained by taking data after stopping the infusion t Slope= -KE/2.303
  • 34. Semi log Plot obtained by taking data during the infusion t Slope= -KE/2.303 Css
  • 35. Infusion plus loading dose It takes very long time for the drugs having longer half lives to reach the steady state concentration. An i.v. loading dose is given to yield the desired steady-state immediately upon injection prior to starting the infusion. It should then be followed immediately by i.v. infusion at a rate enough to maintain this concentration.
  • 36. The equation for the plasma concentration time profile following i.v. loading dose and constant rate i.v. infusion, C =X0,L e-kEt + R0 (1- e-Ket) Vd KEVd
  • 37. Amountofdrug inthebody Half life Start of infusion Exponential decline of loading dose Asymptotic rise of infused drug Loading dose Constant plasma level
  • 38. Apparent Volume of distribution Vd= Clearance=KEVd R0 KE Css
  • 39. EXTRAVASCULAR ADMINISTRATION Absorption is prerequisite for its therapeutic activity. The rate of absorption may be described mathematically as zero-order or first-order process. After e.v. administration (Oral,i.m.,rectal,etc) the rate of change in the amount of drug in the body is given by dx = Rate of absorption – Rate of elimination dt dX = dXev - dXE dt dt dt
  • 40. During absorption phase, the rate of absorption is greater than elimination phase. dXev > dXE dt dt At peak plasma concentration, dXev = dXE dt dt During post absorption phase, dXev < dXE dt dt
  • 42. ZERO-ORDER ABSORPTION MODEL Similar to that for constant rate infusion All equations that explain the plasma concentration time profile are also applicable to this model Elimination KEBlood and other body tissuesZero order absorption R0 Drug at e.v. Site
  • 43. FIRST-ORDER ABSORPTION MODEL Equation. dX = dXev - dXE dt dt dt Differentiating above equation We get, dX = Ka Xa – KEX, Ka= absorption rate const. dt Xa= amt of drug remaining to be absorbed. Drug at e.v. site Blood & other body tissue Elimination Ka First order absorption KE
  • 44. Integrating above equation:  tKTK Ea oa aE ee KK FXK    )( X = Transferring into concentration terms:  tKTK Ea oa aE ee KK FXK    )( C = X= Vd F=Fraction of drug absorbed systematically after e.v. administration  tKTK Ea oa aE ee KK FXK    )( C =  tKTK Ea oa aE ee KK FXK    )( C =  tKTK Ea oa aE ee KK FXK    )( C =  tKTK Ea oa aE ee KK FXK    )( C =C = Vd
  • 45. Assessment of Pharmacokinetic parameters-Extra vascular administration Elimination Rate constant When the absorption is complete the change in plasma concentration depends only on elimination rate In log form: KaFX0 C= Vd (Ka-KE) e -KEt C= KaFX0 Vd (Ka-KE) KEt 2.303 -
  • 46. ABSORPTION RATE CONSTANT This can be calculated by METHOD OF RESIDUALS. Method is also known as Feathering, stripping and peeling.
  • 47. Drug that follows one- compartment kinetics and administered e.v. , the concentration of drug in plasma is expressed by biexponential equation: If KaFX0/Ka-KE=A  tKTK Ea oa aE ee KK FXK    )(  tKTK Ea oa aE ee KK FXK    )( C =  tKTK Ea oa aE ee KK FXK    )(  tKTK Ea oa aE ee KK FXK    )(  tKTK Ea oa aE ee KK FXK    )( C = Vd A  tKTK Ea oa aE ee KK FXK    )( C = A  tKTK Ea oa aE ee KK FXK    )( C = A C = A e-KEt – A e-Kat
  • 48. During the elimination phase, when absorption is most over, Ka >>KE C = A e-KEt In log form above equation is Log C = Log A - Where, C = back extrapolated plasma conc. values 2.303 KEt
  • 49. Subtracting true plasma conc. From extrapolated one: C – C = Cr = A e-Kat log Cr = log A – Cr=Residual conc. Kat 2.303
  • 50. 50
  • 51. Absorption Rate Ka = - Slope of residual curve X 2.303 Elimination Rate KE = - Slope of terminal line X 2.303
  • 52. This method works best when difference between Ka & KE is large (Ka/KE >3) In some cases KE obtained after i.v. bolus of the same drug is very large much larger than the Ka obtained by the method f residuals (eg. Isoprenaline) and If KE/Ka > 3 , the terminal slope estimates Ka and not KE whereas the slope of residuals line gives KE and not Ka. This is called as flip-flop phenomenon since the slopes of the two lines have exchanged their meanings. Flip-Flop Phenomenon
  • 53. Time Lag Ideally the extrapolated and residual lines: Intersect at time=0 No lag in absorption Intersect at time not =0Time lag(t0) Time difference between the drug administration and start of absorption Not onset time
  • 54. Curve Fitting Method Above method for estimation of Ka is a crve fitting method Suited for: Drugs which are rapidly and completely absorbed. Follow on compartment kinetics even when given i.v. Not suited for: If absorption of drug is affected by GI motility or enzymatic degradation Shows multicompartment characteristic after i.v. administration (true for virtually all drugs)