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
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
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
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
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)
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
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
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 =0Time 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)