Pharmacokinetics of Drug absorption
Drug absorption and factors depends on it:
Entry of something in to the blood stream from any part of the body is called systemic
absorption. It may occur from the GIT, muscle, skin and other parts of the body.
The systemic drug absorption from the gastrointestinal tract or from any other extravascular
site is dependent on
a. The physicochemical properties of the drug,
b. The dosage form used,
c. The anatomy and physiology of the absorption site.
For oral dosing, some factors are also considered,
a. surface area of the GI tract,
b. stomach emptying rate,
c. GI mobility,
d. blood flow to the absorption site
These are all affects the rate and the extent of drug absorption.
In pharmacokinetics, the overall rate of drug absorption may be described as either a firstorder or zero-order process. Most pharmacokinetic models assume first-order absorption
unless an assumption of zero-order absorption improves the model significantly or has been
verified experimentally.
𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛
DGI →−−−−−−−→
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛
DBVD →−−−−−−−→
The rate of change in the amount of drug in the body
𝑑𝐷𝐵
𝑑𝑡
DE
is dependent on the relative rates
of drug absorption and elimination. The net rate of drug accumulation in the body at any
time is equal to the rate of drug absorption less the rate of drug elimination, regardless of
whether absorption is zero-order or first order.
𝑑𝐷𝐵 𝑑𝐷𝐺𝐼 𝑑𝐷𝐸
=
−
𝑑𝑡
𝑑𝑡
𝑑𝑡
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1
Where DGI is amount of drug in the gastrointestinal tract and DE is amount of drug
eliminated.
Pharm acokinetics of Drug absorption
Kushal
Plasma-level time curve:
A plasma level time curve showing drug adsorption
and elimination rate processes is given in following
figure.
During the absorption phase of a plasma level time
curve, the rate of drug absorption is greater than
the rate of drug elimination. During the absorption
phase, elimination occurs whenever drug is present
in the plasma, even though absorption
predominates.
𝑑𝐷𝐺𝐼 𝑑𝐷𝐸
>
𝑑𝑡
𝑑𝑡
At the peak drug concentration in the plasma the rate of drug absorption just equals the
rate of drug elimination, and there is no net change in the amount of drug in the body.
𝑑𝐷𝐺𝐼 𝑑𝐷𝐸
=
𝑑𝑡
𝑑𝑡
Immediately after the time of peak drug absorption, some drug may still be at the
absorption site. However, the rate of drug elimination at this time is faster than the rate of
absorption, as represented by the post-absorption phase.
𝑑𝐷𝐺𝐼 𝑑𝐷𝐸
<
𝑑𝑡
𝑑𝑡
When the drug at the absorption site becomes depleted, the rate of drug absorption
approaches zero, i.e.
𝑑𝐷𝘎𝐼
𝑑𝑡
= 0 . The plasma level time curve then represents only the
elimination of drug from the body, usually a first-order process. Therefore, during the
elimination phase the rate of change in the amount of drug in the body is described as a
first-order process,
𝑑𝐷𝐺𝐼
= −𝑘𝐷𝐵
𝑑𝑡
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2
Where, k is the first-order elimination rate constant.
Pharm acokinetics of Drug absorption
Kushal
Zero order absorption models:
In this model, drug in the gastrointestinal tract, DGI, is absorbed systemically at a constant
rate, k0. Drug is simultaneously and immediately eliminated from the body by a first-order
rate process defined by a first-order rate constant, k.
The rate of first-order elimination at any time is equal to DBk. The rate of input is simply k0.
Therefore, the net change per unit time in the body can be expressed as,
𝑑𝐷𝐵
= 𝑘0 − 𝑘𝐷𝐵
𝑑𝑡
𝐾𝑜
𝐾
DGI → DBVD → DE
Integration of this equation with substitution of VDCP for DB produces,
𝑑𝑉 𝐷 𝐶 𝑃
𝑑𝑡
Or,
So,
= 𝑘0 − 𝑘𝑉𝐷 𝐶𝑃
𝑑𝐶 𝑃
𝑑𝑡
=
𝐶𝑃 =
𝑘0
𝑉𝐷
𝑘0
𝑉𝐷𝑘
− 𝑘𝑉𝐷 𝐶𝑃
(1 − 𝑒−𝑘𝑡)
The rate of drug absorption is constant until the amount of drug in the gut, D GI, is depleted.
The time for complete drug absorption to occur is equal to DGI/k0. After this time, the drug is
no longer available for absorption from the gut. The drug concentration in the plasma
subsequently declines in accordance with a first-order elimination rate process.
First order absorption models:
This model assumes a first-order input across the gut wall and first-order elimination from
the body.
𝐾𝑜
𝐾
DGI → DBVD → DE
After administration of a drug it is absorbed from the absorption site by 1st order process. In
the case of a drug given orally, the dosage form first disintegrates if it is given as a solid,
then the drug dissolves into the fluids of the GI tract. Only drug in solution is absorbed into
the body. The rate of disappearance of drug from the gastrointestinal tract is described by
Pharm acokinetics of Drug absorption
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3
𝑑𝐷𝐺𝐼
= −𝑘𝑎𝐷𝐺𝐼 𝐹
𝑑𝑡
Kushal
Where,
ka
= first-order absorption rate constant from the GI tract,
F
= fraction absorbed,
DGI
= amount of drug in solution in the GI tract at any time t.
Integration of the differential equation gives,
𝑑𝐷𝐺𝐼
= 𝐷0 𝑒 −𝑘𝑎 𝑡
𝑑𝑡
Where, D0 is the dose of the drug.
The rate of drug elimination is described by a first-order rate process for most drugs and is
equal to kDB. The rate of drug change in the body, dDB/dt, is therefore the rate of drug in,
minus the rate of drug out as given by the differential equation,
Or,
𝑑𝐷𝐵
𝑑𝑡
𝑑𝐷 𝐵
𝑑𝑡
= 𝑟𝑎𝑡𝑒 𝑖𝑛 − 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡
= 𝐹𝑘𝑎𝐷𝐺𝐼 − 𝑘𝐷𝐵
Where F is the fraction of drug absorbed systemically. Since the drug in the gastrointestinal
tract also follows a first-order, the amount of drug in the gastrointestinal tract at any time t
is equal to 𝐷0𝑒−𝐾𝑎𝑡
𝑑𝐷𝐵
= 𝐹𝑘𝑎 𝐷0 𝑒 −𝐾𝑎𝑡 − 𝑘𝐷𝐵
𝑑𝑡
The value of F may vary from 1 for a fully absorbed drug to 0 for a drug that is completely
unabsorbed.
𝑑𝑉𝐷 𝐶𝑃
= 𝐹𝑘𝑎 𝐷0 𝑒 −𝐾𝑎𝑡 − 𝑘𝑉𝐷 𝐶𝑃
𝑑𝑡
𝑑𝐶𝑃 𝐹𝑘𝑎𝐷0𝑒−𝐾𝑎𝑡
=
− 𝑘𝐶𝑃
𝑑𝑡
𝑉𝐷
Integrating this equation,
𝐶𝑃 =
𝐹𝑘𝑎𝐷0
(𝑒𝑘𝑡 − 𝑒−𝐾𝑎𝑡)
𝑉𝐷(𝑘𝑎 − 𝑘)
Thus, we can calculate 𝐷𝐵from this equation,
𝐹𝑘𝑎𝐷0
(𝑒𝑘𝑡 − 𝑒−𝐾𝑎𝑡)
(𝑘𝑎 − 𝑘)
𝐹𝑘𝑎𝐷0
(𝑒𝑘𝑡 − 𝑒−𝐾𝑎𝑡)
𝐷𝐵 =
(𝑘𝑎 − 𝑘)
Pharm acokinetics of Drug absorption
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4
𝐶𝑃 𝑉𝐷 =
Kushal
Determination of Cmax and tmax:
Prove that tmax is depended on the rate constant Ka and the elimination rate, k of the drug
not the dose. We know,
𝐶𝑃 =
𝐹𝑘𝑎𝐷0
(𝑒𝑘𝑡 − 𝑒−𝐾𝑎𝑡)
𝑉𝐷(𝑘𝑎 − 𝑘)
After administration of a single dose the time needed to reach maximum concentration is
known as tmax. It is depended on the rate constant and elimination rate of the drug from the
body.
𝐶𝑚𝑎𝑥 =
𝐹𝑘𝑎𝐷0
(𝑒 𝑘𝑡 𝑚𝑎𝑥 − 𝑒 −𝐾 𝑎 𝑡 𝑚𝑎𝑥 )
𝑉𝐷(𝑘𝑎 − 𝑘)
At Cmax the rate of drug absorbed is equal to the rate of drug elimination. Therefore the net
rate of concentration change is equal to zero. We know,
𝐹𝑘𝑎𝐷0
(𝑒𝑘𝑡 − 𝑒−𝐾𝑎𝑡)
𝑉𝐷(𝑘𝑎 − 𝑘)
𝑑𝐶𝑃
𝐹𝑘𝑎𝐷0
=
(−𝑘𝑒−𝑘𝑡 + 𝑘 𝑎 𝑒−𝐾𝑎𝑡)
𝑑𝑡
𝑉𝐷(𝑘𝑎 − 𝑘)
𝐶𝑃 =
At Cmax,
(−𝑘𝑒 −𝑘𝑡 + 𝑘𝑎 𝑒 −𝐾𝑎𝑡 ) = 0
Or,
−𝑘𝑒−𝑘𝑡 + 𝑘𝑎𝑒−𝐾𝑎𝑡 = 0
Or,
𝑘𝑒−𝑘𝑡 = 𝑘𝑎𝑒−𝐾𝑎𝑡
Or,
ln 𝑘 − 𝑘𝑡 = ln 𝑘𝑎 − 𝑘𝑎𝑡
Or,
𝑘𝑎 𝑡 − 𝑘𝑡 = ln
Or,
𝑡(𝑘 𝑎 − 𝑘) = ln
Or,
𝑡=
So,
𝑡 = (𝑘
log
𝑘𝑎
So,
𝑡𝑚𝑎𝑥 =
1
ln
(𝑘𝑎−𝑘)
2.303
𝑎−𝑘)
ln
𝑘
𝑘𝑎
𝑘
𝑘𝑎
𝑘
(𝑘𝑎−𝑘)
𝑘
𝑘𝑎
𝑘
[Here t= tmax]
It is clear from this equation that tmax only dependent on rate constant ka and elimination
constant k but not the dose.
Pharm acokinetics of Drug absorption
Kushal
5
1
𝑘𝑎
Page
𝐹𝑘 𝑎 𝐷0
𝑉𝐷(𝑘𝑎−𝑘)
Determination of 1st order elimination constant k from elimination phase of
plasma level time curve:
The first-order elimination rate constant may be determined from the elimination phase of
the plasma level time curve. At later time intervals, when drug absorption has been
completed, i.e. 𝑒−𝑘𝑎𝑡 = 0. The equation from plasma drug concentration at any time t
reduces to,
𝐶𝑃 =
𝐹𝑘𝑎𝐷0
𝑉𝐷(𝑘𝑎−𝑘)
𝑒𝑘𝑡
Taking log we get,
ln 𝐶𝑃 = ln
𝑙𝑜𝑔𝐶𝑃 =
𝐹𝑘𝑎𝐷0
− 𝑘𝑡
𝑉𝐷 (𝑘𝑎 − 𝑘)
𝐹𝑘𝑎𝐷0
𝑘𝑡
−
𝑉𝐷 (𝑘𝑎 − 𝑘) 2.3
𝑙𝑜𝑔𝐶𝑃 = −
𝑘𝑡
𝐹𝑘𝑎𝐷0
+
2.3 𝑉𝐷 (𝑘𝑎 − 𝑘)
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6
With this equation, a graph constructed by plotting log C p versus time will yield a straight
line with a slope of –k/2.3.
Pharm acokinetics of Drug absorption
Kushal
Determination of absorption rate constant or ka from oral absorption data:
Determination of absorption rate constant by residual method:
Determination of absorption rate constant by Feathering technique:
After oral administration of a drug concentration of drug at
any time,
𝐹𝑘𝑎𝐷0
(𝑒𝑘𝑡 − 𝑒−𝐾𝑎𝑡)
𝑉𝐷(𝑘𝑎 − 𝑘)
𝐶𝑃 =
Assuming k a >> k, the value for the second exponential will
become insignificantly small with time (i.e. 𝑒−𝑘𝑎𝑡 = 0) and
can therefore be omitted. When this is the case, drug
absorption is virtually complete. Then the equation becomes,
𝐶𝑃 =
𝐹𝑘𝑎𝐷0
𝑒𝑘𝑡
𝑉𝐷(𝑘𝑎 − 𝑘)
Taking log we get,
ln 𝐶𝑃 = ln
𝑙𝑜𝑔𝐶𝑃 =
𝐹𝑘𝑎𝐷0
𝑉𝐷(𝑘𝑎−𝑘)
𝐹𝑘𝑎𝐷0
𝑉𝐷(𝑘𝑎−𝑘)
𝑙𝑜𝑔𝐶𝑃 = −
𝑘𝑡
2.3
+
−
− 𝑘𝑡
𝑘𝑡
2.3
𝐹𝑘𝑎𝐷0
𝑉𝐷(𝑘𝑎−𝑘)
This is the equation of a straight line. A plot of Cp versus time gives a straight line with slope
− 𝑘𝑡 and intercept
2.3
𝐹𝑘𝑎𝐷0
𝑉𝐷(𝑘𝑎−𝑘)
.
𝑥1´ − 𝑥1 = ∆1
𝑥2´ − 𝑥2 = ∆2
𝑥3´ − 𝑥3 = ∆3
The value of ka can be obtained by using the method of residual or feathering technique.
The value of ka is obtained by the following procedure:
a. Plot the drug concentration versus time on semi-log paper with the concentration values
on the logarithmic axis.
b. Obtain the slope of the terminal phase (line BC) by extrapolation.
c. Take any points on the upper part of line BC (e.g. x'1, x'2, x’3,) and drop vertically to
obtain corresponding points on the curve (e.g. x1, x2, x3,).
d. Plot the concentration values at x1 and x'1, x2 and x'2, x3and x'3 and so on. Plot the values
obtained with a slope of − 𝑘𝑡 . From the slope we can calculate the value of ka.
Pharm acokinetics of Drug absorption
Page
2.3
7
of the differences at the corresponding time points 1, 2, 3 . . . . A straight line will be
Kushal
Precautions:
When using the method of residuals, a minimum of three points should be used to
define the straight line.
Data points occurring shortly after t max may not be accurate, because drug
absorption is still continuing at that time.
Data points from the elimination phase should be used.
Determination of ka by plotting percent of drug unabsorbed Vs time curve:
Determination of ka by Wagner-Nelson method:
After a single oral dose of a drug, the total dose should be completely accounted for in the
amount present in the body, the amount present in the urine, and the amount present in
the GI tract. Therefore, dose Do is expressed as follows:
𝐷𝑜 = 𝐷𝐺𝐼𝑇 + 𝐷𝐵 + 𝐷𝑈
Let Ab = DB + DU = amount of drug absorbed.
let 𝐴𝑏∝= amount of drug absorbed at t = a. At any given time the fraction of drug absorbed
is Ab/𝐴𝑏∝ and the fraction of drug unabsorbed is 1 − 𝐴𝑏/𝐴𝑏∝.
The amount of drug excreted at any time t can be calculated as
𝐷𝑈 = 𝑘𝑉𝐷[𝐴𝑈𝐶]𝑡0
The amount of drug in the body DB at any time, DB = CPVD. At any time t, the amount of drug
absorbed (Ab) is,
𝐴𝑏 = 𝐶𝑃𝑉𝐷 + 𝑘𝑉𝐷 [𝐴𝑈𝐶]𝑡0
At t =∝, 𝐶𝑃∝= 0 and the total amount of drug absorbed is (Ab) is,
𝐴𝑏∝ = 0 + 𝑘𝑉𝐷[𝐴𝑈𝐶]∝0
The fraction of drug absorbed at any time is,
[𝐶𝑃𝑉𝐷 + 𝑘𝑉𝐷 [𝐴𝑈𝐶]𝑡0]
𝐴𝑏
=
𝐴𝑏∝
𝑘𝑉𝐷[𝐴𝑈𝐶]∝0
[𝐶 + 𝑘[𝐴𝑈𝐶]0𝑡 ]
𝐴𝑏
= 𝑃
𝐴𝑏∝
𝑘[𝐴𝑈𝐶]∝0
The fraction unabsorbed at any time t is = 1 − 𝐴𝑏
Pharm acokinetics of Drug absorption
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Page
The fraction unabsorbed at any time t is = 1 −
𝐴𝑏∝
[𝐶𝑃+𝑘[𝐴𝑈𝐶]𝑡0 ]
𝑘[𝐴𝑈𝐶]∝0
Kushal
The drug remaining in the GIT at any time t is, 𝐷𝐺𝐼 = 𝐷0𝑒−𝑘𝑎𝑡
Therefore the fraction of drug remaining is,
𝐷 𝘎𝐼
= 𝑒−𝑘𝑎𝑡
𝐷0
Or,
𝐷𝘎𝐼
𝐷0
𝐴𝑏
=1−
So, 1 −
𝐴𝑏
= 𝑒−𝑘𝑎𝑡
𝐴𝑏∝
Or, 𝑙𝑛[1 −
𝐴𝑏
]
𝐴𝑏∝
𝐴𝑏
So, 𝑙𝑜𝑔[1 −
A plot of 1 −
𝐴𝑏∝
= −𝑘 𝑎 𝑡
]=
∝
𝐴𝑏
𝐴𝑏
𝐴𝑏∝
−𝑘𝑎𝑡
2.3
versus time curve gives –ka/2.3 as the slope.
The following steps should be useful in determination of ka:
a.
b.
c.
d.
e.
f.
Plot log concentration of drug versus time.
Find k from the terminal part of the slope when the slope = -k/2.3.
Find [𝐴𝑈𝐶]𝑡0 by plotting Cp versus t.
Find k[𝐴𝑈𝐶]𝑡 by multiplying each [𝐴𝑈𝐶]𝑡 by k.
0
Find [𝐴𝑈𝐶]∝0
by adding up all the [AUC] pieces, from t = 0 to t = ∝.
Determine the 1 −
g. Plot 1 −
𝐴𝑏
𝐴𝑏∝
0
𝐴𝑏
𝐴𝑏∝
value corresponding to each time point t.
versus time on semi-log paper, with 1 −
If the fraction of the unabsorbed 1 −
𝐴𝑏
,
𝐴𝑏∝
𝐴𝑏
𝐴𝑏∝
on the logarithmic axis.
gives a linear regression line on the semilog
graph, then the rate of drug absorption dDGI/dt is a first order process.
Estimation of ka from urinary excretion data:
The absorption rate constant may also be estimated from urinary excretion data, using a
plot of percent of drug unabsorbed versus time. For a one-compartment model total
amount of drug absorption:
Pharm acokinetics of Drug absorption
Kushal
Page
Ab = total amount of drug absorbed, that is the amount of drug present in the body and
amount of drug excreted,
DB = amount of drug in the body,
DE = amount of unchanged drug excreted in the urine.
9
AB = DB + DE
The differential of equation with respect to time gives,
𝑑𝐴𝑏 𝑑𝐷𝐵 𝑑𝐷𝐸
=
+
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝐴𝑏 𝑑𝑉𝐷𝐶𝑃 𝑑𝐷𝐸
=
+
𝑑𝑡
𝑑𝑡
𝑑𝑡
Assuming first-order elimination kinetics with renal elimination constant ke,
𝑑𝐷 𝑈
𝑑𝑡
= 𝑘𝑒𝐷𝐵 = 𝑘𝑒𝑉𝐷𝐶𝑃
1
𝐶𝑃 =
𝑑𝐶 𝑃
𝑑𝑡
𝑑𝐷𝑈
𝑘 𝑒 𝐷 𝐵 𝑑𝑡
=
𝑑(
𝑑𝐷 𝑈
)
𝑑𝑡
𝑑𝑡 𝑘 𝑒 𝐷 𝐵
Substituting for dCp/dt into equation and kDu/ke for DE,
𝑑𝐷
𝑑( 𝑈)
𝑑𝐴𝑏
𝑑𝑡 + 𝑘 𝑑𝐷𝑈
= 𝑉𝐷
𝑑𝑡
𝑑𝑡 𝑘𝑒𝐷𝐵 𝑘𝑒 𝑑𝑡
This equation is integrated from the zero to time t,
𝐴𝑏𝑡 =
1 𝑑𝐷𝑈 𝑡 + 𝑘 (𝐷 𝑡)
𝑈
𝑘𝑒 𝑑𝑡
𝑘𝑒
At time t = ∝, all of the drug is ultimately absorbed is expressed as 𝐴𝑏∝ and dDu/dt = 0, the
total amount of drug absorbed is ,
𝐴𝑏 ∝ =
𝑘 ∝
𝐷
𝑘𝑒 𝑈
Fraction of the drug absorbed at any time t is equal to the amount of drug absorbed at this
time.
𝐴𝑏∝
Or,
𝐴𝑏𝑡
𝐴𝑏∝
=
=
1 𝑑𝐷 𝑈
𝑡+ 𝑘 (𝐷𝑈 𝑡)
𝑘 𝑒 𝑑𝑡
𝑘𝑒
𝑘 ∝
𝐷
𝑘𝑒 𝑈
𝑑𝐷 𝑈
𝑡+𝑘(𝐷 𝑈 𝑡)
𝑑𝑡
𝑘𝐷∝
𝑈
Fraction of drug unabsorbed,
1−
𝐴𝑏𝑡
𝐴𝑏∝
=1−
𝑑𝐷 𝑈
𝑡+𝑘(𝐷 𝑈 𝑡)
𝑑𝑡
𝑘𝐷∝𝑈
Page
A plot of the fraction of drug unabsorbed, 1 − 𝐴𝑏𝑡
, versus time gives -ka/2.3 as the slope
𝐴𝑏∝
from which the absorption rate constant is obtained.
10
𝐴𝑏𝑡
Pharm acokinetics of Drug absorption
Kushal
When urinary excretion data is used to determine Ke?
Urinary excretion data is used to determine Ke, only when the volume of distribution of drug
is very high than the plasma drug concentration.
Why Wagner-Nelson methods are used?
When the drug is given by oral route, only the Wagner—Nelson method may represents one
compartment model may be used to provide initial estimation of Ka.
Significance of absorption rate constant:
a. The faster the absorption the greater the peak concentration of drug in plasma and
shorter the time required after addition to peak drug level. Thus addition of an equal
dosage of 3 different dosage form results 3 different types of curve.
b. Absorption rate should be higher than the elimination rate and ka is as much as 5 times
greater than ke.
c. If ka is slow or it is lower than ke. The MEC of drug is not attending to the plasma and at
the site of action.
d. The more the rapid the absorption rate faster the onset of action and more intense
pharmacological action.
e. An active drug may appear to be inactive by adding in the form that results in slow or
incomplete absorption. So absorption rate is important in complete absorption.
f. The absorption rate of drug sometimes influences the frequency and severity of side
effect produced by certain orally administered drug such as tetracycline.
g. Rapid absorption decreases the influence of biological variables such as gastric emptying
rates, intestinal motility or GI content.
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11
h. The calculation of ka is useful in designing a multiple dosage regimen.
Pharm acokinetics of Drug absorption
Kushal