kinetics exam 2 - oral absorption Flashcards
Two formulations of the same drug, A and B, have different absorption rates. A has a faster absorption rate than B. B will arrive at peak concentration in a shorter time.
A
True
B
False
false because since A has a faster absorption time it will reach the peak concentration sooner
they are the same drug so they have the same elimination rate tho
ABSORPTION SITES SUCH AS
G-I TRACT
what different variables mean
A—Measured form of Drug
B—Moiety Chemically Appended to A
A + B—Molecular form of Drug
S = A/(A+B)
Dgi = A + B
F = fraction of drug absorbed
for IV, the F is always 1 because all of the drug goes to the blood and is absorbed
Model for Absorption from Extravascular Sites: Two Compartments
Absorption: mostly first-order, but can be zero-order.
central compartment (Cp, VD, DP) k goes out
Tissue compartment (Ct, Vt, Dt)
k12 and k21 goes between the central and tissues compartments
Model for Absorption from Extravascular Sites: One Compartment
- Anatomy and physiology of absorption site
- Physico-chemical properties of drug
- Dosage form used
absorption goes in
DB, VD
elimination goes out
Model for Absorption from Extravascular Sites II
Anatomy and Physiology: factors such as GI motility, and surface area for absorption
- Dosage Form Used: for instance if solid, needs to first disintegrate, then dissolve in GI fluids before absorption is possible
- Physico-chemical properties of drug: e.g. lipid solubility, size etc…
Oral Absorption: Cp against t Curve
absorption Phase: Rate of absorption > rate of elimination
- the graph line goes up
Cmax: peak concentration
- rate of absorption = rate of elimination
Post-Absorption Phase:
Rate of absorption < rate of elimination
- line begins to go down
Elimination Phase: Rate of absorption = 0
- line steadily goes down
Importance of Parameters
ka (the first-order rate constant for absorption) and k make it possible to determine peak and trough plasma concentrations during multiple dosing
- Changes in ka and k affect the Cmax, tmax, and AUC. If ka increases at a steady k, Cmax increases and tmax decreases while AUC remains the same. However at a steady ka, if k increases, Cmax, tmax, and AUC all decrease.
- ka, k, and tmax are useful in determining the bioequivalence of chemically equivalent products
makes sense because ka is absorption so if it increases and k (which is elimination) decreases then Cmax will increase and tmax will decrease because it will take less time to reach the tmax
makes sense because k is elimination so if it increases then Cmax & tmax will decrease
do problem!
slide 11
Zero-Order Absorption Process
The selection of a model and order is typically based on the data gathered
- Zero order absorption occurs when the process is saturable or when a controlled-release system is used.
- For a zero-order absorption with rate constant k0 and a first-order elimination with rate constant k:
First-Order Absorption Process II
CpSFkaD0 (e-kt-e-kat) VD(ka –k)
At time tmax, corresponding to peak plasma concentration, Cmax, dDB/dt = 0 and so
differentiating,:
so at time tmax, peak plasma concentration is 0
do the problem!
slide 15, 16, 17
First-Order Absorption Process: Determining the Elimination rate Constant (k)
- During the elimination phase e-kat 0. Thus:
- Taking natural logarithms and then substituting with common logarithms:
eqn resembles y = mx + b
Thus k is determined from the slope of the log Cp against t curve; other parameters may be determined from the intercept (FkaD0/VD(ka – k))
graph
the y-intercept that is extrapolated from the straight line is this eqn
(FD0ka)/(VD(ka-k)
the slope of the elimination line = -k/2.3
First-Order Absorption Process: Determining the Elimination rate Constant (k) II
Similarly, urinary excretion data may be used. The urinary rate of excretion:
Thus k may be determined from the slope of the log dDu/dt against t curve; other parameters may be determined from the intercept (Fka keD0/ (ka – k))
graph!
slide 21
First-Order Absorption Process: Determining the Elimination rate Constant (k) III
Thus k may be determined from the slope of the log dDu/dt against t curve; other parameters may be determined from the intercept (Fka keD0/ (ka – k))
do the problem!
slide 23
First-Order Absorption Process: Determining the Elimination rate Constant (k) IV
- Since dDU/ dt cannot be determined for any single time point, the value for each time point on the previous plot is first obtained by taking urine samples over a period for assay (DU/ t is plotted against time t). The average rate over each period is determined and the time used is the mid point of each such period.
/k - The cumulative parent drug excreted in the
urine at time t = : D = Fk D Ue0
Determining k and ka
- -Method of Residuals:
The amount of absorbable drug at the absorption site at time t
D D e-kat
-where D0 is the amount of absorbable drug at time zero
Fraction of unabsorbed drug:
DGI e-kat D0
or
log -
D0 2.3
Determining Absorption Rate Constant ka : Method of Residuals
- From equation
a plot of Cp against time may be used to determine ka: the elimination phase slope is –k/2.3 and the y-intercept (FkaD0/VD(ka – k)) = A
Determining Absorption Rate Constant ka : Method of Residuals II
ka can be obtained by using the feathering technique:
by extrapolating the elimination phase line on the semilog plot, A can be obtained
a line representing the absorption phase is obtained by subtracting points on the extrapolated beta line from the corresponding original observed data points
a plot of the values of the differences gives a straight line with slope –ka/2.3
graph
slide 28, 29
Example for Method of Residuals
do the problem
slide 30
when do you use -Wagner-Nelson and -Loo-Riegelman
Wagner-Nelson - used only for 1 compartment
-Loo-Riegelman - used for 2 or multiple compartments
