kinetics exam 1 Flashcards
One-Compartment Model – IV Bolus Administration
IV bolus administration (entering body)
{Db, Vd} then k arrow out (which is the elimination of drug)
picture the diagram
The single IV bolus dose avoids many variability sources (because it bypasses absorption) related to absorption and so provides the most accurate method of determining drug distribution-related parameters.
used to determine drug distribution-related parameters
two pharmacokinetic parameters with IV bolus
- k: how the drug is eliminated from the body
- DB: how the drug distributes in the body
IIV bolus dose:
- large amount of drug administered one time given directly in the vein
- reach therapeutic levels quickly
- does bypass absorption
Assumptions - what we beleive ti be true about the model
Drugs can enter and/or leave the body
- The entire body is one compartment
*The drug is injected all at once into the compartment which is what bolus means
- The drug is distributed instantly and thoroughly throughout the compartment
*Drug elimination from the compartment occurs immediately after injection
- Rapid drug equilibrium between blood and tissue
*Changes in plasma drug levels result in proportional changes in tissue levels i.e. assaying levels in accessible fluid samples (plasma) provides an indirect measure of drug level in the body
- this is important because it allows us to sample plasma and make measurements of the levels of the drug in the body based on knowing plasma measurements in the body
Which of the following is FALSE?
- If the amount of a drug decreases by a constant amount in a constant time interval, the elimination process is zero-order
- For most drugs, the overall rate of elimination follows first-order kinetics
- Phenytoin follows first-order elimination kinetics
I think: Phenytoin follows first-order elimination kinetics
CORRECT - phenytoin follows zero order
- also these drugs follow zero order: salicylate and ethanol
- so phenytoin, salicylate and ethanol follows zero order
True or False? The elimination half-life of a zero-order reaction is NOT constant.
true
false
true, zero order is constant
- Zero-order kinetics undergo constant elimination regardless of the plasma concentration, following a linear elimination phase as the system becomes saturated. A simple analogy would be an athlete signing an autograph on a picture.
but the half-life:
Thus, in zero-order kinetics of elimination the concept of half-life becomes meaningless because the drug concentration does not decrease by half with every predictable time interval. Instead, we talk about drug dose removed over time.
Elimination Rate Constant
Elimination derives from one or more sources of metabolism (biotransformation) and excretion
k = km + ke
Rate dDB/dt = -kDB
integrating: DB = -kt + constant
DB = DB^0 x e^-kt
or
log DB = (-kt/2.3) + log DB^0 which is related to
y = mx + b
we will be dealing with two different sets of constants one with plasma data and one with urine data and the first constant is k
elimination is by metabolism (km) and excretion elimination (ke)
Elimination Rate Constant
- This equation may be depicted on a semi-log graph paper as:
Apparent Volume of Distribution
“Apparent” because no true physiologic or anatomic space being referenced but rather a box :)
- Plasma samples taken at different times are assayed to get Cp at time t=0, then the amount of the drug in the body is calculated as follows:
DB =VDCp
Apparent Volume of Distribution
- Substituting equation into prior equations:
log Cp = (-kt/2.3) + logCp^0
Cp = Cp^0 x e^-kt
- Where Cp is the plasma drug concentration at time t:
VD = (dose/CP^0) = (doseb^0/CP^0)
so we subsitutted DB for Cp
Example #1a
In a study by Brier and Harding a dose of 45 ng was given by IV bolus to rats. Samples of blood were taken at various intervals throughout the length of the study and the following data were obtained:
Find:
(i) k
(ii) t1⁄2
(iii) Cp0
(iv) VD
the data appears to be in first order because the [ ] decreases but is not constant but if it were zero order then it would be constant rate of decrease
(i) k: find slope then solve for k
slope = (log y2 - log y1)/ x2 - x1
(ii) t1⁄2 = 0.693/k
(iii) Cp0 = y intercept so where on the graph dose it hit the y intercept
(iv) VD = (dose/CP^0) = (dose b^0/CP^0)
dose = 45ng
Apparent Volume of Distribution
Similarly, since the dose given in a rapid IV bolus is equivalent to the amount of the drug in the body, the equation
- dDB/ dt = -kDB
- may be re-written as:
dDB/ dt = -kVDCp
dDB= -kVDCpdt
don’t worry too much about this
Apparent Volume of Distribution
Since integral from 0 to infinity C p dt is equivalent to [ AUC]0 to infinity, after integration previous equation 0
becomes:
D0 = kVD[AUC]0 to infinity
Thus for a model-independent way of determining VD…
VD = (D0/k[AUC] zero to infinity)
need to know hwo to use this eqn
the [ ] of the drug at all time is equal to the AUC
Apparent Volume of Distribution
Note:
* VD is dependent on Cp to 0
- A drug with a large VD is concentrated in the extra-vascular tissues, with lower amounts in the blood vessels
*A drug that is highly blood protein bound remains in the vascular compartment and has a low VD, with low amounts in other tissues
- VD may be expressed as a volume or as a % of the body weight
- Is constant for any drug; pathological conditions(such as edema)may alter the values for individuals
Clearance (Cl)
Definition: a reference to the volume of blood plasma from which the drug is completely removed per unit of time,
OR the fraction of the drug removed from the body per unit of time
- It is a representation of drug elimination and the result of both biotransformation (km) and excretion of the drug (ke)
- Clearance is model-independent
Clearance (Cl)
Since the rate of elimination is constant in a zero-order process, it is convenient to express zero-order elimination as
mass per unit time e.g. mg/ min, mg/ hr
because zero order means the same about the drug tis eliminated as time progresses
- The rate of elimination is not constant for first order processes (since it depends on the drug concentration); elimination for such processes is better expressed as volume per unit time e.g. L/ hr, mL/ min
Clearance (Cl)
Mass Approach (Zero Order):
mass eliminated per unit time.
- Volume/Clearance Approach (First Order)”
dDB/ dt = -kVDCp
(dDB/ dt)/ Cp = -kVDCp/ Cp
(dDB/ dt)/ Cp = -kVD = -Cl
Cl, the clearance, is a constant because k and VD are both constants
- Fractional approach: fraction eliminated per unit time - do not focus too much on
Cl = -k/VD
Cl and Vd
Rearranging
(dDB/ dt)/ Cp = -kVD = -Cl
Gives:
k =Cl/ VD …………………….(1) same as Cl = -kVD
- Recall, for a first order process,
A=A e-kt
C = e-kt……………………(2) - Substituting equation (1) in equation (2):
Cp = Cp to 0 x e^-(Cl/VD)t
Example #1b
A study by Brier and Harding a dose of 45 ng was given by IV bolus to rats. Samples of blood were taken at various intervals throughout the length of the study and the following data were obtained:
find Cl
use two earlier answers and use k and VD to find Cl
Cl = -kVD
Cl and Vd
Since concentration = dose / volume, for a one-compartment model, Cp0 = D0/ VD,, thus equation (3) may be re-written as:
Cp = (D0/VD) x e^-(Cl/VD)t - do not worry too much about, nice to know :)
Equation (4) is used in clinical practice to obtain individualized Cl and VD for patients after Cp corresponding time t is obtained following administration of a given dose. This typically involves a computer-aided adjustment of population data.
Non-Renal Clearance
ClT =ClR+ClNR
where ClT is the total clearance and ClR and ClNR, are the renal and non-renal clearance respectively.
do renal clearance + non renal clearance = total clearance
- In the absence of significant drug elimination from sources such as the lung and bile, non-renal clearance is practically equivalent to hepatic clearance (ClH)
ClT = ClR + ClH
Non-Renal Clearance
ClT may also be defined in terms of drug elimination rate and Cp as
follows:
ClT = (dDE/dt)/Cp
so drug eliminated over time over the concentration of the plasma
where DE = the amount of drug eliminated.
i.e. the drug elimination rate
(dDE/dt) = CP ClT
For first order processes, as the plasma drug level Cp decreases, so will dDE/dt (recall why). However, ClT remains constant
less drug in the plasma, less drug gets eliminated and so the total clearance remains constant
Non-Renal Clearance
For some drugs, elimination is so complex that a non-compartmental model has to be used. ClT then has to be calculated directly from the plasma concentration-time curve using:
ClT = (D0)/[AUC]
focus on -kVD mostly for clearance
Urinary Excretion Data
The renal excretion rate (ke) is assumed to be first order and may be used to determine k as follows:
dDU/dt = keDB
where DU = drug quantity excreted via the urine
dDU/dt = keDB^0e^-kt
log dDU/dt = logkeDB to 0 + (-kt/2.3)
y = b + m
y - intercept: ke DB at time zero
slope = -k/2.3
can determine elimination rate by urine
k is total elimination but ke is excretion rate
we plot first order on semi-log plots because on semi log it is straight :)
Urinary Excretion Data
NB: For a rapid IV dose, DB0 = D0. DB0 is known.
The elimination rate constant k as well as ke may be determined
using the following plot
Urinary Excretion Data
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.
- Since k and ke can be determined from the plot (because on the urinary excretion plot ehre is k which is in the slope and ke which is in the y intercept), the rate for non-renal excretion may be determined from:
k - ke = knr
so the ke could also be kr which is renal
