kinetics exam 1

Question 1

One-Compartment Model – IV Bolus Administration

Answer 1

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

Question 2

Assumptions - what we beleive ti be true about the model

Answer 2

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

Question 3

Which of the following is FALSE?

1. If the amount of a drug decreases by a constant amount in a constant time interval, the elimination process is zero-order
2. For most drugs, the overall rate of elimination follows first-order kinetics
3. Phenytoin follows first-order elimination kinetics

Answer 3

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

Question 4

True or False? The elimination half-life of a zero-order reaction is NOT constant.

true
false

Answer 4

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.

Question 5

Elimination Rate Constant

Answer 5

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)

Question 6

Elimination Rate Constant

Answer 6

• This equation may be depicted on a semi-log graph paper as:

Question 7

Apparent Volume of Distribution

Answer 7

“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

Question 8

Apparent Volume of Distribution

Answer 8

• 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

Question 9

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

Answer 9

(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

Question 10

Apparent Volume of Distribution

Answer 10

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

Question 11

Apparent Volume of Distribution

Answer 11

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

Question 12

Apparent Volume of Distribution

Answer 12

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

Question 13

Clearance (Cl)

Answer 13

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

Question 14

Clearance (Cl)

Answer 14

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

Question 15

Clearance (Cl)

Answer 15

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

Question 16

Cl and Vd

Answer 16

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

Question 17

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

Answer 17

use two earlier answers and use k and VD to find Cl
Cl = -kVD

Question 18

Cl and Vd

Answer 18

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.

Question 19

Non-Renal Clearance

Answer 19

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

Question 20

Non-Renal Clearance

Answer 20

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

Question 21

Non-Renal Clearance

Answer 21

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

Question 22

Urinary Excretion Data

Answer 22

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 :)

Question 23

Urinary Excretion Data

Answer 23

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

Question 24

Urinary Excretion Data

Answer 24

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

Question 25

Urinary Excretion Data

Answer 25

knr = km

where km is the rate constant for metabolism

(i.e. for most drugs k is contributed to by renal excretion which is ke and biotransformation which is km) k = ke + km

Question 26

Example #2

Answer 26

A single IV dose of an antibiotic was given to a 50-kg woman at a dose level of 20 mg/kg. Urine and blood samples were removed periodically and assayed for the parent drug. The following urine data were obtained:

Construct a graph on a semilogarithmic scale of Du/ t versus t*.
* The slope of this line should equal -k/2.3. It is usually easier to determine the elimination t1/2 directly from the curve and then calculate k from k=0.693/ t1/2

so plot the raet mg/hr which is equal to Du/t and have t* on the x axis

Question 27

Elimination Rate Constant - The
Sigma-Minus Method

Answer 27

• Also known as the Amount of Drug Remaining to be Excreted Method
• The amount of parent drug in the urine
  DU = (ke/k x. D0) x 1 - e^-kt …(5)

where t = infinity
DU to infintiy = ke x D0/k

Thus equation (5) becomes:
DU = DU to infinity (1–e-kt)
 DU to infinity - DU = DU to infinity .e-kt

log (Du to infinity - DU) = -kt/2.3 + log DU to infinity

Question 28

Elimination Rate Constant - The Sigma-Minus Method

Answer 28

This method of calculating k minimizes fluctuations in the elimination rate.

• However, it is not applicable to zero-order processes and cannot be used to calculate the renal excretion rate constant.

Question 29

Example #3

Answer 29

Enalapril urinary excretion data from 5 mg IV Bolus

Find k by finding k
Notice how DU∞ - DU was calculated for you

first order so put the data on semi log plot
y axis will be DU to infinity - DU and and x axis is time

Question 30

Example #2b

Answer 30

Using the data in the earlier problem, determine the elimination rate constant…

A single IV dose of an antibiotic was given to a 50-kg woman at a dose level of 20 mg/kg. Urine and blood samples were removed periodically and assayed for the parent drug. The following urine data were obtained:

Plot log(Du -Du) versus time
* Use a semilogarithmic scale
for (Du -Du)
Evaluate k and t1/2 from the slope

the hardest part if plotting and finding slope which is not that bad thank God :)

Question 31

Sampling and Related Issues - the issues that make finding the kinetics of the drug in the urine a little difficult

Answer 31

The assay method must be specific for the unchanged drug (i.e. must ignore metabolites)

• Urine samples must be obtained at regular intervals until almost all the drug is eliminated
• The need to obtain complete specimens (bladder must be emptied)
• Alterations in urine pH and volume may affect the excretion rate
• The need for frequent sampling
• A good portion of the drug must be excreted unchanged in the urine