Index
Sr
no |
Experiment |
signature |
1 |
Mathematical Fundamentals in Pharmacokinetics |
|
2 |
Elimination of Clearance by Model Independent Method |
|
3 |
Calculation of Infusion Rate, Time to reach 99% of Steady
State Concentration and Loading Dose for IV Infusion |
|
4 |
Calculation of “K” from Urinary Excretion By Rate Method and Sigma Minus
Method |
|
5 |
Calculation of Various Pharmacokinetic Parameter after IV
Bolus Injection |
|
6 |
Calculation of Various Pharmacokinetic Parameters after IV Infusion |
|
7 |
Determine the Absolute Bioavailability of IV Injection |
|
8 |
Determine the Relative Bioavailability of Oral Tablets and Oral Solution |
|
9 |
Dose Adjustment in Renal Impairment and Hepatic
Impairment Patient |
|
10 |
Dose Adjustment in Renal Impairment and Hepatic Impairment Patient |
|
Practical # 1
Mathematical Fundamentals in
Pharmacokinetics
Exponents and Logarithms
Exponential
functions:
Exponential functions have the form:
f (x) = b x
Where b is the base and x is the exponent (or power).
Logarithmic Functions:
A logarithm is simply an exponent that is written in a special way.
The logarithmic function is
defined as:
f (x) = log b x
The base of
the logarithm is b.
Examples:
1)
ex = 0.44
2)
e-2X-1 = 75
3)
log x = 0.95
4)
ln x = 1.22
5)
e-4k = 25/50
6) Cp = Cpo e-kt where Cpo=35 k=0.15
t1/2=2 hrs.
Graphs:
1-
Examples:
Q) Plot the following data on
semi-log graph and standard rectangular coordinate graph:
Time |
Drug(mg) |
10 |
96 |
20 |
89 |
40 |
73 |
60 |
57 |
90 |
34 |
120 |
10 |
130 |
2.5 |
Curve
fitting:
Curve
Fitting: It is the fitting a curve to the points on a graph sows that there is
some sort of relation between the variables ‘x’ and ‘y’. For e.g. dose of drug
vs. pharmacological effects.
Least Square Method:
The least square method is a useful procedure for obtaining the line of best
fit through a set of data points.
Reasons
for Using Least Square Method: It is used for obtaining a straight line from
which different pharmacokinetic parameters can be calculated more accurately. Ø
It minimizing the deviation between experimental and theoretical line, so we
obtain a straight line on simple rectilinear graph instead of curved one
Regression Line:
The straight line that characterizes the relationship between two variables is
called regression line.
Equation
for Regression Line: The general equation of regression line is:
y
= mx+b
m = slope
b = y-intercept
Equation for (m)
and (b):
m
=
b
=
Graph Method:
m= slope = Y2-Y1 / X2-X1
b= y-intercept on graph.
Determination of Slope:
Slope: is a
measure of the steepness of a line, or a section of a line, connecting two
points.
Slope
of straight line on rectangular coordinate graph:
Zero order
reaction: slope = y2-y1/x2-x1 &
k = slope
Slope
of straight line on Semi-log graph:
First order
reaction: slope =log y2-log y1/x2-x1 &
k = 2.3 (slope)
Rate and Order of
Reaction:
Rate: it
is the velocity with which a reaction occurs.
Order: it
is the way through which a reaction occurs.
Examples:
Q) A pharmacist dissolved
exactly 10g of a drug into 100 mL of water. The solution is kept at room
temperature and samples removed periodically and assayed for drug.
Following date is obtained:
Dug (mg/mL) |
Time |
Log Drug conc. |
100 |
0 |
2 |
50 |
4 |
1.6 |
25 |
8 |
1.3 |
12.5 |
12 |
1.09 |
6.25 |
16 |
0.79 |
3.13 |
20 |
0.49 |
1.56 |
24 |
0.17 |
“Since the log values are plotted
on quadrant graph, thus it’s a First order process”
t1/2
graphically:
t1/2
through formula:
t1/2 = 0.693/k
Q) A pharmacist dissolved few mgs
of antibiotic into exactly 100 mL of distilled water and placed the solution if
refrigerator (5C). at various time intervals, the pharmacist removed a 10 mL
liquid from the solution and measured the amount of drug contained in each
liquid. Following data is obtained:
Time |
Antibiotic (μg/mL) |
0.5 |
84.5 |
1 |
81.2 |
2 |
74.5 |
4 |
61 |
6 |
48 |
8 |
35 |
12 |
8.7 |
a- Is the
decomposition of this antibiotic a first order or zero order process?
b- What is the rate of
decomposition of this antibiotic?
c- How many mgs of
antibiotic were in the original solution prepared by the pharmacist?
d- Give the equation
for the first line that best fit the experimental data.
y=mx+b
Practical #
2
Determination of Clearance by Model Independent Method.
Introduction:
Clearance: Clearance is a measure of drug elimination from the
body without identifying the mechanism or process.
Volume of distribution: The volume of fluid in which the drug appears to be
distributed. (VD)
Physiological model: This model depends upon the blood flow in body and
drug is determined by extraction ratio.
Model
dependant: This model is also compartmental model and body is
considered to have single or two compartments in which the drug is distributed.
Model independent: This type of model is used to perform non
compartmental analysis
Cl = Do/(AUC)o∞
Q-1) 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 parent drug. The following data were obtained:
Time (hr) |
Cp (ug/ml) |
0.25 |
4.2 |
0.5 |
3.5 |
1.0 |
2.5 |
2.0 |
1.25 |
4.0 |
0.31 |
6.0 |
0.08 |
Slope:
Rate constant:
Area under curve (AUC):
Clearance:
Practical #
3
Calculate infusion rate, time to reach 99% of steady
state concentration and loading dose for IV infusion.
Introduction:
Steady state drug concentration: At steady state rate of drug leaving the body is equal
to rate of drug entering the body so at steady state the rate of change in
plasma drug concentration is zero. Represented by Css
dCp/dt=0
Infusion rate: The rate at which the drug is infused directly into
the systemic circulation is known as infusion rate, which is denoted by R.
Loading dose: It is initial bolus dose of a drug that is used to
obtain desired concentration as early as possible. Loading dose is represented by
symbol DL.
Desired steady state or targeted drug concentration: The plasma drug concentration prior to reach the
theoretical steady state is considered the steady state plasma drug
concentration; this concentration is the targeted or desired steady state drug
concentration.
Time to reach 99% of Css
Css=R/Vdk
For Cssto be 99%:
99% (R/Vdk)
Conc. Of drug at any time during
infusion:
Cp=R/Vdk (1-e-kt)
We want:
Cp=99%Css
Putting values
Clearance of IV infusion:
Css=R/vdk
(Cl=Vdk)
Css=R/Cl
Cl=R/ Css
Q1) An antibiotic
has a volume of distribution of 10 L and a k of 0.2 h-1. A
steady-state plasma concentration of 10 μg/mL is desired.
a.
Calculate the
rate of infusion?
b.
Calculate time
required to reach 99% Css?
c.
Calculate
clearance?
Q2)
A patient was given an antibiotic (t1/2=6hr) by constant IV infusion
at a rate of 2mg/hr at the end of 2days the serum concentration was 10mg/L.
calculate total body clearance for this antibiotic?
Q3) A physician
wants to administer an anesthetic agent at a rate of 2 mg/h by IV infusion. The
elimination rate constant is 0.1 h-1 and the volume of distribution is 10 L.
a. What loading dose would be recommended if the doctor
wants drug level to reach 2μg/ml immediately?
b. Calculate
Css?
Practical #
4
Calculate “K” from
Urinary Excretion Data By Using Rate Method and Sigma Minus Method.
Introduction:
If a drug is eliminated unchanged through
urinary route then its appearance In urine will be a reflection of drug
disappearance from plasma. So, the urine data can be used to calculate the
pharmacokinetic parameters. In such case the collection of samples of urine
after administration of drug is an alternative approach to obtain
pharmacokinetics information. In this situation the excretion rate of drug is
assume to be first order. Rate of drug excretion in urine can be given as:
dDu/dt= Ke.Dbo. e-kt
There
are two methods to determine urinary excretion rate constant
·
Rate method
·
Sigma-minus-method
1)
Rate
method:
In this method the urinary excretion rate is plotted against time on
semi-log graph paper. The slope of the curve is –k/2.3 (as it is first order)
and y-intercept= ke.Dbo
As drug’s urinary excretion rate can not be determined experimentally
for any given instant therefore the average rate of urinary drug excretion Du/t
is plotted against T*(where T* is mid points of collection period and t is the
time interval for collection of urine sample).
Ke.Dbo
Slope
= -k/2.3
Du/t
Time T*
Q: A single IV dose of an anti-biotic was given to a 50kg woman at a dose
of 20mg/kg. Urine samples are removed periodically and assayed for parent drug.
The following data was obtained
Data:
Sr. # |
Time (hrs) |
Du (mg) |
Du/t |
T* |
slope |
1 |
0.25 |
160 |
|
|
|
2 |
0.50 |
140 |
|
|
|
3 |
1 |
200 |
|
|
|
4 |
2 |
250 |
|
|
|
5 |
4 |
188 |
|
|
|
6 |
6 |
46 |
|
|
|
Slope
= log y2 – log y1/x2 – x1
Average
slope:
Find
“k”:
Half
life t1/2 = 0.693/K
2)
Sigma-Minus Method:
An
alternative method for the calculation of elimination rate constant K from
urinary excretion data is “sigma-minus” method or amount of drug remaining to
be excreted method.
In
this method the amount of drug remaining to be excreted (Duα-Du) is
plotted against time on semi-log graph. The slope of the curve is –K/2.3 and
y-intercept is maximum amount to be excreted i.e Duα.
Data:
Sr # |
Time |
Du (mg) |
Cumulative Du |
Duα-Du |
1 |
0.25 |
160 |
|
|
2 |
0.50 |
140 |
|
|
3 |
1 |
200 |
|
|
4 |
2 |
250 |
|
|
5 |
4 |
188 |
|
|
6 |
6 |
46 |
|
|
Slope
= log y2 – log y1/x2 – x1
Find ‘K’?
Find ‘ t 1/2 ’?
y-intercept:
Comparison
of Rate and Sigma-minus Method:
Rate
Method |
Sigma-minus
Method |
|||
·
Rate method does not require complete knowledge
of Du* so the loss of one urine sample does not invalidate the study |
·
Sigma-minus method requires an accurate
method of Du* which requires the collection of urine until a urinary
excretion complete |
|||
·
Fluctuation in the rate of drug
elimination and experimental error including incomplete bladder emptying for
a collection and period can affect the linearity in rate method |
·
In this case sigma-minus method is less
effected |
|||
·
Rate method is applicable to zero-order
drug elimination process |
·
It is not applicable to zero-order
processes |
|||
·
Renal excretion rate constant can
be calculated from rate method Ke.Db |
·
Renal excretion rate constant Ke cannot
be calculated from sigma-minus method
Duα Duα
-Du |
Du/t
t* t
Practical #
5
Calculation of various pharmacokinetic parameter after
IV bolus injection.
Introduction:
Clearance: Clearance is a measure of drug elimination from the
body without identifying the mechanism or process.
Volume of distribution: The volume of fluid in which the drug appears to be
distributed. (VD)
Procedure:
1-
Select
cycle according to data.
2-
Plot
concentration on y-axis and time on x-axis.
3-
Mark
conc. For respected time on graph.
4-
Draw
straight line in such a way that it covers maximum points on graph paper.
5-
Y-intercept
is Co
Estimation of half life by graphical method; To estimate half life on graph select conc. (C1)
on straight line, now draw straight line towards the plotted line so that it
intercepts plotted line. From intercepted line draw a perpendicular straight
line on x-axis which gives t1 from y-axis draw a line from point
which is half of C1 , which is C2 and then draw a
straight line on selective points towards the plotted line from intercepted
line draw perpendicular line on x-axis which gives t2
Calculate half
life by measuring difference of t1-t2
Data A
Q) Plasma data
obtained after IV bolus dose of 184mg of and antibiotic is as followed:
Time (hr) |
Cp (g/ml) |
1 |
137 |
6 |
120 |
12 |
103 |
24 |
76 |
48 |
42 |
72 |
23 |
96 |
12 |
Prepare
semi-log graph and estimate t1/2 of drug from this plot?
1-
Slope?
2-
Find k?
3-
Estimate
total AUC?
3-Calculate VD?
(Model dependant)
4-Calculate
total clearance?
Data B
Q) Data given in table below are the plasma concentrations
of cocaine after IV administration of
330mg of cocaine HCl to a subject:
Time (hr) |
Conc. (ug/ml) |
0.16 |
170 |
0.5 |
122 |
1 |
74 |
1.5 |
45 |
2 |
28 |
2.5 |
17 |
3 |
10 |
Prepare semi log
graph and estimate t1/2from plot?
1) Slope?
2) Find
k?
3) Calculate
total AUC?
4 -Calculate VD?
4- Calculate Total Clearance?
Practical #
6
Calculation of Various Pharmacokinetic Parameters
after IV Infusion.
Estimate
the volume of distribution, area under the curve, elimination rate constant,
half-life and clearance from the data in following table obtained on infusing a
drug at the rate of 50 mg/hr for 7.5 h in a dose of 375mg.
x(time) (hrs.) |
y(Cp)(µg/ml) |
|
0 |
0 |
|
2 |
3.4 |
|
4 |
5.4 |
|
6 |
6.5 |
|
7.5 |
7 |
|
9 |
4.6 |
|
12 |
2 |
|
15 |
0.9 |
|
a.
Prepare a semi logarithmic plot and estimate the half-life of drug
b.
Calculate AUC
c.
Calculate volume of distribution
d.
Calculate elimination half-life from formula
e.
Calculate total clearance
Introduction:
Clearance: Clearance is a measure of drug elimination from the
body without identifying the mechanism or process. Calculate
clearance by using following equation.
Cl= kVD
Volume of distribution: The volume of fluid in which the drug appears to be
distributed (VD). Calculate volume of
distribution by using following equation
VD
= Dose/CP
Elimination
half-life: Calculate volume
of distribution by using following equation
t1/2 =
IV
Infusion:
Intravenous therapy is the infusion of liquid substances
directly into a vein. Intravenous (IV) means "within
vein". Intravenous infusions are commonly referred to as drips.
The intravenous route is the fastest way to deliver fluids and
medications throughout the body.
Procedure;
1-
As
the data is first order so we will plot the graph on semi log graph paper.
Select cycle according to data.
2-
Plot
concentration on y-axis and time on x-axis.
3-
Mark
conc. For respected time on graph.
4-
Draw
straight line in such a way that it covers maximum points on graph paper.
5-
Y-intercept
is Co
6-
Determine
slope of the line and calculate elimination rate constant by using this
formula:
Estimation of half-life by graphical method:To estimate half-life on graph select conc. (C1)
on straight line, now draw straight line towards the plotted line so that it
intercepts plotted line. From intercepted line draw a perpendicular straight
line on x-axis which gives t1 from y-axis draw a line from point
which is half of C1 , which is C2 and then draw a
straight line on selective points towards the plotted line from intercepted
line draw perpendicular line on x-axis which gives t2
Calculate
half-life by measuring difference of t1-t2
Estimation of Elimination half-life from formula:
Calculate elimination half-life by using following equation
t1/2=
Calculations:
1-Estimate
total AUC?
=
=
∑+
2-Calculate slope?
3- Find k?
K=Slope*2.3
4- Calculate elimination half-life ?
t1/2=
5-Calculate Volume of Distribution (VD)
(Modeldependant)?
VD=Dose/CP
6- Calculate total clearance?
Cl=KVD
Practical # 7
TO STUDY AND DETERMINE THE ABSOLUTE BIOAVAILABILTY OF
IV INJECTION AND ORAL SOLUTION
BIOAVAILABILITY: It
is defined as the rate and extent to which the drug is available systemically
in an unchanged form at the site of action.
It is further divided into two types
Ø Relative
Bioavailability
The
availability of the drug in formulation is compared to the availability of drug
in standard dosage formulation
Ø Absolute
Bioavailability
It is the systemic availability of drug
after extra vascular administration (oral, rectal, transdermal, subcutaneous)
compared to IV dosing.
It is generally measured by comparing the
respective AUC after extra vascular and IV administration.
Measurement of Absolute
bioavailability:
Absolute bioavailability
F= [AUC]oral/Dose oral
[AUC]iv/Dose iv
It may be expressed as a fraction or as
percentage by multiplying F with 100.
For drugs administered in iv bolus
injection F=1 as the drug is completely absorbed while for for drugs
administered by other routes F may not exceed 100% or F>1
DATA:
The
dose for Iv injection is 2mg/kg while for oral
solution is 10mg/kg
Sr.No 1 2 3 4 5 6 7 8 |
Time 0.5 1.0 1.5 2 2.0 4.0 6.0 8.0 |
Cp
(IV) 5.94 5.30 4.72 4.21 3.34 2.66 1.68 1.06 |
Cp (oral solution) 23.4 26.6 25.2 22.8 18.2 14.6 9.14 5.77 |
Calculation for Oral:
1)
Calculate
of Slope:
Calculation
of ‘K’:
Calculation
of AUC:
Calculation
for IV:
Calculation
of Slope:
Calculation
of ‘K’:
Calculation
of AUC:
Calculation
of Absolute Bioavailability:
Practical # 8
TO STUDY AND DETERMINE THE RELATIVE BIOAVAILABILTY OF ORAL TABLETS AND ORAL
SOLUTION
BIOAVAILABILITY: It is defined as the rate and extent to
which the drug is available systemically in an unchanged form at the site of
action.
It is further
divided into two types
Ø
Relative
Bioavailability
The availability of the drug in formulation is
compared to the availability of drug in standard dosage formulation
Ø
Absolute
Bioavailability
It is the systemic
availability of drug after extravascular administration(oral, rectal,
transdermal, subcutaneous) compared to IV dosing.
Measurement of
Relative Bioavailability: [AUC] test/Dose test
[AUC]
standard/Dose standard
DATA:
The dose for test
and standard is 10mg/kg
Sr.No 1 2 3 4 5 6 7 8 |
Time 0.5 1.0 1.5 2 2.0 4.0 6.0 8.0 |
Cp (oral
tablets) 13.2 18.0 19.0 18.3 15.4 12.5 7.92 5.00 |
Cp (oral solution) 23.4 26.6 25.2 22.8 18.2 14.5 9.14 5.77 |
Calculation
for Oral Tablets:
Calculation
of Slope:
Calculation
of ‘K’:
Calculate
AUC:
Calculation
for Oral Solution:
Calculation
of Slope:
Calculation
of ‘K’:
Calculation
of AUC:
Calculation
of Relative Bioavailability:
Practical
# 9
DOSE ADJUSTMENT IN RENAL
AND HEPATICALLY IMPAIRED PATIENTS.
Introduction:
RENAL IMPAIRMENT:
Acute disease or trauma to the kidney can cause uremia, in which glomerular
filteration is impaired or reduced leading to accumulation of excessive fluid
and blood nitrogenous products in body.
GENERAL APPROACHES FOR DOSE ADJUSTMENT IN RENAL DISEASE;
1. The
required therapeutic plasma drug concentration in uremic patients is similar to
that required in normal renal function patients.
2. Uremic
patients are maintained on the same Cav∞ after multiple oral doses
or multiple IV bolus injection.
3. For
IV infusion same Css is maintained.
DOSE ADJUSTMENT BASED ON DRUG CLEARANCE:
This
method based on drug clearance tries to maintain desired Cav∞ after
multiple oral doses or multiple IV bolus injection.
The
calculation for Cav∞ :
F=
fraction of drug
ClT=
total body clearance
t=
dosing interval
∴ F=1 for IV
To maintain desire Cav∞, ureamic
dose or dose interval must be changed to tu
N=
Normal Condition
U=uremic
condition
By
rearranging equation,
If
dosage interval is kept constant then,
For IV infusion the same desired Css is
maintain both or Patient with normal renal function and for patients with renal
impairment. Therefore the rate of infusion R, must be changed
DOSE ADJUSTMENT BASED ON CHANGES IN ELIMINATION RATE CONSTANT:
The
overall elimination rate constant for many drugs is reduced in uremic patients
.By reducing the normal dose of the drug and keeping the frequency of the
dosing constant or by decreasing the frequency of dosing and keeping dose
constant.
Assuming the VD is same in both
normal and uremic patients and t
is constant, then the uremic dose Dou is a function of the normal dose,
DOSE CALCULATION ON BASIS PF ELIMINATION RATE CONSTANT:
1. The
renal elimination rate constant decrease proportionally as renal function
decreases.
2. The
non renal routes of elimination remain unchanged.
3. Changes
in the renal clearance of the drug are reflected by changes in the creatinine
clearance.
The overall route of elimination rate constant is the sum total
of all routes of elimination in the body, including the renal rate and non
renal rate constant.
+
Non
renal elimination rate constant.
Renal elimination rate constant.
Renal clearance is product of VD
and kR.
By
rearranging,
Assuming
that the apparent VD and non renal route of elimination do not
change in ureamia then,
And
By
substitution:
A
change in renal clearance ClRU due to renal impairment
will be reflected in the change in overall elimination rate constant.
NUMERICAL:
The maintenance dose of
gentamycin is 80mg every 6hrs for a patient with normal renal function.
Calculate the maintenance dose for uremic patient with Clcrof
20mL/min .assume a normal Clcr is of 100mL/min.
SOLUTION:
Practical # 10
DOSE ADJUSTMENT IN RENAL
AND HEPATICALLY IMPAIRED PATIENTS.
Introduction:
RENAL IMPAIRMENT:
Acute disease or trauma to the kidney can cause uremia, in which glomerular
filteration is impaired or reduced leading to accumulation of excessive fluid
and blood nitrogenous products in body.
GENERAL APPROACHES FOR DOSE ADJUSTMENT IN RENAL
DISEASE;
4. The
required therapeutic plasma drug concentration in uremic patients is similar to
that required in normal renal function patients.
5. Uremic
patients are maintained on the same Cav∞ after multiple oral doses
or multiple IV bolus injection.
6. For
IV infusion same Css is maintained.
DOSE ADJUSTMENT BASED ON DRUG CLEARANCE:
This
method based on drug clearance tries to maintain desired Cav∞ after
multiple oral doses or multiple IV bolus injection.
The
calculation for Cav∞ :
F=
fraction of drug
ClT=
total body clearance
t=
dosing interval
∴ F=1 for IV
To maintain desire Cav∞, ureamic
dose or dose interval must be changed to tu
N=
Normal Condition
U=uremic
condition
By
rearranging equation,
If
dosage interval is kept constant then,
For IV infusion the same desired Css is
maintain both or Patient with normal renal function and for patients with renal
impairment. Therefore the rate of infusion R, must be changed
DOSE ADJUSTMENT BASED ON CHANGES IN ELIMINATION RATE CONSTANT:
The
overall elimination rate constant for many drugs is reduced in uremic patients
.By reducing the normal dose of the drug and keeping the frequency of the
dosing constant or by decreasing the frequency of dosing and keeping dose
constant.
Assuming the VD is same in both
normal and uremic patients and t
is constant, then the uremic dose Dou is a function of the normal dose,
DOSE CALCULATION ON BASIS PF ELIMINATION RATE CONSTANT:
4. The
renal elimination rate constant decrease proportionally as renal function
decreases.
5. The
non renal routes of elimination remain unchanged.
6. Changes
in the renal clearance of the drug are reflected by changes in the creatinine
clearance.
The overall route of elimination rate constant is the sum total
of all routes of elimination in the body, including the renal rate and non
renal rate constant.
+
Non
renal elimination rate constant.
Renal elimination rate constant.
Renal clearance is product of VD
and kR.
By
rearranging,
Assuming
that the apparent VD and non renal route of elimination do not
change in ureamia then,
And
By
substitution:
A
change in renal clearance ClRU due to renal impairment
will be reflected in the change in overall elimination rate constant.
Numerical: Lincomycin
is given at 500mg every 6hrs to a 75kg normal patient. What dose would be use :
a) In
complete renal shutdown ,creatinine clearance = 0
b) When
creatinine clearance = 10ml/min
No comments:
Post a Comment