Index
|
Sr
no |
Experiment |
signature |
|
0 |
Units, Equations and Definitions |
|
|
1 |
Determine the
Absolute Bioavailability of an Antibiotic |
|
|
2 |
Evaluation of 1st order and Zero order by
Graphical Method |
|
|
3 |
Find
the AUC for Zero order and 1st order Reaction |
|
|
4 |
Dosing of Drugs in Patients By Using Body Mass Index,
Ideal Body Weight and Body Surface Area |
|
|
5 |
Determine
the Elimination Half-life in a Patient with Normal Kinetics, Renal failure,
Enzyme Induction |
|
|
6 |
Find the AUC by Trapezoidal Rule |
|
|
7 |
Find the Equation of Regression By Using Least Square Method |
|
|
8 |
To constrict a calibration curve between absorbance and
known concentration of drug for determination of unknown concentration
graphically |
|
|
9 |
To determine the dissolution profile using USP dissolution apparatus for
a tablet containing 500 mg of a drug |
|
|
10 |
Calculation of K by urinary excretion Data by using Rate
Method |
|
Units
in Biopharmaceutics
|
Parameter |
Symbol |
Unit |
Example |
|
Drug
dose |
Do |
Mass |
mg |
|
Plasma
Drug Concentration |
Cp |
|
|
|
Rate |
|
|
|
|
Zero
Order Rate Constant |
Ko |
|
|
|
First
Order Rate Constant |
K |
|
1/h, hr-1 |
|
Volume
of Distribution |
VD |
Volume |
ml or L |
|
Area
Under the Drug Concentration Time Curve |
AUC |
Conc*Time |
|
|
Fraction
of Drug Absorbed |
F |
No unit |
Range from 0
to 1(1 to 100%) |
|
Clearance |
Cl |
|
|
|
Half-life |
t1/2 |
Time |
h |
|
Excretion
Rate Constant |
Ke |
|
hr-1 |
|
Metabolism
Rate Constant |
Km |
|
hr-1 |
|
Bilairy
Constant |
Kb |
|
hr-1 |
Basic
Equation Employed in Biopharmaceutics
|
Equation |
Symbol |
Formula |
|
Half-life
equation for zero order reaction |
t1/2 |
|
|
Half-life
equation for first order reaction |
t1/2 |
|
|
Elimination
rate constant for zero order reaction |
K |
|
|
Elimination
rate constant for first order reaction |
K |
|
|
Slope
for zero order reaction |
Slope |
|
|
Slope
for first order reaction |
Slope |
|
|
Plasma
concentration |
Cp |
|
|
Clearance |
CL |
K.VD or |
|
Area
Under the Curve |
AUC |
|
|
Steady
State Concentration |
CSS |
|
|
Equation
of straight line for zero order |
C |
|
|
Equation
for straight line for first order |
log C |
log C = |
|
Metric
BMI |
BMI |
|
|
Imperial
BMI |
BMI |
|
|
Ideal
body weight for males |
IBW |
IBW= 50kg+2.3kg for each inch of patient height
over 5 feet |
|
Ideal
body weight for females |
IBW |
IBW= 45.5kg+2.3kg for each inch of patient height over 5 feet |
|
Professional
Body Surface Area |
BSA |
|
|
Household
Body Surface Area |
BSA |
|
DEFINITIONS
Biopharmaceutics:
Biopharmaceutics
examines the inter relationship of the physical/chemical properties of the
drug, the dosage form in which the drug is given & the rate of
administration on the rate & extent of systemic drug absorption.
Pharmacokinetics:
Science
of kinetics of drug absorption, distribution & elimination (metabolism + excretion)
Pharmacodynamics:
Refer
to the relationship between the drug concentration at the site of action
(receptors) and pharmacologic response including biochemical and physiologic
effects that influence the interaction of drug with receptors.
Clinical
Pharmacokinetics:
It
involves the application of pharmacokinetic methods to drug therapy
Toxicokinetics:
It
is the application of pharmacokinetic principles to the design, conduct &
interpretation of drug safety evaluation studies.
Clinical
toxicology:
It
is the study of adverse effects of drug & toxic substances in the body.
Minimum
effective concentration (MEC):
It
reflects the minimum concentration of drug needed at the receptors to produce
the desired pharmacologic effect.
Minimum
toxic concentration: (MTC)
It
represents the drug concentration needed to just barely produce a toxic effect.
Onset
time:
Time
required for a drug to reach the MEC.
Duration
of Drug:
It
is a difference between the onset time and the time for the drug to decline
back to MEC
Area
under the drug concentration – time curve(AUC):
It
is used as a measure of total amount of unaltered drug that reaches the
systemic circulation.
Apparent
volume of distribution (VD):
The
volume in which the drug seems to be distributed is termed the apparent volume
of distribution.
Elimination
rate constant (K) :
The
rate at which the drugs concentration in the body declines overtime.
Half
life:
The
time duration in which drug concentration becomes one half of its initial
concentration.
Clearance:
Clearance
is a measure of drug elimination from the body without identifying the
mechanism or process.
Steady
state concentration (CSS):
Concentration
at which rate of drug entering the body becomes equal to the rate of drug
leaving the body.
Rate:
Change
in the concentration of drug in body per unit time.
Rate
of reaction:
The
order of reaction refers to the way by which the concentration of drug or
reactant influences the rate of a chemical reaction.
Zero
order reaction:
That
type of reaction in which rate of reaction is independent of initial
concentration of drug
First
order reaction:
That
type of reaction in which rate of reaction is dependent on initial
concentration of drug
One
compartment open model:
This
model assumes that the drug can enter or leave the body and the entire body
acts like a single, uniform compartment.
Two
compartment open model:
In
this model drug distributes into two compartments i.e. the central compartment
(plasma) and tissue compartment. .
Practical no 1:
DETERMINE
THE ABSOLUTE BIOAVAILABILITY OF AN ANTIBIOTIC
Bioavailability:
The
rate and extent of systemic drug absorption or the amount of drug reaches the
systemic circulation.
Absolute
bioavailability:
Systemic
bioavailability of drug after extra cellular administration of oral, rectal etc
compared to IV drug.
F=
Relative
bioavailability:
Comparison
of an unrecognized (a) drug with recognized drug (b).
R
=
Factors:
Following
are the factors on which bioavailability depends:
·
Tissue binding.
·
Person to person
variation
·
Rate of administration
·
Chemical degradation
·
Physical properties of
drug
·
Stomach emptying
·
Patient condition
(renal/hepatic diseases)
·
Food
Data:
Comparison of plasma concentration of an
oral solution
(10mg/kg with IV solution (2mg/kg)
|
S.no |
Time (hrs) |
Plasma
Concentration |
|
|
IV
solution (2mg/kg) |
Oral
solution (10mg/kg) |
||
|
1 |
0.5 |
5.95 |
23.9 |
|
2 |
1.0 |
5.32 |
26.6 |
|
3 |
1.5 |
4.14 |
25.2 |
|
4 |
2.0 |
4.03 |
21.1 |
|
5 |
3.0 |
3.91 |
19.6 |
|
6 |
4.0 |
3.50 |
10.4 |
|
7 |
6.0 |
2.97 |
6.3 |
|
8 |
8.0 |
1.01 |
4.1 |
PRACTICAL NO: 2
Evaluation
of 1st order and Zero order by Graphical Method
Introduction:
Order of Reaction:
Order of reaction is the way in which
concentration of drug effects the rate of reactions.
1st
Order of reaction:
That
type of reaction in which rate of reaction is dependent on initial
concentration of drug.
Zero
Order of reaction:
That type of reaction in which rate of reaction
is independent of initial concentration of drug.
Half Life:
The time duration in which drug
concentration becomes one half of its initial concentration.
Difference between Zero order and 1st order of reactions
|
Zero Order
|
1st Order
|
|
Rate of reaction
is independent on concentration of drug. |
Rate of reaction
is dependent on concentration of drug. |
|
If the amount of
drug is decreasing at a constant rate i.e. zero order. |
If the amount of
drug is decreasing at a rate i.e. proportional to the amount of drug
remaining. |
|
Formula : dA/dt= - Ko Changes in the
amount of drug with change in time is constant. |
Formula: dA/dt =
- Ko A Both time and
concentration are constant to each other. |
|
constant Ko = slope |
Ko =slope (2.303) |
|
A= -Ko t + Ao |
A = Aoe-kt |
|
A graph of A v/s
T yield a curve line on semi log paper. |
A graph of A v/s
T yields straight line on semi-log paper. |
|
A graph of A v/s
T yields a Straight line on Quadrant paper. |
A graph of A v/s
T yields a Straight line on Quadrant paper. |
|
Half-life is
dependent on initial concentration of drug. |
Half-life is
independent on initial concentration of drug |
|
Half-life for
zero order is not constant. |
Half-life for
zero order is constant. |
|
t1/2= 0.5 Ao / K |
t1/2= 0.693/K |
|
Zero order
process comes to an end. |
First order
never comes to an end. |
|
Zero order has
less practical value. |
First order has more practical value. |
Slope of straight line on
rectangular coordinate graph: Slope = 
= 
Slope
of straight line on semi log graph: Slope
= = 
Data:
|
Time (hrs) |
Cp (μ/ml) |
|
Time (hrs) |
Cp (μ/ml) |
|
0.5 |
38.9 |
|
0 |
12 |
|
1 |
30.3 |
|
1 |
10 |
|
2 |
18.4 |
|
2 |
8 |
|
3 |
11.1 |
|
3 |
6 |
|
4 |
6.77 |
|
4 |
4 |
|
5 |
4.1 |
|
5 |
2 |
Data (A)
Order of Reaction: 1st order of reation.
Graphical Method: From the graph it is concluded that this is 1st order reaction. There is
a straight line on semi-log paper and curved line on quadrant paper.
Slope:
Slope = =
Solution:
Result:
As the data is showing straight line on semi-log graph. Thus data A is the
representing first order process.
Equation of Rate Constant/slope:
K=2.303 (slope)
Calculation of Half-life for 1st order of reaction:
Formula: t1/2= 
Data (B)
Order of reaction: Zero order of reaction.
Graphical Method: From the graph it is concluded that this is zero order reaction. There is a
curved line on semi-log paper and straight line on quadrant paper.
Slope:
Slope = =
Solution:
K for Zero order data:
C=CO-Kt
In Zero order Process :
K = Slope
Calculation of Half-life for Zero order:
t1/2= 
Result:
The rate of change of concentration of drug between two intervals is same so
reaction is zero order.
As the data is showing
curved line on semi-log graph. Thus data A is the representing zero order
process.
Result:
For
1st order reaction half –life is ___________.
For
Zero order reaction half-life is _________
PRACTICAL NO: 3
Find
the AUC for Zero order and 1st order Reaction
|
Time (hrs) |
Cp (μ/ml) |
|
Time (hrs) |
Cp (μ/ml) |
|
0.5 |
38.9 |
|
0 |
100 |
|
1 |
30.3 |
|
2 |
95 |
|
2 |
18.4 |
|
4 |
90 |
|
3 |
11.1 |
|
6 |
85 |
|
4 |
6.77 |
|
8 |
80 |
|
5 |
4.1 |
|
10 |
75 |
Data
(A)
Order of Reaction:
Graphical Method: From the graph it is concluded that this is 1st order reaction. There is
a straight line on semi-log paper and curved line on quadrant paper.
Calculation of Half-life GraphicalMethod:
Formula:
t1/2= 
Random Method:
Slope = =
Solution:
Result:
As the values vary it shows that it is first order result.
Equation of Kinetic of Drug or Rate Constant:
K=2.303 
OR 
Calculation of Half-life Random Method:
Formula:
t1/2=
Calculation of Area under Curve (AUC):
Formula:
= 
= ∑
+
Data (B)
Order of reaction:
Random Method:
Slope = =
Solution:
Calculation of Half-life for Random Method:
t1/2=
Result:
The rate of change of concentration of drug between two intervals is same so
reaction is zero order.
Graphical Method:
For Zero order curved line is for semi-log
paper and for quadrant paper it is straight line.
Calculation of Rate Constant:
C=CO-Kt
Calculation of Half-life for Graphical Method:
t1/2=
Calculation of Area under Curve (AUC):
Formula:
=
= ∑
+
Result:
AUC of 1st
Order reaction is ____________.
AUC of Zero Order
reaction is ____________.
Practical No: 4
Dosing of Drugs in Patients By Using Body
Mass Index, Ideal Body Weight and Body Surface Area
Theory:
Obesity has been
associated with increased mortality resulting from increases in the incidence
of hypertension, arthrosclerosis, coronary artery disease, diabetes and other
conditions compared to non-obese patients. A patient is considered obese if
actual body weight exceeds ideal body weight by 20%. Obesity often is defined
by Body Mass Index (BMI) a value that normalizes body weight based on height.
Body Mass Index:
It is expressed as body
weight (kg or lb) divided by a square of the person’s height (meters or inch).
Types of BMI:
There are two types of
BMI:
Metric BMI:
When the body weight is
taken in kilogram and height in meters.
BMI = 
Imperial BMI:
When the body weight is taken in pounds (lb) and height in inches.
BMI = 
Units of BMI:
Metric = kg/cm2
Imperial = lbs/inch2
|
Physical Condition |
Range (metric) |
Range (Imperial) |
|
Very
Severely Underweight |
Less
than15 |
Less
than 0.6 |
|
Severely
Underweight |
15-16 |
0.6-0.64 |
|
Underweight |
16-18.5 |
0.64-0.74 |
|
Normal
(Healthy Weight) |
18.5-25 |
0.74-1 |
|
Overweight |
25-30 |
1-1.2 |
|
Obese
Class I (Moderate) |
30-35 |
1.2-1.4 |
|
Obese
Class II (Severe) |
35-40 |
1.4-1.6 |
|
Obese
Class III (Very Severe) |
Over 40 |
More than 1.6 |
Question no 1:
Calculate the BMI of a patient weighing 68 kg and height is 165cm?
BMI =
Effect of increased BMI
on pK parameters:
·
Increase fat content results in decrease water
proportion or total body water to total body weight, which effects the apparent
volume of distribution.
·
Distribution changes in drug’s pharmacokinetics due
to partioning of drug between lipid and aqueous environment.
·
Fatty infilteration of liver affecting
biotransformation.
·
CVS changes that may affect renal blood flow and
renal excretion.
Ideal Body Weight:
It refers to appropriate or normal weight for a male or female based on
age, height, and weight and frame size.
Dosing by actual bodyweight may result in overdosing of drugs such as
aminoglycosides (gentimicin), which are very polar and are distributed in
extracellular fluids. Dosing of these drugs is based on ideal body weight or
lean body weight.
Following equations have been used for estimation of lean or ideal body
weight, particularly for adjustment of dosage in renally impaired patients.
For Males:
IBW = 50kg+2.3kg for each inch of patient height over 5 feet.
For Females:
IBW = 45.5kg+2.3kg for each inch of patient height over 5 feet.
Question no 1:
Calculate the ideal body weight for a male patient weighing 16lb and measuring
5ft 8 inch in height?
Question no 2:
Calculate the ideal body weight for a female patient weighing 60kg and measuring
5ft 3inch in height?
IBW used in calculating
Loading and Maintenance dose:
“The loading dose (initial dose required to reach a certain drug
concentration) and the maintenance dose needed to maintain the specified
concentration can be calculated”.
To calculate the
loading dose, perform following:
LD = IBW in kg or lb * drug dose per kg or lb.
To calculate the
maintenance dose, perform following:
MD = IBW in kg or lb *drug dose per kg per dosing interval.
Question no 1: determine the loading and maintenance dose of
gentimicin for a 70 year old male patient weighing 190lb with a height of
6feet. The physician desires a loading dose of 1.5mg/kg of ideal body weight to
be administered every 8hr after initial dose?
For ideal weight:
IBW = 50kg+2.3kg for each inch of patient height over 5 feet.
For loading dose:
LD = IBW in kg or lb * drug dose per kg or lb
For maintenance dose:
MD = IBW in kg or lb *drug dose per kg per dosing interval
Body Surface Area
In physiology and
medicine, the body surface area is measured or calculated surface area of human
body. For many clinical purposes BSA is better indicator of metabolic mass than
body weight because it is less affected by abnormal adipose mass.
Uses:
·
Renal
clearance is usually divided by BSA to gain an appreciation of the required
GFR.
·
Cardiac
index is measure of cardiac output divided by BSA. It gives better
approximation of cardiac output.
·
Chemotherapy
is often dosed according to patient BSA.
·
Glucocorticoid
dosing is also expressed in terms of BSA for calculated maintenance dosing on
to compare high dose with maintain requirement.
Average values:
Neonates…………………….…0.25m2
Child of 2 years………………...0.5m2
Child of 9 years………………..1.07m2
Child of 10 years……………….1.14m2
Child of 12-13years…………….1.33m
Women…………………………1.6m2
Men……………………………..1.9m2
Formula:
Professional BSA(m2)
= 
Household BSA (m2) = 
Calculation for does in Children:
Formula 1:
Pediatric dose = 
adult dose
Question: If the adult
dose of a drug is 50mg. What is the dose for a child weighing 25kg and having
height of 126cm?
BSA =
Pediatric dose = 
adult dose
Formula 2:
If dose is given in
Dose/m2.
Dose = BSA * Dose/m2
Question: If the dose
is 2mg per m2. What is the dose of drug per patient having BSA of
0.87m2?
Dose = BSA * Dose/m2
PRACTICAL NO: 5
Determine the
Elimination Half-life in a patient with Normal Kinetics, Renal failure, Enzyme
Induction
Definition:
Elimination Half-life: It is the time duration when half of the drug is eliminated
from the body.
Renal failure:
A pathological condition in which kidney functions are impaired and it can no
more eliminate the drug from the body.
Enzyme Induction: Enzyme
induction is a process in which a molecule (e.g. a drug) induces (i.e.
initiates or enhances) the expression of an enzyme.
Elimination Rate Constant: It is the rate of drug eliminated from the body.
Ke:
It represents the elimination rate constant for drugs that are eliminated
through kidney.
Kb:
It represents the rate constant for drugs that are eliminated through bile.
Data:
A drug fitting one
compartment model was found to be eliminated from plasma with an elimination
rate constant equal to 0.6/hr. The drug was found to be eliminated by three
pathway-metabolism?
Kidney excretion (Ke =
0.25hr-1) Bilairy
excretion (Kb = 0.150hr-1)
Q no 1: What is the elimination t1/2 for drug?
Q no 2: what would be the Half-life of this drug if drug
excreted through kidney is completely impaired?
K= Km+ Ke + Kb
So, Kidney is impaired
Ke = 0
K = Km + Ke + Kb
Calculation of Half-life:
t1/2=
Q no 3: If the drug metabolizing enzymes are induced so that the
rate of metabolism of this drug is doluble what would be the new Half-life?
K = Km + Ke + Kb
So,
Calculation of Half-life:
t1/2
=
Result:
Q no 4: Calculate the %age of total drug metabolized and total
drug of eliminated?
%age of metabolized drug=
*
100
%age of eliminated drug= 
*
100
Conclusion:
In normal person, the t
½ and K are _________ and ___________. But in Kidney impaired
patients, the elimination rate constant decreases to __________ and t ½ increases
to ___________ due to impaired renal function. While by enzyme induction, the
metabolic enzymes double the rate of metabolism which in turn increases the
overall elimination rate constant from __________ to _________. While t ½ decreases
up to __________ due to rapid elimination of drug.
%age of drug
metabolized is _____________.
%age of drug eliminated
is _____________.
Practical
No: 6
Find the AUC by
Trapezoidal Rule
AUC:
It is the area under the plasma drug concentration vs time curve.
Trapezoid
Rule: is a numerical method frequently used in
pharmacokinetics to calculate the area under the plasma drug concentration vs.
time curve. The total area under the plasma drug level vs. time curve is
obtained by summation of each individual area between two consecutive time
intervals using the trapezoidal rule.
Cp 
time
Each
division is known as trapezoid.
Trapezoid:
The area between the time intervals is the area of trapezoid and it can be
calculated by the formula.
= 
Where,
AUC = Area under the curve
Gradually
area = width*length
Where in AUC formula
Length
= 
and Width = 
Data:
|
Time |
Cp |
|
1 |
35 |
|
2 |
30 |
|
3 |
28 |
|
4 |
22.5 |
|
6 |
15.6 |
|
8 |
7.3 |
Solution:
1)
To
calculate [AUC] from t = 0 to t = n :
Formula: =
= = 
As curve does not touch this x-axis. So
we take up to the infinity.
2) [AUC] from tn to t∞ :
The area under the plasma level time
curve is extrapolated to t∞ .
In this case residual area 
is calculated as
Where,
= last observed plasma concentration.
K
= slope obtained from terminal portion of curve. OR K = constant of
elimination.
t
is the time of observation of drug concentration at Cn.
Cp =
7.3µg/ml
Calculation
of ‘K’: (first order process)
Formula:
K = 2.303
Slope
3) To calculate AUC from t=0 to t=∞:
Result:
The total area under the curve (AUC)
obtained by using trapezoidal rule is -----------------
Practical
No: 7
Find the Equation of
Regression Line By Using Least Square Method
Least
Square Method: The least square method is a useful
procedure for obtaining the line of best fit through a set of data points.
Significance
of This Method. It minimizes the deviation between the
experimental and theoretical line. Experimentally, the data which is obtained is
not always straight line. We will drive an equation that will over our
hypothesis (i.e. equal change in equal interval of time) and experimentally
determined values.
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 once.
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.
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
Physiological variables are not always
related linearly. However the data may be arranged or transformed to express
the relationship between variables as straight line.
Equation for ‘m’ and ‘b’
m = 
b = 
Data:
|
x(time) |
y(Cp) |
|
|
1 |
23 |
|
|
2 |
22 |
|
|
3 |
17.9 |
|
|
4 |
15.3 |
|
|
5 |
12.9 |
|
|
6 |
8.7 |
|
|
7 |
6.0 |
|
|
8 |
3.1 |
|
Objectives:
a) Calculate Co, t 1/2, k, by graphical
graphical method.
b)
Obtain the equation of straight line that best fit the data by using least
square method.
c)
Put data on straight line equation.
d) Calculate Co, t 1/2, k, by using least
square method.
Procedure:
1. On X-axis plot the data of time.
2. On Y-axis plot the data of
concentration.
3. Now by plotting the data and joining
the points we obtain a line.
4. Then extrapolate the line towards
Y-axis to obtain Cpº.
5. As some points are deviating from the
line so to find Y-intercept (Cpº) and to obtain a straight line least square
method is applied on the same data.
Observations,
Calculations and Results:
Calculation of Cpº by Graphical Method
Calculation of half-life By Graphical Method:
Least Square Method:
Calculations:
|
X |
y |
Xy |
X2 |
|
1 |
23 |
1*23= |
|
|
2 |
22 |
2*22= |
|
|
3 |
17.9 |
3*17.9= |
|
|
4 |
15.3 |
4*15.3= |
|
|
5 |
12.9 |
5*12.9= |
|
|
6 |
8.7 |
6*8.7= |
|
|
7 |
6 |
7*6= |
|
|
8 |
3.1 |
8*3.1= |
|
|
|
|
|
|
Equation of Straight Line:
Calculation
of Cº By Least Square Method:
Calculation
of Half-life:
Results:
The
value obtained of Cpº after extrapolating the line toward Y-axis is ------------------µg/ml
by graphical method, while the value of Cº obtained from, least square method
is -------------------µg/ml. Half-life obtained by graphical method is ---------------hrs.
While half-life obtained by least square method is ----------------hrs.
Conclusion:
By
plotting a graph with the new values of Cp obtained by applying least square
method a straight line is obtained on a simple rectilinear graph which have
minimize the deviation between the experimental and theoretical line and hence we
have also obtained more accurate pharmacokinetic parameters such that Cpº &
half-life.
Practical No: 8
To constrict a calibration curve between absorbance
and known concentration of drug for determination of unknown concentration
graphically
Calibration
curve, also
known as a standard curve, is a general method for
determining the unknown concentration of a substance in a sample by comparing
the unknown to a set of standard samples of known concentration.
Determination of drug in a formulation or
in blood sample requires construction of a standard or calibration curve.
Standard solutions of known concentrations are prepared and the absorbance by
UV spectrophotometer is taken. A graph
between the absorbance (dependent variable) and concentration (independent
variable) is plotted on ordinary or rectilinear graph paper. The unknown concentration can be directly
estimated from graph. The unknown
concentration can also be calculated from y-intercept and slope of the curve.
The y-intercept is estimated from the extrapolation of curve to y-ordinate.
Data:
Data
for construction of calibration curve is as follow
|
Sr.no |
Concentration
(µg/ml) |
Absorbance |
|
1 |
5 |
0.11 |
|
2 |
10 |
0.21 |
|
3 |
15 |
0.31 |
|
4 |
20 |
0.43 |
|
5 |
25 |
0.54 |
|
6 |
30 |
0.67 |
|
7 |
35 |
0.77 |
|
|
|
|
Absorbance of unknown concentration is
0.35
Procedure:
1. Plot
values of concentration on x-axis and absorbance on y-axis to construct a
calibration curve.
2. Estimate
the unknown concentration of sample directly from calibration curve by using
absorbance of sample.
3. Give
the observation regarding the curve (linear or non-linear)
4. Extrapolate
the curve towards y-axis so that it intercepts y-axis. This value is
y-intercept (b).
5. Calculate
slope value by using following formula:
Slope = =
6. Calculate
un known concentration x by the following formula:
Where m = slope and b= y-intercept
Results:
Practical No: 9
To determine the dissolution profile using
USP dissolution apparatus for a tablet containing 500 mg of a drug
Introduction
In-vitro dissolution is an important
component in drug development and quality control of the unit solid dosage
forms. It is used to assess the bioequivalence of certain drugs and is called
as in vitro bioequivalence testing.
The bioequivalence is assessed using the
USP dissolution apparatus. It consists of 6 vessels and is equipped with the
paddle or basket. The volume of dissolution medium is 900ml in each vessel. A
tablet or solid unit dosage form is added in the each vessel and the paddle or
basket is rotated at a specific speed given in USP. Samples of specified volume
are taken (usually 5 ml) from each vessel at a specified time intervals for
drug analysis. The drug is quantified in the samples by using the UV and HPLC
method.
During experimentation, several dilutions of
the samples are required, therefore, the dilution factor must be considered
carefully to calculate amount of drug. For example, each tablet is added in the
vessel containing 900 ml of dissolution media, thus, 900 is the dilution factor
and must be incorporated in calculation. Similarly, when the concentration of
drug in sample exceeds from the cut-off detection level of the UV the sample is
also diluted appropriately.
Data
The drug concentrations per ml of the
sample at different time intervals along with the dilution factor are given in
table below:
Table:
drug concentration per ml along with dilution factors
|
Time
(hr) |
Concentration
(µg/ml) |
Dilution
factor |
|
0.0 |
0 |
0 |
|
0.5 |
19 |
5 |
|
1.0 |
16.3 |
10 |
|
2.0 |
27.5 |
10 |
|
4.0 |
25.20 |
15 |
|
6.0 |
22.30 |
20 |
|
8.0 |
24.25 |
20 |
|
12.0 |
25.50 |
20 |
Procedure
1. Arrange
the data from the table as shown in table.
2. Calculate
drug concentration in sample by multiplying drug concentration per ml in column
B of table to respective dilution factors given in column C. Put the resulting
values in the column D.
3. Calculate
the total drug amount dissolved by multiplying the drug concentration in sample
with volume of dissolution media (900 ml). To convert the concentration in
µg/ml to milligram, divided the values by 1000 and put them in column E of
table.
4. Calculate
the percentage drug dissolved using following equation and put the values in
column F. Note that the drug was given as 500 mg.
5. Calculate
the amount remaining to be dissolved as in column G of table.
6. Graph
the % drug dissolved against time on rectilinear graph given in figure.
Table:
Treatment of the data given in table
|
time |
Drug
conc. |
Dilution
factor |
Drug
conc in sample |
Total
drug dissolved (mg) |
%
drug dissolved |
|
Column
A |
Column
B |
Column
C |
Column
D= ColumnB× ColumnC |
Column
E= (Column D×900)/1000* |
Column
F= (Column E × 100)/500** |
|
0.0 |
0 |
0 |
0 |
0 |
0 |
|
0.5 |
19 |
5 |
95 |
85.5 |
17.1 |
|
1.0 |
16.3 |
10 |
163 |
146.7 |
29.34 |
|
2.0 |
27.5 |
10 |
275 |
247.5 |
49.5 |
|
4.0 |
25.20 |
15 |
378 |
340.2 |
68.04 |
|
6.0 |
22.30 |
20 |
446 |
401.4 |
80.28 |
|
8.0 |
24.25 |
20 |
485 |
436.5 |
87.3 |
|
12.0 |
25.50 |
20 |
510 |
459 |
91.8 |
*900 is the volume of dissolution media
as the tablet was dissolved in 900ml of dissolution media and 1000 is to
convert µg into mg.
**500 was the drug contained in unit
dosage form (tablet).
Results
The % drug dissolved at last time
interval (12 hr) is------------------------------------------------------------------
Comment on the profile of the % drug
dissolved
(released):------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Practical No: 10
CALCULATION OF K FROM URINARY EXCRETION
DATA BY USING RATE METHOD
Introduction
If a drug is eliminated unchanged from the
plasma through the urinary excretion then its appearance in urine will be a
reflection of drug disappearance from the plasma body. So the urine data can be
used to calculate the pharmacokinetics parameter. In such case the collection
of samples of urine after administration of drug is an alternative approach. To
obtain pharmacokinetic information. In this calculation the excretion rate of
drug is assumed to be first order.
Db = Db e
dDb/dt = kDb
dDb/dt = keDb e
For urine : dDu/dt = Ke Db e
where ke = excretion rate
constant and –k= overall excretion
Rate Method :
In
the drug this method the urinary excretion rate is plotted against time on semi
log graph paper ,the slope of curve is – k/2.3 (first order) and y intercept =
keDb
As the drug urinary excretion rate (
dD/dt) cannot be determined experimentally for any given instant therefore the
average rate for urinary drug excretion (Du/t) is plotted against the average
time t*,for the collection.t* co response to the midpoint (average time) of the collection period
Data
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 are
removed periodically and assayed for parent drug. The following data was
obtained :
|
Time (hr) |
Cp (µg/ml) |
Du (mg) |
|
0.25 |
4.2 |
160 |
|
0.50 |
3.5 |
140 |
|
1.0 |
2.5 |
200 |
|
2.0 |
1.25 |
250 |
|
4.0 |
0.31 |
188 |
|
6.0 |
0.08 |
46 |
Procedure
|
Time |
Du
(mg) |
Du/t |
t* |
Slope |
|
0.25 |
160 |
160/0.25
= 640 |
0.125 |
-0.24 |
|
0.50 |
140 |
140/0.25
= 560 |
0.375 |
-0.37 |
|
1.0 |
200 |
200/0.5
= 400 |
0.750 |
-0.28 |
|
2.0 |
250 |
250/1
= 250 |
1.50 |
-0.28 |
|
4.0 |
188 |
188/2
= 94 |
3.0 |
-0.3 |
|
6.0 |
46 |
46/2
= 23 |
5.0 |
|
1. Set up the following Table :
Where t* = midpoint of collection period;
and t = time interval for collection of urine sample.
2. Construct a graph on a semi log scale of Du/t versus t*. The slope of this
line should equal - k/2.3. calculate the final results
3.Find slope for each value using the following formula and take out the
average of the slope.
slope = log
y2 – log y1 / x2- x1
Observation and Result
average slope =
intercept =
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