Biopharmaceutics Lab Manual (1)

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 1^st 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.

1^st 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 1^st 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 1^st order of reaction:

Formula: t_1/2=  

                                                          

 

 

 

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=C_O-Kt

     In Zero order Process :

                                                                          K = Slope

 

 

 

 

 

Calculation of Half-life for Zero order:

t_1/2=

 

 

 

 

 

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 1^st order reaction half –life is ___________.

For Zero order reaction half-life is _________         

PRACTICAL NO: 3

Find the AUC for Zero order and 1^st 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 1^st order reaction. There is a straight line on semi-log paper and curved line on quadrant paper.

Calculation of Half-life GraphicalMethod:

Formula:

t_1/2=  

 

 

 

 

 

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:

t_1/2=  

 

 

 

 

Calculation of Area under Curve (AUC):

Formula:

=

 

= ∑+

Data (B)

Order of reaction:

Random Method:

Slope =  =

Solution:

 

 

 

 

 

Calculation of Half-life for Random Method:

t_1/2=

 

 

 

 

 

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=C_O-Kt

 

 

 

 

Calculation of Half-life for Graphical Method:

t_1/2=

 

 

 

 

 

 

Calculation of Area under Curve (AUC):

Formula:

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= ∑+

 

 

 

Result:

AUC of 1^st 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.

K_e: It represents the elimination rate constant for drugs that are eliminated through kidney.

K_b: 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 t_1/2 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:

t_1/2=

 

 

 

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:

t_1/2 =

 

 

 

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 t_n 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 =