Lab thesis

(Phytochemistry/Phyto-Pharmacology/Ethno-Pharmacology)
Dr. Md. Golam Sadik
Professor
Department of Pharmacy
University of Rajshahi
Rajshahi-6205
Bangladesh.
Md.Imran Nur Manik
B.Pharm; M.Pharm(Thesis)
Rajshahi-6205
Bangladesh.
manikrupharmacy@gmail.com

DEDICATED TO
MY
BELOVED & RESPECTED
PARENTS
And
Honorable Supervisor

INTRODUCTION: Laboratory Manual for the Thesis Students
Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page i
manikrupharmacy@gmail.com; Department of Pharmacy, University of Rajshahi.
َ‫ن‬‫ِي‬‫م‬َ‫ل‬‫َا‬‫ع‬ ِّ‫َب‬‫ر‬ ِ‫ه‬َّ‫ِل‬‫ل‬ ُ‫د‬ْ‫م‬َ‫ح‬ْ‫ل‬‫ا‬ All the praise goes to ALLAH SUBHANAHU
WA’TALA the most gracious ,the most merciful.
I would like to express my best regards, profound
gratitude, indebtedness and deep appreciation to my
honorable and beloved supervisor Dr. Md. Golam Sadik,
M.Pharm. (D.U.), Ph.D. (Japan), Post Doc. (Japan),
Professor, Department of Pharmacy, University of
Rajshahi, for his consistent supervision, expert guidance,
enthusiastic encouragement and never-ending inspiration
throughout the entire period of this work, as well as to
prepare this Laboratory Manual. I am thankful to ALLAH
SUBHANAHU WA’TALA for giving me such an opportunity
to work in close association with him. May ALLAH
SUBHANAHU WA’TALA give him proper reward
(za-zha-kallohu-khair).
I convey my heartiest thanks to all of my friends and well-wishers specially
Kushal Biswas, Shadid Hossain, M H Setu, Shahid Hossain, Tafserul Alam, Md.
Nur Islam, Md. Najem Uddin, M N Titu, Bayezid O D, M Taimur Rahman & Nasir
Uddin for their encouragement and kind co-operation throughout this research.
The present work was accomplished in the Department of Pharmacy,
University of Rajshahi. In this regard I would like give my grateful thank to
Dr. Mir Imam Ibne Wahed, Professor and Chairman, Department of Pharmacy,
University of Rajshahi, for supporting the work.
Last but not the least, I would like to express my sincere
gratitude and heartfelt obligation to my beloved and respected
parents & sister for their moral and financial support, constant
encouragement and never ending affection, as well as blessings
when they were needed the most.
April, 2015
Author
Md. Imran Nur Manik
E-mail: gsadik2@yahoo.com
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page ii
Table of Contents
SL.NO. Name of the Experiment Page
01 Determination of Total Phenolics ----------------- 01
02 Determination of total flavanoids ----------------- 10
03 DPPH Radical Scavenging Activity --------------- 15
04 Total Antioxidants Assay ---------------------------- 22
05 Iron Reducing PowerAssay ------------------------ 31
06 Hydroxyl Radical Scavenging Assay ------------ 39
07 Lipid Peroxidation Scavenging Assay ---------- 48
08 Acetylcholinesterase inhibitory activity -------- 57
09 Butyrulcholinesterase inhibitory activity ------- 69
10 Determination of Total Proanthocyanidins ----- 75
11 Determination of AChE enzyme inhibition kinetics-- 81
12 Determination of BuChE enzyme inhibition kinetics- 86
Common Laboratory Equipments
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page iii
Tips
Tips for the most accurate test results
WASHING:
1. Wash your test tubes with soft washing materials like LIQUID VIM.
2. During washing make sure that, there is no sign of residual washing
materials like the bubbles/foam of LIQUID VIM, in the equipments.
After being confirmed the absence of bubbles/foam, wash it with
Distilled Water (DW).
DRYING:
3. Completely dry each test tube. Because, as we are dealing with
microgram (μg) of concentration, any residual water even a single drop would
change the desired concentration of sample/STD solution.
4. For drying, use hot air oven. In the oven place the equipments to be dried
‘‘invertly”. This will place the opening/mouth of the equipment towards the
ground & accordingly the residual debris (if any) would be deposited in the
mouth/opening, instead of being in the bottom of that equipment.
5. Never put your hands/figures inside the dried equipments.
MEASURING:
6. Use waxy paper for weighing of the samples, standards, reagents etc.
For an accurate weighing of the solid reagents with normal paper, take 1.2/1.3 mg
to weigh 1 mg of solids. Because 0.4/0.3/0.5 mg may adhere to the paper on
which you are taking your weights.
MAKING SOLUTION:
7. Use large test tubes (16×150) for 5 ml of total final volume, & small test tubes
(12×100) for 3 ml total volume of final solution. Similarly use beakers of various
sizes as per requirement of the volume of the solution to be prepared in it.
8. Always mark all the test tubes with their corresponding concentrations as well
as the beakers with their corresponding solutions to be prepared in it.
9. If your sample is re-dispersible, then dry the samples (bone dry) in small
amount, and re-disperse them after accurate weighing.
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page iv
10. After taking out any samples, standards, reagents etc from the refrigerator,
keep at least 20 minutes in the room temperature before using. Therefore take out
them at least 20 minutes before your need.
TAKING ABSORBANCE:
11. Make sure that the absorbance cell is free from all kinds of contamination
including residual debris.
12. Take the absorbance from the solutions with the lowest concentration to the
solutions with the highest concentration. (i.e. Gradual increment of the test
solution concentrations)
13. After taking each absorbance, before proceeding to the next one with different
concentrations, if possible ALWAYS completely discard the residual solutions;
i.e. solutions remaining in the ABSORBANCE CELL.
14. If you have enough test solutions (e.g. more than 3 ml, which may be
5 ml/6ml/7ml) then, range the ABSORBANCE CELL with small volumes of the test
solutions having different concentrations in each turn before taking the
absorbance of that solution. This will prevent the cross contamination.
15. If Precipitation (PPT) appears, then CENTRIFUGE the solution and use the
supernatant layer for taking the absorbance. You can use micropipettes to
withdraw the supernatant layer more conveniently.
16. Generally 2.5 ml (more than 2 ml) of solution in the absorbance cell is enough
to take the absorbance.
17. Use new reagents and freshly prepared solvents & solution for each tests.
18. Always use the ‘‘Replacement theory’’ whenever applicable, to make the
sample & standard solutions.
pH ADJUSTMENT:
19. If measurements are taken successively in different samples, it is
recommended to rinse the electrode thoroughly with de-ionized water or tap water
and then with some of the next sample in order to prevent cross-contamination.
20. The pH reading is affected by temperature. In order to measure the pH
accurately, the temperature effect must be compensated for.( Detailed description
will be mentioned in case of the particular test where pH adjustment is
mandatory)
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page v
CENTRIFUGATION:
Centrifuge Safety Guide
Centrifugation may present two serious hazards:
i. Mechanical failure and
ii. Dispersion of aerosols.
Therefore this segment would describe safety
and maintenance procedures to minimize
centrifuge hazards.
Safe Procedures for Centrifugation:
Before centrifugation
------------------------------
1. Check tubes, bottles, and rotors for cracks and deformities before each use.
2. Make sure that the rotor, tubes, and spindle are dry and clean.
3. Never overfill centrifuge tubes (don't exceed ¾ full).
4. Always cap tubes before centrifugation.
5. Always balance buckets, tubes, and rotors properly.
6. Check that the rotor is seated on the drive correctly, close the lid on the
centrifuge, and secure it.
7. When using swinging bucket rotors, make sure that all buckets are hooked
correctly and move freely.
During centrifugation
------------------------------
1. Close lids at all times during operation. Never open a centrifuge until the
rotor has stopped.
2. Do not exceed safe rotor speed.
3. The operator should not leave the
centrifuge until full operating speed is
attained and the machine appears to be
running safely without vibration.
4. Stop the centrifuge immediately if an
unusual condition (noise or vibration)
begins and check load balances.
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page vi
After centrifugation
------------------------------
1. Allow the centrifuge to come to a complete stop before opening.
Emergency Procedures:
Emergency Situations
------------------------------
The following events are considered an emergency:
• If there is a spill in the centrifuge
• If centrifuge malfunctions
• If there is rotor failure
• If there is tube breakage
Emergency Procedures
------------------------------
• Turn centrifuge off immediately, close the centrifuge lid.
• Notify others, evacuate the lab,
Overall, steps would be clear to you, during your lab work.
Molecular Weights Essentials
Name of the Atom(Symbol) Mean Relative Atomic Mass
Hydrogen (H
Nitrogen (N)
Oxygen (O)
Sodium (Na)
Phosphorus (P)
Sulphur (S)
Chlorine(Cl)
Potassium (K)
Ferrous(Fe)
1.0079~1
14.0067~14
15.9994~16
22.9897~23
30.9737~31
32.065~32
35.45
39.09983~39.1
55.85
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page vii
In A Nutshell
Name of the reagents Molecular Formula
Molecular weight
(g/mol)
Ammonium molybdate [NH4]6Mo7O24.4H2O. 1235.9
Trisodium phosphate; TSP Na3PO4 163.94
DI-POTASSIUM
PHOSPHATE, Potassium
Hydrogen Phosphate,
Potassium dibasic
phosphate, Potassium
phosphate dibasic.
K2HO4P
174.18
Potassium acid phosphate KH2PO4 136.09
Caustic potash; Potash lye KOH 56.11/56.1056
2-deoxy-L-erythro-
pentose;
(3R,4S)-3,4,5-
trihydroxypentanal;
L-Deoxyribose
C5H10O4 134.13
Edetate disodium salt
dihydrate
C10H14N2Na2O8·2H2O 372.24
Ferric chloride;
Iron trichloride; Molysite
FeCl3 162.20/ 162.204
Oxydol; perhydrol H2O2 34.01/ 34.01468
Vitamin C;
L-Threoascorbic acid;
Antiscorbutic factor
C6H8O6 176.12/
176.12412
2-Amino-2-
(hydroxymethyl)-1,3-
propanediol TRIS BASE
C4H11NO3 121.14
Potassium Chloride KCl 74.55
Iron (III) chloride FeCl3 162.204
Hydrogen chloride HCl 36.46
Sodium Hydroxide, Pellets NaOH 40
5,5′-dithiobis-(2-
nitrobenzoic acid)
C14H8N2O8S2 396.35
Sodium Chloride NaCl 58.44
Magnesium Chloride,
Hexahydrate
MgCl2.6H2O 203.30
Acetylthiocholine iodide C7H16INOS (Linear Formula)
CH3COSCH2CH2N(CH3)3I
289.18/289.17751
S-Butyrylthiocholine
iodide
C9H20INOS (Linear Formula)
(CH3)3N(I)CH2CH2SCOCH2CH2CH3
317.23/
317.23067
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page viii
Test Number Detection Color(Final solution before Absorbance) Result
01 with increasing sample/Std.concentration: WISC) +VE
02 WISC) +ve
03 গ ো WISC) +VE
(04)& 05 & ‘’ +VE
06+07 গ ো WISC) +VE
08+09 WISC) +VE
10 WISC) +VE
Physical properties of Hydrochloric acid
Concentration Density Molarity pH Viscosity
Specific
heat
Vapour
pressure
Boiling
point
Melting
point
kg HCl/kg kg HCl/m3
Baumé kg/L mol/dm3
mPa·s kJ/(kg·K) kPa °C °C
10% 104.80 6.6 1.048 2.87 −0.5 1.16 3.47 1.95 103 −18
20% 219.60 13 1.098 6.02 −0.8 1.37 2.99 1.40 108 −59
30% 344.70 19 1.149 9.45 −1.0 1.70 2.60 2.13 90 −52
32% 370.88 20 1.159 10.17 −1.0 1.80 2.55 3.73 84 −43
34% 397.46 21 1.169 10.90 −1.0 1.90 2.50 7.24 71 −36
36% 424.44 22 1.179 11.64 −1.1 1.99 2.46 14.5 61 −30
38% 451.82 23 1.189 12.39 −1.1 2.10 2.43 28.3 48 −26
The reference temperature and pressure for the above table are 20 °C and 1 atmosphere
(101.325 kPa). Vapour pressure values are taken from the International Critical Tables, and refer
to the total vapour pressure of the solution.
N.B.The Baumé scale is a pair of hydrometer scales developed by French pharmacist Antoine Baumé in
1768 to measure density of various liquids.
The Test Reagents
Name of the Reagents (Formulae) Name of the Reagents (Formulae)
Folin-Ciocalteu Reagent: FCR Phosphate Buffer (K2HPO4+KH2PO4) pH 6.6
Sodium Carbonate (Na2CO3) Potassium Ferricyanide :K3Fe(CN)6
Aluminium Chloride (AlCl3) Trichloro Acetic Acid (TCA)
Sodium Nitrate (NaNO2) Ferric Chloride (FeCl3)
Soium Hydroxide (NaOH) 2-Deoxy -D- ribose, (C5H10O4)
DPPH, Vanillin KH2PO4-KOH Buffer :pH 7.4
Sulfuric Acid (H2SO4) EDTA
Sodium Phosphate (Na3PO4) Gallic Acid & Catechin (Standard)
Ammonium Molybdate Hydrogen Peroxide (H2O2)
Thio-Barbituric Acid (TBA) Ascorbic Acid,(C6H8O6)
Md.
Imran
Nur
Manik

Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page ix
Clearance: In this manual the term ‘‘Replacement Theory’’ represents
‘‘Serial Dilution’’ procedure. The terms ‘‘WHOLE” and ‘‘REPLACE” has been used
only for distinguishing between each other as well as easy understanding.
The Test apparatuses
01. Rotary Evaporator
02. UV machine
03. pH meter
04. Vortex Machine
05. Water Bath
06. Centrifuge Machine
07. Electrical Balance
{3 Digits after the point (0.000g); 4 Digits after the point (0.0000g)}
08. Hot Air Oven
09. Measuring cylinder (25 ml, 100 ml, 500 ml)
10. Beaker (10 ml, 50 ml,100 ml, 250 ml, 500 ml,1000 ml)
11. Test tubes (12×100, 16×150, 16×125)
12. Falcon tubes (15 ml, 50 ml)
13. Eppendrop Tubes
14. Glass rod
15. Spatula
16. Pipettes (1 ml,2 ml,10 ml)
17. Micropipettes (10-100 μl,100-1000 μl)
18. Tips
19. Reagent Bottles
20. Glass Vials
21. Test tube Stands etc.
Please pay your attention
This manual is an experimental step for the simplification of these tests. It is
expected that, this tests would be made very easier in the coming future.
Therefore as a first step it is very common to have mistakes. Thus we hope that,
you would provide us with your valuable suggestions for the improvement of this
manual. May ALLAH SUBHANAHU WA’TALA give you people proper reward.
(with za-zha-kallohu-khair)- Md. Imran Nur Manik  imran.md39@gmail.com(+fb)
Md.
Imran
Nur
Manik

Laboratory Manual For The Thesis Students
Comprehensible Composition By: Md. Imran Nur Manik; M.Pharm. (Thesis) Page 1
Test Number: 01
Title: Determination of Total Phenolics.
Reagents:
1. Folin-Ciocalteu Reagent
: FCR {10 Times diluted with Distilled Water (DW)},
2. 7.5% W/V Sodium Carbonate (Na2CO3) (Solvent: DW),
3. Gallic Acid: GA (Standard) {(Solvent: 70% Methanol)}.
Also known as Folin-Ciocalteu’s Phenol Reagent.
Method: (Total volume=5 ml)
1. Take 0.5 ml (500 μl) of Sample/Standard solution in the test tube,

2. Add 2.5 ml of Folin-Ciocalteu Reagent: FCR (10 Times diluted with DW),

3. Then add 2 ml of 7.5% W/V Na2CO3,

4. Vortex each test tubes for 1 minute,

5. Keep the reaction mixture for 25 (20-30 minutes) minutes at room temperature,

6. Take the absorbance at 760 nm.
Blank:
Take 0.5ml (500 μl ) of the respective solvent,
(e.g. DW/70% Methanol/100% Methanol/Or ,other solvent)

Add 2.5 ml of FCR (10 Times diluted with DW),

Then add 2 ml of 7.5% W/V Na2CO3,

Vortex each test tubes for 1 minute,

Keep the reaction mixture for 25 minutes at room temperature,

Take the absorbance at 760 nm.
Md.
Imran
Nur
Manik

Question: Prepare a standard Gallic Acid curve for the determination of Total
Phenolics having Gallic acid concentrations as follows.
Concentrations of Gallic Acid (μg/ml)
1
2
4
8
16
32
Answer:
Solvent: For this test & for many others, we would use 70% methanol as solvent.
(For dissolving Standards e.g. Gallic Acid for this case, and for the preparation of the
Sample solutions as well as the blank solutions.)
Preparation of 25 ml, 70% methanol (CH3OH) Solvent:
100 ml of 70% CH3OH requires = 70 ml of CH3OH
So, 25 ml ,, ,, ,, = {( 70×25)÷100} ml ,, ,, = 17.5 ml of CH3OH
Now take 17.5 ml of 100% methanol in a 25 ml measuring cylinder & add the
required volume of DW (7.5 ml) into it, to make the final solution volume = 25 ml.
Preparation of Folin-Ciocalteu Reagent (FCR):
Calculate the volume of FCR needed for your test & then prepare FCR solution.
For e.g. Let, 1 set of test solutions, having 7 test tubes, where each test tube
contains 2.5 ml of FCR, would require 7×2.5=17.5 ml of FCR.
Now let us prepare 20 ml 0f FCR:
Take 2 ml of FCR in a 50 ml beaker & 18 ml of DW in a 25 ml measuring cylinder .
Now, add 18 ml of DW into the FCR containing beaker with continuous stirring &
make the final solution volume =20 ml.
Alternatively you can take 2 ml of FCR in 25 ml measuring cylinder & add 18 ml
of DW into the FCR containing measuring cylinder to make the solution
volume=20 ml.
N.B. Here, you may have to roll the measuring cylinder between your two palms to
prepare a homogeneous FCR solution. You can also use a clean stirrer for this
purpose.
Md.
Imran
Nur
Manik

Preparation of 7.5% Na2CO3 Solution:
Like FCR, Calculate the volume of Na2CO3 solution required for your test.
For e.g.1 set of test solution having 7 test tubes, where each test tube contains
2 ml of Na2CO3, would require 7×2=14 ml of Na2CO3.
Now let us prepare 15 ml of 7.5% (W/V) Na2CO3 Solution:
100 ml of 7.5% Na2CO3 requires = 7.5 gm of Na2CO3
So, 15 ml ,, ,, ,, = {( 7.5×15)÷100} gm = 1.125 gm of Na2CO3.
Now, take 1.125 gram of Na2CO3, in a 50 ml beaker and 15 ml of DW in a 25 ml
measuring cylinder.
Then gradually pour a small volume of DW (lowest volume needed for dissolving
the Na2CO3.) from the measuring cylinder with continuous stirring to completely
dissolve Na2CO3.
After the complete dissolution, pour all the DW left in the measuring cylinder to
make final solution volume =15 ml.
N. B. To make our test cost effective, & to use minimum amount of solvents;
before going to prepare the standard Gallic Acid (GA)solution, we would modify
our test to have the final solution volume of 3 ml, instead of 5 ml.
Because, 3 ml is enough to take the absorbance accurately. 
Modified formula (3 ml):
1. Take 0.3 ml (300 μl) of Sample /Standard solution in the test tube,

2. Add 1.5 ml of Folin-Ciocalteu Reagent: FCR (10 Times diluted with DW),

3. Then add 1.2 ml of 7.5% W/V Na2CO3,

4. Vortex each test tubes for 1 minute,

5. Keep the reaction mixture for 25 (20-30 minutes) minutes at room temperature,

Md.
Imran
Nur
Manik

Blank:
Take 0.3ml (300 μl )of the respective Solvent
( DW/70% Methanol/100% Methanol/respective solvent)

Add 1.5 ml of FCR (10 Times diluted with DW)

Then add 1.2 ml of 7.5% W/V Na2CO3,

Vortex each test tube for 1 minute,

Keep the reaction mixture for 25(20-30 minutes) minutes at room temperature,

Take the absorbance at 760 nm.
Preparation of the Standard Gallic Acid (GA) Solution:
Dissolve 1 mg of GA in 4ml of 70% methanol (STOCK SOLUTION). Therefore, the
GA concentration of this STOCK SOLUTION would be = 1mg÷4 ml
or, = 0.25 mg/ml. (Or, 0.25 μg/μl}
When 1 ml of this STOCK SOLUTION is given in a 3 ml of total volume
(e.g. 1 ml of GA solution+ 2 ml other solution=3 ml solution), then it would be
diluted and it’s GA concentration would be changed as follows
( 0.25 mg÷3 ml)=0.0833mg/ml. or, 0.0833µg/μl (DILUTION FACTOR)
Therefore, to make a solution of 1 μg/ml of GA in a 3 ml of final solution volume,
we have to take,
0.0833 μg is present in = 1 μl of solution
Therefore, 1 μg is present in = (1÷0.0833) μl of solution =12 μl of solution
{This Means After Dilution
0.08333 μg of GA would be presented in =1 μl of solution
Therefore, 1 μg of GA would be presented in =12 μl of solution }
Basic Concept:
When you take 12 μl of STOCK SOLUTION, then you actually take (12×0.25)= 3 μg
of GA, and accordingly, when you put this 3 μg of GA in a 3 ml of solution,
you actually put (3 μg÷3 ml)=1 μg/ml of GA in the final 3 ml of solution.
Md.
Imran
Nur
Manik

To make 0.3 ml (300 μl) of standard GA solution with different concentrations you
have to take the following volumes of GA, from the 4 ml STOCK SOLUTION and
add the following volumes of 70% methanol to make the sample solution
volume =300 μl.
Volumes from the
stock solution (μl)
Volumes of 70% methanol
to be added (μl )
Total Volume
of Sample
(μl)
GA Concentration
(μg ) in the 3 ml of
final volume
12 288 300 1
24 276 300 2
48 252 300 4
96 204 300 8
192 108 300 16
384 -84 is not Applicable(N/A) N/A
32
 You cannot take 384 μl of solution as that exceeds the total sample volume of
0.3 ml (300 μl). Therefore, you have to make another solution to full fill
the reaction set.
Another solution:
Dissolve 1 mg of GA in 1ml of 70% methanol ( STOCK SOLUTION ). Therefore, the
GA concentration of this solution would be 1mg÷1ml=1 mg/ml. {Or,1μg/ μl }.
When 1 ml of this STOCK SOLUTION is given in a 3 ml of total volume, then it
would be diluted and it’s GA concentration would be changed as follows
(1mg÷3 ml)=0.333mg/ml. or, 0.333μg/μl (DILUTION FACTOR).
Therefore, to make a solution of 1μg/ml of GA in a 3 ml of final solution volume,
we have to take,
Therefore, 32 μg is present in = {(1÷0.333)×32} μl of solution = 96 μl of solution
{This Means After Dilution 0.333 μg of GA would present in = 1 μl of solution
Therefore, 32 μg of GA would present in = 96 μl of solution }
Basic Concept:
When you take 96 μl of STOCK SOLUTION, then you actually take (96×1)μg =96 μg
of GA, and accordingly, when you put this 96 μg of GA in a 3 ml of solution,
you actually put (96 μg÷3 ml)=32 μg/ml of GA in the final 3 ml of solution.
Md.
Imran
Nur
Manik

To make 0.3 ml (300 μl) of standard GA solution with different concentrations you
have to take the following volumes, from the 1 ml STOCK SOLUTION and add the
following volumes of 70% methanol to make the sample solutions volume= 300 μl.
Volumes from the
Volumes of
70% methanol
to be added (μl )
Total Sample
Volume (μl)
GA Concentration (μg )
in the 3 ml of final
volume
3 297 N/A
(300) 1
6 294 N/A
(300) 2
12 288 300 4
24 276 300 8
48 252 300 16
96 204 300 32
You would not be able to take 3 μl or 6 μl of solutions as they do not matches
with the capacities of our available micropipettes. Therefore, you have to make
both of the solutions to full fill the required set of GA concentrations.
N.B. To have efficient test is would be very helpful if you have 2-20 µl, 10-100 µl
and 100-1000 µl micropipette. 
Simplified Calculation: (Replacement theory.) -1
Dissolve 1 mg of Gallic Acid (GA) in 1ml of 70% methanol. Therefore, the GA
concentration of this STOCK solution would be (1 mg÷1ml) =1 mg/ml.
{or, 1 μg/μl}.
A. Take 96 μl from the 1 ml STOCK solution & add 204 μl of 70% methanol.
{Total volume =0.3 ml (300 μl)}. Let this 0.3 ml solution be named as, ‘‘WHOLE’’.
B. Now, add the ‘‘WHOLE’’ totally in the 3 ml of final solution. Therefore, the GA
concentration of this 3 ml solution would be 32 μg/ml.
(Because, according to Basic Concept, when you take 96 μl of STOCK
solution, then you actually take (96×1)μg =96 μg of GA, and accordingly, when you
put this 96 μg of GA in a 3 ml of solution, you actually put
(96 μg÷3 ml)=32 μg/ml of GA in the final 3 ml of solution.)
C. Take another of 96 μl from the 1 ml STOCK solution & add 504 μl of 70%
methanol. (Total volume =600 μl ). Let this 600 μl solution be named as,
‘‘RECYCLE’’.
D. Now take 300 μl from the 600 μl of ‘‘RECYCLE’’ and add totally in the 3 ml of
final solution. Therefore, the GA concentration of this 3 ml solution would be
16 μg/ml.
Md.
Imran
Nur
Manik

Because, according to Basic Concept, 600 μl of original ‘‘RECYCLE’’ contains,
(96×1) μg= 96 μg of GA. Therefore, when you take 300 μl from the ‘‘RECYCLE’’ you
actually take (96 ÷2) μg =48 μg of GA & addition of this 300 μl of solution in the 3
ml of final solution, makes the GA concentration,
(48 μg÷3 ml)=16 μg/ml.
Remember:
Always replace the amount 300μl, taken from the ‘‘RECYCLE’’ with 300μl of 70%
methanol; and always keep it’s volume = 600 μl .
E. Accordingly, withdrawing off 300 μl from the ‘‘RECYCLE’’ and replacing it with
300μl of 70% methanol (to the ‘‘RECYCLE’’ ) would decrease the GA concentration,
by one half (i.e. Concentration would be divided by two) in each turn as follows.
Volumes of
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
{‘‘RECYCLE’’}
Volumes from
the 0.6 ml
‘‘RECYCLE’’
Concentration
in the Final
3 ml solution
(μg )
300 μl 300 μl 600 μl 300 μl 8
300 μl 300 μl 600 μl 300 μl 4
300 μl 300 μl 600 μl 300 μl 2
300 μl 300 μl 600 μl 300 μl 1
Simplified Calculation: (Replacement theory.)-2
Dissolve 1 mg of Gallic Acid (GA) in 1 ml of 70% methanol. Therefore, the GA
concentration of this STOCK solution would be (1 mg÷ 1ml) = 1 mg/ml.
or, = 1 μg/μl.
{Total volume =0.6 ml (600 μl)}.
Let this 0.6 ml solution be named as, ‘‘RECYCLE’’.
B. Now take 300 μl from the 600 μl of ‘‘RECYCLE’’ and add totally in the 3 ml of
final solution. Therefore, the GA concentration of this 3 ml solution
would be =32 μg/ml.
Because, 600 μl of original ‘‘RECYCLE’’ contains, (192×1) μg= 192 μg of GA.
Therefore, when you take 300 μl from the ‘‘RECYCLE’’ you actually take
(192 ÷2) μg = 96 μg of GA & addition of this 300 μl of solution in the 3 ml of final
solution, makes the GA concentration,( 96 μg÷3 ml)=32 μg/ml.
Md.
Imran
Nur
Manik

C. Replace the 300 μl taken out from the ‘‘RECYCLE’’ with 300 μl of 70%
methanol. {Total volume =0.6 ml (600 μl)}.
D. Again, take 300 μl from the 600 μl of ‘‘RECYCLE’’ and add totally in the 3 ml of
final solution. In this turn, the GA concentration of this 3 ml solution
would be = 16 μg/ml.
Because, in this time, 600 μl of ‘‘RECYCLE’’ contains, the left (96×1) μg= 96 μg
of GA. Therefore, when you take 300 μl from the ‘‘RECYCLE’’ you actually take
(96 ÷2) μg = 48 μg of GA & addition of this 300 μl of solution to the 3 ml of final
solution, makes the GA concentration,( 48 μg÷3 ml)=16 μg/ml.
E. Accordingly, withdrawing off 300 μl from the ‘‘RECYCLE’’ and replacing it with
300μl of 70% methanol (to the ‘‘RECYCLE’’ ) would decrease the GA concentration,
by one half (i.e. Concentration would be divided by two) in each turn as follows.
Volumes of
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
Volumes from
the 0.6 ml
‘‘RECYCLE’’
Concentration
in the Final
3 ml solution
(μg )
300 μl 300 μl 600 μl 300 μl 8
300 μl 300 μl 600 μl 300 μl 4
300 μl 300 μl 600 μl 300 μl 2
300 μl 300 μl 600 μl 300 μl 1
Remember: Always replace the amount 300μl, taken from the ‘‘RECYCLE’’ with
300μl of 70% methanol; and always keep it’s volume = 600 μl .
Result: (Courtesy: Md.Imran Nur Manik.)
Concentrations of
Gallic Acid
(μg/ml)
Absorbance
Mean ± STD
A B C
1 0.192 0.180 0.172 0.181±0.010066
2 0.293 0.272 0.266 0.277±0.014177
4 0.58 0.521 0.513 0.538±0.036592
8 0.957 0.918 0.913 0.929±0.02409
16 1.649 1.61 1.678 1.645±0.034122
32 3.215 2.914 3.135 3.088±0.155907
Md.
Imran
Nur
Manik

Equation for the Determination of Gallic Acid Equivalent (Also applicable for Catechin Equivalent)
Example:
For, y = 0.093x + 0.131
Sample Concentration =25 μg/ml
Sample Volume = 0.5 ml
Sample Absorbance = 1.663
mg/gm

y = 0.0932x + 0.1314
R² = 0.9986
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40
Absorbence
Concentration(μg/ml)
Gallic Acid
Gallic Acid
Linear (Gallic Acid)
Md.
Imran
Nur
Manik

Test Number: 02
Title: Determination of total flavanoids.
Reagents:
1. 10% W/V Aluminium Chloride (AlCl3), (Solvent: DW),
2. 5% W/V Sodium Nitrate (NaNO2), (Solvent: DW),
3. 1mM (0.001 M) Soium Hydroxide (NaOH), (Solvent: DW),
4. Catechin (Standard).
1. Take 0.5 ml (500 μl) of Sample/Standard solution in the test tube,

2. Add 2.5 ml of DW,

3. Then add 0.15 ml (150 μl) of 5%W/V NaNO2,

4. Stand for 5 minutes,

5. Now add 0.3 ml (300 μl) of 10% W/V AlCl3,

6. Stand for another 5 minutes,

7. Add 1 ml of 1mM (0.001 M) NaOH,

8. Afterwards add 0.55 ml (550 μl) of DW,

9. Mix the solution well,

10. If Precipitation (PPT ) appears then, centrifuge the reaction mixtures
at 4000 rpm for 10 minutes.

11. In case of centrifugation; after the completion, carefully withdraw the
supernatants by using a micropipette.

Blank: Use 0.5 ml (500 μl) of 70% methanol instead of Sample/STD.
Rest of the reagents & procedures would be the same as described above.
Md.
Imran
Nur
Manik

Question: Prepare a standard Catechin curve for the determination of Total Flavanoids
having Catechin concentrations as follows.
Concentration of Catechin (μg/ml)
31.25
62.5
125
250
500
Answer Preparation of the Reagents:
5%W/V NaNO2 (5 ml):
100 ml 5% NaNO2 requires = 5 gm of NaNO2
So, 5 ml ,, ,, = {( 5×5)÷100} gm of NaNO2
= 0.25 gm of NaNO2
Now, take 0.25 gram of NaNO2 in a 50 ml beaker (or in a large test tube) & 5 ml of
DW in a measuring cylinder. Then add 5 ml of DW with continuous stirring
(in case of a beaker) or, careful shaking (in case of a test tube) to completely
dissolve the NaNO2.
10% W/V AlCl3 (10 ml):
100 ml 10% AlCl3 requires = 10gm of AlCl3
So, 10 ml ,, ,, = {( 10×10)÷100} gm of AlCl3
= 1 gm of AlCl3
Now, take 1 gram of AlCl3, in a 50 ml beaker and 10 ml of DW in a measuring
cylinder. Then gradually pour small volumes of DW from the measuring cylinder
with continuous stirring to completely dissolve the AlCl3.
After the complete dissolution of AlCl3, add the rest volume of DW to make the final
solution volume = 10 ml.
1mM (0.001M) NaOH (500 ml):
We Know that, W= (SMV÷1000).
Here, S =1mM~0.001 M; // M= (23×1+16×1+1) =40; // V=250 ml
To make 1000 ml solution, we need W= {(0.001 ×40×1000) ÷1000}
or, W=0.020 gm or 20 mg of NaOH
Now, take 0.02gm (20 mg) of NaOH in a 500 ml beaker and 500 ml of DW in a
500 ml measuring cylinder. Then dissolve the NaOH with small volumes of DW
(20-30 ml). When NaOH becomes completely dissolved, and then make the final
solution volume, 500 ml by pouring the left DW from the measuring cylinder.
Md.
Imran
Nur
Manik

If Proper Storage Condition (air tight system) is available, then you can
store 250 ml or 100 ml of this solution in a 250 ml or 100 ml volumetric
flask respectively, for further utilisation.
70% Methanol solution (25 ml):
Take 17.5 ml of 100% methanol in a 25 ml measuring cylinder and make the
solution volume up to 25 ml mark with DW.
Preparation of the Standard Catechin Solution:
My (Imran Nur Manik) calculation:
Dissolve 5 mg of Catechin (CAT) in 1ml of 70% methanol (STOCK SOLUTION ).
Therefore, the CAT concentration of this STOCK SOLUTION would be = 5 mg÷1ml
Or, = 5 mg/ml. {Or, 5 μg/μl}.
When 1 ml of this STOCK SOLUTION is given in a 5 ml of total volume ,then it
would be diluted & it’s CAT concentration would be changed as follows,
(5 mg ÷5 ml)=1mg/ml. or, 1 μg/μl (DILUTION FACTOR).
Therefore, to make a solution of 500μg/ml of Catechin in a 5 ml of final volume,
we have to take,
1 μg is present in = 1 μl of solution
So, 500 μg is present in= (1×500) μl of solution = 500 μl of solution
1 μg of Catechin would be presented in =1 μl of solution
Therefore, 500 μg of Catechin would be presented in =500 μl of solution }
Basic Concept:
When you take 500 μl STOCK SOLUTION, then you actually take,
(500×5) μg = 2500 μg of Catechin, and accordingly when you put this 2500 μg of
Catechin in the 5 ml of solution, then you actually put 2500 μg÷5 ml=500 μg/ml
of Catechin in the final 5 ml of solution. 
To make 0.5 ml (500 μl) of standard Catechin solution with different
Concentrations you have to take the following volumes of Catechin, from the 1ml
STOCK SOLUTION and add the following volumes of 70% methanol to make the
volume 500 μl.
Md.
Imran
Nur
Manik

Volumes from the
Volumes of
70% methanol
to be added (μl )
Total
Volume (μl)
CAT Concentration (μg )
in the 5 ml final volume
31.25
468.75 500 31.25
62.5 437.5 500 62.5
125 375 500 125
250 250 500 250
500 0 500 500
 You cannot take this volume, as it does not match with the capacities of our
available micropipettes. Therefore; you have to use Replacement theory to full fill
the set of catechin concentrations indicated in the question.
Simplified Calculation: (Replacement theory.)
Dissolve 5 mg of Catechin in 1ml of 70% methanol. Therefore, the catechin
concentration of this STOCK solution would be (5 mg÷1ml)=5 mg/ml. {Or, 5 μg/μl}.
A. Take 0.5 ml (500 μl) from the 1 ml STOCK solution add totally in the 5 ml of
final solution.
Therefore, the catechin concentration of this 5 ml solution would be 500μg/ml.
(Because, according to Basic Concept, 500 μl of STOCK solution contains,
(500×5) μg= 2500 μg of Catechin. & addition of this 0.5 ml (500 μl) of solution
in the 5 ml of final volume, makes the catechin concentration = 2500 μg÷5 ml
=500 μg/ml.)
B. Take the rest of 0.5 ml (500 μl) from the 1 ml STOCK solution & add
0.5 ml (500 μl) of 70% methanol. (Total volume =1 ml).
Let this 1 ml solution be named as, ‘‘RECYCLE’’.
C. Now take of 0.5 ml (500 μl) from the 1 ml of ‘‘RECYCLE’’ and add totally in the
5 ml of final solution. Therefore, the catechin concentration of this 5 ml solution
would be 250μg/ml.
Because, according to Basic Concept, 1 ml of original ‘‘RECYCLE’’ contains,
(500×5) μg= 2500 μg of Catechin. Therefore, when you take 0.5 ml (500 μl )from
the ‘‘RECYCLE’’ you actually take (2500 ÷2) μg =1250 μg of catechin & addition of
this 0.5 ml (500 μl )of solution in the 5 ml of final solution, makes the catechin
concentration 1250 μg÷5 ml=250 μg/ml.
Remember: Always replace the amount 0.5 ml (500 μl ),taken from the
‘‘RECYCLE’’ with the 500μl of 70% methanol; and always keep it’s volume = 1 ml.
Md.
Imran
Nur
Manik

D. Accordingly, withdrawing off 0.5 ml (500 μl) from the ‘‘RECYCLE’’ and replacing
it with 0.5 ml (500μl) of 70% methanol (to the ‘‘RECYCLE’’ ) would decrease the
concentration of Catchin, by one half (i.e. Concentration is divided by two) in each
turn as follows.
Volumes of
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
Volume from
the 1ml
‘‘RECYCLE’’
Concentration
in the Final
5 ml solution (μg
)
0.5ml (500 μl) 0.5ml (500 μl) 1 ml 0.5ml (500 μl) 125
0.5ml (500 μl) 0.5ml (500 μl) 1 ml 0.5ml (500 μl) 62.5
0.5ml (500 μl) 0.5ml (500 μl) 1 ml 0.5ml (500 μl) 6.25
0.5ml (500 μl) 0.5ml (500 μl) 1 ml 0.5ml (500 μl) 31.25
Result: (Courtesy: Md.Imran Nur Manik. )
Concentration(µg/ml)
Absorbance
Mean±STD
A B C
31.25 0.244 0.241 0.245 0.243±0.002082
62.5 0.453 0.450 0.452 0.451±0.001528
125 0.915 0.911 0.914 0.913±0.002082
250 1.805 1.802 1.804 1.803±0.001528
500 3.458 3.51 3.487 3.485±0.026058

y = 0.0069x + 0.036
R² = 0.9997
0
0.5
1
1.5
2
2.5
3
3.5
4
0 100 200 300 400 500 600
Absorbence
Catechin
Catechin
Linear (Catechin)
Md.
Imran
Nur
Manik

Test Number: 03
Title: DPPH Radical Scavenging Activity.
Reagents:
1. 0.004% DPPH (1 mg in 25 ml solvent),
2. Catechin (Standard).
Method (Total Volume=5 ml):
1. Take 2 ml of Sample/Standard Solution in the test tube,

2. Add 3ml of 0.004% DPPH solution,

3. Keep the reaction mixture 30 min in a dark place,

4. After 30 minutes measure the absorbance at 517 nm.
Blank: Take 3 ml of Solvent (Here 70% methanol), and auto zero the absorbance
value at 517 nm.
DPPH (Control): Take 3 ml of DPPH solution in the absorbance cell & take it’s
absorbance at 517 nm.
Calculation of the Percentage (%) of Scavenging:
Percentage of (%) Scavenging = {1-(A Sample÷ A DPPH)} ×100
Here,
A Sample = Absorbance of the Sample.
A DPPH = Absorbance of the DPPH.
Question: Prepare a standard Catechin curve for the determination of DPPH
Radical Scavenging Activity, having Catechin concentrations as follows.
Concentration of Catechin (μg/ml) Concentration of Catechin (μg/ml)
100 3.125
50 1.5625
25 0.78125
12.5 0.390625
6.25 0.1953125
Other names of DPPH:
A. 2,2-diphenyl-1-picrylhydrazyl
B. 1,1-diphenyl-2-picrylhydrazyl radical
C. 2,2-diphenyl-1-(2,4,6-trinitrophenyl)hydrazyl
D. Diphenylpicrylhydrazyl
Md.
Imran
Nur
Manik

70% Methanol solution (25 ml): Take 17.5 ml 100% methanol in a 25 ml
measuring cylinder and make the volume up to 25 ml mark with DW.
DPPH Solution: Dissolve 1 mg of DPPH in 25 ml of 70% of methanol.
Dissolve 1 mg of Catechin in 1ml of 70% methanol (STOCK SOLUTION). Therefore,
the concentration of this STOCK SOLUTION would be = 1 mg÷1ml
=1 mg/ml. {Or, 1 μg/µl}.
When 1 ml of this STOCK SOLUTION is given in a 5 ml of total volume ,then it
would be diluted & it’s catechin concentration would be changed as follows,
1 mg ÷5 ml=0.2 mg/ml. or, 0.2 μg/μl (DILUTION FACTOR).
Therefore, to make a solution of 100μg/ml of Catechin in the 5 ml of final volume,
we have to take,
100 μg is present in= {(1×100)÷0.2} μl of solution = 500 μl of solution
0.2 μg of Catechin would be presented in =1 μl of solution
Therefore, 100 μg of Catechin would be presented in =500 μl of solution }
Basic Concept:
When you take 500 μl STOCK SOLUTION, then you actually take (500×1)μg=500 μg
of Catechin, and accordingly when you put this 500 μg of Catechin in the 5 ml of
final solution, then you actually put 500 μg÷5 ml=100 μg/ml of Catechin in the
final 5 ml of solution. 
Md.
Imran
Nur
Manik

To make 2 ml (2000 μl) of standard Catechin solution with different
concentrations you have to take the following volumes of Catechin, from the 1ml
volume 2000 μl.
Volumes from the stock
solution (μl)
Volumes of 70%
methanol
to be added (μl )
Total
Volume (μl)
Concentration(μg) in
the Final
5 ml final volume
500 1500 2000 100
250 1750 2000 50
125 1875 2000 25
62.5 1937.5 2000 12.5
31.25
1968.75 2000 6.25
15.625
1984.375 2000 3.125
7.8125
1992.188 2000 1.5625
3.90625
1996.094 2000 0.78125
1.953125
1998.047 2000 0.390625
0.976563
1999.023 2000 0.1953125
 You cannot take these volumes of solutions, as they do not match with the
capacities of our available micropipettes. Therefore; you have to use
Replacement theory to full fill the set of concentrations dictated in the question.
Dissolve 1mg of Catechin in 1ml of 70% methanol. Therefore, the concentration of
this STOCK solution would be 1 mg÷1ml=1 mg/ml. {Or, 1 μg/μl}.
A. Take 0.5 ml (500 μl) from the 1 ml STOCK solution & add 1.5 ml of 70%
methanol.(Total volume =2 ml). Let this 2 ml solution be named as ‘‘WHOLE’’.
B. Take the rest of 0.5 ml (500 μl) from the 1 ml STOCK solution & add 3.5 ml of
70% Methanol. (Total volume = 4 ml).
Let this 2 ml solution be named as ‘‘RECYCLE’’.
C. Now add the ‘‘WHOLE’’ (500 μl) totally in the 5 ml of final solution.
Therefore, the Catechin concentration of this 5 ml solution would be 100 μg /ml.
(According to Basic Concept, when you put 500 μl of STOCK solution in the
5 ml of solution, then you actually you put (500×1) μg= 500 μg of Catechin,
and Catechin concentration becomes, 500 μg÷5 ml=100 μg/ml. )
Md.
Imran
Nur
Manik

D. Now, take 2 ml from the 4 ml ‘‘RECYCLE’’ and add totally in the 5 ml of final
solution.
Therefore, the Catechin concentration of this 5 ml solution would be 50 μg/ml.
Beacuse according to Basic Concept, 4 ml of original ‘‘RECYCLE’’ contains
(500×1) μg= 500 μg of Catechin. Thus, when you take 2 ml from the 4 ml
‘‘RECYCLE’’ you actually take (500 ÷2) μg =250 μg of catechin, and addition of this
2 ml of solution in the 5 ml of final volume, makes the Catechin concentration
250 μg ÷5 ml=50 μg /ml. 
Remember: Always replace the amount (2 ml), taken from the ‘‘RECYCLE’’
with 2 ml of 70% methanol; and always keep it’s volume= 4 ml.
E. Accordingly, withdrawing off 2 ml from the ‘‘RECYCLE’’ and replacing it with
2 ml of 70% methanol to the ‘‘RECYCLE’’ would decrease the concentration of
Catchin by one half, (i.e.Concentration is divided by two) in each turn as follows.
Volumes of
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
Volumes from
the 2ml
‘‘RECYCLE’’
Concentration
in the Final
5 ml solution
(μg )
2 ml 2 ml 4 ml 2 ml 25
2 ml 2 ml 4 ml 2 ml 12.5
2 ml 2 ml 4 ml 2 ml 6.25
2 ml 2 ml 4 ml 2 ml 3.125
2 ml 2 ml 4 ml 2 ml 1.5625
2 ml 2 ml 4 ml 2 ml 0.78125
2 ml 2 ml 4 ml 2 ml 0.390625
2 ml 2 ml 4 ml 2 ml 0.1953125
Another Simplified Calculation: (Replacement theory.)
Dissolve 1mg of Catechin in 1ml of 70% methanol. Therefore, the concentration of
this STOCK solution would be 1 mg÷1ml=1 mg/ml. {Or, 1 μg/μl}.
A. Take 0.5 ml (500 μl) from the 1 ml STOCK solution & add 0.5 ml of 70%
methanol.(Total volume =1 ml). Let this 1 ml solution be named as ‘‘WHOLE’’.
B. Take the rest of 0.5 ml (500 μl) from the 1 ml STOCK solution & add another
0.5 ml of 70% Methanol. (Total volume = 1 ml).
Let this1 ml solution be named as ‘‘RECYCLE’’.
Md.
Imran
Nur
Manik

C. Now put the 1ml of ‘‘WHOLE’’ totally in that reaction test tube, marked as
100μg/ml. Then add 1 ml of 70 % methanol to make the sample solution
volume=2 ml.
Finally add 3 ml of 0.004% DPPH Solution to make the 5 ml of final solution.
Therefore, the Catechin concentration of this 5 ml solution would be
100 μg /ml.
(According to Basic Concept, the original 1ml of ‘‘WHOLE’’ contains 500 μl of
STOCK solution i.e. (500×1) μg= 500 μg of Catechin .Therefore, when you put this
1ml of ‘‘WHOLE’’ in the 5 ml of final solution, then you actually put
(500×1) μg= 500 μg of Catechin, and accordingly catechin concentration becomes,
500 μg÷5 ml=100 μg/ml. )
D. OK. Now, take 500 μl from the 1 ml ‘‘RECYCLE’’ and add 1.5 ml of 70%
methanol .Total volume = 2 ml. Aaddition this 2 ml sample solution totally in the
5 ml of final solution, in the same way as described in C, would make the
Catechin concentration 50 μg/ml.
Beacuse according to Basic Concept, 1ml of original ‘‘RECYCLE’’ contains
(500×1) μg= 500 μg of Catechin. Thus, when you take 500 μl from the ‘‘RECYCLE’’
you actually take (500 ÷2) μg =250 μg of catechin, and addition of this 500 μl of
solution in the 5 ml of final solution makes the Catechin concentration
250 μg ÷5 ml=50 μg /ml. 
E. In this way put 500 μl from the ‘‘RECYCLE’’ in the reaction test tubes, marked
with their respective concentrations.
While replace the volume, 500 μl taken from the ‘‘RECYCLE’’ with 500 μl of
70% methanol ,to always keep the ‘‘RECYCLE’’ volume = 1 ml.
Then add 1.5 ml of 70 % methanol to make the sample solution volume=2 ml.
Finally add 3 ml of 0.004% DPPH Solution to make the 5 ml of final solution.}
N.B.
To have more convenience, you can put 500 μl of 70% methanol after
putting 500 μl from the ‘‘RECYCLE’’ in the reaction test tubes, marked with their
respective concentrations.
This would make the sample volume = 1 ml.
Then you have to put another 1ml of 70 % methanol, into these test tubes to
make the 2 ml of the sample solution.
Remember:
Always replace the amount 500 μl, taken from the ‘‘RECYCLE’’ with 500 μl of 70%
methanol; and always keep it’s volume= 1 ml.
F. Accordingly, withdrawing off 500 μl from the ‘‘RECYCLE’’ and replacing it with
500 μl of 70% methanol to the ‘‘RECYCLE’’ would decrease the concentration of
Md.
Imran
Nur
Manik

Volumes of
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
Volumes from
the 1ml
‘‘RECYCLE’’
Concentration
in the Final
5 ml solution (μg )
500 μl 500 μl 1 ml 500 μl 25
500 μl 500 μl 1 ml 500 μl 12.5
500 μl 500 μl 1 ml 500 μl 6.25
500 μl 500 μl 1 ml 500 μl 3.125
500 μl 500 μl 1 ml 500 μl 1.5625
500 μl 500 μl 1 ml 500 μl 0.78125
500 μl 500 μl 1 ml 500 μl 0.390625
500 μl 500 μl 1 ml 500 μl 0.1953125
Precautions:
 All the test tubes must be covered (sealed) with cotton.
Because, evaporation of methanol may change the solute
concentration.
 The absorbance should be taken just after 30 minutes.
Because, after 30 minutes the absorbance value decreases.
Catechin is a potent radical scavenger.
Therefore it should be used in small concentration
(highest concentration should not exceed 25 μg/ml)
Absorbance Percentage (%) of
scavengingA B C
0.195 49.08 49.12 49.09 49.10
0.391 61.32 61.35 61.27 61.31
0.781 75.32 75.35 75.34 75.34
1.563 92.1 92.06 92.07 92.08
3.125 94.78 94.81 94.82 94.80
6.25 95.28 95.24 95.23 95.25
12.5 95.22 95.27 95.25 95.25
25 95.5 95.47 95.47 95.48
Md.
Imran
Nur
Manik

Figure: UV-Spectrophotometer

0
20
40
60
80
100
120
0 20 40 60 80 100 120
Catechin
CME
EAF
CLF
PEF
AQF
Md.
Imran
Nur
Manik

Test Number: 04
Title: Total Antioxidants Assay.
Reagents:
NO. Working Concentration
(when prepared)
Desired Concentration
(in the reaction mixture)
1. 1.8M Sulphuric Acid (H2SO4) 0.6M Sulphuric Acid (H2SO4)
2. 84 mM Sodium Phosphate (Na3PO4) 28 mM Sodium Phosphate (Na3PO4)
3.
3% Ammonium Molybdate
[NH4]6Mo7O24.4H2O
1% Ammonium Molybdate
[NH4]6Mo7O24.4H2O
4. Catechin(Standard) Conc. As directed by the Procedure
Solvent: DW for the Reagents & 70%Methanol for the Standard
Method: (Total volume=3.5 ml)
1. Take 0.5 ml (500 μl) of Sample/Standard Solution in the test tube,

2. Add 3 ml of reaction mixture,

3. Incubate the test tubes at 95°C for 10 minutes,

4. Cool the test tubes at room temperature for 10 minutes,

Blank: 0.5 ml (500 μl) of 70% methanol+3 ml reaction mixture (=3.5 ml)
Rest of the procedures & reagents would be the same as described for the samples
& standard.
Question: Prepare a standard Catechin curve for the determination of Total
Antioxidants Assay, having Catechin concentrations as follows.
100
50
25
12.5
6.25
3.125
1.5625
Md.
Imran
Nur
Manik

1.8 M H2SO4 (20 ml): {Specific Gravity (density) 1.84 g/cm3, liquid & Purity 98 %.}
Generally these acids are supplied in form of V/V solution
We know that, W= (SMV÷1000).
Here, S=1.8 M; / M=( 1×2+32+16×4)= 98; / V=20 ml
To make 20 ml solution, we need W={( 1.8 ×98×20)÷1000}//or, W=3.528 gm
But, H2SO4 is in the liquid form & we know that, ρ=(M÷V)
Here, ρ =Specific Gravity =1.84
Therefore, V=(M÷ρ)//or, V=(3.528÷1.84)=1.917ml
Given that,H2SO4 is 98% Pure,(V/V Solution)
That means, 98 ml is present in = 100 ml
Thus, 1.917 ml is present in = {(100×1.917)÷98}=1.956 ml~2 ml
Now, take 2ml of the concentrated H2SO4 in a measuring cylinder and add 18 ml
of DW to make the solution volume = 20 ml.
Another Calculation (More specific):
Where, M=98.079
Therefore, to make 20 ml of solution, we would need
W ={( 1.8 ×98.079×20)÷1000} gram
or, W =3.530844gm~3.531gm
Accordingly, V=(M÷ρ)//or, V=(3.531÷1.84) // or,V=1.9190217ml~1.919 ml
For 98% purity,(V/V solution)
1.917 ml would be presented in ={(100×1.919)÷98} ml /=1.9581633 ml
= (1.9581633 ×1000) μl / =1958.1633 μl~ 1958 μl
Now, take 1958 μl of concentrated H2SO4 in a 25 ml measuring cylinder
(1.9 ml with 10 ml pipette & 58 μl with a micropipette ) and add 18 ml of DW to
make the final solution volume = 20 ml .
Md.
Imran
Nur
Manik

84 mM Na3PO4 (20 ml):
Here, S=84 mM~0.084 M; // M= (23×3+31+16×4) =164; // V=20 ml
To make 20 ml solution, we need W={( 0.084 ×164×20)÷1000}
W=0.27552 gm~0.276gm
Now, take 0.276gm gram of Na3PO4, in a 50 ml beaker and 20 ml of DW in a
measuring cylinder. Then gradually pour small volumes of DW from the measuring
cylinder with continuous stirring to completely dissolve the Na3PO4.
After the complete dissolution add the rest volume of DW to make the final
solution volume = 20 ml.
Where, M=163.94
Therefore, to make a 20 ml solution, we would need
W= {( 0.084×163.94×20)÷1000} gram
or, W= 0.2754192 gm~0.275gm
Now take 0.275gm of Na3PO4 in a 50 ml beaker & make the solution as described
above.
 N.B. you can use Di-Potassium Mono-Hydrogen Phosphate (K2HPO4) instead of
Sodium Phosphate (Na3PO4). In that case calculation would be as follows:
Calculation for K2HPO4:
84 mM K2HPO4 (20 ml):
Here, S=84 mM~0.084 M; // M= (39×2+1+31+16×4) =174; // V=20 ml
To make 20ml solution, we would need W = {(0.084 ×174×20) ÷1000} gram
or, W= 0.29232 gm~0.292gm
Now, take 0.292 gm of K2HPO4, in a 50 ml beaker and 20 ml of DW in a measuring
cylinder.
Then gradually pour small volume of DW from the measuring cylinder with
continuous stirring to completely dissolve K2HPO4.
After the complete dissolution add the rest volume of DW to make the solution
final volume 20 ml.
Md.
Imran
Nur
Manik

Where, M= 174.18
Therefore, to make 20ml solution,we would need W= {(0.084 ×174.18×20)÷1000}gm
or, W=0.2926224 gm~0.293 gm
Now, take 0.293 gm of K2HPO4 in a 50 ml beaker and make the solution as
described above. 
3% Ammonium Molybdate: [NH4]6Mo7O24.4H2O (25 ml) :
100 ml of 3% of Ammonium Molybdate requires = 3 gm of Ammonium Molybdate
So, 25 ml ,, ,, ,, ,, = {( 3×25)÷100} gm
= 0.75 gm of Ammonium Molyb.
Now, take 0.75 gram of Ammonium Molybdate, in a 50 ml beaker and 25 ml of DW
in a measuring cylinder.
Then gradually pour small volums of DW from the measuring cylinder with
continuous stirring to completely dissolve the Ammonium Molybdate.
Afterwards add the rest volume of DW to make the final solution volume 20 ml.
N.B. Ammonium Molybdate does not dissolve completely; thus use it’s
powder & if needed break down the agglomerates into powder.
Try your best to disperse it into the solution (you can use a stirrer).
 As it forms a suspension (milky white in color) therefore, it must be filtered
before adding into the reaction mixture.
 Filtration can be done with cotton filter or filter paper.
 Make 10 ml more solution than your estimated need, in case of cotton
filtration, and 5 ml for the filtration with filter paper.
70% Methanol solution: Take 17.5 ml 100% methanol in a 25 ml measuring
cylinder and make the volume up to 25 ml mark with DW.
Preparation of the Reaction Mixture:
 Reaction mixture should be prepared freshly, just prior to the addition to the
final Sample/Standard Solution. Standing of reaction mixture for a longer period
may result in precipitation (PPT) and thus accurate result would not be achieved.
Reaction mixture = 1ml of 1.8 M H2SO4+ 1 ml of 84 mM(0.084 M) Na3PO4+1 ml of
3% Ammonium Molybdate Solutions=3 ml of Reaction Mixtures.
Md.
Imran
Nur
Manik

Here we have made  20ml of 1.8 M H2SO4+ 20 ml of 84 mM(0.084 M)
Na3PO4+20 ml of 3% Ammonium Molybdate Solutions = 20 ml Reaction Mixtures.
Dissolve 1 mg of Catechin in 1ml of 70% methanol (STOCK SOLUTION). Therefore,
the catechin concentration of this STOCK SOLUTION would be 1mg÷1ml=1 mg/ml.
Or, =1 μg/μl.
When 1 ml of this STOCK SOLUTION is given in a 3.5 ml of total volume ,then it
would be diluted & it’s concentration would be changed as follows,
1 mg ÷3.5 ml=0.2857143mg/ml. or, 0.2857143 μg/μl (DILUTION FACTOR).
Therefore, to make a solution of 100μg/ml of Catechin in the 3.5 ml of final
volume, we have to take,
100 μg is present in= {(1×100)÷0.2857143 } μl of solution
=349.99998~350 μl of solution
0.2857143 μg of Catechin would be present in = 1 μl of solution
Therefore, 100 μg of Catechin would be present in = 350 μl of solution }
Basic Concept:
When you take 350 μl STOCK SOLUTION, then you actually take
(350×1) μg = 350 μg of Catechin, and accordingly when you put this 350 μg of
Catechin in the 3.5 ml of solution, you actually put 350 μg÷3.5 ml=100 μg/ml of
Catechin in the final 3.5 ml of solution. 
To make 0.5 ml (500 μl) of standard Catechin solution with different
concentrations you have to take the following volumes of Catechin, from the 1ml
sample solution volume 500 μl.
Volumes from
the stock solution (μl)
Volumes of
70% methanol
to be added (μl )
Total
Volume (μl)
Concentration(μg)
in the
3.5 ml final volume
350 150 500 100
175 325 500 50
87.5 412.5 500 25
43.75
456.25 500 12.5
21.875
478.125 500 6.25
10.9375
489.0625 500 3.125
5.46875
494.5313 500 1.5625
Md.
Imran
Nur
Manik

You cannot take these volumes of solutions, as they do not match with the
capacities of our available micropipettes. Therefore; you have to use
Replacement theory to full fill the set of concentrations indicated in the question.
Dissolve 1 mg of Catechin (CAT) in 1ml of 70% methanol. Therefore, the CAT
concentration of this solution would be 1 mg÷1ml=1 mg/ml. {Or,1 μg/μl}.
A. Take 0.7 ml or, 700 μl from the above 1 ml solution & add 0.3 ml (300 μl) of
70% methanol. (Total volume =1 ml).
Let this 1 ml solution be named as, the STOCK.
B. Now take 0.5 ml (500 μl) from the 1 ml STOCK solution & add totally in the
3.5 ml of final solution.
Therefore, the catechin concentration of this 3.5 ml solution would be 100μg/ml.
(According to Basic Concept, 1 ml of STOCK solution contains
(700×1)μg = 700μg of catechin. Therefore, 0.5 ml or 500μl would contain
{(700×500) ÷1000} μg =350μg of CAT. Thus addition of this 500μ of STOCK
solution to the 3.5 ml of final solution makes the Catechin concentration of final
solution 350 μg÷3.5 ml=100 μg /ml. )
C. Take the left 0.5 ml (500μl) of STOCK solution & add 0.5 ml of 70% Methanol.
(Total volume =1 ml). Let this 1 ml solution be named as, ‘‘RECYCLE’’.
D. Now take 0.5 ml (500μl) from the 1 ml ‘‘RECYCLE’’ and add totally in the
3.5 ml of final solution. Therefore, the concentration of this 3.5 ml solution would
be 50μg/ml.
350 μg of catechin. Thus, when you take 500μl from the ‘‘RECYCLE’’ you actually
take (350÷2) μg=175 μg of catechin, and addition of this 500μl solution in the
3.5 ml of final solution makes the catechin concentration=175 μg ÷3.5 ml
=50 μg /ml. 
with their respective concentrations . While replace the volume, 500 μl taken from
the ‘‘RECYCLE’’ with 500 μl of 70% methanol ,to always keep the ‘‘RECYCLE’’
volume = 1 ml.
Remember: Always replace the amount 500μl, taken from the ‘‘RECYCLE’’ with
the 500μl of 70% methanol; and always keep it’s volume= 1 ml.
Md.
Imran
Nur
Manik

F. Accordingly, withdrawing off 0.5 ml (500μl) from the ‘‘RECYCLE’’ and replacing
it with 0.5 ml (500μl) of 70% methanol (to ‘‘RECYCLE’’ ) would decrease the
concentration of Catchin, by one half (i.e.Concentration is divided by two) in each
turn as follows.
Volumes of
‘‘RECYCLE’’
Left
70%methanol
(Replacement)
Equals to
‘‘RECYCLE’’
Volumes from
‘‘RECYCLE’’
Concentration
(μg ) in the Final
3.5 ml solution
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 25
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 12.5
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 6.25
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 3.125
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 1.5625
Dissolve 3.5 mg of Catechin (CAT) in 1ml of 70% methanol (STOCK) . Therefore,
concentration of this solution would be 3.5 mg÷1ml=3.5mg/ml. {Or,3.5 μg/μl}.
A. Take 100 μl from the above 1 ml STOCK solution & add 400 μl of 70%
methanol. (Total volume =500 μl).Let this 1 ml solution be named as, the WHOLE.
B. Now add this 500 μl of WHOLE totally in the 3.5 ml of final solution.
Therefore, the catechin concentration of this 3.5 ml solution would be 100μg/ml.
(According to Basic Concept, 0.5 ml (500 μl) of WHOLE contains
(100×3.5)μg = 350 μg of catechin. Thus addition of this 0.5ml (500μl) of WHOLE
solution to the 3.5 ml of final solution makes the Catechin concentration of final
the solution, 350 μg÷3.5 ml=100 μg /ml. )
C. Now take another 100μl from the STOCK solution & add 900 μl of 70%
methanol. (Total volume =1ml). Let this 1 ml solution be named as, ‘‘RECYCLE’’.
D. Now take 500μl from the 1 ml ‘‘RECYCLE’’ and add totally in the 3.5 ml of final
solution. Therefore, the concentration of this 3.5 ml solution would be 50μg/ml.
(100×3.5) μg = 350 μg of catechin. Thus, when you take 0.5 ml (500μl) from the
‘‘RECYCLE’’ you actually take (350÷2) μg=175 μg of catechin, and addition of this
0.5 ml (500μl) solution in the 3.5 ml of final solution makes the catechin
concentration 175 μg ÷3.5 ml=50 μg /ml. 
Md.
Imran
Nur
Manik

E. In this way put 500 μl from the ‘‘RECYCLE’’ to the reaction test tubes, marked
with their respective concentrations . While replace the volume, 500 μl taken from
the ‘‘RECYCLE’’ with 500 μl of 70% methanol ,to always keep the ‘‘RECYCLE’’
volume = 1 ml.
Remember: Always replace the amount 0.5 ml (500μl), taken from the
‘‘RECYCLE’’ with the 0.5 ml (500μl) of 70% methanol; and always keep it’s
volume= 1 ml.
F. Accordingly, withdrawing off 0.5 ml (500μl) from the ‘‘RECYCLE’’ and replacing
it with 0.5 ml (500μl) of 70% methanol (to ‘‘RECYCLE’’ ) would decrease the
concentration of Catchin, by one half (i.e.Concentration is divided by two) in each
turn as follows.
Volumes of
‘‘RECYCLE’’
Left
70%methanol
(Replacement)
Equals to
‘‘RECYCLE’’
Volumes from
‘‘RECYCLE’’
Concentration
(μg ) in the Final
3.5 ml solution
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 25
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 12.5
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 6.25
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 3.125
0.5 ml (500μl) 0.5 ml (500μl) 1 ml 0.5 ml (500μl) 1.5625
Precautions:
 Do not heat the test tubes for more than 10 minutes. Turn on the water bath
earlier and put the test tubes when the temperature is accurately 95°C.
 Do not delay in submerging the test tubes in the 95 °C heated water bath, after
the addition of the reaction mixtures.
 The test should be repeated if PPT appears after heating.
(As it would not give the accurate absorbance.)
 The test tubes should be cooled after withdrawing from the water bath.
But the cooling time should not exceed for more than 10 minutes.
Thus the UV machine should be turned on previously, so that the absorbance
can be taken just after the cooling period.
This test requires 2 set of equal number of test tubes for 1 set of reaction.
Md.
Imran
Nur
Manik

Name of sample Conc. (μg/ml)
Absorbance
Mean
A B C
Catechin
6.25 0.107 0.101 0.103 0.104
12.5 0.202 0.201 0.206 0.203
25 0.381 0.375 0.374 0.377
50 0.69 0.685 0.688 0.688
100 1.255 1.251 1.248 1.251

0
0.5
1
1.5
2
2.5
0 20 40 60 80 100 120
Absorbance
Concentrate (μg/ml)
Total Antioxidant Activity
Catechin
CME
AQF
EAF
CHCl3
PET
Md.
Imran
Nur
Manik

Test Number: 05
Title: Iron Reducing Power Assay.
Reagents:
1. 0.2 M Phosphate Buffer (K2HPO4+KH2PO4): pH 6.6,
2. 1% Potassium Ferricyanide :K3Fe(CN)6 (Solvent :DW),
3. 10% Trichloro Acetic Acid (TCA) (Solvent: DW),
4. 0.1%Ferric Chloride (FeCl3) (Solvent: DW),
5. Catechin (Standard) (Solvent: 70% Methanol).
Method:
1. Take 1 ml of Sample/Standard Solution in the test tube,

2. Add 2.5 ml of 0.2 M Phosphate Buffer & 2.5 ml of 1% Potassium Ferricyanide,

3. Incubate the reaction mixtures at 50°C for 20 minutes,

4. Add 2.5 ml of 10% TCA immediately after withdrawing from the water bath,

5. If Precipitation (PPT) appears, then CENTRIFUGE the reaction mixtures
at 4000 rpm for 10 min,

6. Withdraw 2.5 ml of supernatant (in case of PPT )
Otherwise withdraw the test tube solution
and add 2.5 ml of DW,

7. Now add 0.5 ml of 0.1% FeCl3,

8. Keep the Reaction solutions at room temperature, for 10 minutes,

Blank: Use 1 ml of 70% methanol instead of Sample/Standard.
Rest of the procedures & reagents would be the same.
1 ml of 70% methanol+2.5 ml 0.2 M Phosphate Buffer & 2.5 ml K3Fe (CN)6+ Incubate the
reaction mixture at 50°C for 20 minutes+ Add 2.5 ml of 10% TCA immediately after
withdrawing from water bath.(If PPT appears, then CENTRIFUGE the reaction mixture at
3000 rpm for 5 min)+In case of PPT after CENTRIFUGATION withdraw 2.5 ml of
supernatant; otherwise withdraw 2.5 ml from the heated test tubes and add 2.5 ml of DW+
Add 0.5 ml of 0.1% FeCl3+ Keep the solution for 10 minutes+Take the absorbance at 700 nm.

Md.
Imran
Nur
Manik

Question: Prepare a standard Catechin curve for the determination of Iron Reducing
Power Assay, having Catechin concentrations as follows.
100
50
25
12.5
6.25
3.125
1.5625
0.2M Phosphate Buffer (K2HPO4+KH2PO4): pH 6.6
0.2 M K2HPO4 (100 ml):
Here, S=0.2M; // M= (39×2+1+31+16×4) =174; // V=100 ml
To make 100ml solution, we need W= {(0.2 ×174×100) ÷1000}//or, W=3.480 gm
Where, M= 174.18
To make 100ml solution, we would need W = {(0.2 ×174.18×100)÷1000} grams
or, W =3.4836 gm~3.484 gm of K2HPO4.
0.2 M KH2PO4 (100 ml):
Here, S=0.2M; // M= (39×1+1×2+31+16×4) =136; // V=100 ml
Where, M= 136.09
To make 100ml solution, we would need W= {(0.2 ×136.09×100)÷1000} grams
or, W=2.7218 gm~2.722 gm of KH2PO4.
Now, take 3.48gm of K2HPO4 in a 100 ml beaker and 100 ml of DW in a 100 ml
measuring cylinder. Then gradually pour small volumes of DW (30 ml) from the
measuring cylinder with continuous stirring to completely dissolve K2HPO4.
Finally add the rest volume of DW (70 ml) to make the solution volume 100 ml.
Similarly dissolve 2.72 gm KH2PO4 in another 250 ml beaker.
Md.
Imran
Nur
Manik

Another Calculation (More specific): Take 3.484gm of K2HPO4 in a 100 ml beaker
and make the solution as described above. Similarly dissolve 2.722gm KH2PO4 in
another 250 ml beaker.
pH Adjustment at 6.6:
Place the 250ml beaker with the K2HPO4, beneath the
pH meter stand and Submerge the electrode and the
temperature probe approximately 4 cm (1½”) into it and stir
gently.
Allow time for the electrode to stabilize.
The pH is displayed on the primary LCD and the
temperature on the secondary LCD.
Now, add the KH2PO4 solution and observe the pH meter reading. When pH meter
reading shows 6.6 than your pH is considered to be adjusted. If needed use 10M
KOH or 1M HCl solution to adjust the desired pH at 6.6.
Fig. Use of pH meter.
Storage: After preparation, the buffer can be kept for more than 1 week at cool
temperature. (See Storage condition of Phosphate buffer.) In case of deep fridge storage
/cool/cold temperature storage always the buffer must be brought to the normal
temperature and check for the desired pH.
Md.
Imran
Nur
Manik

1% Potassium Ferricyanide: K3Fe(CN)6 (50 ml):
100 ml 1% K3Fe(CN)6 requires = 1 gm of K3Fe(CN)6
So, 50 ml ,, ,, ,, = {( 1×50)÷100} gm = 0.5 gm of K3Fe(CN)6
Now, take 0.5 gram of K3Fe(CN)6, in a 100 ml beaker & 50 ml of DW in a 100 ml
measuring cylinder. Then gradually pour a small volume of DW (20 ml) from the
measuring cylinder with continuous stirring to completely dissolve the K3Fe(CN)6.
When the K3Fe(CN)6 becomes completely dissolved then, add the rest volume of
DW (30 ml) to make the final solution volume 50 ml.
Generally K3Fe(CN)6 solution has a clear yellowish colour, but with time the
colour becomes deep. Thus always use the freshly prepared solution.
N.B. Potassium Ferricyanide K3Fe(CN)6 is very toxic and
corrosive, therefore handle it with the gloves. 
10% Trichloro Acetic Acid (TCA) (50ml):
For Solid TCA:
100 ml 10 % TCA requires = 10 gm of TCA
So, 50 ml ,, ,, ,, = {( 10×50)÷100} gm = 5 gm of TCA
Now, take 5 gram of TCA, in a 100 ml beaker & 50 ml of DW in a 100 ml
measuring cylinder. Then gradually pour a small volume of DW (20 ml) from the
measuring cylinder with continuous stirring to completely dissolve TCA.
Finally add the rest volume of DW (30 ml) to make the final solution volume 50 ml.
For Liquid TCA:
100 ml 10 % TCA requires = 10 ml of TCA
So, 50 ml ,, ,, ,, = {( 10×50)÷100} ml = 5 ml of TCA
Now, take 5 ml of TCA, in a 100 ml beaker & 50 ml of DW in a 100 ml measuring
cylinder. Then gradually pour a small volume of DW (20 ml) from the measuring
cylinder with continuous stirring to completely dissolve TCA.
N.B. Trichloro Acetic Acid (TCA) is very toxic and corrosive. So, handle with
gloves
Md.
Imran
Nur
Manik

0.1%Ferric Chloride (FeCl3)(25 ml):
100 ml 0.1% FeCl3 requires = 0.1gm of FeCl3
So, 25 ml ,, ,, ,, = {( 0.1×25)÷100} gm = 0.025 gm of FeCl3
Now, take 0.025 gram of FeCl3, in a 50 ml beaker & 25 ml of DW in a measuring
cylinder. Then gradually pour a small volume of DW (10 ml) from the measuring
cylinder with continuous stirring to completely dissolve FeCl3.
Generally FeCl3 solution is faintly yellowish in colour, but with time the
colour becomes deep. Since it is highly reactive (oxidises rapidly in the air),
thus use freshly prepared solution.
N.B. Ferric Chloride (FeCl3) is very toxic and corrosive, thus handle with
gloves.
Preparation of the Standard Catechin Solution: (Replacement theory)
N.B. In case of this experiment, the calculation is not clearly understood.
There was no scope for me (Imran Nur Manik) to understand the calculation in my
own way. Thus I am just giving you what have been provided in the manual written
by Muniya Khatun . 
Dissolve 5 mg of Catechin in 1ml of 70% methanol. Therefore, the concentration
of this solution would be 5 mg÷1ml=5 mg/ml. {Or,5 μg/μl}.
 N.B. The writer of this formula has dictated the ‘‘DILUTION FACTOR’’ for this
test, as 18 
A. Take 0.72 ml (720 μl) from the above1 ml solution & add 1.28 ml of 70%
methanol. (Total volume =2 ml). Let this 2 ml solution be named as, the STOCK.
Md.
Imran
Nur
Manik

B. Take 1 ml from the 2ml STOCK add totally in the 18 ml of final solution.
Therefore, the catechin concentration of this 18 ml solution would be 100μg/ml
Basic Concept:
When you take 720 μl of solution, then you actually take (720×5) μg= 3600 μg of
Catechin. Therefore, 2 ml of original STOCK solution contains 3600 μg of Catechin.
Thus, when you take 1 ml from the 2 ml STOCK solution, you actually take
(3600 ÷2) μg =1800 μg of catechin.
So, addition of this 1 ml STOCK solution to the final 18 ml solution, makes the
Catechin concentration, 1800 μg÷18 ml=100 μg/ml. 
C. Now take the rest of 1 ml from the 2 ml STOCK solution & add 1 ml of 70%
methanol. (Total volume =2 ml). Let this 2 ml solution be named as, ‘‘RECYCLE’’.
D. Now take 1 ml from the 2 ml ‘‘RECYCLE’’ and add totally in the 18 ml of final
solution. Therefore, the Catechin concentration of this 18 ml solution would be
50μg/ml.
Beacuse according to Basic Concept, 2 ml of original ‘‘RECYCLE’’contains
(3600 ÷2) μg =1800 μg of catechin. Thus, when you take 1 ml from the 2 ml
‘‘RECYCLE’’ you actually take (1800÷2) μg=900 μg of catechin, and addition of this
1 ml solution to the 18 ml of final solution makes the Catechin concentration
900 μg ÷18 ml=50 μg /ml. 
Remember: Always replace the amount (1 ml), taken from the ‘‘RECYCLE’’
with1 ml of 70% methanol; and always keep it’s volume= 2 ml.
E. Accordingly, withdrawing off 1 ml from the ‘‘RECYCLE’’ and replacing it with
1 ml of 70% methanol to ‘‘RECYCLE’’ would decrease the concentration of Catchin
by one half, (i.e.Concentration is divided by two) in each turn as follows.
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
‘‘RECYCLE’’
Volume
from the
‘‘RECYCLE’’
Concentration
(μg ) in the final
18 ml(?) solution
1 ml 1 ml 2 ml 1 ml 25
1 ml 1 ml 2 ml 1 ml 12.5
1 ml 1 ml 2 ml 1 ml 6.25
1 ml 1 ml 2 ml 1 ml 3.125
1 ml 1 ml 2 ml 1 ml 1.5625
Md.
Imran
Nur
Manik

Dissolve 5 mg of Catechin in 1ml of 70% methanol. Therefore, the concentration
of this STOCK solution would be 5 mg÷1ml=5 mg/ml. {Or, 5 μg/μl}.
(Total volume =500 μl ). Let this 500 μl solution be named as ‘‘WHOLE’’.
B. Now take another 360 μl from the 1 ml STOCK solution & add 640 μl of 70%
Methanol. (Total volume =1 ml ). Let this 1 ml solution be named as ‘‘RECYCLE’’.
C. Now put the 500 μl of ‘‘WHOLE’’ totally in that reaction test tube, marked as
100μg/ml. Then add 500 μl of 70 % methanol to make the sample solution
volume=1 ml.
Therefore, the Catechin concentration of this solution would be 100 μg /ml.
(According to Basic Concept, the original 500 μl of ‘‘WHOLE’’ contains 360 μl
of STOCK solution i.e. (360×5) μg= 1800 μg of Catechin .Therefore, when you put
this 500 μl of ‘‘WHOLE’’ in the 18 ml of final solution, then you actually put
1800 μg of Catechin, and accordingly catechin concentration becomes,
1800 μg÷18 ml=100 μg/ml. )
D. OK. Now, take 500 μl from the 1 ml ‘‘RECYCLE’’ and add 0.5 ml of 70%
methanol .Total volume =1 ml. Addition this 1 ml sample solution totally in the
18 ml of final solution, in the same way as described in C, would make the
Catechin concentration 50 μg/ml
(360×5) μg= 1800 μg of Catechin. Thus, when you take 500 μl from the 1 ml
‘‘RECYCLE’’ you actually take (1800 ÷2) μg =900 μg of catechin, and addition of
this 500 μl of solution in the 18 ml of final solution makes the Catechin
concentration 900 μg ÷18 ml=50 μg /ml. 
with their respective concentrations.
While replace the volume, 500 μl taken from the ‘‘RECYCLE’’ with 500 μl of
70% methanol ,to always keep the volume of the ‘‘RECYCLE’’ = 1 ml.
Then add 1.5 ml of 70 % methanol to make the sample solution volume=2 ml.
Finally add 3 ml of 0.004% DPPH Solution to make the 5 ml of final solution.}
N.B. To have more convenience, you can put 500 μl of 70% methanol after
putting 500 μl from the ‘‘RECYCLE’’ in the reaction test tubes, marked with their
respective concentrations. This would make the sample volume = 1 ml.
Md.
Imran
Nur
Manik

Remember: Always replace the amount 500 μl , taken from the ‘‘RECYCLE’’
with 500 μl of 70% methanol; and always keep it’s volume= 1 ml.
F. Accordingly, withdrawing off 500 μl from the ‘‘RECYCLE’’ and replacing it with
500 μl of 70% methanol to the ‘‘RECYCLE’’ would decrease the concentration of
Volumes of
‘‘RECYCLE’’
Left
70% methanol
(Replacement)
Equals to
Volumes from
the 1ml
‘‘RECYCLE’’
Concentration
in the Final
5 ml solution (μg )
500 μl 500 μl 1 ml 500 μl 25
500 μl 500 μl 1 ml 500 μl 12.5
500 μl 500 μl 1 ml 500 μl 6.25
500 μl 500 μl 1 ml 500 μl 3.125
500 μl 500 μl 1 ml 500 μl 1.5625
500 μl 500 μl 1 ml 500 μl 0.78125
500 μl 500 μl 1 ml 500 μl 0.390625
500 μl 500 μl 1 ml 500 μl 0.1953125
Name of sample Conc. (μg/ml) % of scavenging Mean
A B C
CATECHIN
1.5625 0.306 0.302 0.305 0.304
3.125 0.503 0.508 0.507 0.506
6.25 0.776 0.772 0.775 0.774
12.5 1.387 1.381 1.401 1.390
25 2.425 2.419 2.421 2.422
50 3.441 3.432 3.434 3.436
100 3.911 3.915 3.914 3.913

0
1
2
3
4
5
0 20 40 60 80 100 120
Absorbance
Concentration(µg/ml)
Reducing Power Capacity
CATECHIN
CME
EAF
CLF
PEF
AQF
Md.
Imran
Nur
Manik

Test Number: 06
Title: Hydroxyl Radical Scavenging Assay.
Reagents:
1. 28 mM (0.028 M) 2-Deoxy -D- ribose, (C5H10O4) {Prepared In Buffer: PIB}
2. 20 mM (0.02 M) KH2PO4-KOH Buffer (pH 7.4),
3. 1.04 mM (0.00104 M) EDTA, C10H14N2Na2O8·2H2O  Edetate disodium salt dehydrate (PIB)
4. 200 μΜ (0.0002 M) Ferric Chloride (FeCl3), (PIB )
5. 1.0 mM (0.001 M) Hydrogen Peroxide (H2O2), (PIB )
6. 1 mM (0.001 M) Ascorbic Acid,(C6H8O6) (Prepared in DW )
7. 1% TBA, (Thio-Barbituric Acid) {1st
dissolve with 0.1 M KOH & then make final solution with Buffer}
8. 10% TCA, (Tri-Chloro Acetic acid) {Prepared in DW}
9. Catechin (Standard). (PIB )
1. 100 μl of 28 mM 2-deoxy2-ribose (2-deoxy D –ribose)
+ 500 μl of Extarct/Standard Solution
+ 100 μl of 1.04 mM EDTA
+ 100 μl of 1.0 mM hydrogen peroxide (H2O2)
+ 100 μl of 1 mM Ascorbic acid
+100 μl of 200 μm FeCl3 = 1 ml Reaction mixture,

2. Incubate the reaction mixture at 37°C for 60 minutes (1 hr),

3. Add 1 ml of 1% TBA & 1ml of 10% TCA in each test tubes,

4. Incubate the reaction mixture at 100°C for 20 minutes,

5. Cool the reaction mixture,

Blank: 2-Deoxy -D- ribose Prepared in buffer except of other reagents and test
samples.Now, auto zero the absorbance value at 532 nm.
The blank 3ml solution
contains only100 μl of 28 mM 2-Deoxy -D- ribose in the buffer solvent(100 μl of 28 mM 2-Deoxy -
D- ribose+ 2.9 ml buffer=3ml), treated in the same manner as Sample/Standard solutions. 
Control: Solution with all the reagents except the sample.
3ml control contains
everything but no Sample/Standard, treated in the same manner as Sample/Standard solutions.
Here 500 μl of buffer solution is given instead of the Sample/STD. 
Md.
Imran
Nur
Manik

Calculation of the Percentage (%) of Scavenging:
Percentage of (%) Scavenging = {1-(A Sample÷ A Control)} ×100
Here,
A Sample = Absorbance of the Sample.
A Control = Absorbance of the Control.
Question: Prepare a standard Catechin curve for the determination Hydroxyl Radical
Scavenging Assay, having Catechin concentrations as follows.
100
50
25
12.5
6.25
3.125
1.5625
0.78125
0.390625
0.1953125
20 mM (0.02 M) KH2PO4–KOH Buffer: pH 7.4
0.02 M KH2PO4 (500 ml):
Here, S=0.02M; // M= (39×1+1×2+31+16×4) =136; // V=500 ml
Where, M= 136.09
To make 500ml solution, we need W= {(0.02 ×136.09×500)÷1000} gram KH2PO4
W=1.3609 gm~1.361 gm KH2PO4
0.02 M KOH (500 ml):
Here, S=0.02M; // M= (39×1+16×1+1) =56; // V=500 ml
Md.
Imran
Nur
Manik

Where, M= 56.1056
To make 500ml solution, we need W= {(0.02 ×56.1056×500)÷1000} gram of KOH
W=0.561056 gm~0.561 gm of KOH
Now, take 1.36 gm of KH2PO4 in a 500 ml of beaker & 500 ml of DW in a
500 ml measuring cylinder. Then gradually pour a small volume of DW (70 ml)
from the measuring cylinder with continuous stirring to completely dissolve the
KH2PO4. After the complete dissolution, add the rest volume of DW (430 ml) to
make the final solution volume 500 ml.
Next dissolve 0.56 gm of KOH in another 500 ml beaker in the same manner as
described for KH2PO4.
Take 1.361 gm of KH2PO4 in a 500 ml of beaker, and make it’s solution as
described above. Similarly dissolve 0.561 gm KOH in another 500 ml beaker.
pH Adjustment at 7.4
 Place a 1000 ml beaker beneath the pH meter stand and pour 500 ml of
KH2PO4 into it followed by 500 ml of KOH.
 Now, submerge the electrode and the temperature probe approximately
4 cm (1½") into the sample and stir gently.
 Allow time for the electrode to stabilize. The pH is displayed on the
primary LCD and the temperature on the secondary LCD.
If needed use 0.02M KOH solution to adjust the desired pH at 7.4.
Storage: After preparation, the buffer can be kept for more than 1 week at cool
temperature. (See Storage condition of Phosphate buffer.) In case of deep fridge
storage /cool/cold temperature storage always the buffer must be brought to the
normal temperature and check for the desired pH.
28 mM (0.0028 M) 2-Deoxy -D- ribose (C5H10O4): (in 10 ml Buffer)( 2DDR)
Here, S=0.028M; // M= (12×5+1×10+16×4) =134; // V=10 ml.
To make 10ml solution, we need W= {(0.028 ×134×10) ÷1000} gram of 2DDR
W=0.03752~0.0375 gm of 2DDR
Now, take 0.0375 gm of 2-Deoxy -D- ribose in a 50 ml of beaker & 10 ml of buffer
Md.
Imran
Nur
Manik

solution in a measuring cylinder. Then gradually pour a small volume of buffer
solution from the measuring cylinder with continuous stirring to completely
dissolve the 2-Deoxy -D- ribose (C5H10O4.)
After the complete dissolution, add the rest volume of buffer solution to make the
solution volume 10 ml.
Where, M= 134.13
To make 10ml solution, we need W = {(0.028 ×134.13×10)÷1000} gram of 2DDR
W=0.0375564gm~0.0376 gm of 2DDR
Now Dissolve 0.0376 gm of 2-Deoxy -D- ribose in 10 ml of Buffer solution in a
50 ml beaker as described above.
1.04 mM (0.00104 M) EDTA Solution (C10H14N2Na2O8·2H2O): (in 10 ml Buffer)
Here, S=0.00104M;
M= (12×10+1×14+14×2+23×2+16×8+2×18) =372gm; & V=10 ml
To make 10ml solution, we need W= {(0.00104 ×372×10) ÷1000} gram of EDTA
or, W=0.0038688~0.0039 gm of EDTA
Now Dissolve 0.0039 gm of EDTA in 10 ml of Buffer solution in a 50 ml beaker as
described above for 2-Deoxy -D- ribose.
If, M=372.24 (More Specific); Then W=0.0038713 gm~0.0039 gm
200 μΜ (0.0002 M) Ferric Chloride (FeCl3):( in 100 ml Buffer Solution)
Here, S=0.0002M; // M= (55.85×1+35.45×3) =162.20; // V=100 ml
To make 100ml solution, we need W= {(0.0002 ×162.20×100) ÷1000} gm of FeCl3
or, W=0.003244 gm~0.0033 gm of FeCl3issolve 0.0
Now, take 0.0033 gm of FeCl3, in a 100 ml of beaker & 100 ml of buffer solution
in a 100 ml measuring cylinder. Then gradually pour small volume of buffer
solution from the measuring cylinder with continuous stirring to completely
dissolve FeCl3.
After complete dissolution, add the rest volume of buffer solution to make the
solution volume 100 ml.
1.0 mM (0.001 M) Hydrogen Peroxide (H2O2): (in 100 ml Buffer)
Here, S=0.001M; // M= (1×2+16×2) =34gm; // V=100 ml
To make 100ml solution, we need W= {(0.001×34×100) ÷1000}//or, W=0.0034gm
Md.
Imran
Nur
Manik

But, we have a 30% Hydrogen Peroxide (H2O2) solution.
For 30% Hydrogen Peroxide (H2O2) solution:
Therefore, to make 100 ml,1mM H2O2 solution, we have to take,
30 gm H2O2 is present in = 100 ml Solution
So,0.0034 gm ,, ,, ,, ,, = {(0.0034×100) ÷30}=0.00113333 ml solution
=(0.00113333×1000) μl
=11.33 μl ~12 μl of Solution
Now Dissolve 12 μl Solution of 30% H2O2 in 100 ml of Buffer solution in a 100 ml
beaker.
For 6% Hydrogen Peroxide (H2O2) solution:
6 gm H2O2 is present in = 100 ml Solution
So, 0.0034 gm ,, ,, ,, ,, = {(0.0034×100) ÷6} = 0.0566667 ml Solution
=(0.0566667×1000)μl
=56.6667μl ~57μl of Solution
Now Dissolve 57 μl Solution of 6% H2O2 in 100 ml of Buffer solution in a 100 ml
beaker.
1 mM (0.001 M) Ascorbic Acid (C6H8O6): ( in 100 ml DW)
Here, S=0.001M; // M= (12×6+1×8+16×6) =176 gm; // V=100 ml
To make 100ml solution, we need W= {(0.001 ×372×100) ÷1000}//or,
W=0.0176gm
Now, take 0.0176 gm of C6H8O6, in a 100 ml of beaker & 100 ml of DW in a 100 ml
measuring cylinder. Then gradually pour small volume of DW from the measuring
cylinder with continuous stirring to completely dissolve C6H8O6.
If, M=176.12412; then W=0.0176124gm ~0.0176 gm
10% Trichloro Acetic Acid (TCA): (in 50 ml DW )
For Solid TCA:
100 ml 10 % TCA requires = 10 gm of TCA
So, 50 ml ,, ,, ,, = {( 10×50)÷100} gm = 5 gm of TCA
Now dissolve 5 gm of Trichloro Acetic Acid (TCA) in 50 ml of DW in a 100 ml
beaker.
Md.
Imran
Nur
Manik

For Liquid TCA(50 ml):
100 ml 10 % TCA requires = 10 ml of TCA
So, 50 ml ,, ,, ,, = {( 10×50)÷100} ml = 5 ml of TCA
Now, dissolve 5 ml of Trichloro Acetic Acid (TCA) in 50ml of DW in a 100 ml
beaker.
N.B. Trichloro Acetic Acid (TCA) is very toxic and corrosive. So, handle with
gloves.
1% Thio-Barbituric Acid (TBA): (in 50 ml Buffer through 0.1 M KOH):
100 ml 1 % TBA requires = 1 gm of TBA
So, 50 ml ,, ,, ,, = {( 1×50)÷100} gm = 0.5 gm of TBA
Now, take 0.5 gm of TBA, in a 100 ml of beaker & gradually pour small volume of
0.1 M KOH with a micropipette (e.g. 500 μl in each turn, for 2 or more times).
When TBA has almost dissolved (formed a suspension) then pour the required
volume of buffer solution (e.g. 49.5 ml) from the measuring cylinder with
continuous stirring to completely dissolve the TBA.
0.1 M KOH (10 ml DW):
Here, S=0.1M; M= 56.1056 ;V=10 ml
To make 10ml solution, we need W= {(0.1 ×56.1056×10)÷1000} gram of KOH
W=0.0561056 gm~0.056 gm of KOH
Now, take 0.056 gm of KOH in a 50 ml of beaker & 10 ml of DW in a measuring
cylinder. Then gradually pour a small volume of DW from the measuring cylinder
with continuous stirring to completely dissolve the KOH. After the complete
dissolution, add the rest volume of DW to make the final solution volume 10 ml.
 N.B. Generally it forms a suspension and hence
must be filtered before addition to the reaction solution.
Remember, always use clear TBA filtrate.
For cotton filter make 10-15 ml more TBA solution.
For filter paper filtration, make 5-8 ml more TBA solution 
Md.
Imran
Nur
Manik

Preparation of the standard Catechin Solution:
Simplified Calculation: (Replacement theory)
Dissolve 1 mg of Catechin in 1ml of Buffer Solution. Therefore, the concentration
of this solution would be 1 mg÷1ml=1 mg/ml. {Or,1 μg/μl}.
When 1 ml of this solution is given in a 3 ml of total volume ,then it would be
diluted & it’s concentration would be changed as follows,
1 mg ÷3 ml=0.333mg/ml. or, or, 0.333 μg/μl (DILUTION FACTOR)
Thus, to make a solution of 100 μg/ml of Catechin in 3 ml of final volume, we
have to take,
0.333 μg is present in= 1 μl of solution
100 μg is present in= ({100×1)÷0.333} μl of solution =300 μl of solution
0.333 μg of Catechin would be presented in = 1 μl of solution
Therefore, 100 μg of Catechin would be presented in = 300 μl of solution }
Basic Concept:
When you take 300 μl of solution, then you actually take (300×1) μg= 300 μg of
Catechin. And accordingly when you put this 300 μg of Catechin in 3 ml of
solution, you actually put 300 μg÷3 ml=100 μg/ml of Catechin in the final 3 ml
solution.
A. Take 0.6 ml (600 μl) from the above 1 ml solution & add 0.4 ml (400μl) of Buffer
Solution. (Total volume =1 ml). Let this 1 ml solution be named as STOCK.
B. Now, take 0.5 ml (500 μl) from the 1 ml STOCK solution & add totally in the
final 3 ml solution. [Through the reaction mixture. (i.e. by adding/mixing with the
reaction mixture.)]
Therefore, the Catechin concentration of in this 3 ml solution would be 100μg/ml.
(According to Basic Concept,1 ml of STOCK solution contains (600×1)μg=600μg
of catechin. Therefore, when you take 0.5 ml (500 μl) from the STOCK solution you
actually take (600÷2) μg =300μg of catechin.
Therefore, addition of this 300 μg of catechin in the 3 ml of final solution, makes
the catechin concentration, 300μg÷3ml=100 μg/ml.
C. Now take the rest of 500 μl from the 1 ml STOCK solution & add 500 μl of Buffer
Solution. (Total volume =1 ml). Let, this 1 ml solution be named as ‘‘RECYCLE’’.
Md.
Imran
Nur
Manik

D. Now take 0.5 ml (500 μl) from the 1 ml ‘‘RECYCLE’’ and add totally in the
3 ml of final solution. Therefore, the Catechin concentration of this 3 ml solution
would be 50μg/ml.
300 μg of catechin. Thus, when you take 0.5 ml (500μl) from the ‘‘RECYCLE’’ you
actually take (300÷2) μg=150 μg of catechin, and addition of this 0.5 ml (500μl) of
solution in the 3 ml of final solution makes the catechin concentration
150 μg ÷3 ml=50 μg /ml. 
Remember: Always replace the amount 500 μl, taken from the ‘‘RECYCLE’’ with
500 μl of 70% methanol; and always keep it’s volume= 1 ml.
E. Accordingly, withdrawing off 500 μl from ‘‘RECYCLE’’ and replacing it with
500 μl of Buffer Solution to ‘‘RECYCLE’’ would decrease the concentration of
‘‘RECYCLE’’
Left
Buffer Solution
(Replacement)
Equals to
‘‘RECYCLE’’
Volumes from
‘‘RECYCLE’’
Concentration (μg ) in
3 ml Final solution
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 25
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 12.5
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 6.25
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 3.125
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 1.5625
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 0.78125
0.5 ml (500 μl) 0.5 ml (500 μl) 1 ml 0.5 ml (500 μl) 0.390625
Name of the sample Conc. (μg/ml)
% of scavenging
Mean ± STD
A B C
Catechin
0.78125 28.989 29.145 29.112 29.082±0.082213
1.5625 54.124 53.293 53.295 53.571±0.479202
3.125 66.958 67.976 68.638 67.857±0.846263
6.25 78.231 77.985 77.967 78.061±0.147499
12.5 81.339 80.993 81.341 81.224±0.200343
25 86.451 86.786 86.968 86.735±0.262246
50 91.717 92.005 91.789 91.837±0.14988
100 92.989 93.429 93.439 93.286±0.25697
Md.
Imran
Nur
Manik

Figure: UV-Spectrophotometer

0
10
20
30
40
50
60
70
80
90
100
0 20 40 60 80 100 120
Catechin
CME
EAF
CLF
PEF
AQF
Md.
Imran
Nur
Manik

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