Practical Manual of Biochemistry
Practical Manual of Biochemistry
Practical Manual of Biochemistry
Dr. K. Saravanan
Principal,
MASS College of Arts and Science,
Kumabkonam.
ACKNOWLEDGMENTS
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Practical Manual of Biochemistry
18. It is necessary to wash the glassware with soap and water, then rinse it with
distilled water.
19. Flame transfer loops, wires, or needles for transferring biological material
before and immediately after use.
20. Do not walk around the laboratory with infectious matter containing
transfer loops, wires, needles, or pipettes.
21. Around Bunsen burners, be careful. It is not always possible to see flames.
22. Turn off the incinerators before the laboratory leaves.
23. Report any broken equipment, report any broken glass, in particular those
containing infectious materials immediately.
24. Contact your course instructor or TA immediately if you are injured in the
laboratory.
25. In the event of further treatment being required, spills, cuts and other
accidents should be reported to the instructor or TA.
26. Familiarize yourself with safety equipment and emergency escape routes in
the laboratory.
27. Before putting it away, always wipe and clean your microscope's lenses. To
this end, use the relevant tissue paper and cleaning solution.
28. With all biological fluids, apply appropriate universal precautions.
29. Without the written permission of the course instructor or TA, do not
remove any materials from the laboratory.
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Principal
A buffer's main purpose is to control the solution's pH. Buffers
can also play secondary roles in a system, such as controlling ionic strength
or species solving, perhaps even affecting the structure or activity of protein
or nucleic acid. Nucleic acids, nucleic acid-protein complexes, proteins, and
biochemical reactions are stabilized by buffers (whose products might be
used in subsequent biochemical reactions). Complex buffer systems in
electrophoretic systems are used to control the pH and to establish the pH
gradient. Weak acids and bases are made up of buffer solutions that make
them comparatively resistant to pH change. Theoretically, buffers offer a
ready source of both acid and base to either supply additional H+ if the
process consumes H+ or if a reaction produces acid, combine it with excess
H+.
Reagents:
a. Acetic acid 0.2 M: glacial acetic acid 1.5 ml is made up to 100 ml by using
distilled water.
b. Citric acid: citric acid 2.10 gm in 100 ml distilled water.
c. Dibasic sodium phosphate: 5.3 gm of disodium hydrogen phosphate in
100 ml distilled water.
e. Monobasic sodium phosphate: 2.78 gm sodium dihydrogen phosphate in
100 ml distilled water.
f. Sodium acetate solution: 0.64 gm of sodium acetate in 100 ml distilled
water.
g. Sodium bicarbonate solution: sodium bicarbonate 1.68 gm in 100 ml
distilled water.
h. Sodium carbonate solution 0.2 M:2.12 gm anhydrous sodium carbonate
in 100 ml distilled water.
i. Sodium citrate solution 0.1 M: sodium citrate 2.94 gm in 100 ml distilled
water.
Procedure:
a. Acetic acid-sodium acetate buffer
Take a 100 ml flask and use a pipette to add 36.2 ml of sodium
acetate solution, and then add 14.8 ml of glacial acetic acid to it. Using
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distilled water to produce a total volume of 100 ml. The resulting acetic
acid-sodium acetate buffer is 0.2 M. With the help of a pH meter, the pH is
measured. With distilled water, the electrode is washed, and then dipped in
the prepared buffer solution. The resultant pH is 4.6.
b. Barbitone buffer
In distilled water, mix 2.85 gm of diethyl barbituric acid and 14.2
gm of sodium diethyl barbititrate and then produce 1000 ml of final
volume. That's the buffer of barbitone. With the help of the pH meter, the
pH is measured, and the final pH comes out as 6.8.
c. Citrate buffer
Using distilled water, mix 46.5 ml of citric acid with 3.5 ml of
sodium citrate solution and add 100 ml to the final volume. This is a citrate
buffer of 0.1 M. With the help of a pH meter, the pH of this buffer is
measured and the pH is 2.5.
d. Carbonate-Bicarbonate buffer
Put 27.5 ml of sodium carbonate solution in a flask and add 22.5
ml of sodium bicarbonate solution to it. Then make 100 ml of the total
volume with the aid of distilled water. This is a buffer of 0.2 M carbonate-
bicarbonate. The pH meter is standardized, and the pH of the buffer
solution that is prepared is measured. The pH would be 10.2.
e. Phosphate buffer
Dihydrogen sodium phosphate (39 ml) is mixed with disodium
hydrogen phosphate (61 ml), and 200 ml of distilled water is added to the
final volume. This solution results in the phosphate buffer being 0.2 M.
Using a pH meter, the pH of the phosphate buffer is measured and 6.8 is
obtained.
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The beaker, stirring rod, and funnel must be carefully rinsed before
adding additional solvent to the flask, and the washes added to the vol flask
to ensure that all remaining traces of the solution have been transferred.
The vol flask is finally diluted to volume (additional solvent is added to the
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flask until the liquid level reaches the calibration mark). As before the flask
is capped and inverted until the contents are mixed thoroughly. The
downside to this technique is that if not thoroughly washed, some of the
solution may stick to the beaker, stirring rod, or funnel. Also, if they have
not been washed carefully, there is a chance of contamination from the
beaker, rod, or funnel.
c. Molar Solution
It consists of one mole of solvent in a solution equal to one liter.
Molar solution = Molecular weight in the solution in grams / liters.
Example: Sodium chloride molar solution I (NaCl).
Sodium atomic weight = 23
Chloride atomic weight = 35.5
Total molecular weight = 58.5 gram / mol
Now dissolve 58.5 grams of NaCl in distilled water and make the solution
to one liter.
d. Normal Solution
The normal solution is defined as the solution's gram equivalent
weight per liter.
Normal solution = gram equivalent solvent weight/solution liter.
These alternatives are expressed as N.
Gram equivalent weight = weight/valence of molecules
Example
1 N sodium chloride solution to make
• The NaCl molecular weight is 58.55.
• Gram equivalent NaCl weight = molecular weight / 1 (valency)
Dissolve 58.5 grams of NaCl in one liter of distilled water and make-up.
e. Percent Solution
1. This is per hundred of the total solution component.
2. A percentage solution has three possibilities.
Weight/weight is a solvent percentage equal to solute + solvent in 100
grams of final solution.
1. Weight/weight is a solvent percentage equal to solute + solvent in 100
grams of final solution.
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f. The Dilution
1. This procedure is very common for the preparation of serum dilution
where, if it is above 300 mg/dL, there is a high concentration of chemicals
such as urea in the blood.
2. If we manufacture a serum dilution like this:
1. Serum = 1 millilitre
2. Diluting 4 mL of fluid
3. This will be a dilution of 1:5 (1+4 =5).
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3. INTRODUCTION OF CARBOHYDRATES
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Practical Manual of Biochemistry
Monosaccharides
Hydrolysis can break disaccharides into two monosaccharide units
and lose one molecule of water.
FDi-saccharides
Oligosaccharides can be broken by hydrolysis into 3-6
monosaccharide units and lose one or more water molecules, see the
example of oligosaccharides below.
Poly-saccharides
Polysaccharides can be broken into 7 or more units of monosaccharides by
hydrolysis and one or more molecules of water can be lost.
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1. Solubility
Due to the polar hydroxyl groups that form H-bonds with water,
monosaccharides and oligosaccharides are readily soluble in water.
However, due to their large molecular weight, the polysaccharides make
translucent colloidal solutions.
a. Molisch's Test
Principle
The response is based on the fact that H2SO4 concentrated
catalyzes the dehydration of sugars to form furfural (from pentoses) or
hydroxymethyl furfural (from hexoses). These furfurals then condense to
give a purple or violet colored product with sulfonated alpha-naphthol.
A+ve reaction is also provided by polysaccharides and glycoproteins. The
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b) Anthrone test
Principle
Another general test for carbohydrates is anthrone reaction. Its
principle is the same as that of Molisch's, except that furfurals and
hydroxyl-methyl furfurals offer bluish green condensation products with
anthrone.
Reagents
Anthrone reagent: in conc.H2SO4, 0.2% (w/v) solution.
Procedure
In a test tube, add about 2 mL of Anthrone reagent to
approximately 0.5-1mL of the test solution and mix thoroughly. Watch if
the colour changes to bluish green. If not, after keeping them in a boiling
water bath for ten minutes, re-examine the tubes. A positive test indicates a
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blue-green color.
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Practical Manual of Biochemistry
absence of sugar reduction, the alkaline copper solution is heated, the black
precipitate of cupric oxide forms:
Heat
Cu (OH)2 ------------ 2OH
A) Fehling's test
In this reaction, Rochelle salt acts as a chelating agent:
Reagents
i) Fehling solution A:
Dissolve 69.38 g of copper sulfate in distilled water and make 1 L
of copper sulfate.
ii) Fehling's solution B:
Dissolve 250 g of NaOH in DW, add 346 g of potassium sodium
tartrate and add up to 1 L of volume.
Just before use, mix equivalent volumes of A & B solutions
because mixing causes deterioration with time.
Procedure
1 mL of Fehlings reagent is added to the remaining 1 mL solution.
Mix the test tubes thoroughly and place them in a boiling water bath. The
formation of yellow or red Cuprous Oxide precipitates indicates the
presence of sugar reduction.
Note
i) Leave the solution to stand for 10-15 minutes in the case of a mild
reduction, then decant the supernatant. It is then possible to see a small
amount of red or yellow precipitates adhering to the inner side of the tube.
ii) The Fehling test is only carried out with an alkaline solution
iii) Cuprous oxide is ammonia-dissolved. Small amounts of sugar reducers
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can therefore not be detected in fluids saturated with ammonium salts, such
as urine.
B) Benedict's Test
Benedict modified the Fehling solution to produce a fairly stable,
enhanced single reagent. As a chelating agent, Sodium Citrate works. It is
very sensitive and sufficient precipitates are produced by even small
amounts of sugar reduction (0.1 percent).
Reaction
Reagents:
Qualitative reaction of Benedict:
Dissolve 173g of Sodium Citrate and 100g of Anhydrous Sod. In
about 800mL of water, carbonate by gently heating the contents. Then,
dissolve 17.3g of copper sulfate in about 100mL DW in a separate beaker.
Slowly pour this solution into the Carbonate-Citrate mixture with constant
stirring, and make up to 1 L with DW.
Procedure:
To add about 2 mL of Benedict's reagent, add 0.5-1 mL of the test
solution. Keep the test tubes in a bath of boiling water. .Note that green,
orange, yellow or red precipitates are formed, indicating the presence of
sugar reduction in the solution.
Note:
i) This test is particularly suitable for urinary sugar reduction detection
because it is more specific than the Fehling test, which is also positive for
non-reducing substances such as urates present in urine.
ii) This is a semi-quantitative test.
C) Barfoed's Test:
This test is performed to differentiate between mono- and
disaccharide reduction. Monosaccharides are more reactive reducing agents
than disaccharides and thus react in about 1-2 minutes, while it takes 7-12
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minutes for the reducing disaccharides to get hydrolysed and then react in
the acidic solution. Therefore it is possible to detect the difference in
property reductions.
Reagents:
Barfoed's reagent:
66.5 g of Cupric acetate dissolved in approximately 900 mL DW.
Add 9 mL of Glacial Acetic Acid and boil. Cool and use DW to cool the
volume to 1 L and filter if necessary.
Procedure:
Take 2- solution: Keep the test tubes for only 1-2 min in a boiling
water bath. Then allow a while for the tubes to cool down. Thin red
precipitates indicate the presence of a reduction of monosaccharide at the
bottom or sides of the tube.
Note:
i) The boiling should not be prolonged beyond 1-2min, otherwise the
disaccharide reduction will respond to this test as well.
ii) This test is not effective in detecting urine sugar reduction due to the
presence of chloride ions.
E) Seliwanoff’s Test
Principle:
This test is a keto hexose-specific timed colour reaction. It is
therefore used for the differentiation of aldoses from ketoses. Dehydration
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Principle:
In an unknown solution, the sucrose present is hydrolyzed to
glucose and fructose by acid. Then Seliwanoff's reagent forms the resulting
fructose in the solution.
Reagents:
i) Conc. of HCL
ii) Seliwanoff's reagent
iii) Sodium carbonate
Procedure:
Add 1-2 drops of conc HCl to approximately 2-3mL of test
solution and boil for about 8-minutes in a water bath. The appearance of
the red color indicates the presence of the hydrolytic sucrose product
fructose.
Note:
Benedict's reagent can be tested for sugar reduction by acid hydrolyzed
sample after cooling and then neutralizing with sodium carbonate.
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Procedure:
In a test tube, take 7-8 ml of carbohydrate solution and add a pinch
of phenylhydrazine to that and double the amount of sodium acetate and 10
drops of acetic acid. Dissolve and allow slow cooling by shaking. Observe
the shape of the crystal under low microscope power (10x).
Observations and conclusion:
Lactose forms crystals in the form of powder puffs, maltose forms
sunflower-shaped or star-shaped crystals, while glucose and fructose form
identical crystals in the form of needles.
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5. INTRODUCTION OF PROTEINS
Dipeptide
They are the products formed by the condensation of two
molecules of alpha-amino acid.
Tripeptide
They are formed by the condensation of three molecules of al-
amino acid.
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Stability to Alkali
Amino acids do not develop NH3 or alkaline vapor when boiled
with alkali, unlike amides and volatile amines. In order to differentiate
amino acids from amines and amides, this method can be used. Apply this
test to the amine or amide provided and to glycine as well.
Procedure:
Pipette 1 mL 1% glycine and the amide or amine solution into
separate test tubes.
Add 1 mL dilute NaOH to each test tube and boil.
Test the vapor from each boiling tube with wet litmus paper.
Specific Reactions for Individual Amino Acids:
a) Xanthoproteic Test:
There are aromatic groups that are derivatives of benzene in some
amino acids. Reactions that are characteristics of benzene and benzene
derivatives can be experienced by these aromatic groups. The nitration of a
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Practical Manual of Biochemistry
benzene ring with nitric acid is one such reaction. The amino acids that
have the benzene ring activated can easily undergo nitration. This nitration
reaction forms a yellow product in the presence of the activated benzene
ring. Apply this test to tyrosine, phenylalanine, tryptophan, and glutamic
acid.
Procedure:
To 2 mL amino acid solution in a boiling test tube, add equal
volume of concentrated HNO3.
Heat over a flame for 2 min and observe the color.
Now COOL THOROUGHLY under the tap and CAUTIOSLY
run in sufficient 40% NaOH to make the solution strongly alkaline.
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Pour 1-2 mL H2SO4 down the side of the sloping test tube to
form a layer underneath the acetic acid.
The development of a purple color at the interface proves a
positive reaction.
c) Lead-Sulfide Test:
When cystine is boiled with 40 percent NaOH, sodium sulfide is
covered with some sulfur in its structure (Na2S). Using a sodium plumbate
solution that causes the precipitation of PbS from an alkaline solution,
Na2S can be detected. In order to carry out this test, the sodium plumbate
solution must be prepared first. This test is applied to cysteine and cystine.
Procedure:
Sodium Plumbate Solution Preparation:
Add 5 mL dilute NaOH to 2 mL dilute lead acetate.
A white precipitate of lead hydroxide forms.
Boil until the precipitate dissolves with the formation of sodium
plumbate.
Boil 2 mL amino acid solution with a few drops of 40% NaOH for
2 min.
Cool and add a few drops of the sodium plumbate solution.
A brown color or precipitate is a positive test for sulfides.
d) Ehrlich Test:
Aromatic amines and many organic compounds (indole and urea)
provide this test with a color complex. This test is applied to tryptophan,
urea and glycine.
Procedure:
Put 0.5 mL of the amino acid solution to a test tube.
Add 2 mL Ehrlich reagent and observe the color changes.
Repeat the test with urea solution.
e) Sakaguchi Test:
To test for a certain amino acid and proteins, the Sakaguchi reagent
is used. The amino acid that this test detects is arginine. As arginine has a
group of guanidine in its side chain, in the presence of an oxidizing agent
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f) Nitroprusside Test:
The nitroprusside test is specific to cysteine, the only sulfhydryl
group amino acid containing cysteine (-SH). In the presence of excess
ammonia, that group reacts with nitroprusside. This test is used for
cysteine, cystine and methionine.
Procedure:
Put 2 mL amino acid solution into the test tube.
Add 0.5 mL nitroprusside solution and shake thoroughly.
Add 0.5 mL ammonium hydroxide.
Observe the color change.
g) Biuret Test:
The Biuret Test recognizes positively the presence of proteins (not less than
two peptides). The reaction in this test involves the complex formation in a
strongly alkaline solution of the proteins with Cu2+ ions. This test is
applied to gelatin, casein and albumin.
Procedure:
To 2 mL protein solution, add 5-6 drops of dilute CuSO4
(Fehling’s solution A diluted 1/10 with water)
Add 3 mL 40% NaOH solution.
Observe the color change.
If the protein tested is insoluble in water, then apply the procedure
given below:
Measure 3 mL acetone and 1.5 mL water into a test tube.
Add 1 drop of dilute NaOH and a little piece of protein to be
tested.
Boil continuously over a small flame for 2 min and cool.
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h) Ninhydrin Test:
This test is given by only amino acids and proteins which contain
free –NH2 groups in their structure. Apply this test for all the proteins
provided.
Precipitation of Proteins:
The precipitation of a protein occurs in a stepwise process. The
addition of a precipitating agent and steady mixing destabilizes the protein
solution. Mixing causes the precipitant and the target product to collide.
Enough mixing time is required for molecules to diffuse accross the fluid.
By Neutral Salts:
The precipitation of a protein by neutral salt is commonly known
as salting-out method. Addition of a neutral salt, such as ammonium sulfate,
compresses the solvation layer and increases the protein-protein interaction.
As the salt concentration of a solution is increased, more of the bulk water
becomes associated with the ions. As a result, less water is available to take
part in the solvation layer around the protein, which exposes hydrophobic
parts on the protein surface. Therefore, proteins can aggregate and form
precipitates from the solution. The amount of neutral salt required to cause
protein precipitation varies with the nature of the protein and the pH of the
solution. Apply this test to all the proteins provided.
Procedure:
Add solid ammonium sulfate to about 5 mL of protein solution in
a test tube (the salt should be added in quantities of approximately 1 g at a
time)
Agitate the solution gently after each addition to dissolve the
ammonium sulfate.
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salt bridges in proteins. The reaction of a heavy metal salt with a protein
usually leads to an insoluble metal protein salt. Apply this test to all the
proteins provided.
Procedure:
Treat 3 mL of the protein solution provided with a few drops of
mercuric nitrate.
A white precipitate formation should be observed.
By Acid Reagents:
The precipitation of a protein in the presence of acid reagents is
probably due to the formation of insoluble salts between the acid anions
and the positively charged protein particles. These precipitants are only
effective in acid solutions. Apply this test to all the proteins provided.
Procedure:
Treat 3 mL of protein solution provided with a few drops of
trichloroacetic acid solution.
Note the protein precipitate formed.
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7. INTRODUCTION OF AMINOACIDS
For life, amino acids are critical. They have important functions in
particular, such as being the building blocks of proteins and the
intermediates in metabolism.
Amino acids are usually classified into four groups according to the
properties of their side chain. The side chain can make a weak acid or a
weak base of an amino acid, and if the side chain is polar, a hydrophobe or
if it is nonpolar, a hydrophile.
Proteins (also known as polypeptides) are organic compounds
arranged in a linear chain made of amino acids. The peptide bonds between
the carboxyl and amino groups of adjacent amino acid residues are joined
together by the amino acids in a polymer.
Proteins are essential parts of organisms, like other biological
macromolecules such as polysaccharides and nucleic acids, and participate
in virtually every process within cells. In: Proteins are significant in:
-catalyzing biochemical reactions (enzymes)
-structural and mechanical functions (actin and myosin)
-cell signaling
-immune responses
-cell adhesion
-cell cycle
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Solubility Tests
The solubility of amino acids and proteins is largely dependent on
the pH of the solution. The structural changes that occur at different pH
values in an amino acid or protein change the molecule's relative solubility.
Amino and carboxylic groups are both protonated in acidic solutions. Both
groups are deprotonated in basic solutions.
In water, amino acids are essentially soluble. Depending on the
structure of its side chains, their solubility in water, dilute alkali and dilute
acid varies from one compound to another. Glycine, tyrosine, glutamic acid
and cysteine will be used for this test.
Procedure:
The solubility of amino acids in water and alcohol should be noted
by placing a small quantity in the test tube, adding a few mL of solvent and
if necessary, warming up.
Use the litmus paper to determine whether the amino acid solution
is acidic or basic, while testing the solubility in water.
Using dilute HCl and dilute NaOH, repeat the solubility test.
WARNING: Avoid spilling ninhydrin solutions on your skin because it is
hard to remove the resulting stains.
a) Millon’s Test:
The test by Millon is specific to phenol that contains structures
(tyrosine is the only common phenolic amino acid). Millon's reagent is a
concentrated HNO3 reagent that dissolves mercury. A red precipitate or a
red solution is regarded as a positive test as a consequence of the reaction.
A yellow HgO precipitate is NOT a positive response, but it usually shows
that the solution is too alkaline. This test can be applied to tyrosine,
phenylalanine.
Procedure:
To 2 mL amino acid solution in a test tube, add 1-2 drops of
Millon2s reagent.
Warm the tube in a boiling water bath for 10 min.
A brick red color is a positive reaction.
Note that this is a test for phenols, and the ninhydrin test should also be
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c) Lead-Sulfide Test:
When cystine is boiled with 40 percent NaOH, sodium sulfide is
covered with some sulfur in its structure (Na2S). Using a sodium plumbate
solution that causes the precipitation of PbS from an alkaline solution,
Na2S can be detected. In order to carry out this test, the sodium plumbate
solution must be prepared first. This test is applied to cysteine and cystine.
Procedure:
Sodium Plumbate Solution Preparation:
Add 5 mL dilute NaOH to 2 mL dilute lead acetate.
A white precipitate of lead hydroxide forms.
Boil until the precipitate dissolves with the formation of sodium
plumbate.
Boil 2 mL amino acid solution with a few drops of 40% NaOH for
2 min.
Cool and add a few drops of the sodium plumbate solution.
A brown color or precipitate is a positive test for sulfides.
d) Ehrlich Test:
Aromatic amines and many organic compounds (indole and urea)
provide this test with a color complex. This test is applied to tryptophan,
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e) Sakaguchi Test:
To test for a certain amino acid and proteins, the Sakaguchi reagent
is used. The amino acid that this test detects is arginine. As arginine has a
group of guanidine in its side chain, in the presence of an oxidizing agent
such as bromine solution, it gives a red color with alpha-naphthol. Apply
Arginine to this test.
Procedure:
1 mL NaOH and 3 mL arginine solution is mixed and 2 drops of α-
naphthol is added.
Mix thoroughly and add 4-5 drops of bromine solution UNDER
THE HOOD!!
Observe the color change.
f) Nitroprusside Test:
The nitroprusside test is specific to cysteine, the only sulfhydryl
group amino acid containing cysteine (-SH). In the presence of excess
ammonia, that group reacts with nitroprusside. This test is used for
cysteine, cystine and methionine.
Procedure:
Put 2 mL amino acid solution into the test tube.
Add 0.5 mL nitroprusside solution and shake thoroughly.
Add 0.5 mL ammonium hydroxide.
Observe the color change.
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Qualitative Tests:
I. Physical Test:
1. Grease spot test:
Take a small amount of oil on a piece of paper, a greasy spot
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penetrating the paper will be formed. This happens because lipid does not
wet paper unlike water.
2. Test for free fatty acids:
Take a few drops of phenolphthalein solution in a test tube and
add to it one or two drops of very dilute alkali solution, just sufficient to
give the solution a pink colour. Now add a few drops of the oil and shake.
The colour will disappear as the alkali is neutralized by the free fatty acids
present in the oil.
3. Emulsification:
Oil or liquid fat becomes finely divided and is dispersed in water
when shaken with water to form emulsification. Emulsification is
permanent and complete in the presence of emulsifying agent. The
important emulsifying agents are bile salts, proteins, soaps, mono- and
diglycerides. Emulsification is important in the processes of fat digestion in
the intestine. Emulsifying agents lower surface tension of the liquid.
Take 2 clean and dry test tubes, in one test tube added 2 ml water and in
other 2ml dilute bile salt solution. Now to each tube added 2 drops of
mustard oil and shaken vigorously for about one minute. Allow the tubes to
stands for two minutes and note that the water, oil is broken in small pieces
and floats on the surface; where as in the bile salt solution, the oil can be
seen in minute droplets suspended in the liquid (permanent emulsification).
4. Saponification test:
Esters can be hydrolysed by alkali to yield the parent alcohol and
salt. When the fatty acid possesses a long chain the salt formed is a soap
which we commonly use. This process is called saponification. Oils and fats
usually contain long chain fatty acids and are, therefore, the starting
materials for the preparation of soap.
Take 1 ml of the oil in a test tube and add an equal amount of
alcoholic KOH solution, mix them thoroughly and keep the mixture during
the course of warming and shake up gently with a little distilled water.
Appearance of some oil drops will indicate the incomplete saponification.
After complete saponification no oil drops will appear.
5. Tests for unsaturation of fatty acids:
Unsaturated fatty acids like oleic acid can react with halogens like
bromine and iodine due to presence of double bonds as shown below.
CH3 (CH2)7CH = CH (CH2)7COOH + Br2 → CH3 (CH2)7CHBr-CHBr
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(CH2)7COOH
The amount of Br2 or I2 taken up will indicate the amount of
unsaturation present in a particular acid. Approximate idea about the
unsaturation in a different oils and fats can be obtained by the following
test. Set up four clean and dry test tubes each containing 5 ml of CCl4.
To the first, add one drop of shark liver oil, to the second, one drop of
coconut oil, to the third, a drop of vegetable ghee and add nothing to the
fourth tube. Now test for the unsaturation of the added oil by adding
bromine water drop by drop to each tube followed by shaking.
Record the number of drops required to obtain a permanent
yellowish red colour in each tube and infer the relative unsaturation in the
three samples used. It may be mentioned here, vegetable ghee is prepared
by hydrogenating vegetable oil. Hydrogenation means saturation of
unsaturated fatty acid by hydrogen.
6. Isolation of free fatty acids from soap:
Take a few ml of 20% H2SO4 in a test tube and gradually add 5 ml
of some soap solution. The fatty acids will separate out in a distinct layer
due to the hydrolysis of the soap.
RCOONa + H2O → RCOOH + NaOH
Cool the solution which will become hot and skim off the surface
layer and wash it several times with water till free from H2SO4. Then
dissolve it in some water and add alkaline phenolphthalein solution and
shake. The pink colour will be discharged indicating the presence of free
fatty acids.
Calcium soap formation:
To a small amount of the soap solution in a test tube add
CaCl2 solution. A white precipitate will be formed. The white precipitate is
due to insoluble calcium salt of fatty acid. This is referred to as calcium
soap.
Lead soap formation:
To a small amount of the soap solution in a test tube add lead
acetate solution, a white precipitate will appear. The white ppt is due to
insoluble lead salt of fatty acids. This is referred to as lead soap.
7. Tests for Glycerol:
I. Acrolein test:
Take pure glycerol in a dry test tube; add to it a few crystals of
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potassium hydrogen sulphate. Warm gently to mix and then heat strongly.
A very pungent odour of acrolein is produced. Acrolein is formed due to
removal of water from glycerol by potassium hydrogen sulphate.
II. Dichromate Test:
Take in a dry test tube 3 or 4 ml of glycerol solution, to it add a few
drops of 5% potassium dichromate solution and 5 ml of conc. HNO3, mix
well and note that the brown colour is changed to blue. This test is given by
the substances containing primary and secondary alcohol groups. The
chromic ions oxidize the glycerol and in this process they are reduced to
chromous ions which give the blue colour. This test is also given by
reducing sugars, so before confirming glycerol be sure that the reducing
sugars are not present.
Quantitative Tests:
1. Determination of Iodine Number:
The iodine number of a fat is the amount in gm. of iodine taken up
by 100 gm. of fat. Not only iodine but also equivalent amounts of other
halogens will add at double bonds; so bromine is often used instead of
iodine because it is more reactive. The halogenating reagent used in this
method is pyridine sulphate di-bromide. This reagent can be prepared by
adding carefully 8.1 ml pyridine in 20 ml glacial acetic acid and making the
volume up to 1 litre with glacial acetic acid.
Weigh the bottle containing sample of oil plus a medicine dropper
and then transfer about 0.1 to 0.3 gm. of oil to a flask. Reweigh the bottle
containing oil and dropper to find out the exact quantity of the sample
transferred. Add 10 ml of chloroform and then 25 ml of the pyridine
sulphate di-bromide reagent.
Shake thoroughly; allow standing for 5 minutes and then
determining the residual bromine. To do this, add 10 ml of 10% KI and
titrate the equivalent amount of iodine liberated by the residual bromine
with the help of 0.1 (N) Na2S2O3 (sodium thiosulphate). The titration can
be done by adding sodium thiosulphate solution through a burette to the
flask.
When the colour of the solution in flask becomes light yellow add 1 ml of
starch solution. It will become blue. Slowly add the thiosulphate solution
again till it becomes colourless. Note the total volume of thiosulphate used.
The total amount of bromine originally added is found by titrating
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Enzymatic Methods:
Assays have been developed in which cholesterol oxidase obtained
from the bacterium Nocardia erythropolis is used to convert cholesterol
into cholest-4-en-3-one with the formation of Hydrogen peroxide. The
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B. COLLECTION OF SAMPLES
1. Tissues (in general):
For the post-mortem examination of the species of animals with
which they work, animal health personnel should be trained in the correct
procedures. The required equipment will depend on the size and species of
the animal, but for the opening of the intestines, a knife, saw and cleaver
will be required, as well as scalpels, forceps and scissors, including scissors
with a rounded tip on one blade.
There must be an abundant supply of containers appropriate to the
nature of the sample required, as well as labels and reporting forms. For
transporting samples from the field, special media may be required.
Protective clothing should be worn by the operator: overalls, rubber gloves
and rubber boots. It is common to detach the animal's head if rabies is
suspected, and the operator should wear a face mask and goggles, gloves,
and a plastic apron.
Tissues may be collected for culture or histopathology and occasionally in
serological tests for use as an antigen. In the post-mortem technique, the
person removing the tissues should be experienced and have sufficient
pathology knowledge to select the right organs and the most promising
lesions for sampling.
Using ordinary instruments, the skin of the dead animal may be
removed, but the body cavities should be opened with sterile instruments,
and a fresh set of sterile instruments should be used to collect the required
pieces of the various organs. Each piece of tissue should be placed in a
separate plastic bag or sterile screw-capped jar, fully labeled with the date,
tissue and identification of the animal. It is necessary to take care not to
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2. Blood
Blood samples can be taken for haematology or culture purposes
and/or for direct examination of bacteria, viruses, or protozoa, in which
case anti-coagulants such as heparin are added to the blood. They may also
be taken for serology, which requires a clotted sample. A blood sample is
taken through venepuncture, as cleanly as possible.
The jugular vein or caudal vein is chosen in most large mammals,
but brachial veins and mammary veins are also used. A wing vein (brachial
vein) is generally selected in birds. Blood can be taken through a syringe
and needle or through a needle and vacuum tube (not easy in delicate veins
but convenient in strong veins). Ideally, the skin should be shaved (plucked)
and swabbed with 70 percent ethyl alcohol at the venepuncture site first and
allowed to dry.
In order to reduce bacterial growth, whole blood samples can have
antibiotics added, taking care that the antibiotics are chosen to avoid
interference with the growth of the pathogens concerned. Thorough mixing
is required for samples with anti-coagulant and/or antibiotics as soon as the
sample has been taken. A smear of fresh blood on the microscope slide
may also be required.
A smear of fresh blood on the microscope slide may also be
required. Blood should be left to stand at room temperature for serum
samples (but protected from excessive heat) until the clot begins to
contract. With a rod, the clot can then be ringed round and the bottles then
placed at 4o C in a refrigerator. Later, after centrifugation, the serum can be
decanted or removed.
The use of chemical preservatives such as boric acid or merthiolate
in sera in virus neutralization tests should be avoided. An alternative
method is to transport on a filter paper disk a drop of dried blood that
contains sufficient material for sensitive antibody assay systems.
3. Faeces:
It is necessary to select freshly voided faeces and send them with or
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without a means of transport. Taking swabs from the rectum (or cloaca),
taking care to swab the mucosal surface, is an alternate and sometimes
preferable method. It is also possible to transport swabs either dry or in
transport media. In order to reduce air and prevent hatching of parasite
eggs, Faeces for Parasitology should fill the container.
4. Skin
Samples are taken from the lesions themselves in diseases which
produce vesicular rashes or where the lesions are exclusively in the skin.
Scrapings of the lesion can be taken, and where unruptured vesicles are
present, the vesicular fluid should also be sampled.
5. Genital tract:
Samples may be taken by washing them vaginally or prepucially, or
by using appropriate swabs. The cervix or urethra is also sometimes
sampled through swabbing.
8. Milk:
Samples of milk should be taken after cleansing the tip of the teat.
The initial stream of milk is discarded and a tube filled with the next
stream(s). In severe mastitis, there may be little fluid present.
9. Environment:
Samples may be taken to monitor hygiene or as part of a disease
inquiry, for example, from litter, ventilation ducts, feed troughs, drains, soil,
hatcheries and slaughter houses.
10. Serum:
Serum samples are the most commonly collected specimens from
live animals for conducting various serological tests. Generally serum
samples early in the course of disease (acute, within 1-4 days) and during
convalescence (convalescent, around 21 days) are collected. Such samples
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are called paired serum samples and are used to demonstrate the rising
antibody titre. In comparing the antibody titres of acute and convalescent
phase sera, a minimum four fold rise is considered significant.
C. SELECTION OF SAMPLES
Considerable skill and care are required to decide on the correct
samples to be sent to the laboratory. Frequently a combination of blood
samples for serology and tissues from dead or culled animals for
microbiological culture will be required. Also, it is usually important to
collect tissues for fixation for histology.
D. SAMPLE SIZE
There are some general statistical rules which should be borne in
mind, particularly when sampling herds or flocks for a health surveillance
scheme. It is possible to calculate how many animals must be sampled
from a herd/flock of a certain size, to achieve a 95% probability infection
assumed to be present in a certain percentage of the animals.
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F. TRANSPORT OF SAMPLES
Samples must be carefully packed, to avoid any possibility of
leakage or cross-contamination. They should be delivered within 48 hours
and must be kept cool during transit. Some samples should not be frozen.
Screw-capped bottles should be used and should be additionally sealed with
adhesive tape or paraffin wax. Samples in individually identified containers
should be placed in larger strong, outer containers and packed with enough
absorbent material to protect from damage. Official shipping regulations
must be consulted. It is advisable to contact the laboratory in advance in the
case of unusual requests. It is essential to do so, where material is sent to a
laboratory in another country. Many countries require a special import
license to be obtained in advance for any biological material, especially for
tissues which could contain animal pathogens. This should accompany the
package and be attached in an envelope to the outside of the parcel.
G. PRESERVATION OF SPECIMENS
Various preservatives are used for different specimens, e.g.
phosphate buffered glycerin for tissues; EDTA, sodium citrate, heparin or
OCG mixture for whole blood and transport media (TPB) for swabs. The
preserved specimens are most frequently transported on ice in a thermos
flask or other suitable containers.
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11. ANTICOAGULANTS
Anticoagulants are used to treat and prevent blood clots that may
occur in your blood vessels. Blood clots can block blood vessels (an artery
or a vein). A blocked artery stops blood and oxygen from getting to a part
of your body (for example, to a part of the heart, brain or lungs). The tissue
supplied by a blocked artery becomes damaged or dies, and this results in
serious problems such as a stroke or heart attack. A blood clot in a large
vein, such as a clot in a leg vein - a deep vein thrombosis (DVT), can lead
to serious problems. For example, it can lead to a clot that travels from a
leg vein to the lungs (a pulmonary embolism). Anticoagulants are used to
prevent blood clots as well - the most common condition for this is atrial
fibrillation (AF).
EDTA
Salt of ethylene diamine tetraacetic acid. Dipotassium (K2),
tripotassium (K3) (41) and disodium (Na2) salts are used; concentrations:
1.2 to 2.0 mg/mL blood (4.1 to 6.8 mmol/L blood) based on anhydrous
EDTA
Citrate
Trisodium citrate with 0.100 to 0.136 mol/L citric acid. Buffered
citrate with pH 5.5 to 5.6: 84 mmol/L tris odium citrate with 21 mmol/L
citric acid. Differences were noticed between 3.2% and 3.8% (v/v) citrate
when reporting results in INR (1, 145, 192, 210). WHO and NCCLS
recommend 0.109 mol/L (3.2%) citric acid. The International Society for
Thrombosis and Haemostasis (ISTH) recommends the use of Hepes
buffered citrate for all investigations of haemostastic functions.
a. A mixture of one part citrate with nine parts blood is
recommended for coagulation tests.
b. One part citrate mixed with four parts blood is recommended to
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Sodium Citrate
Citrate is used as trisodium citrate salt.
It is a white hygroscopic crystalline powder.
Purpose:
Sodium citrate is widely used for coagulation studies.
For PT and PTT.
The sample can be used for ESR by the Westergren method.
Mechanism of action:
it is used in solution form.
This will chelate calcium. Inactivates Ca++ ions.
This will prevent the rapid deterioration of labile coagulation
factors like factor V and factor VII.
Sodium citrate mechanism as an anticoagulant
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Potassium Oxalate
This may be sodium, potassium, ammonium, or lithium oxalic acid
salt used as an anticoagulant.
This form insoluble complex with calcium ions (precipitate with calcium as
a salt).
Potassium oxalate mechanism as an anticoagulant
Sodium Fluoride
This is a weak anticoagulant but used an antiglycolytic agent to
preserve the glucose.
This inhibits the system involved in glycolysis and preserve the
glucose.
This can be used as a dry additive.
Mechanism of action: It acts in two ways:
As an anticoagulant by binding the calcium.
As an enzyme inhibitor which prevents the glycolytic enzyme to destroy the
glucose.
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Sodium Iodoacetate
This is also an antiglycolytic agent at a concentration of 2 g/L.
This may be substituted for sodium fluoride.
This has no effect on urease.
Drawback:
It inhibits creatine kinase.
Adverse effects of the additives:
The additive may contain the substance to be tested like Na+oxalate for the
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estimation of Na+.
The additive may remove the component to be tested like in oxalate,
removes the calcium.
The additive may affect enzymes like Na+flouride. This may destroy many
enzymes.
A small amount of the anticoagulant gives rise to microclots and
this will interfere with cell count.
The additive may distort the cells like oxalate will change the cell
morphology like RBCs and these will become crenated. While WBCs show
vacuoles. Lymphocytes and monocytes will have distorted shapes.
If the excess quantity is used that will dilute the substance to be tested.
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Total ash
The total ash was determined by burning 2g of feed in a pre-
weighed China dish and then samples were placed in a muffle furnace for
ignition at 550 – 600°C till residue was obtained after 4 – 5 hours. Then the
sample residue were placed in desiccators to cool and weight was recorded.
Percentage of ash was obtained by using the following formula:
Wt. of ash
Total ash (%) = ---------------------------× 100
Wt. of feed
Dry matter
Dry matter content was determined by difference between fresh
weight of sample and moisture content. The crude protein, crude lipid,
crude fibre, nitrogen free extract (NFE), ash contents were calculated on %
dry matter basis. Dry matter percentage was calculated by the following.
Dry matter (%) = 100 - moisture (%)
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Principle
Benedict’s quantitative reagent is a modification of qualitative. It
contains copper sulphate, sodium acetate and sodium corbonate. It also
contains potassium thio cyanate and small amount of potassium
ferricyanide. The inclusion of acetate prevents the precipitation of copper
carbonate by chelating Cu3+ion. The thiocyanate causes the precipitation of
white cuprous thio cyanate rather than red cupric oxide. On reduction of
Cu3+ ions which enables the end point of the titration ie., the transition
from blue to white to be readily observable. Methylene blue will be used as
an additional indicator. The small amount of potassium ferricyanide
prevents the re-oxidation of copper. A non-stoicheometric reaction is on
which does not follow a defined pathway and cannot be described by an
equation either quantitatively or qualitatively. The reduction of Cu3+ ions by
sugar is a non-stoicheometric equation and is only constant over a small
range of sugar concentration. To obtain accurate results the volume of
sugar added must be with in 6- 12 ml for 10 ml of benedict’s reagent. If the
preliminary titre value Falls outside this range the sugar solution must be
titrations are repeated.
Reagents Required:
Standard Glucose Solution:
200 mg of glucose was weighed accurately and made upto 100 ml
with distilled water (concentration: 2 mg / ml)
provide the required allcaling with A few porcelain bits and heated to
boiling over a moderate flame.
The standard glucose solution is taken in the burette when the
benedict’s solution Boils, glucose solution is added drop by drop (one
drop per second) till the last trace of blue colour disappears. The volume
of glucose rundown is noted and the titrations are repeated for concordant
value.
The given unknown sugar solution was made upto 100 ml in a
standard flask with distilled water. Then the burette was filled with
unknown sugar solution and the benedict’s reagent was titrated as before.
The volume of sugar solution rundown was noted and titrations are
repeated for concordant values.
Self
TITRATION 2:
Estimation Of Glucose Standardised Benedict’s Reagent Vs Unknown
Glucose
S.No Volume of Burette Readings Volume of Indicator
Benedict’s Intial Final unknown
reagents(ml) ml ml glucose(ml)
Self
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Calculation:
Result:
The amount of glucose present in 100 ml of given unknown
solution is………..
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Reagents Required
Stock Solution:
Bovine serum albumin of 100mg is weighed accurately and
dissolved in 100ml of distilled water in a standard flask.
(Concentration:1mg/ml)
Working Standard:
The stock solution of 10ml is diluted to 100ml with distilled water
in a standard flask. (Concentration:100mg.ml)
Folin;s Phenol Reagent:
Folin’s phenol reagent is mixed with distilled water in a the ratio
1:2
Alkaline CuS04 Reagent: Solution A:
Sodium carbonate of 2% in 0.1N sodium hydroxide.
Solution B: Sodium Potassium tartrate of 1%
Solution C: Copper sulphate of 0.5%
Solutions A, B, C are mixed in the proportion of 50:1:0.5
Unknown Preparation:
The given protein is made upto 100ml with distilled water.
Procedure:
Working standard of 0.2 to 1.0ml is pipetted out into clean test
tubes labelled as S1, to S5 . Test solution of 0.2 and 0.4 ml is taken in test
tubes labelled as T1 and T2. The volume is made upto 1.0ml with distilled
water. Distilled water of 1.0ml serves as blank. To all the test tubes 4.5 ml
of alkaline copper sulphate reagent is added and it is incubated at room
temperature for 10 minutes. To all the test tubes 0.5ml of Folin’s Phenol
reagent is added. The contents are mixed well and the blue colour
developed is read at 640nm after 15 minutes. From the standard graph the
amount of protein in the given unknown solution is calculated.
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S.No Reagents B S1 S2 S3 S4 S5 T1 T2
1 Volume of working - 0.2 0.4 0.6 0.8 1.0 - -
standard (ml)
2 Concentration of working - 20 40 60 80 100 - -
standard (mg)
3 Volume of unknown - - - - - - 0.2 0.4
solution (ml)
4 Volume of distilled water 1.0 0.8 0.6 0.4 0.2 - 0.8 0.6
(ml)
5 Volume of alkaline 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5
copper reagent (ml)
6 Volume of Folin’s phenol 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
reagent
The contents are mixed well and kept at room temperature for 10 minutes. The
blue colour developed is read at 640nm
Calculation:
………of unknown solution corresponds to……..xOD
OD corresponds of ……….. mg of protein
i.e. 0.2ml of unknown solution contains ……….of protein 100ml of
unknown solution contains
100 x ----
Result:
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Reagents Required:
Stock Standard Solution :
About 100 mg of cholesterol was dissolved and made up to 100 ml
with glacial acetic acid (concentration 1 mg / ml).
Working Standard :
About 4 ml of stock solution was made up to 100 ml with ferric
chloride acetic acid reagent (concentration in 40 mg / ml).
Ferric chloride of 0.05% in acetic acid.
Apolar sulphuric acid.
Glacial acetic acid.
Preparation of unknown sample:
20 ml of sample and 40 ml of chloroform was added and
centrifuged. The supernatant was used for estimation.
Procedure:
0.5 ml to 2.5 ml of working standard were Pipetted out into a clean
test tubes. About of 0.5 ml and 1 ml of unknown food sample supernatant
was taken in a test tubes. The volume was made upto 5.0 ml with ferric
chloride and 3.0 ml of concentrated sulphuric acid were added. The test
tubes were kept at room temperature for 15 minutes. The pinkish red
colour formed was measured at 540 nm. Standard graph was drawn for the
values obtained. From the standard graph the amount of cholesterol
present in the food sample can be calculated.
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S.N Reagents B S1 S2 S3 S4 S5 T1 T2
o
1. Volume of standard _ 0.5 1.0 1.5 2.0 2.5 _ _
chloresterol
(ml)
2. Concentration of cholesterol _ 20 40 60 80 100 _ _
(mg)
3. Volume of sample _ _ _ _ _ _ 1 1
supernatant (ml)
4. Volume of 0.05% ferric 5 4.5 4 3.5 3 2.5 _ _
chloride
acetic acid reagent (ml)
5. Volume of cone sulphuric acid 3 3 3 3 3 3 3 3
(ml)
Incubated the tubes for 15 minutes at
room temperature
6. O.D at 540 nm
Calculations
Result:
The amount of cholesterol present in unknown food sample was
found to be……………….
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Principle:
The deoxyribose of DNA in the presence of acid forms hydroxyl
levolinic aldehyde which reacts with diphenylamine to give a blue colour.
But only the deoxy ribose of purine nucleotide react.
Reagents required:
Stock standard solution: Weighed 100 mg of DNA and dissolved in 100 ml
of distilled water.
Working standard solution: 10 ml of stock was diluted to 100 ml using
distilled water.
Diphenyl amine reagent:
It should be prepared freshly by dissolving 1 gm of diphenyl amine
in 100 ml of glacial acetic acid and by adding 2.5 ml of concentrated
sulphuric acid.
Procedure:
Pipetted out 0.2 ml – 1.0 ml of DNA solution into a series of test
tubes and made up the volume to 3.0 ml with distilled water. 0.2 ml and 0.4
ml of unknown is taken and made upto 3.0 ml with water. Added 5.0 ml of
disphenylamine reagent, mixed well and is heated in a boiling water bath for
10 minutes. Cooled and the colour developed is read at 595 nm.
Calculations
Results:
The amount of DNA present in 100 ml of given unknown solution
is……………
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S.No Contents B S1 S2 S3 S4 S5 U1 U2
6. Optical density at
595 mm
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Principle:
Chromatography is a method by which a mixture of substances in
smaller quantities can be separated both qualitatively and quantitatively. In
chromatography there are twophases-the stationary phase and other mobile
phase. When the mobile phase moves along stationary phase, separation of
substances takes place. In thin layer chromatography, the thin layer of gel
functions as an inert supporting material. When the mobile phase moves
along the gel solvent, as the partition coefficient differ for different sugars,
the rate of flow differs and therefore separation occurs.
Materials Required:
Silica gel G 2.Microscopic slides 3.n-butanol
Acetic acid
Spraying reagent (0.3% solution of Ninhydrin in butanol containing
3 ml acetic acid)
Amino acids
Procedure:
A slurry of silicagel G is prepared in 0.02M sodium acetate buffer.
Taken the microscopic slides and keeping them flat, pipetted out about 1-2
ml of the slurry into them. By tilting the slides spread the slurry evenly on
the surface. Lining the edges with Vaseline will be of help. Allowed the
slides to dry completely leaving them flat. 5l samples of amino acid (or
mixture) are spotted and the slide is then dipped in a trough containing n-
butanol-acetic acid-water in the ration 8:2:2. The slide must be handled with
care not to break the surface. After development, that is, when the solvent
has reached the top, the slide is dried and sprayed with the developing
reagent. The slide is then heated in an oven at 1100 C for 10 minutes and
Rf values of the spots are measured.
Distance moved by solute
Rf =-------------------------------------------
Distance moved by solvent
Result:
The given sample Rf value is ……………………
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Materials Required:
Whatmann No :1 filter paper 2.N-butanol
Glacial acetic acid 4.Sugars
Spraying reagent-Alkaline permanganate spraying agent 0.1%
KmnO4 in 2% sodium carbonate
Procedure:
3 strips of whatman No 1 filter paper of size 122 cm are cut. A
line is drawn at a distance of 2.5 cm from the base and a pencil mark made
at the mid point. Sugar solutions of glucose and fructose at a concentration
of 100 mg/10ml is prepared. The chromatography paper is placed on a box
having a slit with lighting arrangment underneath the slit. Spotting is done
on the paper using a micropipette. Care is taken to see that the spot does
not spread beyond a particular diameter. 10ml each glucose and fructose are
spotted on a paper A and B. To the strips a mixture is applied. The three
strips are placed in three different boiling tubes each containing 5ml of n-
butanol acetic acid, water in the ration 4:2:1. The boiling tubes are plugged
with cotton, The paper are kept in a a pre-saturated chromatographic
chamber and the solvent is allowed to rise. When the solvent front has
reached three fours of the paper the strips are removed and air dried. It is
then sprayed with the spraying agent and dried in hot air oven at 1000C.
The sugars appeared as yellow spots in a purple background. The
distances travelled by the solvent are mesured. The distance from the base
line to the centre of each spot are measured, Rf values is then calculated as
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follows.
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Enzyme reagent
Consisted of amino 4 antipyrine (0.125 m mol /L) glucose (30000
m/l ) Peroxidase (10000 m/1) & phosphate butter (100 m mol /L)
Phenol (16 m mol /l)
Standard glucose (5.55 m mol /L )
Preparation of working enzyme reagent
The enzyme reagent was prepared in 500 ml distilled water to
which 5 ml of phenol reagent was added. It was mixed gently & used after 3
hours.
Procedure
Three clean dry test tubes were marked as T for test S for standard
& B for blank & following solutions were added in ml in given order.
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Principle:
This is a general reaction for pentoses and depends on the
formation of furfural when the pentose is heated with concentrated
hydrochloric acid. Orcinol reacts with the furfural in the presence of ferric
chloride as a catalyst to give a green colour, which can be measured at 665
nm.
Requirements:
Standard RNA solution- 200µg/ml in 1 N perchloric acid/buffered
saline.
Orcinol Reagent- Dissolve 0.1g of ferric chloride in 100 ml of
concentrated HCl and add 3.5 ml of 6% w/v orcinol in alcohol.
Buffered Saline- 0.5 mol/litre NaCl; 0.015 mol/litre sodium citrate,
pH 7.
Procedure:
Pipette out 0.0, 0.2, 0.4, 0.6, 0.8 and 1 ml of working standard in to
the series of labeled test tubes.
Pipette out 1 ml of the given sample in another test tube.
Make up the volume to 1 ml in all the test tubes. A tube with 1 ml of
distilled water serves as the blank.
Now add 2 ml of orcinol reagent to all the test tubes including the
test tubes labeled 'blank' and 'unknown'.
Mix the contents of the tubes by vortexing / shaking the tubes and
heat on a boiling water bath for 20 min.
Then cool the contents and record the absorbance at 665 nm
against blank.
Then plot the standard curve by taking concentration of RNA
along X-axis and absorbance at 665 nm along Y-axis.
Then from this standard curve calculate the concentration of RNA
in the given sample.
Result:
The given unknown sample contains ----------µg RNA/ml.
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2. 0.2 0.8 40 2
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Materials
Alkaline Copper tartrate
(A) Dissolve 2.54g anhydrous sodium carbonate, 2g sodium
bicarbonate, 2.5g potassium sodium tartrate and 20g anhydrous sodium
sulphate in 80mL water and make up to 100mL.
(B) Dissolve 15g copper sulphate in a small volume of distilled
water. Add one drop of sulphuric acid and make up to 100mL. Mix 4mL of
B and 96mL of solution A befire use.
Arsenomolybdate Reagent: Dissolve 2.5g ammonium molybdate in
45mL water. Add 2.5mL sulphuric acid and mix well. Then add 0.3g
disodium hydrogen arsenate dissolved in 25mL water. Mix well and
incubate at 37°C for 24 to 48 hours.
Standard Glucose Solution: Stock: 100mg in 100mL distilled water.
Working Standard: 10mL of stock diluted to 100mL with distilled water
[100mg/mL].
Procedure
Weigh 100mg of the sample and extract the sugars with the hot
80% ethanol twice (5mL each time)
Collect the supernatant and evaporate it by keeping it on a water
bath at 80°C.
Add 10mL water and dissolve the sugars.
Pipette out aliquots of 0.1 or 0.2mL to separate test tubes.
Pipette out 0.2, 0.4, 0.6, 0.8 and 1mL of the working standard
solution into a series of test tubes.
Make up the volume in both standard and samp-le tubes to 2mL
with distilled water.
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Calculate
Absorbance corresponds to 0.1mL of test = ‘x’ mg of glucose
10mL contains = (‘x’ ÷ 0.1) × 10mg of glucose = % of reducing sugars
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Procedure
Take beaker containing 15 ml of 1% starch solution + 3 ml of 1%
NaCl solution.
Divide and pour this solution into three test tubes and mark them
as A, B and C.
Maintain the temperature of the beaker containing ice cubes at 5°C.
Take beaker containing ice cubes and keep it on the table.
Take another two beakers containing water and heat over the
Bunsen burner.
Now transfer experimental tube A into a beaker containing ice.
Transfer the second experimental tube B into water bath set at
37°C and third experimental tube C into the beaker maintained at 50°C.
Using a dropper, take 1 ml saliva solution and transfer the solution
into test tube A.
Similarly, add 1 ml saliva solution into test tube B and test tube C.
Immediately, using a dropper, take few drops from experimental
tube A and transfer this into first series of test tubes having iodine solution.
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Similarly, using fresh droppers, do the same procedure for test tube
B and test tube C and transfer the solution into second and third series of
test tubes having iodine solution.
Note this time as zero minute reading.
After an interval of 2 minutes, again take a few drops from each
tube and add to the iodine tubes and note the change in colour of iodine.
Keep on repeating the experiment at an interval of every 2 minutes
till colour of iodine does not change.
Results
It takes less time to reach achromic point at 37°C, as the enzyme is
maximum active at this temperature, while at higher and lower temperatures
more time is taken to reach the achromic point.
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Principle
The optimum pH for the enzymatic activity of salivary amylase
ranges from 6 to 7. Above and below this range, the reaction rate reduces as
enzymes get denaturated. The enzyme salivary amylase is most active at pH
6.8. Our stomach has high level of acidity which causes the salivary amylase
to denature and change its shape. So the salivary amylase does not function
once it enters the stomach.
Materials Required
Three series of test tubes having iodine solution
test tubes, pH tablets of 5, 6.8 and 8,
15 ml 1% starch solution + 3 ml 1% NaCl,
saliva solution, droppers, Bunsen burner and wire gauze.
Procedure
Take a beaker containing 15 ml of 1% starch solution + 3 ml of
1% NaCl solution.
Divide and pour this solution into three test tubes and mark them as A,
B and C.
Add pH tablet 5 into test tube A, pH tablet 6.8 into test tube B and pH
tablet 8 into test tube C.
Now transfer experimental tube A, B and C into a beaker containing
water and a thermometer for recording temperature. Temperature of this
beaker is to be maintained at 37°C.
Using a dropper, take 3 ml saliva solution and add 1 ml of solution to
each of the three test tubes.
Immediately using a dropper, take few drops from experimental tube A
and transfer this into the first series of test tubes having iodine solution.
Similarly, do the same procedure for test tube B and test tube C and
transfer the solution into second and third series of test tubes having iodine
solution.
Note this time as zero minute reading.
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After an interval of 2 minutes, again take a drop from each tube and
add to the iodine tubes and note the change in colour of iodine.
Keep on repeating the experiment at an interval of every 2 minutes till
colour of iodine does not change.
Results
pH 5 is acidic and pH 8 is alkaline, therefore salivary amylase did
not act in these tubes. Whereas, the enzyme acted in the tube with pH 6.8
(i.e., slightly acidic) and digested the starch.
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Equipment required
Hemocytometer which consists of z comparator box which has
brown colored glass on either side
Hb pipette which is marked upto 20mm3(0.02ml blood) z Tube
with markings of Hb on one side z glass rod
Dropper
Reagents required
N/10 HCl Distilled water Sample:
Venous blood collected in EDTA
Procedure
1. Add N/10 HCl into the tube upto mark 2g%
2. Mix the EDTA sample by gentle inversion and fill the pipette with
0.02ml blood. Wipe the external surface of the pipette to remove any excess
blood.
3. Add the blood into the tube containing HCl. Wash out the contents of
the pipette by drawing in and blowing out the acid two to three times. Mix
the blood with the acid thoroughly.
4. Allow to stand undisturbed for 10min.
5. Place the hemoglobinometer tube in the comparator and add distilled
water to the solution drop by drop stirring with the glass rod till it’s color
matches with that of the comparator glass. While matching the color, the
glass rod must be removed from the solution and held vertically in the tube.
6. Remove the stirrer and take the reading directly by noting the height of
the diluted acid hematin and express in g%.
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Principle:
Equipment required
Hb pipette
Spectrophotometer
Reagents required
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Procedure
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Principle
The DNA of plant cells is found in three distinct genomes. First,
there is nuclear DNA familiar as the DNA that makes up the
chromosomes, but mitochondria and chloroplasts each have DNAs of their
own. These genomes are closed circular DNA molecules encoding many of
enzymes necessary for the function of the organelles.
Materials
Extraction buffer (EB):
50 mM Tris pH 8.0
1% CTAB; 50 mM EDTA
1 mM 1,10 o-phenanthroline
0.7 M NaCl
0.1% β-mercaptoethanol
chloroform; isopropyl alcohol
80% ethanol
15 mM ammonium acetate pH 7.5.
TE buffer:
10 mM Tris pH 8.0
1 mM EDTA,
centrifuge tubes, water bath (65°C), centrifuge.
Protocol
Start with the frozen chloroplast preparation. This sample will have a
volume of several mL. For each mL of chloroplasts, add 4 mL EB. If
necessary, transfer the mixture to a capped centrifuge tube of at least twice
the volume of the chloroplasts and EB.
Incubate the mixture at 65°C for 1 h on a water bath.
Remove the tube from the water bath and allow cooling on the bench
for several minutes to proceed further.
Add an approximately equal volume of chloroform to the tube, recap,
and mix by inversion.
Centrifuge at >3500g for 10 min.
The tube contents will be separated into two distinct layers. Transfer
the upper (aqueous) layer into a fresh centrifuge tube with the help of an
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ultra pipette. This tube should be of the same size as that used in the first
step. The lower (organic) layer is hazardous water, dispose as per safety
guidelines.
Add 0.6 mL of isopropanol for each mL of DNA containing extract
in the centrifuge tube and mix by inversion.
Centrifuge at >10,000g for 20 min.
Decant liquid in the tube away from the DNA containing pellet. The
tube’s inside can be wiped carefully to remove liquid, but take care not to
dislodge the DNA pellet.
Result
The yield of chloroplast DNA is expected to be low and will
depend on the quality of the chloroplast preparation produced.
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Principle
Roots of pea seedlings are used to isolate mitochondria by
differential centrifugation and assay the activity of a mitochondrial enzyme
to assess the success of the isolation protocol. Differential centrifugation
separates cell components based on differences in the rate at which they
sediment in a centrifugal field. The first centrifugation step will remove
whole cells, cell wall fragment, nuclei, starch, etc. Mitochondria, because of
their small size, will not be sedimented by this step but will remain
suspended in the supernatant. A subsequent spin at a higher speed will then
be used to pellet the mitochondria to the bottom of the centrifuge tube.
Materials
Pea seedlings, cheesecloth/miracloth, centrifuge tubes,
Pasteur pipettes, small paint brush, ice bucket, blender
centrifuges
UV–Vis spectrophotometer.
Harvest pea root tissue, prepare extract, isolate mitochondria by
differential centrifugation.
Homogenization buffer:
70 mM sucrose, 220 mM mannitol, 0.5 g/L BSA, 1 mM HEPES
pH 7.4 homogenization buffer.
Solutions:
0.1 M potassium phosphate buffer, pH 7.4,
0.8 M ascorbic acid,
4% Triton X-100,
5 mg/mL cytochrome,
sodium dithionate crystals.
Isolation of Mitochondria
Procedure
All solutions and samples should be kept on ice while working.
Harvest 5 g of 7-day-old pea roots, shake off vermiculite in which
they are growing, and rinse in distilled water in a beaker.
Chop roots into small pieces with a razor blade or scissors and put
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Principle
In biological materials, lipids are found as lipoprotein complexes and
these have to be extracted. Lipids being soluble in nonpolar organic solvents and
proteins being soluble in polar aqueous solvents, the efficient lipid extraction
can be achieved only with an aqueous solvent like chloroform and diethyl
ether. This would help in breaking the lipoprotein complexes. Extracted
lipid components can be separated on TLC based on their differential
mobility along the porous stationary phase such as silica gel, and these can
be located by spraying the plates with either 2′,7′-dichlorofluorescein or
50% sulfuric acid.
Materials
TLC tank; glass plates (20 × 20 cm) for TLC; 2′,7′-
dichlorofluorescein (prepare 0.2% solution of 2′,7′3-dichlorofluorescein in
95% v/v ethanol).
glass plates; spreader;
silica gel G;
oven set at 110°C;
UV lamp; solvent system—petroleum ether (b.p. 60°C–70°C).
Procedure
Extraction of lipids from sample
Grind 1 g of the tissue in the extraction solvent (either diethyl
ether/ethanol, 3:1, or chloroform/methanol, 2:1) in a pestle and mortar.
Transfer the homogenate to a separating funnel. Shake the contents
vigorously, and allow it to stand till the two phases have completely
separated. Drain out the lower organic layer that contains the lipids.
Evaporate the solvent under vacuum, and keep the concentrated lipid
extract protected from light under N2 atmosphere.
Prepare the TLC plates using silica gel G as described in the
experiment for sugars.
Activate the TLC plates at 110°C for 30 min, cool them in a
desiccator, and spot the lipid samples, standards, as well as unknown.
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Principle
Sugars are first extracted by treating the finely powdered dried
grain or leaf sample repeatedly with hot 80% (v/v) alcohol. The residue is
then treated with cold perchloric acid to solubilize starch. This derivative is
then condensed with anthrone to give a blue-colored complex that is
determined quantitatively by anthrone sulfuric acid procedure.
Materials
Dry leaf powder, test tubes, water bath, refrigerated centrifuge,
polypropylene tubes, colorimeter, cuvettes, anthrone, sulfuric acid, ethyl
alcohol, perchloric acid, glucose.
Anthrone–sulfuric acid reagent:
Dissolve 2 g of anthrone in cold 95% conc. H2SO4, store at 4°C.
Use freshly prepared reagent.
80% absolute alcohol: Eighty milliliters of ethyl alcohol make up to 100
mL with distilled water.
52% perchloric acid: Add 270 mL of 72% perchloric acid to 100 mL of
water.
Glucose standard:
Dissolve 0.1 g of anhydrous glucose in 100 mL of water containing
0.001% benzoic acid as preservative. The standard solution contains 1 mg
glucose per mL.
Procedure
For extraction of sugars and starch, take 0.2 g of finely ground
sample in a 50 mL centrifuge tube, and add 20 mL of hot 80% alcohol.
Shake the tubes for 5–10 min, and after centrifuging at 3000 rpm for 10
min, decant the supernatant.
Repeat this extraction with 80% hot ethanol five to six times till the
supernatant is free of sugars as judged by a negative test with anthrone
reagent.
Cool the residue in ice water and add 6.5 mL of 52% perchloric
acid while stirring the contents with a glass rod.
Let it stand for 15 min with occasional stirring and then centrifuge
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at 4°C. Collect the supernatant and repeat this extraction Step four to
five times. Combine the supernatant fractions and make up the volume
to 100 mL with water.
Dilute the previous extract so that it contains 5–20 µg of glucose
per mL. Take 5 mL aliquot of this diluted extract and place the tubes in a
cold water bath.
Add 10 mL freshly prepared anthrone reagent. Mix properly and
transfer the tubes to boiling water bath for 5–7 min.
After cooling the tubes under running tap water, note the
absorbance of these solutions at 630 nm.
Prepare a standard curve using 0–100 µg glucose and anthrone
reagent (Step 3). Calculate the amount of glucose in the sample aliquot.
Results
From the standard curve, determine the amount of glucose in 5 mL
of the diluted aliquot of sample extract.
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Principle:
Casein is the main protein found in milk and is present at a
concentration of about 35 g/L. It is actually a heterogeneous mixture of
phosphorus-containing proteins and not a single compound. Most proteins
show minimum solubility at their isoelectric point, and this principle is used
to isolate the casein by adjusting the pH of milk to 4.8, its isoelectric point.
Casein is also insoluble in ethanol, and this property is used to remove
unwanted fat from the preparation.
Materials:
Milk, sodium acetate buffer (0.2 mol/L, pH 4.6)
Ethanol (95% v/v), ether
Procedure:
Place 100 mL of milk in a 500 mL beaker and warm to 40°C; also
warm 100 mL of the acetate buffer and add slowly with stirring.
The final pH of the mixture should be about 4.8, and this can be
checked with a pH meter. Cool the suspension to room temperature and
then leave to stand for a further 5 min before filtering through muslin.
Wash the precipitate several times with a small volume of water,
and then suspend it in about 30 mL of ethanol.
Filter the suspension on a Buchner funnel and wash the precipitate
a second time with a mixture of equal volumes of ethanol and ether.
Finally, wash the precipitate on the filter paper with 50 mL of
ether, and suck dry. Remove the powder and spread out on a watch glass to
allow evaporation of the ether.
Weigh the casein and calculate the percentage yield of the protein.
Result:
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Principle
During storage, fats may become rancid as a result of peroxide
formation at the double bonds by atmospheric oxygen and hydrolysis by
microorganisms with the liberation of free acid. The amount of free acid
present, therefore, gives an indication of age and quality of the fat. The acid
value is the number of milligrams of KOH required to neutralize the free
fatty acid present in 1 g of fat.
Materials
Olive oil, butter and margarine,
fat solvent (equal volumes of 95% v/v alcohol and ether
neutralized to phenolphthalein),
Phenolphthalein (10 g/L in alcohol),
KOH (0.1 mol/L)
Protocol
Accurately weigh out 10 g of the test compound and suspend the
melted fat in about 50 mL of fat solvent.
Add 1 mL of phenolphthalein solution, mix thoroughly, and titrate
with 0.1 mol/L KOH until the faint pink color persists for 20–30 s.
Note the number of milliliters of standard alkali required and
calculate the acid value of the fat.
Results
The acid value present in the given fat sample is ………g/mL.
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Principle
On refluxing with alkali, glyceryl esters are hydrolyzed to give
glycerol and the potassium salts of the fatty acids (soaps). The
saponification value is the number of milligrams of KOH required to
neutralize the fatty acids resulting from the complete hydrolysis of 1 g of
fat. The saponification value gives an indication of the nature of the fatty
acids in the fat since the longer the carbon chain, the less acid is liberated
per gram of fat hydrolyzed.
Materials
Fats and oils (tristearin, coconut oil, corn oil, and butter),
Fat solvent (equal volumes of 95% ethanol and ether),
Alcoholic KOH (0.5 mol/L),
Phenolphthalein (10 g/L in alcohol),
HCl (0.5 mol/L)
Protocol
Weigh 1 g of fat in a tarred beaker and dissolve in about 3 mL of
the fat solvent.
Quantitatively transfer the contents of the beaker to a 250 mL
conical flask by rinsing the beaker three times with a further addition of
solvent; add 25 mL of 0.5 mol/L alcoholic KOH and attach to a reflux
condenser.
Set up another reflux condenser as blank with everything present
except the fat, and heat both flasks on a boiling water bath for 30 min.
Leave to cool to room temperature and titrate with 0.5 mol/L HCl
and phenolphthalein indicator.
The difference between the blank and test reading gives the
number of milliliters of 0.5 mol/L KOH required to saponify 1 g of fat.
Results
The molecular weight of KOH is 56, and since three molecules of
fatty acid are released from a triglyceride
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Principle
Acetic anhydride reacts with cholesterol in a chloroform solution
to produce a characteristic blue-green color. Blood or serum is extracted
with an alcohol–acetone mixture that removes cholesterol and other lipids
and precipitates protein. The organic solvent is removed by evaporation on
a boiling water bath and the dry residue dissolved in chloroform. The
cholesterol is then determined colorimetrically using the Liebermann–
Burchard reaction. Free cholesterol is equally distributed between the cells
and plasma, while the esterifies form occurs only in plasma. It is essential to
use absolutely dry glassware for this estimation.
Materials
Serum or blood, alcohol–acetone mixture (1:1)
Chloroform, acetic anhydride–sulfuric acid mixture (30:1)
Stock cholesterol solution (2 mg/mL in chloroform)
Working cholesterol solution (dilute the previous solution in 1:5
ratios in chloroform to give a strength of 0.4 mg/mL).
Protocol
Place 10 mL of the alcohol–acetone solvent in a centrifuge tube,
and add 0.2 mL of serum or blood.
Immerse the tube in a boiling water bath with shaking until the
solvent begins to boil. Remove the tube and continue shaking the mixture
for a further 5 min. Cool to room temperature and centrifuge.
Decant the supernatant fluid into a test tube and evaporate to
dryness on a boiling water bath.
Cool and dissolve the residue in 2 mL of chloroform. At the same
time, set up a series of standard tubes containing cholesterol and a blank
with 2 mL of chloroform.
Add 2 mL of acetic anhydride–sulfuric acid mixture to all tubes and
thoroughly mix. Leave the tubes in the dark at room temperature and read
the extinction at 680 nm.
Results
The normal serum cholesterol lies within the range of 100–250
mg/100 mL.
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Principle
Total yeast RNA is obtained by extracting a whole-cell homogenate
with phenol. The concentrated solution of phenol disrupts hydrogen
bonding in the macromolecules, causing denaturation of the protein. The
turbid suspension is centrifuged, and two phases appear: the lower phenol
phase contains DNA, and the upper aqueous phase contains carbohydrate
and RNA. Denatured protein, which is present in both phases, is removed
by centrifugation. The RNA is then precipitated with alcohol. The product
obtained is free of DNA but usually contaminated with polysaccharide.
Further purification can be made by treating the preparation with amylase.
Materials
Dried yeast, phenol solution (90%),
Potassium acetate (20%, pH 5),
Absolute ethanol, diethyl ether
Protocol
Suspend 30 g of dried yeast in 120 mL of water previously heated
to 37°C. Leave for 15 min at this temperature and add 160 mL of
concentrated phenol solution.
Stir the suspension mechanically for 30 min at room temperature,
and then centrifuge at 3000g for 15 min in the cold to break the emulsion.
Carefully remove the upper aqueous layer with a Pasteur pipette, and
centrifuge at 10,000g for 5 min in a refrigerated centrifuge to sediment
denatured protein.
Add potassium acetate to the supernatant to a final concentration
of 20 g/L, and precipitate the RNA by adding two volumes of ethanol.
Cool the solution in ice and leave to stand for 1 h.
Collect the precipitate by centrifuging at 2000g for 5 min in the
cold. Wash the RNA with ethanol–water (3:1), ethanol, and, finally, ether;
air-dry and weigh. (Note: Yeast contains about 4% RNA by dry weight.)
Compare with a commercial preparation by measuring the pentose,
phosphorus, and DNA content and by determining the absorption
spectrum. Keep your preparation for use in later experiments.
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Principle
Procedure
Take 1 mL aliquot of the chloroplast suspension into a 10 mL
graduated cylinder, and dilute to 10 mL with 80% acetone. Cover the
cylinder with parafilm and mix by inverting.
Prepare the spectrophotometer to read the absorbance of the diluted
chlorophyll extract.
Adjust the wavelength to read 652 nm (why is this wavelength
chosen?). Without a cuvette in the machine, adjust to 0% transmittance
(left-hand knob).
Blank the spectrophotometer with the reagent blank (80% acetone) to
read 0 absorbance (right-hand knob). Transfer some of diluted chlorophyll
extract to a cuvette and read the absorbance. Record the absorbance value.
Chlorophyll absorbance (A652) = OD
Calculate the chlorophyll content in the diluted sample using the
following equation. Record the chlorophyll concentration of the
diluted sample.
A = ECd where A is the observed absorbance, E is a proportionality
constant (extinction coefficient) (=36 mL/cm), C is the chlorophyll
concentration (mg/mL), and d is the distance of the light path (=1
cm).
Calculate the chlorophyll concentration in the original chloroplast
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Chemicals:
Mercuric chloride : 0.5gm
Sodium chloride : 1.0gm
Sodium sulphate : 5.0 gm
Distilled water : 200ml
Procedure
An improved Neubauer,s counting chamber was used for counting
RBC (Baker and Silverton, 1982).
Using RBC pipette, the blood was drawn upto 0.5 mark and the
diluting fluid to the mark 101.
Although fluid is drawn to the mark 101 but the real dilution is
0.5:100 or 1:200 because the fluid in the capillary tube is discarded before
the count.
Calculations
The number of RBC’s per sq mm was calculated as follows:
Area of a small square : 1/400 sq mm
Depth of the counting chamber : 1/10mm
The volume of Small Squareis : 1/4000cumm
The dilution of the blood is : 1/200
Total RBC = cu mm
N = No of cells in 80 small squares
Result
The total RBC present in the given sample is ……………….
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A white cell count (TLC) estimates the total number of white cells
in a cubic millimetre of blood. WBC diluting fluid or Turk fluid contains a
weak acid to lyse the red blood cells and Gentian violet stain for staining
the nucleus of white blood cells.
Chemicals:
The Turks fluid with following composition was used for TLC:
Glacial acetic acid :1.5ml
1% aqueous solution of Gentian violet :1 ml
Distilled water :100ml
Procedure
Neubauer,s hemocytometer (Baker and Silverton, 1982) was used
for counting of leucocytes.
Using white cell pipette, the blood was drawn upto 0.5 mark and
the diluting fluid to 11 mark, thus the dilution was 1:20.
Calculations
The number of RBC’s per sq mm was calculated as follows:
Area of a small square : 4 sq mm
Depth of the counting chamber : 1/10mm
The dilution of the blood is : 1/20
N x 20/10
Total WBC = ----------------- cu mm
4
Result
The total WBC present in the given sample is ……………….
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Chemicals:
EDTA blood
PCV tube
Procedure
The heparinised blood was filled upto the mark 100 of the
haematocrit tube with the help of Pasteur pipette and centrifuged at 3000
rpm for 30minutes.
The relative volume of the height of the RBC’s packed at the
bottom of the haematocrit tube was recorded as (PCV) in terms of
percentage of total blood column taken in the haematocrit tube.
Result
The PCV or Haematocrit present in the given sample is
……………….
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PCV×10
MCV (fl) =
RBC Count
Result
The MVC present in the given sample is ……………….
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Hemoglobin(g / dL)×10
MCH (pg) =
RBC Count
Result
The MCH present in the given sample is ……………….
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Hemoglobin(g / dL)
MCHC (%) = ×100
PCV
Result
The MCHC present in the given sample is ……………….
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Reagents
Leishman’s stain (Eosin and methylene blue dyes dissolved in
acetone-free absolute methyl alcohol).
Procedure
A blood film was prepared, dried and placed on the staining rack
and was covered with Leishman’s stain, allowed to stand for 2 minutes.
Then equal volume of distilled water was added and mixed by the
slide first one way and then other.
Allowed to stand it for 6 minutes. Drained off diluted stain in a
stream of distilled water from a wash bottle for about 20 seconds and
allowed the slide to remain on the staining rack for 1-2 minutes with the last
wash covering it.
Then kept the slide against a support in an inclined position,
stained smear facing down and allowed it to dry.
Then the stained slides were studied under low and high power
objectives for differential leucocyte count by placing two drops of glycerol
on the stained smear and using oil-immersion lens.
Calculations
Differential leucocyte was expressed as percentage.
Number of type cells
DLC (%) = -------------------------------------------------- x 100
Total number of leukocytes
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Principle:
Serum is incubated with starch substrate. The amylase in the serum
hydrolyses the starch to simpler units with a resulting increase in reducing
groups. In the method presented here iodine is added which reacts with the
starch molecules not hydrolysed by the amylase. The iodine-starch complex
is blue in colour and is measured in the spectrophotometer. The degree of
loss in colour is proportional to the amount of starch hydrolysed and hence
to the activity of the amylase in the serum. A substrate control is carried
though the procedure to give a reference value for the amount of starch
substrate present before hydrolysis.
Procedure
1. Pipette 5 ml of substrate into two 50 ml volumetric flasks.
2. Place the 'test' flask into a 34oC water bath for 5 minute to warm
the contents.
3. Using a pipette that deviloers between two marks add 0.1 ml of
serum to the 'test' flask and mix. Do not use blow out pipettes as the
smallest amount of saliva can give a large error.
4. Time the addition of serum using a stop watch.
5. After exactly 7.5 minutes add 5 ml or working iodine solution, mix
and immediately remove from the water bath.
6. Similarly add 5 ml of the working iodine solution to the flask
containing the 'substrate control', which has not been incubated.
7. Dilute the contents of both flasks to the 50 ml mark with distilled
water and mix the flasks well.
8. Read the absorbance of both against water using the large (19 mm)
cuvettes at 660 nm.
Calculation
Absorbance of substrate control – absorbance of test x 800 Absorbance of
control
units of amylase activity per 100 ml of serum.
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Principle:
Phosphotungstic acid in alkaline medium oxidizes uric acid
to allantion and itself gets reduced to tungsten blue which is
estimated colorimetrically at 700mm.
Reagents :
Sodium tungstate 10%.
2/3 N Sulphuric acid.
Tungstic acid: Add 50ml of 10% sodium tungstate 50ml 2/3 N H2SO4 and
a drop of phosphoric acid with mixing to 800ml water. Discard when
cloudy. Store in brown bottle.
Phosphotungstic acid: Stock-Dissolve 50g sodium tungstate in about 400ml
of water. Add 40ml 85% phosphoric acid and reflux gently for 2 hours,
cool, make volume to 500m. store in brown bottle. Dilute 1 to 1 for use.
Na3CO3 10%.
B T S S2 S3
1. Standard uric acid - - 1.0 2.0 3.0
(1mg%) ml.
2. Supernatant (ml) - 3.0 - - -
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Natural process
This process include sun-drying. In this the plants are kept in the
shades and are air dried in sheds. This process takes few weeks for
complete drying of the moisture. This time depends on the temperature and
humidity.
Artificial drying
Using the help of artificial driers, artificial drying is done. The time
consumed is reduced to a few hours or minutes by this process. Warm-air
drying is the common method used for the drying of medicinal plants. The
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hot air furnace on which hot air is blown is used to do this. For drying
succulent parts of plants and fragile flowers, this method is applicable. To
avoid disintegration of the thermolabile compounds, the drying must be
done at a lower temperature.
Grinding of plant materials
The plant samples are to be powdered for further analysis
following complete drying of moisture. There are various forms of
powdering, including the following,
1. Grinding can be done by grinding in an electric grinder or a spice mill or
in a mortar or pestle as well.
2. Due to the increased surface area of the plants, grinding increases the
effectiveness of extraction. The reduction in the area of the surface can lead
to dense packing of the material.
3. It is always ideal to mill the plants into a fine powder, but if they are too
fine, this impacts the flow of the solvent and also generates more heat that
could degrade some thermolabile compounds.
Choice of Solvent
For the determination of biologically active phytochemicals from
plants, the solvent that is used for the extraction process is very important.
These solvents must be less toxic, easy to evaporate in less heat, preserve
and not dissociate the compounds in them. For extraction, the different
solvents commonly used include:
1. Water:
It is a universal solvent; plant extracts are usually extracted with
water for anti-microbial activity. But when compared to water, the organic
solvents give consistent results in anti-microbial activities. No significant
results can be obtained from water soluble compounds in the extract.
2. Alcohol:
Due to the presence of higher amounts of polyphenols, these
alcoholic extracts from plants show more activity than aqueous extracts.
This is due to the higher alcohol degradation of the cell wall and seed,
which releases the polyphenols that will be degraded if aqueous extracts are
extracted. Ethanol, however is microbicidal rather than water. 70 percent
ethanol is used to extract more bioactive compounds than pure ethanol.
Intracellular ingredients from plant materials are also found to be easier to
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3. Acetone:
Acetone dissolves many of the plants' hydrophilic and lipophilic
compounds and is water-mixable. It is low in toxicity and volatile and is
used to extract antimicrobial activity. Tannins and other phenolic
compounds are extracted using acetone. In addition, they are used to
extract saponins.
4. Chloroform:
Terpenoid lactones are obtained from barks by extraction with
chloroform. Tannins and Terpenoids are treated with less polar solvents.
5. Ether: They are used for the extraction of coumarins and fatty acids.
Methods of Extraction:
Homogenization
One of the most commonly used methods for extraction is this
technique. This is done using either the dried or wet method of extraction.
The dried plant samples are finely powdered in this dried extraction method
and added to the solvent blended for a few minutes and kept for about 24
hours in an orbital shaker. The parts of the plants are cut into small pieces
during the wet extraction process, ground in a mortar and pestle and added
to a solvent and shaken for 24 hours in an orbital shaker and then filtered.
The filtrate can be used for the further analysis.
Serial Exhaustive Extraction
It is done to extract a wide range of polarities of compounds with a
variety of solvents from a non-polar solvent such as hexane to more polar
solvent such as methanol. The drawback is that due to the high heat leading
to the degradation, thermolabile compounds can not be extracted.
Soxhlet Extraction
It is used when the compound in the solvent is less soluble and the
impurities in the solvent are soluble. The impurities can be eliminated by
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Digestion
This is a process where the extraction is done as maceration with a
gentle heat applied. It is used when the elevated temperature do not
interfere the solvent efficiency or the compounds.
Percolation
For this process an instrument called percolator is used which is a
narrow, cone shaped Vessel with open ends. The ingredients are moistened
with an appropriate amount of the specified menstrum and allowed to stand
for approximately 4 h in a well closed container, after which the mass is
packed and the top of the percolator is closed. Additional menstrum is
added to form a shallow layer above the mass, and the mixture is allowed to
macerate in the closed percolator for 24 h. The outlet of the percolator then
is opened and the liquid contained therein is allowed to drip slowly.
Additional menstrum is added as required, until the percolate measures
about three quarters of the required volume of the finished product. The
marc is then pressed and the expressed liquid is added to the percolate.
Sufficient menstrum is added to produce the required volume, and the
mixed liquid is clarified by filtration or by standing followed by decanting.
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Sonication
In this method the ultrasound with higher frequencies of 20 kHz –
2000 kHz are used which will disrupt the cells and releases the constituents.
Although the process is useful in some cases, like extraction of rauwolfi a
root, its large-scale application is limited due to the higher costs. One
disadvantage of the procedure is the occasional but known deleterious
effect of ultrasound energy (more than 20 kHz) on the active constituents
of medicinal plants through formation of free radicals and consequently
undesirable changes in the drug molecules.
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Miscellaneous compounds
Test of resins
1. Precipitation test: about 0.2 g of extract was extracted with 15ml of 95%
ethanol. The alcoholic extract was then poured into a beaker containing
about 20ml of distilled water.
2. 1ml of extract was taken and to this few ml of acetic anhydride was
added to this 1ml of conc.H2SO4 was added. The appearance of orange to
yellow colour indicates the presence of resins
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Carboxylic acids
1. To 1ml of extract a pinch of sodium bicarbonate is added. The
production of effervescence indicates the presence of carboxylic acids
2. 2ml of alcoholic extract was taken in warm water and filtered. The
filtrate was then tested with litmus paper and methyl orange. The
appearance of blue colour.
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3. 2ml of extract is mixed with about 0.4 ml of glacial acetic acid containing
traces of ferric chloride and 0.5 of conc. H2SO4 was added the production
of blue colour is positive for glycosides.
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chalcones
Test for Terpenoids (Salkowski test) 3ml of the extract was taken
and 1ml of chloroform and 1.5 ml of concentrated H2SO4 are added along
the sides of the tube. The reddish brown colour in the interface is
considered positive for the presence of terpenoids
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Gas Chromatography
Volatile compounds are analysed using gas chromatography. In this
method, there is a gas and a liquid phase. The liquid phase is stationary
where the gas phase is a mobile phase. These compounds to be analysed are
also in the mobile phase with a carrier gas which is usually helium, hydrogen
or argon. The chemicals are separated depending on the migration rate into
the liquid phase. Higher percentage of the chemical will lead to faster
migration in the liquid phase. This is widely used in qualitative and
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REFERENCES
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