Transcript of "VK Malhotra - Practical Biochemistry for Students, 4th Edition"
Varun Kumar Malhotra
Department of Biochemistry
Maulana Azad Medical College
MEDICAL PUBLISHERS (P) LTD
Preface to the Fourth Edition
Over the years, the world of Biochemistry has seen many changes for better investigational
techniques for the welfare of the patients. Hence it becomes mandatory for any written
material to show the changes duly. In this edition, my efforts have gone in a direction to
improve the material.
In a modest attempt, additions have been made about the methods of expressing
concentrations, instrumentations, collection and preparation of blood specimen, SGOT,
SGPT, etc. The inquiring mind will certainly benefit from such exposures to manage the
clinical situation in a more creative and challenging manner.
Some modifications became necessary in various chapters and thus the matters have
been updated in a befitting manner to serve the demanding needs of the consumers.
Lastly, I thank many people like Prof B Misra and Publishers for their guidance and
Varun Kumar Malhotra
Preface to the First Edition
In the ever expanding knowledge of Biochemistry it is very difficult on the part of single
individual to go through the various bigger valume of textbooks on practice of biochemical
investigations. This book in a concise but equally satisfactory form will help the user,
students and practitioners either to a great extent. Besides being handy it has been kept up
to its spirit of recent approaches by virtue of which one has the best utility in a busy time.
For every medical practitioner and enlighted patients, Biochemistry has been playing a
significant role. This book will certainly be of significant importances to the practitioners
as well as laboratories.
I thank all my colleagues and friends in contributing to have brought out this book in its
near perfect shape.
My extreme heartful thanks are due to the incessant guidance from Prof B Misra,
Department of Physiology, MAM College in bringing out every chapter in an excellent
way. The contributions that I have received by the constant cooperation of my parents and
wife cannot be ignored.
Lastly my utmost thanks are due to Dr K Chaudhry and my Publisher Mr Jitendar Vij.
I hope the book will bring out greater number of readers keeping in view the worth of
The author always keeps an open eye for suggestions.
Varun Kumar Malhotra
x Practical Biochemistry for Students
24. Urea Clearance ................................................................................................................................... 83
25. Blood Cholesterol .............................................................................................................................. 85
26. Serum Calcium .................................................................................................................................. 87
27. Inorganic Phosphorus ...................................................................................................................... 90
28. Serum Total Proteins and Albumin: Globulin Ratio ................................................................. 92
29. Serum Bilirubin ................................................................................................................................. 95
30. Prothrombin Time ............................................................................................................................. 98
31. Liver Function Tests ........................................................................................................................ 100
32. Demonstrations ............................................................................................................................... 104
Appendix ............................................................................................................................................ 121
Questions Analysis 121
Normal Values 124
Methods of Expressing Concentration
Concentration may be defined as weight per unit volume.
The most common expressions are:
According to Caraway there are three ways of expressing percentage of solution, i.e. W/W,
a. Weight per unit weight (W/W)
A 10% W/W solution contains 10 gm of solute in 90 gm of solvent.
b. Weight per unit volume (W/V)
A 10% W/V solution contains 10 gm of solute dissolved in final volume of 100 ml of
c. A 10% V/V solution contains 10 ml of the concentrate per 100 ml of solution.
A molar solution contains 1 gm mol. wt. (mole) of solute in one litre of solution. Molarity of
solution is expressed as ‘M’
1 mole = 1000 millimoles
gm.mol.wt.(moles) of solute
Molarity = _____________________________________
Volume of solution (L)
Amount of solute (gm)
Molarity = ____________________________________________
Mol. wt. × volume of solution (L)
Molarity of a solution of 2 litres containing
58.5 gm. NaCl dissolved.
Amt. of NaCl (gm)
Moles of NaCl = __________________________
23 + 35.5
2 Practical Biochemistry for Students
∴ Molarity =
Moles of solute
Volume of solution (L)
= 0.5 M
So, Molarity of given solution is 0.5 M.
A normal solution contains 1 gm equivalent Wt. (eq.) of solute dissolved in one litre of solution.
Normality of solution is expressed as ‘N’
1 equivalent = 1000 milliequivalent.
gm. eq. wt. of solute
Normality = __________________________________
Volume of solution (L)
⇒ Normality =
Amount of solute (gm)
Equivalent wt. × Vol. of solution (L)
To evaluate equivalents of solute, we divide the molecular weight by its total valency of
cation or anion.
Eg. equivalent weight:
a. of NaCl = Mol. wt. of NaCl
Mol. wt. of MgCl2
b. of MgCl2 = __________________________
Mol. wt. of AICl3c. of AICl3 = ________________________
Normality of a solution of 2 litres containing 180 gm of glucose.
• Equivalents of glucose will be same as its moles because it does not dissosiate in solution.
Amt. of glucose (gm)
Equivalents of glucose =
Equivalents of solute
Vol. of solution (L)
= 0.5 N
So, Normality of given solution is 0.5 N.
Molality is defined as number of moles of solute dissolved in 1000 gm of solvent (Not in
It is designated as ‘M’.
Moles of solute
Weight of solvent (gm)
Methods of Expressing Concentration 3
Amount of solute (gm)
⇒ Molality =
Mol. wt. × Weight of solvent (gm)
Unlike normality and molarity, this parameter of expressing the concentration is
independent of temperature as there is no volume term in this relation.
Formality parameter is not used nowadays because all of the above parameters are better than
it and are able to express concentration sufficently.
Formality is same as molarity if molecular weight in the formula is replaced by formula
4 Practical Biochemistry for Students
All biochemical reactions are greatly influenced by the hydrogen ion concentration of the
surrounding medium in which the reaction takes place. The most convenient way of expressing
hydrogen ion concentration is by the term pH.
pH is defined as the negative logarithm of the hydrogen ion concentration of the solution.
pH = – log CH
Hence it is both important and useful to know some of the simple methods of pH determi-
pH can be determined both by colorimetric and electrometric methods. Electrometric
method is the most accurate one and is done by using a pH meter whereas colorimetric
determination of pH can be simply done by the following methods:
1. Indicator papers also called narrow range pH papers.
2. Universal indicators.
3. Gillespie’s drop method.
Indicators are substances which change in colour with change in the pH of the solution to
which they are added. Indicators are weak organic acids or bases. Their unionized forms show
a colour while their ionized forms, i.e. cations or anions have different or another colour. The
colour of the solution in presence of an indicator depends upon the relative proportions of
ionized and unionized forms of the indicator which in turn depend upon the hydrogen ion
concentration. For each indicator there is a definite pH range in which it is present as a mixture
of its ionized and unionized forms. In this specific range, variations in the pH of the solution
will bring visible change in the colour of the indicator. It is necessary that the effective pH
range of the indicator includes the pH of the unknown sample.
Selection of Indicator
Place 2 ml each of N/10 acetic acid, N/10 sodium carbonate and unknown solution in three
Add 2 drops of the indicator in each tubes. Mix and observe the colour in all the three test
tubes. Test tube containing acetic acid will display acid colour of the indicator whereas test
tube containing sodium carbonate will display alkali colour of the indicator. If the colour
obtained with the unknown solution lies in between (i.e. intermediate) acid and alkali colour
of the indicator and hence it is a suitable indicator for the colorimetric determination of the
unknown solution. But on the other hand if the unknown solution shows either the full acid or
Physical Chemistry 5
full alkali colour then the indicator used is unsuitable. Repeat the same procedures with the
other indicators till a suitable indicator is selected.
Some common indicators useful for biological pH range are:
Indicator pK pH range Acid colour Alkali colour
1. Thymol blue (acid range) 1.65 1.2-2.8 Red Yellow
2. Methyl yellow (Topfer’s reagent) 2.9-4.0 Red Yellow
3. Methyl orange 3.46 3.1-4.4 Red Yellow-orange
4. Methyl red 5.01 4.3-6.1 Red Yellow
5. Phenol red 7.81 6.7-8.3 Yellow Red
6. Thymol blue (alkaline range) 9.7 8.0-9.6 Yellow Blue
7. Phenolphthalein 9.7 8.2-10.0 Colourless Pink
Indicators are used in:
1. Determining the end point in acid-base titration.
2. Determining the pH of the unknown solutions.
1. Indicator paper
Indicator paper consists of a strip of a sensitized paper and is accompanied by a colour chart
which shows different colour which the indicator exhibits at different pH values.
Take a strip of indicator paper and moisten it or dip it in the solution whose pH is to be
determined. Remove the excess of the fluid adhering to the indicator paper strip by means of
pressing between the folds of filter papers. Compare the colour of the pH paper with the
colour chart on the indicator paper and thus determine the pH of the solution.
2. Universal indicator
Universal indicator is a wide range indicator solution having pH between 0 to 14.
Take 5 ml of the unknown solution. Add to it 0.1 ml of the universal indicator. Mix well
and find out the pH by matching the colour of the solution with the colour chart on the universal
indicator bottle. The one with which it coincides or matches, is the pH of the unknown solution.
3. Gillespie’s drop method
In the determination of pH by this method, the ratio of two forms of the indicator may be
found out by adding a known number of drops of an appropriate indicator to the test solution
and finding out how the same number of drops has to be distributed between an acid and
alkali so that the colour of the test solution matches with that
of the acid and alkali solution. When superimposed, then the
pH can be calculated by using Handerson-Hasselbatch
pH = pK + log
Apparatus used is Cole and Onslow’s comparator.
Place 5 ml of N/10 HCl in tube no. 1, 5 ml of N/10 sodium
carbonate in tube no. 2, 5 ml of unknown solution in tube no.
3 and 5 ml of distilled water in tube no. 4 (this is to equalise
the optical conditions).
6 Practical Biochemistry for Students
Add few drops (count the number of drops) of suitable indicators to the unknown solution
in the tube no. 3. Now distribute the same number of drops in acid solution (because any drop
of indicator solution going into the solution, will give its acid form) and alkali solution (because
any drop of indicator solution going into alkali will give its alkali form). Mix the contents of
each tube and examine the colours by white light. If the colour viewed through tube no. 3 and
4 appears to have more of the alkalis form of the indicator as compared to the colour viewed
through tube nos 1 and 2, add more drops of the indicator to the sodium carbonate tube (alkali
tube) and an equal no. of drops to unknown solution.
Now again observe colour. If the colour viewed through the tube nos 1 and 2 matches with
the colour viewed through the tube nos 3 and 4, then count the number of drops added to acid
tube and alkali tube.
The manner in which matching has been done it can be argued that the pH of the solution
which contains ionized and unionised indicator in the ratio they are present in tube nos 1
pH of unknown solution = pK + log _________________
Number of drops of indicator added to alkaline solution
= pK + log ___________________________________________________________________
Number of drops of indicator added to acid solution
To Study the Phenomenon of Osmotic Pressure, Diffusion and Dialysis
Osmotic flow occurs whenever a semipermeable membrane
separates a solution and its pure solvent or between two solutions
differing in concentration. Water passes through the membrane until
the concentration on both sides becomes same. Such a movement of
solvent molecules from a pure solvent or dilute solution, through a
semipermeable membrane is called osmosis.
Osmotic pressure is the pressure that must be applied on a
solution to keep it in equilibrium with the pure solvent when the
two are separated by semipermeable membrane or osmotic pressure
is the force required to oppose the osmotic flow.
Since osmotic pressure is proportional to the total number of
solute particles in solution so the substances which ionize, will have
the higher osmotic pressure as compared to those which do not
If the solution containing crystalloids and colloids is placed in a cellophane sac and this is
immersed in a jar of distilled water, the crystalloids diffuse across the membrane while the
large colloidal molecules are retained. By repeatedly changing the distilled water outside the
sac, it is possible to free the colloidal material virtually completely from salts and other
crystalloids. This process is known as dialysis.
Cover the top of the thistle funnel by tying a piece of cellophane over it. Then pour a
solution made by mixing 10 ml of saturated ammonium sulphate with 90 ml of 1% starch into
the thistle funnel till the level rises to the middle of the stem. Clamp the funnel on a stand and
suspend this into a breaker of distilled water.
Physical Chemistry 7
After half an hour, take out a portion of solution from the beaker and perform the iodine
test and barium chloride (BaCl2) test.
i. 2 ml of solution (from beaker) add few drops of iodine. No change in colour is observed.
ii. 2 ml of solution (from beaker), add 2 ml of BaCl2. A white precipitate of barium sulphate
The starch test will be negative and sulphate test positive.
To Study the Phenomenon of Adsorption and to
Compare Two Eluters—Acetone and Water
Adsorption is a phenomenon in which a substance is adsorped on the surface of a substance. It
takes place due to the presence of free valencies on the surface of the adsorbate which attracts
and takes up the adsorbant.
Take two test tubes. Add 10 ml of 0.1% methylene blue solution, add 0.5 gm of activated
charcoal in each. Shake the tube vigorously and filter the contents into the separate test tubes.
Both dyes are completely adsorbed by the charcoal and are retained on the filter paper.
Now place these two funnels with filter paper to a separate test tubes. Wash the charcoal
on the filter paper in one funnel with acetone and the other funnel with water.
The colour of the filtrate with acetone washings will be blue while with water it will be
The solution looses its blue colour due to the adsorption of methylene blue particles on the
activated charcoal. This proves that the charcoal is a good adsorbant.
On adding water to it, water cannot elute methylene blue and hence colourless filtrate is
obtained. But acetone dissolves the methylene blue particles and thus the solution regains its
original blue colour.
This acetone is a better eluter than water and a strong adsorbant than the charcoal.
To Study and Compare Surface Tension of Two Liquids—Water and Soap Solution
Surface tension is a phenomenon concerned with the force acting at the surface of a liquid
giving an appearance of a stretched membrane. A liquid or a fluid is made up of molecules.
8 Practical Biochemistry for Students
Molecules present in the interior of fluid are equally attracted in all directions by intermolecular
forces due to surrounding molecules. But molecules on the surface are unequally attracted
because of the absence of forces from above. This leads to certain unbalanced forces on the
surface. Due to these forces, the surface acts as a membrane. Surface tension is defined as the
workdone in ergs in streching the membrane by 1 square centimeter.
Take two test tubes. Add 3 ml of water in one and 3 ml of soap solution in the other. Sprinkle
sulphur powder in both the test tubes.
Sulphur powder sinks in the test tube containing a soap solution whereas it floats in the
test tube containing water.
Due to lower surface tension of soap solution, the surface could not keep the sulphur powder
floating. Therefore, the surface tension of water is more than that of soap solution.
To Study the Reactions of Monosaccharides
Solutions provided are 1% glucose and 1% fructose.
This is a general test for carbohydrates. Carbohydrates on treatment with strong concentrated
sulphuric acid undergo dehydration to give furfural or furfural derivate which on conden-
sation with α-naphthol yields a violet or purple coloured complex whose exact structure is
If oligosaccharides or polysaccharides are present, they are first hydrolysed to the constituent
monosaccharides which are then dehydrated.
Pentoses yield furfural and hexoses yield 5-hydroxymethyl furfural.
Molisch reagent α-naphthol in ethanol (ethanolic α-naphthol).
In a clean and dry test tube, take 2 ml of the carbohydrate solution. Add 2 drops of ethanolic
α-naphthol (Molisch reagent). Mix and incline the test tube and cautiously add 2 ml of concen-
trated sulphuric acid by the side of the test tube so that the acid forms a layer under the
carbohydrate solution. Gently rotate the test tube between the palms of the hands to bring
about slight mixing at the interface. An appearance of violet or purple ring at the interface
(junction) of two solutions indicate the presence of carbohydrates.
– 3 H2O
Molisch reagent: α-naphthol in ethanol (ethanolic a-naphthol)
10 Practical Biochemistry for Students
Test tube for this test should be completely dry.
Benedict’s Qualitative Test
This test is positive for reducing sugars only.
Reducing sugars (mono or disaccharides) by virtue of free aldehydic or ketonic group in
their structure reduce cupric ions in alkaline solutions at high temperature. The alkali present
in the Benedict’s reagent enolises the reducing sugar to form enediols (different forms of
reducing sugar) which are highly reactive and act as strong reducing agent.
Benedict’s qualitative reagent contains:
i. Copper sulphate Furnishes cupric ions (Cu++
) in solution.
ii. Sodium carbonate Makes medium alkaline.
iii. Sodium citrate Prevents the precipitation of cupric ions as cupric hydroxide by forming a
loosely bound cupric-sodium citrate complex which on dissociation gives a continuous
supply of cupric ions.
Benedict’s qualitative reagent is prepared by dissolving 173 gm of sodium citrate, 90 gm of
anhydrous Na2CO3 in 500 ml of distilled water. Slightly heat the contents to dissolve. Filter
the solution and make the volume to 850 ml. Dissolve separately 17.3 gm of CuSO4.5H2O in
150 ml of water. Add this solution slowly and with stirring to the above solution—the mixed
solution is ready for use.
– 3 H2O
Violet or purple
CHOH — CHOH
HOH2C – CHOH CHOH — CHO ____________→
___→ Cu++ +SO4
Cu++ + Sodium Citrate ___→ Cupric: Sodium citrate complex
HO — C — COO
Reducing sugar _________→ Enediols forms of reducing sugar.
Enediols + Cupric: Sodium citrate complex _________→ Cu+ + Mixture of sugar acids.
Cu+ + OH– _______→ CuOH
2CuOH _________→ Cu2O ↓ + H2O
Pipette 5 ml of Benedict’s qualitative reagent in a test tube. Add to it 8 drops of given carbo-
hydrate solution. Boil over a flame or in a boiling water bath for 2 minutes. Cool the solution.
An appearance of green, yellow or red precipitate indicates the presence of reducing sugars.
The colour of the solution or precipitate gives an approximate amount of reducing sugars
present in the solution.
Green color — upto 0.5 g% (+)
Green precipitate — 0.5-1.0 g% (++)
Green to yellow ppt — 1.0-1.5 g% (+++)
Yellow to red ppt — 1.5-2.0 g% (++++)
Brick red ppt — more than 2.0 g%
This is another reduction test to detect the presence of reducing sugars.
It differs from Benedict’s qualitative test in that Fehling reagent contains Rochelle’s salt
(Sodium-potassium tartarate) in place of sodium citrate.
Fehling solution consists of:
Fehling solution A It contains copper sulphate solution. It is prepared by dissolving 34.65 gm
of CuSO4-5H2O in 500 ml of distilled water
Fehling solution B It contains potassium hydroxide and Rochelle salt (Sodium potassium
tartarate). It is prepared by dissolving 125 gm of KOH and 173 gm of Rochelle salt in 500 ml of
Mix, equal volume of Fehling A and Fehling B before use.
Benedict’s reagent is superior to Fehling test. It is semiquantitative and more sensitive.
Sodium citrate in Benedict’s reagent and sodium-potassium tartarate (Rochelle’s salt) in Fehling
solution prevent the precipitation of cupric hydroxide or cupric carbonate by forming a deep
12 Practical Biochemistry for Students
blue soluble, slightly dissociated complex with the cupric ions. These complexes dissociate
sufficiently to provide a continuous supply of ready available cupric ions for oxidation.
To 2 ml of Fehling solution (1 ml of Fehling A + 1 ml of Fehling B), add 2 ml of Carbohydrate
solution. Mix and boil. Appearance of yellow or red precipitate of cupric oxide indicates the
presence of reducing sugars.
This test is used to distinguish monosaccharides from disaccharides by controlling the pH
and the time of heating.
Barfoed’s test is a reduction test carried out in an acidic medium. The acidity makes it a
weaker oxidising reagent. Therefore only monosaccharides, will reduce cupric ions. However
if heating is prolonged, disaccharides may be hydrolysed by the acid and the resulting
monosaccharide will give the test positive.
Cupric acetate in lactic acid. Barfoed’s reagent is prepared by dissolving 24 gm of copper
acetate in 400 ml of boiling water. To this add 25 ml of 8.5% lactic acid solution. Stir cool the
solution and dilute to 500 ml.
To 2 ml of Barfoed’s reagent add 2 ml of carbohydrates solution. Place the test tube in boiling
water bath for 3 minutes. An appearance of brick red precipitate of cuprous oxide indicates
the presence of monosaccharides.
Precautions The solution should be boiled for 3 minutes only. Overheating should be avoi-
ded because on prolonged heating disaccharides will also give this test positive.
This test is positive for ketohexoses only and hence is used in the detection of fructose.
Ketohexoses, i.e. fructose on treatment with hydrochloric acid form 5 hydroxymethyl fur-
fural which on condensation with resorcinol gives a cherry red coloured complex.
Seliwanoff’s test distinguishes between fructose and glucose. Overheating of the solution
is avoided because on continuous boiling, aldoses will also give this test positive because of
their conversion to ketoses by hydrochloric acid.
Sucrose will also give Seliwanoff’s test positive because the acidity of reagent is sufficient
enough to hydrolyse sucrose to glucose and fructose but Benedict’s test will be negative.
– 3 H2O
Resorcinol in concentrated hydrochloric acid (diluted 1:1 with water).
To 3 ml of Seliwanoff’s reagent in a test tube add 3 drops of carbohydrate solution. Heat over
a flame for 30 seconds only. Cool the solution. An appearance of cherry red colour indicates
the presence of fructose.
Phenylhydrazine Test (Osazone Formation Test)
Reducing sugar can be distinguished by phenylhydrazine test when characteristic osazone
crystals are formed. These osazones have definite crystal structure, precipitation time and
melting point and hence help in the identification of reducing sugars.
CH = N — NH —C6H5
H — C — OH
HO — C — H
H — C — OH
H — C — OH
CH = N — NH — C6H5
C = O
HO — C — H
| + NH3
H — C — OH
| + C6H5NH2
H — C — OH
H — C — OH
HO — C — H
H — C — OH
H — C — OH
C = O
HO — C — H
H — C — OH
H — C — OH
Osazones of monosaccharides separate out while in hot.
Examine the shape of crystals under low power of microscope.
The shape of the osazones are as follow:
Glucose: Needle shape osazone.
Fructose: Needle shape osazone.
Osazone formation test
C = N — NH — C6H5
HO — C — H
H — C — OH
H — C — OH
C = N — NH C6H5
HO — C — H
| + NH3
H — C — OH
| + C6H5NH2
H — C — OH
– H 2
_________________→ C6H5 NHNH2C6H5 NHNH2
CH = N — NH — C6H5
C = N — NH C6H5
HO — C — H
H — C — OH
H — C — OH
14 Practical Biochemistry for Students
In contains equal part of phenylhydrazine hydrocholoride and anhydrous sodium acetate. It
is prepared only at the time of reaction.
In a clear and dry test tube, take approximately 0.5 gm. of phenylhydrazine mixture (equal
part of phenylhydrazine hydrochloride and anhydrous sodium acetate). Add 5 ml of
carbohydrate solution and 1-2 drops of glacial acetic acid. Mix and place the test tube in
boiling water bath for 30 minutes.
Bial’s Test for Pentoses
This is a sensitive test for the detection of pentoses. Pentoses on heating with strong acid are
converted to furfural which reacts with the coloured compound produced when orcinol and
ferric chloride react with each other.
Bial’s reagent (0.2% orcinol in concentrated hydrochloric acid).
To 5 ml of Bial’s reagent acid add 10 drops of pentose solution (i.e. Aarabinose). Boil.
Appearance of green colour.
To Detect Galactose
Mucic acid test
Galactose on oxidation with strong acid gives mucic acid which crystallises out and can be
In a test tube take 1 ml of galactose solution followed by 1 ml of concentrated nitric acid.
Evaporate the mixture by using boiling water bath for 1½ hours in a furming cup board.
Keep it overnight. Examine a drop of the crystals under low power of microscope.
– 3 H2O
solution+ Fe Cl3
To Study the Chemical Reactions of Disaccharides
The most common disaccharides are maltose, lactose and sucrose. Maltose and lactose are
reducing disaccharides where as sucrose is a non-reducing disaccharide.
One percent solution of each maltose, lactose and sucrose are provided.
1. Molisch test
Dissaccharides are first hydrolysed to constituent monosaccharides which are then dehydrated.
Test In a clean and dry test tube, take 2 ml of the carbohydrate solution. Add 2 drops of
ethanolic α-naphthol (Molisch reagent). Mix. Incline the test tube and cautiously add 2 ml of
concentrated H2SO4 by the side of the test tube. An appearance of violet or purple ring at the
junction of two solutions indicate the presence of dicarbohydrates.
2. Benedict’s qualitative reagent
Test Pipette 5 ml of Benedicts qualitative reagent in a test tube. Add 8 drops of given
dicarbohydrate solution. Boil for 2 minutes. An appearance of green, yellow or red precipitate
indicates the presence of disaccharides.
Maltose and lactose give Benedict’s qualitative test positive whereas with sucrose the test
is negative, i.e. no reduction is observed.
3. Barfoed’s test
Test To 2 ml of Barfoed’s reagent add 2 ml of disaccharide solution. Place the test tube in
boiling water bath for 3 minutes. No change in colour indicates the presence of disaccharides
in the solution.
Negative for disaccharides.
4. Osazone formation (i.e. phenylhydrazine test)
Osazones of disaccharides separate out on cooling.
Test In a clean and dry test tube, take roughly 0.5 g of phenylhydrazine mixture. Add 5 ml
of disaccharide solution and 2 drops of glacial acetic acid. Mix. Place the test tube in boiling
water bath for 30 minutes.
After 30 minutes, take out the test tube from the boiling water bath and allow it to cool by
itself in a test tube rack (Do not disturb the test tube in between as the osazones of disac-
charides separates out on slow cooling).
Appearance of yellow crystals takes place. Observe the shape of crystals under low power
H — C — OH
H — C — OH
HO — C — H
HO — C — H
HO — C
H — C — OH
HO — C — H
HO — C — H
H — C — OH
16 Practical Biochemistry for Students
The shape of osazones are:
Malatose : Sunflower shape
Lactose : Cotton ball shape
Sucrose It will give Benedict’s qualitative test negative.
Lactose (Cotton ball) Maltose (Sun flower)
Sucrose is Confirmed As
Test Take 5 ml of sucrose solution in a test tube. Add to it 1-2 drops of concentrated hydrochloric
acid. Boil the contents for few minutes (2-5 minutes). Cool the solution.
Divide it in two parts.
Neutralise one part of the solution with sodium carbonate and carry out the Benedict’s
qualitative test. The test will be positive.
Carry out the Barfoed’s test with the other part of the solution. It will be positive now.
Carry out the osazone test with the hydrolysate solution of sucrose. Appearance of needle
To Study the Chemical Reactions of Polysaccharides
Solutions provided are 1% starch and 1% dextrins.
1. Molisch test
2. Iodine test.
This test is used for polysaccharides detection and differentiation.
Iodine forms a coordination complex between the helically coiled polysaccharides chain
and the iodine centrally located with in the helix due to adsorption. The iodine colour obtained
with the polysaccharides depends upon the length of the unbranched or linear (α1, 4 linkage)
chain available for complex formation.
Amylose a linear chain component of starch gives a deep blue colour.
Amylopectin, a branched chain component of starch gives a purple colour.
Glycogen gives a reddish brown colour.
Dextrins, formed from the partial hydrolysis of starch gives colours ranging from brown
red to colourless depending on the size of the molecule.
Cellulose, inulin, disaccharides or monosaccharides gives no colour with iodine.
Polysaccharides Colour with iodine
Starch Blue colour
Amylose Blue colour
Amylopectin Purple colour
Glycogen Brown red colour
Dextrins Brown to colourless
Cellulose or inulin No colour
Disaccharides or monosaccharides No colour
In two ml of carbohydrate solutions, add few drops of hydrochloric acid (to make the medium
acidic) followed by 1 ml of iodine solution. Mix and observe the colour.
No change in colour indicates the absence of polysaccharides.
i. Acid hydrolysis
In a 100 ml conical flask, take 20 ml of 1% starch solution. Add 5 ml of 2N HCl (prepared by
diluting one part of concentrated HCl to 4 parts of water).
Divide the solution in five equal parts (i.e. 5 ml each) in five different tubes and place the
tubes in a boiling water bath. Remove the tube from the boiling water bath at an intervals of 1,
5, 8, 12 and 20 minutes.
Now divide the solution in each tubes in two parts:
i. With one part, perform Benedict’s qualitative test, after making the solution alkaline (i.e.
by neutralising the acidity of the solution with sodium carbonate).
ii. Second part, perform iodine test.
Time Colour with Benedict’s Reduction of hydrolysis Product
iodine test (extend)
1 minute Blue Blue No reduction Starch
5 minute Violet Green Reduction starts(+) Amylodextrins
8 minute Reddish violet Red Initiation of reduction (++) Amylo and erythrodextrin
12 minute Red Red Partialreduction(+++) Achrodextrin
20 minute No colour Red Completely reduced(++++) Glucose
The acidified starch takes about 20 minutes for complete hydrolysis.
ii. Enzymatic hydrolysis
Take a clean test tube and collect some saliva in it.
Take two dry test tubes and label them as blank and experimental. Add 5 ml of 2% starch
solution and 1 ml of Citrate buffer (pH 6.0, prepared in 0.25 M NaCl) in each test tube. Mix
Now add 1 ml of distilled water only in the blank and 1 ml of saliva in the test. Keep the
test tubes for 30 minutes at room remperature.
1. In Blank
a. Iodine test : Blue colour
b. Benedict’s test : Negative.
2. In Test
a. Iodine test : Negative
b. Benedict’s test : Red precipitate.
Blank Test sample
Iodine test Blue colour Negative
Benedict’s test Negative Red precipitate
18 Practical Biochemistry for Students
Starch with saliva shows reduction as starch is converted to glucose which is a reducing
sugar. Where as starch without saliva is not broken up in to smaller molecules because there is
Summary of Qualitative Test
Glucose Fructose Lactose Maltose Sucrose Starch
Identification of Unknown Carbohydrate Solution
1. No need to perform Molisch test, as the unknown solution is carbohydrate in nature.
2. Iodine test Positive for polysaccharides. Depending upon colour, the polysaccharide is
identified. If negative, polysaccharides are absent.
3. Benedict’s test Positive for reducing sugars.
Reducing sugars can be monosaccharides or disaccharides. If the Benedict’s test is nega-
tive, it means reducing sugars are absent.
Absence of Benedict’s test, indicates the presence of non-reducing disaccharide, i.e
4. Barfoed’s test Positive for monosaccharides.
Barfoed’s test differentiates between monosaccharides and disaccharides.
5. Seliwanoff’s test Positive for ketohexoses.
Indicates the presence of fructose.
6. Osazone test For the identification of particular carbohydrates.
Identification of unknown carbohydrate
| | |
No change in colour Red colour Blue colour (starch)
| | |
Clear solution Hazy solution
Positive (Reducing sugar) Negative (sucrose)
Glucose, Fructose, Galactose, (Nonreducing sugar)
Mannose, Lactose, Maltose
(Glucose, Fructose, Galactose) (Lactose, Maltose)
Step IV A
(Fructose) (Glucose, Mannose, Galactose)
Step IV B
| | | |
Precipitate before Yellow insoluble during Small ball like Osazone separates on cooling
heating (Mannose) heating needle shape cluster thronedge |
(Glucose, Fructose, Mannose) (Galactose) |
Cotton ball shape Sunflower shape
20 Practical Biochemistry for Students
Determination of Achromic Point of Your Own Saliva
This is a simple enzymatic hydrolysis of starch. This is different from acid hydrolysis. Enzymatic
hydrolysis gives bigger units and does not break the branched point (i.e. amylopectin chains)
or α-1, 6 linkage.
Achromic point is that point at which no colour is obtained with iodine. Chromic period is
that time period which is required to obtain achromic point when enzymatic hydrolysis is
In an animal, there are enzymes which break only the α-1, 4 linkage, i.e. they break only the
straight chains. Enzymes for breaking the α-linkage are present only in the plants.
Salivary amylase is a β-linkage enzyme which acts randomly breaking the starch into
monosaccharides (i.e. glucose) and maltose units. At places where there is α-1, 6 linkage
oligosaccharide units, the breaking proceeds in the following order.
i. Once enzyme activity starts we get first soluble starch.
ii. The first product hereafter formed is amylodextrins.
iii. Next we get erythrodextins which gives reddish colour with iodine.
iv. No colour with iodine is got when we get achrodextrin. This point is called achromic
Take 5 ml of 1% starch. Add 2 ml of buffer (pH 6.7) to it and 1 ml of 1% NaCI. Mix.Take out 5
ml of it. This is prepared buffer-starch solution.
Rinse your mouth with water. Take 10-15 ml of warm water in mouth and rotate the water
with tongue. Take this in a polythene beaker and now taken 5 ml of this.
Take a tile having grooves and put iodine in equal amount in each groove.
0 sec 30 sec 1 min 1.5 min
2 min 2.5 min
Achromic Point 21
Now mix the saliva and prepared buffer-starch solution, and a drop of it at zero hour and
then at intervals of 30 seconds in iodine till the achromic point is reached. Note the chromic
30 sec. Blue
1 minute Light blue
1½ minute Lighter blue
2 minute Faint blue
2½ minutes Colourless
Result and Conclusion
The achromic point of saliva is 2½ minutes, this means that 2½ minutes are taken till achromic
point is got and formation of achrodextrin takes place. Now if we put the hydrolysed solution
in a drop of Fehling’s very little red precipitate is got because enzyme hydrolysis does not
produce many monomers.
22 Practical Biochemistry for Students
To Study the General Reactions of Proteins
1. Biuret Test
Biuret test is given by all compounds that contains two or more peptide bonds. Since proteins
are polypeptide, hence it is a general tests for proteins.
The name of the reaction is derived from the organic compound, a biuret, obtained by
heating urea at high temperature which gives a positive test.
180°C || ....... .....|| ...
NH2CONH2 + NH2CONH2
_________→ H2N — C — NH — C — NH2
Biuret reagent contains dilute copper sulphate in strong alkali.
Biuret Reagent (Stock) It is prepared by dissolving 22.5 gm of Rochelle salt in 200 ml of 0.2N
NaOH to this is added 7.5 gm of CuSO4 5H2O with constant stirring. Then is added 2.5 gm of
KI and make the volume to 500 ml with 0.2 N NaOH.
Working biuret reagent is prepared by dissolving 50 gm of stock biuret reagent to 250 ml
with 0.2N NaOH containing 5 gm of KI per litre.
The purple or violet colour produced is
believed to be due to a coordinate
complex between the cupric ions and four
nitrogen atoms, two from each of two
adjacent peptide chains.
Take 6 ml of 5% NaOH, in a test tube and
add few drops of 1% CuSO4 solution till
blue colour solution is produced. Divide
the solution, i.e. 3 ml each in two test
tubes marked experimental test ‘A’ and
control test ‘C’.
• To ‘A’ add 3 ml of protein solution.
• To ‘C’ add 3 ml of distilled water.
O == C
R — C
O == C
R — C
H — C == O
CH — R
C == O
N — CH — R
Appearance of purple or violet colour in the tube ‘A’ shows the presence of proteins with
respect to tube ‘C’ which serves as a control for the test.
2. Ninhydrin Test
Proteins containing free α-amino acid radical in the molecule reacts with ninhydrin to give a
blue-violet coloured compound. Ninhydrin test is not given by protein and hydroxy proline
because no free α-amino group is present. They give only a yellow colour.
Ninhydrin dissolved in acetone.
Protein solution heated with ninhydrin leads to the formation of a blue coloured compound
called Ruhemann’s complex.
Take 1 ml of protein solution, add to it 2-3 drops of freshly prepared ninhydrin solution. Heat
the solution. Appearance of blue colour indicates the presence of proteins.
To Study the R-Groups of Proteins
The most important aspect of R-group of proteins is that of their nutritional significance. These
acids which the animal body is unable to synthesise either completely or in amounts sufficient
to normal growth maintainence must be supplied in the food. These so called essential amino
acids can be tested for in various way and can in many cases, be detected and determined by
means of simple colour reactions based upon the qualitative test.
A variety of colour reactions specific to particular functional groups in amino acids are
known. They are useful in both the qualitative and quantitative identification of particular
1. Xanthoproteic test (For benzenoid radical)
The reaction is based upon the nitration of the benzene ring with concentrated HNO3 yielding
yellow derivatives of nitrobenzene which turns to orange in alkaline medium.
+ H2N — C — H ______→
C = N — C
24 Practical Biochemistry for Students
Take 3 ml of test solution (protein solution). Add 1 ml of concentrated HNO3. A white precipitate
due to denaturation of protein is formed. Boil the solution. A yellow solution derivative is
formed due to nitration of benzene ring. Cool the solution and make it alkaline with 20%
NaOH, orange colour is produced.
2. Millon’s test (For hydroxy benzene radical)
This test is specific for tyrosine and is an indication of the presence of tyrosine in the protein
because tyrosine is the only amino acid containing hydroxy phenyl group.
A pink colour is obtained in this test is due to mercury complex of nitrophenol derivatives.
Mercuric nitrate dissolved in concentrated HNO3.
Take 1 ml of test solution in a test tube. Acidify with dilute-H2SO4. Check with litmus. Add 1
ml of Millon’s reagent. Boil the solution. A yellow precipitate adheres to the side of the test
tube. Cool the solution under tap water. Add a drop of 1% sodium nitrite (NaNO2) and gently
warm. The precipitate or solution turns red.
3. Hopkins-Cole test (For indole group)
This test is specific for tryptophan and is an indication of the presence of tryptophan in the
Take 1 ml of test solution and add few drops of 1 : 500 commercial formalin (40% formalde-
hyde). Add 2 drops of mercuric sulphate (i.e. 10% HgSO4 in 10% H2SO4). Mix. Incline the test
tube and add 1 ml of concentrated H2SO4 by the side of the test tube. A purple ring is formed
at the junction of the two layers.
4. Sakaguchi test (For guanidino group)
This test is given by all compounds containing guanidino group and thus is an indication of
the presence of arginine present either in free or in combined form.
Guanidines in alkaline solution give a red colour in the presence of α-naphthol and sodium
Take 3 ml of test solution in a test tube. Add 1 ml of 5% NaOH, 2 drops of ethanolic α-napthol
and 2 drops of 10% sodium hypobromite. Mix well. Wait for 5 minutes. Development of bright
red colour takes place.
Run a control by taking 3 ml of distilled water instead of protein solution and add all other
reagents as in test.
5. Test for cystine or cysteine (–S–S–and–SH radicals)
Sulphur is present in proteins as cystine, cysteine or methionine.
Take 2 ml of test solution and add 2 ml 40% NaOH. Boil for 2 minutes. Cool and add lead
acetate solution. A black precipitate of PbS insoluble in dilute HCl is formed.
6. Test for free—SH radical
Take 1 ml of test solution. Add few crystals of ammonium sulphate. To it add few drops of
freshly prepared solution of sodium nitroprusside and 1 ml of liquor ammonia. Development
of rose red colour takes place.
Colour tests Albumin Globu lin Phenylalanine Tyrosine Arginine Guanidine Tryptophan
Precepitation Reactions of Proteins
1. Precipitation by heavy metals.
2. Precipitation by alkaloidal reagents.
1. Precipitation by heavy metals (10% lead acetate, 10% CuSO4
and 10% ZnSO4
Proteins are precipitated from solutions by salts of heavy metals probably by combination of
the metal ions with the anionic form of the protein. On the alkaline side of the isoelectric point
proteins exits as negative ions.
1. Take 3 ml of protein solution. Add 2 drops of 5% NaOH. Mix, followed by 2 ml of 10%
lead acetate solution. Appearance of white precipitate.
2. Take 3 ml of protein solution. Add 2 drops of 5% NaOH. Mix, followed by 2 ml of 10%
CuSO4 solution. A light blue precipitate appears.
3. Take 3 ml of protein solution. Add 2 drops of 5% NaOH. Mix, followed by 2 ml of 10%
ZnSO4 solution. An intense white precipitate appears.
2. Precipitation by alkaloidal reagents
Proteins are precipitated from the solution by combination between the acid anions and the
positively charged protein molecule by forming all insoluble complex. The alkaloidal reagents
precipitate proteins by combination of the acidic radical of the former with the cationic form
of the protein, which predominates when the solution is on the acidic radical of the former
with the cationic form of the protein, which predominates when the solution is on the acidic
side of the isoelectric point.
26 Practical Biochemistry for Students
a. To 3 ml of protein solution, add few drops of metaphosphoric acid. A dirty white
b. To 3 ml of protein solution, add few drops of 20% sulphosalicyclic acid. White
c. To 3 ml of protein solution, add 3 ml of Esbach reagent. A precipitate appears.
d. To 3 ml of protein solution, add few drops of glacial acetic acid followed by 1 ml of 5%
potassium ferrocyanide. A deep yellow precipitate appears.
3. Precipitation of proteins by heating
Proteins which are precipitated when their solutions are heated are termed as heat coaguable
proteins. This property of proteins is made use of in the detection of albumin in urine simply
by heating the urine.
Solubilities of several proteins in various solvents are shown below:
Distilled water Dilute acids Dilute alkali Ammonium sulphate
Albumin Soluble Soluble Soluble Soluble Insoluble
Globulins Insoluble Soluble Soluble Insoluble Insoluble
Gelatin Soluble Soluble Soluble Insoluble Insoluble
Difference in Albumin and Globulin
1. It is obtained by full saturation of 1. It is obtained by half saturation of
(NH4)2 SO4 ammonium sulphate.
2. It is soluble in water. 2. It is insoluble in water but soluble in dilute
mineral acids and salt solution.
3. It is a smaller molecule having more charge. 3. It is bigger molecule with less charge.
4. In isoelectric point field it travels faster 4. Due to big molecule and less charge it
than globulins. travel slowly in isoelectric field.
Isoelectric Point 27
Determine the isoelectric point of casein, the pK for acetic acid is 4.74.
Solution of casein in 0.1 N sodium acetate
1 N acetic acid
0.1 N acetic acid
0.01 N acetic acid
Isoelectric pH is defined as the pH at which the protein molecule does not migrate to the
cathode or to the anode in an electric field. At this pH, the protein molecule exists as zwitterion.
At the isoelectric point, the precipitation of protein is maximum, i.e. in other words the solubility
of proteins is minimum. At the isoelectric pH, the electrostatic repulsive force which normally
prevents the protein molecules from coming together is minimum as a result of no net change
on the protein molecule and hence give rise to maximum precipitation.
Arrange a series of nine test tubes, clean and dry. Label the test tubes as 1, 2....9.
Add the following solution in given order.
1 2 3 4 5 6 7 8 9
Protein solution 1 ml 1 ml 1 ml 1 ml 1 ml 1 ml 1 ml 1 ml 1 ml
Water (ml) 8.38 7.75 8.75 8.50 8.0 7.0 5.0 1.0 7.40
0.01 N CH8COOH (ml) 0.62 1.25 – – – – – – –
0.01 N CH3COOH (ml) – – 0.25 0.50 1.00 2.00 4.00 8.00 –
1.0 N CH3COOH (ml) – – – – – – – – 1.60
Mix, the contents after every addition.
Observe the tube after 5 minutes and see which tubes has the maximum turbidity. Again
observe after 10 minutes, 30 minutes and note again which tube has the maximum turbidity.
The test tube number 5 shows the maximum turbidity, the corresponding pH is the
isoelectric point of casein.
According to Handerson-Haselbatch equation.
pH = pK + log _____________
pKa for acetic acid is 4.74
28 Practical Biochemistry for Students
[Salt] = Sodium acetate = 0.1 N
[Acid] = Acetic acid = 0.1 N Maximum turbidity
pH = pK + log __________
pH = 4.74
The isoelectric point (pI) for casein is 4.74.
The pH of the respecting tubes will be as
Tube Precipitate formed
No. Zero time Ten minutes Thirty minutes
1. – – – 5.95
2. – – – 5.64
3. + + + 5.34
4. ++ ++ – 5.04
5. ++ +++ ++++ 4.74
6. ++ ++ + 4.44
7. + + + 4.12
8. + + + 3.14
9. – – – 3.54
To Perform General Tests for Lipids
Lipids are defined as a group of fatty nature which are insoluble in water but soluble in nonpolar
solvents like ether, chloroform, etc. Lipids thus include fats, oils, waxes and related compounds.
Lipids are classified as simple and complex.
Take three perfectly dry test tubes. To the first add 2 ml of distilled water, to the second 2 ml
of ethyl alcohol and to the third 2 ml of chloroform. To each of the three test tubes add 3 drops
of the provided oil. Shake gently and observe.
The provided oil is insoluble in water, hence it floats on the surface of water, forming a separate
layer. The oil is fairly soluble in ethyl being heavier than alcohol, some of the oil (undissolves,
settle down at the bottom as minute droplets, whereas it is extremely soluble in chloroform.
The resulting solution is clear.
Oils are insoluble in water sparingly soluble in alcohols but extremely soluble in fat solvents
Take three perfectly clean test tubes.
• To the first add 5 ml of distilled water.
• To the second add 5 ml of bile salt solution.
• To the third add 5 ml of household detergent (Surf solution).
• To each of the three test tubes, add 3 ml of the provided oil.
• Shake vigorously and observe.
In the first tube a temporary emulsion of oil in water is formed. On vigorous shaking, this
emulsion is unstable and so breaks down early. In case of bile salt solution and household
30 Practical Biochemistry for Students
detergent solution, a highly stable emulsion is formed. The emulsion is very fine and breaks
down after long time.
Oils form a coarse and unstable emulsion with water. This readily breaks down. So this con-
cludes that oils do not reduce the surface tension of water. In case of bile salt solution and
household detergent, the emulsion is fine and very stable because bile salts and household
detergents reduce the surface tension of water as a result of which oil fragments into small
droplets which form an emulsion.
3. Acrolein test
Take a clear test tube and add 4 drops of provided oil to it. Then add a pinch of potassium
bisulphite and heat vigorously. Smell the fumes of the gas which come from one of the test
Pungent smelling fumes arise from one of the test tube.
Acrolein is evolved which has a pungent smell.
All the triglycerides give this test.
Take a clean dry test tube and add 0.5 ml of the provided oil then add 2.5 ml of ethanol to it
and mix well. After mixing, add 10 ml of 10% alcoholic sodium hydroxide, shake well and
keep in boiling water bath for 15 minutes. Take the test tube after 15 minutes and add water so
that the resulting volume of the solution is 15-20 ml. Shake well to dissolve. Divide the contents
into 4 equal parts in four different test tubes.
a. To the first part, add 3 ml of conc. HCl and shake well.
b. To the second part, add an equal volume of saturated NaCl solution.
c. To the third part, add 3 drops of CaCl2.
d. To the fourth part, add 3 drops of MgCl2.
A white precipitate of liberated fatty acid is obtained.
Sodium salts of fatty acid rise up and form a pale white layer.
A white precipitate of calcium salt of fatty acid is formed.
A white precipitate of magnesium salt of fatty acid is formed.
The liberated fatty acid being insoluble in water is precipitated.
The sodium salt of fatty acid is salted out.
Calcium salt of fatty acid being insoluble is precipitated.
5. Test for unsaturation
Take a clean dry test tube and add 3 drops of oil in it. Then add 2 ml of ethyl alcohol and mix
well. Then add 0.5% alcoholic bromine solution until bromine solution imparts its own colour.
The colour of the solution was colourless at first but gradually turned pale yellow, i.e. the
colour of the bromine solution itself.
Bromine goes into the solution forming a dibromide, i.e. it add to the double bonds. In other
words, bromine solution is decolourised, but when all the double bonds are saturated the
bromine solution imparts its own colour.
6. Test for cholesterol
a. Libermann-Burchard reaction
Take a perfectly dry test tube and add 2 ml of CHCl3 solution of cholesterol to it. Then add 10
drops of acetic anhydride and mix well. Then add drops of concentrated H2SO4 from the sides
of the test tube. Keep it in dark after mixing well.
A deep green coloured solution is obtained.
This indicates the presence of cholesterol.
b. Salkowski reaction
Take a perfectly clean and dry test tube and add to it 2 ml cholesterol. Solution prepared in
cholesterol. Then add an equal volume of concentrated H2SO4 dropwise along the side of the
Two layers are formed. The upper brown one is formed by CHCl3 and the lower one yellow in
colour formed by concentrated H2SO4. This layer of conc. H2SO4 gives fluorescence.
This indicates the presence of cholesterol.
32 Practical Biochemistry for Students
Determination of Saponification Number of an Oil
Saponification number is defined as the number of milligrams of KOH required to saponify
completely 1 gm of fat.
Since fats are mixture of triglycerides, most of which are of mixed type, so saponification
number is a measure of average moleculer weight of the fatty acids comprising the fats (i.e.
the measure of the average chain length of the fatty acid).
Saponification number is an important constant particularly in distinguishing or identifying
Take a clean and dry 100 ml conical flask. Using a 2 ml pipette, transfer 1.5 ml of the oil sample
provided in the conical flask. Add 15 ml of 0.5 N ethanolic KOH into the flask containing the
oil. Mix the contents well. Place a funnel at the neck of the conical flask (the stem of funnel acts
as a condenser) and place it in the boiling water bath for half an hour till all the oil globules
disappear and a yellow cake is formed by potassium salts of fatty acids. After half an hour,
take out the conical flask, cool it to room temperature. Add 20 ml of distilled water in the flask,
and shake till a clear solution is formed. Now add 1-2 drops of phenolphthalein as an indicator.
Titrate with 0.5 NHCl till the colour is changed from red to colourless. Note the titre value.
Also run a blank titration, without using oil under similar conditions and note the titre
• Volume of HCl required for saponified solution = T ml.
• Volume of HCl required for blank titration = B ml.
• Volume of HCl utilised = (B–T) = Blank test reading value.
According to Normality equation
• 1 ml of 0.5 N HCl = 1 ml of 0.5 N KOH
• (B–T) ml of 0.5 N HCl = (B–T) ml of 0.5 N KOH
• 1 ml of 0.5 N KOH = 28 mg of KOH
• (B–T) ml of 0.5 N KOH = 28 (B–T) mg of KOH
• Weight of oil = Volume × density = 1.5 × 0.9 = 1.35 gm.
• Saponification number of 1.35 gm of oil = 28 × (B–T) mg or KOH
28 × (B – T) mg of KOH
• Saponification number of 1 gm of oil = ____________________________________
Iodine Number 33
To Determine the Iodine Number of the Given Oil
Iodine number is defined as the number of grams of iodine absorbed by 100 gm of the fat.
Halogens, e.g. iodine or bromine are taken up by the fats because of the presence of double
bonds present in the fatty acid part of the fat.
Iodine number is a measure of the degree of unsaturation of a fat. The higher the iodine
number, the more is the unsaturation present in the fat. Iodine number is a useful characteristic
for assessment of both purity and nutritive value of the fat.
Bromine is often used instead of iodine because it is more reactive.
The value is influenced by the percentage of each unsaturated fatty acid, the degree of
unsaturation of each acid and the mean molecular weight of the fat.
The iodine members of some important fats are mentioned below:
Fats Iodine numbers
Butter fat 26-28
Human fat 65-70
Peanut oil 80-90
Soyabean oil 137-143
Linseed oil 170-200
The given amount of fat is treated with a measured excess of Hanus solution.
R — CH = CH — COOH + I — Br _______→ R — CH — CH — COOH + I — Br
Excess of Left
Hanus solution over
To the left over Hanus solution is added potassium iodide solution. The iodine thus
liberated is titrated against standard solution of “hypo” (Na2S2O3) using starch as an indicator.
The colour change is from deep blue-black to white which marks the end point of the titration.
I Br+KI ________→ I2 ↑ + KBr
2Na2S2O3 + I2
At the end point
I2+Starch ______→ Blue colour
34 Practical Biochemistry for Students
In a 250 ml conical flask, add 5 ml of given oil sample (the oil sample is dissolved in CCl4. The
concentration of the oil sample is 5 g%), followed by 10 ml of Hanus solution.
Mix well, cover the mouth of the flask with a paper and keep it for 30 minutes for reaction
to take place.
After 30 minutes, add 5 ml of KI solution into it. Mix well, followed by 25 ml of distilled
water. Add 4-5 drops of starch as indicator. The colour of the solution turn blue-black.
Titrate the contents of the flask with N/10 Na2S2O3 till the colour changes from blue-black
to white, which marks the end point of the titration. Note down the titre value which is x ml.
In 250 ml conical flask add 5 ml of CCl4 only instead of oil sample and repeat the same procedure
as in the test. Note down the titre value which is y ml.
The difference between the two (i.e. blank-test) gives the amount of Na2S2O3 utilised in
titrating the IBr which was used in saturating the unsaturated fatty acid moiety, i.e. the volume
of I Br required to saturate the oil = (Blank-Test) value.
In the test titration the excess of I-Br, i.e. the left over I-Br is titrated against Na2S2O3.
In blank titration the excess of I-Br (as in the first case) and the actual volume of I-Br which
would have used up by an oil to be saturated are together titrated against Na2S2O3.
Titre value obtained for test titration = x ml
Titre value obtained for blank titration = y ml
According to Normality equation
1 ml of N/10 Na2S2O3 solution ≡ 1 ml of N/10 I-Br solution
1 ml of n/10 Na2S2O3 solution ≡ 1 ml of N/10 I-Br solution
≡ 1 ml of N/10 iodine solution
Equivalent weight of iodine = 127.
1 ml of N/10 iodine = ________ × ____ = 0.0127 gm.
(Two molecules of Na2S2O3 are equivalent to one molecule of iodine; thus one molecule of
Na2S2O3 is equivalent to one atom of iodine).
1 ml of N/10 Na2S2O3 solution = 0.127 gm of iodine.
Amount of iodine absorbed by given amount of oil or fat = (y-x) × 0.0127 gm of iodine
The concentration of oil sample is 5 gm%, i.e. 5 ml of oil = 0.25 gm of oil.
0.25 gm of oil or fat consumes (y – x) × 0.0127 gm of iodine.
y – x
∴ 100 gm of oil or fat consumes __________
× 0.0127 gm of iodine.
y – x × 0.0127 × 100
The iodine number is ____________________________ gm of iodine.
Formal Titration 35
Sorensen’ formal titration method is used for the estimation of free carboxyl group in amino
acids and in mixture of amino acids. By this method one can determine the increase in carboxyl
groups which accompanies the enzymatic hydrolysis of proteins.
Amino acids by virtue of zwitterion formation are merely neutral in reaction. If formaldehyde
is added to a solution of amino acids an adduct is formed at the amino group leaving the
carboxyl group free.
In other words, formaldehyde suppresses the basicity of amino group. So that carboxyl
group can exert its maximum acidity and hence can be titrated against standard solution of
alkali using phenolphthalein as an indicator.
Preparation of neutral formation.
Pipette 2 ml of formalin into a test tube and add 2 drops of indicator phenolphthalein to it.
Mix well and then titrate the above solution against N/10 NaOH. Note the amount of NaOH
used up in the titration.
In a 100 ml conical flask, take 10 ml of glycine solution, add to it 2 ml of above neutralised
formalin solution. Then titrate the resulting solution against N/10 NaOH. The end point
in each case is observed when the colour just turns pink.
Volume of NaOH utilised in neutralising glycine solution along with neutralised formalin
= x ml.
10 ml of glycine = x ml of N/10 NaOH
= x ml of N/10 glycine
NH3 | CH2OH
R — CH — COO– + 2HCHO ____________→ R — CH — COO– – + H+
36 Practical Biochemistry for Students
1000 ml of 1 N glycine solution contains 75 gm of glycine, i.e 1000 ml of N/10 glycine
solution contains 7.5 gm of glycine
7.5 × χ
χ ml of N/10 glycine contains _________ gm of glycine
Gastric Analysis 37
Stimuli which cause secretion of gastric juice are of two types viz.
Chemical stimulus—by giving dry toast, oatmeal, gruel and alcohol and humoral stimulus—
by giving 0.5 gm of histamine. The patient is given a light super the previous evening and in
the morning. Ryle’s tube is inserted through the nose to suck the gastric juice. The patient is
given 50 ml of 7% alcohol after every 15 minutes, 10 ml of gastric juice is sucked up. Collect 4
such samples; 15 minutes, 30 minutes, 45 minutes and 1 hour. Then inject 0.5 gm of histamine
intramuscularly, after every 15 minutes, suck 10 ml of gastric juice. Collect four such samples
1 hour 15 minutes, 1 hour 30 minutes, 1 hour 45 minutes and 2 hours.
Histamine has definite advantages over the ordinary test meal. It evokes a maximum response
and often produces acid secretion where the ordinary test meal fails completely. It adds nothing
to the stomach so that a pure juice, undiluted and uncontaminated is obtained and no
neutralisation of the acid can take place through food constituents.
Combined Alcohol and Histamine Tests
Alcohol as a physiological stimulant has several advantages over gruel, charcoal biscuit or
dry toast. It is much less objectionable to take, is easily swallowed and evokes a feeling of
pleasure instead of distaste on the part of the subject. The gastric juice obtained is ideal for
The main content of the gastric juice is hydrochloric acid. In a normal person the pH of the
gastric juice is 1 to 2. The HCl of gastric juice is found in two forms, i.e. the free (uncombined)
HCl and combined acidity, the protein bound HCl (protein hydrochloride).
Total acidity = Free acidity + combined acidity (i.e. protein bound acidity).
A known volume of gastric juice is taken. 2 drops of Topfer’s reagent (P-dimethylamino-
azobenzene in alcohol) as indicator is added. The colour of the solution turns yellow due to
the presence of free HCl. Yellow colour indicates a pH of 2.90. The end point appears when the
colour turns orange. The pH is 3.60 at the end point. The free HCl being liberated. Proceed to
titrate the protein bound HCl. 2 drops of phenolphthalein are added to the solution (which
has been titrated for free HCl) and it is titrated against N/10 NaOH. The end point appears
when the colour becomes pinkish. The pH at this end point is 8.50.
38 Practical Biochemistry for Students
The first titre value with Topfer’s reagent expressed in units of N/10 HCl per 100 ml of
gastric juice gives free acidity. The second titre value with phenolphthalein expressed in units
of N/10 HCl per 100 ml of gastric juice gives combined acidity.
In a conical flask, pipette 1 ml of gastric juice. Then add 1-2 drops of Topfer’s indicator to it. If
an orange colour appears (pH 3.6) then free acidity is zero. In that case add a drop of
phenolphthalein and titrate with N/100 NaOH till the end point is reached (pH 8.50). The titre
value expressed as units of N/10 HCl per 100 ml of gastric juice gives the combined acidity
(=Total acidity because free acidity is zero).
If by adding Topfer’s reagent a yellow colour is obtained (pH is 2.9), titrate with N/100
NaOH till a orange colour is obtained. The titre value expressed in units of N/100 HCl per 100
ml of gastric juice gives free acidity. Now in the same solution, add a drop of two of
phenolphthalein and titrate with N/100 NaOH till a pink colour (pH 8.5) is obtained. This is
the end point. This titre value expressed in units of N/100 HCl per 100 ml of gastric juice gives
1 ml of N/100 NaOH ≡ 1 ml of N/100 HCl
1 ml of N/10 HCl ≡ 10 ml of N/100 HCl
10 ml of N/100 HCl ≡ 10 ml of N/100 NaOH
≡ 1 ml of N/10 HCl
If the titre value is x, then
10 × ml of N/100 HCl = x ml of N/10 HCl
Total acidity in 1 ml of gastric juice = x/10 ml
Total acidity in 100 ml of gastric juice = 10 x
So in order to express the titre value in terms of units of N/10 HCl per 100 ml of gastric juice, it has to
be multiplied by a factor of 10.
The acidity of gastric juice is expressed in terms of units.
One unit is 1 ml of N/10 HCl present per 100 ml of gastric juice.
Achlorhydria : Free HCl absent
Hyperchlorhydria : Free HCl above 60 units
Hypochlorhydria : Free HCl never above 10 units
Absence of free HCl in gastric juice is called achlorohydria. However enzymes such as pepsin
may still be present in gastric juice. The condition is found in pernicious anemia.
It means the absence of free HCl as well as enzymes like pepsin, renin, etc.
e.g. pernicious anemia, very advanced stages of gastric carcinoma and gastritis.
Free HCl is present but in very low concentration, i.e. less than 10 units.
Most of the conditions where there is an increased evidence of achlorhydria may show
Many patients with hypochlorhydria may become achlorhydric subsequently.
Gastric Analysis 39
Hyperchlorhydria (Gastric hyperacidity)
It means that the concentration of free HCl in the gastric juice is more than the upper normal
limit, i.e. values about 60 units, e.g. Duodenal ulcer.
Normal Value Gastric ulcer Duodenal ulcer
of stomach anemia
Free HCl 10-70 Normal 10-45 Normal Zero
Total acidity 5-100 10-100 15-110 3-80 0-40
Volume 10-100 40-100 10-115 10-500 5-50
1. Presence of starch
Iodine test: Take 3 ml of gastric juice in a test tube. Add few drops of iodine solution. The
appearance of blue colour indicates the presence of starch.
2. Presence of Iactic acid
Take 2 ml of gastric juice in a test tube. Add few drops of Maclean’s reagent (contain HgCl2
and Fe Cl3). Development of yellow colour indicates the presence of lactic acid.
Run a control with 2 ml of distilled water instead of gastric juice.
3. Presence of blood
Take 3 ml of saturated solution of benzidine in glacial acetic acid. Add 2 ml of gastric juice
followed by 1 ml of hydrogen peroxide. Wait for few minutes. The appearance of blue or
green colour indicates the presence of blood in gastric juice.
4. Presence of bile
In a test tube take 3 ml of gastric juice. Sprinkle sulphur powder over it. The sinking of sulphur
powder indicates the presence of bile salts.
5. Presence of HCl
A drop of two of Gunzberg reagent is placed in a small porcelein dish. It is carfully evaporated
on a small flame. A glass rod is dipped in the gastric juice and is touched thoroughly with the
residue in the porcelein dish. It is gently heated. The appearance of purple red colour indicates
the presence of free HCl.
6. Presence of mucus
Examination of gastric juice in broad day light will reveal the stingy appearance of juice for
presence of mucus.
40 Practical Biochemistry for Students
The average volume of urine excreted daily is about 1.5 litres. It contains nitrogeneous organic
compounds such as urea, uric acid, creatinine, hippuric acid, indican, purines and amino acids
as well as organic compounds which do not contain nitrogen. The average composition of
urine excreted in 24 hours is as follows:
Urea 20-30 gm
Uric acid 0.7 gm
Amino acid 0.5-1 gm
Ammonia 0.7 gm
Creatinine 1.4 gm
Ascorbic acid 15-20 gm
Guanidine 3-16 gm
Hippuric acid 0.6 gm
Indican 4-20 mg
Iodine 50-250 mg
Lactic acid 50-200 mg
Chloride as NaCl 10-15 mg
Inorganic sulphate as S 60-120 mg
Neutral sulphate as S 80-160 mg
Sodium 3-5 gm
Calcium 0.1-0.3 gm
Phosphate as inorganic phosphates 1.0-1.5 gm
Urine specimens must be collected in the correct way and in a clean container or bottle. In
the hospital, when only one specimen is needed, the best time to collect is first urine in the
morning when it is most concentrated. The specimen should be analysed without delay. If this
is not possible the urine should be stored in a refrigerator.
The ordinary urine is examined under the following heads:
1. Physical examination.
2. Chemical examination.
3. Microscopic examination.
The physical examination of urine includes appearance, colour, pH, specific gravity, etc.
These tests are carried out before the microscopic or chemical examination. Whereas in chemical
examination, abnormal constituents which are normally not present in easily detectable
quantities in urine of healthy beings but are known to occur in urine under certain diseased
conditions. Abnormal or pathological constituents of urine are looked for in urine such as
reducing sugars, proteins, ketone bodies, bile salts, bile pigments, blood, urobilinogen, etc.
Urine Analysis 41
The normal urine is yellow in colour. The intensity of normal urine is dependent on the
concentration of urine. The yellow or amber colour of a normal urine is due to presence of a
yellow pigment urochrome.
The colour of urine changes in many disease conditions because of the presence of pigments
that do not normally occur.
Colour Possible cause
Orange Concentrated urine
Almost colourless Dilute urine
Yellow to yellow brown or greenish colour Bile pigments
Reddish brown colour Hemoglobin
Milky Presence of fats
Black upon standing Alkaptonuria
Brown-black colour on standing Melanin or homogentisic acid
Orange brown Urobilinogen
Cloudy Presence of insoluble calcium and magnesium phosphates
The urine may also assume many different colours following ingestion of various dyes,
foods and drugs.
The normal volume of urine voided by an adult per day ranges from 750 to 2000 ml. The
average volume is 1500 ml.
The amount of urine excreted is directly related to fluid intake, the temperature, climate
and the amount of sweating that occurs.
Polyuria It is the increased excretion of urine.
Polyuria may indicate the loss of concentrating ability by the kidney.
Polyuria occurs in physiological conditions.
1. Excessive fluid intake
2. Ingestions of diurates
1. Diabetes mellitus
2. Diabetes insipidous
3. Chronic renal damage
Oliguria It is decreased in which the urinary output is 500 ml per 24 hours.
Oliguria occurs in
1. Less fluid intake
2. Excessive fluid loss due to vomiting, diarrhoea and sweating
5. Acute nephritis
6. Cardiac failure
Anuria It is a total loss of urine. The ratio of day urine (i.e. 8 AM to 8 PM) to night urine (i.e. 8 PM
to 8 AM) should be at least 2 : 1 and sometimes 3 : 1 or more in healthier individual. In renal
disease, this ratio is reduced and may even be reversed.
42 Practical Biochemistry for Students
3. Specific Gravity
Normal specific gravity of urine varies from 1.005 to 1.030.
Specific gravity is highest in the first morning specimen and is generally greater than 1.020.
A specific gravity of 1.025 or above in a random normal urine sample indicates normal ability.
The specific gravity of urine varies according to kidney function. Concentrated urine has a
high specific gravity while diluted urine has a low specific gravity.
Specific gravity of urine indicates the relative proportions of dissolved components to the
total volume of the urine. It also reflects the relative degree of concentration or dilution of the
Low specific gravity is observed in the following:
1. High fluid intake
2. Diabetes insipidous
High specific gravity urine is observed in
1. Excessive loss of water due to sweating fever, vomiting and diarrhoea.
2. Diabetes mellitus
4. Hepatic diseases
5. Congestive heart failure.
Whereas fixed specific gravity of urine is an indication of several renal damages
with disturbance of both concentrating and diluting abilities of kidney.
Determination of Specific Gravity
The specific gravity of urine is determined by urinometer. Urinometer is a glass
made instrument having a cylinderical stem containing a scale.
The urinometer is floated in a cylinder containing urine. Care should be taken
that it should not touch the sides or bottom of the cylinder. The depth to which it
sinks in the urine, indicates the specific gravity of urine.
The urinometer is calibrated with respect to distilled water at 1.000 at specific
temperature, indicated on the instrument itself. If the temperature of urine is above
or below that temperature, a correction of + 0.001 for each 3°C should be made.
The pH of the normal urine is on an acidic side, i.e. 5.5-6.5. The acidity of the urine is mainly
due to acid phosphates.
In general protein rich diets give rise to acidic urine attributable to the sulphur of the amino
acids which is oxidised to sulphuric acid. Similarly phospholipids and nucleic acids all yield
phosphoric acid. Alkaline urines are excreted where there is a predominance of vegetables
and fruits in the diet. However, the production of ammonia by the kidney also plays an
important part in influencing the pH of the urine since it forms salts with acids and can excreted
i. On a high protein diet.
Urine Analysis 43
ii. Uncontrolled diabetes
i. After meals
ii. Diet high in vegetables, citrus fruits, milk, etc.
iii. Renal tubular acidosis.
5. Total Solids
Total solids can be calculated approximately by multiplying the second and third decimal
figures of specific gravity by 2.6 (Longe’s coefficient). The product represents the number of
gm of solids in 1 litre of urine, e.g. if the specific gravity of urine is 1.025, then 25 × 2.6 = 65 gm
totals solids per litre. From this the output during 24 hours can be calculated.
Chemical Examination of Urine (Qualitative).
1. Reducing Sugars (Usually Glucose)
Normally very small amounts of reducing sugars are excreted in urine, which are not detected
by the reagents used. However, under abnormal conditions following reducing sugars are
excreted in urine.
Reducing sugars Conditions
Glucose Renal glucosuria
Blood glucose will differentiate
Lactose Pregnancy and lactation
After ingestion of large amounts of vegetables or fruitsPentoses
Causes of Glycosuria
1. Diabetes mellitus
2. Non-diabetic glycosuria
A. Glycosuria with hyperglycemia
a. Renal glycosuria—due to low renal threshold
b. Alimentary glycosuria—after ingestion of lot of carbohydrates
c. Glycosuria of pregnancy—in 10-15% of normal pregnancies
B. Non-diabetic glycosuria with hyperglycemia
b. Emotional disturbances
c. Ether anesthesia
d. Increased intracranial pressure—tumours, fractures, encephalitis.
Benedict's Qualitative Test
To 5 ml of Benedict’s qualitative reagent in a test tube, add 8 drops of urine. Boil for 2 minutes
and cool. A change in blue colour to green-yellow-red precipitate indicates the presence of
reducing sugars in urine.
44 Practical Biochemistry for Students
The colour of the precipitate gives an appropriate, i.e. rough amount of sugar in the urine.
Blue colour remains No reducing sugar
Green colour Less than 0.5 gm%
Green precipitate 0.5 to 1 gm%
Green to yellow precipitate 1.0 to 1.5 gm%
Yellow to red precipitate 1.5 to 2.0 gm%
Brick red precipitate More than 2gm%
Mix 1 ml each of Fehling A solution and Fehling B solution. Then add 2 ml of urine. Boil for 2
min. Appearance of either green-yellow-red precipitate indicates the presence of reducing
Tests for Specific Reducing Sugar
The following tests will differentiate the various reducing substances which respond to Fehling
and Benedict’s test. Usually the object in testing for reducing substances in urine is to determine
the presence or absence of glucose.
1. Fermentation test: Glucose and fructose are fermented by the enzyme present in the yeast
and the libration of carbon-dioxide confirms the test.
2. Mucic acid test for lactose and galactose: Evaporate urine in a china dish containing a
mixture of 100 ml of urine and 20 ml of concentrated HNO3. Let the volume is reduced to
20 ml. Allow it to cool. Lactose or galactose yields mucic acid through the oxidising action
of HNO3 and this forms a white precipitate.
3. Seliwanoff’s test for fructose.
4. Pentoses by orcinol test.
5. Identification of carbohydrates by osazones formation.
There is no single test capable of detecting the presence of proteins under all conditons. The
proteins which appears in urine however are usually albumin and globulins of which albumin
almost greatly predominate.
Causes of albuminuria
1. Functional—orthostatic, severe muscular exertion, prolonged exposure to cold, pregnancy.
a. Pre-renal causes—no primary kidney disease
Passive congestion of kidney
Fever and toxaemia
Drugs and chemical poisoning
b. Renal—primary disease of kidney
Urine Analysis 45
c. Post renal causes
Normally proteins are not excreted in urine.
Test for Proteins
1. Heat Coagulation Test
Fill a test tube with two-third (2/3) of urine. Gently heat the upper half of the urine to boiling
without disturbing the lower portion which serves as a control. A white turbidity indicates
the presence of proteins or phosphates. Add few drops of glacial acetic acid. If precipitate
dissolves, it is due to phosphates.
2. Sulphosalicyclic Acid Test
Take 5 ml urine in a test tube, add to it 1 ml of 25% sulphosalicyclic acid. Appearance of white
precipitate indicates the presence of proteins in urine.
3. Nitric Acid Test (Heller’s Test)
In a clean dry test tube take 3 ml of concentrated nitric acid. Put 2 ml of urine over the acid to
form a separate layer. Appearance of white ring at the junction indicates the presence of proteins.
Distinction between albumin and globulin when urine is mixed with an equal volume of
saturated ammonium sulphate, the globulin is precipitated and albumin which remains can
then be precipitated by boiling the filtrate or by addition of ammonium sulphate.
Bence Jones Proteins
Bence Jones proteins are found in urine in cases of multiple myeloma, i.e. cases of malignant
disease involving the bone marrow and of leukemia. Its occurrence is very uncommon in
Test for Detection of Bence Jones Proteins
Heat coagulation test
The urine must be faintly acidic. Acetic acid being added if necessary. It is then
gently heated in a test tube. Bence Jones protein gives a precipitate which appears
about 40°C, it is maximal about 60°C, and then disappears as the temperature rises.
The disappearance may be incomplete since albumin is often present as well. In
that case the boiling urine is filtered rapidly to remove albumin. Bence Jones proteins
reappears as the filtrate cools. The precipitate flocculates and sticks to the side of
Quantitative analysis of proteins
Approximation to the quality of proteins present in the urine can be obtained by the
use of Esbach proteinometer. The graduated tube is filled to the mark U by urine to
be tested. Esbach reagent is added to the mark R. The tube is stoppered and after
mixing or inverting once or twice without shaking is allowed to stand undisturbed overnight.
46 Practical Biochemistry for Students
The height of the precipitate in the tube is then read and the reading divided by ten gives the
amount in gm of proteins in 100 ml of urine.
Esbach reagent : 20 gm citric acid, 10 gm picric acid dissolved in 1 litre of water.
3. Test for Ketone Bodies
Acetone, acetoacetic acid and β-hydroxybutyric acids are collectively known as ketone bodies.
Ketone bodies are obtained as the intermediate products in the oxidation of fatty acids and are
oxidised to carbon dioxide and water under normal conditions. In abnormal conditions, ketone
bodies accumulate in the blood and pass out in urine. This condition is called ketosis. When
the carbohydrate metabolism is defective, increased fat is oxidised for energy purposes giving
rise to increased formation of ketone bodies. Such happens in diabetes mellitus, when glucose
present is not metabolised hence ketone bodies appear in urine. Also during starvation when
glucose supply to the body is restric-ted, ketone bodies appear in urine.
Causes of ketonuria
i. Diabetes mellitus
Saturate 5 ml of urine with Rothera’s mixture (ammonium sulphate and sodium nitroprusside).
Add 2 ml of liquor ammonia by the side of the test tube. Appearance of permanganate ring at
the junction indicates the presence of ketone bodies in the urine.
4. Test for Bile Pigments
Bile pigments are billirubin and biliverdin. Most of the tests for bile pigments depend on the
oxidation of bilirubin to coloured compounds of blue, green, violet, red and yellow. Normal
urine does not contain bilirubin as bilirubin is bound to albumin. Bilirubin is present in the
urine of obstructive jaundice and in some cases of hepatic jaundice due to high level of
Take 5 ml of urine in a test tube. Add to it an equal volume of 10% BaCl2 solution. Filter, dry
the filter paper and add a drop of Fouchet’s reagent. Green colour indicates the presence of
Fouchet’s reagent contains trichloroacetic acid and ferric chloride.
Place 5 ml of fuming nitric acid in a test tube. Add 5 ml of urine. Appearance of blue, green
and violet rings are seen at the junction if bilirubin is present.
Filter some urine through a filter paper. Unfold the filter paper. Dry it. Add a drop of
concentrated HNO3 in the centre of filter paper. Green, blue, violet, and red colours show the
presence of bile pigments.
Urine Analysis 47
5. Bile Salts
Bile salts are sodium and potassium salts of glycocholates and taurocholates. Bile salts are
formed in the liver from where they are excreted in the bile. They are absorbed by the intestine
and passed back to the liver through portal circulation. Bile salts are present in urine in
Sprinkle some sulphur powder in test tube containing urine. Sulphur remains on the surface
in normal urine but sinks down in the presence of bile salts.
Blood appears in urine in hematuria and hemoglobinuria. Hematuria consists of hemoglobin
pigment and unruptured corpusules. It is due to passing of blood through the kidney into
urine because of lession of the kidney. Hematuria occurs in polynephritis and hemoglobinuria
occurs in enteric fever, malaria, wrong blood transfusion and hemolytic poisoning.
The peroxidase activity of hemoglobin decomposes hydrogen peroxide and the liberated oxygen
oxidises the benzidine to give a blue solution.
In a test tube take 3 ml of saturated solution of benzidine in glacial acetic acid. Add 3 ml of
urine followed by 1 ml of hydrogen peroxide. Appearance of green or blue colour within few
minutes indicates the presence of blood.
7. For Urobilinogen
Urobilinogen is normally present in the urine in very small amount. The excretion of
urobilinogen is increased in:
i. Hemolytic jaundice: In hemolytic jaundice, liver is not able to excrete the increased amount
of urobilinogen absorbed from the intestine completely.
ii. Infective hepatitis: The liver cell is less able to excrete so more of bilirubin passes into general
circulation and is excreted in urine.
However it is absent in obstructive jaundice.
In a test tube take 5 ml of urine. Add to an equal volume of Ehrlich reagent. Allow it to wait for
Appearance of faint pink or brown colour indicates the normal amount of urobilinogen,
whereas deep red colour suggests increased amount of urobilinogen. The test is done in
undiluted urine sample and also in serial dilution of the urine. Normally urobilinogen is present
in urine upto a dilution of 1 : 10.
48 Practical Biochemistry for Students
Test for Carbohydrates
Take about 1 gm of given food sample in a test tube. Add 10 ml of distilled water. Slightly
warm the solution. Perform the following tests.
To 2 ml of the above suspension (solution), add 2 drops of ethanolic α-napthol. Mix and incline
the tube and add 2 ml of concentrated sulphuric acid by the side of the test tube. Appearance
of green-violet or pink ring at the interface indicates the presence of carbohydrates.
To 2 ml of suspension, add a drop of dilute HCl, followed by few drops of iodine solution.
Appearance of intense blue colour indicates the presence of starch in the food mixture.
If the polysaccharides/starch is absent, filter the suspension and perform the following
tests in the filtered solution.
To 5 ml of Benedict’s qualitative test, add 8 drops of the above filtered solution. Boil for 3
minutes. Appearance of green-yellow orange-red precipitate indicates the presence of reducing
sugar in the food.
To 2 ml of Barfoed’s reagent, add 2 ml of above filter solution. Boil for 2 minutes. Appearance
of red precipitate indicates the presence of monosaccharides in the food.
To 3 ml Seliwanoff’s reagent, add 3 drops of above filtered solution. Boil for 30 seconds only.
Cool. Appearance of cherry red colour indicates the presence of fructose in the food sample.
In a clean dry test tube, take 0.5 gm of phenylhydrazine mixture (prepared by taking equal
part of phenylhydrazine hydrochloride and anhydrous sodium acetate), 5 ml of filtered solution
and add 2 drops of glacial acetic acid. Mix and place the tube in the boiling water bath for 30
Food Analysis 49
The osazones of monosaccharides separate out while in hot, whereas the osazones of
disaccharides separate out in cold, i.e. by allowing the solution to cool by itself without disturb-
ing the solution in between at room temperature.
Examine the shape of the osazones under low power microscope. Appearance of needle
shaped osazones suggests/indicates the presence of monosaccharides either glucose or fructose
while appearance of sunflower or cotton ball shaped osazones indicates the appearance of
maltose or lactose in the food sample.
Test for Proteins
Take about 1 gm of food sample in a test tube, add 10 ml of distilled water. Warm the solution
(do not boil). Perform the following tests for proteins and R-groups with “suspension” only.
Take 6 ml of 5% NaOH. Add few drops of 1% CuSO4. Divide the solution in two parts and
label one as experimental and other as control.
Experimental tube: 3 ml of above divided solution. Add 3 ml of suspension. Mix. Appearance
of purple or violet colours indicates the presence of proteins in the given food sample.
Control tubes: 3 ml of above divided solution. Add 3 ml of distilled water.
This tube serves as a control for the above experimental test.
To 2 ml of the suspension, add few drops of ninhydrin solution (0.2 g% in acetone). Boil the
solution. Appearance of deep blue colour indicates the presence of proteins in the solution.
Test for R-groups
Perform the following R-group tests for proteins with suspension.
i. X anthoproteic test.
ii. Millon’s test
iii. Hopkin’s-Cole test
iv. Sakaguchi test
v. Test for — S — S — and — SH — group
Test for Fats (i.e. For Cholesterol)
In a clean dry test tube, take about 1 gm of food sample. Add 10 ml of chloroform. Mix. Since
fats are soluble in chloroform, it will go into the solution. Filter the solution. Care should be
taken that funnel and test tube receiving the filtrate should not be wet with water. Perform the
following test for cholesterol.
a. Libermann Burchard test: In a clean and dry test tube take 2 ml of filtered chloroform extract.
Add 10 drop of acetic anhydride and 2 drops of concentrated sulphuric acid. Mix. Keep it
in dark for 10 minutes. Appearance of green colour indicates the presence of cholesterol
in the food sample.
b. Salkowski test: In a clean dry test tube, take 2 ml of the filtered chloroform extract. Add 2
ml of concentrated sulphuric acid by the side of the test tube. Slightly mix and allow to
50 Practical Biochemistry for Students
stand. Appearance of brown red colour in the upper layer (CHCl3 layer) and yellow colour
in the lower layer (H2SO4 layer) indicates the presence of cholesterol in the food sample.
Test for Minerals
Take about 1 gm of food sample in the test tube, add 10 ml of distilled water. Mix thoroughly.
Minerals being water soluble will go into the solution. Filter the solution. Perform the following
tests for minerals.
Test for Iron
i. To 2 ml of the filtrate, add a drop of concentrated nitric acid. Boil the solution. Cool it.
Add 1 ml of 5% ammonium thiocyanate solution. Appearance of red colouration indicates
the presence of iron in the food sample.
ii. To 2 ml of the filtrate, add 2 ml of sodium nitroprusside. Appearance of blue colouration
indicates the presence of iron in the food sample. This is known as Pearl Reaction.
Test for Phosphorus
To 2 ml of the filtrate, add 2 ml of 5% ammonium molybdate. Boil the solution. Appearance of
yellow coloured precipitate indicates the presence of phosphorus in the food sample.
Test for Calcium
To 2 ml of the filtrate, add few drops of glacial acetic acid, followed by 2 ml of saturated
ammonium oxalate solution. Appearance of white precipitate indicates the presence of calcium
in the food sample.
Test for Chloride
To 2 ml of the filtrate, add 2 ml of AgNO3 solution. Appearance of white precipitate insoluble
in dilute hydrochloric indicates the presence of chloride in the food sample.