biochemistry Record book

Certificate
This is certified that Mr/Miss ……………………………… being a
Register No………… of second year Diploma in Pharmacy from SHRI
SANGAMESHWAR COLLEGE OF PHARMACY SINDAGI has
completed the term work satisfactorily in Biochemistry and Clinical
Pathology (ER20-23P) for academic year 20.…..To 20…… as
prescribed in curriculum.
Place:
Date:
Subject Teacher Principal

Biochemistry and Clinical Pathology Expt. Date: .
Shri Sangameshwar College of Pharmacy Sindagi-586128 Page 2
Sl. No Name of the Experiments Page No. Date of Expt Sign
1. Introduction of Carbohydrates
2. To perform chemical identification of Glucose
3. To perform chemical identification of Fructose
4. To perform chemical identification of Lactose
5. To perform chemical identification of Maltose
6. To perform chemical identification of Sucrose
7. To perform chemical identification of Starch
8. Introduction of Proteins
9. To perform chemical identification of Arginine
10. To perform chemical identification of Tyrosine
11. To perform chemical identification of Albumin
12. To perform chemical identification of Globulin
13. To perform chemical identification of Gelatin
14. Introduction of Lipids
15. To perform chemical identification of Triglycerids
16. To perform chemical identification of Cholestrol
17. To identify the physical properties of given urine
sample
18. To identify the normal/ abnormal inorganic constitute
of given urine sample
19. To identify the normal/ abnormal organic constitute of
given urine sample
20. To estimate glucose in given urine sample

21. To estimate creatinine in given Urine sample by a
Folin-Wu tube method
22. To estimation of chlorides in urine by volhard method
23. To estimate glucose in given serum sample
24. To estimate cholesterol in given serum sample
25. To estimate calcium in given serum sample
26. To estimate urea in given blood sample
27. To estimate creatinine in given blood sample
28. To estimate serum glutamate oxalo trans aminase
(SGOT)
29. To estimate serum glutamate pyruvate trans aminase
(SGPT)
30. To study the effect of hydrolysis of starch from acid
31. To study the effect of hydrolysis of starch from
salivary amylase enzyme

EXPERIMENT NO.I
INTRODUCTION OF CARBOHYDRATES
Theory: Carbohydrate is naturally occurring organic compounds containing carbon hydrogen
oxygen elements chemically they are poly hydroxyl aldehyde are ketones compounds that
produce these compounds on hydrolysis. Monosaccharides also called simple sugar are the most
basic unit of carbohydrates. They are fundamental units of carbohydrates and cannot be further
hydrolyzed to simpler compounds. They are simplest form of sugar and usually colorless water
soluble and crystalline solid. Some monosaccharide sweet taste. Monosaccharides are the
building blocks of disaccharides and polysaccharides.
Carbohydrates are the most abundant bimolecular on earth. Oxidation of carbohydrates is the
central energy-yielding pathway in most non-photosynthetic cells.
Definition: Carbohydrates are polyhydroxy aldehyde or ketones, or substances that yield such
compounds on hydrolysis. Carbohydrates have the empirical formula (CH2O)n. There are three
major classes of carbohydrates:
1. Monosaccharides
Monosaccharides, or simple sugars, consist of a single polyhydroxy aldehyde or ketones
unit. The most abundant monosaccharide in nature is the six-carbon sugar Glucose,
sometimes referred to as dextrose.
2. Oligosaccharides
Oligosaccharides consist of short chains of monosaccharide units, or residues, joined by
characteristic linkages called glycoside bonds. The most abundant are the disaccharides,
with two monosaccharide units. Example: sucrose (cane sugar).
3. Polysaccharides
The polysaccharides are sugar polymers containing more than 20 or so monosaccharide
units, and some have hundreds or thousands of units. Example: starch. Polysaccharides
are of two types based on their function and composition. Based on function,
polysaccharides of two types storage and structural.
A. Storage polysaccharide - starch.
B. Structural polysaccharide – cellulose.

General properties of carbohydrates
 Carbohydrates act as energy reserves, also stores fuels, and metabolic intermediates.
 Ribose and deoxyribose sugars forms the structural frame of the genetic material, RNA
and DNA.
 Polysaccharides like cellulose are the structural elements in the cell walls of bacteria and
plants.
 Carbohydrates are linked to proteins and lipids that play important roles in cell
interactions.
 Carbohydrates are organic compounds; they are aldehydes or ketones with many
hydroxyl groups.
Physical Properties of Carbohydrates
 Steroisomerism - Compound shaving same structural formula but they differ in spatial
configuration. Example: Glucose has two isomers with respect to penultimate carbon
atom. They are D-glucose and L-glucose.
 Optical Activity - It is the rotation of plane polarized light forming (+) glucose and (-)
glucose.
 Diastereoisomeers - It the configurational changes with regard to C2, C3, or C4 in
glucose. Example: Mannose, galactose.
 Annomerism - It is the spatial configuration with respect to the first carbon atom in
aldoses and second carbon atom in ketoses.
Biological Importance
 Carbohydrates are chief energy source, in many animals; they are instant source of
energy. Glucose is broken down by glycolysis/ kreb's cycle to yield ATP.
 Glucose is the source of storage of energy. It is stored as glycogen in animals and starch
in plants.
 Stored carbohydrates act as energy source instead of proteins.
 Carbohydrates are intermediates in biosynthesis of fats and proteins.
 Carbohydrates aid in regulation of nerve tissue and are the energy source for brain.

 Carbohydrates get associated with lipids and proteins to form surface antigens, receptor
molecules, vitamins and antibiotics.
 They form structural and protective components, like in cell wall of plants and
microorganisms.
 In animals they are important constituent of connective tissues.
 They participate in biological transport, cell-cell communication and activation of growth
factors.
 Carbohydrates that is rich in fibre content help to prevent constipation.
 Also they help in modulation of immune system.
1. Monosaccharides
 The word “Monosaccharides” derived from the Greek word “Mono” means Single and
“saccharide” means sugar
 Monosaccharides are polyhydroxy aldehydes or ketones which cannot be further
hydrolysed to simple sugar.

 Monosaccharides are simple sugars. They are sweet in taste. They are soluble in water.
They are crystalline in nature.
 They contain 3 to 10 carbon atoms, 2 or more hydroxyl (OH) groups and one aldehyde
(CHO) or one ketone (CO) group. Classification of Monosaccharides
Monosaccharides are classified in two ways. (a) First of all, based on the number of carbon
atoms present in them and (b) secondly based on the presence of carbonyl group. The naturally
occurring monosaccharides contain three to seven carbon atoms per molecule. Monosaccharides
of specific sizes may be indicated by names composed of a stem denoting the number of carbon
atoms and the suffix -ose. For example, the terms triose, tetrose, pentose, andhexose signify
monosaccharides with, respectively, three, four, five, and six carbon atoms. Monosaccharides are
also classified as aldoses or ketoses. Those monosaccharides that contain an aldehyde functional
group are called aldoses; those containing a ketone functional group on the second carbon atom
are ketoses. Combining these classification systems gives general names that indicate both the
type of carbonyl group and the number of carbon atoms in a molecule. Thus, monosaccharides
are described as aldotetroses, aldopentoses, ketopentoses, ketoheptoses, and so forth. Glucose
and fructose are specific examples of an aldohexose and a ketohexose, respectively.

2. Disaccharides
Disaccharides consist of two sugars joined by an O-glycosidic bond. The most abundant
disaccharides are sucrose, lactose and maltose. Other disaccharides include isomaltose,
cellobiose and trehalose.
Sucrose
Sucrose being dextrorotatory in nature gives dextrorotatory glucose as well as laevorotatory
fructose on hydrolysis. The overall mixture is laevorotatory and this is because the laevorotation
of fructose (-92.4) is more than the dextrorotation of glucose (+52.5).
Maltose
Maltose is also one of the disaccharides which have two α -D-glucose units which are connected
by the first carbon of the glucose and also linked to the fourth carbon of another glucose unit. In
the solution, a free aldehyde can be produced at the first carbon of the second glucose of the
solution and it is a reducing sugar as it shows reducing properties.

Lactose
Commonly it is called milk sugar as this disaccharide is found in milk. It is made up of Beta-D-
galactose and β-D-glucose. The bond is between the first carbon of galactose and the fourth
carbon of glucose. This is also a reducing sugar.
3. Polysaccharides
Polysaccharides contain hundreds or thousands of carbohydrate units. • Polysaccharides are
not reducing sugars, since the anomeric carbons are connected through glycosidic linkages. •
Nomenclature: Homopolysaccharide- a polysaccharide is made up of one type of
monosaccharide unit Heteropolysaccharide- a polysaccharide is made up of more than one type
of monosaccharide unit
Starch
 Starch is a polymer consisting of D-glucose units.
 Starches (and other glucose polymers) are usually insoluble in water because of the
high molecular weight, but they can form thick colloidal suspensions with water.
 Starch is a storage compound in plants, and made of glucose units
 It is a homopolysaccharide made up of two components: amylose and amylopectin.
 Most starch is 10-30% amylose and 70-90% amylopectin.
 Amylose – a straight chain structure formed by 1,4 glycosidic bonds between α-D-
glucose molecules.
 The amylose chain forms a helix.
 This causes the blue colour change on reaction with iodine.

 Amylose is poorly soluble in water, but forms micellar suspensions.
 Amylopectin-a glucose polymer with mainly α -(1→4) linkages, but it also has
branches formed by α -(1→6) linkages. Branches are generally longer than shown
above Amylopectin causes a red-violet colour change on reaction with iodine.
 This change is usually masked by the much darker reaction of amylose to iodine.
Glycogen
 Storage polysaccharide in animals • Glycogen constitutes up to 10% of liver mass and
1-2% of muscle mass.
 Glycogen is stored energy for the organism
 Similar in structure to amylopectin, only difference from starch: number of branches
 Alpha(1,6) branches every 8-12 residues
 Like amylopectin, glycogen gives a red-violet color with iodine.

Cellulose
 The β-glucose molecules are joined by condensation, i.e. the removal of water,
forming β-(1,4) glycosidic linkages.
 The glucose units are linked into straight chains each 100-1000 units long.
 Weak hydrogen bonds form between parallel chains binding them into cellulose
microfibrils.
 Cellulose microfibrils arrange themselves into thicker bundles called microfibrils.
(These are usually referred to as fibres.)
 The cellulose fibres are often “glued” together by other compounds such as
hemicelluloses and calcium pectate to form complex structures such as plant cell
walls.
 Because of the β-linkages, cellulose has a different overall shape from amylose,
forming extended straight chains which hydrogen bond to each other, resulting in a
very rigid structure.

 Cellulose is an important structural polysaccharide, and is the single most abundant
organic compound on earth. It is the material in plant cell walls that provides
strength and rigidity; wood is 50% cellulose.
 Most animals lack the enzymes needed to digest cellulose, although it does provide
needed roughage (dietary fiber) to stimulate contraction of the intestines and thus
help pass food along through the digestive system
 Some animals, such as cows, sheep, and horses, can process cellulose through the
use of colonies of bacteria in the digestive system which are capable of breaking
cellulose down to glucose; ruminants use a series of stomachs to allow cellulose a
longer time to digest. Some other animals such as rabbits reprocess digested food to
allow more time for the breakdown of cellulose to occur.
 Cellulose is also important industrially, from its presence in wood, paper, cotton,
cellophane, rayon, linen, nitrocellulose (guncotton), photographic films (cellulose
acetate), etc
IN LABORATORY CARBOHYDRATES CAN BE TESTED BY.

Principle:
1. Molisch’s test: This is a common test for all a carbohydrates larger than tetroses. The test is
on the basis that pentoses and hexoses are dehydrated by concentrated sulphuric acid to form
furfural or hydroxyl methylfurfural, respectively. These products condense with Alpha naphthol
to form purple condensation product.
Procedure: Collect 2ml of sample and 4-5 drops of molish reagent mix well then add
along sides of test tube only sulpuric acid 1ml observe violet coloured complex.
2. Fehling’s test: This forms the reduction test of carbohydrates. Fehling’s solution contain blue
alkaline cupric hydroxide solution, heated with reducing sugars gets reduced to yellow or red
Corpus oxide and is precipitated. Hence formation of the yellow or bruise red colour precipitate
helps in the detection of reducing sugar in the test solution.
Procedure: In a clean tube mix equal volume of Fehlings A & B reagent and boil to this
hot solution add 5-8 drops of sample and heat observed red PPT.
3. Benedict’s test: As in Fehling’s test, free aldehyde or keto groups in the reducing sugar
reduce cupric hydroxide in alkaline medium to red colour cuprous oxide. Depending on the
concentration of sugars, yellow to green color is a developed. All monosaccharides are reducing
sugar as they all have a free reactive carbonyl group. Some dicesaccharides like maltose have
exposed carbonyl groups and are also reducing sugars but less reactive than monosaccharides.
Procedure: Collect 5ml of at Benedict solution in a tube boil to this hot solution add 4-5
drops of sample and again heat observed red PPT quickly.
4. Barfoed’s test: Barfoed's tests are used to detect the presence of Monosachharide sugar in
solution. Barfoed's reagent a mixture of ethanoic acid and copper 2 acetate is combined with the
test solution and boiled. A red copper 2 oxide precipitate is formed will indicates the presence of
reducing sugar. The reaction will be negative in the presence of disaccharide sugar because they
are weaker reducing agents. The test specific for Monosachharide. Due to the weakly acidic
nature of Barfoed's reagent, it is reduced only by Monosachharides.

Procedure: Collect 4ml of barfoed solution add 0.5ml of sample boil for a min then cool
and add phosphomolybdic acid observe light blue colour with disaccharide and dark blue colour
monosaccharide.
5. Iodine Test: Any polysaccharides answer iodine test hence iodine test differentiate
polysaccharide from the rest of carbohydrates. Iodine is trapped in between helical coils of
polysaccharides which give a colour starch give a blue coloue, glycogen gives a red colour &
destrin gives a purple colour in case of polysaccharides if not it will be monosaccharide.
Procedure: Collect 2ml of sample in a test tube add iodine drop by drop observed blue
colour and red purple colour in case of polysaccharides if not it will be monosaccharide.
Test for non-reducing sugar: sucrose is non reducing sugar it doesn’t answer benedicts and
Fehling’s test it can be tested by hydrolysis then performing this test.
6. Seliwanoffs Test: Seliwanoff test distinguish between aldoses and ketose sugar ketose are
distinguish from aldoses their ketones, aldehyde functionality. If the sugar contains a ketone
groups it is a ketose and if it contains an aldehyde groups it is an aldose this test is based on the
fact that when heated ketose are more rapidly dehydrated then aldoses.
Procedure: To dilute aqueous solution add crystals of resorcinol add and equal volume
of conc. HCL and a head on water bath formation of cherry red colour or reddish brown colour
indicates presence of ketoses like fructose and sucrose.
7. Osazone Test: For this the reaction mixture should be between pH 5-6 fructose takes 2min to
form the osazone where as for glucose it is 5min. the disaccharides takes longer time to forms
osazone. Disaccharides form crystals only on cooling.
Procedure: To 0.5gm of Phenyl hydrazine hydrochloride add 0.1gm of sodium acetate and 10
drops of glyceal acetic acid to this mixture add 5ml of sample and on boiling water bath for
about 30min allow the tube for cool slowly and examine the crystals under microscope. glucose,
fructose and maltose produced needle shape yellow osazone crystals where as lactose Is balls of
prickle shaped different osazone shows crystals of different shapes. Maltose produced sun flower
shaped crystals the ketose and aldoses reacts with phenyl hydrazine to produce a phenyl
hydrazine in tern reacts with another 2 molecule of phenyl hydrazine to form osazone.

EXPERIMENT NO. II
Aim: To perform chemical identification of GLUCOSE
Sl.
No.
Test Observation Result
01. Molish Test
2ml of sample + 4 drops of molish
reagent + 1ml conc. H2SO4along the
sides of test tube
02. Fehlings Test
2ml sample + 2ml of Fehling’s A &
B solution mix and boil for 2min
03. Benedict Test
3ml of sample + 5ml of Benedict
regent boil and cool
04. Barfoeds Test
4ml of barfoed solution + 0.5ml
sanple boil & cool then
phosphomobdic acid to get clear
solution
05. Tommers Test
2ml tommers reagent + 3ml of
sample boil and cool for 2min
06. Picric Acid Test
3ml of sample + 1ml of picruc acid
solution + 1ml of NaOH
Result: The given sample identified as.........................

EXPERIMENT NO. III
Aim: To perform chemical identification of FRUCTOSE
Sl.
No.
01. Molish Test
sides of test tube
02. Fehlings Test
03. Benedict Test
04. Barfoeds Test
solution
05. Tommers Test
2ml tommers reagent + 3ml of
sample boil and cool for 2min
06. Picric Acid Test
3ml of sample + 1ml of picruc acid
solution + 1ml of NaOH

EXPERIMENT NO. IV
Aim: To perform chemical identification of LACTOSE
Sl.
No.
01. Molish Test
sides of test tube
02. Fehlings Test
03. Benedict Test
04. Barfoeds Test
solution
05. Osazone Test
0.2gm of sample + 0.4gm of phenyl
hydrazine HCl + 0.6gm sodium
acetate + 4ml H2O heat on water bath
20min cool and allow to
crystallization

EXPERIMENT NO. V
Aim: To perform chemical identification of MALTOSE
Sl.
No.
01. Molish Test
sides of test tube
02. Fehlings Test
03. Benedict Test
04. Barfoeds Test
solution
05. Osazone Test
crystallization

EXPERIMENT NO. VI
Aim: To perform chemical identification of SUCROSE
Sl.
No.
01. Molish Test
sides of test tube
02. Fehlings Test
03. Benedict Test
04. Barfoeds Test
solution
05. Osazone Test
crystallization

EXPERIMENT NO. VII
Aim: To perform chemical identification of STARCH
Sl.
No.
01. Molish Test
sides of test tube
02. Fehlings Test
03. Benedict Test
04. Barfoeds Test
solution
05. Iodine Test
2ml of sample + 3ml NaOH solution
+ add iodine drop by drop

EXPERIMENT NO. VII
INTRODUCTION OF PROTEINS
The term protein is derived from a Greek word proteios, meaning holding the first place.
Berzelius (Swedish chemist) suggested the name proteins to the group of organic compounds
that are utmost important to life. Mulder (Dutch chemist) in 1838 used the term proteins for the
high molecular weight nitrogen-rich and most abundant substances present in animals and plants.
Proteins are the most abundant organic molecules of the living system. They occur in
every part of the cell and constitute about 50% of the cellular dry weight. Proteins form the
fundamental basis of structure and function of life.
Functions of proteins
Proteins perform a great variety of specialized and essential functions in the living cells. These
functions may be broadly grouped as static (structural) and dynamic.
1. Structural functions: Certain proteins perform brick and mortar roles and are primarily
responsible for structure and strength of body.
2. Dynamic functions : The dynamic functions of proteins are acting as enzymes,
hormones, blood clotting factors, immunoglobulin’s, membrane receptors, storage
proteins, besides their function in genetic control, muscle contraction, respiration etc.
Elemental composition of proteins
Proteins are predominantly constituted by five major elements in the following proportion.
Carbon: 50 – 55%
Hydrogen: 6 – 7.3%
Oxygen: 19 – 24%
Nitrogen: 13 – 19%
Sulfur: 0 – 4%
Structure of proteins
Proteins are the polymers of L-D-amino acids.
The structure of proteins is rather complex which can be divided into 4 levels of organization
1. Primary structure: The linear sequence of amino acids forming the backbone of proteins
(Polypeptides).
2. Secondary structure: The spatial arrangement of protein by twisting of the Polypeptide
chain.

3. Tertiary structure: The three dimensional structure of a functional protein.
4. Quaternary structure: Some of the proteins are composed of two or more polypeptide
chains referred to as subunits. The spatial arrangement of these subunits is known as
quaternary structure.
[The structural hierarchy of proteins is comparable with the structure of a building. The
amino acids may be considered as the bricks, the wall as the primary structure, the twists in a
wall as the secondary structure, a full-fledged self-contained room as the tertiary structure. A
building with similar and dissimilar rooms will be the quaternary structure].
PROPERTIES OF PROTEINS
1. Solubility: Proteins form colloidal solutions instead of true solutions in water. This is due
to huge size of protein molecules.
2. Molecular weight: The proteins vary in their molecular weights, which, in turn, is
dependent on the number of amino acid residues. Each amino acid on an average
contributes to a molecular weight of about 110. Majority of proteins/polypeptides may be
composed of 40 to 4,000 amino acids with a molecular weight ranging from 4,000 to
440,000.
3. Shape: There is a wide variation in the protein shape. It may be globular (insulin), oval
(albumin) fibrous or elongated (fibrinogen).

4. Isoelectric pH: Isoelectric pH (pI) as a property of amino acids has been described. The
nature of the amino acids (particularly their ionizable groups) determines the pI of a
protein.
5. Acidic and basic proteins: Proteins in which the ratio (H Lys + H Arg)/(H Glu + H Asp)
is greater than 1 are referred to as basic proteins. For acidic proteins, the ratio is less than
1.
6. Precipitation of proteins: Proteins exist in colloidal solution due to hydration of polar
groups (COO–, NH3 +, OH). Proteins can be precipitated by dehydration or neutralization
of polar groups.
DENATURATION
The phenomenon of disorganization of native protein structure is known as denaturation.
Denaturation results in the loss of secondary, tertiary and quaternary structure of proteins. This
involves a change in physical, chemical and biological properties of protein molecules.
Agents of denaturation
 Physical agents: Heat, violent shaking, X-rays, UV radiation.
 Chemical agents: Acids, alkalies, organic solvents (ether, alcohol), salts of heavy metals
(Pb, Hg), urea, salicylate, detergents (e.g. sodium dodecyl sulfate).
Characteristics of denaturation
1) The native helical structure of protein is lost.
2) The primary structure of a protein with peptide linkages remains intact i.e., peptide bonds
are not hydrolysed.
3) The protein loses its biological activity.
4) Denatured protein becomes insoluble in the solvent in which it was originally soluble.
5) 5 The viscosity of denatured protein (solution) increases while its surface tension
decreases.
6) Denaturation is associated with increase in ionizable and sulfhydryl groups of protein.
This is due to loss of hydrogen and disulfide bonds.
7) Denatured protein is more easily digested. This is due to increased exposure of peptide
bonds to enzymes. Cooking causes protein denaturation and, therefore, cooked food
(protein) is more easily digested. Further, denaturation of dietary protein by gastric HCl
enchances protein digestion by pepsin.

8) Denaturation is usually irreversible. For instance, omelet can be prepared from an egg
(protein-albumin) but the reversal is not possible.
9) Careful denaturation is sometimes reversible (known as renaturation). Hemoglobin
undergoes denaturation in the presence of salicylate. By removal of salicylate,
hemoglobin is renatured.
10) Denatured protein cannot be crystallized.
Coagulation: The term ‘coagulum’ refers to a semi-solid viscous precipitate of
protein. Irreversible denaturation results in coagulation. Coagulation is optimum and
requires lowest temperature at isoelectric pH. Albumins and globulins (to a lesser extent)
are coagulable proteins. Heat coagulation test is commonly used to detect the presence of
albumin in urine.
Functional classification of proteins
Based on the functions they perform, proteins are classified into the following groups (with
examples)
1. Structural proteins: Keratin of hair and nails, collagen of bone.
2. Enzymes or catalytic proteins: Hexokinase, pepsin.
3. Transport proteins: Hemoglobin, serum albumin.
4. Hormonal proteins: Insulin, growth hormone.
5. Contractile proteins: Actin, myosin.
6. Storage proteins: Ovalbumin, glutelin.
7. Genetic proteins: Nucleoproteins.
8. Defense proteins: Snake venoms, Immunoglobulins.
9. Receptor proteins: hormones, viruses.
Protein classification based on chemical nature and solubility
This is a more comprehensive and popular classification of proteins. It is based on the amino
acid composition, structure, shape and solubility properties. Proteins are broadly classified into 3
major groups
1. Simple proteins: They are composed of only amino acid residues.
2. Conjugated proteins: Besides the amino acids, these proteins contain a non-protein
moiety known as prosthetic group or conjugating group.

3. Derived proteins: These are the denatured or degraded products of simple and conjugated
proteins.
1. Simple proteins
(a) Globular proteins: These are spherical or oval in shape, soluble in water or other solvents
and digestible.
(i) Albumins: Soluble in water and dilute salt solutions and coagulated by heat. e.g.
serum albumin, ovalbumin (egg), lactalbumin (milk).
(ii) Globulins: Soluble in neutral and dilute salt solutions e.g. serum globulins, vitelline
(egg yolk).
(iii) Histones: Strongly basic proteins, soluble in water and dilute acids but insoluble in
dilute ammonium hydroxide e.g. thymus histones.
(iv) Globins: These are generally considered along with histones. However, globins are
not basic proteins and are not precipitated by NH4OH.
(v) Lectins are carbohydrate-binding proteins, and are involved in the interaction
between cells and proteins. They help to maintain tissue and organ structures. In the
laboratory, lectins are useful for the purification of carbohydrates by affinity
chromatography e.g. concanavalin A, agglutinin.
(b) Fibrous proteins: These are fiber like in shape, insoluble in water and resistant to digestion.
Albuminoids or scleroproteins are predominant group of fibrous proteins.
(i) Collagens are connective tissue proteins lacking tryptophan. Collagens, on boiling
with water or dilute acids, yield gelatin which is soluble and digestible.

(ii) Elastins: These proteins are found in elastic tissues such as tendons and arteries.
2. Conjugated proteins
(a) Nucleoproteins: Nucleic acid (DNA or RNA) is the prosthetic group e.g. nucleohistones,
nucleoprotamines.
(b) Glycoproteins: The prosthetic group is carbohydrate, which is less than 4% of protein.
The term mucoprotein is used if the carbohydrate content is more than 4%. e.g. mucin
(saliva), ovomucoid (egg white).
(c) Lipoproteins: Protein found in combination with lipids as the prosthetic group e.g. serum
lipoproteins.
(d) Phosphoproteins: Phosphoric acid is the prosthetic group e.g. casein (milk), vitelline (egg
yolk).
3. Derived proteins
The derived proteins are of two types. The primary derived is the denatured or coagulated
or first hydrolysed products of proteins. The secondary derived are the degraded (due to
breakdown of peptide bonds) products of proteins.
(a) Primary derived proteins
(i) Coagulated proteins: These are the denatured proteins produced by agents such as
heat, acids, alkalies etc. e.g. cooked proteins, coagulated albumin (egg white).
(ii) Proteans: These are the earliest products of protein hydrolysis by enzymes, dilute
acids, alkalies etc. which are insoluble in water. e.g. fibrin formed from fibrinogen.
(b) Secondary derived proteins: These are the progressive hydrolytic products of protein
hydrolysis. These include proteoses, peptones, polypeptides and peptides.
Biologically important proteins
Regulatory
Proteins A number of proteins do not perform any obvious chemical transformation but
nevertheless can regulate the ability of other proteins to carry out their physiological functions.
Such proteins are referred to as regulatory proteins. A well-known example is insulin, the
hormone regulating glucose metabolism in animals.

Transport Proteins
A third class of proteins is the transport proteins. These proteins function to transport specific
substances from one place to another. One type of transport is exemplified by the transport of
oxygen from the lungs to the tissues by haemoglobin.
Storage Proteins
Proteins whose biological function is to provide a reservoir of an essential nutrient are called
storage proteins. Because proteins are amino acid polymers and because nitrogen is commonly a
limiting nutrient for growth, organisms have exploited proteins as a means to provide sufficient
nitrogen in times of need. For example, ovalbumin, the protein of egg white.
Contractile and Motile Proteins
Certain proteins endow cells with unique capabilities for movement. Cell division, muscle
contraction, and cell motility represent some of the ways in which cells execute motion. The
contractile and motile proteins underlying these motions share a common property: they are
filamentous or polymerize to form filaments. Examples include actin and myosin.
Structural Proteins
An apparently passive but very important role of proteins is their function in creating and
maintaining biological structures. Structural proteins provide strength and protection to cells and
tissues. Monomeric units of structural proteins typically polymerize to generate long fibers (as in
hair) or protective sheets of fibrous arrays, as in cowhide (leather). - Keratins are insoluble
fibrous proteins making up hair, horns, and fingernails. Collagen, another insoluble fibrous
protein, is found in bone, connective tissue, tendons, cartilage, and hide, where it forms inelastic
fibrils of great strength.
Exotic Proteins
Some proteins display rather exotic functions that do not quite fit the previous classifications.
Monellin, a protein found in an African plant, has a very sweet taste and is being considered as
an artificial sweetener for human consumption.
Glycoproteins
Glycoproteins are proteins that contain carbohydrate. Proteins destined for an extracellular
location are characteristically glycoproteins. For example, fibronectin and proteoglycans are
important components of the extracellular matrix that surrounds the cells of most tissues in

animals. Immunoglobulin G molecules are the principal antibody species found circulating free
in the blood plasma. Many membrane proteins are glycosylated on their extracellular segments.
Lipoproteins
Blood plasma lipoproteins are prominent examples of the class of proteins conjugated with lipid.
The plasma lipoproteins function primarily in the transport of lipids to sites of active membrane
synthesis. Serum levels of low density lipoproteins (LDLs) are often used as a clinical index of
susceptibility to vascular disease.
Nucleoproteins
Nucleoprotein conjugates have many roles in the storage and transmission of genetic
information. Ribosomes are the sites of protein synthesis. Virus particles and even chromosomes
are protein–nucleic acid complexes.
Phosphoproteins
These proteins have phosphate groups esterified to the hydroxyls of serine, threonine, or tyrosine
residues. Casein, the major protein of milk, contains many phosphates and serves to bring
essential phosphorus to the growing infant.
Metalloproteins
Metalloproteins are either metal storage forms, as in the case of ferritin, or enzymes in which the
metal atom participates in a catalytically important manner. We encounter many examples
throughout this book of the vital metabolic functions served by metalloenzymes.
Hemoproteins
These proteins are actually a subclass of metalloproteins because their prosthetic group is heme,
the name given to iron protoporphyrin IX. Because heme-containing proteins enjoy so many
prominent biological functions, they are considered a class by themselves.
Flavoproteins
Flavin is an essential substance for the activity of a number of important oxidoreductases. We
discuss the chemistry of flavin and its derivatives, FMN and FAD, in the chapter on electron
transport and oxidative phosphorylation.
IN LABORATORY PROTEINS CAN BE TESTED BY.

1) Biuret test: it is a very common identification test for proteins. In basic medium CuSO4
reacts with peptide bond and proteins to give violet coloured complex.
Procedure:
 Take 2 mL of protein solution (milk, albumin of egg or gram seed extract) in a test
tube.
 Add 1 mL of 40% NaOH solution and 1 or 2 drops of 1% CuSO4 solution.
 A violet colour indicates the presence of proteins. Care must be taken that excess of
copper sulphate is not added otherwise there will be blue colour instead of violet
colour.
2) Xanthoproteic test: This test is a identification test for the presence of aromatic amino acids.
Principle: In this reaction aromatic amino acids (tyrosine or tryptophan or proteins with
aromatic amino acids) undergo nitration in the presence of nitric acid to produce nitro-
derivatives that are yellow in colour. At the end of the reaction addition of NaOH (alkaline
pH), the colour changes to orange due to the ionization of the phenolic group.
Procedure:
 Add carefully 1mL of concentrated HNO3 to 2mL of protein solution (albumin of
egg, milk or gram seed extract).
 A white precipitate is formed.
 Boil the solution and the colour changes to yellow.
 Cool the test tube and add 2mL of 20% NaOH (or ammonia solution) to make it
alkaline.
 The colour changes to orange indicating the presence of proteins
3) Lead Acetate test: This test is a identification test for sulphur containing amino acids.
Principle: This test mainly depends on formation of inorganic sulphide from organic
sulphur. The sulphur containing amino acids, (cysteine and cystine) upon boiling with
sodium hydroxide (hot alkali) produce sodium sulphide. This can be detected by
precipitating inorganic sulphide to lead sulphide(black), using lead acetate solution.
Procedure:
 Take 1ml of protein sample
 Add equal volume of 40% NaOH and boil for 5min

 Cool the mixture then Add lead acetate observe black to brown colour.
4) Sakaguchi test:
Principle: α-naphthol (1-hydroxy naphthalene) reacts with a guanidine group containing
amino acid like arginine under alkaline condition,, which upon treatment with hypobromite
or hypochlorite, produces a characteristic red colour.
Procedure:
 Take 1 mL of amino acid solution in a test tube and chill it in ice box.
 Add few drops of sodium hydroxide solution, followed by 1 mL αnaphthol
reagent and mix well. After 2 min add few drops of 5% urea solution
 Add freshly prepared hypobromite reagent by drop wise observe red colour.
5) Millon’s test:
Principal: This test is specific to phenolic group containing amino acid such as tyrosine.
Tyrosine reacts with mercuric ions in acidic condition in the presence of sodium nitrite, to
give a red colour complex (Millon’s red).
Procedure:
 Take 1 mL of amino acid solution in a test tube
 Add few drops of Millon’s reagent and mix well.
 Boil the contents over a bunsen flame for 3-5 min. Later cool the contents under
running tap water and add few drops of sodium nitrite solution.
 Observe bright red colour.
6) Hopkins-Cole test: This test is a confirmatory test for the presence of amino acid
tryptophan.
Principle: In this reaction indole moiety of tryptophan condenses with aldehydes under
acidic environment to yield purple or violet coloured compounds.
Tryptophan + aldehydes Condences / acidic conditions Purple colored compound
Procedure:
 Take 1 mL of amino acid solution and add 1 mL of acetic acidglyoxylic acid
reagent, in a test tube and mix.

 Then carefully, add (use pipette) conc. H2SO4along the side of the test tube,
keeping the tube in an inclined position (do not shake the test tube, while adding
the acid).
 A purple- violet ring appears at the junction of the amino acid solution and the
conc. sulphuric acid.
7) Ninhydrin test: This test is a identification test for the presence of amino acids.
Principle: In this reaction amino acids react with ninhydrin (a powerful oxidizing agent)
reagent to give a purple coloured complex (Ruhemann’s purple) while, imino acids (proline
and hydroxyproline) react with ninhydrin to produce yellow colour.
Producer:
 Take 5ml of protein solution
 Add freshly prepared ninhydrin solution reagent
 Place test tube on water bath for 5min
 Remove and cool at room temperature
 Appearance of purple to blue color presence of alpa-aminoacid
 Appearance of yellow color presence of imino acid
8) Sodium nitroprusside test:
Principle: Sodium nitroprusside reacts with the thiol group of the cysteine under alkaline
condition to yield an intense purple coloured compound, which fades after few minutes.
Procedure:
 Take 1 mL of amino acid solution
 Add few drops of sodium nitroprusside reagent and mix well.
 Add few drops of liquor ammonia and mix well.
 Appearance of intense purple colour
9) Heat coagulation test:
Principal: Protein when heat undergo denaturation such a denatured proteins are insoluble
in water hence they form coagulation.

Procedure:
 Collect half of the test tube sample
 Hold the bottom of test tube and heat upper portion of solution
 Add about 1-2 drops of acetic acid observe thick coagulation
10) Pauly’s diazo test:
Principle: This test is based on the formation of diazonium salt where, sulphanilic acid
upon diazotization in the presence of sodium nitrite and HCl produces diazonium salt. The
diazonium salt formed, can combine with either tyrosine or histidine in alkaline medium to
give a red coloured product (azo dye).
Procedure:
 Take 1 mL of sulphanilic acid reagent in a test tube and chill the contents in a ice
box. Add few drops of pre-cooled sodium nitrite solution and mix. Quickly add
few drops of pre-cooled amino acid solution and mix well. (Later add sodium
carbonate solution drop by drop).
11) Neumanns Test:
Principal: When boiled with HNO3 organic phosphorus present in a casein is converted to
inorganic phosphorous and that gives yellow precipitate which is ammonium
phaspomolybdate.
Procedure:
 Take 5ml of protein solution add 3-4 drops od sulphuric acid add 10-12 drops of
HNO3 heat until mixture turns colourless allow to cool few ml of ammonium
molybdate solution observe canary yellow colour.

EXPERIMENT NO. IX
Aim: To perform chemical identification of ARGININE
Sl.
No.
01. Ninhydrine Test
2ml of sample + 0.5ml of ninhydrine
reagent boil for 2min and cool.
02. Xanthoproteic Test
2ml sample + 1ml of conc. HNO3
boil and cool for 2min + 40% NaOH
drop by drop
03. Sakaguchi Test
2ml of sample + 1ml of 10% NaOH
solution + 5 drops of 1% alpha-
naphtol regent mix + 5drop of freshly
prepared sodium hypobromite soln
04. Lead Acetate Test
1ml of sample + few drop NaOH boil
tube for 5min & cool + few drops
10% lead acetate solution
05. Millions Test
2ml of sample + 2ml millions reagent
+ boil for 2min, cool + few drops of
NaNO2 solution

EXPERIMENT NO. X
Aim: To perform chemical identification of TYROSINE
Sl.
No.
01. Ninhydrine Test
drop by drop
03. Sakaguchi Test
1ml of sample + drop NaOH boil for
5min & cool + few drops 10% lead
acetate
05. Millions Test
NaNO2 solution
06. Pauly’s diazo test:
1 mL of H2SO4reagent + precool few
drops of sodium nitrite + Quickly add
few drops of pre-cooled amino acid
solution and mix well + sodium
carbonate drop by drop

EXPERIMENT NO. XI
Aim: To perform chemical identification of ALBUMIN
Sl.
No.
01. Ninhydrine Test
drop by drop
03. Sakaguchi Test
04. Coagulation Test
5ml of sample + 1drop acetic acid
heat upper portion
05. Neumanns Test
5ml of sample + 0.5ml of 40% NaOH
heat for 1min and cool add 0.5ml of
conc. HNO3 to mix, add 1ml of
ammonium molybdate

EXPERIMENT NO. XII
Aim: To perform chemical identification of GLOBULIN
Sl.
No.
01. Ninhydrine Test
drop by drop
03. Sakaguchi Test
04. Half saturation Test
5ml of sample + 5ml ammonium
sulphate till soln saturated ppt keep
for 5min filer the soln.
05. Millions Test
NaNO2 solution

EXPERIMENT NO. XIII
Aim: To perform chemical identification of GELATIN
Sl.
No.
01. Ninhydrine Test
drop by drop
03. Aldehyde Test
2ml of sample + 5 drops of millions
reagent + 5 drops of formaline mix +
2ml of H2SO4 from along side
1ml of sample + drop NaOH boil for
5min & cool + few drops 10% lead
acetate
05. Millions Test
NaNO2 solution
06. Half saturation Test
5ml of sample + 5ml ammonium
sulphate till soln saturated ppt keep
for 5min filer the soln.

(a) Standard creatinine solution -0.5 ml
(b) Sodium hydroxide solution - 1.0 ml (NaOH 10%)
(c) Picric acid (1%) – 10.0 ml.
Dilute by adding distilled water to 50 ml.
Mix and keep for 15 minutes.
Step 3. Preparation of unknown:
In unknown flask (U), add following...
(a) Given urine sample - 0.5 ml.
(b) Sodium hydroxide solution 1.0 ml. (NaOH 10%)
(c) Picric acid (1%) – 100 ml.
Dilute by adding distilled water to make final volume 50 ml.
Mix and keep for 15 minutes.
Step 4. Record the colors (i.e., optical density) obtained to that of standard and unknown by
using photoelectric colorimeter.
Result: creatine present in given urine sample is ………. mg %.

EXPERIMENT NO. XXII
TO ESTIMATION OF CHLORIDES IN URINE BY VOLHARD METHOD
Principle: This method is also known as argentometric determination of chlorides. It involves
the precipitation of chlorides present in urine as silver chloride (AgCl) by adding a known
additional amount of standard silver nitrate (AgNO,) solution and the unused AgNO, (excess) is
determined by titrating against standard potassium thiocyanate (KSCN) using ferric alum
(Fe,(SO), (NH),SO,) as indicator. From this the amount of AgNO, used for precipitation of
chlorides is found out.
Each ml of AgNO3 is equivalent to 10 ml of NaCl.
Cl + AgNO3 AgCl + NO3
AgNO3 + KSCN AgSCN + KNO3
6 KSCN + Fe2(SO4)3 (NH4)2SO4 Fe(SCN3) + (NH4)2,SO4 + 3 K2SO4
Clinical significance:
An adult person on an average diet excretes 8-15 g (or 170 to 250 meq) of chlorides in
the form of NaCl. Vomiting and diarrhoea result in low excretion of chlorides in urine. Chlorides
excretion is also diminished when retention of chlorides by body fluids occurs, as in some cases
of nephrites and inflammations, in Cushing's syndrome and steroid therapy. When tubular
reabsorption in impaired (Addison's disease), urinary excretion of chlorides becomes
appreciable.
Procedure:
Pipette 5 ml urine into a conical flask and add 5 ml of Conc. HNO3, (to Prevent
precipitation of urates). Then add 10 ml of standard AgNO3, using a pipette, slowly with constant
stirring. Add 0,5 ml of ferric alum indicator and allow to stand for 15 min. Make up the volume
to 50 ml with distilled water and filter. Pipette out 25 ml filtrate and titrate against standard
KSCN from a microburette to a reddish brown end point.

Normal Range: 8 g to 15 g per day.
Reagents:
1) Standard AgNO3; 1 ml = 10 mg of NaCl.
Dissolve 29.061 g of crystalline AgNO3, in water and make the volume to 1000 ml.
2) Standard KSCN: 1 ml AgNO3, = 10 mg of NaCl.
Prepare 2% solution of KSCN and titrate against standard AgNO3, (10 ml) using
ferric alum as indicator. Dilute the solution of KSCN to make 1 ml KSCN = 1 ml
AgNO3.
3) Ferric Alum indicator: 5 % solution of ferric ammonium sulphate in water
Fe2{(SO4)3,(NH4,)2 SO4}.
4) Conc. HNO3
Result: The amount of chloride present in given urine sample is ………. mg %.

EXPERIMENT NO. XXIII
TO ESTIMATION OF GLUCOSE IN BLOOD BY FOLIN-WU METHOD
Principle: In this method, protein-free filtrate is obtained [Folin-Wu filtrate) so that 10 ml of
filtrate corresponds to 1 ml of blood sample. Protein-free filtrate is obtained by precipitating
proteins of blood by tungstic acid. Then this protein-free filtrate containing glucose is heated
with alkaline copper sulphate solution. Thus glucose reduces copper sulphate to form equivalent
quantity of cuprous oxide.
This cuprous oxide formed is reduced with phosphomolybdic acid to produce
corresponding equivalent quantity of molybdenum blue. The molybdenum blue gives intense
blue colour, the intensity of which is directly proportional to cuprous oxide which corresponds to
the amount of glucose present in the given sample of "Folin-Wu" filtrate.
Reactions:
C6H12O6 + Cu** Cu* + Oxidation
glucose cupric products of glucose
Cu*Na2PO4 : 12MoO3 Cu** + 12 MoO
Cuprous cupric molybdenum blue
The blue colour obtained with test blood sample is compared with standard solution by
similar procedure and by using photoelectric colorimeter. The optical density of test and standard
is measured and concentration of glucose in blood can be calculated using colorimetric principle.
Clinical significance: The chief end product of carbohydrate digestion blood is monosaccharide
like glucose and others like fructose, galactose etc. Glucose metabolism supplies...
Major amount of energy for body activities
Reserve fat depots
Tissue glycolipids
Amino acids
Thus it appears that glucose metabolism plays a central role in carbohydrate metabolism
which is closely associated with metabolism of protein and fat.

Availability of glucose from various dietary sources and its management/utilization is regulated
by hormones of islets of Langerhans. This can be explained schematically as follows...
Glucose B cells of Langerhans Glycogen (stored)
(Excess) Insulin
Energy
Fat depots
Tissue repair
Thus blood sugar level is maintained and it is normally expressed as Fasting. Post meal.
Hyperglycemia i.e. increase in blood sugar level is a real characteristic sign of "diabetes
mellitus". Excess free glucose appears in blood due to lack of insulin or in functioning of ẞ cells
to secrete insulin.
In diabetes mellitus, high values for fasting blood sugar are obtained and vary from
normal to 500 mg/100 ml and over according to severity of condition.
Increase in blood sugar level above 500 mg/100 ml indicates the increasing possibility of
coma. Hyperactivity of thyroid, pituitary, aderenal glands which include states of emotion stress
increases blood sugar level about 150 mg %.
Similar increase in blood sugar level is observed in convulsions and in terminal stages of
many diseases.
A moderate increase in blood sugar level L.e. hyperglycemia can occur in sepsis and
number of infectious diseases.
Increase in blood sugar level is also found in some intracranial diseases such as meningitis,
encephalities, tumours and haemorrhages.
Method used: Folin-Wu (modified)
Procedure:
1. Wash clean, label three Folin-Wu tubes as...
 Unknown ... 'U'
 Standard I-Std I
 Standard II - Std II

2. To the Folin-Wu tube labelled as "U" take 2 ml of "Folin-Wu filtrate".
3. In a Folin-Wu tube labelled as "Std 1" take 1 ml of standard sugar solution 1 (0.1 mg
sugar).
4. In a Folin-Wu tube labelled as "Std II' take 1 ml of standard sugar solution II (0.2 mg
sugar).
5. To all above tubes, add 1 ml of alkaline copper sulphate solution.
6. Keep the tubes in boiling water bath for 6 to 8 minutes.
7. Remove from the water bath and add 1 ml of phosphomolybdic acid to all tubes. 8.
8. Keep the tubes again in boiling water bath for 2 minutes and after 2 minutes cool to
room temperature.
9. Add 25 ml of distilled water to each tube, mix well and record. Compare the optical
densities by using photoelectric colorimeter by using tube filter 420 mµ.
Result: Glucose present in given blood sample is ..………… mg/100 ml.

EXPERIMENT NO. XXIV
TO ESTIMATION OF CHOLESTROL IN SERUM SAMPLE BY FOLIN-WU METHOD
Principle: This is a direct method used for estimation of cholesterol in serum by Ferro and Ham
In this method, a mixture of acetic anhydride, glacial acetic acid, and sulphuric acid in
appropriate proportion are used.
The colour reagent gives bluish colour with cholesterol.
In this method, the colour is developed directly without extraction of lipids.
In the "standard" preparation, two drops of distilled water addition is advised, as it
hastens the reaction and develops colour.
This "standard" colour with which "unknown" colour is compared by using photo-
electric colorimeter.
Clinical significance: Cholesterol is a complex monohydric secondary alcohol and is an
important member of sterol class.
It is present in all cells, all body fluids (except cerebrospinal fluid), brain, blood, muscles
etc. Body normal cholesterol serves following functions:
1. It is the essential constituent of cell.
2. It controls cell permeability.
3. It prevents haemolysis.
4. It is defensive in action.
5. It controls cell division.
Clinically, increase and decrease in serum cholesterol is observed in following conditions.
(a) In pregnancy, serum cholesterol level increases about 20 to 25% than the normal value.
(b) Hypercholestermia ie. Increase in cholesterol level is observed specifically in nephrosis,
diabetes mellitus, obstructive jaundice, myoxedema, xanthomatosis, gout and
atherosclerosis. In nephrosis, value of 600 to 700 mg percent is common.
(c) Values upto 400 mg percent cholesterol is found in diabetes mellius when the treatment is
inadequate.
(d) In coronary thrombosis and angina pectoris, serum cholesterol values range between 300
mg % to 400 mg %.

(e) A decrease in cholesterol value is not so well defined. In hyperthyroidism, the serum
cholesterol is reduced and values as low as 80-100 mg % are reported.
(f) In pernicious anaemia, low values of serum cholesterol are frequently observed.
(g) Values below 100 mg percent are seen in malabsorbtion syndrome, in severe wasting,
acute infections and in number of terminal states.
Normal serum cholesterol value:
In healthy young adult... 150 mg % to 270 mg %, cholesterol levels are not affected by ordinary
dietary changes.
Procedure:
[A] Preparation of Unknown Sample:
1. In a test tube labelled as "U" pipette out 0.2 ml of serum.
2. Add 5 ml freshly prepared colour reagent.
3. Mix well by shaking and keep the tube in dark for 10 minutes.
4. Obtain a optical density for unknown by using photoelectric colorimeter at 660 mµ.
Record and note it as "Eu".
[B] Preparation of standard:
1. In a tube labelled as 'S' take 0.2 ml of standard cholesterol solution.
2. Add 2 drops of distilled water and 5 ml of colour reagent.
3. Mix well by shaking and keep the tube in dark for 10 minutes.
4. Obtain a optical density for standard by using photoelectric colorimeter at 660 mµ.
Record and note it as "Es".
Result: Cholesterol present in given serum sample is...………… mg %

EXPERIMENT NO. XXV
TO ESTIMATION OF CALCIUM IN BLOOD SAMPLE BY FOLIN-WU METHOD
Principle: Calcium from serum is precipitated with 4% ammonium oxalate as calcium oxalate.
This calcium oxalate precipitate is washed for several times with washing solution.
Then calcium oxalate precipitate is decomposed by sulphuric acid 3% to oxalic acid. This
oxalic acid liberated is titrated against standard potassium permanganate solution 0.01 N in hot
(70°C to 80°C) till a faint colour appears as end point of titration.
Reaction:
Ca** + COONH4 COO Ca
COONH4 COO
Serum calcium Ammonium Oxalate Calcium oxalate
COO Ca + H2SO4 COOH + CaSO4
COO Sulphuric acid COOH
Calcium oxalate Oxalic acid Calcium sulphate
This formed oxalic acid is titrated against 0.01 N potassium permanganate.
COOH
+ 3H2SO4 + 2[KMnO4] 70°C K2SO4 + 2MnSO4 + 8H₂O + 10CO₂ ↑
COOH 80°C
Oxalic acid
Clinical significance: Calcium is the most chief cation essential for bone, teeth formation. It also
plays an important role in all body tissues as an intracellular messenger.
 It helps to regulate activity of skeletal muscle, heart and many other tissues.
 Low values of serum calcium are found in hypoparathyroidism about 6 mg %.
 In rickets, calcium levels slightly lowered to about 8 mg % to 9 mg %. A low serum
calcium level is a characteristic sign of osteomalacia.

 In steatorrhea, the serum calcium level observed is slightly low.
 In advanced renal failure, decrease in serum calcium levels upto 6 mg % is observed.
 A diagnostic decrease of calcium is found in acute pancreatitis.
 Increase in serum calcium is rare. The highest values are observed in
hyperparathyroidism (upto 20 mg %).
 Excessive administration of vitamin D may also raise the serum calcium (about 15 mg %
at times).
 In cases like multiple myeloma and in carcinoma mestatic to bone, rise in level is
observed.
Normal value: 9 mg % - 11 mg % (slightly higher in young children).
Procedure:
[A] Unknown Titration:
1. In a 15 ml conical centrifuge tube labelled as "U", pipette 2 ml of clear serum.
(Supernatant clear serum is obtained by centrifuging the blood sample for15 minutes).
2. Add to the above "U" tube distilled water 4 ml.
3. Add to the above "U" tube, 2 ml of ammonium oxalate. Mix by rotating in betwe palms.
4. Allow to stand for 30 minutes in an ice bath or refrigerator.
5. After 30 minutes centrifuge tube for 5 minutes at about 1000 r.p.m.
6. Pour off supernatant fluid. Drain off fluid completely.
7. Wash the precipitate twice with 5 ml of washing solution. Centrifuge again as before
Pour off supernatant fluid. Finally wash the precipitate with 5 ml water and pour of
supernatant fluid.
8. To the precipitate in above centrifuge tube, add 2 ml of 3 % sulphuric acid. Heat in water
bath at 70°C to 80°C the precipitate of calcium oxalate dissolves to give oxal acid.
Note: Do not heat above 80°C as oxalic acid which is formed decomposes at high temperature.
9. By 1 ml graduated pipette, add 0.01 N KMnO, drop by drop till faint pink colour which
persists at least one minute is obtained.
Note: Shake well after each drop addition of KMnO Record the reading as unknown_X

[B] Blank titration:
1. Label conical centrifuge tube as "B" and take 6 ml water.
2. Add 2 ml of ammonium oxalate and mix by rotating in between palms. Allow to stand in
ice bath or refrigerator for 30 minutes.
3. After 30 minutes, centrifuge the tube for 5 minutes at 1000 r.p.m. and pour supernatant
fluid. Wash the precipitate as stated previously.
4. To the precipitate, add 2 ml of 3% sulphuric acid and heat in water bath at 70°C to 80°C.
5. Titrate against 0.01 N KMnO, as before till a faint pink colour develops indicating end
point. Record the reading as blank.. "Y".
Note: If the first drop of KMnO, gives a faint/deep pink colour with blank then the blank reading
should be taken as zero.
Result: Calcium present in given serum sample is...………… mg/100 ml.

EXPERIMENT NO. XXVI
TO ESTIMATION OF UREA IN BLOOD SAMPLE BY FOLIN-WU METHOD
Principle: Urea is compleely diffusible substance.its percentage remains constant in plasma
serum and cells. Hence whole blood is preferred. In this method blood is collected in chemistry
bulb with potassium oxalate as an anticoagulant.
The blood is digested with enzyme urease.thus by chemical reaction blood urea is converted
into ammonium salt.
This formed ammonium salt is then treated wih nesslers reagent. This gives typical colour
which is compared colorimetrically against standard urea solution which is prepared in similar
way as unknown.
Clinical significance: Urea is formed in the liver mainly by the breakdown of amino acids. It is
also the chief excretory product of protein metabolism.
The concentration of urea in the blood is the balance between the urea formation from
metabolism and the urea excretion by the kidney. A small amount of urea is lost the protein in
faces and sweat.
The main function of urea in the body is to maintain the reaction of blood constant. The fact
is that it contains carbonic acid and two molecules of ammonia remain neutralized. Main as there
are so many pathological conditions in which blood urea may be raised, the estimation has little
value if performed as random diagnostic procedure. However, the estimation is a valuable guide
to observe the progress in particular diseased case.
The various pathological conditions in which blood urea is raised are:
1. Primarily in renal diseases like acute and chronic nephritis, acute tubular nephrosis,
hepato renal syndrome, uramia due to fall in glomerular filtration rate.
2. Obstruction due to enlarged prostate gland to the flow or urine.
3. In fever, wasting disease, thyrotoxicosis, diabetic coma, leukemia or after a major
operation due to increased protein catabolism.
A low blood urea sometimes occurs in late pregnancy and in acute hepato cellular necrosis
and intensive testosterone treatment.

Normal value: 20-40 mgm/100 ml blood.
Procedure:
Note: When urea is to be estimated from the whole blood proceed for the Folin-Wu filtrate
preparation as follows.
[A] Filtrate Preperation
1. In a conical flask take 14 ml distilled water.
2. Add 2 ml oxalated blood and mix.
3. Add urease powder about and mix
4. Mix and allow to stand for 15 minutes at room temperature. Shake frequently.
5. After 15 minutes incubation add 2 ml of 10% sodium tungstate followed by 2 ml of 0.66
N H₂SO4.
6. Mix well, keep for five minutes and filter by using funnel.
7. Use this filtrate for colorimetric estimations.
[B] Preparation of unknown:
1. Take 1 ml of above filtrate in a tube labelled as "U".
2. Add 5 ml water and cool in ice bath (below 10°C).
3. When sufficiently cooled add 2.5 ml of Nessler's reagent (cool the reagent also in ice
cold bath).
4. Add water to make final volume about 25 ml.
5. Read the optical density by using photoelectric colorimeter and blue filter Note it as
"Eu".
[C] Preparation of standard:
1. In a tube labelled as "std" take ml of standard urea solution.
2. Add water to make the volume about 5 ml and cool in ice cold bath (below 10
3. Add 2.5 ml of Nessler's reagent (which is also cooled in ice cold bath)
4. .Add water to make final volume about 25 ml.
5. Obtain the optical density for standard by using photoelectric colorimeter and filter (420
mp). Note it as "Es".

[D] Preparation of blank:
Take 3 ml water. Add Nessler's reagent 2.5 ml. Obtain optical density for blank (as
possible adjust to 100% transmission so that Eu and Es can be used directly)
Result: Urea present in given blood sample is...………… mg/100 ml

EXPERIMENT NO. XXVII
TO ESTIMATION OF CREATININE IN BLOOD SAMPLE BY FOLIN-WU METHOD
Principle: Creatinine in blood is estimated by Folin modified method using photoelect
colorimeter. In this Folin-Wu ie protein-free blood filtrate is used.
This creatinine (unknown) contain in filtrate is treated with picric acid in alkaline
medium to obtain red coloured creatinine picrate. Optical density of this red coloured creatinine
thus obtained is compared with that d standard solution, similarly converted by picric acid to
creatinine picrate By using colorimetry principle, concentration of creatinine in the given blood
sample can be
Calculated
Clinical significance:
Creatinine represents the waste products of creatine metabolic C and it arises in the body
from the spontaneous breakdown of creatine phosphate. It is N non threshold substance. It is
normally filtered by the glomerulli. As its excretion is related with food protein, so its variations
in the excretion indicate some of the metabol disorders.
Appearance of creatinine in urine is known as creatinuria
 Its excretion increases in fevers, starvation, in a carbohydrate free diet and in diabete
mellitus
 It may increase due to excessive tissue destruction releasing creatine or due to failure of
creatine being properly phosphorylated
 So creatinine excretion is independent of food proteins and is to be considered as an
index of endogenous protein metabolism.
 Endogenous creatinine clearance is a rough measure of the glomerular filtration rate and
is normally 100-130 ml/minute in adult of normal size.
 Values below 90 ml/minute are indicative of diminished glomerular filtration rate
Normal value: 0.7 to 2.0 mg/100 ml blood

Procedure:
[A] Preparation of unknown sample:
1. In a flask labelled as "U", pipette out 5 ml of Folin Wu filtrate.
2. Add 2 ml of 1% picric acid, mix well
3. Add 0.5 ml of sodium hydroxide solution 10%
4. Allow to keep for 15 minutes and obtain the optical density by using green filter (530
mp). Note it as "E"
[B] Preparation of standard sample:
1. In a flask labelled as "S", pipette out 5 ml of standard creatinine solution
2. Add 2 ml of 1% picric acid, mix well.
3. Add 0.5 ml of sodium hydroxide solution 10%
4. Allow to keep for 15 minutes and obtain the optical density by using photoelectric
colorimeter using green filter (530 mp). Note it as "Es
[C] Preparation of blank sample:
1. In a flask labelled as "B" pipette out 5 ml of distilled water.
2. To it add 2 ml of 1% picric acid and 0.5 ml of 10 % sodium hydroxide solution.
3. Allow to keep for 15 minutes and compare in a colorimeter by using green filter. (530
mp).
Result: Creatinine present in given blood sample is...………… mg/100 ml

EXPERIMENT NO. XXVIII
TO ESTIMATE SERUM GLUTAMATE OXALO TRANS AMINASE (SGOT)
Principle: Blood Analysis (Quantitative) Transaminases are enzymes which promote the process
of removal of a amino groups of most of Lamino acids to an a keto acid. As a result number of
alpha amino acids and alpha keto acids are formed.
One of these are serum aspartate transminase ie which catalyses the reaction of glutamate
oxalo acetate transaminase ie GOT
L – alpha - axaglutarate + L - aspartate L-glutamate + L-oxaloacetate.
This oxaloacetate formed in the reaction with glutamate oxaloacetate transaminase (GOT)
decarboxylates spontaneously to pyruvate which is again measured by hydrazone formation.
The colour obtained is measured in colorimeter at 510 mp (filter).
Clinical significance: Serum glutamate oxaloacetate transaminase (GOT) is widely distributed
in human tissues. The liver, kidney, heart contain large amount of glutamate oxaloacetate
transaminase (GOT) Clinically it is observed that serum glutamate oxaloacetate transaminase
rises rapidly after myocardial infarction. High levels of "SGOT" are observed in hepatocellular
damage due to hepato toxic drug, infective hepatitis and primary or secondary liver cancer
Normal value: 2-201 IU per litre
Procedure:
1. In a tube labelled as "U" take aspartate substrate 0.5 ml.
2. Add 0.1 ml serum sample.
3. Incubate the tube at 37°C for 30 minutes.
4. Remove the tube and add 0.5 ml DNPH solution, keep 20 minutes temperature.
5. Add 5 ml of 0.4 N NaOH in the tube.
6. Obtain the optical density for unknown by comparing the colours by using photoelectric
colorimeter with green filter (520 mµ). Note it as "Eu".

[B] Preparation of Control:
1. In a tube labelled as "C" take 0.5 ml of aspartate substrate, 0.5 ml DNPH solution and 0.1
ml serum.
3. After 30 minutes, remove from water bath and keep 20 minutes at room temperature.
4. Add 5 ml of 04 N NaOH to the tube, compare the colour by using green 5. Note it as "Ec
[C] Preparation of Standard:
1. In a tube labelled as "S" take 0.5 ml of aspartate substrate and solution Add 0.1 ml of
standard pyruvate
3. Remove after 30 minutes and keep at room temperature for 20
4. Add 5 ml of 04 N NaOH solution,
5. Compare the colour by using green filter Note it as "Es
[D] Preparation of Blank Sample:
1. In a tube labelled as "B" take 0.5 ml aspartate substrate 0.5 ml. DNPH solution 0.1 ml
distilled water. Incubate at 37°C for 30 minutes.
2. After 30 minutes keep at room temperature for 20 minutes.
3. Add 5 ml NaOH (0.4 N) solution.
4. Compare the colour using green filter. Note it as "E".
5. Prepare unknown control standard, blank as before [SGPT experiment).
6. Incubate the tube for 60 minutes at 37°C.
Result: SGOT present in given blood sample is...………… IU/lit.

EXPERIMENT NO. XXIX
TO ESTIMATE SERUM GLUTAMATE OXALO TRANS AMINASE (SGPT)
Principle: Transaminase are the enzymes which promote the process of removal of a amino
groups of most of L amino acids to an a keto acid. As a result, number of alpha amino acids and
alpha keto acids are formed.
One of these are serum alanine transminase (S.G.P.T.) This catalyses the reaction as
follows.
L - alpha oxoglutamate + Lalanine L - glutamate + L - pyruvate.
This pyruvate produced by "glutamate-pyruvate-transminase" reacts with dinitrophenyl
hydralazine (DNPH solution) in an alkaline medium which is measured at 510 mu filter.
Note: In the estimation, the concentration of substrate is suboptimal, to reduce background
colour produced by ketoglutarate in the reaction with dinitro-phenylhydrazine. (DNPH).
Clinical significance: Serum glutamate pyruvate transaminase is widely distributed in human
tissues. The liver, kidneys, heart and skeletal muscle contain large amount of glutamate pyruvate
transaminase (GPT).
Clinically it is observed that serum glutamate pyruvate transaminase level increases
disorder like myocardial infarction with large lessions and associated with liver damage.
Serum GPT levels are raised in heart conditions without infarction, such as a pectoris,
pericarditis, and in patients with pulmonary embolism.
Serum G.P.T. shows high levels in hepatocellular damage due to hepato-toxic angle drug
infective hepatitis and primary or secondary liver cancer Serum GPT levels are increased in
certain muscular disorders like progressive muscular dystrophy, muscular trauma etc..
Normal value: 2 to 15 IU/litre
Procedure:
1. In a tube labelled as "U" take alanine substrate 0.5 ml
2. Add 01 ml serum sample.
3. Incubate the tube at 37°C for 30 minutes

4. Remove the tube and add 05 ml DNPH solution, keep 20 minutes at room temperature.
5. Add 5 ml of 04 N NaOH in the tube.
6. Obtain the optical density for unknown by comparing the colours by using photoelectric
colorimeter with green filter (520 mg) Note it as "Eu".
[B] Preparation of Control:
1. In a tube labelled as "C" take 05 ml of alanine substrate, 0.5 ml DNPH solution and 0.1
ml serum.
3. After 30 minutes, remove from water bath and keep 20 minutes at room temperature
4. Add 5 ml of 04 N NaOH to the tube, compare the colour by using green filter. 5. Note it
as "Ec".
[C] Preparation of Standard:
1. In a tube labelled as "S" take 0.5 ml of alanine substrate and 0.5 ml of DNPH solution
Add 01 ml of standard pyruvate.
2. Incubate the tube at 37°C for 30 minutes
3. Remove after 30 minutes and keep at room temperature for 20 minutes.
4. Add 5 ml of 0.4 N NaOH solutions.
5. Compare the colour by using green filter Note it as "Es"
[D] Preparation of Blank Sample:
1. In a tube labelled as "B" take
 0.5 ml ... alanine substrate
 0.5 ml ... DNPH solution
 0.1ml ... distilled water
 Incubate at 37°C for 30 minutes.
2. After 30 minutes keep at room temperature for 20 minutes.
3. Add ... 5 ml NaOH (0.4 N) solution.
4. Compare the colour using green filter. Note it as "Eg".
Result: SGPT present in given blood sample is...………… IU/lit.

EXPERIMENT NO. XXX
TO STUDY THE EFFECT OF HYDROLYSIS OF STARCH FROM ACID
Principle: Starch is a polymer of glucose units with two structural components, namely amylose
and amylopectin. Amylose is a linear polymer of alpha D glucose with alpha (1-4) glycosidic
linkages and amylopectin is a branched polymer of glucose with (1-6) glycosidic linkages.
Hydrochloric acid hydrolyses starch by converting it into number of units of glucose. Starch is a
polysaccharide, so it do not give positive test with Benedict's reagent. But on hydrolysis starch
yields reducing sugar glucose and so gives positive test with Benedict's reagent. The rate of
hydrolysis of starch depends on time for which starch is boiled with hydrochloric acid. So
Benedict's reagent test on hydrolysis produces more dark colour on 10 minutes and 15 minutes
boiling accordingly.
Procedure:
1. Test tube No. 1: 5 ml of 3% starch solution + 3 ml of Benedict's reagent, boil, 5 minutes
cool + saturated solution of sodium bicarbonate till alkaline, boil.
→No brick red/yellow ppt/colour

2. Test tube No. 2 : 5 ml of 3% starch solution + 3 ml dilute HCI. Boil for 5 minutes in
water bath, cool + 3 ml of Benedict's reagent, boil, cool + saturated solution of sodium
bicarbonate till alkaline
→ Brick red/Yellow ppt / colour.
3. Test tube No. 3: 5 ml of 3% starch solution + 3 ml of dil. HCI. Boil for 10 minutes in hot
water bath. Cool + 3 ml of Benedict's reagent, boil, cool + saturated solution of sodium
bicarbonate till alkaline.
→ Brick red / Yellow ppt/colour (Darker than test tube No. 2)
4. Test tube No. 4: 5 ml of 3% starch solution + 3 ml dil HCI, boil for 15 minutes in hot
water bath cool + 3 ml Benedict's reagent, boil, cool and add saturated solution of sodium
bicarbonate till alkaline.
→Brick red/Yellow ppt/colour (Darker than test tube No. 3)
Conclusions:
1. Starch is hydrolysed by HCI producing reducing sugar glucose which gives pos
Benedict's test.
2. Rate/Extent of hydrolysis of starch by HCI depends on time for which it is boiled.
Result:

EXPERIMENT NO. XXXI
TO STUDY THE EFFECT OF HYDROLYSIS OF STARCH FROM SALIVARY
AMYLASE ENZYME
Principle: Enzyme amylase present in salivary juice catalyses hydrolysis of alpha (1-4)
glycosidic linkages of amylose and amylopectin components of starch. Hydrolysis proceeds with
the formation of polysaccharide dextrin, disaccharide maltose and then finally monosaccharide
reducing sugar glucose. Hydrolysis is tested by appearance of product glucose and disappearance
of starch over period of time. Disappearance of starch is observed by using iodine solution and
appearance of reducing sugar glucose is observed by using Benedict's reagent. Concentration of
enzyme amylase in saliva varies from person to person. So result also varies.
Procedure:
1. Collect approximately 5 ml. of saliva in a test tube.
2. Place 5 ml each of 1% starch solution in 5 test tube.
3. Place the test tube in water bath at 37°C. Do not allow to increase the temperature above
40° C which may inactivate the enzyme amylase.
4. After 5 minutes add saliva to each test tube as follows and mix each solution and keep in
water bath at 37° C.
5. Place one drop of iodine reagent in each of 5 depression of spot plate.
6. After 5 minutes place one drop of reaction mixture to 5 depression of spot plate and mix
iodine solution and observe the colour. Repeat the procedure after every five minutes by
washing the spot plate every time and keeping fresh one drop of iodine solution every
time. Violet colour indicates starch is not yet completely hydrolysed while red yellow
colour indicates complete hydrolysis of starch.
Conclusions:
1. Starch is hydrolysed to reducing sugar glucose by salivary amylase enzyme.
2. The rate of hydrolysis is directly proportional to the concentration of salivary amylase
enzyme.
Result:

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