Introduction
of Proteins
Definition:
Proteins
are polypeptides, which are made up of many amino acids linked together as a
linear chain. The structure of an amino acid contains a amino group, a carboxyl
group, and a R group which is usually carbon based and gives the amino acid its
specific properties.
Classification:
Simple proteins:Only yield amino acids or
their derivatives on hydrolysis, e.g. albumins,histones,globins etc.
Compound or conjugated proteins: Protein is
attached to some non-protein groups which are called prosthetic groups e.g.
nucleoproteins,phosphoproteins etc.
Derived
proteins: Proteins which are derived from simple or conjugated proteins e.g.
primary derived proteins,secondary derived proteins etc.
Practical#1:
To detect protein in
the given solution by salt saturation method
Proteins which are colloidal in nature
are kept in nature by two factors:
1.
Electric
charges
2.
Shell
of hydration
Electric
Charges:
A large no. of electric charges are
present on the surface of protein molecules.Similar charges repel each other
and prevent their shell of hydration.
Shell
of hydration:
Each molecule is surrounded by a film of
water, known as shell of hydration.This prevents the coalescence.The particles are ppt. if these factors are
removed.This can be done y addition of salts such as ammonium sulphate.This
process is known as salting out.This is of two types:
1.
Half
saturation test
2.
Full
saturation test
Reagents:
1.
Solid
ammonium sulphate
2.
Sat.
solution of ammonium sulphate
3.
40%
NaOH
4.
1%
CuSO4 sol.
Half Saturation Test
Procedure:
1.
Take
ammonium sulphate in a test tube containing sample.
2.
Shake
it,allow to stand for 5 mint.
3.
Filter
and perform biuret test with filtrate using equal vol. of 40% NaOH and 2 drops
of CuSO4.
Interpretation:
White
ppt. are formed in the case of globulin, casein and gelatin .If no ppt.
are formed it indicates the presence of albumin, peptones or polypeptides.
Full Saturation
Test
Procedure:
1.
Take
5 ml of given solution and add solid ammonium sulphate while mixing until the
solution is saturated.
2.
Shake
it,allow to stand for 5 mint.
3.
Filter
and perform biuret test with filtrate using equal vol. of 40% NaOH and 2 drops
of CuSO4.
Interpretation:
White ppt. are formed in the case of
albumin.If ppt. are not formed then peptones or polypeptides are present.
Principle:
When salt is added to protein solution
the effective conc. Of water available for protein is decreased.Amount of
ammonium sulphate required to ppt. a collide depends upon the surface area of
particles.Small molecules like albumin having a relatively large surface area
are ppt. by full saturation test method.Gelatin and casein mol. Having small surface
area are ppt. by half saturation method.
Practical#2:
To detect sulphur containing protein (cysteine) by lead
sulphide method
Reagents:
1.
NaOH…..40%
2.
Lead
acetate…..2%
3.
Protein
solution
Procedure:
1.
Take
2ml of prtein solution in a test tube.
2.
Add
2ml of 40%NaOH solution.
3.
Boil
for 2 mint.
4.
Add
1ml of lead acetate solution and warm.
Interpretation:
Brown or black ppt. are formed if cysteine or
protein containing cysteine is present in solution.
Principle:
When cysteine or protein containing
cysteine is boiled with strong alkali e.g. NaOH, the sulphur present in protein
is liberated as sodium sulphide and then react with lead acetate to form brown
or black ppt. of lead sulphide.Cysteine containing protein respond to this test
but the sulphur present in methionine is not released by this
procedure,therefore methionine does not respond to this test.Casein and gelatin
which contain methionine but negligible amount of cysteine that is why casein
and gelatin give negligible test.
R-SH + NaOH à
R-OH + Na2S + H2O
Na2S +
(CH3COO)2Pb à PbS
+ 2CH3COONa+
Practical#3:
To detect protein by heat
coagulation test
Reagents:
1.
Egg
white solution
2.
Phenol
red (indicator)
3.
Acetic
acid…..1%
Procedure:
1.
Fill
2/3rd of test tube with original solution
2.
Add
4 to 5 drops of phenol red drop by dropand mix
3.
When
purple red colour develop, add 1% acetic acid drop by drop until colour changes
to faint pink
4.
Hold
the test tube near its bottom, incline it slightly and heat the upper portion
of tube
Principle:
Dense coagulant will be formed on the
upper part as compare to lower part which serve as a control. This is due to
heat coagulable protein like
albumin.Test will be negative in case of non-coagulable proteins such as
gelatin and agar or peptones and polypeptides.
Introduction
of lipids
Lipids:
Lipids
are a group of naturally occurring molecules that include fats, waxes, sterols,
fat-soluble vitamins (such as vitamins A, D, E, and K), monoglycerides,
diglycerides, triglycerides, phospholipids, and others.
Classification
of lipids:
Simple
Lipids: Esters of fatty acids with various alcohols. a. Fats: Esters of fatty
acids with glycerol. Oils are fats in the liquid state. b. Waxes: Esters of
fatty acids with higher molecular weight monohydric alcohols.
Compound
lipids:These contain esters of fatty acids containing groups in addition to
alcohol and fatty acid.Further divided into: phospholipids,glycolipids,gangliolipids
etc.
Derived
lipids:These are hydrolytic products of complex lipids.
Practical#4:
To test glycerol in given
sample
Reagent:
1.
Solid
potassium hydrogen sulphate
2.
Pure
glycerol
Procedure:
1.
Take a clean and dry test tube
2.
Add
two drops of glycerol and a pinch of sodium potassium hydrogen sulphate
3.
Heat
gently and strongly over low flame until a pungent smell is produced.
Glycerol becomes dehydrated to
form an unsaturated aldehyde called acrolein which has pungent odour.Potassium
hydrogen sulphate is dehydrating agent.
Practical#5
To demonstrate the
emulsification of natural fat in water and solution of sodium carbonate and
soap
Reagent:
1.
Water
2.
Sodium
carbonate
3.
Soap
solution
Procedure:
1.
Take
3 test tubes
2.
Add
5 ml of water, sodium carbonate solution and soap solution in separate test
tubes and name these as A,B,C
3.
Add
2 drops of oil in each test tube, shake well all the test tubes then stand
these for 10 mints and observe
4.
Oil
and water separate quickly which indicates that emulsion is unstable
5.
Sodium
carbonate separate slowly which indicates that emulsion is more stable than
water
6.
In
soap solution separation takes longer time than other two solutions, this
indicates the emulsion is stable
Principle:
When oil and water which are immiscible
are shaken together, oil is broken into tiny droplets which are dispersed in
water.This is known as oil in water emulsion.The water molecules due to their
higher surface tension have a tendency to come together and form a different
layer,therefore oil and water emulsion is unstable.In presence of substance
that lower the surface tension of water like sodium carbonate or soap solution,
the tendency of water to come together is decreased and emulsion become stable.
TO
DEMONSTRATE THE ACTION OF BUFFER
BUFFER:
A
solution which can resist change in its pH when small amount of acid or base is
added is called as Buffer.
A
buffer solution contains conjugated acid-base pair so that it can both uptake
and release hydrogen ion.Buffers are mainly of two types.
1.
A mixture of weak acid with salt of its strong conjugated base.For example
CH3COOH and CH3COONa act as a buffer. Because acetic acid is aweak acid and
sodium acetate is a strong conjugated base, Similarly NaHCO3 and H2CO3 can form
a buffer solution.
2.
A mixture of weak base with salt of its strong conjugate base.For example
Sodium Monohydrogen phosphate ( Na2Hpo4) n sodium dihydrogen phosphate can act
as a buffer.
it
is important that when acid is weak its conjugate base should be strong and
when base is weak its conjugate acid should be strong.
Reagents:
1.
Distilled water
2.
phosphate buffer solution
3.
0.4% Hcl Solution
4.
METHYL ORANGE
PHOSPHATE
BUFFER SALINE:
potassium
chloride...kCl.. .02g
NaCl….
o.8g
Disodiumhydrogen
phosphate.... 0.178g
Dihydrogen
potassium phosphate... 0.027g
Preparation:
Weigh
all the reagents and mix them all together in a beaker.
Procedure:
1.
Take
two test tubes, Label them as A & B. Add 10ml of water in a test tube A and
10ml buffer solution in a test tube B.To
both test tubes add two drops of methyl orange as indicator. The colour will be
orange in test tube
2.
Take
10ml pipette, fill it with 0.4% HCL . Now add this acid drop wise to test tube
A and shake it after each addition until the color will be change to red and
then note the amount of hcl use to change the color of test tube from orange to
red.
3.
Take
10ml pipette, fill it with 0.4% aaHCL and add this acid drop wise to test tube
B and shake the test tube after each Addition. ANd note the amount of hcl
required to change the color of test tube B
from orange to red.
INTERPRETATION:
The amount of hCl use to change
the orange color to red will be greater in test tube B As it contains the buffer
solution and has buffering capacity. A buffer can resist change in pH and take
most of acid to reach the same pH.
PRINCIPLE:
A
buffer solution is a conjugate acid base pair and it can both uptake and
release the hydrogen ion.
In
this case buffer solution is made by dissolving conjugate acid base pair of
Na2HPO4/ KH2PO4 in water.
It
can buffer addition of acid when HCl is added in this phosphate buffer saline.
Practical#7
To demonstrate the effects of
solutions of different tonicities on RBCs
Reagents:
1.
0.9%
NaCl solution in water
2.
0.5%
NaCl solution in water
3.
0.3%
NaCl solution in water
4.
3.0%
NaCl solution in water
5.
Blood
6.
Microscope
Procedure:
1.
Take
4 dry and clean test tubes and mark them as A,B,C and D
2.
Take
5ml of 0.9% NaCl solution in test tube A
3.
Take
5ml of 0.5% NaCl solution in test tube B
4.
Take
5ml of 0.3% NaCl solution in test tube C
5.
Take
5ml of 3.0% NaCl solution in test tube D
6.
To
each test tube take 5 drops of fresh human blood and mix, let the test tubes
stand for an hour
7.
Take
small drop from each of test tubes on separate glass slides apply cover slips
and observe under microscope
Interpretation:
Cells taken from test tube A will show no
change from normal.Cells from test tube B will swell.There will be no intact cell
in test tube C due to hemolysis, this is because the liberated hemoglobin goes
into solution.Cells fron test tube D will appear shrunken due to loss of water.
Principle:
When
human red blood cells are placed in isotonic solution,they remain intact and
maintain their original shape and volume because in this case the amount of
water entering in the cell is equal to that of leaving the cell.When human RBCs
are placed in hypotonic solution of NaCl then water will enter in the cells
making them to swell and if solution is very dilute the cells will rupture,
releasing their content in the solution.When human RBCs are suspended in
hypertonic solution of NaCl,water will leave the cells and the cells will
shrink and become crenated.
Practical#8
To
prepare,observe and draw the shape of haemin crystals
Reagents:
1.
Distilled
water
2.
0.1%
KCl solution in glacial acetic acid
3.
Microscope
Procedure:
1.
Take
a drop of blood on a glass slide and add 2-3 drops of distilled water
2.
Hemolyze
the RBCs with glass rod and heat the slide over low flame gently till it dries
3.
Add
one drop of 0.5% KCl in glacial acetic acid
4.
Apply
cover slip
5.
Again
heat it gently over a low flame until gas bubbles form
6.
Then
cool the slide and observe it under microscope
Interpretation:
Under microscope haemin crystals have
rhombic shape and are dark brown in colour
Principle:
On heating with acid the haemoglobin is
denatured and haem is oxidized ti haematin.In presence of chloride,haematin is
finally converted to haematin chloride
Practical#9
To demonstrate the process of adsorption
and show that adsorption is reversible process
Reagents:
1.
Eosin
dye solution
2.
Charcoal
3.
Filter
paper
4.
Acetone
solution
Procedure:
1.
Take
two test tubes and label them as A and B
2.
Take
10 ml of 10:10000 solution of red dye eosin in both test tubes
3.
Now
only in test tube A add a knife point of charcoal and shake the tube for some
time
4.
Filter
the contents of both test tubes and observe the colour of filtrate
Interpretation:
The filtrate of test tube A will be
colourless due to the adsorption of the dye by the charcoal. The filtrate of
test tube B will be red in colour
Adsorption
is reversible process:
To show that adsorption is loose
combination and is reversible process,place the funnel containing charcoal dye
complex over another test tube
Take 5 ml of acetone in another tube
and warm it gently in a boiling water bath, then pour this acetone into funnel.
The filtrate obtained will be red showing that the dye has been separated from
charcoal
Principle:
The exact mechanism of adsorption is
still unknown because the molecules adsorbed on the surface are arranged in a
definite manner, it has been purposed that surfaces of all kinds have free
valencies which have the power of attracting and holding other molecules upon
the surface.Adsorption is a reversible process, because there is no strong
covalent bonding between the adsorbing and adsorbed substances
Practical#10
Liebermann Burchard’s test
Reagent:
1.
Cholesterol
in chloroform solution
2.
Acetic
anhydride
3.
Conc.
Sulphuric acid
Procedure:
1.
Take
1 ml of cholesterol solution ina test tube
2.
Add
10 drops of acetic anhydride and 3 drops of concentrated sulphuric acid
3.
Observe
colour changes.This solution first turn red then blue.
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