Biochem Journal 2015

1

Student’s name: Harshavardhan Mahalakshmi Ganesan

School/College

name:

D. Y. PATIL UNIVERSITY, NAVI MUMBAI,

SCHOOL OF BIOTECHNOLOGY AND

BIOINFORMATICS

Class: BBT-2

Roll number: BBT-2-14017

Subject: Practical – II: Biochemistry/Analytical Techniques

Index

Exp

no. Date Experiment title

Pg

no Signature

1 06.08.2015 Study of pH meter 4

2 13.08.2015 Preparation of buffer solutions 6

3 20.08.2015 Qualitative tests for carbohydrates 10

4 27.08.2015 Lipid solubility test 14

5 24.09.2015 Lipid TLC 16

6 08.10.2015 λ max & Beer-Lambert’s law 18

7 13.10.2015 Paper chromatography of A.A 22

8 17.10.2015 Qualitative tests for proteins and A.A 24

9 17.10.2015 Starch extraction 26

10 20.10.2015 Reducing sugar estimn - DNS method 28

11 29.10.2015 Protein estimation by Biuret method 30

12 05.11.2015 Extraction of casein from milk 32

2

3

Certificate

Class: BBT-2 Year: 2015

This is to certify that the work entered in this

journal is the bonafide work of

Mr. HARSHAVARDHAN M GANESAN,

Roll number: BBT-2-14017,

who has worked for III semester

in D.Y.Patil University, Navi Mumbai,

School of Biotechnology & Bioinformatics

during the academic year: 2015-16

__________ __________ __________

Head of the External Internal

Department examiner examiner

Date: 16.11.2015 School of Biotechnology

& Bioinformatics

4

D. Y. PATIL UNIVERSITY, NAVI MUMBAI

SCHOOL OF BIOTECHNOLOGY AND BIOINFORMATICS

Practical – II: Biochemistry/Analytical Techniques

Programme: B.Tech Biotechnology Semester: III

Practical

Title: Study of pH meter

Experiment. No.: 1 Date: 06.08.2015

Aim: To Study the construction & working of pH Meter.

Principle: • Acids dissolve in water forming H+ ions. The greater the concentration

of H+ ions, stronger is the acid, the converse applies to bases (w.r.t OH-

ions). pH can be defined as ‘negative log to the base10 of hydrogen ion

concentration’. pH= -log10 [ H+]

• pH is the degree of acidity or alkalinity of a solution on a scale of 1 to

14. The e.m.f developed between the 2 elements depends on the

concentration of hydrogen ions & hence the pH. In order to measure this

e.m.f no current flow should occur in the electrode flow pair, otherwise

chemical reactions at the cell boundaries with results in polarization of

electrodes.

• To avoid this, a high impendence (reaction) circuit is used to detect the

e.m.f (potential). This converts the voltage to current which can be

amplified & measured & an electric current circuit which measures e.m.f

developed across the electrode. These two electrodes used in the pH

meter are two half cells.

• The glass electrode contains a glass bulb, coated with silica & mineral

salts, that is permeable to hydrogen ions. As a result, potential develops

across the glass membrane. The potential developed is linearly

proportional to the pH of the solution in which the electrode is immersed.

• The relationship is expressed by Nernst’s equation as:

E=E0-(2.303RT)/nf *log a

5

Where,

E= Electrode potential at specified concentrations; E0= Standard electrode

potential; R= Gas constant; T= absolute temperature; F= faraday’s

constant & a= activity of ions.

• Working: On placing the probe into the solution, the H+ ions will move

towards the glass electrode replacing some of the metal ions on the glass.

This causes a potential difference to be developed which is picked up by

the silver wire & the same is passed on to the voltmeter where it is

amplified and displayed as pH units. Increase in concentration of H+ ions

increases the voltage and hence displays lower pH.

Requirement: 1. pH meter

2. standard buffer solution(s)

Procedure: 1. STANDARDISATION OF pH METER: Sureties on the pH

meter & allow it to warm up for 15mins with electrodes properly

dipped in distilled water. This prevents any fluctuations while

measuring the voltage or pH.

2. All the test solution & buffer solution added should be brought to

ambient temperature of room in which pH meter is kept. Check

the position of pointer on scale & it should show zero or set zero if

required.

3. The pH meter should be adjusted to measure the pH at room

temperature by adjusting the temperature knob.

4. The standardisation of the pH with 3 standard buffers is important

i.e. pH 4, pH 7 & pH 9 to ensure that the pH is functioning

properly over the entire range. The electrodes should be washed

thoroughly with D/W & immersed in a beaker containing fresh

distilled water. This protects the electrode from dying & increases

its lasting period.

Result: The construction and working of pH meter was observed and studied.

Precautions: 1. pH meter bulb should not be allowed to dry, so it should be kept immersed

in water when not in use.

2. After each operation bulb should be washed with distilled water.

3. pH meter should be calibrated using standard buffer solution of pH 4,7 & 9.

6





Practical

Title: Preparation of buffer solutions


Aim: To prepare solutions and buffers

Principle: A buffer system is one that can resist a change in pH on addition

of acid or alkalis. It consists of a conjugate acid-base pair. The

buffer system consists of a mixture of weak acid & its

conjugate/base buffer of acetic acid, sodium acetate. A

physiological buffer is carbonate, bicarbonate system in blood.

Buffers are extensively used in biochemical studies since they aid

in maintaining a rear constant pH of media while performing

various lab operations viz. extraction of various bio molecules,

their isolation, purification, selection of an appropriate buffer with

optimal pH is important as it may have an influence on

extractability stability & biological functioning of cell

constituents. For e.g. the activity & stability of extracted enzyme

is dependent on the pH of the system. The pH may vary during an

enzyme reaction due to release or acceptance of proteins which

may affect the activity of the enzyme. Therefore buffers are

necessary to stabilize the H+ ion concentration & remove adverse

effects due to pH change on enzyme under investigation.

Henderson Hasselbach equation can be used to express the

dissociation constant of an acid.

HA↔H+ + A-

K= [H+] [A-]/ [HA]

Where,

7

K= dissociation constant of the acid by rearranging,

[H+] = K [HA]/ [A-]

Taking negative logarithm on both the sides;

-log [H+] = -logK – (log [HA])/ [A-]

But

-log [H+] = pH

& -log K = pKa

Therefore pH = pKa – (log [HA]) / [A-]

Therefore pH= pKa + (log [A-]) / [HA]

This can be written in the general form as:

pH= pKa + log[conjugate base]/[conjugate acid]

Theory: Solution: Any non-reactive substances mixed together homogenously

form a solution.

One mole is its molecular weight in grams (gram equivalent)

Ex: NaOH : 40 mol wt

1 mol of NaOH = 40 gm of NaOH

1 mol of NaOH in 1L water = 1M NaOH

Q1. How many grams of NaOH would be required to make 50ml of 0.5M

solution?

A. Molarity = (amt of solvent/ mol w.t)

1M = 40gm of NaOH 0.5M = 20gm of NaOH

If 20gm in 1L, then ‘x’gm in 50ml

x = 1gm

1gm of NaOH is to be mixed in water, and brought up to 50ml, to

make 0.5M solution.

Q2. Prepare 0.1N & 0.5N Na2CaO3 whose equivalent weight is 53gm.

A. 1N = 53gm in 1L 0.1N = ‘y’gm y = 5.3gm

1N = 53gm 0.5N = ‘z’gm z = 26.5gm

5.3gm & 26.5 gm of Na2CaO3 is to be mixed in 1L of water to make 0.1N

& 0.5N Na2CaO3 respectively.

8

Q3. Prepare 150ml of 70% alcohol from 95% stock.

A. N1V1 = N2V2

0.95 x ‘x’ = 70% x 150ml

‘x’ = 110.5ml

Or

Req concn x total vol/given concn = 110.5ml

150ml – ‘x’gm = amount of water to be added

Water to be added = 39.5ml

Q4. Prepare 100ml of 0.1M NaOH

A. 1mol = 40gm of NaOH

0.1M in 1L water = 4gm

0.1M in 100ml water = 0.4gm

0.4gm of NaOH is to be added to 100ml of water to make a 0.1M

solution.

Q5. Prepare 50ml of 70% alcohol from 100% stock.

A. N1V1 = N2V2

0.1 x ‘x’ = 70% x 50ml

‘x’ = 35ml

Or

Req concn x total vol/given concn = 35ml

50ml – ‘x’gm = amount of water to be added

Water to be added = 15ml

Q6. What volume of HCl do you need to prepare 0.1N? If concn = 37.5%,

sp.gravity = 1.84 & eq.wt = 49gm.

A. V1 = eq.wt of acid x V2 x N x 100

1000 x sp.gravity x purity

V1 = 36.5gm x 100 x 0.1 x 100

1000 x 1.84 x 37.5%

V1 = 52.9ml

Q7. Prepare a buffer solution of Acetic acid = 0.2M & Sodium acetate =

0.2M with total volume 50ml. Find out the resultant pH.

A. 46.3ml of acetic acid & 3.7ml of sodium acetate for 100ml

Hence, for 50ml 23.15ml of acetic acid & 1.85ml of

sodium acetate. pH = 2.17

9

10





Practical

Title:

Qualitative tests for carbohydrates


Aim: To detect the presence and type of carbohydrate(s), in the given

environmental sample, using qualitative tests

Theory: Carbohydrates are compounds consisting of carbon, hydrogen and

oxygen; with hydrogen and oxygen in the ratio of 2:1. Carbohydrates

include sugars and starches and can be divided into three types namely –

monosaccharides and disaccharides and polysaccharides formed by

condensation of monosaccharides. Monosaccharides are polyhydroxy

aldehydes and ketones.

Principle: Molisch’s test:- (General test for Carbohydrates):

Conc. H2SO4 hydrolyses the glycosidic bonds of polysaccharides which

are then dehydrated to furfural and its derivatives. These products then

combine with sulphonated α – naphthol to give a purple colored complex.

This reaction is given generally by all carbohydrates.

Iodine test for Polysaccharides:

Iodine forms colored absorption complexes with polysaccharides. Starch

gives blue color with Iodine, while Glycogen reacts to form reddish

brown complex. Hence it is a useful, convenient and rapid test for

detection of amylase, amylopectin and glycogen.

Benedict’s test (for reducing sugars):

It is a test for detecting reducing compounds. It is a Modified form of

Fehling’s test where alkaline CuSO4 containing Cu(OH) is converted to a

reddish precipitate of Cu2O when heated in the presence of a reducing

11

agent such as Glucose or Maltose.

Na2CO3 + 2H2O → NaOH + H2CO3

2NaOH + CuSO4 → Cu(OH)2 + Na2SO4

Cu(OH)2 → CuO + H2O (Cupric Oxide)

Glucose + 2CuO → (Gluconic acid oxidised) + Cu2O (reduced red ppt

cuprous oxide)

Requirement: 1. Conc. H2SO4

2. Molisch’s reagent

3. Benedict’s reagent

4. Barfoed’s reagent

5. Fehling’s A & B

6. Seliwanoff’s reagent

7. Bial’s reagent

8. Sample solution to test for presence of carbohydrates

9. Iodine solution: 0.005N I2 in 3% KI solution

Procedure: Molisch’s Test: (for carbohydrates)

1. Take 2ml of test solution in a test tube.

2. Add 2 drops of Molisch’s reagent α – naphthol reagent

mix well and wait for 2-3 minutes.

3. Take 1ml of conc. H2SO4 solution allowing it to flow

from the side of the tube.

4. At the junction of the 2 reddish violet purple ring appears

in the presence of the sugar.

Iodine Test: (for polysaccharides)

To 1ml of sample extract or test solution in a test tube add 4-5

drops of iodine solution to it and mix the contents and observe for

color formation.

Benedict ’s test: (for reducing sugars)

1. Take 2ml of Benedict’s solution in the test tube

2. Add 1ml of sugar sample and mix the solution well

3. Boil the solution in boiling water bath for 5 to 10 minutes

and allow to cool

4. Observe the color of precipitate.

12

Barfoed’s test: (for mono/disaccharides)

1. Take 2ml of Barfoed’s solution in the test tube


3. Place the solution in boiling water bath for 5 minutes and

allow to cool


Fehling’s test: (for reducing sugars)

1. Take 1ml of Fehling’s A & B solution in the test tube


3. Place the solution in boiling water bath for 5 to 10 minutes

and allow to cool


Seliwanoff’s test: (for mono/disaccharide & aldose/ketose)

1. Take 2ml of Seliwanoff’s reagent in the test tube


3. Place the solution in boiling water bath for 1 to 2 minutes

and allow to cool


Bial’s test: (for pentose/hexose)

1. Take 2ml of Bial’s reagent in the test tube

2. Add 4 to 5 drops of sugar sample and mix the solution well

3. Place the solution in boiling water bath for 2 minutes and

allow to cool


13

Observation:

Type Test Observation Inference

GT

Molisch’s test:

α-naphthol + conc H2SO4 +

sample

Purple ring

Carbohydrat

es may be

present

GT Iodine test:

Iodine solution + sample

No change in

colour

Polysacchari

des - absent

GT

Benedict’s test:

Reagent + sample {boiling

water bath for 5-10 minutes}

Red ppt

Reducing

sugar may

be present

CT

Barfoed’s test:


water bath for 5 minutes}

Red ppt

within 2-5

minutes

Monosaccha

rides present

CT

Bial’s test:


water bath for 2 minutes}

Colour

change but

not blue

green

Hexose

present

CT

Seliwanoff’s test:


water bath for 1-2 minutes}

Deep red Keto hexose

present

Result: The given sample (B) is a ketohexose monosaccharide reducing sugar.

14





Practical

Title: Lipid solubility test


Aim: To identify the solvent in which butter, ghee, palmitic acid, oleic acid,

vegetable oil & coconut oil gets dissolved

Requirement: Fatty acid samples, distilled water, acetone, chloroform & ethylether

Procedure: 1. Add 0.5 ml of each of the seven solvents in seven test tubes.

2. Add 0.5 ml of a particular lipid sample (say butter) to all the seven

test tubes containing solvents. Mix well and observe for solubility.

3. Incubate the tubes at 55oC for about 10 minutes & observe the

solubility again & record the observations.

4. Repeat the same with other lipid samples.

Observation & Result:

Fatty acids

Solvents

Distilled

water Acetone Chloroform Ethyl ehter

RT IT RT IT RT IT RT IT

Butter I I I M M M M M

Ghee I I M M M M M M

Palmitic acid I M* M M M M M M

Oleic acid I I M M M M M M

Veg. oil I I M M I I M M

Coc. Oil I I M M M M M M

Stearic acid I M* M M M M M M

Key: {RT: Room temperature , IT: Incubated temperature (55 oC), I: Immiscible, M:

Miscible, M*: Sparingly Miscible}

15

16





Practical

Title: Separation of various components in different lipid factions by TLC


Aim: To identify the components of a lipid mixture using TLC

Principle: In Chromatography techniques, the separation of components of a

mixture takes place because of the partition co-efficient of these

components between two immune-soluble phases, separation is due to

small variation in physical and chemical properties which result from

structural differences of chemically related group of compounds under

investigation. They therefore have a relatively lesser or greater affinity for

mobile or stationary phase of chromatographic system and the separation

is based on the difference in there affinities for stationary and mobile

phase. These differences are due to the two physical phenomenon

exhibited by compounds.

1) Adsorption

2) Partition

On heat activated layer of silica gel a mixture of unknown lipids is

spotted and run to determine its constituent lipid by their comparison with

the standard lipid spots on the basis of Rf.

Rf = Distance travelled by solute (lipid)

Distance travelled by solvent front

Requirement: 1. Activated TLC plates.

2. Mixture of unknown lipids.

3. Standard lipid samples.

4. Capillary/micropipette.

5. Solvent system: Hexane: di-ethyl-ether: g.a.a (80:20:1)

6. Saturated solvent chamber.

7. Oven at 110 C.

17

Procedure: 1. A clean grease-free plate is placed on top of a tray and slurry of

silica gel is prepared and spread evenly over it by means of an

applicator.

2. The plates are dried and activated in a hot air oven at 110C for 45

minutes.

3. The plates are then spotted with small amounts of standard lipid

solutions and unknown samples using capillaries.

4. The plates are then placed in a glass chamber saturated with

solvent system.

5. When the solvent front reaches at least two third of silica gel

layer, the plates are then removed and solvent front is marked.

After drying for 5 minutes, they are then sprayed with 50%

H2SO4 solution (to develop colour) and heated at 120oC for 30

minutes.

6. Spots are marked and the Rf value is measured and compared with

the standard.

Observation table:

Sample Distance

travel by

sample(cm)

Distance travel

by solvent (cm)

Rf = (dist

solute)/(dist

solvent)

Colour

of

sample

Oleic Acid 2.6 7 0.37 Purple

Palmitic

acid

- 7 - -

Steric acid - 7 - -

Unknown

sample 2

2.5 7 0.35 Purple

Result: Unknown sample 2 contains Oleic acid among other chemicals.

Conclusion: Only Oleic acid, among the three standards, was visible. Hence, we

cannot comment on the presence/absence, of Palmitic acid or Steric acid,

in the given sample.

18





Practical

Title: Determination of λ max & verification of Beer – Lambert’s law


Aim: To determine λmax of a given coloured solution and to verify Beer-

Lambert’s law.

Principle:

Lambert’s law: Talks about the proportional-relationship between the

path length (L) of light and absorbance (A). A α L ……………………(1)

Beer’s law: Talks about the inverse-proportional-relationship between

amount of transmitted light (T) and concentration (C). T α 1/C

It is noteworthy that absorbance decreases with increase in amount of

transmitted light, hence A α 1/T A α C ……………………….........(2)

From (1) & (2) A α CL

Beer Lambert Law:

It states that when a monochromatic light passes through a coloured

solution, the amount of light transmitted decreases exponentially:

With the increase in concentration of the coloured substance.

With the increase in length of medium through which light passes.

The following quantitative relationship can be drawn:

i) If Io, is the intensity of incident light and I is the intensity of

transmitted light then I/Io is called the Transmittance.

ii) If Io is taken as 100 then I is represented as percent transmission.

Absorbance A α CL A = εCL

Where C is the concentration (g/l), L is length in cm through which light

19

passes & ε is called extinction coefficient.

When C=1 mol/L and L=1cm, then measured Extinction ε = A

This means absorbance is directly proportional to concentration of

solution. Therefore the graph of absorbance against concentration gives a

straight line passing through origin.

Requirement: Colorimeter, 1% K2Cr2O7, 1% CuSO4, D/W, cuvets, filters: 450nm,

470nm, 510nm, 520nm, 540nm, 570nm, 600nm and 670nm.

Procedure: A) Determination of λmax:

1. Take 1% K2Cr2O7 and 1% CuSO4.

2. Adjust the colorimeter to100% transmittance / zero absorbance with

distilled water as a blank.

3. Read the absorbance of K2Cr2O7 & 1% CuSO4 over entire range of

wavelength from 450nm to 670nm using colorimeter.

4. For every wavelength repeat the blank adjustment using D/W.

5. Note down the absorbance at the corresponding wavelength.

6. Plot a graph of absorbance against wavelength and join the

consecutive points. The wavelength showing maximum absorbance or

optical density is λmax.

B) Verification of Beer’s Law:

1. Prepare a series of dilution of K2Cr2O7 and measure the absorbance of

the K2Cr2O7 dilutions of different concentrations at λmax.

2. Plot the graph of absorbance against concentration & obtain a straight

line passing through origin.

Observation table:

Sr.

No.

Concentra

tion of

K2Cr2O7

(%)

Volume of

K2Cr2O7

(ml)

Volume of

D/W (ml)

Diluent

(ml)

Total

volume

(ml)

Absorban

ce

1 0.2 0.6 2.4 10 13 0.30

2 0.4 1.2 1.8 10 13 0.75

3 0.6 1.8 1.2 10 13 0.87

4 0.8 2.4 0.6 10 13 1.67

5 1.0 3 0 10 13 1.80

20

Sr.No. K2Cr2O7 CuSO4

Wavelength Absorbance Wavelength Absorbance

1 450 1.49 450 0.01

2 470 λmax 1.58 470 0.02

3 510 0.65 510 0.02

4 520 0.21 520 0.01

5 540 0.07 540 0.02

6 570 0.00 570 0.06

7 600 0.00 600 0.10

8 670 0.00 670 λmax 0.21

Result: λmax value of K2Cr2O7 & CuSO4 is 470nm & 670nm respectively

Conclusion: It was observed that the absorbance increases with increase in

concentration of K2Cr2O7 and hence the Beer-Lambert’s law stands

verified.

21

22





Practical

Title: Paper chromatography of amino acids


Aim: To identify the presence of proline, glyceine and alanine in the given

environmental sample using paper chromatography.

Principle: Chromatography is a technique by which substances in a mixture are separated

based on their differential partition coefficients in two immiscible solvents (a

polar and a nonpolar solvent). In ascending paper chromatography the Whatman

filter paper serves as a support for the stationary hydrophilic polar solvent. The

hydrophobic solvent serves as the mobile phase which ascends up the paper

chromatogram by capillary action. Thus a mixture of amino acids spotted on the

filter paper is separated based on their partition coefficients in the stationary and

mobile phases. Those amino acids which have a higher affinity for the

hydrophilic stationary phase move slower (polar amino acids) than those which

have a higher affinity for the mobile hydrophobic phase. The separated amino

acids can be detected using Ninhydrin reagent and identified by their unique Rf

(Relative to front) values. This technique can be used to determine the

composition of amino acids in a protein.

Rf = Distance travelled by solute (amino acid)

Distance travelled by solvent front

Requirement: 1. Whatman’s filter paper no. 1,

2. Standard amino acids, protein hydrolysate containing mixture of amino acids

(glyceine, alanine & proline).

3. Solvent system butanol acetic acid, water ( 4:1:5) in solvent chamber.

4. 0.2 % Ninhydrin

5. Oven set at 80oC

23

Procedure:

1. Add the required amount of solvent system in the solvent chamber and allow

the chamber to saturate.

2. Spot the standard amino acids in W1 equidistant from each other and 2 cm

from the bottom of the filter paper. Also spot the mixture of amino acids

whose components have to be identified.

3. The spots can be concentrated by spotting for a second time after sufficient

time is given for the initial spot to dry.

4. Transfer the chromatography paper into the solvent chamber such that the

bottom of the Whatman paper just touches the solvent system and the amino

acid spots do not go into the solvent.

5. Allow the solvent system to rise up by capillary action over the amino acid

spots. The amino acids in the mixture will separate as per their partition

coefficients. Remove chromatogram from chamber and mark the solvent

front. Allow the filter to dry. And spray it with Ninhydrin.

6. Keep the filter paper in an oven for 5 minutes.

7. Determine distance travelled by solvent (y cm) and distance travelled by

standard amino acids and amino acids in mixture (x cm) using a scale. Record

results in tabular form.

8. Identify the amino acids in the mixture by comparing Rf values and color of

spots in mixture and Rf values and color of standard amino acids (Proline

and hydroxyproline are yellow in color) and calculate Rf values as x/y.

Observation table:

Sample Distance travelled by

Sample (A)

Distance travel by

Solute(B)

Rf=(A)/(B)

Proline 1.5cm 6cm 0.25

Glyceine 0.7cm 6cm 0.11

Alanine 1.3cm 6cm 0.21

Unknown 0.8cm & 1.6cm 6cm 0.13 & 0.26

Result: The unknown sample contains proline & glceine.

Conclusion Only proline and glyceine were found in the unknown sample as their Rf values

corresponded with that of standards and hence alanine was absent.

24





Practical

Title:

Qualitative tests for proteins and amino acids


Aim: To test for the presence of amino acids and proteins using Ninhydrin

and Biuret reagents.

Principle: Proteins are polymers of amino acids linked together by peptide bonds

(CO-NH). There are 20 different amino acids having the common general

structure, where R can vary. Based on their R groups, amino acids are

classified as Polar, Hydrophilic (acidic/ basic/ or uncharged polar) amino

acids and nonpolar amino acids. During protein synthesis in the

ribosomes amino acids are linked together by condensation between the

amino group of one amino acid and the carboxyl group of another amino

acid.

Ninhydrin test for amino acids:

To 1ml of sample solution (adjusted to neutrality) 5 drops of Ninhydrin

solution is added. The solution is placed in a boiling water bath for 2

minutes and observed for appearance of purple, violet (cysteine) or

yellow (proline and hydroxyl proline) color.

Biuret test for polypeptides: A general test where Alkaline copper sulphate reacts with compounds

containing two or more peptide bonds, to give a violet/pink product. This

is due to the formation of co-ordination complex of cupric ions with

unshared pair of electrons of peptide nitrogen as oxygen of water.

Requirement: 1. Test solution

2. 0.2% ninhydrin

3. biuret’s reagent

4. boiling water bath

25

Procedure:

Ninhydrin test for amino acids:

To 1ml of sample solution (adjusted to neutrality) 5 drops of Ninhydrin

solution is added. The solution is placed in a boiling water bath for 2

minutes and observed for appearance of purple, violet (cysteine) or

yellow (proline and hydroxyl proline) color.

Biuret test for polypeptides:

To 1ml of sample solution (adjusted to neutrality) 0.1ml of NaOH

solution is added. To this, 2 to 5 drops of copper sulphate is added and

observed for appearance of violet/pink color.

Observation: Ninhydrin test: Blue-violet color.

Biuret test: Purple

Result: The environmental sample was found to contain proteins as both

ninhydrin & biuret tests yielded positive results.

26





Practical

Title: Starch extraction


Aim: To quantify the amount of starch present in 1gm of potato sample.

Requirement: 1. Finely chopped potato (5gm)

2. Water

3. Pestle-mortar

4. Muslin cloth

5. W1

Procedure: 1. Take potato sample (5gm) in pestle-mortar and add 10ml water.

2. Crush & grind the sample.

3. Filter the contents using 2 layers of muslin cloth.

4. Collect the filtrate in a 100ml beaker.

5. Allow the starch to settle (15mins).

6. Weigh the petri-plate with W1.

7. Collect the starch in W1 .

8. Air dry the starch for 16 hours and calculate the weight.

Calculation:

W1 Weight of petri-plate & W1 = 97gm

W2 Weight of petri-plate, W1 & starch = 97.320gm

W3 = W2 – W1 = 0.32gm

If W3 gm in 5gm, then X gm in 1gm

Xgm = 0.064gm per gram of potato

Result: 0.064gm of starch was to be present in 1gm sample of potato.

27

28





Practical

Title: Estimation of reducing sugar by DNS method


Aim: To estimate the amount of sugar preset in the sample by DNSA

method.

Principle: Maltose is a reducing sugar which will reduce 3,5 - dinitro salicylic acid

(DNSA) to 3-amino, 5-nitro salicylic acid (ANSA) which is an orange

colored solution and can be estimated at 540nm.

Requirement: 20 mg % of maltose stock solution, unknown sugar, DNSA reagent &

D/w.

Procedure: 1. Prepare varying concentrations of standard maltose solution ranging

from 40 µg/ml to 200 µg/ml with an interval of 40 µg/ml.

2. In the test tube labelled “UK”, pipette out 2ml of maltose solution of

unknown concentration. Add DNSA (1ml) and mix.

3. Mix contents of each tube and heat for 10 minutes in a water bath.

4. Cool the tubes and add 7ml of D/W in all tubes. Mix contents of each

tube by ‘vortexing’.

5. Record the absorbance at 540 nm against the blank.

6. Draw a standard graph by plotting the concentration (µg/ml of

maltose on X- axis against the absorbance on the Y-axis.

7. 7) Determine concentration of maltose in given sample.

Calculation:

Given concentration: 200 µg/ml

Required concentration: 40 µg/ml

Total volume required: 2ml

(RxT)/G = 40 µg/ml x 2ml = 0.4ml

200 µg/ml

29

Observation table

Sr

No.

Standard Conc.

(µg/ml)

Amount

of stock

(ml)

Amount

of D/W

(ml)

Amount

of DNSA

(ml)

D/W

(ml)

O.D at

540nm

1 0 0.00 2.0 1 7 0.00

2 40 0.4 0.8 1 7 0.07

3 80 0.8 0.6 1 7 0.09

4 120 1.2 0.4 1 7 0.14

5 160 1.6 0.2 1 7 0.16

6 200 2.0 0.0 1 7 0.21

7 Unknown 2.0 0.0 1 7 0.20

If 120µg/ml 0.14 OD then ‘z’ 0.20 OD ‘z’ = 171.4µg/ml

Concentration of ‘z’ by graph = 170µg/ml

& by calculation = 171.4µg/ml

Result: The concentration of maltose, in the given sugar sample, by graph &

calculation, comes out to be 170µg/ml & 171.4µg/ml respectively.

30





Practical

Title: Estimation of protein by Biuret method


Aim: To quantify the amount of proteins present in the given

environmental sample using Biuret test

Principle: The test is a general test for compounds having a peptide bond. Alkaline

CuSO4 solution reacts with compounds containing two or more peptide

bonds to give a violet colored complex. The intensity of colour contained

is directly proportional to number of peptide bonds present in protein. The

concentration of an unknown sample can be estimated using a series of

standard concentrations.

Requirement: Biuert reagent, standard stock solution of Bovine Serum Albumin (BSA)

of 10 mg/ml concentration, unknown sample, colorimeter & D/W.

Procedure: 1. Prepare a set of known standard varying concentrations of BSA from

0 to 5mg/ml with the interval of 1mg/ml, using the stock BSA and

distilled water given in table as given in the observation table.

2. In other test tubes labelled ‘x’ & ‘y’, pipette out 2ml of given protein

sample and add 3ml Biuret reagent (for ‘x’ and dilute it 1:1 with D/W

for ‘y’)

3. Incubate at room temperature for 10 minutes and then read against

blank at 540 nm.

4. Plot a graph of O.D (Y- axis) versus standard concentration of protein

(X-axis) and calculate concentration of protein in given sample by

extrapolating OD of sample onto X-axis.

31

Observation table:

Sr

No.

Required

Standard

concn(mg/ml)

Amount of

Stock BSA

(ml)

Distilled

water (ml)

Biuret

Reagent

(ml)

O.D at

540nm

1 0 0 2 3 0.00

2 1 0.2 1.8 3 0.07

3 2 0.4 1.6 3 0.15

4 3 0.6 1.4 3 0.23

5 4 0.8 1.2 3 0.33

6 5 1 1 3 0.38

7 x 2 0 3 0.19

8 y 1 1 3 0.10

Calculation:

For ‘x’:

3mg/ml BSA 0.23 OD & ‘x’ 0.19 OD

‘x’ = 2.47mg/ml

Concentration of ‘x’ by graph = 2.5mg/ml

& by calculation = 2.47mg/ml

For ‘y’:

3mg/ml BSA 0.23 OD & ‘y’ 0.10 OD

‘y’ = 1.30mg/ml

Concentration of ‘y’ by graph = 1.3mg/ml

& by calculation = 1.3mg/ml

Result: The protein concentration in the unknown sample ‘x’ & ‘y’ was found to

be 2.47mg/ml & 1.30mg/ml respectively.

32





Practical

Title: Extraction of casein from milk


Aim: To calculate the percentage of casein in the given milk sample

Principle:

Requirement:

Procedure: 1. Add 50ml of milk in 100ml beaker and heat it to 55oC.

2. Add acetate buffer to the milk, while stirring, till it gets turbid.

3. Collect the precipitate by filtering the contents using a muslic-

clothe & funnel.

4. Wash with 25ml 1:1 ethylalcohol: ethylether.

5. Collect the precipitate by filtering the contents using a W1 &

funnel and air dry.

6. Measure the weight & calculate the percentage of casein.

Calculation:

Weight of petri-plate & W1 : 93.18gm & Weight of petri-plate, casein & W1 : 102.59gm

Casein % = Weight of casein x 100 (102.59gm - 93.18gm) x 100 18.82%

Total volume 50gm

Result: Total protein content in percentage is 18.82%

33

Documents

Biochem Journal 2015