HAND BOOK FOR PRACTICAL BIOCHEMISTRY (MEDICAL &DENTAL SCIENCE GRADUATES )

HAND BOOK FOR PRACTICAL BIOCHEMISTRY

(For Dental & Biochemistry Graduates ) Mrs.R.Janani Dr.K.M.Tulaanidi

Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi e-ISBN-Pending UI Media Publications-INDIA No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work.

Product liability: The Publisher cannot guarantee the accuracy of any information about biochemical medical values or text or any other related terms contained in this book. In every individual case the user must check such information by consulting the relavant literature.

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1

Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

DEDICATED TO MY MOTHER, FATHER, HUSBAND AND DAUGHTER

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

PREFACE As an Author I internally have tenacity to publish this book. This Book solves the purpose of referring the experiments and clarifying the doubts then and there. All the experiments in this book are designed to help the Dental as well as the Biochemistry graduates. Hope it serves the purpose.

My uncle Dr.J.Kumar, Principle, Sasuri Engineering College, Vijayamangalam, who is a great mentor, is a prodigious inspiration in my life for my Career. I thank him for his support.

I thank my Principal Dr.G.S.Kumar, for his support in publishing this book.

I express my gratitude to my Head of the Department Dr. G.Rajeswari for her encouragement.

I thank my students who have been stimulation for writing this book.

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CONTENTS TITLE

PAGE NO.

QUANTITATIVE ANALYSIS Determination of Acid number

06

Determination of Saponification number

08

Determination of Iodine number

10

Estimation of Lipid content in Egg yolk

12

Estimation of Amino acids by Sarrenson’s Method

14

Estimation of Glucose by Glucose by Ortho Toluidine Method

16

Estimation of Creatinine by Jaff’s Alkaline Picrate Method

18

Estimation of Blood Urea by DAM’s Method

20

Estimation of Protein by Biuret Method

22

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

QUALITATIVE ANALYSIS Analysis of Carbohydrates

24

Hydrolysis of Sucrose

27

Hydrolysis of Starch

28

Reactions of Protein

29

Analysis of Amino Acids

31

Precipitation reactions of Protein

33

Analysis of Lipid

34

Analysis of Normal Urine

35

Analysis of Abnormal Urine

37

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

QUANTITATIVE ANALYSIS LIPID ANALYSIS DETERMINATION OF ACID NUMBER AIM: To determine the acid number of the unknown oil.

PRINCILPE: Acid number is the number of milligram of potassium hydroxide required to neutralize the free fatty acids, present in one gram of oil.

The experiment consists of procedure of taking a known weight of fat, dissolving it in alcohol and titrates with standard alkali using phenolphthalein as indicator.

PROCEDURE: Weigh 2g of the oil into 250ml conical flask. Pipette out 25ml of 90%alcohol and add it to the conical flask.

Add 4 to 6 drops of phenolphthalein to it. Titrate the solution against 0.1N potassium hydroxide. The end point will be appearance of pink colour which persists for 20 to 40 second. The conical flask containing 25ml of alcohol serves as a blank.

RESULT: The acid number of the unknown oil is _______________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

DETERMINATION OF ACID NUMBER STANDARDISATION OF POTASSIUM HYDROXIDE AGAINST FREE FATTY ACID

BURETTE READING VOLUME OF 0.1N (ml) S.No

SOLUTION

POTASSIUM HYDROXIDE INITIAL

1.

BLANK

0

2.

SAMPLE I

0

3.

SAMLE II

0

INDICATOR

(ml)

FINAL

PHENOLPHTHALEIN

CALCULATION: Weight of the oil

= 2g

Volume of potassium hydroxide require by 2g of oil

= Blank volume – sample volume =A

1g of oil require

= Equivalent weight of KOH x A x Blank volume x 1000 2 x 1000

Acid number of given oil

= ________________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

DETERMINATION OF SAPONIFICATION NUMBER AIM: To determine the saponification number of the unknown oil.

PRINCIPLE: The saponofication number is the number of mg of Potassium hydroxide required to saponify 1g of fat.

A known weight of lipid is reacted with excess of alcoholic potassium hydroxide for 30 minutes and is titrated against hydrochloric acid using phenolphthalein as indicator.

REAGENTS: 1. Alcoholic Potassium hydroxide 2. Phenolphthalein 3. Standard hydrochloric acid.

PROCEDURE: Weigh 1g of fat in about 3ml of fat solvent (equal volume of 95% alcohol and ether) in a beaker. Quickly transfer the contents of the beaker into 250ml conical flask. To this add 25ml of 0.5N alcoholic KOH keeping a funnel at the mouth of the flask. Keep the flask in boiling water bath for 30 minutes. Cool the flask and titrate it against 0.5N hydrochloric acid using phenolphthalein as indicator. Simultaneously heat another flask without oil and titrate it against 0.5N hydrochloric acid. This serves as the Blank. Calculate the amount of KOH in mg used by 1g of oil from the difference between the titration value of blank and the sample. The calculation value gives the saponification number of the unknown oil.

RESULT: The Saponification number of the given oil is _____________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

DETERMINATION OF SAPONIFICATION NUMBER STANDARDISATION OF HYDROCHLORIC ACID AGAINST POTASSIUM HYDROXIDE

BURETTE READING VOLUME OF 0.1N (ml) S.No

SOLUTION

HYDROCHLORIC ACID INITIAL

1.

BLANK

0

2.

SAMPLE I

0

3.

SAMLE II

0

INDICATOR

(ml)

FINAL

PHENOLPHTHALEIN

CALCULATION: Weight of the oil

= 2g

Volume of Hydrochloric acid require by 2g of oil

= Blank volume – sample volume =A

1g of oil require

= Equivalent weight of Hcl x A x Blank volume x 1000 2 x 1000

Saponification number of given oil

= ________________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

DETERMINATION OF IODINE NUMBER AIM To determine the Iodine number of the unknown oil.

PRINCIPLE: The Iodine number of oil is the grams of iodine absorbed by 100g of oil.

A known weight of the oil is dissolved in chloroform and Hanu’s Idoine solution is added to it. The unreacted iodine titrated against sodiumthiosulphate (0.1N) using starch as indicator.

REAGENTS: 1. HANU’S IODINE SOLUTION Dissolve 13.2g of the iodine in 1 litre of glacial acetic acid and add 3ml of liquid bromine to increase halogen content.

2. STANDARD SODIUMTHIOSULPHATE SOLUTION Dissolve 24.82g of sodiumthiosulphate in 1 litre of water.

3. 15% POTASSIUM IODINE SOLUTION 4. FRESHLY PREPARED 1% STARCH SOLUTION 5. Chloroform PROCEDURE: Accurately weigh about 1g of the given oil in a dry clear iodine flask. Dissolve the sample in 10cc chloroform.

Then add 30ml of Hanu’s iodine solution to the flask. Allow it to stand in dark for 30 minutes with occasional shaking.

Add 10 of 15% potassium iodine solution. Shake well and then add 100ml of water. Titrate the contents against sodiumthiosulphate till yellow colour disappears. Add starch and continue the titration till the end point, which is the disappearance of blue colour. Repeat the same with the flask without oil which serves as the blank.

RESULT: The Iodine number of the given oil is ____________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

DETERMINATION OF IODINE NUMBER STANDARDISATION OF SODIUMTHIOSULPHATE AGAINST OIL

BURETTE READING VOLUME OF 0.1N (ml) S.No

SOLUTION

SODIUMTHIOSULPHATE INITIAL

1.

BLANK

0

2.

SAMPLE I

0

3.

SAMLE II

0

INDICATOR

(ml)

FINAL

STARCH

CALCULATION: Weight of the oil

= 2g

Volume of Sodiumthiosulphate require by 2g of oil

= Blank volume – sample volume =A

1g of oil require

= Equivalent weight of Hcl x A x Blank volume x 1000 2 x 1000 =B

100g of oil requires Iodine number of given oil

= B x 100 = ________________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF LIPID CONTENT IN EGG YOLK AIM: To determine the lipid content in the egg yolk.

PRINCIPLE: Lipids are ethers of fatty acids. Chemically they are widely spread in nature being found in all vegetables and animal matters. Some of the groups of lipids such as phosphotides and steroids are found in all living cells. They form an essential part of colloidal complex of cytoplasm. Complex lipids are also found in large quantities in brain and nervous tissues. Other lipids such as fats and oil represent the chief form in which the excess nutrients are stores in animal body. Lipids act as heat insulators and reserve supply of energy.

REAGENTS: 1. CHLOROFORM METHANOL REAGENT 2 parts of chloroform is mixed with 1 part of methanol volume by volume 2. EGG YOLK 3. CENTRIFUGE TUBES 4. DISTILLED WATER PROCEDURE: Weigh 1g of egg yolk and take it in the centrifuge tube. Add 3.5ml of distilled water and 10ml of 2:1 chloroform methanol reagent. Mix the solution with a glass rod. Centrifuge the tubes at 2000rpm for 15 minutes. The mixture will separate into two phases. The top most layers will be methanol and water which contains all nutrients. The lower layer is chloroform which contains lipid. Remove the top layer by using an aspirator. Transfer the lower layer into 50ml previously weighed conical flask. Evaporate the contents of flask in a boiling water bath and dry it in an oven at 60â °c and weigh it.

RESULT: 100g Egg yolk contains approximately _____________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF LIPID CONTENT IN EGG YOLK CALCULATION: Weight of the flask

=

A

Weight of the Egg yolk

=

1g

Weight of the flask and lipid

=

B

Weight of the lipid

=

B–A

=

C

1g of the Egg yolk contains

=

C grams

100 g Egg yolk contains

=

C x 100

=

D

=

D grams

100g of Egg yolk contains

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF AMINOACID BY SARRENSON’S METHOD AIM: To estimate the amount of glysine present in the whole of the given sample.

PROCEDURE: Carboxyl group of d- amino acids cannot be accurately titrated in water solution against alkali because it forms zwitter ions by reacting with basic amino acids. Sarrenson observer that the amino acid solution are neutralised when titrated against excess of formaldehyde solution with phenolphthalein as indicator. The mixture becomes acidic and can be titrated sharply against the standard alkali solution. Under the condition formaldehyde helps the amino group to form dimethyl amino acid. Titration completes when the amino group is formed. The reaction is reversible.

REAGENTS: 1. 0.1N STANDARN SODIUM HYDROXIDE 2. ACIDIFIED FORMALDEHYDE Add 2 – 3 drops of phenolphthalein indicator to 50ml formaldehyde and then add 10ml 0.1N hydrochloric acid.

3. STANDARD GLYCINE SOLUTION Add 75mg of glycine is dissolved in 100ml of 0.1N hydrochloric acid.

PROCEDURE: Pipette out 10ml of given glycine solution into a conical flask. To this add 5ml of acidified formaldehyde and titrate against standard sodium hydroxide solution in the burette. The end point is the appearance of pale permanent pink colour. Repeat the experiment for concordant values. A blank with formaldehyde alone is titrated against alkali. The difference in the titre value of the blank and sample gives the amount of alkali required to neutralize the amino acids.

RESULT: The amount of glycine present in the given solution is ____________.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF AMINOACID BY SARRENSON’S METHOD BURETTE READING VOLUME OF 0.1N VOLUME OF

(ml)

S.No

SODIUM HYDROXIDE SOLUTION INITIAL

FINAL

(ml)

0 1.

BLANK

0 0 0

2.

SAMPLE I

0 0

CALCULATION: Volume of NaOH required to neutralize 10ml of amino acid and 5ml of formaldehyde

=

A

Volume of NaOH required to neutralize 5ml of formaldehyde

=

B

Volume of NaOH required to neutralize 10ml of amino acid

=

A–B=C

=

7.5 x C

10ml of amino acid require C ml of 0.1N Sodium hydroxide 100ml of glycine solution contains

x 100

10

The amount of glycine present in the given solution

=

D mg

=

D mg

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF GLUCOSE BY ORTHO – TOULIDINE METHOD AIM To Estimate the amount of glucose present in 100ml of the given sample.

PRINCIPLE Proteins are precipitated by Tricholoro acetic acid. Glucose in protein free filtrate is treated with Ortho Toluidine reagent in glacial acetic acid to form N- Glycosylamine derivatives which forms the colour. The intensity of the colour developed is directly proportional to the amount of Glucose present which is compared with the standard similarly treated.

REAGENTS 1. Ortho Toluidine reagent 2. Stock Standard glucose solution 3. Working standard glucose solution PROCEDURE Development of colours     

Label 3 test tubes as T, S, and B. Pipette out 1.0ml of the given glucose solution in T, 1.0ml of standard in S and 1.0ml of water in B. Add 5.0ml of Ortho Toluidine reagent to all the tubes.

Mix thoroughly.

Keep them in boiling water bath for 10 minutes.

Cool the test tubes and read the colour developed at 620nm.

RESULT The amount of glucose present in 100ml of the given sample is ___________________________.

CLINICAL SIGNIFICANCE     

Blood glucose level is increased in Diabetes Mellitus.

Increased levels are also seen due to hyper activity of the pituitary gland and Adrenal cortex medulla.

Fear, Anxiety and other emotional status cause elevation of blood glucose level.

Decrease in level of circulating glucose is seen in Insulin secreting tumors.

Prolonged starvation may cause a small decrease in blood glucose level.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF GLUCOSE BY ORTHO – TOULIDINE METHOD REAGENTS

BLANK (ml) -

STANDARD (ml) -

TEST (ml) 1.0

-

1.0

-

WATER

1.0

-

-

O – TOLUDINE

5.0

5.0

5.0

GLUCOSE SOLUTION (GIVEN) STANDARD SOLUTION

OPTICAL DENSITY (O.D) AT 620nm

CALCULATION O.D of the Test – O.D of the Blank The amount of the glucose present in 100ml of the given sample = O.D of the Standard – O.D of the Blank

= ______________________ mg/dl.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF CREATININEBY JAFF’S ALKALINE PICRATE METHOD AIM To estimate the amount of creatinine present in 100ml of the given solution.

PRINCIPLE Serum proteins are precipitated by Tungstic acid. Creatinine reacts with picric acid to give orange colour. The intensity of the colour developed is directly proportional to the amount of creatinine present. It is compared with the standard similarly treated in photometer at 520nm.

REAGENTS 1.

2.

3.

4.

5.

2/3 N Sulphuric acid 0.7 N Sodium hydroxide Picric acid Stock Standard creatinine solution Working Standard creatinine solution

PROCEDURE Colour Development     

Label 3 test tubes as T, S, and B. Pipette out 3.0ml of the given creatinine solution in T, 3.0ml of standard in S and 3.0ml of water in B. Add 1ml picric acid and 1ml of 0.7N Sodium hydroxide to all the tubes.

Mix thoroughly.

Allow it to stand for 15 minutes at room temperature.

Read the colour developed at 520nm.

RESULT The amount of Creatinine present in 100ml of the given solution is = ________________________mg/100ml.

CLINICAL SIGNIFICANCE  Serum Creatinine level is 0.5 -1.4mg/dl.  Daily excretion of creatinine is 1.0 – 2.0g. Excretion is more in men.  The level of excretion is fairly constant in an individual and is not influenced by diet.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF CREATININEBY JAFF’S ALKALINE PICRATE METHOD REAGENTS

BLANK (ml) -

STANDARD (ml) -

TEST (ml) 3.0

-

3.0

-

WATER

3.0

-

-

PICRIC ACID

1.0

1.0

1.0

0.75N SODIUM HYDROXIDE

1.0

1.0

1.0

CREATININE SOLUTION (GIVEN) STANDARD SOLUTION

OPTICAL DENSITY (O.D) AT 520nm

CALCULATION O.D of the Test – O.D of the Blank The amount of the Creatinine present in 100ml of the given sample = O.D of the Standard – O.D of the Blank

= ______________________ mg/dl.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF BLOOD UREA BY DAM’S METHOD AIM To estimate the amount of urea present in 100ml of the given sample.

PRINCIPLE Proteins present in blood are precipitated by Tungstic acid. Urea in protein free precipitate reacts with Diacetyl Monoxime to form yellow colour. The intensity of the colour developed is directly proportional to the amount of creatinine present and it is compared with standard similarly treated at 480nm.

REAGENTS 1.

2.

3.

4.

5.

6.

10% Sodium Tungstate 2/3 N Sulphuric acid Diacetyl Monoxime Acid mix (Sulphuric acid and phosphoric acid) Stock standard urea solution Working standard urea solution.

DEPROTEINATION OF PROTEIN To 1.0ml of blood add 3.3ml water, 3ml of 10% sodium Tungstate and 0.3ml of 2/3N Sulphuric acid. Mix and filter with Whatman no.1 filter paper. 1.0ml of protein free filtrate contains 0.25mg of blood (0.1/4=0.25mg).

PROCEDURE Colour development     

Label 3 test tubes as T, S, and B. Take 3.0ml of water in B, 2.0ml of water in S and T also. . Pipette out 1.0ml of the given urea solution in T, 1.0ml of standard in S To all the tubes add 0.4ml DAM reagent and 1.6ml of acid mix.

Keep all the tubes in boiling water bath for 15 minutes.

Cool the tubes and read them at 480nm.

RESULT The amount of Urea present in 100ml of the given solution = ______________________mg/100ml.

CLINICAL SIGNIFICANCE    

Normal blood Urea is <40 mg/dl.

When protein intake is high the level of urea is in upper range.

Blood urea concentration increases with age.

It is increased in malignant hypertension, vomiting, diahorrea.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF BLOOD UREA BY DAM’S METHOD REAGENTS

BLANK (ml) 3.0

STANDARD (ml) 2.0

TEST (ml) 2.0

STANDARD SOLUTION

-

1.0

-

UREA SOLUTION (GIVEN)

-

-

1.0

DAM

0.4

0.4

0.4

ACID MIX

1.6

1.6

1.6

WATER

OPTICAL DENSITY (O.D) AT 480nm

CALCULATION O.D of the Test – O.D of the Blank The amount of the Urea present in 100ml of the given sample = O.D of the Standard – O.D of the Blank

= ______________________ mg/dl.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF TOTAL PROTEIN BY BIURET METHOD AIM To estimate the amount of total protein present in the given solution.

PRINCIPLE The peptide bond present in the protein reacts with copper sulphate in alkaline medium to give violet colour, which is compared with the standard similarly treated at 540nm.

REAGENTS 1. Serum – unhemolysed.

2. Stock standard 3. Working standard biuret reagent PROCEDURE      

Mark three tubes as B, S AND T. Take 3.0ml of water in B and 2.9ml water in S and T. Pipette out 0.1ml standard to S and 0.1ml Serum to T. Add 3.0ml of Biuret reagent to all the tubes.

Mix thoroughly and incubate them at room temperature for 15 minutes.

The violet colour developed is read at 540nm.

RESULT The amount of protein present in serum sample = _______________________________.

CLINICAL SIGNIFICANCE      

Total protein concentration in serum = 63.75g/dl Albumin level = 3.6 – 5.2 g/dl.

Globulin level = 2-5 – 3.5g/dl.

Increased level is seen in multiple myeloma. Due to increase in globulin level.

Decreased because of lower albumin level.

Kwashiorkor is due to inadequate supply of dietary protein.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ESTIMATION OF TOTAL PROTEIN BY BIURET METHOD REAGENTS

BLANK (ml) 3.0

STANDARD (ml) 2.9

TEST (ml) 2.9

STANDARD SOLUTION

-

0.1

-

SERUM

-

-

0.1

3.0

3.0

3.0

WATER

BIRUET REAGENT OPTICAL DENSITY (O.D) AT 540nm

CALCULATION

The amount of the Urea present in

O.D of the Test – O.D of the Blank =

100ml of the given sample

Std. concentration x

O.D of the Standard – O.D of the Blank

x 100 Volume

= ______________________ mg/dl.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ANALYSIS OF CARBOHYDRATES S.No 1.

2.

3.

4.

5.

6.

EXPERIMENT Solubility test: To a little amount of sugar add a few drops of distilled water and mixed well.

OBSERVATION Soluble

Insoluble

INFERENCE Presence of Monosaccharides or Disaccharides Presence of Polysaccharides

PRINCIPLE POLYSACCHARIDES, BECAUSE OF THEIR SIZE, THESE ARE OFTEN TIMES NOT SOLUBLE IN WATER. Molish Test: A purple colour develops at the Presence of carbohydrates To 2ml of the sugar solution interface add 2 drops of molish reagent mixed well and then add 1ml of concentrated sulphuric acid along the sides of the test tube.

PRINCIPLE CARBOHYDRATES WHEN TREATED WITH CONCENTRATED SUPLHURIC ACID UNDERGO HYDRATION REACTION TO GIVE FURFURAL DERIVATIVES, WHICH CONDENSES WITH PHENOLIC COMPOUNDS LIKE α –NAPTHOL TO FORM COLOURED COMPOUNDS. Iodine Test: Blue colour precipitate develops Presence of Polysaccharides To a small amount of sugar No Characteristic change Absence of Polysaccharides solution add a few drops of iodine solution.

PRINCIPLE A COORDINATED COMPLEX IS FORMED BETWEEN THE HELICALLY COILED POLYSACCHARIDES AND IODINE. THE COLOUR OBTAINED DEPENDS ON THE LENGTH OF THE CHAIN. Fehling’s Test: A brick red colour precipitate Presence of Reducing sugar To 5ml of Fehling’s reagent add develops a small amount of the sugar No Characteristic change Absence of Reducing sugar solution and heat in the boiling water bath PRINCIPLE THE TEST DEPENDS ON THE FREE ALDO OR KETO GROUP OF THE REACTING COMPOUND. WHEN AN ALKALINE SOLUTION OF CUPROUS HYDROXIDE IS HEATED IN THE PRESENCE OF REDUCING SUGAR, IT IS REDUSED TO CUPROS OXIDE, WHICH GIVES A RED COLOUR. Benedict’s Test: Reddish brown precipitate develops Presence of Reducing sugar To 5ml of Benedict’s reagent add 8 drops of sugar solution No Characteristic change Absence of Reducing sugar and heat in the boiling water bath for 3 minutes and allow to cool spontaneously PRINCIPLE REDUCING SUGAR UNDER ALKALINE CONDITION TAUTOMERIZES TO FORM ENEDIOLS. THESE ENEDIOLS ARE POWERFUL REDUCING AGENTS WHICH REDUCES THE CUPRIC IONS TO CUPROUS IONS. CUPROUS HYDROXIE IS CONVERTED TO RED PRECIPITE CUPROUS OXIDE ON HEATING. Barfoed’s Test: Red colour precipitate develops at Presence of To 5ml of Barfoed’s reagent the bottom of the test tube monosaccharides add 1ml of sugar solution and No Characteristic change Presence of disaccharides heat for few minutes in boiling water bath.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

7.

8.

9.

PRINCIPLE THIS IS ALSO A COPPER RED TEST IN ACIDIC CONDITION. MONOSACCHARIDES REACTS VERY FAST WHERE AS REACTION IN DISACCHARIDES IS SLOW. HOWEVER PROLONGED HEATING OF DISACCHARIDES WILL GIVE A POSITIVE RESULT FOR THEM. Seliwanoff’s Test: Cherry red colour precipitate Presence of Keto sugar To 5 drops of sugar solutions develops add 5ml of seliwanoff’s reagent No Characteristic change Absence of Keto sugar and boil it for half a minute in direct flame.

PRINCIPLE KETO – HEXOSES ON TREATMENT WITH HYDROCHLORIC ACID FORM 5 HYDROXY METHYL FURFURALS WHICH ON CONDENSES WITH RESORSINOL TO GIVE A CHERRY RED COLOUR. Picric Acid Test: A mahogany red colour solution Presence of Reducing sugar To 5ml of the solution add 5ml develops of saturated picric acid and 1ml No Characteristic change Absence of Reducing sugar of 10% sodium carbonate and heat in a boiling water bath.

PRINCIPLE REDUCING SUGARS REACT WITH PRCRIC ACID TO FORM PICRAMIC ACID WHICH GIVES A RED COLOUR. Tollen’s Test: Red colour solution develops Presence of Pentose (or) To equal amount of sugar Galactose solution add concentrated No Characteristic change Absence of Pentose (or) hydrochloric acid and a pinch Galactose of phloroglucinol and heat in a boiling water bath.

PRINCIPLE THIS TEST RELIES ON REACTION OF THE FURFURAL WITH PHLOROGLUCINOL TO PRODUCE A COLORED COMPOUND WITH HIGH MOLAR ABSORPTIVITY.

10.

11.

Bial’s Test: Green colour develops Presence of Pentose To 5ml of the Bial’s reagent No Characteristic change Absence of Pentose add 2ml of sugar solution heat in a boiling water bath.

PRINCIPLE SUGARS FORM FURFURALS IN ACID MEDIUM WHICH CONDENSES WITH ORICINOL IN THE PRESENCE OF FERRIC IONS TO GIVE A GREEN COLOURED COMPLEX. Phenyl Hydrazine Test (or) Ozazone Test: Presence of Glucose, To 5ml of the sugar solution Fructose. add one spatula of phenyl hydazinehydrochloride, 2 Presence of Sucrose, Starch. spatulas full of sodium acetate Formation of sheaves of corn shaped (After hydrolysis) and 1ml of glacial acetic acid crystals. and mix well. Keep the solution in boiling water bath and cool it at room temperature. View the crystals under the microscope.

Presence of Lactose Formation of Yellow puff shaped crystals.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

Presence of Galactose Formation of Yellow sunflower shaped crystals.

Principle COMPOUNDS CONTAINING CARBONIC GROUP REACT WITH PHENYL HYDRAZINE TO GIVE OSAZONE CRYSTALS. THESE CRYSTALS HAVE CHARACTERISTIC SHAPES AND MELTING POINTS WHICH ARE USED FOR IDENTIFICATION OF REDUCING SUGARS.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

HYDROLYSIS OF SUCROSE S.NO

PROCEDURE To 2ml of sucrose solution 2 drops of concentrated hydrochloric acid was added and boiled for 1-3 minutes, in a boiling test tube and cooled.

To the cooled solution 20% sodium Carbonate and sodium hydroxide was added to neutralize the solution.

OBSERVATION

A pale yellow liquid is formed

INFERRENCE

Sucrose is converted to its constituents glucose and fructose.

Effervescence due to excess hydrochloric acid present in the hydrolyzed sucrose solution.

Divide the solution equally to perform Benedict’s test, Seliwanoff’s test and Osazone Tests.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

HYDROLYSIS OF STARCH 25ml of 1% of Starch solution and 10% concentrated hydrochloric acid were heated in a 100ml beaker. A the end of each minute, few drops of the solution was test with N/50 Iodine and Benedict’s solution and the colour changes were noted.

S.NO PROCEDURE 1. IODINE TEST To 2ml of the starch solution Few drops of Iodine solution was added

2. BENEDICT’S TEST To 2ml of Benedict’s reagent 8 drops of the hydrolyzed starch solution was added and heated to boil and the colour changes were observed.

OBSERVATION i.

Bluish colour appears. ii.

Red colour appears iii.

Partial colour is present. iv.

Colour disappears.

i. ii. iii. iv.

No change in colour Green colour appears Orange colour appears Brick red coloured precipitate Is formed.

INFERENCE Amylodextrin is present Erythrodextrin is present Acrodextrin is present Maltose is present.

Starch is present Amylodextrin is present Erythrodextrin is present Maltose is present.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

REACTION OF PROTEINS S.No EXPERIMENT 1.

Solubility Test: Cold water

2.

3.

4.

5.

OBSERVATION

INFERENCE

Insoluble

Hot water

Soluble

Dilute alkali

Soluble

Dilute Acid

Soluble

Proteins are insoluble in cold water and soluble in hot water and alkali

Molish Test: Formation of a violet Confirms the presence of amino To 2ml of the given solution add 2 drops colour ring at the group forming Ruhemann’s purple. of molish reagent mixed well and then junction of two liquids add 1ml of concentrated sulphuric acid along the sides of the test tube.

PRINCIPLE GLYCOPROTEINS WHEN TREATED WITH CONCENTRATED SUPLHURIC ACID UNDERGO HYDRATION REACTION TO GIVE FURFURAL DERIVATIVES, WHICH CONDENSES WITH PHENOLIC COMPOUNDS LIKE α –NAPTHOL TO FORM COLOURED COMPOUNDS. Biuret Test: A violet colour develops This is due to the presence of To 5ml of the given solution add 2ml of carbamino group present in protein.

Biuret reagent.

PRINCIPLE THIS IS A GENERAL TEST FOR THE COMPOUNDA HAVING PEPTIDE BOND. ALKALINE COPPER SULPHATE REACTS WITH COMPOUNDS CONTAINING 2 OR MORE PEPTIDE BOND TO GIVE A VIOLET COLOUR. THE COLOUR FORMATION IS DUE TO THE COORDINATION COMPLEX OF CUPRIC IONS WITH THE UNSHARDED ELECTRON PAIR OF NITROGEN AND OXYGEN F WATER. Ninhydrine Test: Formation of a blue Confirms the presence of amino To 2ml of the given solution add 2ml of colour solution group forming Ruhemann’s purple.

0.1% Ninhydrine solution heat and cool the solution.

PRINCIPLE NINHYDRIN REACTS WITH α AMINO GROUP TO FORM PURPLE COLOUR IT IS A POWERFUL OXIDISING AGENT AND CAUSE OXIDATIVE DECARBOXYLATION OF L AMINO ACID PRODUCING AN ALDEHYDE. THE REDUCED NINHYDRIN REACTS WITH AMMONIA TO FORM PURPLE COLOUR. PROLEIN AND HYDROXY PROLEIN GIVE YELLOW COLOUR. Xanthoproteic Test: Formation of a yellow Presence of aromatic group in To 2ml of the given solution add 1ml of precipitate which protein. concentrated nitric acid. Boil the solution disappears by the and cool it. To that solution add 40% addition of 40% sodium sodium hydroxide by drops. hydroxide to give a orange colour PRINCIPLE NITROGEN IN THE AROMATIC AMINO ACID IS NITRATED ON HEATING WITH NITRIC ACID AND GIVES A An online product of www.ebooks.elearninfinity.com,Unique Intellectuals Media Publications-India

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

6.

7.

8.

9.

10.

YELLOW COLOURED NITROPHENYL COMPOUND. THIS GETS IODISED IN ALKALINE CONDITION TO GIVES YELLOW COLOUR. Million Test: Formation of red colour Presence of phenyl hydroxyl group of To 5ml of the given solution add 1ml of solution Tyrosine. the sulphuric acid & 1ml of 10%mercuric sulphate (10%mercuric sulphate in 10N sulphuric acid). Place the solution in water bath for 10 minutes, cool it and add 1ml of 1% sodium nitrate solution.

PRINCIPLE SODIUM REACTS WITH SULPHURIC ACID TO FORM NITROUS ACIDS. MERCURIC ACID PRECIPITATES PROTEINS, WHICH ON BOILING EXPOSES THE REACTING GROUP. THIS GROUP REACTS WITH NITROUS ACID TO GIVE RED COLOUR PRECIPITATE. Hopkin’s Test: Formation of a violet Presence of Tryptophan To 2ml of the given solution add 2ml of colour ring at the Hopkin’s reagent. Mix thoroughly and add junction of two liquids 2ml of concentrated sulphuric acid slowly along the sides of the test tube.

PRINCIPLE SEVERAL ALDEHYDE REACTS WITH THE OXIDISED PRODUCT OF THE INDOL GROUP OF TRYPTOPHAN TO GIVE VIOLET COLOUR. SULPHURIC ACID WITH MERCURIC SULPHATE ACTS OXIDISING AGENT IN THE REACTION SAKAGUCHI TEST To 3ml of the solution 5 drops of 40% Caramel red colour is Arginine is present sodium hydroxide, 1% of α napthol are obtained added and 2- 3 drops of bromine water was added and cooled.

PRINCIPLE FREE ARGININE IN PROTEIN REACTS WITH α NAPTHOLAND ALKALINE HYPOBROMIDE TO GIVE BRIGHT RED COLOUR COMPLEX, THIS REACTION IS SPECIFIC FOR GUANIDIO GROUP. Sulphur Test To 1ml of Protein solution 1ml 0f 40% Brownish black colour is Sulphur containing amino acids are Sodium Hydroxide is added and heated for obtained present 1 minute and cooled. 1ml of 1% lead acetate is added and warmed PRINCIPLE THE ORGANIC SULPHUR PRESENT IN SULPHAR CONTAINING PROTEIN IN A ALKALINE MEDIUM REACTS WITH LEAD ACETATE TO FORM BLACK PRECIPITATE OF LEAD SULPHATE Molish Test: A purple colour develops Presence of carbohydrates To 2ml of the sugar solution add 2 drops at the interface of molish reagent mixed well and then add 1ml of concentrated sulphuric acid along the sides of the test tube.

PRINCIPLE CARBOHYDRATES WHEN TREATED WITH CONCENTRATED SUPLHURIC ACID UNDERGO HYDRATION REACTION TO GIVE FURFURAL DERIVATIVES, WHICH CONDENSES WITH PHENOLIC COMPOUNDS LIKE α –NAPTHOL TO FORM COLOURED COMPOUNDS.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ANALYSIS OF AMINOACIDS S.No EXPERIMENTS 1.

Solubility Test: Cold water Hot water Dilute potassium hydroxide Dilute ammonium hydroxide Dilute hydrochloric acid Alcohol 2.

Ninhydrine Test: To 2ml of the given solution add 2ml of 0.1% Ninhydrine solution heat and cool the solution.

3.

Pauli’s Test: To 2ml of the given solution add 1ml of 1% sulphanilic acid and cool it. Then add 1ml of 5%sodium nitrate and leave for 3 minutes. Make the solution alkaline by the addition of 2ml of 1% sodium carbonate and observe the change.

4.

Test for Histidine: To 2ml of the given solution add 1% bromine in 33% acetic acid until the yellow colour is formed. After 10 minutes add 2ml of 5% ammonium carbonate and place it in a water bath for 10 minutes.

5.

Millon’s Test: To 2ml of the given solution add few drops of million’s reagent and place it in the water bath for 15 minutes. Then add 1ml of sodium Nitrate solution.

6.

Folin’s Test: To 2ml of the given solution add 1ml of Folin’s reagent and add 1ml of sodium carbonate solution.

7.

Hopkin’s cole test: To 2ml of the given add 2ml of Hopkin’s cole reagent and mix thoroughly. Then add 2ml of concentrated sulphuric acid along the sides of the test tube.

8.

Sakaguchi Test: To 2ml of the given solution add 15%

OBSERVATION Soluble Soluble Soluble Soluble Soluble Insoluble Formation of a violet colour solution

Formation of red colour solution

No Characteristic change

INFERENCE The given solution is solution is soluble in all given solution except alcohol.

Confirms the presence of amino group forming Ruhemann’s purple.

Amino acid combines with diazotised sulphate in an alkaline medium and forms the coloured azo compounds.

(Tyrosine and Histidine) Absence of Tyrosine and Histidine

Formation of red colour solution

Presence of Histidine.

No Characteristic change

Absence of Histidine.

Formation of deep red colour solution

Presence of Tyrosine.

No Characteristic change

Absence of Tyrosine.

Formation of blue colour solution No Characteristic change Formation of a violet colour ring at the junction of two liquids No Characteristic change

Presence of tyrosine and Tryptophan (aromatic amino acids) Absence of tyrosine and Tryptophan

Formation of red colour solution

Presence of Arginine

Presence of Tryptophan

Absence of Tryptophan

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

9.

10.

12.

sodium hydroxide and 2 drops of α Napthol. Mix thoroughly and add 4-5 drops of Bromine water.

Lead acetate test: To 2ml of the given solution, add few drops of 10% lead acetate until the formation of a white precipitate. Then add 0.5ml of 40% sodium hydroxide and boil for 1 minute.

Mccarthy Sullivan’s Test: To 2ml of the given reagent add the following reagent and mix after each addition 1. 1 drop of 20% sodium hydroxide 2. 1ml of 1% glysine 3. 0.3ml of 10% sodium nitro prusside Place the solution in water bath at 40⁰C for 15 minutes and then add 1ml of 5N hydrochloric acid drop by drop and cool the contents at room temperature.

Nitro prusside Test: Treat 2ml of the given solution with 2% sodium nitro prusside and a drop of 20% sodium hydroxide.

No Characteristic change

Absence of Arginine

Formation of black colour precipitate

Presence of cysteine

No Characteristic change

Absence of cysteine

Formation of red colour solution

Presence of Methionine is confirmed by the red colouration in which the methyl group is split off to form homocysteine and reacts with sodium nitro prusside.

Absence of Methionine

No Characteristic change

Formation of red colour solution which fades after two minutes No Characteristic change

It is due to the presence of thiol group in cysteine.

Absence of Cysteine.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

PRECIPITATION REACTION OF PROTEINS S.NO PROCEDURE

OBSERVATION

INFERENCE

1. PRECIPITATION OF HEAVY METALS To 3ml of protein solution 5% mercuric nitrate White precipitate is formed due Protein is present or sulphate solution is added drop by drop To heavy metal complex PRINCIPLE IN HIGHER pH PROTEIN EXISTS AS NEGATIVELY CHARGED ANION. TO THIS SOLUTION IF A SALT OF HEAVY METAL IS ADDED, POSITIVELY CHARGED COMPLEX REACTS NEGATIVELY CHARGED PROTEIN TO FORM A PRECIPITATE. 2. PRECIPITATION BY ALCOHOL To 3ml of protein solution 2ml of alcohol solution White precipitate is formed due Alcohol precipitates Is added to dehydrogenation protein Due to dehydrogenation PRINCIPLE ALCOHOL REMOVES THE CHARGES OF PROTEIN BY DRHYDRATION AND DENATURATION. 3. PRECIPITATION BY ALKALOIDS To 3ml of protein solution 2ml of White precipitate is formed Proteins can be phosphomolybdic acid or precipitated sulphosalicylic acid is added By alkaloid solution PRINCIPLE ALKALOIDS HAVE NEGATIVE CHARGE WHICH REACTS WITH THE POSITIVE CHARGE OF PROTEIN TO FORM A PRECIPITATE. 4. HALF AND FULL SATURATIONBY AMMONIUM SULPHATE To 5ml of protein solution add 5 ml of ammonium The white turbidity is formed due Globulin is present Sulphate. to precipitation of the solution, this is half saturation.

The content of the tube is allowed to stand for A white precipitate is obtained in Albumin is present 5- 10 minutes and filtered by using Whatman full saturation no.1 filter paper. The filtrate contains only albumin. This is precipitated by saturated ammonium sulphate PRINCIPLE AMMONIUM SALT PRECIPITATE PROTEINS BY NEUTRALISATION OF CHARGES ON THE PROTEIN BY DEHYDRATION. An online product of www.ebooks.elearninfinity.com,Unique Intellectuals Media Publications-India

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ANALYSIS OF LIPID S.No EXPERIMENTS 1.

Solubility Test: Water Ether Ethyl alcohol 2.

Grease spot Test: Put a drop of Oil over a piece of ordinary writing paper 3.

Oil spot Test: Mix a drop of oil with 2ml ether and transfer the solution to a china dish and subject it to evaporation.

4.

Emulsification Test: To 2ml of water in a test tube add one drop of oil and shake vigorously and add few drops of soap solution and mix well 5.

Halogenation Test (Unsaturation in fats): In a series of two test tubes add the following 1. 3ml of chloroform 2. Bromine water 6.

Test with Cholesterol: 1. Microscopic Examination

2. Appearance 7. a.

b.

Colour reaction of sterol Salkawoski Reaction: To5ml of a solution in a dry test tube add equal volume of concentrated sulphuric acid along the sides of the test tube Libermann Burchard Reaction: To 2ml of the solution in a dry test tube add chloroform and 2ml of acetic anhydride and 2-3 drops of concentrated sulphuric acid. Mix well and stand for few minutes in dark

OBSERVATION Insoluble Soluble Soluble Formation of greasy appearance

INFERENCE

Presence of Lipid Presence of Fat

Formation of an oil spot

Presence of Lipids

Formation of minute droplets which floats on the surface

Presence of Fat

Solution discharge bromine colour (orange)

Presence of unsaturated fat

Presence of Cholesterol

Colourless and Rhombic in Shape A change of colour from bluish red to violet.

Formation of emerald green colour solution

Presence of Cholesterol

Presence of Cholesterol

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ANALYSIS OF NORMAL URINE S.NO PROCEDURE OBSERVATION RESULT 1.

TEST FOR CHLORIDE To 5ml of urine few drops of concentrated Formation of curdy Normal urine Nitric acid and 3ml of silver nitrate white precipitate contains chloride were added PRINCIPLE SLIVER IN SILVER NITRATE REACTS WITH NITRIC ACID TO FORM A WHITE PRECIPITE OF SILVER CHLORIDE 2.

TEST FOR CALCIUM AND PHSPHROUS To 10ml of urine concentrated Ammonium hydroxide is added and heated to boil and cooled. A flasky precipitate of calcium phosphate is formed which is filtered. To the precipitate 2ml of 10% acetic acid is added through the sides of the paper and divide into two parts.

TEST FOR CALCIUM To one part of the filtrate 1ml of 2% potassium oxalate was Formation of white Normal urine added. precipitate contains calcium TEST FOR PHOSPHOROUS To another part of the filtrate 3 drops of ammonium Formation of canary Normal urine Molybdate and concentrated nitric acid were added and yellow precipitate contain warmed. phosphorous PRINCIPLE WHITE PRECIPITATE IN CASE OF CALCIUM IS DUE TO THE FORMATION OF CALCIUM OXALATE. AND CANARY YELLOW COLOUR IS DUE TO THE FORMATION OF AMMONIUM PHOSPHATE. 3.

TEST FOR SULPHATE To 5ml of urine 1ml of concentrated hydrochloric acid and White precipitate is Normal urine 5ml of barium chloride were added formed contains sulphate PRINCIPLE WHITE PRECIPITATE IS DUE TO THE FORMATION OF BARIUM SULPHATE 4.

TEST FOR AMMONIA To 2.5ml of urine one drop of phenolphthalein and 2% Formation of Pink Normal urine sodium carbonate drops were added and boiled. colour at the end of contain Ammonia Glass rod dipped in phenolphthalein is held at the mouth of glass rod the test tube.

5.

TEST FOR THE ORGANIC CONTENTS IN URINE a.

JAFF’S TEST FOR CREATININE To 5ml of urine 1ml of picric acid and 10 drops of 10% sodium Formation of Reddish Normal urine hydroxide were added orange colour contains creatinine PRINCIPLE CREATINIE REACTS WITH PICRIC ACID IN AN ALKALINE MEDIUM TO GIVE REDDISH ORANGE COLOUR. b.

SCHIFF’S TEST FOR URIC ACID To 5 drops of urine 2 drops of 2% sodium carbonate was Formation of Metallic Normal urine added and filtered. The filter paper is moisture with silver colour contains uric acid ammonium silver nitrate.

PRINCIPLE URIC ACID REDUCES AMMONICAL SILVER NITRATE TO METALLIC SILVER An online product of www.ebooks.elearninfinity.com,Unique Intellectuals Media Publications-India

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi c.

TEST FOR UREA To 2ml of urine, add 1ml of 40% sodium hydroxide. Heat the Red litmus paper Normal urine contents. Note the smell of ammonia. Test the vapour by change to blue contain urea holding a piece of red litmus paper at the mouth of the test tube.

PRINCIPLE UREA DECOMPOSES TO GIVE AMMONIA AND SODIM CARBONATE ON HEATING.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

ANALYSIS OF ABNORMAL URINE S.NO PROCEDURE OBSERVATION INFERENCE 1.

TEST FOR PROTEIN HEAT AND COAUGLATION TEST To a test tube fill 3/4th of urine. Add 2% acetic acid The upper part of the Abnormal Urine and heat the upper portion of the test tube by keeping it in a tube gets coagulated contains protein slanting position PRINCIPLE URINE CONTAINS MAILNLY ALBUMIN WHICH IS A HEAT COAGUBLE PROTEIN 2.

SULPHOSALICYLIC TEST To 5ml of urine 1ml of sulphosalicylic acid was added.

Formation of white Abnormal Urine precipitate contains protein PRINCIPLE SULPHOSALICYLIC ACID REMOVES THE CHARGES OF THE PROTEIN TO FORM A PRECIPITATE 3.

HELLER’S TEST To3ml of concentrated nitric acid, 2ml of urine was added along Formation of a white Abnormal Urine the sides of the test tube ring at the junction of contains protein two layers PRINCIPLE NITRIC ACID CAUSES THE PRECIPITATION OF PROTEIN 4.

TEST FOR BLOOD UREA AND HEMATUREA BENZIDINE TEST Warm 1ml of urine. Add 2ml of glacial acetic acid and benzidine Formation of blue or Abnormal Urine power. green colour contains blood PRINCIPLE THE HEME PART OF THE HEMOGLOBIN ACT AS A CATALYST AND DECOMPOSES HYDROGEN PEROXIDE. THE NACENT OXYGEN OXIDISES BENZIDINE TO GIVE A GREEN OR BLUE COLOUR 5.

TEST FOR SUGAR To 5ml of Benedict’s reagent add 8 drops of urine is added and Formation of red Abnormal Urine heated for 2 minutes and cooled. precipitate contains reducing sugar PRINCIPLE REDUCING SUGAR UNDER ALKALINE CONDITION TAUTOMERIZES TO FORM ENEDIOLS. THESE ENEDIOLS ARE POWERFUL REDUCING AGENTS WHICH REDUCES THE CUPRIC IONS TO CUPROUS IONS. CUPROUS HYDROXIE IS CONVERTED TO RED PRECIPITE CUPROUS OXIDE ON HEATING. 6.

TEST FOR KETONE BODIES ROTHERA’S TEST Saturate 5ml of urine with ammonium sulphate and add 5 drops Formation of purple Abnormal Urine of 5% sodium nitropruside and few drops of ammonia colour contains acetone PRINCIPLE NITROPRUSIDE IN ALKALINE MEDIUM REACTS WITH KETONE GROUP TO FROM PERMANGANATE COLOUR. 7.

GERHARDT’S TEST To 5ml of urine add 10% ferric chloride drop by drop. The yellow Formation of red Abnormal urine precipitate formed is filtered and 2 drops of ferric chloride is colour contains acetic added acid PRINCIPLE ACETIC ACID IN URINE REACTS FERRIC CHLORIDE TO GIVE A RED COLOUR An online product of www.ebooks.elearninfinity.com,Unique Intellectuals Media Publications-India

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi 8.

9.

TEST FOR BILE SALT HAY’S TEST Take ½ test tube urine and add sulphur power gently.

Sulphur will not sink

Abnormal urine contains bile salt

PRINCIPLE HAY’S TEST IS BASED ON THE FACT THAT BILE SALT LOWERS THE SURFACE TENSION AND ALLOWS THE SULPHUR POWDER TO SINK. TEST FOR BILE PIGMENT FOUCHET’S TEST To 5ml of urine add a few crystals of magnesium sulphate. Shake Formation of green Abnormal urine the tube till it dissolves and add 10% barium chloride. Filter the colour contains bile contents of the tube and discard the filter. Remove the moisture pigments by placing another filter paper. Now add Fouchet’s reagent to the dry filter paper.

PRINCIPLE THE FORMATION GREEN COLOUR IS DUE TO THE OXIDATION OF BILIRUBIN TO BILIVERDIN.

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Mrs. R. Janani HAND BOOK FOR PRACTICAL BIOCHEMISTRY Dr.K.M.Tulaanidi

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