Practical Biochemistry PDF

Practical Biochemistry  Practical Biochemistry Geetha Damodaran K MD Associate Professor Department of Biochemistry Government Medical College,Thrissur, Kerala, India...

Practical Biochemistry  Practical Biochemistry Geetha Damodaran K MD Associate Professor Department of Biochemistry Government Medical College,Thrissur, Kerala, India ® JAYPEE BROTHERS MEDICAL PUBLISHERS (P) LTD Kochi St Louis (USA) Panama City (Panama) London (UK) New Delhi Ahmedabad Bengaluru Chennai Hyderabad Kolkata Lucknow Mumbai Nagpur Published by Jitendar P Vij Jaypee Brothers Medical Publishers (P) Ltd Corporate Office 4838/24 Ansari Road, Daryaganj, New Delhi - 110002, India, Phone: +91-11-43574357 Fax: +91-11-43574314 Registered Office B-3 EMCA House, 23/23B Ansari Road, Daryaganj, New Delhi - 110 002, India Phones: +91-11-23272143, +91-11-23272703, +91-11-23282021 +91-11-23245672, Rel: +91-11-32558559, Fax: +91-11-23276490, +91-11-23245683 e-mail: [email protected], Website: www.jaypeebrothers.com Offices in India Ahmedabad, Phone: Rel: +91-79-32988717, e-mail: [email protected] Bengaluru, Phone: Rel: +91-80-32714073, e-mail: [email protected] Chennai, Phone: Rel: +91-44-32972089, e-mail: [email protected] Hyderabad, Phone: Rel:+91-40-32940929, e-mail: [email protected] Kochi, Phone: +91-484-2395740, e-mail: [email protected] Kolkata, Phone: +91-33-22276415, e-mail: [email protected] Lucknow, Phone: +91-522-3040554, e-mail: [email protected] Mumbai, Phone: Rel: +91-22-32926896, e-mail: [email protected] Nagpur, Phone: Rel: +91-712-3245220, e-mail: [email protected] Overseas Offices North America Office, USA, Ph: 001-636-6279734 e-mail: [email protected], [email protected] Central America Office, Panama City, Panama, Ph: 001-507-317-0160 e-mail: [email protected] Website: www.jphmedical.com Europe Office, UK, Ph: +44 (0) 2031708910, e-mail: [email protected] Practical Biochemistry © 2011, Jaypee Brothers Medical Publishers (P) Ltd. All rights reserved. No part of this publication should be reproduced, stored in a retrieval system, or transmitted in any form or by any means: electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author and the publisher. This book has been published in good faith that the material provided by author is original. Every effort is made to ensure accuracy of material, but the publisher, printer and author will not be held responsible for any inadvertent error (s). In case of any dispute, all legal matters are to be settled under Delhi jurisdiction only. First Edition: 2011 ISBN 978-93-5025-141-6 Typeset at JPBMP typesetting unit Printed at  To my father (late) Sri KV Damodaran  Preface Nearly two decades of teaching experience have driven me to write this book. I realized that if an illustrated book is available, students will be able to recollect the experiments done earlier, to face the different types of questions during practical examinations. Hence all the items in this book are illustrated. The contents of this book are structured in the practical examination-oriented manner. The major sections are qualitative experiments, quantitative experiments, charts, spotters and objective structured practical examination questions. All the tests are provided with diagrams and interpretations. This will help the students to understand each concept thoroughly and enable them to use it as an instant doubt clearing book. I hope it will be very useful for day-to-day studies and exam preparations. Details of reagent preparations given along with the respective chapters are useful for the staff involved in the laboratory preparation of practical sessions. This part will also help to improve the level of understanding of students about the reagents they are using for various experiments in the laboratory. Questions provided with the chapters are useful for having better clarity and grasp of the topic. Moreover, it will definitely boost the confidence of students to face the examination. Chapters on charts and spotting and OSPE questions are useful for self-training of such type of evaluation methods. I warmly welcome the views of those using the book and I shall be grateful to the readers for bringing to my notice of mistakes for corrections, in future editions of the book. Geetha Damodaran K  Acknowledgments I would like to thank God for enabling me to do this work. I thank my parents, teachers for molding me to reach this level. I extend my gratitude to my colleagues for their support. I should thank my husband Dr PK Balachandran for constantly persuading me to write.  Contents SECTION ONE: QUALITATIVE ANALYSIS 1. Reactions of Carbohydrates........................................................................................................... 3 2. Reactions of Proteins..................................................................................................................... 18 3. Reactions of Lipids........................................................................................................................ 37 4. Reactions of Urea........................................................................................................................... 42 5. Reactions of Creatinine................................................................................................................. 45 6. Uric Acid......................................................................................................................................... 47 7. Scheme for Identification of Biologically Important Substance in a Given Solution........... 50 8. Urine Analysis............................................................................................................................... 51 9. Spectroscopy.................................................................................................................................. 65 10. Reactions of Milk........................................................................................................................... 72 SECTION TWO: QUANTITATIVE ANALYSIS 11. Principles of Colorimetry............................................................................................................. 77 12. Determination of Blood Sugar..................................................................................................... 83 13. Determination of Urea.................................................................................................................. 90 14. Determination of Creatinine........................................................................................................ 94 15. Determination of Total Protein and Albumin........................................................................... 98 16. Determination of Cholesterol.................................................................................................... 102 17. Determination of Uric Acid....................................................................................................... 105 18. Determination of Bilirubin......................................................................................................... 109 19. Determination of Transaminases.............................................................................................. 113 20. Determination of Alkaline Phosphatase.................................................................................. 119 21. Determination of Calcium.......................................................................................................... 122 22. Determination of Phosphorus................................................................................................... 127 23. Determination of Titrable Acidity and Ammonia in Urine.................................................. 131 24. Determination of Urine Chloride.............................................................................................. 135 SECTION THREE: CHARTS 25. Charts............................................................................................................................................ 141  Practical Biochemistry SECTION FOUR: SPOTTERS 26. Spotters......................................................................................................................................... 173 SECTION FIVE: OSPE (OBJECTIVE STRUCTURED PRACTICAL EXAMINATION) QUESTIONS 27. OSPE (Objective Structured Practical Examination) Questions........................................... 213 Index............................................................................................................................................................ 225 xii SECTION ONE Qualitative Analysis  Reactions of Carbohydrates 1 1A. REACTIONS OF in urine in diabetes mellitus, fructose in urine in MONOSACCHARIDES fructosuria , galactose in urine in galactosemia). Hence, it is essential to understand the tests for their detection. INTRODUCTION The classification of carbohydrates will be Carbohydrates are aldehyde or ketone useful for the detection of various types of derivatives of polyhydric alcohols. They are carbohydrates by different chemical tests. widely distributed in plants and animals. Plants synthesize glucose by photosynthesis and it is CLASSIFICATION converted mainly to storage form, the starch and structural frame work form, the cellulose. 1. Monosaccharides: Cannot be hydrolyzed into Animals largely depend on plant source to simpler carbohydrates. They are classified into obtain carbohydrates though they can synthesize trioses,tetroses,pentoses, hexoses, heptoses based carbohydrates from non carbohydrates sources on the number of carbon atoms present in them. like glycerol and amino acids in their body They are again divided into aldoses and ketoses (gluconeogenesis). based on the functional group present in them The glucose is the major form of carbohydrate (see Table 1A-1). absorbed from the gut in humans. Table 1A-1: Classification of Monosaccharides According to the metabolic status it has different fates– Monosaccharides Aldoses Ketoses catabolized to release energy Trioses Glycerose Dihydroxyacetone polymerized to form the storage fuel—the Tetroses Erythrose Erythrulose glycogen Pentoses Ribose Ribulose sometimes converted to other sugars like Hexoses Glucose Fructose fructose and galactose. 2. Disaccharides: Give rise to two monosac- Different carbohydrates are present in charide units upon hydrolysis intracellular and extracellular fluids and are E.g.: Sucrose (glucose + fructose) excreted in urine when the concentration of them Lactose (glucose + galactose) rises in the blood as in certain diseases (glucose Maltose (glucose + glucose)  Qualitative Analysis 3. Oligosaccharides: Yields less than ten thoroughly. Add 3 ml of concentrated sulphuric monosaccharides. acid along the sides of the test tube by slightly E.g.: Maltotriose (3 glucose units), inclining the tube, thus forming a layer of acid Raffinose (glucose + fructose + galactose) (acid being heavier goes down beneath the sugar 4. Polysaccharides: Contain more than ten solution) in the lower part. Observation: A reddish violet ring appears monosaccharide units at the junction of two liquids. (i) Homopolysaccharides (consisting of same Inference: Indicates presence of a type of monomeric units) carbohydrate and hence the presence of Polymer of glucose: Starch, glycogen, cellulose monosaccharide. Polymer of fructose: Inulin Principle: Concentrated acid dehydrates the (ii) Heteropolysaccharides (consisting of sugar to form furfural (in the case of pentoses) different types of monomeric units) or furfural derivatives (hexoses and heptoses ) Proteoglycans, e.g. Heparin (D-glucosamine which then condense with α-naphthol to give a sulfate + D-sulfated iduronic acid) reddish violet colored complex Hyaluronic acid (D-β glucuronic acid + N- Application of the test: Used as a general test acetylglucosamine). to detect carbohydrates. REACTIONS OF MONOSACCHARIDES Aberrant Observations Monosaccharides possess one or more hydroxyl 1. Instead of a violet ring in the Molisch test, groups and an aldehyde or keto group. Therefore appearance of dark brown color indicates many reactions of monosaccharides are the charring of sugar due to the heat generated known reactions of alcohols,aldehydes or during the addition of acid (acid water ketones. Many of the reactions shown by interaction generates heat). It will become monosaccharides are exhibited by higher obvious when the concentration of the sugar carbohydrates also. Differences in the structures solution is high. To avoid charring, dilute the of sugars often affect the rate of a reaction and sugar sample solution with water as depicted in sometimes the ability to react. figure 1A-2 and repeat the Molisch test. The reactions described below, are applied in 2. Appearance of a green color while doing the test, the identification of sugars. which persist even after completion of the test The reactions due to hydroxyl group: suggest excess use of Molisch reagent than required – Dehydration (e.g. Molisch test, Rapid furfural or due to the presence impurities in the reagent. test, Seliwanoff’s test ) The reactions due to carbonyl group: 2. Benedict’s Test (Fig. 1A-3) – Reduction (e.g. Benedict’s test, Barfoed’s test) Procedure: To 5 ml of Benedict’s reagent in a test – Condensation (e.g. Osazone test) tube add exactly 8 drops of the sugar solution. Mix well. Boil the solution vigorously for two α -Naphthol Reaction) 1. Molisch Test (α minutes or place in a boiling water bath for three (Fig. 1A-1) minutes. Allow the contents to cool by keeping Procedure: To 5 ml of sugar solution in a test in a test tube rack. Do not hasten cooling by tube add two drops of Molisch reagent. Mix immersion in cold water. 4  Reactions of Carbohydrates 1 Furfural and hydroxymethylfurfural condense with phenolic (alpha naphthol in Molisch test) compounds to give rise to colored products. Fig. 1A-1: Chemistry of Molisch test Fig. 1A-2: Method to avoid charring Observation: The entire body of the solution In the absence of reducing substance, blue will be filled with a precipitate, the color of which color of the Benedict’s reagent remains as such. varies with the concentration of the sugar The test is sensitive up to 0.1-0.15 gm% of sugar solution—green, yellow, orange or red. solution (that is Benedict’s test will not be 5  Qualitative Analysis positive with solutions containing less than reduce various metallic ions. In this test cupric 0.1-0.15 gm% of sugar). ions are reduced to cuprous ions by the enediols Inference: Reducing monosaccharides, formed from sugars in the alkaline medium of glucose, fructose, galactose and mannose give a Benedict’s reagent. positive reaction with Benedict’s reagent. Benedict’s reagent contains copper sulphate, The color of the precipitate give an idea about sodium citrate and sodium carbonate. the concentration of the sugar solution as shown Copper sulphate dissociate to give sufficient below. cupric ions (in the form of cupric hydroxide) for Blue – absence of reducing sugar the reduction reactions to occur. Green – up to 0.5 gm% Sodium citrate keeps the cupric hydroxide in Yellow – > 0.5 to 1.0 gm% solution without getting precipitated. Orange – > 1.0 to 2.0 gm% Sodium carbonate (Na2CO3 ) make the pH of Brick red – ≥ 2 gm% the medium alkaline. Thus, Benedict’s test is described as a semi- In the alkaline medium sugars form enediols quantitative test. which are powerful reducing agents. They Principle: (see Fig. 1A-4) Carbohydrates with reduce blue cupric hydroxide to insoluble yellow a free aldehyde or keto group have the ability to to red cuprous oxide. Application of the test: To detect reducing sugars. It is widely used in detecting glucose in urine even though not specific for glucose. 3. Barfoed’s Test (Fig. 1A-5) Procedure: To 5 ml of Barfoed’s reagent in a test tube add 0.5 ml of sugar solution. Mix well. Keep in a boiling water bath for 2 minutes. Keep the Fig. 1A-3: Benedict’s test at different sugar tube in a test tube rack and examine for concentrations precipitate after 10-15 minutes. Fig. 1A-4: Chemistry of Benedict’s test 6  Reactions of Carbohydrates 1 Unlike the Benedict’s test, Barfoed’s test is unsuitable for testing sugars in urine or any fluids containing chloride. The red precipitate is formed at the bottom of the tube. To see the precipitate, lift the tube to the eye level, otherwise the precipitate formed adhering to the bottom most part of the tube may escape notice. Application of the test: Useful to distinguish between monosaccharides and disaccharides. Chemistry of the test: Reduction reaction as shown under Benedict’s test. Fig. 1A-5: Barfoed’s test 4. Rapid Furfural Test Observation: A red precipitate clinging to the bottom most part of the test tube forms, in the Procedure: To 2 ml of sugar solution add 6 drops presence of a monosaccharide. of Molisch reagent and 3 ml of concentrated Inference: The test is answered by monosac- HCl. Boil for 30 seconds only. charides only, e.g. glucose, fructose, galactose, Observation: Positive reaction is indicated by the development of violet color (Fig. 1A-6). mannose. Inference: Development of violet color within Principle: It is a reduction test. Reducing 30 seconds of boiling indicates presence of a keto property owes to the carbonyl group (aldehyde sugar, e.g. fructose. or keto group). Barfoed’s reagent is copper Principle: A dehydration reaction which owe acetate in acetic acid. to the hydroxyl groups of the sugar. Concentrated Difference between Barfoed’s test and HCl being weaker than concentrated sulphuric Benedict’s test: Barfoed’s test differs from acid, dehydrate ketoses (e.g. fructose) more readily Benedict’s test with respect to the pH of the than aldoses to form hydroxymethyl furfural, medium. It is alkaline in the case of Benedict’s which then condenses with α-naphthol to form a and acidic in the case of Barfoed’s test. In the acid violet colored complex. medium monosaccharides enolize much more readily than disaccharides and these enediols reduce cupric ions released by copper acetate of Barfoed’s reagent to produce a red precipitate. Points to Ponder It is important to keep the time limit (2 minutes) prescribed for Barfoed’s test otherwise disaccharides will also respond to the test positively. Disaccharides when present in high concentrations (> 5 gm%) also will give positive response Fig. 1A-6: Positive rapid furfural test 7  Qualitative Analysis Chemistry of the test: Dehydration reaction Principle: A dehydration reaction due to the as shown under Molisch test. hydroxyl groups of the sugar. Selivanoff’s Aberrant reaction: If red color develops reagent is resorcinol in dilute hydrochloric acid. instead of violet color due to charring action of Ketoses (e.g. fructose) are more readily acid, dilute the sugar sample with water and dehydrated by HCl than the aldoses to form conduct the test with diluted sugar solution (Fig. hydroxymethyl furfural which then condenses 1A-7). with resorcinol of Seliwanoff’s reagent to form a red colored complex. Application of the Test Points to Ponder For the detection of ketoses. Useful for differentiating ketoses from aldoses. The test is sensitive up to 0.1 gm% of fructose in the absence of glucose. In the presence of glucose, the test becomes less sensitive to fructose. Large amounts of glucose gives the same color. If the boiling is prolonged a positive reaction may occur with glucose because of Lobry de Bruyn-van Ekenstein transformation of glucose into fructose in the presence of acid. The precautions to be followed to get a positive test for fructose are given below: Fig. 1A-7: Method to avoid charring response in rapid furfural test 1. Concentration of HCl used must be less than 12%. 5. Seliwanoff’s Test 2. The reaction must be observed within 20 to 30 seconds of performing the test. Procedure: To 5 ml of Seliwanoff’s reagent in a 3. Those reactions occurring after 20 -30 seconds, test tube add 5 drops of fructose solution and must not be taken into account. heat the contents to just boiling. 4. Glucose must not be present in amounts more Observation: Positive reaction gives a red than 2% or else it will interfere with the test. color within half a minute (Fig. 1A-8). Inference: This test is given by ketoses. 6. Osazone Test e.g. fructose. Procedure: To 5 ml of sugar solution in a test tube add 300 mg (one or two scoopfuls) of phenyl hydrazine mixture. Shake well. Heat in a boiling water bath for 15 minutes. Then take the tube out of the water bath and allow cooling at room temperature by placing it in the test tube rack. Avoid showing under the tap water because rapid cooling disturbs crystallization where as slow cooling ensures crystallization (ideally within the water bath itself). Fig. 1A-8: Positive Seliwanoff’s test 8  Reactions of Carbohydrates 1 Observation: Crystals are formed readily Inference: Glucose, fructose, mannose yield (within 1-5 minutes) at the room temperature in the same shaped phenyl osazone crystals because the case of mannose. For other sugars minimum of the elimination of differences in configuration time required in minutes in the water bath for about the carbon atoms 1 and 2 during osazone the formation of insoluble yellow osazone is formation. given in the Table 1A-2. Principle: The reaction involves the carbonyl Look under the microscope to view the carbon (either aldehyde or ketone as the case may crystals (see Fig. 1A-9). be) and the adjacent carbon. One molecule of sugar reacts with one molecule of phenyl- Table 1A-2: Time of Formation of Osazones hydrazine to form phenylhydrazone which then Sugar Time (minutes) reacts with two additional phenyl hydrazine Glucose 5 molecules to form the osazones as shown in the Fructose 2 figure 1A-10. Galactose 20 Points to Ponder If the solution appears red after heating process, it indicates that the solution has become concentrated in the boiling process and no crystals will separate in the concentrated form. So dilute with water for the separation of Fig. 1A-9: Osazone crystals crystals. Fig. 1A-10: Chemistry of Osazone test 9  Qualitative Analysis 1B. REACTIONS OF DISACCHARIDES galactose is involved in the β galactoside bond with the carbon number 4 of glucose. Sucrose (α α -D-glucopyranosyl-β β -D fructo- INTRODUCTION furanoside): (see Fig. 1B-3). Disaccharides are glycosides in which both Hydrolysis of sucrose yields one molecule of components are monosaccharides. The general glucose and one molecule of fructose. Sucrose is formula of common disaccharides is C12H22O 11. dextrorotatory. After hydrolysis by enzymes or The common disaccharides studied are detailed weak acids , it becomes levorotatory. This is below. because of the formation of fructose upon Maltose (α α -D-glucopyranosyl-(1→ → 4) α -D- hydrolysis, which is strongly levorotatory than glucopyranose) (Fig. 1B-1): Maltose yield 2 the glucose.Thus the change of optical rotation glucose molecules upon hydrolysis. Maltose is of sucrose solution from dextro to levo rotation formed from the hydrolysis of starch by the upon hydrolysis is known as inversion and the action of the enzyme maltase. It is also produced mixture of glucose and fructose obtained is called as an intermediate product of mineral acid invert sugar. hydrolysis of starch. It is dextrorotatory, exhibits Sucrose do not reduce metallic ions (do not mutarotation, reduces metallic ions in alkaline answer Benedict’s and Barfoed’s tests) and also solutions. Like other disaccharides maltose is do not form osazone with phenylhydrazine. hydrolyzed by dilute acid leading to the But prolonged boiling with phenylhydrazine in formation of two molecules of glucose. With acid medium will form osazone due to the phenyl hydrazine maltose forms maltosazone. reaction of products of hydrolysis of sucrose with Examples for other disaccharides that produce only glucose upon hydrolysis: – Cellobiose a β glucoside with 1,4 linkage derived from partial hydrolysis of cellulose. – Gentiobiose, a β glucoside with 1,6 linkage derived from roots of Gentiana lutea. – Trehalose, α glucoside with 1,1 linkage obtained from yeast and mushrooms. – Isomaltose, α glucoside with 1,6 linkage formed as a side product of hydrolysis of starch by Fig. 1B-1: Maltose (α-D-glucopyranosyl-(1→4) α-D- amylase enzyme. glucopyranose) Lactose (ββ -D-galactopyranosyl-(1→ → 4) β -D- glucopyranose) (Fig. 1B-2): Lactose give rise to one molecule of glucose and galactose upon enzymatic (lactase) or acid hydrolysis. Lactose is normally present in milk and in the urine of women during later half of pregnancy and during lactation. It is dextrorotatory, shows mutarotation in solution. It reduces metallic ions, forms lactosazone with phenylhydrazine. It is a Fig. 1B-2: Lactose (β-D-galactopyranosyl-(1→4) β-D- galactoside since the carbon number 1 of glucopyranose) 10  Reactions of Carbohydrates 1 phenylhydrazine and not due to the reaction of 2. Nonreducing disaccharides intact sucrose molecules with phenylhydrazine. E.g. Sucrose, Trehalose. These are the disac- charides in which the functional groups of constituent monaosaccharides are in linkage. 3. Barfoed’s Test Procedure: ⎫ ⎪ Same as given under Observation: ⎬ Inference: ⎪⎭ monosaccharides Fig. 1B-3: Sucrose – (α-D-glucopyranosyl – β-D fructofuranoside) Principle: Response of the disaccharides: Disaccharides will not reduce cupric ions in the REACTIONS OF DISACCHARIDES weak acid medium within the prescribed keeping time of 2 minutes in the boiling water bath and 1. Molisch Test do not give a positive response to the test. Application: Useful to differentiate monosac- Principle: Response of the disaccharides: All the charides from disaccharides. disaccharides that are experimented routinely give the positive reaction – reddish violet ring as Points to Ponder this is a general test to detect the presence of – If the heating time is prolonged disaccharides carbohydrate. will also give a positive response to Barfoed’s Procedure: ⎫ test. ⎪ Same as given under Observation: ⎬ – If the concentration of disaccharide solution is Inference: ⎪⎭ monosaccharides high, Barfoed’s test tends to become positive. 2. Benedict’s Test 4. Osazone Test Procedure: ⎫ Procedure: Same as given under monosac- ⎪ Same as given under Observation: ⎬ charides except for the period for which the Inference: ⎪⎭ monosaccharides reaction tube to be placed in the boiling water bath – it is 45 minutes for disaccharides. Response of the Disaccharides Lactose gives a characteristic yellow puff Based on Benedict’s test disaccharides are shaped lactosazone crystals (see Fig. 1B-4). classified into: 1. Reducing disaccharides, e.g. Lactose, Maltose. These disaccharides have a free carbonyl (keto/ aldehyde) group which is not involved in glycosidic linkage will reduce cupric ions in the alkaline medium as explained under the monosaccharides, E.g. Lactose, maltose. Fig. 1B-4: Lactosazone (Puff shaped) 11  Qualitative Analysis Maltose: Individual crystals of maltosazone Principle: The disaccharide sucrose contains looks like a yellow colored petal and when glucose and fructose. Fructose formed from grouped looks like a sun flower (see Fig. 1B-5). sucrose upon acid hydrolysis by the HCl of Seliwanoff’s reagent, is dehydrated by the acid HCl to form hydroxymethyl furfural which then condenses with the resorcinol of Seliwanoff’s reagent to form a red colored complex. 6. Rapid Furfural Test Procedure: Same as given under monosac- Fig. 1B-5: Maltosazone (Petal shaped) charides. Observation: Sucrose gives violet color (see Inference Fig. 1A-6) whereas lactose and maltose do not Lactose → Puff shaped lactosazone crystals give violet color. Maltose → Petal shaped or sunflower shaped Inference: Sucrose upon acid hydrolysis by maltosazone crystals the HCl added in the test yields a keto sugar Sucrose → Will not form osazone fructose. Fructose being a keto sugar gives positive response to Rapid furfural test as Principle: Reducing disaccharides with a reactive described under monosaccharides. Where as carbonyl group condense with phenyl hydrazine lactose (galactose + glucose) and maltose (glucose to form respective osazone crystals with + glucose) contain no keto sugar and cannot give characteristic shapes as detailed above. positive response to this test. Application: Useful to differentiate disac- Principle: The disaccharide sucrose contains charides. glucose and fructose. Fructose formed from sucrose upon acid hydrolysis by the HCl, is 5. Seliwanoff’s Test dehydrated by the same HCl to form Procedure: Same as given under monosac- hydroxymethyl furfural which then condenses charides. with the α -naphthol of Molisch reagent to form Observation: Sucrose gives bright red color a violet colored complex. (see Fig. 1A-8) whereas lactose and maltose do not give red color. 7. Specific Sucrose Test (Fig. 1B-6) Inference: Sucrose upon acid hydrolysis by the HCl in the Seliwanoff’s reagent yields a keto Procedure: It is done in two steps. sugar, fructose. Fructose being a keto sugar gives Step 1: Hydrolysis positive response to Seliwanoff’s test as described under monosaccharides. Whereas lactose To 5 ml of sucrose solution add 1 drop of thymol (galactose + glucose) and maltose (glucose + blue indicator and one or two drops of dilute HCl glucose) contain no keto sugar and cannot give to make the solution acidic as shown by the positive response to this test upon acid development of pink color. Divide it into two hydrolysis by the HCl present in the Seliwanoff’s equal parts. Boil one part for 1 minute and the reagent. other part is kept as control. Neutralize both parts 12  Reactions of Carbohydrates 1 by adding 2% sodium carbonate drop by drop needed for getting correct reaction in the until a blue color develops. second step. Step 2: Benedict’s Test Thymol blue indicator contains two components that work at acid range (pH range 1.2-2.8; Perform Benedict’s test with each portions. color change – red to yellow) and at alkaline Observation: Unboiled sucrose solution will range (pH range 8.0-9.6; color change – yellow not give a positive response to Benedict’s test to blue). where as boiled portion gives a positive response. Inference: Sucrose is hydrolyzed by HCl in the first step to form glucose and fructose and 1C. REACTIONS OF the medium is neutralized by the 2% sodium POLYSACCHARIDES carbonate. In the second step, products of acid hydrolysis INTRODUCTION reduce cupric ions to red cuprous oxide. The polysaccharides are complex carbohydrates Precautions of high molecular weight, which on hydrolysis 1. Avoid adding excess acid because it will yields monosaccharides or products related to dehydrate sugar to form furfural derivatives monosaccharides. The various polysaccharides and that will interfere the test. differ from one another with respect to their 2. Always remember to add alkali as per the test constituent monosaccharide composition, procedure since neutralization of acidic pH is molecular weight and other structural features. Fig. 1B-6: Specific sucrose test 13  Qualitative Analysis In all types the linkage between the Observation: Deep blue color appears which monosaccharide units is the glycosidic bond. then disappears on heating and then reappears This may be α or β which join the respective on cooling (see Fig. 1C-1). units through 1 → 2, 1 → 3, 1 → 4 or 1 → 4 linkages Inference: Starch forms a adsorption complex in the linear sequence or at branch points in the with iodine to give a blue color. The blue color polymer. disappears on heating due to the breaking of the Polysaccharides are classified based on the Iodine starch adsorption complex and appears type of monosaccharide units present in them. on cooling due to reformation of the adsorption complex. 1. Homopolysaccharide 3. Benedict’s Test Contains only one type of monosaccharide, Procedure: Same as given with monosaccharides. e.g. Starch, Glycogen. Observation: No colored precipitate. Inference: Starch is a nonreducing carbohydrate. 2. Heteropolysaccharide Contains more than one type of monosaccharide 4. Starch Hydrolysis Test units, e.g. Glycosaminoglycans (heparin, Procedure: Take 25 ml of starch solution in a hyaluronic acid). beaker. Add 10 drops of concentrated HCl and We will discuss the reactions of starch in this boil gently. At the end of each minute , transfer a chapter in order to understand the chemical drop (using glass tube) of the solution on to a properties of polysaccharides in general. plate for doing the iodine test and 3 drops to 5 ml of Benedicts solution (Set tubes containing 5 REACTIONS OF STARCH ml of Benedict’s reagent in series). Continue until the iodine test becomes negative. Then place the 1. Molisch Test tubes for the Benedict’s test in the boiling water bath for 3 minutes. Procedure: Observation: See Table 1C-1. Observation and Inference: Same as given under Inference: Starch upon hydrolysis by HCl gives monosaccharides. the following products. Principle: The test is answered by all furfural yielding substances and hence all the carbohydrates. 2. Iodine Test Procedure: To 2-3 ml of starch solution add 2 drops of dilute (0.05 N ) iodine solution. Observe the changes on heating and on subsequent cooling. Fig. 1C-1: Iodine test 14  Reactions of Carbohydrates 1 Table 1C-1: Response of Starch Hydrolysis Test Time in minutes Color with I2 Benedict’s test Hydrolysis Product 1 Blue Blue No reduction Starch 5 Violet Green Reduction starts Amylodextrins 8 Reddish violet Red Initiation of reduction Amylo and erythrodextrins 12 No color Red Partial reduction Achrodextrins 20 No color Red Complete reduction Glucose Starch → Soluble starch → Amylodextrins → the level of formation of maltose and glucose Erythrodextrins → Achrodextrins → Maltose → iodine test becomes negative and Benedict’s test Glucose. When the hydrolytic stage reaches to becomes positive. 15  Qualitative Analysis 1D. IDENTIFICATION OF UNKNOWN CARBOHYDRATES 16  Reactions of Carbohydrates 1 1E. QUESTIONS 1F. REAGENT PREPARATION 1. Name the following: 1. Molisch’s Reagent: Dissolve 5 g of α-naphthol a. General test for detecting carbohydrates in 100 ml of 95% of alcohol. b. Reduction test for monosaccharides 2. Benedict’s Qualitative Reagent: Heat to c. Sugars giving positive response for Rapid dissolve 173 g sodium citrate and 100 g sodium furfural test and Seliwanoff’s test carbonate in about 800 ml of water in a conical d. The disaccharide yielding puff shaped flask. Transfer to a graduated cylinder through osazone crystals a folded filter paper placed in a funnel or beaker e. Tests based on reduction property of sugars of 1L capacity. Dissolve 17.3 g copper sulfate in f. The test used to detect sugar in urine about 100 ml of water. Add the copper sulfate g. Reducing disaccharides solution slowly with constant stirring to the h. Nonreducing disaccharides carbonatecitrate solution and make up to 1L. 2. Give the principle of the following tests: 3. Barfoed’s Reagent: Dissolve 13.3 g neutral a. Molisch test copper acetate crystals in 200 ml water. Pass b. Benedict’s test through a filter paper placed in a funnel to c. Barfoed’s test remove the particles if present to another d. Osazone test graduated beaker. Then add 1.8 ml glacial acetic e. Iodine test for starch acid. f. Rapid furfural test 4. Seliwanoff’s Reagent: Dissolve 0.05 g g. Seliwanoff’s test resorcinol in 100 ml dilute HCl. 3. Give the ingredients of following reagents: 5. Phenylhydrazine Mixture: Mix 2 parts phenyl- a. Molisch’s reagent hydrazine hydrochloride and 3 parts sodium b. Benedict’s reagent acetate by weight thoroughly in a mortar c. Barfoed’s reagent (Mixture with longer shelf life may be prepared d. Seliwanoff’s reagent by using equal weights of phenylhydrazine 4. Benedict’s test is described as a semiquantitive hydrochloride and anhydrous sodium acetate). test. Explain. 6. 0.1 N iodine Solution: Dissolve 1.27 g iodine 5. Unlike Benedict’s test, Barfoed’s test is not and 3 g pure KI (potassium iodide) crystals in suitable for testing glucose in urine. Why? 100 ml distilled water. Dilute 1:10 in distilled 6. Give the difference between Benedict’s and water before use. Barfoed’s test. 7. Why do glucose, mannose and fructose give 7. Glucose, Fructose, Lactose, Maltose, Sucrose, similar osazone crystals? Starch Solutions: 1% solutions -Weigh 1 gm of respective sugars and dissolve in 100 ml of water. 8. Sucrose do not form osazone crystals with osazone test. Why? 9. Make a scheme for the detection of an unknown carbohydrate solution. 17  Reactions of Proteins 2 2A. GENERAL REACTIONS OF troponin, collagen, myosin and globular proteins PROTEINS (offer mainly dynamic functions), e.g. Hb, enzymes, peptide hormones, plasma proteins. Proteins are present in all types of body fluids. INTRODUCTION Proteins have to be studied in different ways. During routine analytical laboratory work, two Proteins are the most abundant organic types of reactions are practiced. molecules (carbon containing) in the living 1. Precipitation reactions system. They offer structural and dynamic 2. Color reactions function.They are polymers of amino acids linked by covalent peptide bonds. Proteins ingested undergo digestion and get absorbed as 1. PRECIPITATION REACTIONS OF amino acids into the portal vein and reaches liver PROTEINS and then to other tissues. They are used mainly for protein synthesis as Proteins have to be precipitated for different dictated by the genes of respective tissues purposes during routine laboratory work. Two (differential expression). Some amino acids such situations are described below: undergo specific metabolic reactions to produce For its own identification and estimation, specialized compounds.eg; epinephrine and nor e.g. Proteins are excreted in urine in various epinephrine formed from Tyrosine, Serotonin forms of kidney dysfunction. According to the from Tryptophan. degree of kidney damage different proteins House keeping proteins like aldolase have are excreted in urine. In the early stages low longer half life where as regulatory proteins like molecular weight albumin is excreted. As the HMG CoA reductase have shorter half lives. disease progresses high molecular weight After their life span proteins are catabolized to globulin starts excreting. release nitrogen which ultimately get converted For the analysis of other compounds in the into urea and excreted in the urine where as the specimen, proteins are first precipitated out. carbon skeletons may be utilized for other Proteins form emulsoid colloidal solutions purposes like gluconeogenesis. (colloid solutions are formed by particles with a Proteins are classified into fibrous (offer diameter ranging from 1 μm to 200 μm). mainly structural function) eg: fibrinogen, Emulsoids (here proteins) in general possess two  Reactions of Proteins 2 stability factors – charge and water of hydration the filter. Perform Biuret test with the filtrate either of these prevent aggregation and using an equal volume of 40% sodium hydroxide precipitation of proteins. The electrical charges and 2 drops of 1% CuSO4.) carried by the proteins may be changed in sign Observation: Upon doing Biuret test with the or magnitude by changing the acidity or filtrate, violet color forms. alkalinity of the solution causing them to precipitate. The inorganic salts like ammonium Inference sulfate act as dehydrating agent, there by Albumin is not precipitated by half saturation removing the shell of hydration of the proteins. with ammonium sulfate. The dehydration is also carried out by organic Globulins are precipitated solvents like alcohol and ether. Principle: The molecular weight of the albumin is much less than the globulin so albumin is not (i) Precipitation by Salts precipitated by half saturation (see Fig. 2A-1) Inorganic salts when added to the protein whereas high molecular weight globulins are solutions, water of hydration around the protein precipitated. molecules is removed causing aggregation of Points to Ponder: Use 40% sodium hydroxide protein molecules leading to their precipitation. for doing Biuret test. (In the routine Biuret test Proteins are lyophilic colloids as they have much 5% sodium hydroxide is used) Here the filtrate affinity for the dispersion medium. contains ammonium sulfate. Ammonium (a) Half saturation test with saturated ions form a deep blue cuprammonium ion, ammonium sulfate solution: [Cu(NH3)4++] which will mask the violet color of Procedure: To 3 ml of protein solution add an Biuret test. To avoid this 40% NaOH is used. equal volume of saturated ammonium sulfate (b) Full saturation test with ammonium sulfate solution. Mix and allow to stand for 5 minutes. crystals: Filter (for this take a round filter paper of 5 cm Procedure: To 5 ml of protein solution, keep on radius and fold it to form a cone to make it fit adding ammonium sulfate crystals and at the same into a funnel. Then place this funnel over a test- time shaking the tube till a few crystals remain at tube and pour the contents of the tube through the bottom of the test tube. Filter (for this take a round filter paper of 5 cm radius and fold it to form a cone so as to fit it in a funnel. Then place this funnel over a test-tube and pour the test tube contents through the filter. Perform Biuret test with the filtrate using an equal volume of 40% sodium hydroxide and 2 drops of 1% CuSO4.) Observation: Upon doing Biuret test with the filtrate no purple or violet color develops. Inference: The protein (e.g. Albumin) is completely precipitated by full saturation with ammonium sulfate. Upon filtration no protein passes into the filtrate, to be detected by the Fig. 2A-1: Half saturation test—Albumin Biuret test. 19  Qualitative Analysis Principle: Neutral salt, (e.g. ammonium acidic the proteins acquire net positive charge. sulfate) precipitate proteins by salting out which In this test upon adding alkali proteins gain involves the removal of the shell of hydration negative charge and they form ionic bond with causing precipitation of proteins. Higher the positively charged metal ions leading to molecular weight lesser will be salt required for precipitation of proteins. the precipitation. Here globulins have much higher molecular weight than albumin so that albumin require only saturated solution where as globulins require addition of further amount of salt for complete precipitation (see Fig. 2A-2). Fig. 2A-3: Mechanism of precipitation of proteins by metal ions (b) Precipitation by 10% CuSO4 solution: Procedure: To 3 ml of protein solution add 2 drops of 5% NaOH. Mix well and add 10% Fig. 2A-2: Salting out CuSO4. Observation: A light blue precipitate forms (ii) Precipitation by Heavy Metals Inference: Proteins are precipitated by positively charged copper ions. (a) Precipitation by 10% Lead acetate: Principle: The same as that of precipitation Procedure: Take 3 ml protein solution; add 2 by lead acetate. drops of 5% NaOH. Mix well and add 2 ml of (c) Precipitation by 10% ZnSO4 solution: 10% lead acetate solution. Procedure: Take 3 ml of protein solution, add Observation: White precipitate forms. 2 drops of 5% NaOH. Mix well and add 2 ml of Inference: Proteins are precipitated by 10% ZnSO4 solution. positively charged lead ions. Observation: An intense white precipitate Principle: (see Fig. 2A-3.) The isoelectric point forms. of a protein is that pH at which the net charge on Inference: Proteins are precipitated by the protein is zero. If the pH of the medium is positively charged zinc ions. made alkaline the proteins acquire net negative Principle: The same as that of precipitation charge and if the pH of the medium is made by lead acetate (Fig. 2A-4). 20  Reactions of Proteins 2 (iii) Precipitation by Anionic Reagents Principle: (Fig. 2A-5) Alkaloids when (Alkaloids) dissolved lower the pH of the medium and they themselves form anions. Proteins in this acidic (a) Precipitation by metaphosphoric acid: medium acquire positive charge and they Procedure: Take 3 ml of protein solution in a complex with negatively charged ions in the test tube and add a few drops of metaphosphoric medium. These complexes are insoluble and they acid. are precipitated. (iv) Precipitation by Organic Solvents (a) Precipitation by ethanol: Procedure: Add 2 ml of ethanol to 1 ml of protein solution taken in a test tube and mix well. Observation: Cloudy precipitate forms Inference: Proteins are precipitated due to removal of water of hydration. Principle: Organic solvents cause Fig. 2A-4: Precipitation of proteins by different precipitation of proteins by the removal shell of metal ions hydration surrounding the proteins (Fig. 2A-6). Fig. 2A-5: Mechanism of precipitation of proteins by anionic agents (alkaloids) Fig. 2A-6: Precipitation by organic solvents Inference: Metaphosphoric acid in solution (v) Precipitation by Heat forms acid anion. Proteins become positively charged. Hence positively charged protein ions Procedure: Take a test tube and fill protein and negatively charged acid anions derived from (albumin) solution up to two thirds. Heat the metaphosphoric acid combine to form insoluble upper one third portion of protein solution complex leading to precipitation. column. Note whether any precipitate has 21  Qualitative Analysis appeared. Irrespective of the presence or absence albumin–4.9, human globulin–6.4, casein–4.6). of the development of the precipitate, add 2% At pI proteins are least soluble. So the denatured acetic acid drop by drop. Note whether the protein get precipitated upon adding acetic acid. precipitate formed earlier (if any) became intensified or appeared upon adding acetic acid. (vi) Precipitation by Strong Mineral Acids Observation: White coagulum formed on initial heating intensifies on adding acetic acid Procedure: Take 2 ml of concentrated HNO3 or (see Fig. 2A-7). concentrated HCl in a test tube. Add 2 ml of Inference: Albumin is denatured by heating protein solution along the sides of the test tube and is precipitated by acetic acid. slowly. Principle: Heating caused denaturation. Observation: White ring forms at the junction Disruption of secondary, tertiary, quaternary of two liquids. structures maintained by noncovalent forces Inference: Albumin as well as globulins are (hydrogen bonds, ionic interactions, van der precipitated by strong mineral acids. waals forces, hydrophobic interactions) causes Principle: Strong acids causes denaturation denaturation. and precipitation of proteins. Aggregation of denatured protein is referred Points to Ponder: Precipitation by HNO3 is to as coagulum. Denaturation may be reversible named as Heller’s test. It is used as a test for in some cases (not always).But coagulation is detecting protein in urine or other body fluids. always irreversible. Addition of acetic acid lowers the pH of the medium towards the TEST TO DEMONSTRATE DENATURATION isoelectric pH (pI) of the albumin (IEP of AND COAGULATION different proteins: Human albumin – 4.7, egg Procedure: Step 1 (See Fig. 2A-8) Take 3 test tubes and add 9 ml of a clear salt free albumin solution in them.Mark A,B and C on them. To the test tube marked A add 1 ml of 0.1 N HCl, to the test tube B add acetate buffer (pH – 4.7) and to the test tube C add 1 ml of 0.1 N NaOH. – Heat tube B in a boiling water bath for 15 minutes. – Cool Step 2 (See Fig. 2A-9) To the tubes A and C add 10 ml of acetate buffer solution (pH 4.7) – Filter of the precipitates in each tube – Wash the precipitate obtained in the filter Fig. 2A-7: Precipitation by heating and influence of pI paper with distilled water. 22  Reactions of Proteins 2 – Precipitate in tubes A and C are denatured egg These are conferred by noncovalent forces albumin. Precipitate in the tube B is coagulated hydrogen bonds, hydrophobic interactions, protein. electrostatic interactions and van der waals interactions. Step 3 (See Fig. 2A-10) Changes in higher orders of protein structure – Suspend each of the precipitates in 10 ml of leading to loss of protein function is caused by distilled water and divide each suspension denaturation. Denaturing agents can be chemicals into 3 parts. (mineral acids, alkalies urea) or physical (heat, uv – To the first part add dilute HCl drop by drop, radiation, ultrasonic waves, shaking, stirring). to the second add dilute NaOH and heat the Denatured proteins flocculate at or near the third part of the suspensions drawn from pI which is reversible at room temperature. But tubes A and B in a boiling water bath for 15 if it is heated, the floccules form large tenacious minutes. masses of coagulated protein.The coagulated – Cool and check the solubility of the precipitate proteins are not redissolved by treatment with in dilute acid and alkali. dilute acids or alkalies. Observation: Precipitate forms at pI brought Denaturation is the primary change – floc- about by the addition of acid or alkali. This culation (reversible sometimes) and coagulation precipitate dissolves readily in a few drops of (irreversible) are visible manifestations of dilute acid or alkali. The coagulated protein denaturation. obtained from tube B remains insoluble upon adding dilute acid or alkali. 2. COLOR REACTIONS OF PROTEINS Inference: Proteins have got a primary AND AMINO ACIDS structure as dictated by amino acid sequence bonded by peptide bonds and location of Proteins react with a variety of reagents to form disulfide bonds if any. Functional form of colored products because of their constituent proteins are achieved by higher orders of protein peptide bonds and amino acids.These reactions structure (secondary, tertiary and quaternary). are useful for quantitative and qualitative studies Figs 2A-8 and 2A-9: Test to demonstrate coagulation and denaturation—step 1 and step 2 23  Qualitative Analysis Filter the precipitate and wash the precipitate Suspend the washed precipitate from each tube in 10 ml distilled water and divide into 3 tubes Fig. 2A-10: Test to demonstrate coagulation and denaturation—step 3 of proteins. By quantitative studies the pathways and the deficiency of any enzyme of concentration of the proteins are estimated. these pathways lead to accumulation of Qualitative studies help to know the presence of compounds proximal to the defective step proteins or specific amino acids present in the causing disorders called aminoacidurias. For protein. They are useful mainly in the following instance phenylketonuria due to phenylalanine situations hydroxylase deficiency causes elevated blood 1. For the diagnosis of aminoacidurias: levels of phenylalanine in the blood and urine. Individual amino acids undergo unique catabolic Study of aminoacidurias need identification of 24  Reactions of Proteins 2 abnormally elevated specific amino acids in the Inference: The biuret reaction is given by body fluids. Study of color reactions of amino substances which contain two carbamyl groups acids are useful in the diagnosis of (–CONH 2) joined either directly together or aminoacidurias. through a single atom of nitrogen or carbon. 2. For the nutritional assessment: Out of Positive reaction indicates that the given protein twenty standard amino acids, only eight are solution contains at least two peptide bonds. essential and the rest of the twelve amino acids Chemistry of the reaction: The biuret test is are nonessential in adults. Those proteins given by those substances containing two containing all the essential amino acids are carbamyl groups (–CONH2) joined either directly considered to be good quality proteins eg; egg or by a single nitrogen or carbon atom. The albumin. Hence for the making nutritional purplish violet colour is due to the formation of assessment (roughly) of proteins also the study a copper coordination complex (Fig. 2A-12). of reactions of amino acids are helpful. 3. To detect the presence of proteins or amino acids in biological fluids or in fluids with unknown composition: This chapter deals with different color reactions of amino acids.The color reactions are due to reaction between constituent radical or groups of the amino acids and the chemical reagents used in the test. Amino acid composition of different proteins is different. Fig. 2A-12: Copper coordination complex Depending on the nature of amino acids contained in a protein, the response and the The molecule should have a minimum of two intensity of the color reactions varies. peptide bonds to give copper coordination complex that impart violet color to test mixture. (a) Biuret Test (Fig. 2A-11): [This reaction is first carried out with the compound Procedure: To 2-3 ml of protein solution add an biuret formed by the condensation of 2 molecules of equal volume of 10% sodium hydroxide solution, urea upon heating. This compound contains two mix thoroughly. Then add 0.5% copper sulfate peptide bonds as shown below.] solution drop by drop, mixing between drops until a purplish violet color is obtained. Observation: Purplish violet color develops. Fig. 2A-13: Biuret Biuret (a nonprotein, formed from urea on heating; biuret formed gives violet color with copper sulfate solution in the alkaline medium) (See Fig. 2A-13). Proteins give violet color with biuret test since there are several pairs of CONH groups in the molecule (see Fig. 2A-14). Fig. 2A-11: Biuret test 25  Qualitative Analysis Fig. 2A-14: CONH groups in the peptide Points to Ponder If too much copper sulfate solution is added blue colored copper hydroxide will be formed and that will mask the violet color. If magnesium sulfate is present in the test solution it forms magnesium hydroxide and interferes with the test If much ammonium sulfate is present excess of alkali have to be used. The color depends on the nature of the protein. – Proteoses - Purple – Peptones - Pink Fig. 2A-15: Chemistry of ninhydrin reaction (b) Ninhydrin Test: Procedure: To 1 ml of protein solution (pH must SPECIFIC COLOR REACTIONS USED TO be between 5 and 7) in a test tube add 2-3 drops IDENTIFY THE SIDE CHAIN (R) GROUPS of freshly prepared 0.1% ninhydrin OF AMINO ACIDS (triketohydrindene hydrate) solution. Heat the (c) Xanthoproteic Reaction (Fig. 2A-16): solution to boil for 2 minutes and allow to cool. Procedure: Add 1 ml of concentrated nitric acid Observation: Blue color with α-aminoacids to 2-3 ml of test protein solution. Heat to boil. and yellow color with iminoacid – proline. Cool and pour half of the solution into another Inference: Blue color is due to the formation tube. of a complex - Ruhemann’s purple formed One tube is kept as control and the other as between N-terminal nitrogen and ninhydrin. test, so as to understand the development of even Chemistry of the reaction (see Fig. 2A-15): faint color. To one tube add 40% NaOH or liquor α-amino acids reacts with ninhydrin to form ammonia (ammonium hydroxide) in excess. aldehyde, hydrindantin, ammonia and carbon Observation: A white precipitate forms on dioxide. Then one molecule of hydrindantin, adding nitric acid, which on heating turns yellow ninhydrin and ammonia complex to form and then dissolves to impart yellow color to the Ruhemann’s purple. solution. Upon adding alkali the color deepens Proteins give a faint blue color. In the case of to attain orange colour. proteins amino terminal nitrogen participate in Interpretation: Addition of nitric acid causes the action. denaturation of proteins to get white precipitate. Application: Staining of amino acids in paper Yellow color due to nitration of benzene ring chromatography. of amino acids – tryptophan and tyrosine 26  Reactions of Proteins 2 Fig. 2A-16: Xanthoproteic test (Fig. 2A-17). Addition of alkali increases the Fig. 2A-18: Millon’s test ionization of compounds hence the color deepens to get final orange color. Observation: Red precipitate forms and Points to Ponder solution turns red. Amino acid solutions gives red color without a precipitate (see Fig. 2A-18). This test cannot be employed for urine testing Inference: Protein contains the amino acid as the final color of the test and the natural Tyrosine which contains a phenolic radical. color of urine are similar. Principle: The protein precipitated by The aromatic amino acid Phenylalanine will mercuric sulfate in acidic medium to form not give a positive response to the test even mercury – protein complex (metalloprotein though it contains benzene ring. complex). Nitrous acid is formed by the reaction (d) Millon’s Test: between sodium nitrite and sulphuric acid. This Procedure: To 2 ml of protein solution in a test nitrous acid causes nitration of phenolic groups tube add 2 ml of 10% mercuric sulfate (HgSO4) of tyrosine.Warming enhances nitration process in 10% sulphuric acid. Boil for 30 seconds. A and intensifies the color. precipitate may form at this stage. Add a few (e) Aldehyde Test (Glyoxylic Acid , Hopkins – drops of 1% NaNO2 and gently warm. Cole Reaction): Procedure: Take 2-3 ml of test solution add 2 drops of 1/500 formaldehyde (HCHO) and 1 drop of 10% mercuric sulfate in sulfuric acid. Mix well. Add 3 ml of concentrated sulfuric acid through the sides of the test tube. Observation: A purple ring develops at the junction of two layers (see Fig. 2A-19). Inference: The purple color is due to the indole ring (see Fig. 2A-20) of the amino acid tryptophan. Principle: Mercuric sulfate in sulphuric acid act as an oxidizing agent and it oxidizes the indole ring of tryptophan. Then formaldehyde react with the oxidized indole ring to form purple Fig. 2A-17: Tyrosine and Tryptophan colored complex. 27  Qualitative Analysis Principle: Molisch reagent is α-naphthol in alcohol. Sodium hydroxide provides alkaline pH. At the alkaline pH guanidino group of arginine combines with α-naphthol to form bright red color. (g) Sulfur Test (Lead Blackening Test): Procedure: To 3 ml of protein solution add 3 ml of 40% NaOH and boil for 3 minutes. Cool, add 1 ml of lead acetate solution. Observation: Solution turns dark brown (see Fig. 2A-22). Fig. 2A-19: Aldehyde test Inference: This test is answered by “S” containing aminoacids – cysteine and cystine but not methionine because of the placement of S in the thio ether linkage (Fig. 2A-23). Principle: Upon boiling with strong alkali the organic sulfur in the cystine and cysteine is converted into sulfide (here Na2S). The sodium sulphide react with lead acetate to form black lead sulphide (PbS) and solution turns brownish Fig. 2A-20: Tryptophan black. (f) Sakaguchi’s Test: Procedure: Add 5 drops of 5% sodium hydroxide to 5 ml of protein solution. Shake well. Add 2-4 drops of Molisch’s reagent. Observation: A bright red (see Fig. 2A-21) color develops. Inference: The protein solution contains aminoacid with guanidino group (Arginine). Fig. 2A-22: Sulfur test and cysteine and cystine Fig. 2A-21: Sakaguchi’s test Fig. 2A-23: Methionine 28  Reactions of Proteins 2 Points to Ponder: Casein and gelatin gives a negative response due to the deficiency of cysteine in them. (h) Pauly’s Test: Procedure: To 0.5 ml of 0.5% sulfanilic acid add an equal volume of 0.5% freshly prepared sodium nitrite. Allow to stand for 1 minute and add 1 ml of protein solution. Mix well and Fig. 2A-25: Orange red color of tyrosine add 1 ml of 10% Na2CO3 to make the solution alkaline. Observation: Cherry red or orange red color may be observed. 2B. REACTIONS OF ALBUMIN (TABLE 2B-1) Inference: Cherry red color indicates presence or predominance of histidine and orange red color shows the presence or predominance of Albumins are compact roughly spherical in shape tyrosine in the solution (Fig. 2A-24). and have axial ratios not more than 3 (that is the Principle: Diazotized sulfanilic acid when ratio of their shortest to longest dimensions). complexes with imidazole ring of histidine gives Hence albumins come under globular proteins. cherry red colored complex and when it They are soluble with definite molecular weight. complexes with phenolic group of tyrosine yields Albumins of interest are serum albumin of blood, orange red color (Fig. 2A-25). lactalbumin of milk and ovalbumin of egg. It is also present in pulses. They are soluble in solute free water and coagulable on heating. They are not precipitated by half saturation with salts away from isoelectric pH. Egg albumin has a molecular weight of 45,000 Kda and it’s isoelectric point (pI) is 4.55-4.9. It is commonly employed in the laboratory to carry out experiments to study the properties of albumin in general. Human serum albumin has got a molecular weight of 69,000 Kda and its pI is 4.7. Fig. 2A-24: Cherry red color of histidine 29  Qualitative Analysis Table 2B-1: Reactions of Albumin Reaction Observation Inference PRECIPITATION REACTIONS 1. Isoelectric precipitation: Take 10 ml of protein White coagulum At pI, albumin is solution in a test tube. Add 2-3 drops of denatured. Upon heating Chlorophenol red (pH range – 5.0-6.6; color denatured protein range – yellow to red).The purpose of adding the aggregate to form visible indicator is to get pH around 5.0. Look at the change named color change. coagulation. 2% Na2 CO3 2% acetic acid With chlorophenol red, the yellow color denotes pH either equal to 5 or less than 5. So even if a yellow color is observed the pH may not be 5, it may be less than 5 also. In order to make sure of the required pH 5 add 2% Na2CO3 in drops until a pink color forms and then add 2% acetic acid in drops till the solution turns just yellow. If it is red or pink, we could infer that the prevailing pH is either 6.6 or more than that as indicated by the indicator. Add 2% acetic acid in drops till the yellow color just develops. Boil the above solution 2. Heller’s test: To 2 ml of concentrated nitric acid A white ring at Stratification of Albumin in a test tube (keep the mouth of test tube away the junction of solution over strong from your face and others) add gently equal two liquids mineral acid causes amounts of albumin solution forms denaturation and precipitation of albumin at the point of contact that is at the junction between two layers. 3. Half saturation test: To 5 ml of albumin solution Violet color Albumin being relatively add equal volume of saturated ammonium small in size (MW 69000 sulfate solution.Shake vigorously for 2 minutes. Kda) is not completely Keep it for 5 more minutes. Filter and collect the precipitated by saturated filtrate.Perform Biuret test with the filtrate.To 2ml solution of ammonium of the above filtrate taken in a test tube add 2ml sulfate and hence go into 40 % NaOH and 1% CuSO4 drop by drop. the filtrate. (Contd.) 30  Reactions of Proteins 2 (Contd.) 4. Full saturation test: No violet color Albumin is completely To 5 ml of albumin solution add ammonium precipitated by full sulfate crystals and shake well till some crystals saturation with remain at the bottom of the tube. Keep it for 5 ammonium sulfate minutes and filter. Collect the filtrate.Do Biuret crystals and so all the test with the filtrate.To 2ml of the above filtrate albumins get retained in taken in a test tube add 2ml 40 % NaOH and 1% the filter paper without CuSO4 drop by drop. going into the filtrate. Hence the filtrate devoid of albumin do not produce a positive Biuret reaction. 5. Heat and acetic acid test: Precipitate formed Heating caused Take a test tube and fill protein (albumin) on heating become coagulation of albumin solution up to two thirds.Heat the upper one denser on adding and the addition of acetic third portion of protein solution column. Note acetic acid acid lowered the pH of whether any precipitate has appeared. the medium towards the Irrespective of the presence or absence of the isoelectric pH (pI) of the appearance of the precipitate add 2% acetic acid albumin and enhanced drop by drop. Note whether the precipitate the precipitation formed earlier (if any ) has intensified or appeared upon adding acetic acid. COLOR REACTIONS 1. Biuret reaction: (see Section 2A) Violet color Albumin contains more than 2 peptide bonds 2. Xanthoproteic reaction: Deepening of Albumin contains Tryptophan and (see Section 2A) yellow color Tyrosine. Benzene ring of these amino acids are giving the reaction. 3. Millon’s test: (see Section 2A) Red color Albumin contains the phenolic group containing amino acid, Tyrosine 4. Aldehyde test: (see Section 2A) Violet or purple Indole ring containing amino acid ring at the junction Tryptophan is present in albumin of two liquids 5. Sakaguchi’s test: (see Section 2A) Bright red color Albumin contains the guanidino group 6. Sulfur test: (see Section 2A) Brownish black Indicates the presence of cysteine in solution albumin 7. Pauly’s test: (see Section 2A) Orange red Indicates predominance of Tyrosine in albumin 31  Qualitative Analysis 2C. REACTIONS OF CASEIN Contains all the essential amino acids. (TABLE 2C-1) It is less soluble and is made soluble at acid or alkaline pH and precipitate when the pH is Casein is the main protein present in the milk. It brought to isoelectric point (4.6). It is also is a phosphoprotein and constitute one-third of precipitated by half saturation with ammonium proteins of human milk , five sixth of proteins of sulfate and it is not coagulated by heat. Casein cow’s milk and three fourths of proteins of goat’s act like a suspensoid, the particles of it flocculate milk. Casein is secreted by mammary gland only. when their charges are neutralized. Table 2C-1: Reactions of Casein Reaction Observation Inference PRECIPITATION REACTIONS 1. Isoelectric precipitation: Take 4 ml of protein solution in a test tube. Add 2-3 drops of Bromocresol green (pH range 4.0 – -5.6; color range – yellow to blue).The purpose of adding the indicator is to get pH around 4.6 Look at the color change. 2% sodium carbonate 2% acetic acid If it is yellow that means pH is ≤ 4.0. Then add 2% sodium carbonate drop by drop till the solution turns light green. (pH around 4.6) If it is blue that means pH is ≥ 5.6. Then add 2% Precipitate seen When the pH reaches 4.6 acetic acid drop by drop till the solution turns (pI of casein) casein get light green. (pH around 4.6) precipitated 2. Half saturation test: To 5 ml of Casein solution No violet color Casein is completely precipita- add equal volume of saturated ammonium with Biuret test. ted by saturated solution of sulfate solution. Shake vigorously for 2 minutes. ammonium sulfate. So the Keep it for 5 more minutes. Filter and collect the filtrate do not contain any filtrate. casein as it is completely Perform Biuret test. To 2 ml of the above filtrate filtered off due to precipitation. taken in a test tube add 2 ml 40% NaOH and Hence the Biuret test done 32 1% CuSO4 drop by drop. with filtrate becomes negative. (Contd.)  Reactions of Proteins 2 (Contd.) Color Reactions: All the color reactions will be positive except the Sulfur test. Sulfur test will be faintly positive because only 0.3 g of cysteine/cystine is present in 100 gms of casein where as in egg albumin about 2.5g cysteine/cystine present in 100 gm of albumin. Specific Tests for Casein 3. Neumann’s test (detec