VK Malhotra - Practical Biochemistry for Students, 4th Edition_2

PRACTICAL BIOCHEMISTRY FOR STUDENTS

PRACTICAL BIOCHEMISTRY FOR STUDENTS 4th Edition

Varun Kumar Malhotra PhD/Gold Medalist

Department of Biochemistry Maulana Azad Medical College New Delhi

JAYPEE BROTHERS MEDICAL PUBLISHERS (P) LTD New Delhi

Published by Jitendar P Vij Jaypee Brothers Medical Publishers (P) Ltd EMCA House, 23/23B Ansari Road, Daryaganj New Delhi 110 002, India Phones: 23272143, 23272703, 23282021, 23245672, 23245683 Fax: 011-23276490 e-Mail: [email protected] Visit our website: http://www.jpbros.20m.com Branches • 202 Batavia Chambers, 8 Kumara Kruppa Road Kumara Park East, Bangalore 560 001, Phones: 2285971, 2382956 Tele Fax: 2281761, e-mail: [email protected] • 282 IIIrd Floor, Khaleel Shirazi Estate, Fountain Plaza Pantheon Road, Chennai 600 008, Phone: 28262665 Fax: 28262331 e-mail: [email protected] • 4-2-1067/1-3, Ist Floor, Balaji Building Ramkote Cross Road Hyderabad 500 095, Phones: 55610020, 24758498 Fax: 24758499 e-mail: [email protected] • 1A Indian Mirror Street, Wellington Square Kolkata 700 013, Phone: 22451926 Fax: 22456075 e-mail: [email protected] • 106 Amit Industrial Estate, 61 Dr SS Rao Road Near MGM Hospital, Parel, Mumbai 400 012 Phones: 24124863, 24104532 Fax: 24160828 e-Mail: [email protected] Practical Biochemistry for Students © 2003, Varun Kumar Malhotra 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 the 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 to be settled under Delhi jurisdiction only. First Edition: 1984 Second Edition: 1986 Third Edition: 1989 Fourth Edition: 2003 Publishing Director: RK Yadav ISBN 81-8061-109-4 Typeset at JPBMP typesetting unit Printed at Gopsons Papers Ltd., A-14, Sector 60, Noida

Preface to the Fourth Edition Over the years, the world of Biochemistry has seen many changes for better investigational techniques for the welfare of the patients. Hence it becomes mandatory for any written material to show the changes duly. In this edition, my efforts have gone in a direction to improve the material. In a modest attempt, additions have been made about the methods of expressing concentrations, instrumentations, collection and preparation of blood specimen, SGOT, SGPT, etc. The inquiring mind will certainly benefit from such exposures to manage the clinical situation in a more creative and challenging manner. Some modifications became necessary in various chapters and thus the matters have been updated in a befitting manner to serve the demanding needs of the consumers. Lastly, I thank many people like Prof B Misra and Publishers for their guidance and assistance respectively. Varun Kumar Malhotra

Preface to the First Edition In the ever expanding knowledge of Biochemistry it is very difficult on the part of single individual to go through the various bigger valume of textbooks on practice of biochemical investigations. This book in a concise but equally satisfactory form will help the user, students and practitioners either to a great extent. Besides being handy it has been kept up to its spirit of recent approaches by virtue of which one has the best utility in a busy time. For every medical practitioner and enlighted patients, Biochemistry has been playing a significant role. This book will certainly be of significant importances to the practitioners as well as laboratories. I thank all my colleagues and friends in contributing to have brought out this book in its near perfect shape. My extreme heartful thanks are due to the incessant guidance from Prof B Misra, Department of Physiology, MAM College in bringing out every chapter in an excellent way. The contributions that I have received by the constant cooperation of my parents and wife cannot be ignored. Lastly my utmost thanks are due to Dr K Chaudhry and my Publisher Mr Jitendar Vij. I hope the book will bring out greater number of readers keeping in view the worth of this book. The author always keeps an open eye for suggestions. Varun Kumar Malhotra

Contents 1. Methods of Expressing Concentration ............................................................................................ 1 2. Physical Chemistry ............................................................................................................................. 4 3. Carbohydrates ...................................................................................................................................... 9 4. Achromic Point .................................................................................................................................. 20 5. Proteins ................................................................................................................................................ 22 6. Isoelectric Point ................................................................................................................................. 27 7. Lipids ................................................................................................................................................... 29 8. Saponification Number ................................................................................................................... 32 9. Iodine Number .................................................................................................................................. 33 10. Formal Titration ................................................................................................................................. 35 11. Gastric Analysis ................................................................................................................................. 37 12. Urine Analysis ................................................................................................................................... 40 13. Food Analysis ..................................................................................................................................... 48 14. Collection and Preparation of Blood Specimen .......................................................................... 51 15. Urinary Reducing Sugars ................................................................................................................. 56 16. Urinary Chlorides.............................................................................................................................. 60 17. Urinary Creatinine ............................................................................................................................ 63 18. Ascorbic Acid in Urine ..................................................................................................................... 66 19. Serum Uric Acid ................................................................................................................................ 68 20. Colorimetry ......................................................................................................................................... 70 21. Blood Sugar ........................................................................................................................................ 74 22. Glucose Tolerance Test (GTT) ......................................................................................................... 77 23. Blood Urea .......................................................................................................................................... 80

x Practical Biochemistry for Students 24. Urea Clearance ................................................................................................................................... 83 25. Blood Cholesterol .............................................................................................................................. 85 26. Serum Calcium .................................................................................................................................. 87 27. Inorganic Phosphorus ...................................................................................................................... 90 28. Serum Total Proteins and Albumin: Globulin Ratio ................................................................. 92 29. Serum Bilirubin ................................................................................................................................. 95 30. Prothrombin Time ............................................................................................................................. 98 31. Liver Function Tests ........................................................................................................................ 100 32. Demonstrations ............................................................................................................................... 104 Appendix ............................................................................................................................................ 121 Questions Analysis 121 Normal Values 124 Index .................................................................................................................................................... 127

1 Methods of Expressing Concentration Concentration may be defined as weight per unit volume. The most common expressions are: 1. Percent 2. Molarity 3. Normality 4. Molality 5. Formality. 1. Percent

According to Caraway there are three ways of expressing percentage of solution, i.e. W/W, W/V, V/V. a. Weight per unit weight (W/W) A 10% W/W solution contains 10 gm of solute in 90 gm of solvent. b. Weight per unit volume (W/V) A 10% W/V solution contains 10 gm of solute dissolved in final volume of 100 ml of solution c. A 10% V/V solution contains 10 ml of the concentrate per 100 ml of solution. 2. Molarity

A molar solution contains 1 gm mol. wt. (mole) of solute in one litre of solution. Molarity of solution is expressed as ‘M’ 1 mole = 1000 millimoles Molarity =

gm.mol.wt.(moles) of solute _____________________________________

Molarity =

Amount of solute (gm) ____________________________________________

Volume of solution (L)

Mol. wt. × volume of solution (L)

For example: Molarity of a solution of 2 litres containing 58.5 gm. NaCl dissolved. Moles of NaCl =

Amt. of NaCl (gm) __________________________ Mol. wt

=

58.5 23 + 35.5

=

58.5 =1 58.5

(1)

2 Practical Biochemistry for Students Moles of solute Volume of solution (L) 1 = = 0.5 M 2

∴ Molarity =

So, Molarity of given solution is 0.5 M. 3. Normality

A normal solution contains 1 gm equivalent Wt. (eq.) of solute dissolved in one litre of solution. Normality of solution is expressed as ‘N’ 1 equivalent = 1000 milliequivalent. Normality =

gm. eq. wt. of solute __________________________________ Volume of solution (L)

⇒ Normality =

Amount of solute (gm) Equivalent wt. × Vol. of solution (L)

To evaluate equivalents of solute, we divide the molecular weight by its total valency of cation or anion. Eg. equivalent weight: a. of NaCl = Mol. wt. of NaCl b. of MgCl2 = c. of AICl3 =

Mol. wt. of MgCl2 __________________________ 2 Mol. wt. of AICl3 ________________________ 3

For example: Normality of a solution of 2 litres containing 180 gm of glucose. • Equivalents of glucose will be same as its moles because it does not dissosiate in solution. Amt. of glucose (gm)

Equivalents of glucose =

equivalent weight 180

=

= 1

180 ∴Normality = 1 = 2

Equivalents of solute Vol. of solution (L) = 0.5 N

So, Normality of given solution is 0.5 N. 4. Molality

Molality is defined as number of moles of solute dissolved in 1000 gm of solvent (Not in solution). It is designated as ‘M’. Molality =

Moles of solute Weight of solvent (gm)

Methods of Expressing Concentration 3 ⇒

Molality =

Amount of solute (gm) Mol. wt. × Weight of solvent (gm)

Unlike normality and molarity, this parameter of expressing the concentration is independent of temperature as there is no volume term in this relation. 5. Formality

Formality parameter is not used nowadays because all of the above parameters are better than it and are able to express concentration sufficently. Formality is same as molarity if molecular weight in the formula is replaced by formula weight.

4 Practical Biochemistry for Students

2

Physical Chemistry pH Determination

All biochemical reactions are greatly influenced by the hydrogen ion concentration of the surrounding medium in which the reaction takes place. The most convenient way of expressing hydrogen ion concentration is by the term pH. pH is defined as the negative logarithm of the hydrogen ion concentration of the solution. pH = – log CH+ = log

1 CH+

Hence it is both important and useful to know some of the simple methods of pH determination. pH can be determined both by colorimetric and electrometric methods. Electrometric method is the most accurate one and is done by using a pH meter whereas colorimetric determination of pH can be simply done by the following methods: 1. Indicator papers also called narrow range pH papers. 2. Universal indicators. 3. Gillespie’s drop method. Indicators are substances which change in colour with change in the pH of the solution to which they are added. Indicators are weak organic acids or bases. Their unionized forms show a colour while their ionized forms, i.e. cations or anions have different or another colour. The colour of the solution in presence of an indicator depends upon the relative proportions of ionized and unionized forms of the indicator which in turn depend upon the hydrogen ion concentration. For each indicator there is a definite pH range in which it is present as a mixture of its ionized and unionized forms. In this specific range, variations in the pH of the solution will bring visible change in the colour of the indicator. It is necessary that the effective pH range of the indicator includes the pH of the unknown sample. Selection of Indicator

Place 2 ml each of N/10 acetic acid, N/10 sodium carbonate and unknown solution in three different tubes. Add 2 drops of the indicator in each tubes. Mix and observe the colour in all the three test tubes. Test tube containing acetic acid will display acid colour of the indicator whereas test tube containing sodium carbonate will display alkali colour of the indicator. If the colour obtained with the unknown solution lies in between (i.e. intermediate) acid and alkali colour of the indicator and hence it is a suitable indicator for the colorimetric determination of the unknown solution. But on the other hand if the unknown solution shows either the full acid or (4)

Physical Chemistry 5

full alkali colour then the indicator used is unsuitable. Repeat the same procedures with the other indicators till a suitable indicator is selected. Some common indicators useful for biological pH range are: Indicator 1. 2. 3. 4. 5. 6. 7.

Thymol blue (acid range) Methyl yellow (Topfer’s reagent) Methyl orange Methyl red Phenol red Thymol blue (alkaline range) Phenolphthalein

pK

pH range

1.65

1.2-2.8 2.9-4.0 3.1-4.4 4.3-6.1 6.7-8.3 8.0-9.6 8.2-10.0

3.46 5.01 7.81 9.7 9.7

Acid colour Red Red Red Red Yellow Yellow Colourless

Alkali colour Yellow Yellow Yellow-orange Yellow Red Blue Pink

Indicators are used in: 1. Determining the end point in acid-base titration. 2. Determining the pH of the unknown solutions. 1. Indicator paper

Indicator paper consists of a strip of a sensitized paper and is accompanied by a colour chart which shows different colour which the indicator exhibits at different pH values. Take a strip of indicator paper and moisten it or dip it in the solution whose pH is to be determined. Remove the excess of the fluid adhering to the indicator paper strip by means of pressing between the folds of filter papers. Compare the colour of the pH paper with the colour chart on the indicator paper and thus determine the pH of the solution. 2. Universal indicator

Universal indicator is a wide range indicator solution having pH between 0 to 14. Take 5 ml of the unknown solution. Add to it 0.1 ml of the universal indicator. Mix well and find out the pH by matching the colour of the solution with the colour chart on the universal indicator bottle. The one with which it coincides or matches, is the pH of the unknown solution. 3. Gillespie’s drop method

In the determination of pH by this method, the ratio of two forms of the indicator may be found out by adding a known number of drops of an appropriate indicator to the test solution and finding out how the same number of drops has to be distributed between an acid and alkali so that the colour of the test solution matches with that 4 2 of the acid and alkali solution. When superimposed, then the pH can be calculated by using Handerson-Hasselbatch equation. pH = pK + log

Alkaline form Acid form

Apparatus used is Cole and Onslow’s comparator. Place 5 ml of N/10 HCl in tube no. 1, 5 ml of N/10 sodium carbonate in tube no. 2, 5 ml of unknown solution in tube no. 3 and 5 ml of distilled water in tube no. 4 (this is to equalise the optical conditions).

1

3


Add few drops (count the number of drops) of suitable indicators to the unknown solution in the tube no. 3. Now distribute the same number of drops in acid solution (because any drop of indicator solution going into the solution, will give its acid form) and alkali solution (because any drop of indicator solution going into alkali will give its alkali form). Mix the contents of each tube and examine the colours by white light. If the colour viewed through tube no. 3 and 4 appears to have more of the alkalis form of the indicator as compared to the colour viewed through tube nos 1 and 2, add more drops of the indicator to the sodium carbonate tube (alkali tube) and an equal no. of drops to unknown solution. Now again observe colour. If the colour viewed through the tube nos 1 and 2 matches with the colour viewed through the tube nos 3 and 4, then count the number of drops added to acid tube and alkali tube. The manner in which matching has been done it can be argued that the pH of the solution which contains ionized and unionised indicator in the ratio they are present in tube nos 1 and 2. pH of unknown solution = pK + log

= pK + log

Alkali form _________________ Acid form

Number of drops of indicator added to alkaline solution ___________________________________________________________________ Number of drops of indicator added to acid solution

To Study the Phenomenon of Osmotic Pressure, Diffusion and Dialysis

Osmotic flow occurs whenever a semipermeable membrane separates a solution and its pure solvent or between two solutions differing in concentration. Water passes through the membrane until the concentration on both sides becomes same. Such a movement of solvent molecules from a pure solvent or dilute solution, through a semipermeable membrane is called osmosis. Osmotic pressure is the pressure that must be applied on a solution to keep it in equilibrium with the pure solvent when the two are separated by semipermeable membrane or osmotic pressure is the force required to oppose the osmotic flow. Since osmotic pressure is proportional to the total number of solute particles in solution so the substances which ionize, will have the higher osmotic pressure as compared to those which do not ionize. If the solution containing crystalloids and colloids is placed in a cellophane sac and this is immersed in a jar of distilled water, the crystalloids diffuse across the membrane while the large colloidal molecules are retained. By repeatedly changing the distilled water outside the sac, it is possible to free the colloidal material virtually completely from salts and other crystalloids. This process is known as dialysis. Cover the top of the thistle funnel by tying a piece of cellophane over it. Then pour a solution made by mixing 10 ml of saturated ammonium sulphate with 90 ml of 1% starch into the thistle funnel till the level rises to the middle of the stem. Clamp the funnel on a stand and suspend this into a breaker of distilled water.

Physical Chemistry 7

After half an hour, take out a portion of solution from the beaker and perform the iodine test and barium chloride (BaCl2) test. i. 2 ml of solution (from beaker) add few drops of iodine. No change in colour is observed. ii. 2 ml of solution (from beaker), add 2 ml of BaCl2. A white precipitate of barium sulphate is obtained. The starch test will be negative and sulphate test positive. To Study the Phenomenon of Adsorption and to Compare Two Eluters—Acetone and Water

Adsorption is a phenomenon in which a substance is adsorped on the surface of a substance. It takes place due to the presence of free valencies on the surface of the adsorbate which attracts and takes up the adsorbant. Procedure

Take two test tubes. Add 10 ml of 0.1% methylene blue solution, add 0.5 gm of activated charcoal in each. Shake the tube vigorously and filter the contents into the separate test tubes. Both dyes are completely adsorbed by the charcoal and are retained on the filter paper. Acetone





Water

Charcoal with adsorbent

Stand

Colourless solution

Coloured solution

Now place these two funnels with filter paper to a separate test tubes. Wash the charcoal on the filter paper in one funnel with acetone and the other funnel with water. The colour of the filtrate with acetone washings will be blue while with water it will be colourless. Interpretation

The solution looses its blue colour due to the adsorption of methylene blue particles on the activated charcoal. This proves that the charcoal is a good adsorbant. On adding water to it, water cannot elute methylene blue and hence colourless filtrate is obtained. But acetone dissolves the methylene blue particles and thus the solution regains its original blue colour. This acetone is a better eluter than water and a strong adsorbant than the charcoal. To Study and Compare Surface Tension of Two Liquids—Water and Soap Solution

Surface tension is a phenomenon concerned with the force acting at the surface of a liquid giving an appearance of a stretched membrane. A liquid or a fluid is made up of molecules.


Molecules present in the interior of fluid are equally attracted in all directions by intermolecular forces due to surrounding molecules. But molecules on the surface are unequally attracted because of the absence of forces from above. This leads to certain unbalanced forces on the surface. Due to these forces, the surface acts as a membrane. Surface tension is defined as the workdone in ergs in streching the membrane by 1 square centimeter. Procedure

Take two test tubes. Add 3 ml of water in one and 3 ml of soap solution in the other. Sprinkle sulphur powder in both the test tubes.





Sulphur yellow colour Soap solution



Water Sulphur powder

Sulphur powder sinks in the test tube containing a soap solution whereas it floats in the test tube containing water. Interpretation

Due to lower surface tension of soap solution, the surface could not keep the sulphur powder floating. Therefore, the surface tension of water is more than that of soap solution.

Carbohydrates 9

3

Carbohydrates To Study the Reactions of Monosaccharides

Solutions provided are 1% glucose and 1% fructose. Molisch Test

This is a general test for carbohydrates. Carbohydrates on treatment with strong concentrated sulphuric acid undergo dehydration to give furfural or furfural derivate which on condensation with α-naphthol yields a violet or purple coloured complex whose exact structure is unknown. If oligosaccharides or polysaccharides are present, they are first hydrolysed to the constituent monosaccharides which are then dehydrated. Pentoses yield furfural and hexoses yield 5-hydroxymethyl furfural. Reagent

Molisch reagent α-naphthol in ethanol (ethanolic α-naphthol). O ||

OH | 2

CONC H2SO4 ________________ → HEXOSES – 3 H2 O H2C O | OH

_______________→

CHO CH2 | OH

O

|| C |

| OH

SO3H

Molisch reagent: α-naphthol in ethanol (ethanolic a-naphthol)

Test

In a clean and dry test tube, take 2 ml of the carbohydrate solution. Add 2 drops of ethanolic α-naphthol (Molisch reagent). Mix and incline the test tube and cautiously add 2 ml of concentrated sulphuric acid by the side of the test tube so that the acid forms a layer under the carbohydrate solution. Gently rotate the test tube between the palms of the hands to bring about slight mixing at the interface. An appearance of violet or purple ring at the interface (junction) of two solutions indicate the presence of carbohydrates. (9)

10 Practical Biochemistry for Students O ||

OH | 2

CONC. H2SO4 ________________ → PENTOSES – 3 H2 O

_______________→

O

CHO O

|| C |

+ 2H2O

| OH CH

CHOH — CHOH

|

|

HOH2C – CHOH

CHOH — CHO

SO3 H

CONC. ____________ → H2SO4

D-Glucose (Aldohexose) OH |

CH

||

||

C

C

HOCH2

O

CHO

5-Hydroxymethyl furfural

Furfural or Violet or purple 5-Hydroxymethyl ____________________→ colour condensed α-Naphthol furfural product

Precaution

Test tube for this test should be completely dry. Benedict’s Qualitative Test

This test is positive for reducing sugars only. Reducing sugars (mono or disaccharides) by virtue of free aldehydic or ketonic group in their structure reduce cupric ions in alkaline solutions at high temperature. The alkali present in the Benedict’s reagent enolises the reducing sugar to form enediols (different forms of reducing sugar) which are highly reactive and act as strong reducing agent. Benedict’s qualitative reagent contains: i. Copper sulphate Furnishes cupric ions (Cu++) in solution. ii. Sodium carbonate Makes medium alkaline. iii. Sodium citrate Prevents the precipitation of cupric ions as cupric hydroxide by forming a loosely bound cupric-sodium citrate complex which on dissociation gives a continuous supply of cupric ions. Benedict’s qualitative reagent is prepared by dissolving 173 gm of sodium citrate, 90 gm of anhydrous Na2CO3 in 500 ml of distilled water. Slightly heat the contents to dissolve. Filter the solution and make the volume to 850 ml. Dissolve separately 17.3 gm of CuSO4.5H2O in 150 ml of water. Add this solution slowly and with stirring to the above solution—the mixed solution is ready for use.

Carbohydrates 11 Reactions CuSO4 ___→ Cu++ +SO4– – Cu++ + Sodium Citrate ___→ Cupric: Sodium citrate complex CH2COONa HO — C — COO Cu++ CH2 COO Na2CO3 Reducing sugar _________→ Enediols forms of reducing sugar. Boiling Enediols + Cupric: Sodium citrate complex _________→ Cu+ + Mixture of sugar acids. Cu+ + OH– _______→ CuOH Heat 2CuOH _________→ Cu2O ↓ + H2O Cuprous oxide.

Tests

Pipette 5 ml of Benedict’s qualitative reagent in a test tube. Add to it 8 drops of given carbohydrate solution. Boil over a flame or in a boiling water bath for 2 minutes. Cool the solution. An appearance of green, yellow or red precipitate indicates the presence of reducing sugars. The colour of the solution or precipitate gives an approximate amount of reducing sugars present in the solution. Green color — upto 0.5 g% (+) Green precipitate — 0.5-1.0 g% (++) Green to yellow ppt — 1.0-1.5 g% (+++) Yellow to red ppt — 1.5-2.0 g% (++++) Brick red ppt — more than 2.0 g% Fehling Test

This is another reduction test to detect the presence of reducing sugars. It differs from Benedict’s qualitative test in that Fehling reagent contains Rochelle’s salt (Sodium-potassium tartarate) in place of sodium citrate. Fehling solution consists of: Fehling solution A It contains copper sulphate solution. It is prepared by dissolving 34.65 gm of CuSO4-5H2O in 500 ml of distilled water Fehling solution B It contains potassium hydroxide and Rochelle salt (Sodium potassium tartarate). It is prepared by dissolving 125 gm of KOH and 173 gm of Rochelle salt in 500 ml of distilled water. Mix, equal volume of Fehling A and Fehling B before use. Benedict’s reagent is superior to Fehling test. It is semiquantitative and more sensitive. Sodium citrate in Benedict’s reagent and sodium-potassium tartarate (Rochelle’s salt) in Fehling solution prevent the precipitation of cupric hydroxide or cupric carbonate by forming a deep


blue soluble, slightly dissociated complex with the cupric ions. These complexes dissociate sufficiently to provide a continuous supply of ready available cupric ions for oxidation. Test

To 2 ml of Fehling solution (1 ml of Fehling A + 1 ml of Fehling B), add 2 ml of Carbohydrate solution. Mix and boil. Appearance of yellow or red precipitate of cupric oxide indicates the presence of reducing sugars. Barfoed’s Test

This test is used to distinguish monosaccharides from disaccharides by controlling the pH and the time of heating. Barfoed’s test is a reduction test carried out in an acidic medium. The acidity makes it a weaker oxidising reagent. Therefore only monosaccharides, will reduce cupric ions. However if heating is prolonged, disaccharides may be hydrolysed by the acid and the resulting monosaccharide will give the test positive. Reagent

Cupric acetate in lactic acid. Barfoed’s reagent is prepared by dissolving 24 gm of copper acetate in 400 ml of boiling water. To this add 25 ml of 8.5% lactic acid solution. Stir cool the solution and dilute to 500 ml. Test

To 2 ml of Barfoed’s reagent add 2 ml of carbohydrates solution. Place the test tube in boiling water bath for 3 minutes. An appearance of brick red precipitate of cuprous oxide indicates the presence of monosaccharides. Precautions The solution should be boiled for 3 minutes only. Overheating should be avoided because on prolonged heating disaccharides will also give this test positive. Seliwanoff’s Test

This test is positive for ketohexoses only and hence is used in the detection of fructose. Ketohexoses, i.e. fructose on treatment with hydrochloric acid form 5 hydroxymethyl furfural which on condensation with resorcinol gives a cherry red coloured complex. Seliwanoff’s test distinguishes between fructose and glucose. Overheating of the solution is avoided because on continuous boiling, aldoses will also give this test positive because of their conversion to ketoses by hydrochloric acid. Sucrose will also give Seliwanoff’s test positive because the acidity of reagent is sufficient enough to hydrolyse sucrose to glucose and fructose but Benedict’s test will be negative. OH | CONC. HCl FRUCTOSE _____________________→ H2C – 3 H2 O | OH

O

CHO

Cherry red

OH ______________________ → coloured

compound

Carbohydrates 13 Reagent

Resorcinol in concentrated hydrochloric acid (diluted 1:1 with water). Test

To 3 ml of Seliwanoff’s reagent in a test tube add 3 drops of carbohydrate solution. Heat over a flame for 30 seconds only. Cool the solution. An appearance of cherry red colour indicates the presence of fructose. Phenylhydrazine Test (Osazone Formation Test)

Reducing sugar can be distinguished by phenylhydrazine test when characteristic osazone crystals are formed. These osazones have definite crystal structure, precipitation time and melting point and hence help in the identification of reducing sugars.

C6H5 NHNH2 CH = N — NH — C6H5 | C6H5 NHNH2 C=O _________________ → | HO — C — H | + NH3 H — C — OH | + C6H5NH2 H — C — OH | CH2OH

G–Glucose

D–Fructose OSAZONE ____→

CH = N — NH —C6H5 | H — C — OH | HO — C — H | H — C — OH | H — C — OH | CH2OH

CH2OH | C=O | HO — C — H | H — C — OH | H — C — OH | CH2OH

CH = N — NH — C6H5 | C = N — NH C6H5 | HO — C — H | H — C — OH | H — C — OH | CH2OH

Osazones of monosaccharides separate out while in hot. Examine the shape of crystals under low power of microscope. The shape of the osazones are as follow: Glucose: Needle shape osazone. Fructose: Needle shape osazone. Osazone formation test

C

6H 5

NH NH

2

CH2OH | C = N — NH — C6H5 | HO — C — H | H — C — OH | H — C — OH | CH2OH C6H5 NHNH2

CHO | C6H5 NHNH2 C = N — NH C6H5 | HO — C — H | + NH3 H — C — OH | + C6H5NH2 H — C — OH | CH2OH

_________________→

2 NH NH O H5 2 C6 –H

CHO | H — C — OH | HO — C — H | H — C — OH | H — C — OH | CH 2OH

14 Practical Biochemistry for Students Phenylhydrazine reagent

In contains equal part of phenylhydrazine hydrocholoride and anhydrous sodium acetate. It is prepared only at the time of reaction. Test

In a clear and dry test tube, take approximately 0.5 gm. of phenylhydrazine mixture (equal part of phenylhydrazine hydrochloride and anhydrous sodium acetate). Add 5 ml of carbohydrate solution and 1-2 drops of glacial acetic acid. Mix and place the test tube in boiling water bath for 30 minutes.

Needle shaped

Bial’s Test for Pentoses

This is a sensitive test for the detection of pentoses. Pentoses on heating with strong acid are converted to furfural which reacts with the coloured compound produced when orcinol and ferric chloride react with each other. PENTOSE

CONC. HCl ___________ → – 3 H2 O

OH |

CHO

→ Blue, Green compound

+ Fe Cl3 H3C

O

_→

Coloured solution

OH

Bial’s reagent (0.2% orcinol in concentrated hydrochloric acid). To 5 ml of Bial’s reagent acid add 10 drops of pentose solution (i.e. Aarabinose). Boil. Appearance of green colour. To Detect Galactose Mucic acid test

Galactose on oxidation with strong acid gives mucic acid which crystallises out and can be observed microscopically. Test

In a test tube take 1 ml of galactose solution followed by 1 ml of concentrated nitric acid. Evaporate the mixture by using boiling water bath for 1½ hours in a furming cup board. Keep it overnight. Examine a drop of the crystals under low power of microscope.

Carbohydrates 15

| H — C — OH | H — C — OH | HO — C — H | HO — C — H | HO — C | CH2OH Galactose

O

Δ ______ → OXn

COOH | H — C — OH | HO — C — H | HO — C — H | H — C — OH | COOH

+ HOH

Mucic acid

To Study the Chemical Reactions of Disaccharides

The most common disaccharides are maltose, lactose and sucrose. Maltose and lactose are reducing disaccharides where as sucrose is a non-reducing disaccharide. One percent solution of each maltose, lactose and sucrose are provided. 1. Molisch test Dissaccharides are first hydrolysed to constituent monosaccharides which are then dehydrated. Test In a clean and dry test tube, take 2 ml of the carbohydrate solution. Add 2 drops of ethanolic α-naphthol (Molisch reagent). Mix. Incline the test tube and cautiously add 2 ml of concentrated H2SO4 by the side of the test tube. An appearance of violet or purple ring at the junction of two solutions indicate the presence of dicarbohydrates. 2. Benedict’s qualitative reagent Test Pipette 5 ml of Benedicts qualitative reagent in a test tube. Add 8 drops of given dicarbohydrate solution. Boil for 2 minutes. An appearance of green, yellow or red precipitate indicates the presence of disaccharides. Maltose and lactose give Benedict’s qualitative test positive whereas with sucrose the test is negative, i.e. no reduction is observed. 3. Barfoed’s test Test To 2 ml of Barfoed’s reagent add 2 ml of disaccharide solution. Place the test tube in boiling water bath for 3 minutes. No change in colour indicates the presence of disaccharides in the solution. Negative for disaccharides. 4. Osazone formation (i.e. phenylhydrazine test) Osazones of disaccharides separate out on cooling. Test In a clean and dry test tube, take roughly 0.5 g of phenylhydrazine mixture. Add 5 ml of disaccharide solution and 2 drops of glacial acetic acid. Mix. Place the test tube in boiling water bath for 30 minutes. After 30 minutes, take out the test tube from the boiling water bath and allow it to cool by itself in a test tube rack (Do not disturb the test tube in between as the osazones of disaccharides separates out on slow cooling). Appearance of yellow crystals takes place. Observe the shape of crystals under low power microscope.


The shape of osazones are: Malatose : Sunflower shape Lactose : Cotton ball shape Sucrose It will give Benedict’s qualitative test negative.

Lactose (Cotton ball)

Maltose (Sun flower)

Sucrose is Confirmed As

Test Take 5 ml of sucrose solution in a test tube. Add to it 1-2 drops of concentrated hydrochloric acid. Boil the contents for few minutes (2-5 minutes). Cool the solution. Divide it in two parts. Neutralise one part of the solution with sodium carbonate and carry out the Benedict’s qualitative test. The test will be positive. Carry out the Barfoed’s test with the other part of the solution. It will be positive now. Osazone Formation

Carry out the osazone test with the hydrolysate solution of sucrose. Appearance of needle shaped crystals. To Study the Chemical Reactions of Polysaccharides

Solutions provided are 1% starch and 1% dextrins. 1. Molisch test 2. Iodine test. This test is used for polysaccharides detection and differentiation. Iodine forms a coordination complex between the helically coiled polysaccharides chain and the iodine centrally located with in the helix due to adsorption. The iodine colour obtained with the polysaccharides depends upon the length of the unbranched or linear (α1, 4 linkage) chain available for complex formation. Amylose a linear chain component of starch gives a deep blue colour. Amylopectin, a branched chain component of starch gives a purple colour. Glycogen gives a reddish brown colour. Dextrins, formed from the partial hydrolysis of starch gives colours ranging from brown red to colourless depending on the size of the molecule. Cellulose, inulin, disaccharides or monosaccharides gives no colour with iodine. Polysaccharides

Colour with iodine

Starch Amylose Amylopectin Glycogen Dextrins Cellulose or inulin Disaccharides or monosaccharides

Blue colour Blue colour Purple colour Brown red colour Brown to colourless No colour No colour

Carbohydrates 17 Test

In two ml of carbohydrate solutions, add few drops of hydrochloric acid (to make the medium acidic) followed by 1 ml of iodine solution. Mix and observe the colour. No change in colour indicates the absence of polysaccharides. Hydrolysis i. Acid hydrolysis

In a 100 ml conical flask, take 20 ml of 1% starch solution. Add 5 ml of 2N HCl (prepared by diluting one part of concentrated HCl to 4 parts of water). Divide the solution in five equal parts (i.e. 5 ml each) in five different tubes and place the tubes in a boiling water bath. Remove the tube from the boiling water bath at an intervals of 1, 5, 8, 12 and 20 minutes. Now divide the solution in each tubes in two parts: i. With one part, perform Benedict’s qualitative test, after making the solution alkaline (i.e. by neutralising the acidity of the solution with sodium carbonate). ii. Second part, perform iodine test. Time

Colour with iodine

Benedict’s test

Reduction of hydrolysis (extend)

Product

1 minute 5 minute 8 minute 12 minute 20 minute

Blue Violet Reddish violet Red No colour

Blue Green Red Red Red

No reduction Reduction starts(+) Initiation of reduction (++) Partialreduction(+++) Completely reduced(++++)

Starch Amylodextrins Amylo and erythrodextrin Achrodextrin Glucose

The acidified starch takes about 20 minutes for complete hydrolysis.

ii. Enzymatic hydrolysis

Take a clean test tube and collect some saliva in it. Take two dry test tubes and label them as blank and experimental. Add 5 ml of 2% starch solution and 1 ml of Citrate buffer (pH 6.0, prepared in 0.25 M NaCl) in each test tube. Mix well. Now add 1 ml of distilled water only in the blank and 1 ml of saliva in the test. Keep the test tubes for 30 minutes at room remperature. 1. In Blank a. Iodine test : Blue colour b. Benedict’s test : Negative. 2. In Test a. Iodine test : Negative b. Benedict’s test : Red precipitate. Iodine test Benedict’s test

Blank Blue colour Negative

Test sample Negative Red precipitate


Starch with saliva shows reduction as starch is converted to glucose which is a reducing sugar. Where as starch without saliva is not broken up in to smaller molecules because there is no hydrolysis. Summary of Qualitative Test

Glucose

Fructose

Lactose

Maltose

Sucrose

Starch

Molisch test Iodine test Benedict’s test Barfoed’s test Seliwanoff’s test Phenylhydrazine test Identification of Unknown Carbohydrate Solution

1. No need to perform Molisch test, as the unknown solution is carbohydrate in nature. 2. Iodine test Positive for polysaccharides. Depending upon colour, the polysaccharide is identified. If negative, polysaccharides are absent. 3. Benedict’s test Positive for reducing sugars. Reducing sugars can be monosaccharides or disaccharides. If the Benedict’s test is negative, it means reducing sugars are absent. Absence of Benedict’s test, indicates the presence of non-reducing disaccharide, i.e sucrose. 4. Barfoed’s test Positive for monosaccharides. Barfoed’s test differentiates between monosaccharides and disaccharides. 5. Seliwanoff’s test Positive for ketohexoses. Indicates the presence of fructose. 6. Osazone test For the identification of particular carbohydrates. Step I Identification of unknown carbohydrate Iodine test | No change in colour |

| | Red colour | |

Clear solution (Dextrin)

| Blue colour (starch) | Hazy solution (Glycogen)

Step II

| Positive (Reducing sugar) Glucose, Fructose, Galactose, Mannose, Lactose, Maltose

Benedict’s test |

| Negative (sucrose) (Nonreducing sugar)

Carbohydrates 19 Step III

| Positive (Glucose, Fructose, Galactose)

Barfoed’s test |

| Negative (Lactose, Maltose)

Step IV A

| Positive (Fructose)

Seliwanoff’s test |

| Negative (Glucose, Mannose, Galactose)

Step IV B Osazone test | Precipitate before heating (Mannose)

| Yellow insoluble during heating needle shape (Glucose, Fructose, Mannose)

|

| Small ball like cluster thronedge (Galactose)

| Osazone separates on cooling

| Cotton ball shape (Lactose)

| |

| Sunflower shape (Maltose)


4

Achromic Point Determination of Achromic Point of Your Own Saliva

This is a simple enzymatic hydrolysis of starch. This is different from acid hydrolysis. Enzymatic hydrolysis gives bigger units and does not break the branched point (i.e. amylopectin chains) or α-1, 6 linkage. Achromic point is that point at which no colour is obtained with iodine. Chromic period is that time period which is required to obtain achromic point when enzymatic hydrolysis is being performed. In an animal, there are enzymes which break only the α-1, 4 linkage, i.e. they break only the straight chains. Enzymes for breaking the α-linkage are present only in the plants. Salivary amylase is a β-linkage enzyme which acts randomly breaking the starch into monosaccharides (i.e. glucose) and maltose units. At places where there is α-1, 6 linkage oligosaccharide units, the breaking proceeds in the following order. i. Once enzyme activity starts we get first soluble starch. ii. The first product hereafter formed is amylodextrins. iii. Next we get erythrodextins which gives reddish colour with iodine. iv. No colour with iodine is got when we get achrodextrin. This point is called achromic point. Reagents required 1% starch Buffer—pH 6.7 1% NaCl. Procedure

Take 5 ml of 1% starch. Add 2 ml of buffer (pH 6.7) to it and 1 ml of 1% NaCI. Mix.Take out 5 ml of it. This is prepared buffer-starch solution. Rinse your mouth with water. Take 10-15 ml of warm water in mouth and rotate the water with tongue. Take this in a polythene beaker and now taken 5 ml of this. Take a tile having grooves and put iodine in equal amount in each groove.

0 sec

30 sec

2 min

2.5 min

1 min

(20)

1.5 min

Achromic Point 21

Now mix the saliva and prepared buffer-starch solution, and a drop of it at zero hour and then at intervals of 30 seconds in iodine till the achromic point is reached. Note the chromic period. Observations Time

Colour

30 sec. 1 minute 1½ minute 2 minute 2½ minutes

Blue Light blue Lighter blue Faint blue Colourless

Result and Conclusion

The achromic point of saliva is 2½ minutes, this means that 2½ minutes are taken till achromic point is got and formation of achrodextrin takes place. Now if we put the hydrolysed solution in a drop of Fehling’s very little red precipitate is got because enzyme hydrolysis does not produce many monomers.


5 Proteins

To Study the General Reactions of Proteins 1. Biuret Test

Biuret test is given by all compounds that contains two or more peptide bonds. Since proteins are polypeptide, hence it is a general tests for proteins. The name of the reaction is derived from the organic compound, a biuret, obtained by heating urea at high temperature which gives a positive test. ......

......

NH2CONH2 + NH2CONH2

O O || ....... .....|| ... → H N — C — NH — C — NH 2 2 .......... ............. Biuret

180°C _________

Reagent

Biuret reagent contains dilute copper sulphate in strong alkali. Biuret Reagent (Stock) It is prepared by dissolving 22.5 gm of Rochelle salt in 200 ml of 0.2N NaOH to this is added 7.5 gm of CuSO4 5H2O with constant stirring. Then is added 2.5 gm of KI and make the volume to 500 ml with 0.2 N NaOH. Working biuret reagent is prepared by dissolving 50 gm of stock biuret reagent to 250 ml with 0.2N NaOH containing 5 gm of KI per litre. Reaction

The purple or violet colour produced is believed to be due to a coordinate complex between the cupric ions and four nitrogen atoms, two from each of two adjacent peptide chains. Test

Take 6 ml of 5% NaOH, in a test tube and add few drops of 1% CuSO4 solution till blue colour solution is produced. Divide the solution, i.e. 3 ml each in two test tubes marked experimental test ‘A’ and control test ‘C’. • To ‘A’ add 3 ml of protein solution. • To ‘C’ add 3 ml of distilled water.

H | O == C

H | H — C == O N

O H N

H R — C | |

CH — R | |

Cu++

C == O | | N — CH — R H |

O == C N H R — C |

O

H

(22)

H

Proteins 23

Appearance of purple or violet colour in the tube ‘A’ shows the presence of proteins with respect to tube ‘C’ which serves as a control for the test. 2. Ninhydrin Test

Proteins containing free α-amino acid radical in the molecule reacts with ninhydrin to give a blue-violet coloured compound. Ninhydrin test is not given by protein and hydroxy proline because no free α-amino group is present. They give only a yellow colour. H || C 2

COOH | Δ + H2N — C — H ______→ | OH R

OH C

C || O

O || C

O || C C = N — C C || OH

C || O Purple coloured complex

Ninhydrin

Δ

NH3

Blue colour (RUHEMANN'S PURPLE) Ninhydrin test

Reagent

Ninhydrin dissolved in acetone. Reaction

Protein solution heated with ninhydrin leads to the formation of a blue coloured compound called Ruhemann’s complex. Test

Take 1 ml of protein solution, add to it 2-3 drops of freshly prepared ninhydrin solution. Heat the solution. Appearance of blue colour indicates the presence of proteins. To Study the R-Groups of Proteins

The most important aspect of R-group of proteins is that of their nutritional significance. These acids which the animal body is unable to synthesise either completely or in amounts sufficient to normal growth maintainence must be supplied in the food. These so called essential amino acids can be tested for in various way and can in many cases, be detected and determined by means of simple colour reactions based upon the qualitative test. A variety of colour reactions specific to particular functional groups in amino acids are known. They are useful in both the qualitative and quantitative identification of particular amino acids. 1. Xanthoproteic test (For benzenoid radical) The reaction is based upon the nitration of the benzene ring with concentrated HNO3 yielding yellow derivatives of nitrobenzene which turns to orange in alkaline medium.

24 Practical Biochemistry for Students Test

Take 3 ml of test solution (protein solution). Add 1 ml of concentrated HNO3. A white precipitate due to denaturation of protein is formed. Boil the solution. A yellow solution derivative is formed due to nitration of benzene ring. Cool the solution and make it alkaline with 20% NaOH, orange colour is produced. 2. Millon’s test (For hydroxy benzene radical) This test is specific for tyrosine and is an indication of the presence of tyrosine in the protein because tyrosine is the only amino acid containing hydroxy phenyl group. A pink colour is obtained in this test is due to mercury complex of nitrophenol derivatives. Reagent

Mercuric nitrate dissolved in concentrated HNO3. Test

Take 1 ml of test solution in a test tube. Acidify with dilute-H2SO4. Check with litmus. Add 1 ml of Millon’s reagent. Boil the solution. A yellow precipitate adheres to the side of the test tube. Cool the solution under tap water. Add a drop of 1% sodium nitrite (NaNO2) and gently warm. The precipitate or solution turns red. 3. Hopkins-Cole test (For indole group) This test is specific for tryptophan and is an indication of the presence of tryptophan in the protein. Test

Take 1 ml of test solution and add few drops of 1 : 500 commercial formalin (40% formaldehyde). Add 2 drops of mercuric sulphate (i.e. 10% HgSO4 in 10% H2SO4). Mix. Incline the test tube and add 1 ml of concentrated H2SO4 by the side of the test tube. A purple ring is formed at the junction of the two layers. 4. Sakaguchi test (For guanidino group) This test is given by all compounds containing guanidino group and thus is an indication of the presence of arginine present either in free or in combined form. Guanidines in alkaline solution give a red colour in the presence of α-naphthol and sodium hypobromite. Test

Take 3 ml of test solution in a test tube. Add 1 ml of 5% NaOH, 2 drops of ethanolic α-napthol and 2 drops of 10% sodium hypobromite. Mix well. Wait for 5 minutes. Development of bright red colour takes place. Run a control by taking 3 ml of distilled water instead of protein solution and add all other reagents as in test. 5. Test for cystine or cysteine (–S–S–and–SH radicals) Sulphur is present in proteins as cystine, cysteine or methionine.

Proteins 25 Test

Take 2 ml of test solution and add 2 ml 40% NaOH. Boil for 2 minutes. Cool and add lead acetate solution. A black precipitate of PbS insoluble in dilute HCl is formed. 6. Test for free—SH radical Take 1 ml of test solution. Add few crystals of ammonium sulphate. To it add few drops of freshly prepared solution of sodium nitroprusside and 1 ml of liquor ammonia. Development of rose red colour takes place. Colour tests

Albumin

Globu lin

Phenylalanine

Tyrosine

Arginine

Guanidine

Tryptophan

Biuret test Ninhydrin test Xanthoprotic test Millon’s test Hopkin’sCole test Sakaguchi test Cystine-cysteine sulphur test

Precepitation Reactions of Proteins

1. Precipitation by heavy metals. 2. Precipitation by alkaloidal reagents. 3. Heat. 1. Precipitation by heavy metals (10% lead acetate, 10% CuSO4 and 10% ZnSO4)

Proteins are precipitated from solutions by salts of heavy metals probably by combination of the metal ions with the anionic form of the protein. On the alkaline side of the isoelectric point proteins exits as negative ions. Test

1. Take 3 ml of protein solution. Add 2 drops of 5% NaOH. Mix, followed by 2 ml of 10% lead acetate solution. Appearance of white precipitate. 2. Take 3 ml of protein solution. Add 2 drops of 5% NaOH. Mix, followed by 2 ml of 10% CuSO4 solution. A light blue precipitate appears. 3. Take 3 ml of protein solution. Add 2 drops of 5% NaOH. Mix, followed by 2 ml of 10% ZnSO4 solution. An intense white precipitate appears. 2. Precipitation by alkaloidal reagents

Proteins are precipitated from the solution by combination between the acid anions and the positively charged protein molecule by forming all insoluble complex. The alkaloidal reagents precipitate proteins by combination of the acidic radical of the former with the cationic form of the protein, which predominates when the solution is on the acidic radical of the former with the cationic form of the protein, which predominates when the solution is on the acidic side of the isoelectric point.


a. To 3 ml of protein solution, add few drops of metaphosphoric acid. A dirty white precipitate appears. b. To 3 ml of protein solution, add few drops of 20% sulphosalicyclic acid. White precipitate appears. c. To 3 ml of protein solution, add 3 ml of Esbach reagent. A precipitate appears. d. To 3 ml of protein solution, add few drops of glacial acetic acid followed by 1 ml of 5% potassium ferrocyanide. A deep yellow precipitate appears. 3. Precipitation of proteins by heating

Proteins which are precipitated when their solutions are heated are termed as heat coaguable proteins. This property of proteins is made use of in the detection of albumin in urine simply by heating the urine. Solubilities of several proteins in various solvents are shown below:

Albumin

Distilled water

Dilute acids

Dilute alkali

Soluble

Soluble

Soluble

Ammonium sulphate Soluble

Insoluble

Globulins

Insoluble

Soluble

Soluble

Insoluble

Insoluble

Gelatin

Soluble

Soluble

Soluble

Insoluble

Insoluble

Difference in Albumin and Globulin Albumin 1. It is obtained by full saturation of (NH4)2 SO4 2. It is soluble in water. 3. It is a smaller molecule having more charge. 4. In isoelectric point field it travels faster than globulins.

Globulin 1. It is obtained by half saturation of ammonium sulphate. 2. It is insoluble in water but soluble in dilute mineral acids and salt solution. 3. It is bigger molecule with less charge. 4. Due to big molecule and less charge it travel slowly in isoelectric field.

Isoelectric Point 27

6

Isoelectric Point Determine the isoelectric point of casein, the pK for acetic acid is 4.74. Things required

Solution of casein in 0.1 N sodium acetate 1 N acetic acid 0.1 N acetic acid 0.01 N acetic acid Principle

Isoelectric pH is defined as the pH at which the protein molecule does not migrate to the cathode or to the anode in an electric field. At this pH, the protein molecule exists as zwitterion. At the isoelectric point, the precipitation of protein is maximum, i.e. in other words the solubility of proteins is minimum. At the isoelectric pH, the electrostatic repulsive force which normally prevents the protein molecules from coming together is minimum as a result of no net change on the protein molecule and hence give rise to maximum precipitation. Procedure

Arrange a series of nine test tubes, clean and dry. Label the test tubes as 1, 2....9. Add the following solution in given order. 1

2

3

4

5

6

7

8

9

Protein solution

1 ml

1 ml

1 ml

1 ml

1 ml

1 ml

1 ml

1 ml

1 ml

Water (ml)

8.38

7.75

8.75

8.50

8.0

7.0

5.0

1.0

7.40

0.01 N CH8COOH (ml)

0.62

1.25

–

–

–

–

–

–

–

0.01 N CH3COOH (ml)

–

–

0.25

0.50

1.00

2.00

4.00

8.00

–

1.0 N CH3COOH (ml)

–

–

–

–

–

–

–

–

1.60

Mix, the contents after every addition. Observe the tube after 5 minutes and see which tubes has the maximum turbidity. Again observe after 10 minutes, 30 minutes and note again which tube has the maximum turbidity. The test tube number 5 shows the maximum turbidity, the corresponding pH is the isoelectric point of casein. According to Handerson-Haselbatch equation. [Salt] pH = pK + log _____________ [Acid] pKa for acetic acid is 4.74 (27)


[Salt] = Sodium acetate = 0.1 N [Acid] = Acetic acid = 0.1 N Maximum turbidity 0.1 pH = pK + log __________ 0.1 pH = 4.74 The isoelectric point (pI) for casein is 4.74. The pH of the respecting tubes will be as Tube No. 1. 2. 3. 4. 5. 6. 7. 8. 9.

Zero time

Precipitate formed Ten minutes

Thirty minutes

– – + ++ ++ ++ + + –

– – + ++ +++ ++ + + –

– – + – ++++ + + + –

pH 5.95 5.64 5.34 5.04 4.74 4.44 4.12 3.14 3.54

Lipids 29

7

Lipids To Perform General Tests for Lipids

Lipids are defined as a group of fatty nature which are insoluble in water but soluble in nonpolar solvents like ether, chloroform, etc. Lipids thus include fats, oils, waxes and related compounds. Lipids are classified as simple and complex. Experiment 1. Solubility

Take three perfectly dry test tubes. To the first add 2 ml of distilled water, to the second 2 ml of ethyl alcohol and to the third 2 ml of chloroform. To each of the three test tubes add 3 drops of the provided oil. Shake gently and observe. Observation

The provided oil is insoluble in water, hence it floats on the surface of water, forming a separate layer. The oil is fairly soluble in ethyl being heavier than alcohol, some of the oil (undissolves, settle down at the bottom as minute droplets, whereas it is extremely soluble in chloroform. The resulting solution is clear. Inference

Oils are insoluble in water sparingly soluble in alcohols but extremely soluble in fat solvents like chloroform. Experiment 2. Emulsification

Take three perfectly clean test tubes. • To the first add 5 ml of distilled water. • To the second add 5 ml of bile salt solution. • To the third add 5 ml of household detergent (Surf solution). • To each of the three test tubes, add 3 ml of the provided oil. • Shake vigorously and observe. Observation

In the first tube a temporary emulsion of oil in water is formed. On vigorous shaking, this emulsion is unstable and so breaks down early. In case of bile salt solution and household (29)


detergent solution, a highly stable emulsion is formed. The emulsion is very fine and breaks down after long time. Inference

Oils form a coarse and unstable emulsion with water. This readily breaks down. So this concludes that oils do not reduce the surface tension of water. In case of bile salt solution and household detergent, the emulsion is fine and very stable because bile salts and household detergents reduce the surface tension of water as a result of which oil fragments into small droplets which form an emulsion. Experiment 3. Acrolein test

Take a clear test tube and add 4 drops of provided oil to it. Then add a pinch of potassium bisulphite and heat vigorously. Smell the fumes of the gas which come from one of the test tube. Observation

Pungent smelling fumes arise from one of the test tube. Inference

Acrolein is evolved which has a pungent smell. All the triglycerides give this test. Experiment 4. Saponification

Take a clean dry test tube and add 0.5 ml of the provided oil then add 2.5 ml of ethanol to it and mix well. After mixing, add 10 ml of 10% alcoholic sodium hydroxide, shake well and keep in boiling water bath for 15 minutes. Take the test tube after 15 minutes and add water so that the resulting volume of the solution is 15-20 ml. Shake well to dissolve. Divide the contents into 4 equal parts in four different test tubes. a. To the first part, add 3 ml of conc. HCl and shake well. b. To the second part, add an equal volume of saturated NaCl solution. c. To the third part, add 3 drops of CaCl2. d. To the fourth part, add 3 drops of MgCl2. Observation

A white precipitate of liberated fatty acid is obtained. Sodium salts of fatty acid rise up and form a pale white layer. A white precipitate of calcium salt of fatty acid is formed. A white precipitate of magnesium salt of fatty acid is formed.

Lipids 31 Inference

The liberated fatty acid being insoluble in water is precipitated. The sodium salt of fatty acid is salted out. Calcium salt of fatty acid being insoluble is precipitated. Experiment 5. Test for unsaturation

Take a clean dry test tube and add 3 drops of oil in it. Then add 2 ml of ethyl alcohol and mix well. Then add 0.5% alcoholic bromine solution until bromine solution imparts its own colour. Observation

The colour of the solution was colourless at first but gradually turned pale yellow, i.e. the colour of the bromine solution itself. Inference

Bromine goes into the solution forming a dibromide, i.e. it add to the double bonds. In other words, bromine solution is decolourised, but when all the double bonds are saturated the bromine solution imparts its own colour. Experiment 6. Test for cholesterol

a. Libermann-Burchard reaction Take a perfectly dry test tube and add 2 ml of CHCl3 solution of cholesterol to it. Then add 10 drops of acetic anhydride and mix well. Then add drops of concentrated H2SO4 from the sides of the test tube. Keep it in dark after mixing well. Observation

A deep green coloured solution is obtained. Inference

This indicates the presence of cholesterol. Experiment

b. Salkowski reaction Take a perfectly clean and dry test tube and add to it 2 ml cholesterol. Solution prepared in cholesterol. Then add an equal volume of concentrated H2SO4 dropwise along the side of the test tube. Observation

Two layers are formed. The upper brown one is formed by CHCl3 and the lower one yellow in colour formed by concentrated H2SO4. This layer of conc. H2SO4 gives fluorescence. Inference

This indicates the presence of cholesterol.


8

Saponification Number Determination of Saponification Number of an Oil

Saponification number is defined as the number of milligrams of KOH required to saponify completely 1 gm of fat. Since fats are mixture of triglycerides, most of which are of mixed type, so saponification number is a measure of average moleculer weight of the fatty acids comprising the fats (i.e. the measure of the average chain length of the fatty acid). Saponification number is an important constant particularly in distinguishing or identifying certain oils. Procedure

Take a clean and dry 100 ml conical flask. Using a 2 ml pipette, transfer 1.5 ml of the oil sample provided in the conical flask. Add 15 ml of 0.5 N ethanolic KOH into the flask containing the oil. Mix the contents well. Place a funnel at the neck of the conical flask (the stem of funnel acts as a condenser) and place it in the boiling water bath for half an hour till all the oil globules disappear and a yellow cake is formed by potassium salts of fatty acids. After half an hour, take out the conical flask, cool it to room temperature. Add 20 ml of distilled water in the flask, and shake till a clear solution is formed. Now add 1-2 drops of phenolphthalein as an indicator. Titrate with 0.5 NHCl till the colour is changed from red to colourless. Note the titre value. Also run a blank titration, without using oil under similar conditions and note the titre valve. Observation

• Volume of HCl required for saponified solution = T ml. • Volume of HCl required for blank titration = B ml. • Volume of HCl utilised = (B–T) = Blank test reading value. According to Normality equation • 1 ml of 0.5 N HCl = 1 ml of 0.5 N KOH • (B–T) ml of 0.5 N HCl = (B–T) ml of 0.5 N KOH • 1 ml of 0.5 N KOH = 28 mg of KOH • (B–T) ml of 0.5 N KOH = 28 (B–T) mg of KOH • Weight of oil = Volume × density = 1.5 × 0.9 = 1.35 gm. • Saponification number of 1.35 gm of oil = 28 × (B–T) mg or KOH 28 × (B – T) mg of KOH • Saponification number of 1 gm of oil = ____________________________________ 1.35 (32)

Iodine Number 33

9

Iodine Number To Determine the Iodine Number of the Given Oil

Iodine number is defined as the number of grams of iodine absorbed by 100 gm of the fat. Halogens, e.g. iodine or bromine are taken up by the fats because of the presence of double bonds present in the fatty acid part of the fat. Iodine number is a measure of the degree of unsaturation of a fat. The higher the iodine number, the more is the unsaturation present in the fat. Iodine number is a useful characteristic for assessment of both purity and nutritive value of the fat. Bromine is often used instead of iodine because it is more reactive. The value is influenced by the percentage of each unsaturated fatty acid, the degree of unsaturation of each acid and the mean molecular weight of the fat. The iodine members of some important fats are mentioned below: Fats

Iodine numbers

Butter fat Human fat Peanut oil Cornoil Soyabean oil Linseed oil

26-28 65-70 80-90 110-125 137-143 170-200

Principle

The given amount of fat is treated with a measured excess of Hanus solution. I

Br | | R — CH = CH — COOH + I — Br _______→ R — CH — CH — COOH + I — Br Excess of Left Hanus solution over

To the left over Hanus solution is added potassium iodide solution. The iodine thus liberated is titrated against standard solution of “hypo” (Na2S2O3) using starch as an indicator. The colour change is from deep blue-black to white which marks the end point of the titration. I Br+KI ________→ 2Na2S2O3 + I2 ______→ At the end point

I2+Starch ______→

I2 ↑ + KBr 2Na2S4O6+2NaI Blue colour

(33)

34 Practical Biochemistry for Students Procedure Test

In a 250 ml conical flask, add 5 ml of given oil sample (the oil sample is dissolved in CCl4. The concentration of the oil sample is 5 g%), followed by 10 ml of Hanus solution. Mix well, cover the mouth of the flask with a paper and keep it for 30 minutes for reaction to take place. After 30 minutes, add 5 ml of KI solution into it. Mix well, followed by 25 ml of distilled water. Add 4-5 drops of starch as indicator. The colour of the solution turn blue-black. Titrate the contents of the flask with N/10 Na2S2O3 till the colour changes from blue-black to white, which marks the end point of the titration. Note down the titre value which is x ml. Blank

In 250 ml conical flask add 5 ml of CCl4 only instead of oil sample and repeat the same procedure as in the test. Note down the titre value which is y ml. The difference between the two (i.e. blank-test) gives the amount of Na2S2O3 utilised in titrating the IBr which was used in saturating the unsaturated fatty acid moiety, i.e. the volume of I Br required to saturate the oil = (Blank-Test) value. In the test titration the excess of I-Br, i.e. the left over I-Br is titrated against Na2S2O3. In blank titration the excess of I-Br (as in the first case) and the actual volume of I-Br which would have used up by an oil to be saturated are together titrated against Na2S2O3. Calculation

Titre value obtained for test titration = x ml Titre value obtained for blank titration = y ml According to Normality equation 1 ml of N/10 Na2S2O3 solution ≡ 1 ml of N/10 I-Br solution 1 ml of N/10 I-Br solution 1 ml of n/10 Na2S2O3 solution ≡ ≡ 1 ml of N/10 iodine solution Equivalent weight of iodine = 127. 127 1 1 ml of N/10 iodine = ________ × ____ = 0.0127 gm. 1000 10 (Two molecules of Na2S2O3 are equivalent to one molecule of iodine; thus one molecule of Na2S2O3 is equivalent to one atom of iodine). 1 ml of N/10 Na2S2O3 solution = 0.127 gm of iodine. Amount of iodine absorbed by given amount of oil or fat = (y-x) × 0.0127 gm of iodine The concentration of oil sample is 5 gm%, i.e. 5 ml of oil = 0.25 gm of oil. 0.25 gm of oil or fat consumes (y – x) × 0.0127 gm of iodine. y–x ∴ 100 gm of oil or fat consumes __________ × 0.0127 gm of iodine. 0.25 y – x × 0.0127 × 100 The iodine number is ____________________________ gm of iodine. 0.25

Formal Titration 35

10

Formal Titration Formal Titration

Sorensen’ formal titration method is used for the estimation of free carboxyl group in amino acids and in mixture of amino acids. By this method one can determine the increase in carboxyl groups which accompanies the enzymatic hydrolysis of proteins. Principle

Amino acids by virtue of zwitterion formation are merely neutral in reaction. If formaldehyde is added to a solution of amino acids an adduct is formed at the amino group leaving the carboxyl group free. + NH3 | R — CH — COO–

+

2HCHO

____________→

CH2OH N CH2OH | | R — CH — COO– – +

H+

In other words, formaldehyde suppresses the basicity of amino group. So that carboxyl group can exert its maximum acidity and hence can be titrated against standard solution of alkali using phenolphthalein as an indicator. Procedure

Preparation of neutral formation. Pipette 2 ml of formalin into a test tube and add 2 drops of indicator phenolphthalein to it. Mix well and then titrate the above solution against N/10 NaOH. Note the amount of NaOH used up in the titration. In a 100 ml conical flask, take 10 ml of glycine solution, add to it 2 ml of above neutralised formalin solution. Then titrate the resulting solution against N/10 NaOH. The end point in each case is observed when the colour just turns pink. Observation

Volume of NaOH utilised in neutralising glycine solution along with neutralised formalin = x ml. Calculation

10 ml of glycine = x ml of N/10 NaOH = x ml of N/10 glycine (35)


1000 ml of 1 N glycine solution contains 75 gm of glycine, i.e 1000 ml of N/10 glycine solution contains 7.5 gm of glycine 7.5 × χ χ ml of N/10 glycine contains _________ gm of glycine 1000

Gastric Analysis 37

11

Gastric Analysis Stimuli which cause secretion of gastric juice are of two types viz. Chemical stimulus—by giving dry toast, oatmeal, gruel and alcohol and humoral stimulus— by giving 0.5 gm of histamine. The patient is given a light super the previous evening and in the morning. Ryle’s tube is inserted through the nose to suck the gastric juice. The patient is given 50 ml of 7% alcohol after every 15 minutes, 10 ml of gastric juice is sucked up. Collect 4 such samples; 15 minutes, 30 minutes, 45 minutes and 1 hour. Then inject 0.5 gm of histamine intramuscularly, after every 15 minutes, suck 10 ml of gastric juice. Collect four such samples 1 hour 15 minutes, 1 hour 30 minutes, 1 hour 45 minutes and 2 hours. Histamine Test

Histamine has definite advantages over the ordinary test meal. It evokes a maximum response and often produces acid secretion where the ordinary test meal fails completely. It adds nothing to the stomach so that a pure juice, undiluted and uncontaminated is obtained and no neutralisation of the acid can take place through food constituents. Combined Alcohol and Histamine Tests

Alcohol as a physiological stimulant has several advantages over gruel, charcoal biscuit or dry toast. It is much less objectionable to take, is easily swallowed and evokes a feeling of pleasure instead of distaste on the part of the subject. The gastric juice obtained is ideal for analysis. The main content of the gastric juice is hydrochloric acid. In a normal person the pH of the gastric juice is 1 to 2. The HCl of gastric juice is found in two forms, i.e. the free (uncombined) HCl and combined acidity, the protein bound HCl (protein hydrochloride). Total acidity = Free acidity + combined acidity (i.e. protein bound acidity). Principle

A known volume of gastric juice is taken. 2 drops of Topfer’s reagent (P-dimethylaminoazobenzene in alcohol) as indicator is added. The colour of the solution turns yellow due to the presence of free HCl. Yellow colour indicates a pH of 2.90. The end point appears when the colour turns orange. The pH is 3.60 at the end point. The free HCl being liberated. Proceed to titrate the protein bound HCl. 2 drops of phenolphthalein are added to the solution (which has been titrated for free HCl) and it is titrated against N/10 NaOH. The end point appears when the colour becomes pinkish. The pH at this end point is 8.50. (37)


The first titre value with Topfer’s reagent expressed in units of N/10 HCl per 100 ml of gastric juice gives free acidity. The second titre value with phenolphthalein expressed in units of N/10 HCl per 100 ml of gastric juice gives combined acidity. Procedure

In a conical flask, pipette 1 ml of gastric juice. Then add 1-2 drops of Topfer’s indicator to it. If an orange colour appears (pH 3.6) then free acidity is zero. In that case add a drop of phenolphthalein and titrate with N/100 NaOH till the end point is reached (pH 8.50). The titre value expressed as units of N/10 HCl per 100 ml of gastric juice gives the combined acidity (=Total acidity because free acidity is zero). If by adding Topfer’s reagent a yellow colour is obtained (pH is 2.9), titrate with N/100 NaOH till a orange colour is obtained. The titre value expressed in units of N/100 HCl per 100 ml of gastric juice gives free acidity. Now in the same solution, add a drop of two of phenolphthalein and titrate with N/100 NaOH till a pink colour (pH 8.5) is obtained. This is the end point. This titre value expressed in units of N/100 HCl per 100 ml of gastric juice gives combined acidity.

1 ml of N/100 NaOH ≡ 1 ml of N/100 HCl 1 ml of N/10 HCl ≡ 10 ml of N/100 HCl 10 ml of N/100 HCl ≡ 10 ml of N/100 NaOH ≡ 1 ml of N/10 HCl If the titre value is x, then 10 × ml of N/100 HCl = x ml of N/10 HCl Total acidity in 1 ml of gastric juice = x/10 ml Total acidity in 100 ml of gastric juice = 10 x So in order to express the titre value in terms of units of N/10 HCl per 100 ml of gastric juice, it has to be multiplied by a factor of 10. The acidity of gastric juice is expressed in terms of units. One unit is 1 ml of N/10 HCl present per 100 ml of gastric juice. Achlorhydria : Free HCl absent Hyperchlorhydria : Free HCl above 60 units Hypochlorhydria : Free HCl never above 10 units

Achlorhydria

Absence of free HCl in gastric juice is called achlorohydria. However enzymes such as pepsin may still be present in gastric juice. The condition is found in pernicious anemia. Achylia Gastrica

It means the absence of free HCl as well as enzymes like pepsin, renin, etc. e.g. pernicious anemia, very advanced stages of gastric carcinoma and gastritis. Hyperchlorhydria

Free HCl is present but in very low concentration, i.e. less than 10 units. Most of the conditions where there is an increased evidence of achlorhydria may show hypochlorhydria. Many patients with hypochlorhydria may become achlorhydric subsequently.

Gastric Analysis 39 Hyperchlorhydria (Gastric hyperacidity)

It means that the concentration of free HCl in the gastric juice is more than the upper normal limit, i.e. values about 60 units, e.g. Duodenal ulcer.

Free HCl Total acidity Volume

Normal Value

Gastric ulcer

Duodenal ulcer

Carcinoma of stomach

Pernicious anemia

10-70 5-100 10-100

Normal 10-100 40-100

10-45 15-110 10-115

Normal 3-80 10-500

Zero 0-40 5-50

Qualitative Tests 1. Presence of starch

Iodine test: Take 3 ml of gastric juice in a test tube. Add few drops of iodine solution. The appearance of blue colour indicates the presence of starch. 2. Presence of Iactic acid Maclean’s test

Take 2 ml of gastric juice in a test tube. Add few drops of Maclean’s reagent (contain HgCl2 and Fe Cl3). Development of yellow colour indicates the presence of lactic acid. Run a control with 2 ml of distilled water instead of gastric juice. 3. Presence of blood Benzidine test

Take 3 ml of saturated solution of benzidine in glacial acetic acid. Add 2 ml of gastric juice followed by 1 ml of hydrogen peroxide. Wait for few minutes. The appearance of blue or green colour indicates the presence of blood in gastric juice. 4. Presence of bile Hay’s test

In a test tube take 3 ml of gastric juice. Sprinkle sulphur powder over it. The sinking of sulphur powder indicates the presence of bile salts. 5. Presence of HCl Gunzberg test

A drop of two of Gunzberg reagent is placed in a small porcelein dish. It is carfully evaporated on a small flame. A glass rod is dipped in the gastric juice and is touched thoroughly with the residue in the porcelein dish. It is gently heated. The appearance of purple red colour indicates the presence of free HCl. 6. Presence of mucus

Examination of gastric juice in broad day light will reveal the stingy appearance of juice for presence of mucus.


12

Urine Analysis Urine Analysis

The average volume of urine excreted daily is about 1.5 litres. It contains nitrogeneous organic compounds such as urea, uric acid, creatinine, hippuric acid, indican, purines and amino acids as well as organic compounds which do not contain nitrogen. The average composition of urine excreted in 24 hours is as follows: Urea Uric acid Amino acid Ammonia Creatinine Ascorbic acid Guanidine Hippuric acid Indican Iodine Lactic acid Chloride as NaCl Inorganic sulphate as S Neutral sulphate as S Sodium Calcium Phosphate as inorganic phosphates

20-30 0.7 0.5-1 0.7 1.4 15-20 3-16 0.6 4-20 50-250 50-200 10-15 60-120 80-160 3-5 0.1-0.3 1.0-1.5

gm gm gm gm gm gm gm gm mg mg mg mg mg mg gm gm gm

Urine specimens must be collected in the correct way and in a clean container or bottle. In the hospital, when only one specimen is needed, the best time to collect is first urine in the morning when it is most concentrated. The specimen should be analysed without delay. If this is not possible the urine should be stored in a refrigerator. The ordinary urine is examined under the following heads: 1. Physical examination. 2. Chemical examination. 3. Microscopic examination. The physical examination of urine includes appearance, colour, pH, specific gravity, etc. These tests are carried out before the microscopic or chemical examination. Whereas in chemical examination, abnormal constituents which are normally not present in easily detectable quantities in urine of healthy beings but are known to occur in urine under certain diseased conditions. Abnormal or pathological constituents of urine are looked for in urine such as reducing sugars, proteins, ketone bodies, bile salts, bile pigments, blood, urobilinogen, etc. (40)

Urine Analysis 41 1. Colour

The normal urine is yellow in colour. The intensity of normal urine is dependent on the concentration of urine. The yellow or amber colour of a normal urine is due to presence of a yellow pigment urochrome. The colour of urine changes in many disease conditions because of the presence of pigments that do not normally occur. Colour

Possible cause

Orange Almost colourless Yellow to yellow brown or greenish colour Reddish brown colour Milky Black upon standing Brown-black colour on standing Orange brown Cloudy

Concentrated urine Dilute urine Bile pigments Hemoglobin Presence of fats Alkaptonuria Melanin or homogentisic acid Urobilinogen Presence of insoluble calcium and magnesium phosphates

The urine may also assume many different colours following ingestion of various dyes, foods and drugs. 2. Volume

The normal volume of urine voided by an adult per day ranges from 750 to 2000 ml. The average volume is 1500 ml. The amount of urine excreted is directly related to fluid intake, the temperature, climate and the amount of sweating that occurs. Polyuria It is the increased excretion of urine. Polyuria may indicate the loss of concentrating ability by the kidney. Polyuria occurs in physiological conditions. 1. Excessive fluid intake 2. Ingestions of diurates Pathological conditions 1. Diabetes mellitus 2. Diabetes insipidous 3. Chronic renal damage Oliguria It is decreased in which the urinary output is 500 ml per 24 hours. Oliguria occurs in 1. Less fluid intake 2. Excessive fluid loss due to vomiting, diarrhoea and sweating 3. Fever 4. Shock 5. Acute nephritis 6. Cardiac failure Anuria It is a total loss of urine. The ratio of day urine (i.e. 8 AM to 8 PM) to night urine (i.e. 8 PM to 8 AM) should be at least 2 : 1 and sometimes 3 : 1 or more in healthier individual. In renal disease, this ratio is reduced and may even be reversed.

42 Practical Biochemistry for Students 3. Specific Gravity

Normal specific gravity of urine varies from 1.005 to 1.030. Specific gravity is highest in the first morning specimen and is generally greater than 1.020. A specific gravity of 1.025 or above in a random normal urine sample indicates normal ability. The specific gravity of urine varies according to kidney function. Concentrated urine has a high specific gravity while diluted urine has a low specific gravity. Specific gravity of urine indicates the relative proportions of dissolved components to the total volume of the urine. It also reflects the relative degree of concentration or dilution of the urine sample. Low specific gravity is observed in the following: 1. High fluid intake 2. Diabetes insipidous 3. Glomerulonephritis 4. Pyelonephritis High specific gravity urine is observed in 1. Excessive loss of water due to sweating fever, vomiting and diarrhoea. 2. Diabetes mellitus 3. Nephrosis 4. Hepatic diseases 5. Congestive heart failure. Whereas fixed specific gravity of urine is an indication of several renal damages 1000 with disturbance of both concentrating and diluting abilities of kidney. 1010

Determination of Specific Gravity

1020

The specific gravity of urine is determined by urinometer. Urinometer is a glass made instrument having a cylinderical stem containing a scale. The urinometer is floated in a cylinder containing urine. Care should be taken that it should not touch the sides or bottom of the cylinder. The depth to which it sinks in the urine, indicates the specific gravity of urine. The urinometer is calibrated with respect to distilled water at 1.000 at specific temperature, indicated on the instrument itself. If the temperature of urine is above or below that temperature, a correction of + 0.001 for each 3°C should be made.

1030 1040 1050

4. pH

The pH of the normal urine is on an acidic side, i.e. 5.5-6.5. The acidity of the urine is mainly due to acid phosphates. In general protein rich diets give rise to acidic urine attributable to the sulphur of the amino acids which is oxidised to sulphuric acid. Similarly phospholipids and nucleic acids all yield phosphoric acid. Alkaline urines are excreted where there is a predominance of vegetables and fruits in the diet. However, the production of ammonia by the kidney also plays an important part in influencing the pH of the urine since it forms salts with acids and can excreted as such. Acid Urine

i. On a high protein diet.

Urine Analysis 43

ii. Uncontrolled diabetes iii. Acidosis iv. Fever. Alkaline Urine

i. After meals ii. Diet high in vegetables, citrus fruits, milk, etc. iii. Renal tubular acidosis. 5. Total Solids

Total solids can be calculated approximately by multiplying the second and third decimal figures of specific gravity by 2.6 (Longe’s coefficient). The product represents the number of gm of solids in 1 litre of urine, e.g. if the specific gravity of urine is 1.025, then 25 × 2.6 = 65 gm totals solids per litre. From this the output during 24 hours can be calculated. Chemical Examination of Urine (Qualitative). 1. Reducing Sugars (Usually Glucose)

Normally very small amounts of reducing sugars are excreted in urine, which are not detected by the reagents used. However, under abnormal conditions following reducing sugars are excreted in urine. Reducing sugars

Conditions

Glucose

Renal glucosuria Blood glucose will differentiate Diabetes mellitus Galactosemia Pregnancy and lactation

}

Galactose Lactose Fructose Pentoses

}

After ingestion of large amounts of vegetables or fruits

Causes of Glycosuria

1. Diabetes mellitus 2. Non-diabetic glycosuria A. Glycosuria with hyperglycemia a. Renal glycosuria—due to low renal threshold b. Alimentary glycosuria—after ingestion of lot of carbohydrates c. Glycosuria of pregnancy—in 10-15% of normal pregnancies B. Non-diabetic glycosuria with hyperglycemia a. Hyperthyroidisim b. Emotional disturbances c. Ether anesthesia d. Increased intracranial pressure—tumours, fractures, encephalitis. Benedict's Qualitative Test

To 5 ml of Benedict’s qualitative reagent in a test tube, add 8 drops of urine. Boil for 2 minutes and cool. A change in blue colour to green-yellow-red precipitate indicates the presence of reducing sugars in urine.


The colour of the precipitate gives an appropriate, i.e. rough amount of sugar in the urine. Colour/Precipitate

Result

Blue colour remains Green colour Green precipitate Green to yellow precipitate Yellow to red precipitate Brick red precipitate

No reducing sugar Less than 0.5 gm% 0.5 to 1 gm% 1.0 to 1.5 gm% 1.5 to 2.0 gm% More than 2gm%

Fehling Test

Mix 1 ml each of Fehling A solution and Fehling B solution. Then add 2 ml of urine. Boil for 2 min. Appearance of either green-yellow-red precipitate indicates the presence of reducing sugars. Tests for Specific Reducing Sugar

The following tests will differentiate the various reducing substances which respond to Fehling and Benedict’s test. Usually the object in testing for reducing substances in urine is to determine the presence or absence of glucose. 1. Fermentation test: Glucose and fructose are fermented by the enzyme present in the yeast and the libration of carbon-dioxide confirms the test. 2. Mucic acid test for lactose and galactose: Evaporate urine in a china dish containing a mixture of 100 ml of urine and 20 ml of concentrated HNO3. Let the volume is reduced to 20 ml. Allow it to cool. Lactose or galactose yields mucic acid through the oxidising action of HNO3 and this forms a white precipitate. 3. Seliwanoff’s test for fructose. 4. Pentoses by orcinol test. 5. Identification of carbohydrates by osazones formation. 2. Proteins

There is no single test capable of detecting the presence of proteins under all conditons. The proteins which appears in urine however are usually albumin and globulins of which albumin almost greatly predominate. Causes of albuminuria 1. Functional—orthostatic, severe muscular exertion, prolonged exposure to cold, pregnancy. 2. Organic a. Pre-renal causes—no primary kidney disease Passive congestion of kidney Fever and toxaemia Intra-abdominal tumours Drugs and chemical poisoning b. Renal—primary disease of kidney Nephritis Nephrosis

Urine Analysis 45

c. Post renal causes Pyetitis Cystitis Urethritis, prostatitis Normally proteins are not excreted in urine. Test for Proteins 1. Heat Coagulation Test

Fill a test tube with two-third (2/3) of urine. Gently heat the upper half of the urine to boiling without disturbing the lower portion which serves as a control. A white turbidity indicates the presence of proteins or phosphates. Add few drops of glacial acetic acid. If precipitate dissolves, it is due to phosphates. 2. Sulphosalicyclic Acid Test

Take 5 ml urine in a test tube, add to it 1 ml of 25% sulphosalicyclic acid. Appearance of white precipitate indicates the presence of proteins in urine. 3. Nitric Acid Test (Heller’s Test)

In a clean dry test tube take 3 ml of concentrated nitric acid. Put 2 ml of urine over the acid to form a separate layer. Appearance of white ring at the junction indicates the presence of proteins. Distinction between albumin and globulin when urine is mixed with an equal volume of saturated ammonium sulphate, the globulin is precipitated and albumin which remains can then be precipitated by boiling the filtrate or by addition of ammonium sulphate. Bence Jones Proteins

Bence Jones proteins are found in urine in cases of multiple myeloma, i.e. cases of malignant disease involving the bone marrow and of leukemia. Its occurrence is very uncommon in urine. Test for Detection of Bence Jones Proteins

Heat coagulation test The urine must be faintly acidic. Acetic acid being added if necessary. It is then gently heated in a test tube. Bence Jones protein gives a precipitate which appears about 40°C, it is maximal about 60°C, and then disappears as the temperature rises. The disappearance may be incomplete since albumin is often present as well. In that case the boiling urine is filtered rapidly to remove albumin. Bence Jones proteins reappears as the filtrate cools. The precipitate flocculates and sticks to the side of test tube.

R U 10 9 8 7 6 5 4 3 2

Quantitative analysis of proteins 1 ½ Approximation to the quality of proteins present in the urine can be obtained by the use of Esbach proteinometer. The graduated tube is filled to the mark U by urine to be tested. Esbach reagent is added to the mark R. The tube is stoppered and after mixing or inverting once or twice without shaking is allowed to stand undisturbed overnight.


The height of the precipitate in the tube is then read and the reading divided by ten gives the amount in gm of proteins in 100 ml of urine. Esbach reagent : 20 gm citric acid, 10 gm picric acid dissolved in 1 litre of water. 3. Test for Ketone Bodies

Acetone, acetoacetic acid and β-hydroxybutyric acids are collectively known as ketone bodies. Ketone bodies are obtained as the intermediate products in the oxidation of fatty acids and are oxidised to carbon dioxide and water under normal conditions. In abnormal conditions, ketone bodies accumulate in the blood and pass out in urine. This condition is called ketosis. When the carbohydrate metabolism is defective, increased fat is oxidised for energy purposes giving rise to increased formation of ketone bodies. Such happens in diabetes mellitus, when glucose present is not metabolised hence ketone bodies appear in urine. Also during starvation when glucose supply to the body is restric-ted, ketone bodies appear in urine. Causes of ketonuria i. Diabetes mellitus ii. Starvation. Rothera’s test

Saturate 5 ml of urine with Rothera’s mixture (ammonium sulphate and sodium nitroprusside). Add 2 ml of liquor ammonia by the side of the test tube. Appearance of permanganate ring at the junction indicates the presence of ketone bodies in the urine. 4. Test for Bile Pigments

Bile pigments are billirubin and biliverdin. Most of the tests for bile pigments depend on the oxidation of bilirubin to coloured compounds of blue, green, violet, red and yellow. Normal urine does not contain bilirubin as bilirubin is bound to albumin. Bilirubin is present in the urine of obstructive jaundice and in some cases of hepatic jaundice due to high level of conjugated bilirubin. Fouchet’s Test

Take 5 ml of urine in a test tube. Add to it an equal volume of 10% BaCl2 solution. Filter, dry the filter paper and add a drop of Fouchet’s reagent. Green colour indicates the presence of bilirubin. Fouchet’s reagent contains trichloroacetic acid and ferric chloride. Gmelin’s Test

Place 5 ml of fuming nitric acid in a test tube. Add 5 ml of urine. Appearance of blue, green and violet rings are seen at the junction if bilirubin is present. Rosenbach Test

Filter some urine through a filter paper. Unfold the filter paper. Dry it. Add a drop of concentrated HNO3 in the centre of filter paper. Green, blue, violet, and red colours show the presence of bile pigments.

Urine Analysis 47 5. Bile Salts

Bile salts are sodium and potassium salts of glycocholates and taurocholates. Bile salts are formed in the liver from where they are excreted in the bile. They are absorbed by the intestine and passed back to the liver through portal circulation. Bile salts are present in urine in obstructive jaundice. Hay Test

Sprinkle some sulphur powder in test tube containing urine. Sulphur remains on the surface in normal urine but sinks down in the presence of bile salts. 6. Blood

Blood appears in urine in hematuria and hemoglobinuria. Hematuria consists of hemoglobin pigment and unruptured corpusules. It is due to passing of blood through the kidney into urine because of lession of the kidney. Hematuria occurs in polynephritis and hemoglobinuria occurs in enteric fever, malaria, wrong blood transfusion and hemolytic poisoning. Benzidine Test

The peroxidase activity of hemoglobin decomposes hydrogen peroxide and the liberated oxygen oxidises the benzidine to give a blue solution. Test

In a test tube take 3 ml of saturated solution of benzidine in glacial acetic acid. Add 3 ml of urine followed by 1 ml of hydrogen peroxide. Appearance of green or blue colour within few minutes indicates the presence of blood. 7. For Urobilinogen

Urobilinogen is normally present in the urine in very small amount. The excretion of urobilinogen is increased in: i. Hemolytic jaundice: In hemolytic jaundice, liver is not able to excrete the increased amount of urobilinogen absorbed from the intestine completely. ii. Infective hepatitis: The liver cell is less able to excrete so more of bilirubin passes into general circulation and is excreted in urine. However it is absent in obstructive jaundice. Ehrlich’s Test

In a test tube take 5 ml of urine. Add to an equal volume of Ehrlich reagent. Allow it to wait for five minutes. Appearance of faint pink or brown colour indicates the normal amount of urobilinogen, whereas deep red colour suggests increased amount of urobilinogen. The test is done in undiluted urine sample and also in serial dilution of the urine. Normally urobilinogen is present in urine upto a dilution of 1 : 10.


13

Food Analysis Test for Carbohydrates

Take about 1 gm of given food sample in a test tube. Add 10 ml of distilled water. Slightly warm the solution. Perform the following tests. Molisch Test

To 2 ml of the above suspension (solution), add 2 drops of ethanolic α-napthol. Mix and incline the tube and add 2 ml of concentrated sulphuric acid by the side of the test tube. Appearance of green-violet or pink ring at the interface indicates the presence of carbohydrates. Indine Test

To 2 ml of suspension, add a drop of dilute HCl, followed by few drops of iodine solution. Appearance of intense blue colour indicates the presence of starch in the food mixture. If the polysaccharides/starch is absent, filter the suspension and perform the following tests in the filtered solution. Benedict’s Test

To 5 ml of Benedict’s qualitative test, add 8 drops of the above filtered solution. Boil for 3 minutes. Appearance of green-yellow orange-red precipitate indicates the presence of reducing sugar in the food. Barfoed’s Test

To 2 ml of Barfoed’s reagent, add 2 ml of above filter solution. Boil for 2 minutes. Appearance of red precipitate indicates the presence of monosaccharides in the food. Seliwonoff’s Test

To 3 ml Seliwanoff’s reagent, add 3 drops of above filtered solution. Boil for 30 seconds only. Cool. Appearance of cherry red colour indicates the presence of fructose in the food sample. Osazone Test

In a clean dry test tube, take 0.5 gm of phenylhydrazine mixture (prepared by taking equal part of phenylhydrazine hydrochloride and anhydrous sodium acetate), 5 ml of filtered solution and add 2 drops of glacial acetic acid. Mix and place the tube in the boiling water bath for 30 minutes. (48)

Food Analysis 49

The osazones of monosaccharides separate out while in hot, whereas the osazones of disaccharides separate out in cold, i.e. by allowing the solution to cool by itself without disturbing the solution in between at room temperature. Examine the shape of the osazones under low power microscope. Appearance of needle shaped osazones suggests/indicates the presence of monosaccharides either glucose or fructose while appearance of sunflower or cotton ball shaped osazones indicates the appearance of maltose or lactose in the food sample. Test for Proteins

Take about 1 gm of food sample in a test tube, add 10 ml of distilled water. Warm the solution (do not boil). Perform the following tests for proteins and R-groups with “suspension” only. Biuret Test

Take 6 ml of 5% NaOH. Add few drops of 1% CuSO4. Divide the solution in two parts and label one as experimental and other as control. Experimental tube: 3 ml of above divided solution. Add 3 ml of suspension. Mix. Appearance of purple or violet colours indicates the presence of proteins in the given food sample. Control tubes: 3 ml of above divided solution. Add 3 ml of distilled water. This tube serves as a control for the above experimental test. Ninhydrin Test

To 2 ml of the suspension, add few drops of ninhydrin solution (0.2 g% in acetone). Boil the solution. Appearance of deep blue colour indicates the presence of proteins in the solution. Test for R-groups

Perform the following R-group tests for proteins with suspension. i. X anthoproteic test. ii. Millon’s test iii. Hopkin’s-Cole test iv. Sakaguchi test v. Test for — S — S — and — SH — group Test for Fats (i.e. For Cholesterol)

In a clean dry test tube, take about 1 gm of food sample. Add 10 ml of chloroform. Mix. Since fats are soluble in chloroform, it will go into the solution. Filter the solution. Care should be taken that funnel and test tube receiving the filtrate should not be wet with water. Perform the following test for cholesterol. a. Libermann Burchard test: In a clean and dry test tube take 2 ml of filtered chloroform extract. Add 10 drop of acetic anhydride and 2 drops of concentrated sulphuric acid. Mix. Keep it in dark for 10 minutes. Appearance of green colour indicates the presence of cholesterol in the food sample. b. Salkowski test: In a clean dry test tube, take 2 ml of the filtered chloroform extract. Add 2 ml of concentrated sulphuric acid by the side of the test tube. Slightly mix and allow to


stand. Appearance of brown red colour in the upper layer (CHCl3 layer) and yellow colour in the lower layer (H2SO4 layer) indicates the presence of cholesterol in the food sample. Test for Minerals

Take about 1 gm of food sample in the test tube, add 10 ml of distilled water. Mix thoroughly. Minerals being water soluble will go into the solution. Filter the solution. Perform the following tests for minerals. Test for Iron

i. To 2 ml of the filtrate, add a drop of concentrated nitric acid. Boil the solution. Cool it. Add 1 ml of 5% ammonium thiocyanate solution. Appearance of red colouration indicates the presence of iron in the food sample. ii. To 2 ml of the filtrate, add 2 ml of sodium nitroprusside. Appearance of blue colouration indicates the presence of iron in the food sample. This is known as Pearl Reaction. Test for Phosphorus

To 2 ml of the filtrate, add 2 ml of 5% ammonium molybdate. Boil the solution. Appearance of yellow coloured precipitate indicates the presence of phosphorus in the food sample. Test for Calcium

To 2 ml of the filtrate, add few drops of glacial acetic acid, followed by 2 ml of saturated ammonium oxalate solution. Appearance of white precipitate indicates the presence of calcium in the food sample. Test for Chloride

To 2 ml of the filtrate, add 2 ml of AgNO3 solution. Appearance of white precipitate insoluble in dilute hydrochloric indicates the presence of chloride in the food sample.

Collection and Preparation of Blood Specimen 51

14

Collection and Preparation of Blood Specimen Blood is one of the most common specimens studied in biochemical laboratories in search of blood disorders, metabolic disorders and infection. Blood clots within a few minutes after it has been removed from the body, unless an anticoagulant is used which stops the process of clotting. Anticoagulated blood is also known as whole blood. Plasma (Fluid portion of unclotted blood) is obtained from the anticoagulated blood. Serum (Fluid portion of clotted blood) is obtained from clotted blood. Usually, blood is obtained by vein puncture with the help of a sterilized dry syringe. For clinical biochemistry, mostly serum/plasma and occasionally whole blood is required. Preparation of Specimen Bottles Using Different Anticoagulants

Anticoagulants are used when either whole blood or plasma is required. Most of the anticoagulants remove calcium which is one of the factors required in coagulation process. Heparin, however, directly interferes in the coagulation process by destroying thrombin and thromboplastin. Use of an anticoagulant largely depends upon the nature of clinical investigation, e.g. heparin is necessary for determination of blood gases (and pH) while sodium fluoride in addition to oxalate, is necessary for glucose estimation. Specific amount of anticoagulant is also recommended. Containers of blood collection must be thoroughly cleaned with no trace of detergent. Specific amount of anticoagulant is added (preferably in solution form) and dried oven inside the container. For oxalate anticoagulant (which is most common in biochemistry lab), a high temperature is avoided as it changes oxalate into carbonate. All previously prepared containers can be stored at room temperature. For easy identification, the containers are labelled differently with coloured markers. EDTA (Ethylene Diamine Tetraacetate)

10% solution of dipotassium salt of EDTA, 0.l ml of this is used/5 ml of blood (Conc. 2 mg/ml of whole blood). Double Oxalate

1.2% of ammonium oxalate + 0.8% of potassium oxalate. Use 0.5 ml for 5 ml blood (Conc. 2 mg/ml of whole blood). (51)

52 Practical Biochemistry for Students Sodium Citrate

0.106 M trisodium citrate in distilled water, sterilize it. This is used in solution form and is not dried inside the container. 1 ml of this solution is used for 9 ml blood (The ratio of anticoagulant solution and whole blood comes to 1 : 9, which is ideal for coagulation studies). Not very common for biochemical studies/purposes. Sodium Fluoride-Potassium Oxalate

Used for blood sugar estimation (except in enzymatic method of sugar estimation). 1.2% sodium fluoride + 5% potassium oxalate. 0.25 ml is used/5 ml of blood (=3 mg sodium fluoride + 15 mg potassium oxalate). [Caution: Fluoride is a poison]. Heparin

Limited to the determination of blood gases and electrolytes, 0.1-0.2 mg/ml of the blood. Collection of Blood by Vein Puncture

Before proceeding to collect the blood from the patient, assemble all the necessary equipment which includes the container for blood collection, a needle, syringe, tourniquet, disinfectant (methylated spirit or 70% alcohol), swab and a tray of cold water to rinse the needle and syringe after drawing blood. Needle Sterilized sharp needle of bore size 18 to 20 gauge for adults and 23 gauge for children is used. Use of disposable needles is highly recommended. In case of reusable ones, the needle point should be kept sharp. Keep a stock of sterile needles in small glass tubes and the point should be on a pad of non-adsorbent cotton wool and the tube should be plugged with the same material. A sterile needle should be attached to the syringe under asceptic conditions. Syringes Of different capacities, 2 ml, 5 ml, 10 ml and 20 ml are available. The size depends upon the amount of blood needed. Check that the end of each syringe fits into the needle. At any stage of assembly of the needle and the syringe, do not touch the tip of the needle and keep the asembled syringe and needle inside a sterilized tube or covered with sterile gauze. With reusable glass syringes, the fit of the plunger and the barrel and the integrity of the syringe tip should be checked. Disposable plastic syringes come in a sterilized wrap and they have sharpened needles for single time use. Following use, syringe should be dipped in cold water to remove the blood. All glass syringes should be properly sterilized and perfectly dry. A wet syringe, an improperly washed syringe. Use of a too fine needle, emptying the syringe without removing the needle are some of the causes that result in haemolysis. Tourniquet It is a soft rubber tubing of 2.5 mm bore and 30-40 cm in length. A flat elastic rubber strip can also be used. This is applied to the arm to slow the blood flow and make the veins more prominent. This helps to select the puncture site.

Collection and Preparation of Blood Specimen 53 Procedure of Blood Collection

Venous blood is frequently collected while the blood may be taken from any prominent vein, a vein on the front of the elbow or forarm is almost universally employed. The arm should be warm. This improves the circulation. The arm is extended and a rubber torniquets firmly applied a few inches above the elbow. The skin over the vein is cleaned by rubbing over with spirit or ether. A well sharpened sterile hypodermic needle fixed onto a syringe can be held steady by a thump of the other hand of the operator. When the needle enters the vein, the plunger is withdrawn slightly. If blood appears, the tourniquet is released. When the desired amount of blood has been drawn into the syringe, a small pad of wodsoaked with spirit or ether is placed on the arm where the needle was inserted, and the needle is withdrawn. This pad is held on firmly for a few minutes until bleeding stops. The needle is removed from the syringe and the blood transferred to an appropriate container, using minimum amount of pressure. The needle and syringe are immediately washed out with cold water to remove any remaining blood. The needle to prevent any infection must be stressed. Capillary Puncture

A capillary is a small blood vessel connecting the small arteries (arterioles) to the small veins (venules). Capillary blood is obtained by skin puncture which is recommended for babies and where vein puncture is difficult (old, prolonged illness). In adults and older children, the tip of the finger is punctured and in infants, the heel is chosen. Arterial Blood

It is rearely examined. It is taken for blood gas determinations. It is most commonly obtained by inserting a needle into radial, brachial or femoral artery usually under a local anesthesia. Separation of Serum and Plasma Serum

Non-anticoagulated blood yields serum, the fluid portion of clotted blood in order to obtain serum is collected in a plain tube without anticoagulant and allowed to clot for 30 min at room temp. The tube can preferable be kept slanted. When the clot separates out, it is centrifuged (2500 rpm for 10 min) and serum be separated. Separation must not be delayed much because prolonged contact of the cells with the serum will cause the release of the red cells constituents into the serum composition. If the centrifuge is not available, refrigerate the blood for several hours, the clot will separate from the serum which can be removed with the help of a pasteur pipette. Plasma

Appropriate aanticoagulant should be mixed with the blood by gentle rotation. Excessive amount of anticoagulant should not be used. Separation of plasma should be done by centrifuging at low to moderate speed. If the plasma is not required immediately, the cells can be allowed to sediment out. The upper fluid is removed and finally cleared of any remaining cells by centrifugation.

54 Practical Biochemistry for Students Preparation of Protein-free Filterate Use of Different Protein Precipitating Agents

The first step in majority of biochemical determination made on blood is to remove the proteins as these interfere in many ways in the actual investigation. For this purpose, many substances are used, chiefly acids, heavy metal, ion in presence of suitable alkali, organic solvents, etc. As acids, e.g. tungstic acid and trichloroacetic acid are mostly used, others include perchloric, tungstomolybdic, phosphotungstic and sulphosalicylic acid. Choice of method depends on nature of estimation. Few Common Methods are: a. Folin method Tungstic acid is prepared by adding 10% sodium tungstate and 2/3 NH2SO4. To prepare the protein free filterate (PFF), mix 1 volume of blood with volume of D/W, add 1 volume of 10% sodium tungstate and 1 volume of 2/3 NH2SO4. Shaking well during the addition of H2SO4. Allow to stand for 10 minutes and centrifuge at 2000 rpm for 5 min. b. Modified folin method/Maden method Mix 1 volume of whole blood to 1 volume of 10% sodium tungstate and 8 volume of N/12 H2SO4. Mix well. Keep for 10 min and centrifuge. c. Somogyi’s method In its original form, the additions are as follows: 4.2 ml isotonic sodium sulphate + 0.2 ml blood + 0.3 ml 10% zinc sulphate + 0.3 ml (N/S) sodium hydroxide. Mix well. Keep for 10 min and centrifuge. In addition to proteins saccharides (non-sugar reducing substances) are also precipitated. This method or its modification is particularly used for true sugar estimation. COLLECTION AND PRESERVATION OF URINE SPECIMENS Collection of Urine Specimens

For most qualitative tests, any fresh specimen is suitable. Because composition of urine varies during the days, quantitative estimation is usually carried out on 24 hr collection. While the specimen is being collected and during the period before it is analysed, it is essential either to keep the specimen refrigerated or to have a preservative in the collection container, so that unstable substances may be preserved and the growth of bacteria prevented. The preservative used will depend on the substance which is to be determined. Preservatives Hydrochloric Acid

50 ml of 2 M HCl is adequate for a 24-hr specimen suitable in particular for the determination of Ca, P, N, ammonia, VMA estimation. It can also be used when sodium, potassium, urea estimations are required but it is not suitable for uric acid and protein estimation as these may be precipitated.

Collection and Preparation of Blood Specimen 55 Toluene

10 ml of toluene is commonly used for 24 hr collection for Na, K, uric acid, creatinine and protein analysis but is not suitable for determination of Ca, P, VMA, ammonia, etc. Hibitane (Chlorhexidine diacetate)

It is a preservative when glucose is to be estimated in urine since it inhibits the action of bacteria but does not interfere with the analysis either by enzyme or reduction methods. 5 ml of a 5% aq solution is added to the container. Chloroform, light petroleum, thymol and formalin are used as urine preservatives. A few and one of these is put into a 2.5 l container and given for collection. Urine Analysis

A routine biochemical examination of urine consists of the following tests: Physical Tests

Colour Urine is normally clear and pale yellow in colour due to a pigment, urochrome, said to be a compound of urobilin, urobilinogen and a peptide substance. Odour Freshly passed urine has a characteristic of aromatic odour, said to be due to volatile organic acids. When the urine is allowed to stand, decomposition of urea by bacteria occurs, acid ammonia is evolved. Reaction Freshly passed normal urine is usually slightly acidic with a pH about 6.0 and range of 4.8-6.8. The pH is determined by simply dipping the pH indicator strips into the urine and comparing the colour change of the strip against the colour guide supplied with the indicator papers. Volume Volume of urine passed by a normal person an average ranges from 1200-1500 ml/ day. Specific Gravity

Specific gravity of normal urine ranges from 1.008 to 1.030. It is determined by using urinometer as follows: Mix urine well. Take urine in specific gravity cylinder. Then put the urinometer gently in the urine. Take care it does not touch the wall of the cylinder. Take reading of the lower meniscus of the urine touching the calibration viewed at eye-level. If the volume of urine is not sufficient to the cylinder, report as q.n.s. (quantity not sufficient). The physical test of urine is valuable if fresh urine is used. Upon standing, almost all the physical properties are changed.


15

Urinary Reducing Sugars Estimation of Reducing Sugars in Urine Principle

Reducing sugars by virtue of free aldehydic or ketonic groups in their structure possess the property of reducing certain heavy metallic cations in an alkaline medium and in the process get oxidised to a mixture of sugar acids. This principle of oxidation-reduction is used in the estimation of reducing sugars. Reducing sugars undergo enolisation when placed in alkaline solution. These enediol forms of reducing sugars are very reactive and are easily oxidised by the oxidised agents. The reagent used is Benedict’s quantitative reagent. In Benedict’s qualitative reagent the colour change is from blue to different colour of the precipitates depending upon the concentration of reducing sugars in the solution, whereas in Benedict’s quantitative reagent the colour change is from blue to colourless, which is sharp and easy to detect. The composition of Benedict’s qualitative reagent. 1. CuSO4 : Furnishes cupric ions (Cu++) in the solution. 2. Na2CO3 : To make the medium alkaline as Benedict’s quantitative reagent is positive in alkaline medium. 3. Sodium citrate : To prevent the precipitation of cupric ions (Cu++) as CuCO3 or Cu(OH)2 by forming a loosely bound complex with Cu++ions (i.e. Sodium citrate: cupric complex) which on dissociation gives a continuous supply of Cu++ions. 4. Potassium ferrocyanide : To prevent the precipitation of Cu2 O in the solution, i.e. K4Fe(CN)6 5. Potassium thiocyanate : Precipitates Cuprous ions (Cu+) as CuCNS, a white colour precipitates. Benedict’s quantitative reagent is prepared as follows: Cooper sulphate 18 gm, sodium carbonate 200 gm, sodium or potassium cirtrate 200 gm, potassium thiocyanate 125 gm, potassium ferrocyanide (5%) 5 ml. With the aid of heat dissolve carbonate, citrate and thiocyanate in enough water to make about 800 ml of the mixture. Dissolve copper sulphate in about 100 ml of distilled water with constant stirring. Add ferrocyanide solution, cool and make the volume to 1 litre. (56)

Urinary Reducing Sugars 57 Reaction Reducing sugars OH– KCNS Cu++ ________________________→ Cu++ _______________→ Cu(OH) ________________→ CuCNS High temperature K4Fe(CN)6

↑

↑

CuSO4

Cu2O

Procedure

In a 100 ml conical flask, pipette 10 ml of Benedict’s quantitative reagent. Add 2-3 gm of anhydrous sodium carbonate (i.e. to ensure the alkaline medium), a pinch of pumic power (to avoid bumping) and 10-15 ml of distilled water (to prevent the loss of water due to evaporation). 0

Burette 25

Conical flask Wire gauge Tripot stand Burner

Place the conical flask on the burner and boil the contents. Once the solution starts boiling, add urine solution from the burrette drop by drop till the blue colour disappears. The colour change is from blue to colourless which marks the end point of the titration. Note the volume of urine consumed. Observation

Sl no

Initial reading of burrette

Final reading of burrette

Volume of urine consumed

1. 2. 3. 4. The volume of urine used = x ml. Calculation

Benedict’s quantitative reagent is prepared in such a way that 25 ml of it is completely reduced by 50 mg of glucose. Therefore 10 ml of Benedict’s quantitative reagent will be reduced by 20 mg of glucose, i.e. 10 ml of Benedict’s quantitative reagent = 20 mg of glucose.


Let x ml of urine reduces 10 ml of Benedict’s quantitative reagent, which is equal to 20 mg of glucose. So x ml of urine contains 20 mg of glucose ∴ 100 ml of urine will contain =

20 × 100 ____________ x

2

____

x

mg of glucose

gm of glucose

Result 2 ___ gm of glucose per 100 ml of urine, depending upon the volume of urine excreted per day x the results are expressed accordingly.

Whatever may be the reducing sugar excreted in the urine, the results are expressed in terms of glucose because of Benedicts’s quantitative reagent relation with glucose. The final confirmation regarding the nature of reducing sugar excreted comes through osazone formation where shape of osazone crystals reveals the identity of reducing sugars. Interpretation

Normally reducing sugars are not excreted in urine even if they are excreted, their amount is too small to be detected by Benedict’s qualitative reagent. The various reducing sugars that can appear in urine are. 1. Glucose

The excretion of reducing sugar in urine is called glycosuria and if the reducing sugar excreted in glucose then the condition is called glucosuria. a. Diabetes mellitus b. Renal glucosuria Diabetes mellitus

In diabetes mellitus, due to high level of glucose in the blood, glucose appears in urine. Renal glucosuria

Due to low renal thereshold, glucose appears in urine even though blood sugar is normal. This may be seen in pregnancy or an inherited characteristic. The condition is usually benign. Differentiation from diabetes millitus is made by simultaneous blood and urine glucose estimation or if necessary glucose tolerance test. 2. Lactose

Lactosuria occurs in: a. Pregnancy: During later stages of pregnancy and lactation. b. Congenital lactase deficiency. 3. Fructose

Fructose appears in urine due to consumption of lot of fruits containing fructose such as grapes, cherry, plums, etc. or oral ingestion.

Urinary Reducing Sugars 59

But in general fructosuria occurs in two rare inborn errors of metabolism. a. Essential fructosuria: Observed in persons of Jewish origin and is a harmless condition. b. Heredity fructose intolerance: The basic defects is a deficiency of fructose-l-phosphate aldolase and accumulation of fructose-l-phosphate. 4. Galactose

Galactose appears in urine in galactosemia due to the deficiency or absence of enzyme galactosel-phosphate uridyl transferase. Galactose conversion to glucose does not take place. Hence it appears in urine. The diagnosis is made by identifying the urinary sugar as galactose by paper chromatography. 5. Pentoses

Pentosuria is very rare. It may occur in: a. Alimentary pentosuria: Xylose and arabinose reappear in urine after consumption of lot of fruits such as cherries, prunes, plums, grapes, etc. b. Essential pentosuria: It may occur as a congential defect, characterised by excretion of xylulose due to a block in the metabolism of glucuronic acid.


16

Urinary Chlorides Estimation of Chlorides in Urine Principle

Urine is acidified with concentrated nitric acid (concentrated HNO3 prevents the precipitation of urates, phosphates and carbonates as silver salts) and the chlorides present in urine are precipited as silver chloride by adding a measured excess of standard solution of silver nitrate solution. AgCl is filtered and the left over AgNO3 in the filtrate is back titrated with standard solution of ammonium thiocyanate using ferric alum (ferric ammonium sulphate) as an indicator. The formation of reddish brown colour due to the formation of ferric thiocyanate marks the end point of the titration. Reagents

1. 2. 3. 4.

N/10 AgNO3 Concentrated HNO3 Ferric alum indicator N/10 NH4CNS

Reaction XCl + excess AgNO3 → AgCl↓ + left over AgNO3 Left over AgNO3 + NH4CNS → AgCNS↓ + NH4NO3 white ppt. FE2 (SO4)3 (NH4)2 SO4 24H2O Ferric alum. ↓ Fe2(SO4)3 + 2NH4CNS → 2Fe(CNS)3 + (NH4)2 SO4 Red colour End point colour

Procedure

In a 100 ml volumetric flask, pipette 10 ml of urine. Add 5 ml of concentrated HNO3, followed by 20 ml of N/10 AgNO3. Make the volume to 100 ml with distilled water (do not use tap water). Mix well and keep it for 5 minutes. Filter the solution. Take a portion of filtrate say 50 ml for titration in a conical flask. Add 1 ml of ferric alum as an indicator. Titrate with N/10 NH4CNS solution from burette till the light brown colour appears which is the end point of the titration. (60)

Urinary Chlorides 61

Note the volume of NH4CNS solution used: End point: The appearance of the light brown colour. The end point colour is due to ferric thiocyanate. Observations

Volume of filtrate taken = 50 ml Sl no.

Initial reading of NH4CNS

final reading of NH4CNS

Volume of NH4CNS used

1. 2. 3. 4. Volume of NH4CNS used = x ml Calculation

The output of chloride is expressed in gm of sodium chloride per day. According to Normality equation: 1 ml of N/10 AgNO3 = 1 ml N/10 NH4CNS Since 50 ml of filtrate consumes x ml of N/10 NH4CNS Which means that 50 ml of filtrate contains x ml of leftover N/10 AgNO3 So 100 ml of filtrate will contain 2 x of leftover N/10 AgNO3 Therefore, the volume of N/10 AgNO3 consumed by 10 ml of urine is (20 – 2x) ml. Again applying the Normality equation (20 – 2x) ml of N/10 AgNO3 = (20 – 2x) ml of N/10 NaCl 1 litre of Normal NaCl = 58.5 gm of NaCl or 1 ml of Normal NaCl = 5.85 gm of NaCl or 1 ml of N/10 NaCl = 58.5 mg of NaCl ∴ (20 – 2x) ml of N/10 AgNO3 = (20 – 2x) ml of NaCl (20 – 2x) ml of N/10 AgNO3 = (20 – 2x) × 5.85 mg of NaCl ∴ 10 ml of urine contains (20 – 2x) × 5.85 mg of chloride as sodium chloride. (20 – 2x) × 5.95 ∴ 100 ml of urine contains _________________________ × 100 mg of chloride as sodium chloride. 10 = 2(10 – x) × 58.5 mg of chloride as NaCl = (10 – x) × 117 mg of chloride as NaCl. Depending upon the volume of urine excreted per day the results are expressed accordingly. Since chloride excreted is an anion, the result can also be expressed in terms of milliequivalent.


mEq/L of chloride =

mg% × 10

___________________________________________

Equivalent weight of chloride

Interpretation

Normally an adult excretes 10 to 15 gm of chloride as sodium chloride per day. The rate of excretion of sodium and chloride ions is controlled by the mineralocorticoid aldosterone. It helps in the reabsorption of Na+ and Cl– ions. In addison’s disease, production of aldosterone is less. Less reabsorption of Na+ and hence more excretion. Low chloride excretion in urine. 1. When chloride intake is less, i.e. in fasting. 2. Any condition where water is retained in the body. a. Edema b. Nephrotic syndrome. c. Heart disease. d. Malnutrition. 3. Any conditions in which water is lost through other sources except kidney. a. Excessive sweating. b. Vomiting. c. Diarrhoea. 4. In Cushing's syndrome of Cushing’s disease where aldosterone production is more, more reabsorption of Na+ and Cl– ions takes place, hence output is decreased. 5. During pneumonia and other infectious disease, hypochloremia results from withdrawal of blood chlorides, i.e. exudates and excretion of chloride in urine falls. 6. Extremely low in severe diabetes mellitus. More Chloride Excretion in Urine

1. In Addison’s disease, where aldosterone, production is more.

Urinary Creatinine 63

17

Urinary Creatinine Estimation of Creatinine in Urine Principle

Creatinine is estimated by Jaffe’s method. Creatinine present in urine on treatment with picric acid in presence of alkali yields red colour due to the formation of creatinine picrate, the colour obtained is compared with standard solution of creatinine. Reaction

Strong alkali Creatinine + Picric acid ______________________→ Creatinine picrate (Red colour) Reagents

1. Saturated solution of picric acid (about 1.5 g%) 2. NaOH (10 g%) 3. Standard creatinine solution (1 mg/ml) Dissolve 161.1 mg of creatinine zinc chloride in some amount of distilled water (say 50 ml). Add 10 ml of dilute HCl. Make the volume to 100 ml with distilled water. The resulting solution contains 1 mg of creatinine per ml of solution. Store in brown bottle at cool place. Procedure

Mark three 100 ml volumetric flask as test, standard and blank. Add the following reagents in each flask as. Reagent Urine Creatinine std. Picric acid soln. NaOH (10 g%)

Test 1 ml — 2 ml 1 ml

Standard — 1 ml 2 ml 1 ml

Blank — — 2 ml 1 ml

Wait for 10 minutes, then make the volume to 100 ml with distilled water in each flask. Read the optical density of the test solution and standard solution at 520 mm setting blank to zero. Observation

Optical density of test solution = Optical density of standard solution = (63)

64 Practical Biochemistry for Students Calculation

Concentration of creatinine excreted per day in urine =

Optical density of test

__________________________________

Optical density of std.

×

Conc. of std.

_______________________

Vol. of urine used

× Volume of urine excreted per day

Interpretation

The normal daily excretion of creatinine in adults is from 0.4 g to 1.8 g. The amount of creatinine excreted varies with the muscle mass and is nearly constant for each individual. Creatinine is derived from creatinine and is a waste product. Increased excretion of creatinine occurs in the following conditions. Increased tissue catabolism, i.e. fever. Decreased excretion of creatinine occurs in the following conditions. 1. Starvation. 2. Muscle atrophy. 3. Muscular weakness. Creatinine Clearance

Clearance is defined as the amount of plasma in ml which would have to be completely cleared of the substance, each minute by both kidneys in order to account for its rate if excretion. Ux V C = ________ Px Where,

Ux = Conc. of concerned substance in urine Px = Conc. of that substance in plasma V = Outflow of urine in ml/min.

The clearance is a parameter for assessing the glomerular function of kidneys. The clearance of that substance is measured which is neither reabsorbed nor secreted by tubules, gives glomerular filteration rate if following conditions are fulfilled. 1. It is not metabolized by body. 2. It is not found to plasma proteins. 3. It is not attached by tubular cells. Ideal substance which is administered exogeneously, is instilled for practical purposes, creatinine clearance is used as it is endogenously synthesized. This is preferred over urea clearance as it is not affected by diet or by rate of urinary flow. Procedure

24 hours urine specimen is collected and blood is taken during day serum and urine concentrations are estimated. Urine volume/ml is calculated by dividing the urine volume (ml) by the urine of collection (min.)

Urinary Creatinine 65 Interpretation

Normal values of creatinine clearance are: Males : 95-140 ml/min. Females : 85-125 ml/min. Clearance values are decreased in impaired renal function and so provide a rough impression of glomerular damage.


18

Ascorbic Acid in Urine Principle

Vitamin C in urine is estimated by direct titration of the acid urine (in presence of acetic acid which inhibits the aerobic oxidation of ascorbic acid) with oxidation-reduction indicator 2, 6 dichlorophenol indophenol, vitamin C reducing the indicator, which becomes decolourised in the process. Vitamin C saturation test: The excretion of vitamin C following a large dose gives an indication of the state of body tissues in respect of vitamin C content. If the tissues are saturated, most of the ingested vitamin will appear in the urine but if there is relatively little vitamin C in the tissues most of the ingested vitamin will be retained by the tissues and little or none appearing in the urine. In acid (pH approximately 3.0), the dye 2,6 dichlorophenol indophenol may exist in two forms, as a red oxidised form or as a leuco (colourless) reduced form Cl | O=

= N —

Cl |

+

+ 2H + 2e — OH

HO —

Red colour Oxidant at pH 3.0

— N —

Colourless Reductant at pH 3.0

At the same pH of 3.0, ascorbic acid exists in two forms O=C— | HO – C || O HO – C | H–C— | HO – C – H | CH2OH

+

– 2H , – 2e

L-Ascorbic acid (Reductant)

O=C— | O–C | O O–C | H–C— | HO – C – H | CH2OH L-Dehydroascorbic acid (Oxidant)

(66)

— OH

Ascorbic Acid in Urine 67 Reaction

O=C— | HO – C || O HO – C | H–C— | HO – C – H | CH2OH

L-Ascorbic acid

OH | Cl — — Cl

| N ||

+

_______________→

|| O 2,6 dichlorophenol indophenol (Oxidised form)

O=C— | O–C | O O–C | H–C— | HO – C – H | CH2OH

OH | Cl — — Cl

| NH |

+

|| OH L-Dehydro Ascorbic acid

Colourless Leucobase (Reduced form)

Collection of Urine Sample

Urine is collected over acetic acid (10 ml of glacial acetic acid per litre of urine). This is done to prevent oxidation of ascorbic acid present by atmospheric oxygen which would take place at more alkaline pH. Procedure

In a clean dry test tube, take 0.5 ml of 2,6 dichlorophenol indophenol indicator dye. Add a drop of glacial acetic acid. The colour of the solution which is blue turns to pink by the glacial acetic acid. Titrate it with urine drop by drop till the pink colour disappears. Note the volume of urine consumed. Let it be x ml. Calculation

The dye is prepared in such a way that 0.5 ml of dye contains 0.04 mg of the dye which is completely reduced by 0.02 mg of ascorbic acid. x ml of urine contains 0.02 mg of ascorbic acid. ∴ 100 ml of urine contains

0.02 × 100 x

__________________

mg of ascorbic acid

2 = ________ mg of ascorbic acid. x

Interpretation

The normal excretion of vitamin C per day is roughly half of the dietary intake. The amount excreted per day is 20-30 mg. Excretion of vitamin C less than 50% indicates unsaturation. In severe deficiency, the vitamin C excretion is nil.


19

Serum Uric Acid Uric acid is major product of purine nucleotide metabolism. Dietary nucleotides get converted to uric acid. In birds, uric acid is converted to allantoin. Uric acid acts as an antioxidant. However, bulk of purine ultimately excreted as in urine as uric acid, derived from endogenous degradation of nucleo acids. In lower primates and mammals other than humans—they carry purine metabolism one step further, i.e. uric acid to allantoin. It requires uricase. Approximately 75% of uric acid is lost in urine and 25% is secreted by GIT. Degradation of purine nucleotide occurs in tissues which contain xanthine oxidase such as liver and small intestine. Uric acid is as such nonionized but at pH 7.4 exist as urate. In plasma, ECF and other fluids have 98% monosodium urate. At 6.8 mg/dl plasma saturates with uric acid and potential for urate crystal to accumulate. It has been seen that even at much higher concentration, precipitation may not occur because plasma contains factors which prevent precipitation. pH of plasma and urine determines the solubility. As pH decreases greater precipitation of uric acid occurs. Uric acid is 100% filtered into glomerulus. Most of it, is absorbed in PCT and around 25% secretion occur in DCT. Determination of Serum Urate by Caraway Method Principle

Phosphotungstic acid in alkaline medium oxidizes uric acid to allantoin and itself gets reduced to tungusten blue which is estimated colorimetrically at 710 nm. Reagents

1. 2. 3. 4. 5.

10% sodium tungstate 2/3 N sulphuric acid Tungstic acid Phosphotungstic acid 10% Na2CO3 Working uric acid solution concentration is 1 mg%.

Procedure

In a centrifuge tube pipette out 0.6 ml of serum and add 5.4 ml of tungstic acid while shaking centrifuge and mix well and wait for 30 min. Read coloured solution within the next 30 min in a colorimeter at 710 nm or using red filter. (68)

Serum Uric Acid 69

1. 2. 3. 4. 5.

Distilled water Standard(1 mg%) Filterate Na2CO3 Phosphotungstic acid (PTA)

Blank 3.0 ml – – 0.6 ml 0.6 ml

Std. – 3.0 ml – 0.6 ml 0.6 ml

Test – – 3.0 ml 0.6 ml 0.6 ml

Mix and keep for 10 min at room temp. Take OD at 710 nm. Normal values

Children Male Female

– – –

2-4 mg/dl 4.4-7.6 mg/dl 2.3-6.6 mg/dl

Other method of estimation: Uricase method Uric acid + 2H2O + O2

Uricase _______________ →

Allantoin + H2O2 + CO2

Urate ion has characteristic absorption peak in UV range. The reaction products allantoin and CO2 has little absorbance at this wavelength so that decrease after uricase activing. Clinical Notes

1. Hyperuricemic value when reaches 77 mg% the term gout is used. Gout is not single disease but is used to describe a number of disorders with crystals of monosodium urate monohydrate, derived from hyperuricemic body fluids give use to inflammatory arthritis and renal disorders. 2. Hypouricemia is reached when uric acid level reaches to less than 2 mg/dl because of decreased production of urate and is indicative of renal diseases, neoplastic diseases, diabetes mellitus and deficiency of xanthine oxidase.


20

Colorimetry Colorimetric procedures are widely used in the hospital laboratory because they are easy to perform, require small amount of blood and give results of high order accuracy. In principle, these methods depend upon the measurements of the amount of colour, i.e. intensity of colour produced during a chemical reaction in which the substance being estimated takes part quantitatively. Within certain limits the colour intensity of a solution is proportional to the concentration of the reacting substances and it is possible to obtain a measure of the concentration of the substance by determining the depth of the colour. The basic laws governing the absorption of light are formulated by Lambert’s and Beer’s which are as follows. Lambert’s (Bouguer) Law

It states that the proportion of light absorbed by a substance is independent of the intensity of the incident light. Beer’s Law

It states that the absorption depends only on the number of absorbing molecules through which the light passes. Mathematical Derivation of Lambert’s and Beer’s Law

Suppose a beam of monochromatic light of intensity I, passes into a small thickness of a solution of an absorbing substance of concentration C (expressed as gm mole/litre), then if the amount of light absorbed be dI. =

dI ______

I

– KC dl

where K is constant for a given wavelength. In order to determine the amount of light absorbed by layer of thickness I where the intensity of incident light is Io and that of emergent light is I. log

I _______

or log

I _______

Io

Io

= – KCI = KCI

Optical density = KCI

From the above equation the optical density of a solution is directly proportional to the concentration of the substance and the depth of solution through which the light passes. (70)

Colorimetry 71 Optical Density (OD) Absorbance

Optical density also called extinction is the logarithmic ratio of the incident light to that of emergent light. OD = log

Io _______ I

Transmission (T)

Transmission is defined as the ratio of the intensity of transmitted light to that of incident light. T =

I _______ Io

Relationship between optical density and transmission: OD

OD

=

log

Io _________

=

log

I ________

=

log

100 _________

=

2 – log %T

I

T

%T

When transmission is 100% the optical density is O. Since optical density is a logarithmic scale, the accuracy of reading is decreased if it is more than 0.60 and reading below 0.10 is also not satisfactory. Hence care should be taken to have the reading between 0.10 to 0.60 by selecting those suitable methods of investigations which give the deflection on the OD scale between 0.10 to 0.60, which is the most sensitive region of the OD scale. The plotting of optical density-concentration of unknown is linear whereas plotting of transmittance-concentration is not. Theory

A beam of light passing through a coloured solution is partly absorbed. The amount of absorption will depend upon the number of coloured ions or molecules in its path. If the number of coloured particles in the solution is doubled, all other things being equal, twice as much light will be absorbed. If this increase in colour is due to double the concentration of the substance it will be seen that there is a direct relationship between the light absorption and the concentration of coloured substance dissolved. This relationship is expressed mathematically by combination of two laws discussed above. The intensity of light emerging from a solution can be determined by: i. Visual colorimetry ii. Photoelectric colorimetry. 1. Visual Colorimetry

In visual colorimetry, the intensity of light entering the two solutions, i.e. standard solution and the unknown solution is the same. The length of the solution through which the white


light passes is kept constant in the standard solution but is varied in the unknown until the intensities of light emerging from two solutions are same, i.e. the two colours match each other. So according to Lambert-Beer Law, a comparison of the concentration is then possible. Io is the intensity of light entering the solution (equal in both cases). I is the intensity of light emerging from solution (made equal by adjusting length of solutions transversed by light). Cs is the concentration of standard solution. Cu is the concentration of unknown or test solution. Rs and Ru are the respective lengths of solution transversed by light (in mm). log

Io ________ I

=

KCs Rs

=

Cu.Rs = Cu =

KCu.Ru Cu.Ru Rs × Cs ______________ Ru

In the visual colorimeter, the length of the absorbing fluids is the distance between the bottom of the cup and the bottom of the plunger. The distance can be varied for each cup and the exact distance (in mm) is measured on a Vernier scale attached to the side of the instrument. EYE

s

s

s

s

LENS

s

s

1234567 1234567 1234567 1234567 1234567 1234567 1234567 1234567 s

12345678 12345678 12345678 12345678 12345678 12345678 12345678 s

12345678901234567890123456789012 12345678901234567890123456789012 The eyepiece is an optical device that brings the two coloured fields next to each other for easy comparison. In making a measurement the length of one column is set and the length of the unknown is varied until the intensities of colour transmittance as viewed in the eyepiece are equal, i.e. the two fields match into a single colour. The length of the unknown column is noted. Since the concentration of the standard solution is known, the concentration of unknown can be calculated. Cu =

Rs × Cs ______________ Ru

This gives the concentration in the final volume of the unknown coloured solution.

Colorimetry 73 2. Photoelectric Colorimetry Principle

In the photoelectric colorimeter (photometer) a photoelectric cell is used instead of eye to measure the intensity of light transmitted. The depth of the solution is kept constant, the intensity of light transmitted through the solution being measured directly in terms of electric current produced by the photoelectric cell. Since the voltage of this current is directly proportional to the intensity of light falling on the cell, the output of current is inversely proportional to the concentration of he coloured substances in the solution. In photoelectric instruments, the light is first allowed to fall upon a filter (filters are made of glass or gelatin). The light transmitted by the filter is allowed to fall upon the coloured solution when some light is absorbed, the amount of light absorbed is proportional to the intensity of the colour. The fraction of light not absorbed is transmitted through the solution and is made to fall upon a photo cell in which photoelectric current is generated which is proportional to the light transmitted through the coloured solution. The current falls on a sensitive galvanometer and although deflection is due to the current generated by the light transmitted by the solution, the scale is calibrated in such a way that the reading represents the light absorbed by the solution and the readings are proportional to the intensity of all colour. The intensity of colour is proportional to the concentration of the unknown substance. Role of Blank in Colorimetric Estimation

The function of blank is to eliminate the effect of light absorption by the reagents used. If the reagents used are coloured they will also absorb light and give rise to high value of optical density and hence falsely higher values for the concentration of the substance to be analysed. This is avoided by running a blank in which distilled water is used instead of unknown and then subjected to similar conditions as in test standard.

–

Ú

Mains Tungsten filament

Ú Ú Voltage stabilizer

F I L T E R Ú

+

Cuvette

Role of Different Filters in Colorimetric Estimations

The choice of filter is made so that its maximum transmission is at a wavelength corresponding to the maximum absorption of the coloured solution. This means that in general the most suitable filter should have a colour complementary to that of a solution. For example, a purple solution requires a green filter, orange red solution a blue or blue-green filter, a yellow solution a blue filter and blue solution a red filter. The accuracy of visual colorimeter is limited by the inability of the eye to detect small difference in the intensity of light because matching of the colour is done by eye. This can be eliminated in the photoelectric colorimeter since the photo cell is affected only by the intensity of light that strikes the cell and the current generated by the cell is directly proportional to the intensity of light.


21

Blood Sugar The normal blood sugar level is 80-100 mg per 100 ml of blood. But the normal level of blood sugar varies depending upon the type of methods used for estimation. Blood sugar can be estimated by a variety of ways. 1. Reduction Method

This method utilises the reducing property of sugar but certain reducing substances other than sugars such as glutathione, ascorbic acid, uric acid, creatine, etc. are also estimated by this method giving slightly higher values upto 20-30 mg% than the normal blood sugar values. 2. Enzymatic Method

By this method, only true blood glucose is estimated. The enzyme used is glucose oxidase. This method is highly specific for glucose and does not involve any other sugar or non-reducing substance. Enzyme glucose oxidase oxidises glucose to glucuronic acid. Glucose + H2O +O2 Glucuronic acid + H2O2

The hydrogen peroxide formed is broken down to water and oxygen by peroxidase of the reagent which in the presence of an oxygen acceptor is converted into a coloured compound. The most common method used for the determination of blood sugar in the laboratories is the reduction method of King and Asatoor. Principle

Cupric ions furnished by copper sulphate are reduced by reducing sugar in alkaline medium to cuprous ions at high temperature. Cuprous ions react with phosphomolybdic acid to give a molybdenum blue colour, the intensity of which is proportional to the concentration of reducing sugar which is measured colorimetrically at 650 nm. Collection of Blood Sample

Blood sample for sugar estimation is collected in a sodium fluoride: potassium oxalate container. Sodium fluoride prevents glycolysis by inhibiting the enzyme enolase of glycolytic pathway where as potassium oxalate is an anticoagulant. Reagents

1. Alkaline copper sulphate solution consists of two solutions. Sugar A : Prepared by dissolving 1.3 gm of copper sulphate in 100 ml of distilled water. (74)

Blood Sugar 75

Sugar B

2. 3. 4. 5.

6. 7.

:

It is prepared by dissolving i. Sodium bicarbonate 50 gm ii. Sodium carbonate 40 gm iii. Potassium oxalate 36.8 gm In 400 ml of distilled water warm the solution then add 24 gm of sodium-potassium tartarate and make the volume to 1 litre with distilled water. Working solution is prepared by mixing equal volume of sugar A and sugar B. Isotonic copper sulphate Sodium tungstate (10g%) Sulphuric acid (2/3N) Phosphomolybdic acid solution: Dissolve 70 gm molybdic acid, 10 gm sodium tungstate, 40 gm sodium hydroxide in 1 litre of distilled water. Boil till the ammonia is off. Then cool and add 250 ml of phosphoric acid. Make the volume to 1 litre with distilled water. Stock glucose standard (100 mg%): Prepared by dissolving 100 mg of glucose in saturated solution of benzoic acid and making the volume to 100 ml. Working glucose standard (10 mg%): Prepared from stock glucose solution by 1:10 dilution.

Procedure

i. Precipitation step In a centrifuge tube, pipette 7.4 ml isotonic CuSO4 (to prevent haemolysis of blood) 0.2 ml blood 0.4 ml Sodium tungstate (to precipitate proteins) Mix well and centrifuge for 15 minutes. Use 2 ml of supernatant for the test i.e. for estimation. ii. Reduction step Label the test tubes in the order given below: Test (T)

S1

S2

Standard S3

S4

Blank (B)

2 ml

—

—

—

—

—

Working glucose std.

—

0.25

0.50

0.75

1.00

—

Distilled water

—

1.75

1.50

1.25

1.00

2 ml

Supernatant

Add 2 ml of alkaline copper sulphate in all the test tubes. Plug the mouth of the test tubes with cotton and place them in the boiling water bath for 10 minutes. After 10 minutes take out the test tubes and cool them at room temperature. iii. Colour development step: Add 3 ml of phosphomolybdic acid in all the test tubes. Then add 8 ml of distilled water. Mix again. Wait for 10 minutes. Read optical density at 650 nm setting blank to zero.

76 Practical Biochemistry for Students Observation

Optical density of test sample (T) Optical density of standard (S1) Optical density of standard (S2) Optical density of standard (S3) Optical density of standard (S4)

= = = = =

, , , ,

conc. (S1) conc. (S2) conc. (S3) conc. (S4)

= = = =

0.025 mg. 0.05 mg. 0.075 mg. 0.10 mg.

Calculation Optical density of test Optical density of standard

Conc. of standard Effective volume of blood used

________________________________________ × ________________________________________

× 100 mg%

Effective volume of blood for colour development 7.4 + 0.4 + 0.2 = 8 ml of the solution contains 0.2 ml blood 2 ml of the solution will contains 0.05 ml blood So effective volume of blood used is 0.05 ml. Interpretation

The normal blood sugar is 80-100 mg%. Increased level of blood sugar is termed as hyperglycemia. Hyperglycemia is observed in: 1. Diabetes mellitus 2. Hyperactivity of thyroid, pituitary and adrenal glands 3. Infectious diseases 4. Meningitis, encephalitis 5. Shock 6. Severe hemorrhage. Decreased level of blood sugar is termed as hypoglycemia. Values below 60 mg% indicate hypoglycemia. Hypoglycemia is observed in 1. Overdosage of insulin in the treatment of diabetes. 2. Hypothyroidism, hyperpituitarism and hyperadrenalism. 3. Severe liver disease.

Glucose Tolerance Test (GTT) 77

22

Glucose Tolerance Test (GTT) Glucose tolerance test determines the degree and duration of hyperglycemia after an oral intake of known quantity of glucose. After the ingestion of food there is a temporary rise in blood sugar level upto 160-180 mg% which then returns to normal fasting level within 2-3 hours after food. This effect can be studied under standard conditions by means of glucose tolerance test which is helpful in investigating abnormalities of carbohydrate metabolism, and cases in which glycosuria has been found. Method

1. 2. 3. 4.

The patient should be in a postabsorptive state, i.e at least 12 hours after last meal. Fasting blood sample and urine specimen is collected. The patient is given 50 gm of glucose dissolved in 100-200 ml of water. Blood samples and urine specimen are collected at half hour interval, i.e. ½ hour, 1 hour, 1½ hour, 2 hour.

Normal Glucose Tolerance Test Curve

The blood glucose level at 2 hours interval is below 120 mg%. It is not necessary that is should return to normal. The zero hour fasting blood glucose value is within normal range and the maximum blood glucose value is reached 1 hour after taking glucose and returns to the near fasting valves after 2 hours always. At no stage the blood glucose values exceed 180 mg% (renal threshold for glucose). All the urine specimens show no reduction and hence are glucose free at all times. Renal Glycosuria

The tolerance curve is normal though the average maximum blood glucose is reached a little lower than in completely normal person. The blood glucose follows the normal pattern, but glucose is found in some or all the urine specimen. In normal GTT, the blood glucose level is below 180 mg%, the renal threshold of glucose. In these cases, this threshold is reduced such persons pass sugar in urine whenever sufficient carbohydrate is taken to raise their blood glucose temporarily above their lowered renal threshold. Carbohydrate metabolism is normal but because of some abnormality in the tubular reabsorption of glucose, an appreciable amount of glucose escapes in the urine. This condition is called renal glycosuria. This type of curve is harmless and patients are not likely to develop into diabetes later on. (77)

78 Practical Biochemistry for Students Normal Curve 200 —

Renal Glycosuria Renal Threshold

200 — •

•

150 —

150 —

Renal Threshold

• •

•

•

100 •—

100 — •

•

50 —

Urine Glucose

0 –

•

50 — | 0.5 –

| 1.0 –

| 1.5 –

| 2.0 –

Urine Glucose

0 –

| 0.5 –

| 1.0 +

| 1.5 –

| 2.0 –

Diabetic Curves •

400 — Mild

•

300 —

200 —

Severe

•

•

150 —

• •

• •

200 —

100 •—

•

150 — •

50 —

•

100 — Urine Glucose

0 –

| 0.5 –

| 1.0 –

| 1.5 –

50 —

Urine Glucose 200 —

0 +

| 0.5 +

| 1.0 +

| 1.5 +

| 2.0 + Flat Curve

Lag Curve

•

•

•

| 2.0 –

200 — •

150 —

150 —

• •

100 —

•

•

50 — Urine Glucose

0 –

• •

| 1.0 +

| 1.5 –

•

•

• •

•

•

| 1.5 –

| 2.0 –

50 —

•

| 0.5 –

100 —

•

| 2.0 –

Urine Glucose

0 –

| 0.5 –

| 1.0 –

Diabetes GTT Curve (Diminished Glucose Tolerance)

Glucose tolerance curve diminished to a greater extent in diabetes mellitus. The most significant finding in the diagnosis of diabetes is the failure of the blood glucose levels to fall below 120 mg% even by 2 hours. The peak level is frequently above normal curve

Glucose Tolerance Test (GTT) 79

and fasting level may or may not be raised. A true diabetic will show a fasting level higher than normal, a rise much above renal threshold and fall to fasting level may be achieved even after 5 hours. They may or may not be glucosuria in the fasting urine but the postglucose urine definitely contains sugar. Lag Type of Tolerance Curve

In this type of curve, the peak blood glucose level may be higher than normal but the 2-hour value is within normal limits or often low. This implies a delay in early compensatory mechanisms (lag) without necessarily impairment of insulin response. The increase in blood glucose level is due to delay in insulin mechanism coming into action. It is more probably due to an increased rate of absorption of glucose into the intestines. If the blood glucose level at peak of the curve is above the renal threshold level, glucose appears in the next urine specimen. Such a curve may be seen in: i. Normal individuals ii. After gastrectomy where rapid entry of glucose into the intestines leads to rapid absorption and sudden outpouring of insulins into overswing. iii. In severe liver disease, probably due to decreased glycogenesis. iv. Rarely in thyrotoxicosis probably due to rapid absorption of glucose. Flat GTT Curve

Blood glucose level does not rise normally following a glucose load. It may however be found in completely healthy individuals and interpretation is difficult. This type of curve indicates malabsorption, hypothyroidism, hypoadrenalism, hypopituitarism.


23

Blood Urea Urea is a major end product of protein metabolism in human body. Urea is synthesised in the liver by urea cycle and is excreted by the kidney. Urea constitutes the major non protein nitrogen (NPN) of the blood and it represents 45-50% of NPN of the blood. It is also the major NPN substance excreted in the urine. The normal blood urea level is 15-40 mg%. It is slightly higher in men than in women. Urea content over a period is influenced by the amount of protein and tends to be lower on a low protein diet. In some countries, blood urea is represented as blood urea nitrogen (BUN). The mg% urea can be convered into BUN and vice versa by the following factor: BUN = mg% urea × 0.467 and mg% urea = BUN × 2.14 Urea diffuses freely in and out of the red blood cells so its concentration both in cells and plasma is nearly the same. It is immaterial whether whole blood, plasma or serum is used for blood urea estimation. Moreover, the collection of the blood for urea estimation depends upon the method used. Urea can be estimated by the following methods: 1. Urease-nesslerization method. 2. Diacetyl monoxime method. 3. Autoanalyser method. In urease-nesslerisation method, urea is converted into ammonium carbonate by the enzyme urease (found in soya beans and jack beans) followed by colour development with Nesslaer reagent. Whereas in diacetyl monoxime method, urea reacts directly with diacetyl monoxime under strongly acidic conditions to give a yellow coloured product. The reaction is intensified by the presence of ferric ions and thiosemicarbazide which gives red/pink coloured complex which is more linear than the yellow one. Reagents

1. Sodium tungstate (10g%): prepared by dissolving 10 gm of sodium tungstate in water and make the volume to 100 ml with distilled water. 2. 2/3 NH2SO4: Prepared by dilution from a stock solution of 1N-H2SO4. 3. Diacetyl monoxime reagent: Prepared by dessolving 2 gm of diacetyl monoxime in a beaker in about 60 ml of water. Add 2 ml of glacial acetic acid, warm if necessary and make the volume to 100 ml with distilled water. (80)

Blood Urea 81

4. Sulphuric acid—phosphoric acid reagent: 150 ml of 85% phosphoric acid is added to 140 ml of water. Mix well. Carefully add 50 ml of concentrated H2SO4 to the mixture. 5. Stock urea standard 250 mg%. 6. Working urea standard 2.5 mg%. Dilute 1 ml of stock standard to 100 ml with water. Procedure

In a centrifuge tube pipette the followings: • 0.1 ml blood • 3.3 ml water • 0.3 ml sodium tungstate • 0.3 ml 2/3 N-H2SO4 • Mix and centrifuge Arrange the test tubes and mark them blank ‘B’, test ‘T’and standards (S1) and (S2). B Supernatant (from test) Working urea standard Distilled water Diacetyl monoxime H2SO4-H3PO4 mixture

— — 2 ml 0.4 ml 1.6 ml

S1 — 0.5 ml 1.5 ml 0.4 ml 1.6 ml

S2

T

— 1.0 ml 1.0 ml 0.4 ml 1.6 ml

1 ml — 1 ml 0.4 ml 1.6 ml

Mix well and place all the test tubes in the boiling water for minutes. Cool it and Read optical density at 490 nm. Observation

• Optical density of test (T) = • Optical density of standard (S1) = • Optical density of standard (S2) = Calculation

The concentration of unknown Optical density of test Concentration of standard = _____________________________________ × __________________________________________ × 100 Optical density of standard Effective volume of blood used Effective volume of blood used = 0.025 ml. (0.1 + 3.3 + 0.3 + 0.3 ml = 4 ml contains 0.1 ml of blood so 1 ml ≡ .025 ml of blood. Interpretation

Increased in blood urea can be considered under the following: Increased tissue protein catabolism associated with negative nitrogen balance such as fever, wasting disease, thyrotoxicosis, leukemia, etc.

82 Practical Biochemistry for Students Prerenal

• • • • • •

Prerenal failure due to low blood supply Vomiting Diarhoea Diabetic coma Addison’s disease In shock due to severe burns.

Renal

• • • • • • •

Blood urea increases in all the kidney diseases Acute and chronic glomerulonephritis Pyelonephritis Nephrotic syndrome Malignant hypertension Heavy metal poisoning Deposition of calcium in kidney due to hypervitaminosis D and hyperparathyroidism.

Postrenal

Postrenal diseases which lead to the increase in blood urea are those in which there is an obstruction to the flow of urine and reduction in the effective filteration at the glomeruli. • Enlargement of the prostrate • Stones in urinary tract • Stricture in the urethra • Tumours of the bladders affecting urethra • Low level of blood urea is found in the following conditions: – Growing infants – Later stages of pregnancy – Low protein and high carbohydrate diet – Severe liver disease.

Urea Clearance 83

24

Urea Clearance Clearance of any substance is defined as the number of ml of plasma which contains the amount of that substance excreted in a minute by the kidneys. mg of substance excreted per minute

=

_____________________________________________________

=

____________

mg of substance per ml of plasma

UV P

where U = Concentration of the substance in urine V = ml of urine excreted per minute P = Concentration of the substance in plasma.

If the concentration of the substance in cells and plasma is not very different, then whole blood can be used. There are substances which are equally distributed between cells and plasma, e.g. urea, than whole blood, can be used. clearance =

UV

____________

B

where B = Concentration of the substance in whole blood.

Urea clearance is defined as the number of ml blood which contain the urea excreted in a minute by the kidneys. Urea clearance = =

mg of urea excreted per minute

______________________________________________

mg of urea per ml of blood

UV

____________

B

U = mg urea per 100 ml of urine B = mg urea per 100 ml of blood V = ml of urine excreted per minute.

The normal urea clearance was determined on a number of normal persons and value found to an average 75 ml/min, when the rate of excretion of urine was 2 ml or more per minute. Urea clearance was found to decrease with V, fell below 2 ml/min. With a urine volume of 1 ml/min the clearance was found to average 54 min/ml and the decrease with decreasing urine volume was seen to be proportional to the square root of V. Hence for value between 0.5 to 2 ml per minute; the average normal clearance is given by 54√V. (83)


For value of V equal to 2 ml or more than the term is called maximum urea clearance. Maximum urea clearance = (% of the average normal)

Observed urea clearance × 100 Average number maximum urea clearance

________________________________________________________

UV

=

______________

=

_________

B 75

100 75

= 1.33

× 100 = 100 UV B

_________

UV B

_________

For values of V less than 2, then the term is called standard urea clearance. Standared urea clearance UV = ___________ × 100 (% of the average normal) B 54√ V U√ V = 1.85 _______________ B

Urea clearance is expressed as a percentage of the average normal. Procedure

The patient is given light breakfast avoiding high protein and then the test is carried out in a two- hour period and can be completed before 12 noon. During the period of the experiment the patient should be at rest in bed and urine collected directly in a clean container. Time should be recorded correct to the minute at the completion of micturition every time. The patient is given about two tumblers of water to drink to ensure adequate urine flow. He then voids urine which is then discarded. Exactly one hour later the patient voids a specimen of urine which is collected. At the same time a blood sample is also collected. One hour later a second urine sample is also collected. Also blood sample is collected. The volume of each urine sample is measured and the volume per minute is calculated. Interpretation

Urea clearance test is a fairly reliable test for assaying the overall functional efficiency of the kidneys. Urea clearance Above 70 40-70 20-40 Below 20 Below 5

Renal function Normal Mild deficit Moderate deficit Severe deficit Coma

Blood Cholesterol 85

25

Blood Cholesterol Cholesterol is present in blood both as free and as esters. Free cholesterol normally forms about 30 per cent of the total cholesterol and the ester fraction forms 70 per cent. Principle

Cholesterol and cholesterol esters from the serum are extracted into an alcohol-ether mixture (Alcohol precipitates the proteins; either solubilises the cholesterol part). The contents are centrifuged. The protein free extract is evaporated to dryness. The cholesterol residue is dissolved in chloroform and is measured colorimetrically by Liebermann-Burchard reaction. Reagents

1. 2. 3. 4. 5.

}

Absolute alcohol Alocohol-either mixture 3.1 Diethyl ether Chloroform Acetic anhydride Standard solution of cholesterol (100 mg%) Dissolve 100 mg of cholesterol in some amount of alcohol. Slightly warm (in water bath), if required. Make the volume to 100 ml with alcohol). Procedure Test

In a centrifuge tube, take 8 ml of alcohol, 2 ml of ether. Mix it. Add 0.2 ml of blood. Mix overall. Keep it in slanting position for half an hour. Centrifuge it. The supernatant is collected (which contains cholesterol) in another tube. This test tube is kept in a boiling water for the evaporation of solvent and the residue, i.e. cholesterol sticks to the bottom of the flask is left behind. Chloroform, acetic anhydride and sulphuric acid must be of highest quality. It is particularly important that chloroform be especially anhydrous. Ordinary chloroform or old stock will lead to weak and uncertain colour. Standard

In different tubes marked S1, S2, S3, and S4, add 0.2, 0.4, 0.6 and 0.8 ml of standard cholesterol solution. All should be kept in boiling water bath for evaporation of solvent till the solvent evaporates and residue is left behind. Blank: Clean dry test tube. (85)


Now add the following reagents in test, different standard, and blank. Standard Blank

S1

S2

S3

S4

Test

CHCl3

5 ml

5 ml

5 ml

5 ml

5 ml

5 ml

Acetic anhydride

2 ml

2 ml

2 ml

2 ml

2 ml

2 ml

0.1 ml

0.1 ml

0.1 ml

0.1 ml

0.1 ml

0.1 ml

Concentrated H2SO4

Mix well, keep it in dark for 10 minutes. Read optical density at 610 nm. Observations

Optical density of test Optical density of standard S1 Optical density of standard S2 Optical density of standard S3 Optical density of standard S4

= = = = =

, , , ,

conc. = conc. = conc. = conc. =

.02 mg .04 mg .06 mg .08 mg

Conclusion

The concentration of cholesterol. Optical density of test Conc of standard = ________________________________________ × _________________________________________ × 100 Optical density of standard Volume of blood/serum used Interpretation

Normal serum cholesterol level is 150-250 mg%. Hypercholesterolemia is observed in the following conditions 1. Diabetes mellitus. 2. Obstructive jaundice. 3. Nephrotic syndrome. 4. Cirrhosis of liver. 5. Hypoparathyroidism. 6. Xanthomatosis. Hypocholesterolemia is observed in the following 1. Hyperthyroidism 2. Pernicious anaemia 3. Haemolytic anaemia 4. Malabsorption syndrome.

Serum Calcium 87

26

Serum Calcium Calcium is present in the body in larger amounts than any other cation. Almost all of it is present in bones and teeth. Very small quantity is present in the body fluid. The blood cells contain very little calcium. Most of the blood calcium is therefore present in the plasma. It is present in three forms in plasma and serum. i. Protein bound fraction (45-50%) ii. Ionized fraction (45-50%) iii. In combination with citrate (5%) Protein bound calcium is non-diffusible whereas the other two are diffusible. Ionised calcium is of great importance in blood coagulation in the function of heart, muscles, nerves and in the permeability of membranes. In the usual determination all the three forms are measured together. Reagents

1. Ammonium oxalate (saturated solution) 2. Ammonia solution (2% v/v). Dilute 2 ml of ammonia to 100 ml with water. 3. Normal sulphuric acid. 4.

N

________

100

KMnO4 solution.

Principle

Total calcium present in the serum is precipitated directly as calcium oxalate by using saturated ammonium oxalate solution. The precipitate is centrifuged, washed with dilute ammonia and is dissolved in sulphuric acid which converts it into oxalic acid. This is titrated with standared solution of potassium permanganate. Collection of Sample

Blood is collected in a plain container (i.e. without any anticoagulant) and allowed to clot and serum is collected. No anticoagulant is used as most of the anticoagulants react with calcium. Fasting blood should be collected as lipaemic serum could lead to erroneous results. (87)

88 Practical Biochemistry for Students Reaction ___________________ → Serum calcium + COONH4 COO | | Ca COO COONH4 (saturated solution Calcium oxalate COO COOH + CaSO4 | Ca + H2SO4 _______________→ | COO COOH _______________→ 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O COOH _______________→ 2CO2 + H2O | +O COOH

Procedure

In a centrifuge tube take, 2 ml of serum, 2 ml of distilled water and 1 ml of saturated ammonium oxalate. Mix and keep it for 30 minutes with frequent mixing in between. Centrifuge for 15 minutes and decant off the supernatant fluid without disturbing the precipitate. After doing so, allow the tubes to stand inverted on a filter paper and allow to drain for five minutes. Wipe the mouth of the tube. Now wash the precipitate with 4 ml of dilute ammonia. Centrifuge and discard the supernatant as before. Repeat the washing twice. After washing dissolve the precipitate in 2 ml of normal sulphuric acid. Warm by placing the tube in a beaker having the temperature between 75° to 85°C so that the precipitate goes into the solution. Titrate it with N/100 KMnO4 while hot till a definite faint pink colour persists for at least 1 minute. Also carry out a blank titration, with 2 ml of sulphuric acid to the same above end point. The difference between the two titrations gives the volume of N/100 KMnO4 required to titrate the calcium oxalate precipitate. Calculation

1 ml of N/100 KMnO4 = 0.2 mg calcium. Volume of serum used = 2 ml Volume of N/100 KMnO4 used to titrate the calcium oxalate precipitate = x ml. This x ml represents the difference of the test titration minus blank titration. x = titration of test – Titration of blank. x ml of N/100 KMnO4 = 0.2 × x mg calcium. Hence 2 ml of serum has 0.2× x mg calcium. 0.2 × x So 100 ml of serum has ____________ × 100 mg of calcium. 2 The normal total serum calcium value is 9-11 mg% (i.e. 4.5 to 5.5 mEq/L). It is slightly higher in young children. The level of serum calcium is affected by deficient calcium absorption from the intestines, by alterations in the amount of parathyroid hormone secreted, by changes in serum inorganic phosphorus and serum proteins.

Serum Calcium 89

Low serum calcium is observed in the following: i. Hypoparathyroidism ii. Osteomalacia iii. Rickets iv. Renal failure v. Tetany vi. During an attack of acute pancreatitis vii. Steatorrhoea Low serum calcium is often associated with high serum inorganic phosphorus. High serum calcium is observed in the following: i. Hyperparathyroidism ii. Hypervitaminosis D iii. Multiple myeloma iv. Neoplastic diseases of bones. Serum calcium is often increased following prolonged excessive intake of milk, alkali and antacids, the treatment for peptic ulcer.


27

Inorganic Phosphorus Phosphorus is present in blood as: i. Inorganic phosphorus ii. Ester phosphorus, i.e. organic phosphorus iii. Lipid phosphorus. Of these inorganic phosphorus is determined most frequently. Principle

Blood sample is treated with trichloroacetic acid to remove protein and lipid phosphorus. The inorganic phosphorus present in the supernatant filtrate reacts with molybdate in acid solution to form phosphomolybdate. Phosphomolybdate so formed is reduced by 1,2,4 aminonaphthol sulphonic acid into molybdenum blue, a blue coloured complex. The intensity of blue colour is proportional to the amount of inorganic phosphorus present in the sample. Reactions

Inorganic phosphorus + Molybdate ______→ Phosphomolybdate Phosphomolybdate + Reducing agent ______→ Μolybdenum blue Reagent

1. 2. 3. 4.

Trichloroacetic acid 10 gm% Ammonium molybdate 5 g% Perchloric acid Reducing agent, i.e. 1, 2, 4, α-aminonaphthol sulphonic acid (ANSA) contains sodium bisulphite and sodium sulphite. It is prepared by dissolving 200 mg of 1, 2, 4, αaminonaphthol sulphonic acid, 12 gm of sodium metabisulphite, 2.4 gm of sodium sulphite in distilled water and make the volume to 100 ml with distilled water. Store the reagent in dark at cool place. 5. Standard phosphorus solution (1 mg/ml): Dissolve 438 mg of potassium dihydrogen phosphate in 100 ml of water, working standard (0.004 mg/ml). Prepared by diluting 1 ml of stock phosphate solution to 250 ml with double distilled water. Procedure

In a test tube, take 9 ml of trichloroacetic acid. Add 1 ml of serum while shaking constantly. Mix well. Stand for 5 minutes and filter the solution. (90)

Inorganic Phosphorus 91

Filtrate Working phosphorus standard Distilled water

Test

S1

S2

S3

S4

S5

Blank

5 ml — —

— 1 ml 4 ml

— 2 ml 3 ml

— 3 ml 2 ml

— 4 ml 1 ml

— 5 ml —

— — 5 ml

To each tube add 0.4 ml perchloric acid, 0.4 ml ammonium molybdate, 0.2 ml of reducing agent. Mix gently after every addition. Wait for 10 minutes. Read optical density at 610 nm setting blank to zero. Precaution

All the reagents should be prepared in double distilled water. All the glass wares used, i.e. test tubes, pipettes, etc. should be also washed with double distilled water. Observation

Optical density of test = Optical density of standard S1 = Optical density of standard S2 = Optical density of standard S3 = Optical density of standard S4 = Optical density of standard S5 =

, , , , ,

conc. conc. conc. conc. conc.

= = = = =

0.004 mg. 0.008 mg. 0.012 mg. 0.016 mg. 0.02 mg.

Calculation

The concentration of inorganic phosphorus is given by Optical density of test

Concentration of blood × ____________________________________________ × 100 mg Optical density of standard Effective volume of blood used

_____________________________________

Effective volume of blood used = 0.5 ml. Interpretation

The normal serum inorganic phosphorus is between 2.5 and 4.5 mg%. It is higher in children whose normal range is from 4 to 6 mg%. Increased serum phosphorus levels may be found in: i. Hypervitaminosis D-↑Ca ↑P ii. Hypoparathyroidism-↓Ca2+ ↑P iii. Renal failure-↓Ca2+ ↑P Low serum phosphorous levels may be found in:‘ i. Rickets-↓Ca2+ ↓P ii. Osteomalacia-↓Ca2+ ↓P iii. Hyperparathyroidism-↑Ca2+ ↓P iv. Fanconi's syndrome


28

Serum Total Proteins and Albumin: Globulin Ratio Serum Proteins

It can be estimated by the number of methods such as a. Colorimetrically b. Precipitation method c. Microkjeldahl method d. Electrophoretically. The normal total serum proteins level is 5.5 to 7.5 gm%, albumin 3.5-5.5% and globulin 22.5g%. Liver is the sole organ chiefly responsible for the formation of plasma albumin and at least 80% of the plasma globulins, i.e. α-and β-globulins. Principle

Proteins and peptide bonds react with alkaline copper sulphate (Biuret reagent) to give a violet coloured coplex. The intensity of violet colour is measured colorimetrically and is proportional to the concentration of the total proteins in the sample. Similarly globulin is precipitated by addition of sodium sulphate. The albumin left in the supernatant is treated with Biuret reagent and violet colour produced is measured colorimetrically as above. Reagents

1. 2. 3. 4.

Normal saline Biuret reagent 28% sodium sulphite Ether.

Procedure

For total proteins: Serum for total protein estimation is 1:10 diluted, i.e. 0.2 ml serum and 1.8 ml normal saline. Serum (1:10 diluted) Bovine albumin standard Normal saline

Test

S1

S2

S3

S4

S5

Blank

0.5 ml — 2.5

— 0.2 ml 2.8

— 0.4 2.6

— 0.6 2.4

— 0.8 2.2

— 1.0 2.0

— — 3

(92)

Serum Total Proteins and Albumin: Globulin Ratio 93

• To each tube, add 3 ml of Biuret reagent. Mix it. • Incubate at 37°C for 10 minutes. • Red optical density at 520 mm. Observation

Optical density of test = Optical density of standard S1= Optical density of standard S2= Optical density of standard S3= Optical density of standard S4= Optical density of standard S5=

, , , , ,

conc, conc, conc, conc, conc,

= = = = =

1.6 mg. 3.2 mg. 4.8 mg. 6.4 mg. 8 mg.

Calculation

The concentration of serum total proteins is given by Optical density of test Concentration of standard _______________________________________ × __________________________________________ × 100 Optical density of standard Effective volume of serum used Effective volume of serum used = 0.05 ml (Because 0.5 ml of 1: 10 diluted serum was taken). A/G Ratio

In a centrifuge tube, take 5.8 ml of 28% sodium sulphite. Add 0.2 ml of serum (undiluted) as such. Gently rotate the centrifuge tube, so that the serum is evently distributed in the solution. Add 2 ml of ether. Place the thumb, and gently shake it up side down slowly 20 times, and keep it for 30 minutes. Centrifuge for 10 minutes. Carefully detach the globulin button formed at the interphase of water: either layer and using a capillary dropper, take out the supernatant and transfer it into another tube marked albumin. Supernatant Normal saline

Test

Standard

Blank

3 ml —

As in total protein

— 3 ml

• Mix and add 3 ml of Biuret reagent. • Keep at 37°C for 10 minutes • Read optical density at 520 nm. Calculation

The concentration of albumin is given by Optical density of test

________________________________________

Optical density of standard

×

Concentration of standard

_____________________________________________

Effective volume of serum used

× 100


Effective volume of seum used = 0.1 ml (5.8 + 0.2 ml solution contains 0.2 ml of serum. Therefore 3 ml will contain 0.1 ml). Interpretation

Total proteins are decreased in i. Nephrotic syndrome ii. Malnutrition iii. Kwashiorkor iv. Cirrhosis of liver v. Liver disorders High protein values are observed in: i. Multiple myeloma ii. Conditions associated with high globulin concentration The normal protein value for serum albumin varies with the technique employed. It ranges from 3-5-5.5 gm% and constitutes about 50-65% of the total serum proteins. Decreased levels of albumin are found in i. Cirrhosis of liver ii. Nephrotic syndrome iii. Malnutrition iv. Malignancies. Pathological conditions associated with increased levels of serum albumin are not known. An albuminaemia is the condition in which plasma albumin is completely absent. All globulin fractions present in increased concentration. Albumin/Globulin ratio i.e. A/G ratio

The normal A/G ratio is 2:1. An alteration in the ratio and reversal may occur due to the reduction in albumin and or elevation of globulin. The ratio is reduced and often reversed in cirrhosis with jaundice. However, the ratio may be increased in some cases of xanthomatosis or biliary cirrhosis.

Serum Bilirubin 95

29

Serum Bilirubin Determination of Total and Conjugated Bilirubin in Serum

Any increase of bilirubin in the blood is an indication of jaundice (derived from French word jaune meaning—yellow), an estimation of bilirubin is a must in liver and biliary tract diseases. Bilirubin is derived from the destruction of red cells in reticuloendothelial system. It is excreted in urine as urobilinogen and in feces as stereobilinogen. Serum bilirubin is of two types: 1. Conjugated or direct bilirubin: It is bilirubin glucuronide. It is water soluble. Van den Bergh called the form of bilirubin which reacted without alcohol as direct (conjugated). 2. Unconjugated or indirect bilirubin: It is albumin bound bilirubin (i.e. protein bound). It is insoluble in water. It is soluble in alcohol that is why van den Bergh called the form of bilirubin which reacted in the presence of alcohol is indirect (Unconjugated). Total bilirubin is measured by adding methanol which permits reaction of direct and indirect bilirubin with diazo reagent. Following different responses are observed. Direct Reaction

a. Immediate direct reaction: A violent colour due to the formation of diazo-bilirubin in 10-30 seconds is observed, i.e. immediate colour development place. b. Delayed direct reaction: No change in appearance of colour take place for 5 to 15 minutes. Then reddish colour develops which gradually becomes violet. c. Biphasic: A red colour appears promptly but takes a longer time to change to violet. Indirect Reaction

The development of colour takes place after the addition of alcohol. Normal Range

• Total bilirubin in adults : • Total bilirubin in infants may go upto: • Conjugated bilirubin :

0.2 to 0.6 mg/100 ml 1.2 mg/100 ml 0.0 to 0.2 mg/100 ml

Serum Collection

Collect blood in a clean container without any anticoagulant and allowed to clot. Separate serum from the clot as soon as possible taking care to avoid hemolysis. Protect specimen from light. If the test cannot be carried out immediately the sample should be frozen. (95)

96 Practical Biochemistry for Students Principles

Direct Conjugated bilirubin couples with diazo reagent (diazotised sulphuric acid forming azobilirubin, a red-purple coloured product in acidic medium. Indirect Unconjugated bilirubin is diazotised only in the presence of methanol. Thus the redpurple coloured azobilirubin produced in presence of methanol originates from both direct and indirect fractions and thus total bilirubin concentration can be found out. The difference of total and direct bilirubin gives indirect (conjugated) bilirubin result. The intensity of red colour so developed above is measured colorimetrically and is proportional to the concentration of the appropriate fraction of bilirubin. The reaction can be represented as. Reaction H+ Bilirubin + Diazotised sulphanilic acid _________________→ Azobilirubin ___ →

H+

Red purple colour

Reagents

1. Diazo A: It contains sulphalinic acid dissolved in concentrated hydrochloric acid. Diazo A is prepared by dissolving 1 gm of sulphanilic acid in 15 ml of concentrated hydrochloric acid and make the volume to 1 litre with distilled water. 2. Diazo B: It contains sodium nitrite in water. Diazo B is prepared by dissolving 500 mg in 100 ml of water. This should be prepared freshly. 3. Diazo reagent: This reagent is prepared freshly before use by adding 10 ml of diazo A to 0.3 ml of diazo B. 4. Hydrochloric acid (1.5% v/v). 5. Methanol. 6. Standard bilirubin (10 mg%). Dissolve 10 mg of bilirubin in chloroform. Reflex the solution to completely dissolve the bilirubin. Cool the solution and make the volume to 100 ml with chloroform. 7. Working bilirubin standard (2 mg%). Dilute the above bilirubin standard solution 1:5 with alcohol, i.e. 1 ml of bilirubin standard add 4 ml of alcohol. Procedure

Arrange the test tubes and add the reagents in the following order. Mix after each additions.

Serum Working bilirubin standard distilled water Diazo reagent Hydrochloric acid Methanol

Blank

Standard

Test (For direct bilirubin)

Test (For folic bilirubin)

— — 2 ml — 0.5 ml 2.5 ml

— 0.2 ml 1.8 ml 0.5 ml — 2.5 ml

0.2 ml — 1.8 ml 0.5 ml — 2.5 ml

0.2 ml — 1.8 ml 0.5 ml — 2.5 ml

Serum Bilirubin 97

Mix and wait for 20 minutes. Read the optical density at 520 mm setting blank to zero. A blank should be run with every specimen tested. Calculation Concentration of bilirubin (total) = (mg%) Optical density of test (total)

___________________________________________

Optical density of standard

×

Concentration of standard

___________________________________________

Volume of serum used

× 100

Same formula can be used determining the concentration of conjugated bilirubin for conjugated test. Concentration of total bilirubin = Conjugated bilirubin + Unconjugated bilirubin. Hence the concentration of unconjugated can be calculated. Interpretation

Serum bilirubin is increased in prehepatic (hemolytic), hepatic (hepatocellular) and posthepatic (obstructive) jaundice. In hemolytic jaundice, the unconjugated bilirubin is increased, the conjugated bilirubin is less than 20% of the total bilirubin. • Indirect/delayed direct reaction—hemolytic jaundice. In hepatic jaundice, the amounts of unconjugated and conjugated bilirubin are variable. Defective transport of unconjugated bilirubin acrosss the hepatocyte or inability of the hepatocyte to conjugate bilirubin results in elevated levels of unconjugated bilirubin. • Direct reaction—hepatic jaundice. In posthepatic (obstructive) jaundice, conjugated bilirubin is increased. Prompt direct reaction—in obstructive jaundice.


30

Prothrombin Time Prothrombin time is the time required for clotting of citrated plasma to which optimum amounts of brain thromboplastin and calcium has been added (Normal values 11-16 sec). One stage prothrombin time measures the adequacy of factors involved in 2nd and 3rd stages of coagulation in extrinsic clotting system. The coagulation in external clotting system is represented as: Stage 1 : Formation of tissue thromoboplastin. Stage 2 : Prothrombin Factor II | | | | Factor V | — — — — — — — — Thrombin | | Factor VII | | | | Factor X | | | | Calcium | | Stage 3 : Fibrinogen — — — — — — — — — — — — — — — — — — Fibrin From the above it is apparent that deficiency of any of the factors involved in 2nd and 3rd stages of extrinsic clotting system (Factors II (Prothrombin), V,VII, X or fibrinogen) will prolong the plasma clotting time. Reagents

1. Sodium citrate 3.8 g%. 2. Thromboplastin suspension: Tablets of thromboplastin containing calcium ions in appropriate amount supplied by Geigy are used. One tablet is dissolved in 2.5 ml of distilled water. Collection of blood

Blood is collected in 3.8% sodium citrate solution as anticoagulant. The ratio of anticoagulant to blood should be 1: 4, i.e. 0.4 ml of sodium citrate and 1.6 ml of blood. Also collect the blood sample from a normal, healthy individuals in the above manner for control. (98)

Prothrombin Time 99 Procedure

Take 0.1 ml of plasma into a test tube and keep it as 37°C in a water bath for 2 minutes. Add 0.2 ml of thromboplastin suspension in the above tube and simultaneously start the stop watch. The contents of the test tube are mixed by gentle rotation. Stop the watch as soon as the fibrin clot is seen and record the time taken for coagulation of plasma. Also find out the prothrombin time of the control plasma. Results

Prothrombin time is expressed in seconds with the control value given. It is also expressed as prothrombin index and prothrombin activity. Prothrombin index (%) =

Prothrombin ratio =

Prothrombin time of control plasma

___________________________________________________

Prothrombin time of test plasma Prothrombin time of control plasma

________________________________________________________

Prothrombin time of test plasma


31

Liver Function Tests The functions of liver are many, and the term liver function tests is something of a misnomer since there is no single test of liver function. The best test of liver function depends on what function of the liver is being tested. Several liver function tests should be performed as a single test may show false-positive or false-negative results. The importances of liver function tests are: 1. To assess severity of liver damage. 2. To differentiate different types of jaundice. 3. To find out the presence of latent liver diseases. There are hosts of tests to evaluate the functions of liver but those that are commonly employed have got significance for assessing the conditions of patients. The first group of tests regarding the secretory, excretory and enzymatic functions are serum bilirubin estimation, bilirubin and urobilinogen in urine, BSP excretion test, serum alkaline phosphatase estimation and SGPT. The second group meant for assessing the protein synthetic function are total proteins estimation, A/G ratio and prothrombin time. The final and third group that are meant for liver metabolic functions are; estimation of serum cholesterol and determination of free and esterified cholesterol ratio. Serum Aminotransferases (Transaminases)

Transaminases are involved in transfer of amino group from α-amino acid to α-ketoacid. In this process X-amino acid is converted to its corresponding X-ketoacid. α-keto- ALT or Pyruvate + Alanine + gluteric _________→ (SGPT) acid glutamate α-keto- AST or Oxaloacetate + Aspartic + gluteric _________→ (SGOT) acid acid glutamate Transamination reactions are important in intermediary metabolism because of their function in synthesis and degradation of amino acid. SGPT is found in hepatocytes in highest concentration. It is considered to be more liver specific. SGOT is found in cardiac tissues, skeletal muscles, pancreas and kidney. Normal values 1. SGPT—3-15 IU/L at 25°C 2. SGOT—4-17 IU/L at 25°C (100)

Liver Function Tests 101 Diagnostic Significance

If the level is raised in serum, it indicates damage to tissues, so clinically its use is in evaluation of damage to tissues conditions in which SGPT is raised are hepatic disorders (like viral hepatitis). Other conditions are malignancies cholestasis and hepatotoxic drugs (like lithium) are used. Procedure

Since enzymes are present in very small quantity in biological fluids and it is difficult to isolate them, so, a convenient way to quantitate them is to measure their catalytic activities and relate this acitivity to concentration. Serum is allowed to react with substrate and product is pyruvate. Pyruvate reacts with 2, 3-dinitro-phenyl hydrazine (DNPH) and gives greenish-yellow colour and intensity of this colour is measured at 550 nm. The intensity of colour is directly proportional to activity of SGPT, NaOH is added to stop the reaction. Calculation

ODT—ODC Plot this OD on standard curve and extrapolate the enzyme activity. If the activity is high and OD is above 0.29, then dilute the serum and repeat the test. B Standard pyurvate Distilled water DNPH

2.0 ml

S1 0.5 ml 1.5 ml 0.5 ml

S2 1.0 ml 1.0 ml 0.5 ml

S3 1.5 ml 0.5 ml 0.5 ml

S4 2.0 ml – 0.5 ml

0.5 ml

Keep at room temperature for 20 min. 4N

NaOH

0.5 ml

0.5 ml

0.5 ml

0.5 ml

0.5 ml

General methods for measuring enzymatic activity 1. End point 2. Kinetic — multipoint — continuous monitoring 1. End Point Reactants are combined and reaction proceeds for designed time (SGPT—30 min.) and then reaction is stopped. 2. Kinetics a. Multipoint measurement made after specifc interval of time, e.g. 30 sec. b. Continuous monitoring In it continuous measurement is done. It is preferred because any deviation from reality can easily be determined. The amount of enzyme that will convert 1μ mole to its product per minute under specific conditions of pH, temperature, time, etc. is one International Unit per minute. Enzyme concentration is expressed in units per litre. Alkaline Phosphatase

Enzyme activity is measured in International Unit. One IU the amount of enzyme that will convert 1μ mole of reactant into product per minute per litre.


Alkaline phosphatase is present in most tissues but is present in high concentration in liver, bones, intestines, spleen, placenta and kidney. It is involved in transport of phosphate across cell membrane. It has a hydrolytic and phosphate transferase activity. Reagents

In Kindlking method following reagents are used. 1. Substrate : Disodium phenyl phosphate 2. Buffer : NaHCO3 + Na2CO3 (pH = 10) 3. Phenol standard : 0.01 mg/ml 4. NaOH stops the reaction 5. 4-amino antipyrene 6. Potassium ferricyanide oxidising agents Test 1. 2. 3. 4. 5. 6. 7. 8. 9.

Substrate 1.0 ml Buffer 1.0 ml Phenol std – Distilled water – Serum 0.1 ml Incubate at 37°C for exactly 15 min NaOH 0.8 ml Serum – 4-amino antipyrene 1.0 ml Potassium ferricyanide 1.0 ml

Control

Std.

Buffer

1.0 ml 1.0 ml – – –

– 1.1 ml 1.0 ml – –

– 1.1 ml – 1.0 ml –

0.8 ml. 0.1 ml 0.8 ml 0.8 ml

0.8 ml. – 1.0 ml 1.0 ml

0.8 ml – 1.0 ml 1.0 ml

Take OD at 540 nm after 10 min

Chemistry

Disodium pH = 10 phenyl ___________________→ Phenol Phosphate AlkPhos | 4-amino antipyrene ↓ Red coloured quinone. Calculations =

=

ODT – ODC

× 0.01 ×

ODS – ODB ODT – ODC ODS – ODB

100 0.1

×

0.10

= ODT – ODC × 60 KA

KAU = Liberation of 1 mg of phenol in 15 min for 100 ml serum KAU × 7.1 = IV = 3-13 KAU Normal value is 3-13 KAU is 2.5 times more in children.

Liver Function Tests 103

Causes of increased activity a. Physiological 1. Infants 2. Pregnancy b. Pathological 1. Bone disorders e.g. osteomalacia rickets Paget's disease 2. Liver diseases Obstructive jaundice Intrahepatic cholestasis Liver cancer 3. Certain drugs e.g. steroids antidioxics Causes of decreased activity 1. Anaemia 2. Kwashiorkor 3. Scurvy


32

Demonstrations ELECTROMETRIC DETERMINATION OF pH

The pH of solutions can be determined more accurately by potential measurements of certain electrodes than by the use of indicators. They give rapid and accurate results. Common electrical methods for pH determination depend upon the use of hydrogen or glass electrode. Solution Tension and Electrode Potentials

According to Nerst theory, all metallic elements and hydrogen have a tendency to pass into solution in the forms of positive ions. When a strip of silver, copper or silver is placed in water or aqueous solutions, an electrical double layer is thus set up at its surface. After a time the rate at which Zn++ ions from the solution combine with electrons on the surface is equal to the rate at which Zn atoms, lose electrons to form Zn++ ions. At equilibrium, the strip has a negative charge and the concentration of zinc ions in solution has a fixed value. The magnitude of the electrical attraction between the two, determines the potential of the metal in the solution and the tendency for zinc to zinc ions in solution is known as solution tension of the metal. Potential developed depends upto the metal and also upon the concentration of metal ions in solution. Electrode potential magnitude is in the order of their arrangement in the electrochemical series, e.g. Na has the high electrode potential as compared to Zn or Cu. Hydrogen Electrode

Its principle is similar to that of Zn electrode. It consists of a small Pt (Platinum) strip coated with patinum black absorb H2 gas. A Pt wire welded to the electrode makes contact with the outer circuit, the Pt strips is surrounding by glass tube with inlets and outlets for H2, which is admitted at 1 atmosphere. Presume while the electrode is in a dilute solution of an acid. H2 (On Pt surface)

_______→

2H (Remain on Pt)

_______→

2e– + 2H+ (Pass into solution from electrode)

Since H2 is admitted at constant pressure, the solution tension of hydrogen atoms has a constant value. Thus the reaction to the right is fixed at a constant rate and the equilibrium will shift accordingly to the H+ ions concentration of the surrounding medium so that it can be used to measure pH. If electrode 1 is maintained constant by immersion 1 N H+ solutions is called normal hydrogen electrode and is arbitrarily assigned a potential, Eno, of zero under all conditions and is used as a standard reference for other electrodes. emf = En + Eno

(104)

Demonstrations 105

If potential difference between the normal H2 electrode and the electrode in the unknown solution is known for a given temperature, pH of the solution can be calculated from the formula. emf pH = _____________________ , where T is absolute temperature. 0.00019837 T

Calomel electrode

It consists of metallic Hg in contact with Hg2Cl2 in KCl solution (0.1 M, 1K saturated KCl). Potential varies with the saturation of KCl solution but for a given temperature and KCl solution but for a given temperature and KCl concentration, the potential of the calomel electrode against the normal hydrogen electrode is constant, e.g. at 25°C potential of saturated calomel electrode is +0.2458 volts. Now if a hydrogen electrode is placed in the solution of unknown pH and connected with a saturated calomel electrode, the potential difference registered will be more than would have been obtained against a normal hydrogen electrode by 0.2458 volts. The pH of unknown solution can therefore, be calculated as: emf – En calomel pH = ____________________________ 0.00019837 T

Where En is amount by which voltage of calomel electrode exceeds that of an hydrogen electrode. The hydrogen electrode method is applicable over the whole pH scale and is very accurate. But it is poisoned and rendered inaccurate by a variety of substances, e.g. O2, NH3, H2S. So it cannot be used in oxidising solution. Glass Electrode

The method determining pH is rapidly replacing the H2 electrode procedure. It is not effected by oxidising or reducing agents. It is based on the principle that when a glass membrane separates two different solution differing in pH, a potential difference is found to exist between the two surfaces of the glass. It consists of a thin walled glass bulb made out of a special type of low melting point. It is filled with normal HCl solution in contact with Ag/AgCl electrodes. The platinum wire dipping in the electrolyte passes out of the glass tube at the bulb is placed in the solution of which pH is to be measured. The potential is measured against a standard calomel electrodes. pH Meter

A glass electrode is made up of a bulb containing solution of known pH into which is dipping a Ag/AgCl electrode. The bulb is fragile as it is made up of a thin layer of glass. The glass electrode and calomel electrode both dip in a solution of unknown pH. The electrodes are connected by potentiometer. Switch on the instruments and let it stabilise for 10 minutes. Take double distilled water and read the pH. It should be zero or else adjust to zero. Then adjust the pH of the standard buffers of pH 4.1 and 6.55. After dipping the electrodes when they are taken out, they should


be washed with distilled water, and then wiped clean. Then put the electrodes in the solution where pH is to be found out. Take the reading as the instrument is directly calibrated in pH units. No connector has to be applied. ESTIMATION OF NITROGEN CONTENT BY MICROKJELDAHL METHOD Principle

Any nitrogen containing substance on digestion with concentrated sulphuric acid is converted into ammonium sulphate. From the ammonium sulphate so formed, the ammonia is distilled off by treating with strong alkali. The evolved ammonia is trapped in a suitable indicator, which on titration with standard acid gives the amount of ammonia trapped. Thus by back calculation the nitrogen content is found out. Reaction

Boric acid Tashiro’s indicator __________________________

Boric acid: NH3 complex ← (Green colour)

(Violet colour)

_______________→

Digestion Nitrogen containing substance ______________________________→ (NH4)2SO4 50% H2SO4 Distillation 40% NaOH

NH3

N/70 H2SO4 Titrate ↓

Violet colour Procedure

The whole procedure for nitrogen estimation is divided into three parts: a. Digestion b. Distillation c. Estimation a. Digestion

A known weight or the volume of the organic compound is taken in a small microKjeldahl flask, followed by 2 ml of concentrated sulphuric acid. To this a pinch of CuSO4 (which acts as a catalyst) and K2SO4 (it raises the boiling point and prevents bumping) are added. The mixture now looks black. The flask is placed on a microburners and is heated slowly over a small flame. In the first stage, a low flame is used and later on a strong flame is used till the material is completely digested and aquired a slight blue tinge. The time of digestion depends upon the nature of nitrogen in the unknown substance (i.e. the complexity of the nitrogen). If the nitrogen present is in the easily extractable form, the

Demonstrations 107

period of digestion should be provided to convert all the nitrogen present into ammonium sulphate. Under similar conditions run a blank which contains all the reagents except test material. b. Distillation

The contents of the digested material are transferred quantitatively into a distillation jacket through a funnel. The digestion flask is washed with distilled water and the contents poured into the distillation jacket followed by 10 ml of 40% NaOH. A steady stream of steam is bubbled into the distillation jacket. The ammonia evolved from ammonium sulphate is trapped into a boric acid-Tashiaro’s indicator. The conical flask containing 5 ml of boric acid-Tashiaro’s indicator is placed at the end of the water condenser in such a way that the tip of condenser is dipped in the indicator. The bubbling of steam in the distillation jacket is continued till the volume of the solution in the conical flask becomes approximate double of the initial volume (i.e. 5 ml the volume of indicator).

 Clamp

12345 12345 12345

12345 12345 12345

12345 12345 12345 Trap

Steam generation

Vacuum

 Water jacket

Burner

→ Outlet

→ Inlet

Distillation jacket Boric acid indicator

The boric acid-Tashiaro’s indicator contains boric acid, methyl red and methylene blue. In acidic pH, it is violet in colour. In this the main indicator is methyl red. In alkaline pH, it is


yellowish orange in colour and the end point is not sharp so methylene is also added to the (boric acid + methyl red) solution. Violet colour is produced. In alkaline solution the yellowish orange colour formed by methyl red combine with methylene blue and a green colour results and the end point is sharp. c. Estimation

The trapped ammonia is estimated by the following ways: i. Colorimetry (using Nesslar’s reagent). ii. Titrimetry (by titrating against standard sulphuric acid). In the microKjedahl method, the trapped ammonia is estimated by titrating against N/70 H2SO4. The colour of the indicator becomes green after the ammonia has been absorbed into it. This is titrated against N/70 H2SO4 until the solution becomes violet again (i.e. acidic pH). Note the amount of N/70 consumed. Reaction NH3 + Boric acid _________→ NH3 : Boric acid complex green colour

Indicator Alkaline Violet

Green Acidic

Calculations

Proteins contain 16% nitrogen. i.e. 16 gm of nitrogen is present in 100 gm of protein i.e. 16 mg of nitrogen is present in 100 mg of protein 1 mg of nitrogen is present in 6.25 mg of protein. 1 litre of 1 N-H2SO4 ≡ 1 litre of 1N-NH3 1 litre of 1 N-H2SO4 ≡ 14 gm of nitrogen 1 ml of 1 N-H2SO4 ≡ 14 mg of nitrogen 1 ml of N/79 H2SO4 ≡ 0.2 mg of nitrogen Let x be the difference between the titre value for test solution and blank. If 0.2 ml of the serum is digested initially then 0.2 ml of serum produces 0.2× x mg of nitrogen. 0.2 × x 1 ml of serum produces ________________ mg of nitrogen. 0.2 0.2× x ∴ 100 ml of serum produces ______________ × 100 mg of nitrogen. 0.2 The results can be converted into proteins by multiplying by factor 6.25.

Demonstrations 109 Uses

1. 2. 3. 4.

Only process by which nitrogen in a mixture can be estimated. Percentage purity of nitrogen containing substance can be determined. Useful for the estimation of nitrogen in fertiliser, drugs and foodstuffs. Nitrogen balance of patients can be calculated.

FLAME PHOTOMETRY

Na+ and K+ are present in all the cells in the body. They are also present in blood. The consistency of pH has to maintained within very narrow limits in our body. Even the slightest change can lead to fatal conditions. This is done by electrolytes. The important anions are chloride and bicarbonates and cations are sodium and potassium. Normal blood level of Na+ is 130-142 mEq/L of plasma. Normal blood level of K+ is 3-4.5 mEq/L of plasma. Since even slightest changes in these levels can causes serious effects, it is very important to estimate the level of Na+ and K+ in blood in case of diseased persons. This is done accurately by flame photometry. Principle

A stream of the sample passes over the flame as it burns, a colour of particular wavelength is formed which after passing through a set of filters, falls on a selenium cell and is changed into electricity. It is measured by potentiometer. The value is directly proportional to the concentration of salt in the sample. Different models of flame photometers are available. The instrument work on 220 volts AC. Various parts of apparatus. 1. Suction needle

It is a hollow syringe and it dips in the beaker containing the solution whose Na+ and K+ contents has to be determined. 2. Atomiser

The needle is attached to atomisor, which breaks up the particles of the solution into very fine particles. Atomiser opens into a spray chamber. 3. Spray Chamber

It is a big chamber which has a rubber tube which serves the function of drainage. There are two other openings, one for gas supply and the other for compressed air supply. The burner has 10-12 holes and enclosed in is a chimney. 4. Compressor

It is connected to spray chamber. It compresses as well as dehumidifies a pressure of 10-20 lbs/sq. inch. The mixture of gas and dry air burns in a nonluminous flame. The burner is arranged in two rows to make the flame strong and stable the light emitted falls on a reflector.


From where it is reflected on a convex lens, a gelation filter and finally on a selenium photo cell, with an oscillating of galvanometer. Procedure

Seven different standard solutions are prepared as shown below: Stock sodium standard (200 mEq/litre) Prepared by dissolving 11.69 gm of pure dry sodium chloride in one litre of double distilled water. Stock potassium standard (10 mEq/litre) Prepared by dissolving 746 mg of pure dry potassium chloride in one litre of double distilled water. Working standards Combined working standards which are used for the estimation of both sodium and potassium are prepared. Into 100 ml volumetric flasks are measured the volumes of stock sodium and potassium standard which are given below. The volume is made upto 100 ml with double distilled water and preserved in polythene bottle. Stock sodium (ml) 0.55 0.60 0.65 0.70 0.75 0.80 0.85

Stock patassium (ml)

Sodium content (mEq/L)

0.20 0.30 0.40 0.50 0.60 0.70 0.80

1.1 1.2 1.3 1.4 1.5 1.6 1.7

Patassium content (mEq/L) 0.02 0.03 0.04 0.05 0.06 0.07 0.08

The compressed air is started and its pressure is adjusted. Burner is lighted and the flame is made nonluminous. First the beaker (sampling up) is filled in the distilled water. It is sucked in and burn. The reading in the galvanometer is adjusted to zero with the help of a knob. Now the beaker with maximum strength of standard is placed under the suction needle. The sensitivity knob is adjusted so that the deflection of galvanometer (maximum) is within the scale. Scale is graduated from 0-100. Now the different solutions are placed one by one under the suction needle and the defections of galvanometer noted. The same procedure is repeated with the test sample. If this gives the same defection as one of the standards, then its Na+ and K+ concentration is that of the standard. If Na+ and K+ content of the serum is to be estimated, the serum is diluted 1: 100 with double distilled water. Plasma, urine, gastric juice and sweat can also be analysed this way. The filter used for Na+ is orange (776 wavelenegth) and for K+ is red (589 wavelength). Pathological Variations of Na+ and K+

Serum Na+ and K+ level decreases in vomiting, diarrhoea, dehydration, diabetic coma, uremia, toxemia of pregnancy serum Na+ decreases and K+ level increases as in Addison’s disease. Serum K+ increases in acute nephritis. ELECTROPHORESIS OF PLASMA PROTEINS

Electrophoresis is defined as the migration of charged particles in the solution under the influence of electric field.

Demonstrations 111

This rate of migration is directly proportional to the number of charges present on a component. Proteins are colloidal particles and have charge either positive or negative which depend on the pH of the solution. In acidic medium, it acts as cation and in alkaline medium as anion. If uncharged particles are changed to charged ones, then they can be separated. If a potential difference is applied across them, current will flow and cations move towards cathode and anions towards anode. Proteins are amphoteric in nature and have a particular pH (isoelectric pH) of which net charge is zero and no migration takes place. At this stage, the ion is called zwitterion. H | R — C — COOH | +NH3

______ → _______

At pH = 1 Net charge= + 1

H | R — C — COO | +NH3 Isoelectric pH Charge = zero

_____→ ←_____

H | R — C — COO | NH2 At pH = 8 Charge = – 1

Separation is best at pH = 8.6. Since all plasma proteins have different isolectric pH, their rate of migration is different. Migration depends upon: 1. pH of buffer 2. Net charge of amphoteric particle 3. Temperature 4. Voltage and current. This distance α time taken α potential gradient. ∴x α T α Σ x or u = _______ sq. m/volts/sec. T.Σ Where u = Electrophoretic mobility (a constant) Potential gradient depends upon, 1. 1 Current E α ______ 1 2. r = Conductance of the solution 3. q = Cross-sectional area. r.q So E = _______ 1 x ∴ μ = ________ sq.cm/volts/sec. Tr q 1

112 Practical Biochemistry for Students Apparatus

Two main types of apparatus are used for separation of proteins. They are: i. Free or moving boundry electrophoresis ii. Zone electrophoresis. 1. Free or moving bounding electrophoresis: In 1933, this apparatus was first introduced, by Tiselius. It consists of rectangular or U-shaped cell. Protein in buffer (pH = 8.6) is taken in it. Proteins has two components. • Albumin 55% • Globulin 45% Globulins are further divided into four fractions: α1, α2, β and γ. Two electrodes are fitted in the two limbs of cell and joined to the battery when electric current flows through the U tubes, the protein being negatively charged migrate towards anode and their rate of migration is: Albumin > α1 globulin > β globulin > α2 globulin > r globulin

When current has flowed through for 16 hours, front edge of albumin have travelled considerably ahead of other proteins in above order. The light rays are passed through these proteins. By appropriate optical apparatus, the degree of refraction can be photographed in the form of so-called electrophoretic pattern. The proportion of each tube of protein is proportional to the area under its refractive peak in the pattern. It is called moving or free boundry because proteins move freely in solution. 2. Zone electrophoresis It is more important. Proteins are supported on various types of inertporous media. Media

1. 2. 3. 4. 5.

Paper Agar-gel Starch Cellulose acetate Silica gel. Generally paper electrophoresis is used because it is easier to operate and is cheaper.

Theory

Albumin is tightly negatively changed. Amount of charge depends upon pH of buffer in which it is dissolved. Apparatus

Apparatus consists of two compartments filled with buffer solutions. Each is divided into a lateral and a medical compartments connected holes plugged with cotton. In the two medical compartments filter paper is dipped and it rests on a purpose bridge. Whole apparatus is covered by perspex cover. In the actual experiment, we use constant voltage then the potential difference is usually 3-4 volts/cm of length of paper and in constant current is 1-15 milliampere per single strip (2.5 cm).

Demonstrations 113

For this purpose we use filter paper strip of 30 cm long and 2.5 cm wide of Whatman paper no 1. Wet the paper in the buffer. Soak it in buffer and then fix on perspex bridge. Buffer used is diethyl barbituric acid and sodium diethyl barbitol. Now cover the compartment and switch current at voltage 110 for 5 minutes to saturate paper with current. After 5 minutes, take out cover and put a drop of serum sample on the paper. Cover it again and switch on the current for 16-18 hours. Afterwards take it out, remove strip and next step is fixation of proteins on this paper. Fixation

Fixation is done by two methods. i. Thermal method. ii. Chemical method. i. Thermal method

Dry strip in oven at 120-150°C for half an hour. These proteins are denatured and get fixed on supporting method. ii. Chemical method

By dipping strip in mixture of ethanol, ether and methanol, proteins are fixed on paper. Location

To locate different proteins in paper electrophoresis or to make them visible, staining is done by bromophenol blue dye. Proteins gets boundry to the dye. Binding capacity of dye is proportional to the normal of protein present. To remove unbound dye: Strip is rinsed with 0.5% acetic acid. Concentration of different fractions or quantitation is done by i. Densitometer ii. Elution. 1. Densitometer: Stand strip in denstiometer and take optical density. By this we can calculate concentration of different patterns. 2. Elution: Here 0.05 N.NaOH is used. Cut different strips of paper and put them in solution. Different colours of different fractions are produced. One can be used as blank. Take different optical density of different fractions. Wavelength is 520 nm. Uses: Changes in electrophoretic pattern of serum proteins are indicated in many chemical condidtion. 1. It is useful in finding out Albumin : Globulin ratio. 2. In diagnosis of certain diseases. Macroglobulinemia: Large quantity of r-globulin. Agamma globulinemia : No r-globulin Cirrhosis of liver : Increase in r-globulin Nephrosis : Increase in α2fraction Multiple myeloma : Increase in r-fraction.

114 Practical Biochemistry for Students CHROMATOGRAPHY

For separation of constituents of a mixture one of the properties of the constituents of the mixture is made use of. In biological mixture separation difficulties are i. Mixture is available in very small quantity. ii. Mixture is usually made up of proteins, hence cannot be subjected to high temperature, high or low pH because they get denatured. iii. Substances have melting and boiling points very close to each other. Their solubilities are also very closely related. To avoid these difficulties, certain methods have been devised. a. Electrophoresis. b. Chromatography. Principle

In any chromatography, there is one stationary phase and one moving phase. Stationary phase is fixed and stationary and the moving phase flows past it. The mixture to be separated is put in the stationary phase and the moving phase carries along with it the different constituents of the mixture. As it does so, certain resistance is offered by stationary phase. Resistance offered is different by different substances. It is therefore the resistance offered by the different constituents of the stationary phase which helps in the separation of the constituents by chromatography. Different types of chromatography are used for separation of different types of substances. 1. Column Chromatography

Stationary phase which can be in solid or liquid state is in the form of a column. If stationary phase is solid like silicon, it can be packed in the phase, then the mixture to be separated is placed on top and on appropriate moving phase which may be a liquid is allowed to run through it. Separation is effected by the resistance offered by stationary phase and different constituents of mixture reach the bottom of tube after different intervals of time. The moving phase is allowed to flow down the tube very very slowly, so that 10-15 drops fall down the column per minute. As the rate is slow, separation is more effective. The moving phase which comes down can be collected in a series of test tube (i test tube for about 5 minutes). Substance which moves fastest reaches the test tube first and one which moves slowest reaches last the process is called elution. If we do not want to separate the mixture, we can allow the moving phase to run till the end of the column and close the tap. This can then be detected by fluoroscopy. It is used for a variety of substances especially whenever particular matter is involved, e.g. antigens. 2. Gas Chromatography

It is usually used for separation of fatty acids. In this case, gas is the moving phase usually an inert gas is used, e.g. argon or sometimes nitrogen gas. The gas should be such that it should not react with fatty acids.

Demonstrations 115 3. Partition Chromatography

Here both the stationary and moving phase are liquids, e.g. if there are two substances, A and B and two solvents C and D and supposing C dissolves 90% of A and 10% of B and D dissolves 90% of B and 10% of A. The two liquids A and B (mixture) should be immiscible. A and B dissolved in C and D are mixed together. It is allowed to stand in funnel with a stop clock, i.e. separating funnel. Two liquids stand at different levels. Each can be removed separately. D contains 90% of B of 10% of A. Then C is added to the funnel. If separated again 9% of it comes to B of 10% of A. So B is 99% pure. The process can be repeated many times. Every time, the purity increases but separation is never 100%. The method depends upon the difference in solubility of solvents. 4. Paper Chromatography

Solvent is employed to flow past the substance so that it takes the different constituents through different distances. Solvent consists of two liquids. As the liquids move up the paper, separation is effected by virtue of different solubilities of two solvents and rate of migration of two solvents is different this is called ascending chromatography, since the solvent moves up against the force of gravity. Descending Chromatography

The paper hangs from the moving phase and the tip of the paper dips into the solvent at the top. By capillary action solvent goes to highest point of paper and then desends due to gravity. Spot is present at the descending past. As the solvent descends down, separation is effected. Two-dimensional Chromatography

The mixture to be analysed is placed below on the paper and solvent is allowed to run upwards. The paper is dried. Then a different solvent is allowed to run on a direction at right angles to the previous one. More spots are obtained in this direction, the constituents which could not be separated by one dimensional chromatography are separated now. To study isotopes and their applications in medicine Isotopes

Isotopes are atomic species having similar atomic number but different in the nucleus of atom. As the atomic number is same, isotopes have the same chemical properties but different physical properties, e.g. Isotopes are of two typesA. Stable or non-radioactive isotopes. Stable isotopes—these do not emit any radioactive radiations, B. Unstable or radioactive isotopes. Unstable isotopes—these emit radioactive radiations viz, α, β or r rays, Radioactivity

It is the phenomenon where radioactive substance emit α, β or r-rays on disintegration.


α-rays—consist of doubly charged Helium atom. They have least penetration power of all the three particles while greatest ionisation power, because they have heavy particles, they cannot penetrate much. β-rays—are fast moving electrons having the ionisation power less than α-particles and less than r-rays. r-rays—are electromagnetic waves with maximum penetration power but least ionisation power of all the three particles. Methods for Counting Radioactivity

Energetic α or β-particles can convert atoms or molecules of a gas to positive ions by displacing electrons from them. Each positive ion so formed is paired either with the free detached electron or with a negative ion formed by the union of the electron with another neutral atom or molecule α or β particles— The number of such ion pairs formed per cm of path of a particle is called specific ionisation. Gases can be ionised by r-rays detaching high velocity electrons from atoms or molecules in their path. These secondary electrons produce ion pairs. Measurement of Radioctivity

1. Geiger-Müller counter The radiations entering the gas mixture produce ions when α, β or r-rays collide with the gas atoms or molecules. The cations and anions move to cathode and anode respectively and produce electric impulse which is proportional to the activity of the radioactive substance. Drawbacks 1. It cannot differentiate between α, β or r rays. 2. All radiations are not taken into account as some of them are lost. 2. Proportional counter: It is same as that of Geiger-Müller counter except the gas is not a mixture but a single mono-atomic gas. It can differentiate between α, β, and r-rays. However, it is less sensitive. 3. Scintilation counter: In this method no gas or mixture of gas is used. The radiations are allowed to fall over fluorescent substance, e.g. crystals or on a liquid organic solvent (liquid scintillation counter). The photons emitted are allowed to pass through a photo multiplier tube that converts this light into electricity and amplified using 10 to 15 diodes, whose strength is proportional to the radio-activity of the substance. For β-rays zns and/ or r-rays zn are used. Units of Radioactivity

Radiation absorption dose: Since the radioactive rays can produce ions inside the tissues also, long exposure to these rays can be dangerous. However limited amount of radiations will not produce any hazardous effect. 1 rad = 100 ergs of energy/gm of tissue. 1 roentgen = 1 rad (practically). Maximum permissible dose = 300 mill r/weak, r = roentgen as rad is difficult to measure, but roentgen is a unit with electromagnetic proceedings, hence can be measured easily.

Demonstrations 117

Roentgen is the amount of r-rays which must be absorbed by Ig of air to cause it to gain 83.8 ergs of energy. Application of Isotopes

A. In biochemistry Isotopes are used in working out metabolic pathways, e.g. cholesterol synthesis from acetic acid, purine synthesis from glycine (4th, 5th and 7th carbon of purine skeleton is derived from glycine). Some commonly used stable isotopes are used in the determination of turnover of different metabolic in the body. B. In physiology Determination of total plasma volume and total blood volume in body. Labelled albumin I131 is administered after the activity of the labelled element has been studied outside the body. When the albumin has been distributed evenly, the activity of the element is studied in a measured volume taken. For measuring the volume of blood or emiscible pool or any other metabolic, the following formula is used: V=v

( coc

_______

)

=1

V = Volume of blood V = Volume of labelled substance administered co = Specific activity in v c = Specific activity in V. Determination of average life of RBC Glycine labelled is used which gets incorporated in the RBC. By determining the rate of decrease of activity over a certain period of time, average life can be calculated. In iodine metabolism We can study iodine metabolism in the body. Iodine is essential for the synthesis of thyroxine. Its distribution can be studied by the administration of radioactive iodine I131. About 1/3 is taken up by thyroid and 2/3 is excreted by the kidneys. In hyperthyroidism, there is increased uptake of I131 while in hypothyroidism there is decreased uptake. Hazards of Radioactivity

Radioactivity is responsible for genetic mutations. When an individual receives a massive dose 500 rad then that individual is susceptible to damage of germinal epithelium and damage to gonads. These result in a permanent genital defect. Hemopoietic system is affected and leukemia is produced. Precautions

1. Body must be covered by lead sheets (particularly gonads and chest) while handling rrays. 2. Gloves must be used. 3. Must get their blood count done every three weeks.


4. 5. 6. 7. 8.

Must carry plates which record radiations received by the body. Isotopes must not be disposed off in running water supply. Eating and drinking should be avoided. Radioactive solutions must not be sucked by mouth. Working should be done in ventilated hood if gas is to be used.

CENTRIFUGE

Centrifuge are used to hasten the deposition of substances suspended in liquids. The suspended matter is deposited in order of weight, the heaviest element being the first to settle. There are many types of centrifuge but the basic principle is the same, that is, the use of ‘centrifugal force’. When this exceeds the force of gravity, heavier elements are thrown to the bottom of the centrifuge tube. For the purpose of comparison, centrifugal force is referred to as RCF, that is relative centrifugal force. The RCF is taken as a guide to the separating capacity of a centrifuge. The RCF value for any centrifuge may be calculated from the following equation: RCF = 118 × r × (rpm)2 × 10-7 (m‘g’)

Where ‘r’ is the radius in cm from the centre of the centrifuge shaft to the external tip of centrifuge tube. ‘rpm’ is the number of revolution per minute of the centrifuge rotor. Small models are designed at speed of 3000-5000 rpm for general clinical chemistry. Constructions

Centrifuge consists of a central shaft or spindle which rotates at high speed. A rotor (head) fixed to the shaft with buckets for holding the centrifuge tubes. Tubes containing the liquid to the centrifugal which are fixed to the rotor. The earlier centrifuge were operated by hands but modern laboratories are equipped with electrically driven models. Centrifuges are used with Z types of rotor—‘Spring out rotor’ and ‘angle rotor’ based on the principles of horizontal or angle sedimentation. The spring-out rotor is safer than angle rotor. Operation

Select two centrifuge tubes of identical size. Place liquid to the centrifugal, water in other to within 2 cm of the tap. Place the tubes in paired centrifuge buckets in exact opposite direction. Start the motor and gradually increase the speed until the required number of revolutions per minute is reached. When the tubes have been centrifugated sufficiently, switch off the motor and allow the centrifuge to stop. Precautions

1. Ensure rubber cushions are present in all metal cups (sockets). 2. Check the balancing carefully; improperly balanced tubes will cause head wobble—a wear out the bearings. 3. Click that the balanced tubes are really opposite to one another. 4. Never start or stop the machine with a jerk.

Demonstrations 119

5. Length of centrifuge tube should not touch surface. 6. Observe the manufacturer’s instructions about the speed limit for various loads. pH METER

This is an instrument by which the pH of a solution can be adjusted. This is an electrical method depending upon the use of hydrogen electrode or glass electrode. The glass electrode method is most commonly used. The principle of the method is that a small potential is set up between the glass and the solution which is directly proportional to the hydrogen ion concentration of the solution. This very small potential difference is indicated by a sensitive measuring devices. This method is particularly useful when great accuracy is required. Glass electrode pH meter covers the pH range 0-14 units. It is ideal for use with biological fluids. The combination electrode system consists of a reference electrode and a measuring electrode. How to Operate

Keep function switch at pH positions. Set the temperature component control to read the solution temperature combination electrode system. Correct it to the connectors marked glass. For a separate pair of glass and reference electrodes, connect the electrodes in their respective sockets. Connect the instrument to the main supply. Put on the main switch, clean the electrodes with distilled water and dip them in buffer solution (prepared from the supplied buffer standard pH value of which is known). Being the READ switch at READ position. Adjust the zero and STD control till the unit reads the correct pH value of supplied buffer solution. Bring the read switch back in position. Take out the electrode from the buffer solution. Wash with distilled water (solution) and dip them in the solution of unknown pH. Read the pH value displayed. Precautions

Temperature of the supplied buffer solution must be very close to that of the test solution (at least within 10°C). In standardization a pH meter, it is advisable to use the supplied buffer solution. The pH value of which approx most closely to the test solution. A check with a second buffer solution is suggested to ensure proper functioning of the equipment. Electrode should be revised with distilled water and wiped with tissue paper gently before immersing in a solution. Measurements must be made in an insulated container such as glass beaker. ELISA

ELISA is short form of enzyme-linked immunosorbent assay. Enzyme can act as labels because their catalytic property also the detection of extremely small quantity of labelled immune reactant. Principle

An enzyme conjugated to an antibody reacts with colourless substrate to generate coloured reactant. A standard curve is obtained plotting absorbance on Y axis and corresponding concentration on X-axis. Generally used enzymes are alk.phosphatase, horse reddish peroxidase.


Glucose-6-phosphate dehydrogenase choice of enzyme depends upon 1. Specificity 2. Availability 3. Satisfactory Nowadays. ELISA is replacing RIA (Radio Immuno Assay) because it is cheaper, longer shelf life, safer and lacks radioactive hazards of RIA and can be used in small labs. Types of ELISA

There are three types of ELISA, which are named below: 1. Indirect 2. Direct sandwich type 3. Competitive. 1. Indirect ELISA It quantitates antibody. Sample containing primary antibody added to antigen coated microtitre wells, which reacts with antigen found to solid phase. Free antibody washed away. Presence of antibody bound to antigen is detected by adding an enzyme conjugated second antibody which ends to primary antibody. Washing is done to remove free antibody. Then substrate is added to develop colour which is measured spectrophotometrically, e.g. detection of HIV antibodies. 2. Sandwich ELISA (Double antibody type) It quantitates antigen specific antibody and is adsorbed on solid phase sample with antigen added. Incubate and washing is done. Second enzyme linked antibody is directed against a different epidope added. Complex is formed. Then enzyme substrate is added and is incubated. The colour developed is measured under specific wavelength. Uses are: 1. All hormones can be detected and assayed 2. Detection of viruses, pregnancy, hepatitis B 3. Competitive ELISA It is inhibition type assay for quantitating antigens. Competetive binding of enzyme labelled antigen and unlabelled antigens to a high affinity antibody. The labelled antigen mixed with antibody at a concentration that just saturates the binding sites of antibody molecule. Then increasing amount of unlabelled antizen of unknown concentration is added. With increasing concentration of unlabelled antigen more and more labelled antigens will be displaced. By measuring the amount of labelled antigen we can determine that of labelled. Concentration of antigens is inversely proportional to colour produced, e.g. HBs antigen, Australian Antigen.

Appendix 121

Appendix QUESTIONS ANALYSIS 1. Identify the glassware laboratory item. 2. Mention one use of it in the biochemistry laboratory. 3. Calculate the Rf value of the amino acid marked ‘A’. 4. Mention one clinical use of this type of chromatography. 5. Which amino acid could the spot marked ‘B’ be? 6. State the principle of heat coagulation test. 7. Why do peptones not answer this test? 8. Mention one common clinical application of this test. 9. Explain the principle of ninhydrin test. 10. Why do amino acids generally not answer biuret test? 11. Mention one clinical application of ninhydrin test. 12. Does gelatin answer Heller’s test? Explain. 13. Name the protein from which gelatin is derived. 14. Why does a solution of gelatin not absorb light significantly at a wavelength of 280 nm? 15. Identify the test (performed for protein). 16. State the principle of this test. 17. Mention one clinical significance of this test. 18. Identify the test (performed for abnormal constituents of urine). 19. State the principle of this test. 20. Mention one clinical significance of this test. 21. Identify the test (performed for abnormal constituents of urine). 22. State the principle of this test. 23. List two non-carbohydrate sources which can answer this test. 24. State the principle of Barfoed’s test. 25. Name one disaccharide which does not answer this test. 26. Mention one clinical application of this test. 27. Explain the basis of starch iodine reaction. 28. List the intermediate products of hydrolysis of starch by amylase in order. 29. Name the co-factor activator of salivary amylase. 30. Identify the test. 31. State the principle of this test. 32. List three different types of crystals formed. (121)

122 Practical Biochemistry for Students A known case of homogentisic oxidase deficiency gave greenish—brown colour with urinary Benedict’s test: 33. What is the likely diagnosis? 34. What is the principle of this test? 35. List two other compounds which answer this test. A patient presented with colicky pain was diagnosed a case of ureteric stone. Biochemical investigations revealed massive aminoacidurias: 36. What is the defect/likely cause of this massive aminoaciduria? 37. Name the different amino acids passing in this aminoaciduria. 38. What will you advise to the patient to prevent further stone formation? A mentally retarded child having past history of frequent seizures was brought to the paediatrician. Biochemical investigations revealed hyperphenylalaninemia: 39. The urine of this patient gave ferric chloride test positive. Name the compound (s) responsible for it. 40. What is the classical name of type I hyperphenylalaninemia. 41. Name the defective enzyme causing this condition. A 30-year-old executive on routine check-up revealed high serum cholesterol and other parameters were within normal range. Similar findings were present in his parents: 42. Name the type of hyperlipidemia in this case. 43. What is the likely cause of this condition? 44. This condition is an alarming signal for which disease in future? Abnormal haemoglobins arise by random mutation and they persist if the functional impairment is compatible with life. Sometimes they show beneficial effect, e.g. increased resistance to malaria: 45. Mention the point mutation present in persons showing increased resistance to malaria. 46. If person is heterozygous for this abnormal Hb what will be status of the disease. 47. Name two abnormal haemoglobin molecules. Denaturants disrupt the different levels of organization of proteins: 48. Name two commonly used denaturants. 49. What will be the effect on protein structure after denaturation? 50. Alteration of levels of organization occurs at what levels in prion diseases?

Appendix 123 51. Identify the type of inhibition. 52. Name one competitive and one non-competitive inhibitor. 53. List two uses of Km value. A 34-year-old African visiting India, presented to the emergency with complaint of fever with chills from past 3 days. Suspecting a case of malaria he was prescribed primaquine tablets but he developed haematuria after 4 days. Laboratory investigations revealed fall in Hb from 11 gm% to 7.4 gm% and rise in total bilirubin from 1.2 mg% to 4.3 mg%: 54. Name the enzyme deficiency causing haemolysis in this patient after primaquine treatment. 55. If the person is heterozygous for this abnormal Hb what will be status of the disease? 56. Name two abnormal Hb molecules. A 25-year-old male brought to the emergency in unconscious state with rapid and deep breathing. Biochemical investigation revealed: Blood pH : 7.1 HCO3– : 5 mEq/L Glucose : 350 mg/dL Urinary sugar and ketone bodies ++++ 57. What is the likely diagnosis? 58. Comment on the pH. What is the explanation of this pH value? 59. List three complications of prolonged diabetes mellitus. Diabetic individuals frequently have type IV hyperlipidemia. Which is characterized by elevated concentrations of very low-density lipoprotein (VLDL): 60. Name other three lipoproteins. 61. What is the likely explanation of this condition? 62. This condition is an alarming signal for which disease in future? 63. List three lipid-metabolic changes in diabetics. 64. What is the criteria to label a person diabetic by doing fasting and post-prandial blood sugar levels? 65. List three complications of prolonged diabetes mellitus.


NORMAL VALUES

Blood Blood glucose (true) Blood glucose (folin) Blood urea Blood cholesterol (total) Blood cholesterol (ester form)

60-80 mg% 80-100 mg% 20-40 mg% 150-250 mg% 2/3rd of total cholesterol Blood uric acid 2-6 mg% Blood creatinine 0.8-1.6 mg% Blood lactic acid 4-16 mg% Blood ascorbic acid 0.4-15 mg% Blood ammonia 40-70 μg% Blood nonprotein nitrogen (NPN) 8-20 mg% Blood amino acid nitrogen 3-5.5 mg% Blood triglycerides 100-250 mg% Blood free fatty acids 8-20 mg%

Serum alkaline phosphatase 3-13 King-Armstrong units (KAU) Serum acid phosphatase 0-3 King-Armstrong units (KAU) SGOT 10-40 IU/L SGPT 9-32 IU/L Serum amylase 80-180 Somogyi unit % Carbon dioxide combining power 55-65 vol % Prothrombin time 14-17 seconds Coagulation time 5-10 seconds Bleeding time 1-3 seconds CSF Glucose Protein Chlorides

45-80 mg% 15-80 mg% 700-750 mg%

Serum Serum total proteins Albumin fraction Globulin fraction A/G ratio Serum calcium (total) or Serum bilirubin Serum iron Serum iron binding capacity Serum iodine (BEI) Serum protein bound iodine Serum sodium or Serum potassium Serum chloride or

5.5-7.5 3.5-5.0 2-2.5 1.2 : 1 9-11 4.5-5.5 0.2-0.6 65-175 250-410 3-6.5 4-8 130-145 310-340 3.5-5.0 14-20 100-126 350-375

gm% gm% g% mg% mEq/L μg% mg% μg% μg% μ% mEq/L mg% mEq/L mg% mEq/L mg%

Urine Reducing sugars Urea Uric acid Creatinine Chlorides as NaCl Urobilinogen Ketone bodies Phosphates Titrable acidity

100 30 0.8 0.4-1.8 10-15 0.4 1 10 200-400

mg/day gm/day mg/day mg/day gm/day gm/day mg/day gm/day ml N/10 acid/day

Conversion of mg% to mEq/L mEq/L =

mg% ×10

_________________________

Equivalent weight

mg% × 10 millimole/litre = _____________________________ molecular weight (mmole/L)

Appendix 125 Molecular weight of some commonly used substances in medical practice. Substance Molecular weight Alcohol Bilirubin Calcium Cholesterol Creatinine Glucose Iron Phosphorus Triglyceride (Triolein) Urea Uric acid

46 585 40 387 113 180 56 31 885 60 168

Tables of prefixes denoting decimal factor Factor Prefix Symbol 10-3 10-6 10-9 10-12

milli Micro nano pico

m μ n p

List of recommended units Molar concentration

Mass concentration

mol/litre m mol/litre μmol/litre n mol/litre

gm/litre (g/l) mg/l μg/l ng/l

Conversion factors of old units to SI units Substance Albumin Bilirubin Calcium Cholesterol Creatinine Glucose Iron Phosphorus Triglyceride Urea Uric acid

Old unit

Conversion factor

New unit

g/100 ml mg/100 ml mg/100 ml mg/100 ml mg/100 ml mg/100 ml μg/100 ml mg/100 ml mg/100 ml mg/100 ml mg/100 ml

10 104/585 1/4 10/387 10/113 1/18 10/56 10/31 10/885 1/6 10/168

gm/l μmol/l mmol/l mmol/l mmol/l mmol/l μmol/l mmol/l mmol/l mmol/l mmol/l

Relation of biochemical tests with disease Blood sugar Blood urea Serum creatinine Serum uric acid Serum cholesterol Serum bilirubin Serum proteins

Diabetes mellitus Kidney disease Gout Heart disease, high blood pressure Jaundice Malnutrition, liver disease Chronic infections Rickets, liver and bone disease Heart attack Liver disease

Serum alkaline phosphatase SGOT SGPT Serum Electrolytes Sodium Potassium Chloride Calcium Phosphorus

Normal values 130-145 mEq/L) 2.5-5 mEq/L) 100-106 mEq/L) 9-11 mg% 3-4.5 mg%

Dehydration or loss of fluids Acid-base balance Vitamin D deficiency Recurrent renal stones

Index A A/G ratio 93, 94 Achlorhydria 38 Achromic point 20 Achylia gastrica 38 Adsorption 7 Albumin 26 Alkaline phosphatase 101 Anticoagulants ethylene diamine tetraacetate 51 heparin 52 oxalate 51 sodium citrate 52 sodium fluoride-potassium oxalate 52 Anuria 41 Arterial blood 53 Ascorbic acid in urine 66 vitamin C saturation test 66

B Beer’s law 70 Bence Jones proteins 45 Bilirubin conjugated or direct 95 reactions 95 reagents 96 unconjugated or indirect 95 Blood collection 52, 53 Blood specimen 51 Blood sugar estimation enzymatic method 74 reduction method 74 Blood urea 80 postrenal 82 prerenal 82 renal 82

C Calcium 87 Capillary puncture 53 Caraway method 68 Centrifugal force 118

Centrifuge 118 Chlorides in urine 60 Cholesterol standard 85 Chromatography column 114 descending 115 gas 114 paper 115 partition 115 two-dimensional 115 Clearance 83 Colorimetry 70, 71 photoelectric choice of filter 73 function of blank 73 visual 71 Creatinine clearance 64 Creatinine in urine 63

D Diabetes GTT curve 78 Diabetes mellitus 58 Dialysis 6 Diffusion 6 Disaccharide reactions 15 Barfoed’s test 15 Benedict’s qualitative test 15 Molisch test 15 osazone formation (phenylhydrazine test) 15

E Electrode 104 calomel 105 glass 105 hydrogen 104 Electrode potentials 104 Electrophoresis of plasma proteins 110 apparatus 112 free or moving bounding 112 zone 112 fixation 113 location 113

ELISA 119 competitive 120 indirect 120 sandwich 120 Eluters 7 Enzymatic hydrolysis of starch 20 Estimation of nitrogen 106 procedure digestion 106 distillation 107 estimation 108

F Flame photometry 109 Food analysis carbohydrate test 48 fats (cholesterol) test 49 mineral test 50 test for proteins 49 Formal titration 35 Formality 3

G Gastric juice 37 Globulin 26 Glucose tolerance curve 78 flat GTT 79 lag type 79 Glycosuria 43

H Histamine test 37 Hydrolysis acid 17 enzymatic 17 Hyperchlorhydria 38 Hyperchlorhydria (gastric hyperacidity) 39

I Identification of carbohydrate 18 Inorganic phosphorus reactions 90 Iodine number 33 Isoelectric pH 111 Isoelectric point 27 Isotopes 115

128 Practical Biochemistry for Students J Jaundice 95

L Lambert’s (Bouguer) law 70 Liver function tests 100

M Microkjeldahl method 106 Migration 111 Molality 1, 2 Monosaccharide reactions 9 Barfoed’s test 12 Benedict’s qualitative test 10 Bial’s test for pentoses 14 Fehling test 11 Molisch test 9 mucic acid test of galactose 14 phenylhydrazine test (osazone formation test) 13 Seliwanoff’s test 12

N Needle 52 Normal values 124 blood 124 CSF 124 serum 124 serum electrolytes 125 urine 124 Normality 2

O Oliguria 41 Optical density 71 Osmosis 6 Osmotic pressure 6

P Pentosuria 59 Percent 1 pH electrometric determinanation 104 pH determination selection of indicator 4 Gillespie’s drop method 5 indicator paper 5 universal indicator 5 pH meter 105, 119

Plasma 53 Polysaccharide reactions 16 Polyuria 41 Protein-free filterate 54 Proteins reactions Biuret test 22 Ninhydrin test 23 precepitation reactions 25 alkaloidal reagents 25 by heating 26 heavy metals 25 study the R-groups of proteins 23 cystine or cysteine test 24 Hopkins-Cole test 24 Millon’s test 24 Sakaguchi test 24 Xanthoproteic test 23 Prothrombin time 98

Q Questions analysis 121 abnormal haemoglobins 122 colicky pain 122 denaturants 122 fever with chills 123 high serum cholesterol 122 homogentisic oxidase 122 seizures 122 type IV hyperlipidemia 123 unconscious state with rapid and deep breathing 123

R Radiation absorption dose 116 Radioactivity 115 application 117 hazards 117 measurement 116 units 116 Reducing sugars 43 Reducing sugars in urine 56 fructose 58 galactose 59 glucose 58 lactose 58 pentoses 59 Renal glucosuria 58

S Saponification number 32 Serum 53 Serum aminotransferases (transaminases) 100 Serum calcium 87 calculation 88 collection of sample 87 Serum proteins 92 Serum urate 68 Serum uric acid 68 Solution tension 104 Sucrose test 16 Surface tension 7 Syringes 52

T Test for proteins 45 Tests for lipids 29 acrolein test 30 emulsification 29 saponification 30 solubility 29 test for cholesterol 31 unsaturation 31 Tourniquet 52 Transmission 71

U Urea 80 Urea clearance 83 Uric acid 68 Urine analysis 40, 55 colour 41 pH 42 specific gravity 42 total solids 43 bile pigments 46 bile salts 47 blood 47 ketone bodies 46 proteins 44 urobilinogen 47 reducing sugars usually glucose 43 volume 41 Urine specimens collection 54 preservatives 54 hibitane 55 hydrochloric acid 54 toluene 55

V Vein puncture 52

Index 129

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