VK Malhotra - Practical Biochemistry for Students, 4th Edition_2

PRACTICALBIOCHEMISTRY

FORSTUDENTS

Varun Kumar MalhotraPhD/Gold Medalist

Department of BiochemistryMaulana Azad Medical College

New Delhi

PRACTICALBIOCHEMISTRY

FORSTUDENTS

4th Edition

JAYPEE BROTHERSMEDICAL PUBLISHERS (P) LTD

New Delhi

Published by

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Practical Biochemistry for Students

© 2003, Varun Kumar Malhotra

All rights reserved. No part of this publication should be reproduced, stored in a retrieval system, ortransmitted 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. Everyeffort is made to ensure accuracy of material, but the publisher, printer and author will not be heldresponsible for any inadvertent error(s). In case of any dispute, all legal matters to be settled under Delhijurisdiction only.

First Edition: 1984Second Edition: 1986Third Edition: 1989Fourth Edition: 2003

Publishing Director: RK Yadav

ISBN 81-8061-109-4

Typeset at JPBMP typesetting unitPrinted 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 investigationaltechniques for the welfare of the patients. Hence it becomes mandatory for any writtenmaterial to show the changes duly. In this edition, my efforts have gone in a direction toimprove the material.

In a modest attempt, additions have been made about the methods of expressingconcentrations, instrumentations, collection and preparation of blood specimen, SGOT,SGPT, etc. The inquiring mind will certainly benefit from such exposures to manage theclinical situation in a more creative and challenging manner.

Some modifications became necessary in various chapters and thus the matters havebeen 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 andassistance respectively.

Varun Kumar Malhotra

Preface to the First Edition

In the ever expanding knowledge of Biochemistry it is very difficult on the part of singleindividual to go through the various bigger valume of textbooks on practice of biochemicalinvestigations. 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 upto 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 asignificant role. This book will certainly be of significant importances to the practitionersas well as laboratories.

I thank all my colleagues and friends in contributing to have brought out this book in itsnear 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 excellentway. The contributions that I have received by the constant cooperation of my parents andwife 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

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1Methods of Expressing Concentration

Concentration may be defined as weight per unit volume.The most common expressions are:

1. Percent2. Molarity3. Normality4. Molality5. Formality.

1. PercentAccording 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

solutionc. A 10% V/V solution contains 10 ml of the concentrate per 100 ml of solution.

2. MolarityA molar solution contains 1 gm mol. wt. (mole) of solute in one litre of solution. Molarity ofsolution is expressed as ‘M’

1 mole = 1000 millimoles

gm.mol.wt.(moles) of soluteMolarity = _____________________________________

Volume of solution (L)

Amount of solute (gm)Molarity = ____________________________________________

Mol. wt. × volume of solution (L)

For example:Molarity of a solution of 2 litres containing58.5 gm. NaCl dissolved.

Amt. of NaCl (gm)Moles of NaCl = __________________________

Mol. wt

= 58.523 + 35.5

= 58.5 = 158.5

2 Practical Biochemistry for Students

∴ Molarity = Moles of soluteVolume of solution (L)

= 1 = 0.5 M2

So, Molarity of given solution is 0.5 M.

3. NormalityA 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.

gm. eq. wt. of soluteNormality = __________________________________

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 ofcation or anion.Eg. equivalent weight:

a. of NaCl = Mol. wt. of NaCl

Mol. wt. of MgCl2b. of MgCl2 = __________________________

2

Mol. wt. of AICl3c. 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 = 1180

Equivalents of solute∴Normality =

Vol. of solution (L)

= 1 = 0.5 N2

So, Normality of given solution is 0.5 N.

4. MolalityMolality is defined as number of moles of solute dissolved in 1000 gm of solvent (Not insolution).

It is designated as ‘M’.

Moles of soluteMolality =

Weight of solvent (gm)

Methods of Expressing Concentration 3

Amount of solute (gm)⇒ Molality =

Mol. wt. × Weight of solvent (gm)

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

5. FormalityFormality parameter is not used nowadays because all of the above parameters are better thanit and are able to express concentration sufficently.

Formality is same as molarity if molecular weight in the formula is replaced by formulaweight.


(4)

2Physical Chemistry

pH Determination

All biochemical reactions are greatly influenced by the hydrogen ion concentration of thesurrounding medium in which the reaction takes place. The most convenient way of expressinghydrogen 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

+

= log1CH+

Hence it is both important and useful to know some of the simple methods of pH determi-nation.

pH can be determined both by colorimetric and electrometric methods. Electrometricmethod is the most accurate one and is done by using a pH meter whereas colorimetricdetermination 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 towhich they are added. Indicators are weak organic acids or bases. Their unionized forms showa colour while their ionized forms, i.e. cations or anions have different or another colour. Thecolour of the solution in presence of an indicator depends upon the relative proportions ofionized and unionized forms of the indicator which in turn depend upon the hydrogen ionconcentration. For each indicator there is a definite pH range in which it is present as a mixtureof its ionized and unionized forms. In this specific range, variations in the pH of the solutionwill bring visible change in the colour of the indicator. It is necessary that the effective pHrange 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 threedifferent tubes.

Add 2 drops of the indicator in each tubes. Mix and observe the colour in all the three testtubes. Test tube containing acetic acid will display acid colour of the indicator whereas testtube containing sodium carbonate will display alkali colour of the indicator. If the colourobtained with the unknown solution lies in between (i.e. intermediate) acid and alkali colourof the indicator and hence it is a suitable indicator for the colorimetric determination of theunknown solution. But on the other hand if the unknown solution shows either the full acid or

Physical Chemistry 5

full alkali colour then the indicator used is unsuitable. Repeat the same procedures with theother indicators till a suitable indicator is selected.

Some common indicators useful for biological pH range are:

Indicator pK pH range Acid colour Alkali colour

1. Thymol blue (acid range) 1.65 1.2-2.8 Red Yellow2. Methyl yellow (Topfer’s reagent) 2.9-4.0 Red Yellow3. Methyl orange 3.46 3.1-4.4 Red Yellow-orange4. Methyl red 5.01 4.3-6.1 Red Yellow5. Phenol red 7.81 6.7-8.3 Yellow Red6. Thymol blue (alkaline range) 9.7 8.0-9.6 Yellow Blue7. Phenolphthalein 9.7 8.2-10.0 Colourless Pink

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

1. Indicator paperIndicator paper consists of a strip of a sensitized paper and is accompanied by a colour chartwhich 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 bedetermined. Remove the excess of the fluid adhering to the indicator paper strip by means ofpressing between the folds of filter papers. Compare the colour of the pH paper with thecolour chart on the indicator paper and thus determine the pH of the solution.

2. Universal indicatorUniversal 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 welland find out the pH by matching the colour of the solution with the colour chart on the universalindicator bottle. The one with which it coincides or matches, is the pH of the unknown solution.

3. Gillespie’s drop methodIn the determination of pH by this method, the ratio of two forms of the indicator may befound out by adding a known number of drops of an appropriate indicator to the test solutionand finding out how the same number of drops has to be distributed between an acid andalkali so that the colour of the test solution matches with thatof the acid and alkali solution. When superimposed, then thepH can be calculated by using Handerson-Hasselbatchequation.

Alkaline formpH = pK + log

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 equalisethe optical conditions).

1

2 4

3


Add few drops (count the number of drops) of suitable indicators to the unknown solutionin the tube no. 3. Now distribute the same number of drops in acid solution (because any dropof indicator solution going into the solution, will give its acid form) and alkali solution (becauseany drop of indicator solution going into alkali will give its alkali form). Mix the contents ofeach tube and examine the colours by white light. If the colour viewed through tube no. 3 and4 appears to have more of the alkalis form of the indicator as compared to the colour viewedthrough tube nos 1 and 2, add more drops of the indicator to the sodium carbonate tube (alkalitube) 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 withthe colour viewed through the tube nos 3 and 4, then count the number of drops added to acidtube and alkali tube.

The manner in which matching has been done it can be argued that the pH of the solutionwhich contains ionized and unionised indicator in the ratio they are present in tube nos 1and 2.

Alkali formpH of unknown solution = pK + log _________________

Acid form

Number of drops of indicator added to alkaline solution= pK + log ___________________________________________________________________

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 membraneseparates a solution and its pure solvent or between two solutionsdiffering in concentration. Water passes through the membrane untilthe concentration on both sides becomes same. Such a movement ofsolvent molecules from a pure solvent or dilute solution, through asemipermeable membrane is called osmosis.

Osmotic pressure is the pressure that must be applied on asolution to keep it in equilibrium with the pure solvent when thetwo are separated by semipermeable membrane or osmotic pressureis the force required to oppose the osmotic flow.

Since osmotic pressure is proportional to the total number ofsolute particles in solution so the substances which ionize, will havethe higher osmotic pressure as compared to those which do notionize.

If the solution containing crystalloids and colloids is placed in a cellophane sac and this isimmersed in a jar of distilled water, the crystalloids diffuse across the membrane while thelarge colloidal molecules are retained. By repeatedly changing the distilled water outside thesac, it is possible to free the colloidal material virtually completely from salts and othercrystalloids. This process is known as dialysis.

Cover the top of the thistle funnel by tying a piece of cellophane over it. Then pour asolution made by mixing 10 ml of saturated ammonium sulphate with 90 ml of 1% starch intothe thistle funnel till the level rises to the middle of the stem. Clamp the funnel on a stand andsuspend 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 iodinetest 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 toCompare Two Eluters—Acetone and Water

Adsorption is a phenomenon in which a substance is adsorped on the surface of a substance. Ittakes place due to the presence of free valencies on the surface of the adsorbate which attractsand takes up the adsorbant.

Procedure

Take two test tubes. Add 10 ml of 0.1% methylene blue solution, add 0.5 gm of activatedcharcoal 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.

Now place these two funnels with filter paper to a separate test tubes. Wash the charcoalon 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 becolourless.

Interpretation

The solution looses its blue colour due to the adsorption of methylene blue particles on theactivated charcoal. This proves that the charcoal is a good adsorbant.

On adding water to it, water cannot elute methylene blue and hence colourless filtrate isobtained. But acetone dissolves the methylene blue particles and thus the solution regains itsoriginal 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 liquidgiving an appearance of a stretched membrane. A liquid or a fluid is made up of molecules.

Colouredsolution

Water Acetone

Charcoal withadsorbent

Stand Colourlesssolution


Molecules present in the interior of fluid are equally attracted in all directions by intermolecularforces due to surrounding molecules. But molecules on the surface are unequally attractedbecause of the absence of forces from above. This leads to certain unbalanced forces on thesurface. Due to these forces, the surface acts as a membrane. Surface tension is defined as theworkdone 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. Sprinklesulphur powder in both the test tubes.

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

Interpretation

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

Sulphuryellowcolour

WaterSulphurpowder

Soapsolution

Carbohydrates 9

(9)

3Carbohydrates

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 concentratedsulphuric acid undergo dehydration to give furfural or furfural derivate which on conden-sation with α-naphthol yields a violet or purple coloured complex whose exact structure isunknown.

If oligosaccharides or polysaccharides are present, they are first hydrolysed to the constituentmonosaccharides which are then dehydrated.

Pentoses yield furfural and hexoses yield 5-hydroxymethyl furfural.

Reagent

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

Test

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

________________→CONC H2SO4

– 3 H2O_______________→HEXOSES

O||

||C|

|OH

O

OCH2|OH

CHO

SO3H

H2C|OH

2

OH|

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


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 presentin the Benedict’s reagent enolises the reducing sugar to form enediols (different forms ofreducing 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 continuoussupply of cupric ions.

Benedict’s qualitative reagent is prepared by dissolving 173 gm of sodium citrate, 90 gm ofanhydrous Na2CO3 in 500 ml of distilled water. Slightly heat the contents to dissolve. Filterthe solution and make the volume to 850 ml. Dissolve separately 17.3 gm of CuSO4.5H2O in150 ml of water. Add this solution slowly and with stirring to the above solution—the mixedsolution is ready for use.

________________→CONC. H2SO4

– 3 H2O_______________→PENTOSES

O||

||C|

|OH

O

O

CHO

SO3H

2

OH|

+ 2H2O

Furfuralor

5-Hydroxymethylfurfural

Violet or purplecolour condensed

product

5-Hydroxymethylfurfural

D-Glucose (Aldohexose)

CH CH|| ||C C

O CHOHOCH2

____________________→

OH|

α-Naphthol

CHOH — CHOH| |

HOH2C – CHOH CHOH — CHO ____________→CONC. H2SO4

Carbohydrates 11

ReactionsCuSO4 ___→ Cu++ +SO4

– –

Cu++ + Sodium Citrate ___→ Cupric: Sodium citrate complexCH2COONa

HO — C — COOCu++

CH2 COO

Na2CO3Reducing sugar _________→ Enediols forms of reducing sugar.

Enediols + Cupric: Sodium citrate complex _________→ Cu+ + Mixture of sugar acids.

Cu+ + OH– _______→ CuOH

Heat2CuOH _________→ Cu2O ↓ + H2O

Cuprous oxide.

Tests

Pipette 5 ml of Benedict’s qualitative reagent in a test tube. Add to it 8 drops of given carbo-hydrate 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 sugarspresent 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 gmof CuSO4-5H2O in 500 ml of distilled water

Fehling solution B It contains potassium hydroxide and Rochelle salt (Sodium potassiumtartarate). It is prepared by dissolving 125 gm of KOH and 173 gm of Rochelle salt in 500 ml ofdistilled 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 Fehlingsolution prevent the precipitation of cupric hydroxide or cupric carbonate by forming a deep

Boiling


blue soluble, slightly dissociated complex with the cupric ions. These complexes dissociatesufficiently 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 Carbohydratesolution. Mix and boil. Appearance of yellow or red precipitate of cupric oxide indicates thepresence of reducing sugars.

Barfoed’s Test

This test is used to distinguish monosaccharides from disaccharides by controlling the pHand the time of heating.

Barfoed’s test is a reduction test carried out in an acidic medium. The acidity makes it aweaker oxidising reagent. Therefore only monosaccharides, will reduce cupric ions. Howeverif heating is prolonged, disaccharides may be hydrolysed by the acid and the resultingmonosaccharide will give the test positive.

Reagent

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

Test

To 2 ml of Barfoed’s reagent add 2 ml of carbohydrates solution. Place the test tube in boilingwater bath for 3 minutes. An appearance of brick red precipitate of cuprous oxide indicatesthe presence of monosaccharides.

Precautions The solution should be boiled for 3 minutes only. Overheating should be avoi-ded 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 fur-

fural 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 oftheir conversion to ketoses by hydrochloric acid.

Sucrose will also give Seliwanoff’s test positive because the acidity of reagent is sufficientenough to hydrolyse sucrose to glucose and fructose but Benedict’s test will be negative.

O CHOH2C|OH

______________________→

OH|

OHCherry redcoloured

compound

_____________________→CONC. HCl

– 3 H2OFRUCTOSE

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 overa flame for 30 seconds only. Cool the solution. An appearance of cherry red colour indicatesthe presence of fructose.

Phenylhydrazine Test (Osazone Formation Test)

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

CH = N — NH —C6H5|

H — C — OH|

HO — C — H|

H — C — OH|

H — C — OH|

CH2OH

C6H5 NHNH2

CH = N — NH — C6H5|C = O|

HO — C — H| + NH3

H — C — OH| + C6H5NH2

H — C — OH|CH2OH

CHO|

H — C — OH|

HO — C — H|

H — C — OH|


G–Glucose

CH2OH|C = O|

HO — C — H|

H — C — OH|


D–Fructose

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

CH2OH|

C = N — NH — C6H5|

HO — C — H|

H — C — OH|

H — C — OH|

CH2OH

C6H5 NHNH2

CHO|C = N — NH C6H5|

HO — C — H| + NH3

H — C — OH| + C6H5NH2


C6 H

5 NHNH2C 6

H 5 NHNH 2

– H 2O

_________________→

_________________→ C6H5 NHNH2C6H5 NHNH2

CH = N — NH — C6H5|C = N — NH C6H5|

HO — C — H|

H — C — OH|


OSAZONE

____→


Phenylhydrazine reagentIn contains equal part of phenylhydrazine hydrocholoride and anhydrous sodium acetate. Itis prepared only at the time of reaction.

Test

In a clear and dry test tube, take approximately 0.5 gm. of phenylhydrazine mixture (equalpart of phenylhydrazine hydrochloride and anhydrous sodium acetate). Add 5 ml ofcarbohydrate solution and 1-2 drops of glacial acetic acid. Mix and place the test tube inboiling 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 areconverted to furfural which reacts with the coloured compound produced when orcinol andferric chloride react with each other.

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 beobserved 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.

___________→CONC. HCl

– 3 H2OPENTOSE

O CHOOH|

OHH3C

Blue, Greencompound

→

Colouredsolution+ Fe Cl3

_→

Carbohydrates 15

To Study the Chemical Reactions of Disaccharides

The most common disaccharides are maltose, lactose and sucrose. Maltose and lactose arereducing disaccharides where as sucrose is a non-reducing disaccharide.

One percent solution of each maltose, lactose and sucrose are provided.

1. Molisch testDissaccharides 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 ofethanolic α-naphthol (Molisch reagent). Mix. Incline the test tube and cautiously add 2 ml ofconcentrated H2SO4 by the side of the test tube. An appearance of violet or purple ring at thejunction of two solutions indicate the presence of dicarbohydrates.

2. Benedict’s qualitative reagentTest Pipette 5 ml of Benedicts qualitative reagent in a test tube. Add 8 drops of givendicarbohydrate solution. Boil for 2 minutes. An appearance of green, yellow or red precipitateindicates the presence of disaccharides.

Maltose and lactose give Benedict’s qualitative test positive whereas with sucrose the testis negative, i.e. no reduction is observed.

3. Barfoed’s testTest To 2 ml of Barfoed’s reagent add 2 ml of disaccharide solution. Place the test tube inboiling water bath for 3 minutes. No change in colour indicates the presence of disaccharidesin 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 mlof disaccharide solution and 2 drops of glacial acetic acid. Mix. Place the test tube in boilingwater bath for 30 minutes.

After 30 minutes, take out the test tube from the boiling water bath and allow it to cool byitself in a test tube rack (Do not disturb the test tube in between as the osazones of disac-charides separates out on slow cooling).

Appearance of yellow crystals takes place. Observe the shape of crystals under low powermicroscope.

|H — C — OH

|H — C — OH

|HO — C — H

|HO — C — H

|HO — C

|CH2OH

Galactose

COOH|

H — C — OH|

HO — C — H|

HO — C — H|

H — C — OH|COOH

Mucic acid

+ HOHO

______→Δ

OXn


The shape of osazones are:Malatose : Sunflower shapeLactose : Cotton ball shape

Sucrose It will give Benedict’s qualitative test negative.

Lactose (Cotton ball) Maltose (Sun flower)

Sucrose is Confirmed AsTest Take 5 ml of sucrose solution in a test tube. Add to it 1-2 drops of concentrated hydrochloricacid. 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 needleshaped crystals.

To Study the Chemical Reactions of PolysaccharidesSolutions provided are 1% starch and 1% dextrins.

1. Molisch test2. 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 obtainedwith 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 Blue colourAmylose Blue colourAmylopectin Purple colourGlycogen Brown red colourDextrins Brown to colourlessCellulose or inulin No colourDisaccharides or monosaccharides No colour

Carbohydrates 17

Test

In two ml of carbohydrate solutions, add few drops of hydrochloric acid (to make the mediumacidic) followed by 1 ml of iodine solution. Mix and observe the colour.

No change in colour indicates the absence of polysaccharides.

Hydrolysis

i. Acid hydrolysisIn a 100 ml conical flask, take 20 ml of 1% starch solution. Add 5 ml of 2N HCl (prepared bydiluting 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 thetubes 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 Benedict’s Reduction of hydrolysis Productiodine test (extend)

1 minute Blue Blue No reduction Starch5 minute Violet Green Reduction starts(+) Amylodextrins8 minute Reddish violet Red Initiation of reduction (++) Amylo and erythrodextrin12 minute Red Red Partialreduction(+++) Achrodextrin20 minute No colour Red Completely reduced(++++) 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. Mixwell.

Now add 1 ml of distilled water only in the blank and 1 ml of saliva in the test. Keep thetest tubes for 30 minutes at room remperature.

1. In Blanka. Iodine test : Blue colourb. Benedict’s test : Negative.

2. In Testa. Iodine test : Negativeb. Benedict’s test : Red precipitate.

Blank Test sample

Iodine test Blue colour NegativeBenedict’s test Negative Red precipitate


Starch with saliva shows reduction as starch is converted to glucose which is a reducingsugar. Where as starch without saliva is not broken up in to smaller molecules because there isno hydrolysis.

Summary of Qualitative Test

Glucose Fructose Lactose Maltose Sucrose Starch

Molisch testIodine testBenedict’s testBarfoed’s testSeliwanoff’s testPhenylhydrazine test

Identification of Unknown Carbohydrate Solution1. 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 nega-tive, it means reducing sugars are absent.

Absence of Benedict’s test, indicates the presence of non-reducing disaccharide, i.esucrose.

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 IIdentification of unknown carbohydrate

Iodine test

|| | |

No change in colour Red colour Blue colour (starch)|

| | |

Clear solution Hazy solution(Dextrin) (Glycogen)

Step IIBenedict’s test

|| |

Positive (Reducing sugar) Negative (sucrose)Glucose, Fructose, Galactose, (Nonreducing sugar)Mannose, Lactose, Maltose

Carbohydrates 19

Step IIIBarfoed’s test

|| |

Positive Negative(Glucose, Fructose, Galactose) (Lactose, Maltose)

Step IV ASeliwanoff’s test

|| |

Positive Negative(Fructose) (Glucose, Mannose, Galactose)

Step IV BOsazone test

|| | | |

Precipitate before Yellow insoluble during Small ball like Osazone separates on coolingheating (Mannose) heating needle shape cluster thronedge |

(Glucose, Fructose, Mannose) (Galactose) || |

Cotton ball shape Sunflower shape(Lactose) (Maltose)

20 Practical Biochemistry for Students 4Achromic Point

Determination of Achromic Point of Your Own Saliva

This is a simple enzymatic hydrolysis of starch. This is different from acid hydrolysis. Enzymatichydrolysis 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 isthat time period which is required to obtain achromic point when enzymatic hydrolysis isbeing performed.

In an animal, there are enzymes which break only the α-1, 4 linkage, i.e. they break only thestraight 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 intomonosaccharides (i.e. glucose) and maltose units. At places where there is α-1, 6 linkageoligosaccharide 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 required1% starchBuffer—pH 6.71% 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 5ml 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 waterwith 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 1 min 1.5 min

2 min 2.5 min

(20)

Achromic Point 21

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

Observations

Time Colour

30 sec. Blue1 minute Light blue1½ minute Lighter blue2 minute Faint blue2½ minutes Colourless

Result and Conclusion

The achromic point of saliva is 2½ minutes, this means that 2½ minutes are taken till achromicpoint is got and formation of achrodextrin takes place. Now if we put the hydrolysed solutionin a drop of Fehling’s very little red precipitate is got because enzyme hydrolysis does notproduce many monomers.

22 Practical Biochemistry for Students 5Proteins

To Study the General Reactions of Proteins

1. Biuret TestBiuret test is given by all compounds that contains two or more peptide bonds. Since proteinsare polypeptide, hence it is a general tests for proteins.

The name of the reaction is derived from the organic compound, a biuret, obtained byheating urea at high temperature which gives a positive test.

O O

180°C || ....... .....|| ...NH2CONH2 + NH2CONH2

_________→ H2N — C — NH — C — NH2Biuret

.......... .............

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 0.2NNaOH to this is added 7.5 gm of CuSO4 5H2O with constant stirring. Then is added 2.5 gm ofKI 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 mlwith 0.2N NaOH containing 5 gm of KI per litre.

Reaction

The purple or violet colour produced isbelieved to be due to a coordinatecomplex between the cupric ions and fournitrogen atoms, two from each of twoadjacent peptide chains.

Test

Take 6 ml of 5% NaOH, in a test tube andadd few drops of 1% CuSO4 solution tillblue colour solution is produced. Dividethe solution, i.e. 3 ml each in two testtubes marked experimental test ‘A’ andcontrol test ‘C’.• To ‘A’ add 3 ml of protein solution.• To ‘C’ add 3 ml of distilled water.

H H

O

Cu++

O

H H

|O == C

HN

HR — C

||

O == C

N

HR — C

|

|H — C == ON

CH — R||

C == O||

N — CH — RH |

(22)

............

Proteins 23

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

2. Ninhydrin TestProteins containing free α-amino acid radical in the molecule reacts with ninhydrin to give ablue-violet coloured compound. Ninhydrin test is not given by protein and hydroxy prolinebecause no free α-amino group is present. They give only a yellow colour.

Reagent

Ninhydrin dissolved in acetone.

Reaction

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

Test

Take 1 ml of protein solution, add to it 2-3 drops of freshly prepared ninhydrin solution. Heatthe 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. Theseacids which the animal body is unable to synthesise either completely or in amounts sufficientto normal growth maintainence must be supplied in the food. These so called essential aminoacids can be tested for in various way and can in many cases, be detected and determined bymeans of simple colour reactions based upon the qualitative test.

A variety of colour reactions specific to particular functional groups in amino acids areknown. They are useful in both the qualitative and quantitative identification of particularamino acids.1. Xanthoproteic test (For benzenoid radical)The reaction is based upon the nitration of the benzene ring with concentrated HNO3 yieldingyellow derivatives of nitrobenzene which turns to orange in alkaline medium.

Ninhydrin test

2

H||C

C||O

C

OH

OH

Ninhydrin

COOH| Δ

+ H2N — C — H ______→

|R

O||C

C||O

C = N — C

O||C

C||OH

Purple colouredcomplex

Δ NH3

Blue colour(RUHEMANN'S PURPLE)


Test

Take 3 ml of test solution (protein solution). Add 1 ml of concentrated HNO3. A white precipitatedue to denaturation of protein is formed. Boil the solution. A yellow solution derivative isformed 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 proteinbecause 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 1ml of Millon’s reagent. Boil the solution. A yellow precipitate adheres to the side of the testtube. Cool the solution under tap water. Add a drop of 1% sodium nitrite (NaNO2) and gentlywarm. 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 theprotein.

Test

Take 1 ml of test solution and add few drops of 1 : 500 commercial formalin (40% formalde-hyde). Add 2 drops of mercuric sulphate (i.e. 10% HgSO4 in 10% H2SO4). Mix. Incline the testtube and add 1 ml of concentrated H2SO4 by the side of the test tube. A purple ring is formedat 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 ofthe 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 sodiumhypobromite.

Test

Take 3 ml of test solution in a test tube. Add 1 ml of 5% NaOH, 2 drops of ethanolic α-naptholand 2 drops of 10% sodium hypobromite. Mix well. Wait for 5 minutes. Development of brightred colour takes place.

Run a control by taking 3 ml of distilled water instead of protein solution and add all otherreagents 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 leadacetate solution. A black precipitate of PbS insoluble in dilute HCl is formed.6. Test for free—SH radicalTake 1 ml of test solution. Add few crystals of ammonium sulphate. To it add few drops offreshly prepared solution of sodium nitroprusside and 1 ml of liquor ammonia. Developmentof rose red colour takes place.

Colour tests Albumin Globu lin Phenylalanine Tyrosine Arginine Guanidine Tryptophan

Biuret testNinhydrin testXanthoprotic testMillon’s testHopkin’s-

Cole testSakaguchi testCystine-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 ofthe metal ions with the anionic form of the protein. On the alkaline side of the isoelectric pointproteins 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 reagentsProteins are precipitated from the solution by combination between the acid anions and thepositively charged protein molecule by forming all insoluble complex. The alkaloidal reagentsprecipitate proteins by combination of the acidic radical of the former with the cationic formof the protein, which predominates when the solution is on the acidic radical of the formerwith the cationic form of the protein, which predominates when the solution is on the acidicside of the isoelectric point.


a. To 3 ml of protein solution, add few drops of metaphosphoric acid. A dirty whiteprecipitate appears.

b. To 3 ml of protein solution, add few drops of 20% sulphosalicyclic acid. Whiteprecipitate 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 heatingProteins which are precipitated when their solutions are heated are termed as heat coaguableproteins. This property of proteins is made use of in the detection of albumin in urine simplyby heating the urine.

Solubilities of several proteins in various solvents are shown below:

Distilled water Dilute acids Dilute alkali Ammonium sulphate

Albumin Soluble Soluble Soluble Soluble Insoluble

Globulins Insoluble Soluble Soluble Insoluble Insoluble

Gelatin Soluble Soluble Soluble Insoluble Insoluble

Difference in Albumin and GlobulinAlbumin Globulin

1. It is obtained by full saturation of 1. It is obtained by half saturation of(NH4)2 SO4 ammonium sulphate.

2. It is soluble in water. 2. It is insoluble in water but soluble in dilutemineral acids and salt solution.

3. It is a smaller molecule having more charge. 3. It is bigger molecule with less charge.4. In isoelectric point field it travels faster 4. Due to big molecule and less charge it

than globulins. travel slowly in isoelectric field.

Isoelectric Point 276Isoelectric 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 acetate1 N acetic acid0.1 N acetic acid0.01 N acetic acid

Principle

Isoelectric pH is defined as the pH at which the protein molecule does not migrate to thecathode 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 solubilityof proteins is minimum. At the isoelectric pH, the electrostatic repulsive force which normallyprevents the protein molecules from coming together is minimum as a result of no net changeon 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 _____________

pKa for acetic acid is 4.74[Acid]

(27)


[Salt] = Sodium acetate = 0.1 N[Acid] = Acetic acid = 0.1 N Maximum turbidity

0.1pH = pK + log __________

pH = 4.740.1

The isoelectric point (pI) for casein is 4.74.The pH of the respecting tubes will be as

Tube Precipitate formedpH

No. Zero time Ten minutes Thirty minutes

1. – – – 5.952. – – – 5.643. + + + 5.344. ++ ++ – 5.045. ++ +++ ++++ 4.746. ++ ++ + 4.447. + + + 4.128. + + + 3.149. – – – 3.54

Lipids 297Lipids

To Perform General Tests for Lipids

Lipids are defined as a group of fatty nature which are insoluble in water but soluble in nonpolarsolvents 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 mlof ethyl alcohol and to the third 2 ml of chloroform. To each of the three test tubes add 3 dropsof 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 separatelayer. 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 solventslike 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, thisemulsion 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 breaksdown after long time.

Inference

Oils form a coarse and unstable emulsion with water. This readily breaks down. So this con-cludes that oils do not reduce the surface tension of water. In case of bile salt solution andhousehold detergent, the emulsion is fine and very stable because bile salts and householddetergents reduce the surface tension of water as a result of which oil fragments into smalldroplets 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 potassiumbisulphite and heat vigorously. Smell the fumes of the gas which come from one of the testtube.

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 itand mix well. After mixing, add 10 ml of 10% alcoholic sodium hydroxide, shake well andkeep in boiling water bath for 15 minutes. Take the test tube after 15 minutes and add water sothat the resulting volume of the solution is 15-20 ml. Shake well to dissolve. Divide the contentsinto 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 mixwell. 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. thecolour of the bromine solution itself.

Inference

Bromine goes into the solution forming a dibromide, i.e. it add to the double bonds. In otherwords, bromine solution is decolourised, but when all the double bonds are saturated thebromine solution imparts its own colour.

Experiment

6. Test for cholesterol

a. Libermann-Burchard reactionTake a perfectly dry test tube and add 2 ml of CHCl3 solution of cholesterol to it. Then add 10drops of acetic anhydride and mix well. Then add drops of concentrated H2SO4 from the sidesof 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 reactionTake a perfectly clean and dry test tube and add to it 2 ml cholesterol. Solution prepared incholesterol. Then add an equal volume of concentrated H2SO4 dropwise along the side of thetest tube.

Observation

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

Inference

This indicates the presence of cholesterol.

32 Practical Biochemistry for Students 8Saponification Number

Determination of Saponification Number of an Oil

Saponification number is defined as the number of milligrams of KOH required to saponifycompletely 1 gm of fat.

Since fats are mixture of triglycerides, most of which are of mixed type, so saponificationnumber 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 identifyingcertain oils.

Procedure

Take a clean and dry 100 ml conical flask. Using a 2 ml pipette, transfer 1.5 ml of the oil sampleprovided in the conical flask. Add 15 ml of 0.5 N ethanolic KOH into the flask containing theoil. Mix the contents well. Place a funnel at the neck of the conical flask (the stem of funnel actsas a condenser) and place it in the boiling water bath for half an hour till all the oil globulesdisappear 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 titrevalve.

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 339Iodine 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 characteristicfor 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 26-28Human fat 65-70Peanut oil 80-90Cornoil 110-125Soyabean oil 137-143Linseed oil 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 — BrExcess of LeftHanus solution over

To the left over Hanus solution is added potassium iodide solution. The iodine thusliberated 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 ________→ I2 ↑ + KBr

2Na2S2O3 + I2______→ 2Na2S4O6+2NaI

At the end point

I2+Starch ______→ Blue colour

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ProcedureTest

In a 250 ml conical flask, add 5 ml of given oil sample (the oil sample is dissolved in CCl4. Theconcentration 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 reactionto take place.

After 30 minutes, add 5 ml of KI solution into it. Mix well, followed by 25 ml of distilledwater. 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-blackto 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 procedureas 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 intitrating the IBr which was used in saturating the unsaturated fatty acid moiety, i.e. the volumeof 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 mlTitre value obtained for blank titration = y mlAccording to Normality equation1 ml of N/10 Na2S2O3 solution ≡ 1 ml of N/10 I-Br solution1 ml of n/10 Na2S2O3 solution ≡ 1 ml of N/10 I-Br solution

≡ 1 ml of N/10 iodine solutionEquivalent weight of iodine = 127.

127 11 ml of N/10 iodine = ________ × ____ = 0.0127 gm.

1000 10

(Two molecules of Na2S2O3 are equivalent to one molecule of iodine; thus one molecule ofNa2S2O3 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 iodineThe 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 × 100The iodine number is ____________________________ gm of iodine.

0.25

Formal Titration 3510Formal Titration

Formal Titration

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

Principle

Amino acids by virtue of zwitterion formation are merely neutral in reaction. If formaldehydeis added to a solution of amino acids an adduct is formed at the amino group leaving thecarboxyl group free.

In other words, formaldehyde suppresses the basicity of amino group. So that carboxylgroup can exert its maximum acidity and hence can be titrated against standard solution ofalkali 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 NaOHused up in the titration.

In a 100 ml conical flask, take 10 ml of glycine solution, add to it 2 ml of above neutralisedformalin solution. Then titrate the resulting solution against N/10 NaOH. The end pointin 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

CH2OH+ NNH3 | CH2OH| |

R — CH — COO– + 2HCHO ____________→ R — CH — COO– – + H+

(35)


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

7.5 × χχ ml of N/10 glycine contains _________ gm of glycine1000

Gastric Analysis 3711Gastric Analysis

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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 inthe morning. Ryle’s tube is inserted through the nose to suck the gastric juice. The patient isgiven 50 ml of 7% alcohol after every 15 minutes, 10 ml of gastric juice is sucked up. Collect 4such samples; 15 minutes, 30 minutes, 45 minutes and 1 hour. Then inject 0.5 gm of histamineintramuscularly, after every 15 minutes, suck 10 ml of gastric juice. Collect four such samples1 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 responseand often produces acid secretion where the ordinary test meal fails completely. It adds nothingto the stomach so that a pure juice, undiluted and uncontaminated is obtained and noneutralisation 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 ordry toast. It is much less objectionable to take, is easily swallowed and evokes a feeling ofpleasure instead of distaste on the part of the subject. The gastric juice obtained is ideal foranalysis.

The main content of the gastric juice is hydrochloric acid. In a normal person the pH of thegastric 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-dimethylamino-azobenzene in alcohol) as indicator is added. The colour of the solution turns yellow due tothe presence of free HCl. Yellow colour indicates a pH of 2.90. The end point appears when thecolour turns orange. The pH is 3.60 at the end point. The free HCl being liberated. Proceed totitrate the protein bound HCl. 2 drops of phenolphthalein are added to the solution (whichhas been titrated for free HCl) and it is titrated against N/10 NaOH. The end point appearswhen the colour becomes pinkish. The pH at this end point is 8.50.


The first titre value with Topfer’s reagent expressed in units of N/10 HCl per 100 ml ofgastric juice gives free acidity. The second titre value with phenolphthalein expressed in unitsof 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. Ifan orange colour appears (pH 3.6) then free acidity is zero. In that case add a drop ofphenolphthalein and titrate with N/100 NaOH till the end point is reached (pH 8.50). The titrevalue 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/100NaOH till a orange colour is obtained. The titre value expressed in units of N/100 HCl per 100ml of gastric juice gives free acidity. Now in the same solution, add a drop of two ofphenolphthalein and titrate with N/100 NaOH till a pink colour (pH 8.5) is obtained. This isthe end point. This titre value expressed in units of N/100 HCl per 100 ml of gastric juice givescombined acidity.

1 ml of N/100 NaOH ≡ 1 ml of N/100 HCl1 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, then10 × ml of N/100 HCl = x ml of N/10 HClTotal acidity in 1 ml of gastric juice = x/10 mlTotal acidity in 100 ml of gastric juice = 10 xSo 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 absentHyperchlorhydria : Free HCl above 60 unitsHypochlorhydria : Free HCl never above 10 units

Achlorhydria

Absence of free HCl in gastric juice is called achlorohydria. However enzymes such as pepsinmay 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 normallimit, i.e. values about 60 units, e.g. Duodenal ulcer.

Normal Value Gastric ulcer Duodenal ulcerCarcinoma Pernicious

of stomach anemia

Free HCl 10-70 Normal 10-45 Normal ZeroTotal acidity 5-100 10-100 15-110 3-80 0-40Volume 10-100 40-100 10-115 10-500 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. Theappearance of blue colour indicates the presence of starch.

2. Presence of Iactic acidMaclean’s test

Take 2 ml of gastric juice in a test tube. Add few drops of Maclean’s reagent (contain HgCl2and 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 bloodBenzidine test

Take 3 ml of saturated solution of benzidine in glacial acetic acid. Add 2 ml of gastric juicefollowed by 1 ml of hydrogen peroxide. Wait for few minutes. The appearance of blue orgreen colour indicates the presence of blood in gastric juice.

4. Presence of bileHay’s test

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

5. Presence of HClGunzberg test

A drop of two of Gunzberg reagent is placed in a small porcelein dish. It is carfully evaporatedon a small flame. A glass rod is dipped in the gastric juice and is touched thoroughly with theresidue in the porcelein dish. It is gently heated. The appearance of purple red colour indicatesthe presence of free HCl.

6. Presence of mucusExamination of gastric juice in broad day light will reveal the stingy appearance of juice forpresence of mucus.

40 Practical Biochemistry for Students 12Urine Analysis

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Urine Analysis

The average volume of urine excreted daily is about 1.5 litres. It contains nitrogeneous organiccompounds such as urea, uric acid, creatinine, hippuric acid, indican, purines and amino acidsas well as organic compounds which do not contain nitrogen. The average composition ofurine excreted in 24 hours is as follows:

Urea 20-30 gmUric acid 0.7 gmAmino acid 0.5-1 gmAmmonia 0.7 gmCreatinine 1.4 gmAscorbic acid 15-20 gmGuanidine 3-16 gmHippuric acid 0.6 gmIndican 4-20 mgIodine 50-250 mgLactic acid 50-200 mgChloride as NaCl 10-15 mgInorganic sulphate as S 60-120 mgNeutral sulphate as S 80-160 mgSodium 3-5 gmCalcium 0.1-0.3 gmPhosphate as inorganic phosphates 1.0-1.5 gm

Urine specimens must be collected in the correct way and in a clean container or bottle. Inthe hospital, when only one specimen is needed, the best time to collect is first urine in themorning when it is most concentrated. The specimen should be analysed without delay. If thisis 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 chemicalexamination, abnormal constituents which are normally not present in easily detectablequantities in urine of healthy beings but are known to occur in urine under certain diseasedconditions. Abnormal or pathological constituents of urine are looked for in urine such asreducing sugars, proteins, ketone bodies, bile salts, bile pigments, blood, urobilinogen, etc.

Urine Analysis 41

1. ColourThe normal urine is yellow in colour. The intensity of normal urine is dependent on theconcentration of urine. The yellow or amber colour of a normal urine is due to presence of ayellow pigment urochrome.

The colour of urine changes in many disease conditions because of the presence of pigmentsthat do not normally occur.

Colour Possible cause

Orange Concentrated urineAlmost colourless Dilute urineYellow to yellow brown or greenish colour Bile pigmentsReddish brown colour HemoglobinMilky Presence of fatsBlack upon standing AlkaptonuriaBrown-black colour on standing Melanin or homogentisic acidOrange brown UrobilinogenCloudy Presence of insoluble calcium and magnesium phosphates

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

2. VolumeThe normal volume of urine voided by an adult per day ranges from 750 to 2000 ml. Theaverage volume is 1500 ml.

The amount of urine excreted is directly related to fluid intake, the temperature, climateand 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 intake2. Ingestions of diurates

Pathological conditions1. Diabetes mellitus2. Diabetes insipidous3. Chronic renal damage

Oliguria It is decreased in which the urinary output is 500 ml per 24 hours.Oliguria occurs in

1. Less fluid intake2. Excessive fluid loss due to vomiting, diarrhoea and sweating3. Fever4. Shock5. Acute nephritis6. 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 renaldisease, this ratio is reduced and may even be reversed.


3. Specific GravityNormal 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 ahigh specific gravity while diluted urine has a low specific gravity.

Specific gravity of urine indicates the relative proportions of dissolved components to thetotal volume of the urine. It also reflects the relative degree of concentration or dilution of theurine sample.

Low specific gravity is observed in the following:1. High fluid intake2. Diabetes insipidous3. Glomerulonephritis4. Pyelonephritis

High specific gravity urine is observed in1. Excessive loss of water due to sweating fever, vomiting and diarrhoea.2. Diabetes mellitus3. Nephrosis4. Hepatic diseases5. Congestive heart failure.

Whereas fixed specific gravity of urine is an indication of several renal damageswith disturbance of both concentrating and diluting abilities of kidney.

Determination of Specific Gravity

The specific gravity of urine is determined by urinometer. Urinometer is a glassmade instrument having a cylinderical stem containing a scale.

The urinometer is floated in a cylinder containing urine. Care should be takenthat it should not touch the sides or bottom of the cylinder. The depth to which itsinks in the urine, indicates the specific gravity of urine.

The urinometer is calibrated with respect to distilled water at 1.000 at specifictemperature, indicated on the instrument itself. If the temperature of urine is aboveor below that temperature, a correction of + 0.001 for each 3°C should be made.

4. pHThe pH of the normal urine is on an acidic side, i.e. 5.5-6.5. The acidity of the urine is mainlydue to acid phosphates.

In general protein rich diets give rise to acidic urine attributable to the sulphur of the aminoacids which is oxidised to sulphuric acid. Similarly phospholipids and nucleic acids all yieldphosphoric acid. Alkaline urines are excreted where there is a predominance of vegetablesand fruits in the diet. However, the production of ammonia by the kidney also plays animportant part in influencing the pH of the urine since it forms salts with acids and can excretedas such.

Acid Urine

i. On a high protein diet.

1000

1010

1020

1030

1040

1050

Urine Analysis 43

ii. Uncontrolled diabetesiii. Acidosisiv. Fever.

Alkaline Urine

i. After mealsii. Diet high in vegetables, citrus fruits, milk, etc.

iii. Renal tubular acidosis.

5. Total SolidsTotal solids can be calculated approximately by multiplying the second and third decimalfigures of specific gravity by 2.6 (Longe’s coefficient). The product represents the number ofgm of solids in 1 litre of urine, e.g. if the specific gravity of urine is 1.025, then 25 × 2.6 = 65 gmtotals 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 detectedby the reagents used. However, under abnormal conditions following reducing sugars areexcreted in urine.

Reducing sugars Conditions

Glucose Renal glucosuriaDiabetes mellitus

Blood glucose will differentiate

Galactose GalactosemiaLactose Pregnancy and lactationFructose

After ingestion of large amounts of vegetables or fruitsPentoses

Causes of Glycosuria

1. Diabetes mellitus2. Non-diabetic glycosuria

A. Glycosuria with hyperglycemiaa. Renal glycosuria—due to low renal thresholdb. Alimentary glycosuria—after ingestion of lot of carbohydratesc. Glycosuria of pregnancy—in 10-15% of normal pregnancies

B. Non-diabetic glycosuria with hyperglycemiaa. Hyperthyroidisimb. Emotional disturbancesc. Ether anesthesiad. 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 minutesand cool. A change in blue colour to green-yellow-red precipitate indicates the presence ofreducing 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 No reducing sugarGreen colour Less than 0.5 gm%Green precipitate 0.5 to 1 gm%Green to yellow precipitate 1.0 to 1.5 gm%Yellow to red precipitate 1.5 to 2.0 gm%Brick red precipitate 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 2min. Appearance of either green-yellow-red precipitate indicates the presence of reducingsugars.

Tests for Specific Reducing Sugar

The following tests will differentiate the various reducing substances which respond to Fehlingand Benedict’s test. Usually the object in testing for reducing substances in urine is to determinethe presence or absence of glucose.

1. Fermentation test: Glucose and fructose are fermented by the enzyme present in the yeastand the libration of carbon-dioxide confirms the test.

2. Mucic acid test for lactose and galactose: Evaporate urine in a china dish containing amixture of 100 ml of urine and 20 ml of concentrated HNO3. Let the volume is reduced to20 ml. Allow it to cool. Lactose or galactose yields mucic acid through the oxidising actionof 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. ProteinsThere is no single test capable of detecting the presence of proteins under all conditons. Theproteins which appears in urine however are usually albumin and globulins of which albuminalmost greatly predominate.

Causes of albuminuria1. Functional—orthostatic, severe muscular exertion, prolonged exposure to cold, pregnancy.2. Organic

a. Pre-renal causes—no primary kidney diseasePassive congestion of kidneyFever and toxaemiaIntra-abdominal tumoursDrugs and chemical poisoning

b. Renal—primary disease of kidneyNephritisNephrosis

Urine Analysis 45

c. Post renal causesPyetitisCystitisUrethritis, 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 boilingwithout disturbing the lower portion which serves as a control. A white turbidity indicatesthe presence of proteins or phosphates. Add few drops of glacial acetic acid. If precipitatedissolves, 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 whiteprecipitate 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 toform 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 ofsaturated ammonium sulphate, the globulin is precipitated and albumin which remains canthen 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 malignantdisease involving the bone marrow and of leukemia. Its occurrence is very uncommon inurine.

Test for Detection of Bence Jones Proteins

Heat coagulation testThe urine must be faintly acidic. Acetic acid being added if necessary. It is thengently heated in a test tube. Bence Jones protein gives a precipitate which appearsabout 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. Inthat case the boiling urine is filtered rapidly to remove albumin. Bence Jones proteinsreappears as the filtrate cools. The precipitate flocculates and sticks to the side oftest tube.

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

1098765432

1½

R

U


The height of the precipitate in the tube is then read and the reading divided by ten gives theamount 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 areoxidised to carbon dioxide and water under normal conditions. In abnormal conditions, ketonebodies accumulate in the blood and pass out in urine. This condition is called ketosis. Whenthe carbohydrate metabolism is defective, increased fat is oxidised for energy purposes givingrise to increased formation of ketone bodies. Such happens in diabetes mellitus, when glucosepresent is not metabolised hence ketone bodies appear in urine. Also during starvation whenglucose supply to the body is restric-ted, ketone bodies appear in urine.

Causes of ketonuriai. 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 atthe 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 theoxidation of bilirubin to coloured compounds of blue, green, violet, red and yellow. Normalurine does not contain bilirubin as bilirubin is bound to albumin. Bilirubin is present in theurine of obstructive jaundice and in some cases of hepatic jaundice due to high level ofconjugated bilirubin.

Fouchet’s Test

Take 5 ml of urine in a test tube. Add to it an equal volume of 10% BaCl2 solution. Filter, drythe filter paper and add a drop of Fouchet’s reagent. Green colour indicates the presence ofbilirubin.

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, greenand 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 ofconcentrated HNO3 in the centre of filter paper. Green, blue, violet, and red colours show thepresence of bile pigments.

Urine Analysis 47

5. Bile Salts

Bile salts are sodium and potassium salts of glycocholates and taurocholates. Bile salts areformed in the liver from where they are excreted in the bile. They are absorbed by the intestineand passed back to the liver through portal circulation. Bile salts are present in urine inobstructive jaundice.

Hay Test

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

6. Blood

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

Benzidine Test

The peroxidase activity of hemoglobin decomposes hydrogen peroxide and the liberated oxygenoxidises 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 ofurine followed by 1 ml of hydrogen peroxide. Appearance of green or blue colour within fewminutes indicates the presence of blood.

7. For Urobilinogen

Urobilinogen is normally present in the urine in very small amount. The excretion ofurobilinogen is increased in:

i. Hemolytic jaundice: In hemolytic jaundice, liver is not able to excrete the increased amountof urobilinogen absorbed from the intestine completely.

ii. Infective hepatitis: The liver cell is less able to excrete so more of bilirubin passes into generalcirculation 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 forfive 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 inundiluted urine sample and also in serial dilution of the urine. Normally urobilinogen is presentin urine upto a dilution of 1 : 10.

48 Practical Biochemistry for Students 13Food Analysis

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Test for Carbohydrates

Take about 1 gm of given food sample in a test tube. Add 10 ml of distilled water. Slightlywarm 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 inclinethe tube and add 2 ml of concentrated sulphuric acid by the side of the test tube. Appearanceof 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 followingtests 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 3minutes. Appearance of green-yellow orange-red precipitate indicates the presence of reducingsugar in the food.

Barfoed’s Test

To 2 ml of Barfoed’s reagent, add 2 ml of above filter solution. Boil for 2 minutes. Appearanceof 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 equalpart of phenylhydrazine hydrochloride and anhydrous sodium acetate), 5 ml of filtered solutionand add 2 drops of glacial acetic acid. Mix and place the tube in the boiling water bath for 30minutes.

Food Analysis 49

The osazones of monosaccharides separate out while in hot, whereas the osazones ofdisaccharides separate out in cold, i.e. by allowing the solution to cool by itself without disturb-ing the solution in between at room temperature.

Examine the shape of the osazones under low power microscope. Appearance of needleshaped osazones suggests/indicates the presence of monosaccharides either glucose or fructosewhile appearance of sunflower or cotton ball shaped osazones indicates the appearance ofmaltose 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 andlabel one as experimental and other as control.

Experimental tube: 3 ml of above divided solution. Add 3 ml of suspension. Mix. Appearanceof 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 thesolution. 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 testiii. Hopkin’s-Cole testiv. Sakaguchi testv. 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. Sincefats are soluble in chloroform, it will go into the solution. Filter the solution. Care should betaken that funnel and test tube receiving the filtrate should not be wet with water. Perform thefollowing 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 itin dark for 10 minutes. Appearance of green colour indicates the presence of cholesterolin the food sample.

b. Salkowski test: In a clean dry test tube, take 2 ml of the filtered chloroform extract. Add 2ml 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 colourin 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 followingtests 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 indicatesthe presence of iron in the food sample.

ii. To 2 ml of the filtrate, add 2 ml of sodium nitroprusside. Appearance of blue colourationindicates 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 ofyellow 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 saturatedammonium oxalate solution. Appearance of white precipitate indicates the presence of calciumin the food sample.

Test for Chloride

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

Collection and Preparation of Blood Specimen 5114Collection and Preparation of

Blood Specimen

(51)

Blood is one of the most common specimens studied in biochemical laboratories in search ofblood disorders, metabolic disorders and infection.

Blood clots within a few minutes after it has been removed from the body, unless ananticoagulant is used which stops the process of clotting. Anticoagulated blood is also knownas whole blood. Plasma (Fluid portion of unclotted blood) is obtained from the anticoagulatedblood. 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. Forclinical 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 theanticoagulants remove calcium which is one of the factors required in coagulation process.Heparin, however, directly interferes in the coagulation process by destroying thrombin andthromboplastin. Use of an anticoagulant largely depends upon the nature of clinicalinvestigation, e.g. heparin is necessary for determination of blood gases (and pH) while sodiumfluoride in addition to oxalate, is necessary for glucose estimation. Specific amount ofanticoagulant 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 insidethe container. For oxalate anticoagulant (which is most common in biochemistry lab), a hightemperature is avoided as it changes oxalate into carbonate. All previously prepared containerscan be stored at room temperature. For easy identification, the containers are labelled differentlywith 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/mlof whole blood).

Double Oxalate

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


Sodium Citrate

0.106 M trisodium citrate in distilled water, sterilize it. This is used in solution form and is notdried inside the container. 1 ml of this solution is used for 9 ml blood (The ratio of anticoagulantsolution and whole blood comes to 1 : 9, which is ideal for coagulation studies). Not verycommon 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 equipmentwhich 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 andsyringe after drawing blood.

Needle Sterilized sharp needle of bore size 18 to 20 gauge for adults and 23 gauge for childrenis used. Use of disposable needles is highly recommended. In case of reusable ones, the needlepoint should be kept sharp. Keep a stock of sterile needles in small glass tubes and the pointshould be on a pad of non-adsorbent cotton wool and the tube should be plugged with thesame 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 dependsupon the amount of blood needed. Check that the end of each syringe fits into the needle. Atany stage of assembly of the needle and the syringe, do not touch the tip of the needle andkeep 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 thesyringe tip should be checked. Disposable plastic syringes come in a sterilized wrap and theyhave sharpened needles for single time use.

Following use, syringe should be dipped in cold water to remove the blood. All glasssyringes should be properly sterilized and perfectly dry. A wet syringe, an improperly washedsyringe. Use of a too fine needle, emptying the syringe without removing the needle are someof 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 elasticrubber strip can also be used. This is applied to the arm to slow the blood flow and make theveins 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 bewarm. This improves the circulation. The arm is extended and a rubber torniquets firmlyapplied a few inches above the elbow. The skin over the vein is cleaned by rubbing over withspirit or ether. A well sharpened sterile hypodermic needle fixed onto a syringe can be heldsteady by a thump of the other hand of the operator. When the needle enters the vein, theplunger is withdrawn slightly. If blood appears, the tourniquet is released. When the desiredamount of blood has been drawn into the syringe, a small pad of wodsoaked with spirit orether is placed on the arm where the needle was inserted, and the needle is withdrawn. Thispad is held on firmly for a few minutes until bleeding stops. The needle is removed from thesyringe and the blood transferred to an appropriate container, using minimum amount ofpressure. The needle and syringe are immediately washed out with cold water to remove anyremaining 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 andwhere vein puncture is difficult (old, prolonged illness). In adults and older children, the tipof 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 obtainedby 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 obtainserum is collected in a plain tube without anticoagulant and allowed to clot for 30 min atroom 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 becauseprolonged contact of the cells with the serum will cause the release of the red cells constituentsinto the serum composition. If the centrifuge is not available, refrigerate the blood for severalhours, the clot will separate from the serum which can be removed with the help of a pasteurpipette.

Plasma

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


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 proteinsas these interfere in many ways in the actual investigation. For this purpose, many substancesare used, chiefly acids, heavy metal, ion in presence of suitable alkali, organic solvents, etc. Asacids, e.g. tungstic acid and trichloroacetic acid are mostly used, others include perchloric,tungstomolybdic, phosphotungstic and sulphosalicylic acid. Choice of method depends onnature 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 duringthe addition of H2SO4. Allow to stand for 10 minutes and centrifuge at 2000 rpm for 5min.

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 sodiumsulphate + 0.2 ml blood + 0.3 ml 10% zinc sulphate + 0.3 ml (N/S) sodium hydroxide. Mixwell. Keep for 10 min and centrifuge. In addition to proteins saccharides (non-sugarreducing substances) are also precipitated. This method or its modification is particularlyused 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 periodbefore it is analysed, it is essential either to keep the specimen refrigerated or to have apreservative in the collection container, so that unstable substances may be preserved and thegrowth of bacteria prevented. The preservative used will depend on the substance which is tobe determined.

Preservatives

Hydrochloric Acid

50 ml of 2 M HCl is adequate for a 24-hr specimen suitable in particular for the determinationof Ca, P, N, ammonia, VMA estimation. It can also be used when sodium, potassium, ureaestimations are required but it is not suitable for uric acid and protein estimation as these maybe 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 andprotein 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 ofbacteria but does not interfere with the analysis either by enzyme or reduction methods. 5 mlof a 5% aq solution is added to the container. Chloroform, light petroleum, thymol and formalinare used as urine preservatives. A few and one of these is put into a 2.5 l container and givenfor 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 tobe a compound of urobilin, urobilinogen and a peptide substance.

Odour Freshly passed urine has a characteristic of aromatic odour, said to be due to volatileorganic 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 rangeof 4.8-6.8. The pH is determined by simply dipping the pH indicator strips into the urine andcomparing the colour change of the strip against the colour guide supplied with the indicatorpapers.

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 urinometeras follows:

Mix urine well. Take urine in specific gravity cylinder. Then put the urinometer gently inthe urine. Take care it does not touch the wall of the cylinder. Take reading of the lower meniscusof 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.

56 Practical Biochemistry for Students 15Urinary Reducing Sugars

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Estimation of Reducing Sugars in Urine

Principle

Reducing sugars by virtue of free aldehydic or ketonic groups in their structure possess theproperty of reducing certain heavy metallic cations in an alkaline medium and in the processget oxidised to a mixture of sugar acids. This principle of oxidation-reduction is used in theestimation of reducing sugars.

Reducing sugars undergo enolisation when placed in alkaline solution. These enediol formsof 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 inBenedict’s quantitative reagent the colour change is from blue to colourless, which is sharpand 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 givesa continuous supply of Cu++ions.

4. Potassium ferrocyanide : To prevent the precipitation of Cu2O in the solution, i.e.K4Fe(CN)6

5. Potassium thiocyanate : Precipitates Cuprous ions (Cu+) as CuCNS, a white colourprecipitates.

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 dissolvecarbonate, citrate and thiocyanate in enough water to make about 800 ml of the mixture. Dissolvecopper sulphate in about 100 ml of distilled water with constant stirring. Add ferrocyanidesolution, cool and make the volume to 1 litre.

Urinary Reducing Sugars 57

Reaction

Reducing sugars OH– KCNSCu++ ________________________→ Cu++ _______________→ Cu(OH) ________________→ CuCNS

High temperatureK4Fe(CN)6

CuSO4 Cu2O

Procedure

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

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 colourchange is from blue to colourless which marks the end point of the titration. Note the volumeof 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 reducedby 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.

Burette

Conical flask

Wire gauge

Tripot stand

Burner

0

25

↑

↑


Let x ml of urine reduces 10 ml of Benedict’s quantitative reagent, which is equal to 20 mgof glucose.

So x ml of urine contains 20 mg of glucose

20 × 100∴ 100 ml of urine will contain ____________ mg of glucose

2x

= ____ gm of glucosex

Result2___ gm of glucose per 100 ml of urine, depending upon the volume of urine excreted per dayx the results are expressed accordingly.

Whatever may be the reducing sugar excreted in the urine, the results are expressed interms of glucose because of Benedicts’s quantitative reagent relation with glucose. The finalconfirmation regarding the nature of reducing sugar excreted comes through osazone formationwhere 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 istoo 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 excretedin glucose then the condition is called glucosuria.

a. Diabetes mellitusb. 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 glucoseestimation 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 galactose-l-phosphate uridyl transferase. Galactose conversion to glucose does not take place. Hence itappears in urine. The diagnosis is made by identifying the urinary sugar as galactose by paperchromatography.

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.

60 Practical Biochemistry for Students 16Urinary Chlorides

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Estimation of Chlorides in Urine

Principle

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

Reagents

1. N/10 AgNO32. Concentrated HNO33. Ferric alum indicator4. N/10 NH4CNS

Reaction

XCl + excess AgNO3 → AgCl↓ + left over AgNO3

Left over AgNO3 + NH4CNS → AgCNS↓ + NH4NO3white ppt.

FE2 (SO4)3 (NH4)2 SO4 24H2O Ferric alum. ↓

Fe2(SO4)3 + 2NH4CNS → 2Fe(CNS)3 + (NH4)2 SO4Red colourEnd point colour

Procedure

In a 100 ml volumetric flask, pipette 10 ml of urine. Add 5 ml of concentrated HNO3, followedby 20 ml of N/10 AgNO3. Make the volume to 100 ml with distilled water (do not use tapwater). 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. Add1 ml of ferric alum as an indicator. Titrate with N/10 NH4CNS solution from burette till thelight brown colour appears which is the end point of the titration.

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 NH4CNSSince 50 ml of filtrate consumes x ml of N/10 NH4CNSWhich 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 NaCl1 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 expressedaccordingly.

Since chloride excreted is an anion, the result can also be expressed in terms ofmilliequivalent.


mg% × 10mEq/L of chloride = ___________________________________________

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. Edemab. 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, morereabsorption of Na+ and Cl– ions takes place, hence output is decreased.

5. During pneumonia and other infectious disease, hypochloremia results from withdrawalof 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 6317Urinary Creatinine

(63)

Estimation of Creatinine in Urine

Principle

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

Reaction

Strong alkaliCreatinine + 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 solutioncontains 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 ineach flask as.

Reagent Test Standard BlankUrine 1 ml — —Creatinine std. — 1 ml —Picric acid soln. 2 ml 2 ml 2 mlNaOH (10 g%) 1 ml 1 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 tozero.

Observation

Optical density of test solution =Optical density of standard solution =


Calculation

Concentration of creatinine excreted per day in urine

Optical density of test Conc. of std.= __________________________________ × _______________________ × Volume of urine excreted per day

Optical density of std. Vol. of urine used

Interpretation

The normal daily excretion of creatinine in adults is from 0.4 g to 1.8 g. The amount of creatinineexcreted 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 clearedof the substance, each minute by both kidneys in order to account for its rate if excretion.

Ux VC = ________

Px

Where,Ux = Conc. of concerned

substance in urinePx = Conc. of that

substance in plasmaV = Outflow of urine in ml/min.

The clearance is a parameter for assessing the glomerular function of kidneys. The clearanceof that substance is measured which is neither reabsorbed nor secreted by tubules, givesglomerular 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 ureaclearance 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 urineconcentrations 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 roughimpression of glomerular damage.

66 Practical Biochemistry for Students 18Ascorbic Acid in Urine

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Principle

Vitamin C in urine is estimated by direct titration of the acid urine (in presence of acetic acidwhich inhibits the aerobic oxidation of ascorbic acid) with oxidation-reduction indicator 2, 6dichlorophenol indophenol, vitamin C reducing the indicator, which becomes decolourisedin the process.

Vitamin C saturation test: The excretion of vitamin C following a large dose gives an indicationof the state of body tissues in respect of vitamin C content. If the tissues are saturated, most ofthe ingested vitamin will appear in the urine but if there is relatively little vitamin C in thetissues most of the ingested vitamin will be retained by the tissues and little or none appearingin the urine.

In acid (pH approximately 3.0), the dye 2,6 dichlorophenol indophenol may exist in twoforms, as a red oxidised form or as a leuco (colourless) reduced form

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

O = C —|

O – C| O

O – C|

H – C —|

HO – C – H|

CH2OH

L-Ascorbic acid(Reductant)

L-Dehydroascorbic acid(Oxidant)

– 2H+, – 2e

O = = N — — OH HO — — N — — OH

Cl|

Cl|

Red colourOxidant at pH 3.0

ColourlessReductant at pH 3.0

+ 2H+ + 2e

Ascorbic Acid in Urine 67

Reaction

Collection of Urine Sample

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

Procedure

In a clean dry test tube, take 0.5 ml of 2,6 dichlorophenol indophenol indicator dye. Add adrop of glacial acetic acid. The colour of the solution which is blue turns to pink by the glacialacetic 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 iscompletely reduced by 0.02 mg of ascorbic acid.

x ml of urine contains 0.02 mg of ascorbic acid.

0.02 × 100∴ 100 ml of urine contains __________________ mg of ascorbic acid

2x

= ________ mg of ascorbic acid.x

Interpretation

The normal excretion of vitamin C per day is roughly half of the dietary intake. The amountexcreted per day is 20-30 mg.

Excretion of vitamin C less than 50% indicates unsaturation. In severe deficiency, the vitaminC excretion is nil.

O = C —|

HO – C|| O

HO – C|

H – C —|

HO – C – H|

CH2OH

O = C —|

O – C| O

O – C|

H – C —|

HO – C – H|

CH2OH

|N||

OH|

Cl — — Cl

||O

|NH

|

OH|

Cl — — Cl

||OH

_______________→+ +

L-Ascorbic acid 2,6 dichlorophenol L-Dehydro Colourless Leucobaseindophenol (Oxidised form) Ascorbic acid (Reduced form)

68 Practical Biochemistry for Students 19Serum Uric Acid

(68)

Uric acid is major product of purine nucleotide metabolism. Dietary nucleotides get convertedto 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 endogenousdegradation of nucleo acids.

In lower primates and mammals other than humans—they carry purine metabolism onestep further, i.e. uric acid to allantoin. It requires uricase. Approximately 75% of uric acid islost in urine and 25% is secreted by GIT. Degradation of purine nucleotide occurs in tissueswhich contain xanthine oxidase such as liver and small intestine. Uric acid is as such nonionizedbut at pH 7.4 exist as urate. In plasma, ECF and other fluids have 98% monosodium urate. At6.8 mg/dl plasma saturates with uric acid and potential for urate crystal to accumulate. It hasbeen seen that even at much higher concentration, precipitation may not occur because plasmacontains 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 intoglomerulus. 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 reducedto tungusten blue which is estimated colorimetrically at 710 nm.

Reagents

1. 10% sodium tungstate2. 2/3 N sulphuric acid3. Tungstic acid4. Phosphotungstic acid5. 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 shakingcentrifuge and mix well and wait for 30 min. Read coloured solution within the next 30 min ina colorimeter at 710 nm or using red filter.

Serum Uric Acid 69

Blank Std. Test1. Distilled water 3.0 ml – –2. Standard(1 mg%) – 3.0 ml –3. Filterate – – 3.0 ml4. Na2CO3 0.6 ml 0.6 ml 0.6 ml5. Phosphotungstic acid (PTA) 0.6 ml 0.6 ml 0.6 ml

Mix and keep for 10 minat room temp.Take OD at 710 nm.

Normal values

Children – 2-4 mg/dlMale – 4.4-7.6 mg/dlFemale – 2.3-6.6 mg/dl

Other method of estimation: Uricase method

UricaseUric acid + 2H2O + O2

_______________→ Allantoin + H2O2 + CO2

Urate ion has characteristic absorption peak in UV range. The reaction products allantoinand 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 singledisease but is used to describe a number of disorders with crystals of monosodium uratemonohydrate, derived from hyperuricemic body fluids give use to inflammatory arthritisand renal disorders.

2. Hypouricemia is reached when uric acid level reaches to less than 2 mg/dl because ofdecreased production of urate and is indicative of renal diseases, neoplastic diseases,diabetes mellitus and deficiency of xanthine oxidase.

70 Practical Biochemistry for Students 20Colorimetry

(70)

Colorimetric procedures are widely used in the hospital laboratory because they are easy toperform, 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 estimatedtakes part quantitatively. Within certain limits the colour intensity of a solution is proportionalto the concentration of the reacting substances and it is possible to obtain a measure of theconcentration 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’swhich are as follows.

Lambert’s (Bouguer) Law

It states that the proportion of light absorbed by a substance is independent of the intensity ofthe incident light.

Beer’s Law

It states that the absorption depends only on the number of absorbing molecules throughwhich 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 solutionof an absorbing substance of concentration C (expressed as gm mole/litre), then if the amountof light absorbed be dI.

dI= ______ – KC dlI

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.

Ilog _______ = – KCI

Io

Ior log _______ = KCI

Io

Optical density = KCI

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

Colorimetry 71

Optical Density (OD) Absorbance

Optical density also called extinction is the logarithmic ratio of the incident light to that ofemergent light.

IoOD = log _______

I

Transmission (T)

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

IT = _______

Io

Relationship between optical density and transmission:

IoOD = log _________

I

I= log ________

T

100= log _________

%T

OD = 2 – log %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 havethe reading between 0.10 to 0.60 by selecting those suitable methods of investigations whichgive the deflection on the OD scale between 0.10 to 0.60, which is the most sensitive region ofthe OD scale.

The plotting of optical density-concentration of unknown is linear whereas plotting oftransmittance-concentration is not.

Theory

A beam of light passing through a coloured solution is partly absorbed. The amount ofabsorption will depend upon the number of coloured ions or molecules in its path. If thenumber of coloured particles in the solution is doubled, all other things being equal, twice asmuch light will be absorbed. If this increase in colour is due to double the concentration of thesubstance it will be seen that there is a direct relationship between the light absorption and theconcentration of coloured substance dissolved. This relationship is expressed mathematicallyby 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 ColorimetryIn visual colorimetry, the intensity of light entering the two solutions, i.e. standard solutionand 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 theintensities of light emerging from two solutions are same, i.e. the two colours match eachother. 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).

Iolog ________ = KCs Rs = KCu.RuI

Cu.Rs = Cu.Ru

Rs × CsCu = ______________Ru

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

The eyepiece is an optical device that brings the two coloured fields next to each other foreasy comparison. In making a measurement the length of one column is set and the length ofthe unknown is varied until the intensities of colour transmittance as viewed in the eyepieceare equal, i.e. the two fields match into a single colour. The length of the unknown column isnoted. Since the concentration of the standard solution is known, the concentration of unknowncan be calculated.

Rs × CsCu = ______________Ru

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

12345678123456781234567812345678123456781234567812345678

12345671234567123456712345671234567123456712345671234567

1234567890123456789012345678901212345678901234567890123456789012

LENS

EYE

s s

s s

s s

s

s

Colorimetry 73

2. Photoelectric Colorimetry

Principle

In the photoelectric colorimeter (photometer) a photoelectric cell is used instead of eye tomeasure the intensity of light transmitted. The depth of the solution is kept constant, theintensity of light transmitted through the solution being measured directly in terms of electriccurrent produced by the photoelectric cell. Since the voltage of this current is directlyproportional to the intensity of light falling on the cell, the output of current is inverselyproportional 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 madeof glass or gelatin). The light transmitted by the filter is allowed to fall upon the colouredsolution when some light is absorbed, the amount of light absorbed is proportional to theintensity of the colour. The fraction of light not absorbed is transmitted through the solutionand is made to fall upon a photo cell in which photoelectric current is generated which isproportional to the light transmitted through the coloured solution. The current falls on asensitive galvanometer and although deflection is due to the current generated by the lighttransmitted by the solution, the scale is calibrated in such a way that the reading representsthe light absorbed by the solution and the readings are proportional to the intensity of allcolour. 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 thereagents used are coloured they will also absorb light and give rise to high value of opticaldensity 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 andthen subjected to similar conditions as in test standard.

Role of Different Filters in Colorimetric Estimations

The choice of filter is made so that its maximum transmission is at a wavelength correspondingto the maximum absorption of the coloured solution. This means that in general the mostsuitable filter should have a colour complementary to that of a solution. For example, a purplesolution requires a green filter, orange red solution a blue or blue-green filter, a yellow solutiona blue filter and blue solution a red filter.

The accuracy of visual colorimeter is limited by the inability of the eye to detect smalldifference in the intensity of light because matching of the colour is done by eye. This can beeliminated in the photoelectric colorimeter since the photo cell is affected only by the intensityof light that strikes the cell and the current generated by the cell is directly proportional to theintensity of light.

Voltage stabilizer

+

Mains

–

Ú

Tungstenfilament

Ú

Ú Cuvette

Ú

FILTER

74 Practical Biochemistry for Students 21Blood Sugar

(74)

The normal blood sugar level is 80-100 mg per 100 ml of blood. But the normal level of bloodsugar 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 otherthan sugars such as glutathione, ascorbic acid, uric acid, creatine, etc. are also estimated bythis 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-reducingsubstance. 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 thereagent 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 isthe reduction method of King and Asatoor.

Principle

Cupric ions furnished by copper sulphate are reduced by reducing sugar in alkaline mediumto cuprous ions at high temperature. Cuprous ions react with phosphomolybdic acid to give amolybdenum blue colour, the intensity of which is proportional to the concentration of reducingsugar 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 pathwaywhere 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.

Blood Sugar 75

Sugar B : It is prepared by dissolvingi. Sodium bicarbonate 50 gm

ii. Sodium carbonate 40 gmiii. Potassium oxalate 36.8 gm

In 400 ml of distilled water warm the solution then add 24 gm of sodium-potassiumtartarate and make the volume to 1 litre with distilled water.

Working solution is prepared by mixing equal volume of sugar A and sugar B.2. Isotonic copper sulphate3. Sodium tungstate (10g%)4. Sulphuric acid (2/3N)5. 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 cooland add 250 ml of phosphoric acid. Make the volume to 1 litre with distilled water.

6. Stock glucose standard (100 mg%): Prepared by dissolving 100 mg of glucose in saturatedsolution of benzoic acid and making the volume to 100 ml.

7. Working glucose standard (10 mg%): Prepared from stock glucose solution by 1:10 dilution.

Procedure

i. Precipitation stepIn a centrifuge tube, pipette7.4 ml isotonic CuSO4 (to prevent haemolysis of blood)0.2 ml blood0.4 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 stepLabel the test tubes in the order given below:

Standard

Test (T) S1 S2 S3 S4 Blank (B)

Supernatant 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

Add 2 ml of alkaline copper sulphate in all the test tubes. Plug the mouth of the testtubes with cotton and place them in the boiling water bath for 10 minutes. After 10 minutestake 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 add8 ml of distilled water.Mix again. Wait for 10 minutes.

Read optical density at 650 nm setting blank to zero.


Observation

Optical density of test sample (T) =Optical density of standard (S1) = , conc. (S1) = 0.025 mg.Optical density of standard (S2) = , conc. (S2) = 0.05 mg.Optical density of standard (S3) = , conc. (S3) = 0.075 mg.Optical density of standard (S4) = , conc. (S4) = 0.10 mg.

Calculation

Optical density of test Conc. of standard________________________________________ × ________________________________________ × 100 mg%Optical density of standard Effective volume of blood used

Effective volume of blood for colour development7.4 + 0.4 + 0.2 = 8 ml of the solution contains 0.2 ml blood

2 ml of the solution will contains 0.05 ml bloodSo 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 mellitus2. Hyperactivity of thyroid, pituitary and adrenal glands3. Infectious diseases4. Meningitis, encephalitis5. Shock6. Severe hemorrhage.

Decreased level of blood sugar is termed as hypoglycemia. Values below 60 mg% indicatehypoglycemia. 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) 7722Glucose Tolerance Test (GTT)

(77)

Glucose tolerance test determines the degree and duration of hyperglycemia after an oralintake of known quantity of glucose. After the ingestion of food there is a temporary rise inblood sugar level upto 160-180 mg% which then returns to normal fasting level within 2-3hours after food.

This effect can be studied under standard conditions by means of glucose tolerance testwhich is helpful in investigating abnormalities of carbohydrate metabolism, and cases in whichglycosuria has been found.

Method

1. The patient should be in a postabsorptive state, i.e at least 12 hours after last meal.2. Fasting blood sample and urine specimen is collected.3. The patient is given 50 gm of glucose dissolved in 100-200 ml of water.4. 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 shouldreturn to normal.

The zero hour fasting blood glucose value is within normal range and the maximum bloodglucose value is reached 1 hour after taking glucose and returns to the near fasting valves after2 hours always. At no stage the blood glucose values exceed 180 mg% (renal threshold forglucose).

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 littlelower than in completely normal person. The blood glucose follows the normal pattern, butglucose 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 sufficientcarbohydrate is taken to raise their blood glucose temporarily above their lowered renalthreshold.

Carbohydrate metabolism is normal but because of some abnormality in the tubularreabsorption of glucose, an appreciable amount of glucose escapes in the urine. This conditionis called renal glycosuria.

This type of curve is harmless and patients are not likely to develop into diabetes later on.


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

Diabetic Curves

UrineGlucose + + + + +

400 —

300 —

200 —

150 —

100 —

50 —

| | | |0 0.5 1.0 1.5 2.0

Severe

UrineGlucose – – – – –

200 —

150 —

100 —

50 —

| | | |0 0.5 1.0 1.5 2.0

Normal Curve

RenalThreshold


200 —

150 —

100 —

50 —

| | | |0 0.5 1.0 1.5 2.0

Mild

UrineGlucose – – + – –

200 —

150 —

100 —

50 —

| | | |0 0.5 1.0 1.5 2.0

Lag Curve


200 —

150 —

100 —

50 —

| | | |0 0.5 1.0 1.5 2.0

Flat Curve

•

•

• •

•

200 —

150 —

100 —

50 —

| | | |0 0.5 1.0 1.5 2.0

Renal Glycosuria

RenalThreshold

UrineGlucose – – + – –

•

••

•

•

•

••

•

•

••

•

•

•

•

•

• ••

•

•

• • •

•

•

•

•

•

•

•

•

•

•

Glucose Tolerance Test (GTT) 79

and fasting level may or may not be raised. A true diabetic will show a fasting level higherthan normal, a rise much above renal threshold and fall to fasting level may be achieved evenafter 5 hours. They may or may not be glucosuria in the fasting urine but the postglucose urinedefinitely 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-hourvalue is within normal limits or often low. This implies a delay in early compensatorymechanisms (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 theblood glucose level at peak of the curve is above the renal threshold level, glucose appears inthe 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 absorptionand 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 foundin completely healthy individuals and interpretation is difficult.

This type of curve indicates malabsorption, hypothyroidism, hypoadrenalism,hypopituitarism.

80 Practical Biochemistry for Students 23Blood Urea

(80)

Urea is a major end product of protein metabolism in human body. Urea is synthesised in theliver 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 NPNsubstance excreted in the urine.

The normal blood urea level is 15-40 mg%. It is slightly higher in men than in women. Ureacontent over a period is influenced by the amount of protein and tends to be lower on a lowprotein 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 × 2.14Urea 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 forblood 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 enzymeurease (found in soya beans and jack beans) followed by colour development with Nesslaerreagent.

Whereas in diacetyl monoxime method, urea reacts directly with diacetyl monoxime understrongly acidic conditions to give a yellow coloured product. The reaction is intensified by thepresence of ferric ions and thiosemicarbazide which gives red/pink coloured complex whichis more linear than the yellow one.

Reagents

1. Sodium tungstate (10g%): prepared by dissolving 10 gm of sodium tungstate in waterand 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 thevolume to 100 ml with distilled water.

Blood Urea 81

4. Sulphuric acid—phosphoric acid reagent: 150 ml of 85% phosphoric acid is added to 140ml 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 centrifugeArrange the test tubes and mark them blank ‘B’, test ‘T’and standards (S1) and (S2).

B S1 S2 T

Supernatant (from test) — — — 1 mlWorking urea standard — 0.5 ml 1.0 ml —Distilled water 2 ml 1.5 ml 1.0 ml 1 mlDiacetyl monoxime 0.4 ml 0.4 ml 0.4 ml 0.4 mlH2SO4-H3PO4 mixture 1.6 ml 1.6 ml 1.6 ml 1.6 ml

Mix well and place all the test tubes in the boiling water for minutes. Cool it and Readoptical 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.


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 anobstruction 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 8324Urea Clearance

(83)

Clearance of any substance is defined as the number of ml of plasma which contains the amountof 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 urineV = ml of urine excreted per minuteP = Concentration of the substance in plasma.

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

UVclearance = ____________

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 aminute by the kidneys.

mg of urea excreted per minuteUrea clearance = ______________________________________________

mg of urea per ml of blood

UV= ____________

B

U = mg urea per 100 ml of urineB = mg urea per 100 ml of bloodV = ml of urine excreted per minute.

The normal urea clearance was determined on a number of normal persons and valuefound to an average 75 ml/min, when the rate of excretion of urine was 2 ml or more perminute.

Urea clearance was found to decrease with V, fell below 2 ml/min. With a urine volume of1 ml/min the clearance was found to average 54 min/ml and the decrease with decreasingurine volume was seen to be proportional to the square root of V. Hence for value between 0.5to 2 ml per minute; the average normal clearance is given by 54√V.


For value of V equal to 2 ml or more than the term is called maximum urea clearance.

Maximum urea clearance Observed urea clearance= ________________________________________________________ × 100(% of the average normal) Average number maximum urea clearance

UV= ______________ × 100

B= 100

75

100 UV= _________ _________

75 B

UV= 1.33 _________

B

For values of V less than 2, then the term is called standard urea clearance.Standared urea clearance UV

(% of the average normal) = ___________ × 100B

54√ VU√ 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 atwo- hour period and can be completed before 12 noon. During the period of the experimentthe patient should be at rest in bed and urine collected directly in a clean container. Timeshould 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 specimenof urine which is collected. At the same time a blood sample is also collected. One hour latera second urine sample is also collected. Also blood sample is collected. The volume of eachurine 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 thekidneys.

Urea clearance Renal function

Above 70 Normal40-70 Mild deficit20-40 Moderate deficitBelow 20 Severe deficitBelow 5 Coma

Blood Cholesterol 8525Blood Cholesterol

(85)

Cholesterol is present in blood both as free and as esters. Free cholesterol normally formsabout 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 arecentrifuged. The protein free extract is evaporated to dryness. The cholesterol residue isdissolved in chloroform and is measured colorimetrically by Liebermann-Burchard reaction.

Reagents

1. Absolute alcohol2. Diethyl ether

Alocohol-either mixture 3.1

3. Chloroform4. Acetic anhydride5. Standard solution of cholesterol (100 mg%)

Dissolve 100 mg of cholesterol in some amount of alcohol. Slightly warm (in water bath), ifrequired. 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 keptin a boiling water for the evaporation of solvent and the residue, i.e. cholesterol sticks to thebottom of the flask is left behind. Chloroform, acetic anhydride and sulphuric acid must be ofhighest quality. It is particularly important that chloroform be especially anhydrous. Ordinarychloroform 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 cholesterolsolution. All should be kept in boiling water bath for evaporation of solvent till the solventevaporates and residue is left behind.

Blank: Clean dry test tube.

}


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

Concentrated H2SO4 0.1 ml 0.1 ml 0.1 ml 0.1 ml 0.1 ml 0.1 ml

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 = , conc. = .02 mgOptical density of standard S2 = , conc. = .04 mgOptical density of standard S3 = , conc. = .06 mgOptical density of standard S4 = , conc. = .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 conditions1. Diabetes mellitus.2. Obstructive jaundice.3. Nephrotic syndrome.4. Cirrhosis of liver.5. Hypoparathyroidism.6. Xanthomatosis.Hypocholesterolemia is observed in the following1. Hyperthyroidism2. Pernicious anaemia3. Haemolytic anaemia4. Malabsorption syndrome.

Serum Calcium 8726Serum Calcium

(87)

Calcium is present in the body in larger amounts than any other cation. Almost all of it ispresent 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 inthe 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.

N4. ________ KMnO4 solution.

100

Principle

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

Collection of Sample

Blood is collected in a plain container (i.e. without any anticoagulant) and allowed to clot andserum 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.


Reaction

Serum calcium + COONH4___________________→ COO

| | CaCOONH4 COO

(saturated solution Calcium oxalateCOO COOH| Ca + H2SO4

_______________→ | + CaSO4COO COOH2KMnO4 + 3H2SO4

_______________→ K2SO4 + 2MnSO4 + 3H2O + 5OCOOH| + O _______________→ 2CO2 + H2OCOOH

Procedure

In a centrifuge tube take, 2 ml of serum, 2 ml of distilled water and 1 ml of saturated ammoniumoxalate. Mix and keep it for 30 minutes with frequent mixing in between.

Centrifuge for 15 minutes and decant off the supernatant fluid without disturbing theprecipitate. After doing so, allow the tubes to stand inverted on a filter paper and allow todrain for five minutes. Wipe the mouth of the tube.

Now wash the precipitate with 4 ml of dilute ammonia. Centrifuge and discard thesupernatant as before. Repeat the washing twice. After washing dissolve the precipitate in 2ml of normal sulphuric acid. Warm by placing the tube in a beaker having the temperaturebetween 75° to 85°C so that the precipitate goes into the solution. Titrate it with N/100 KMnO4while 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 mlVolume 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 × xSo 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 slightlyhigher 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 inorganicphosphorus and serum proteins.

Serum Calcium 89

Low serum calcium is observed in the following:i. Hypoparathyroidism

ii. Osteomalaciaiii. Ricketsiv. Renal failurev. Tetany

vi. During an attack of acute pancreatitisvii. Steatorrhoea

Low serum calcium is often associated with high serum inorganic phosphorus.High serum calcium is observed in the following:

i. Hyperparathyroidismii. Hypervitaminosis D

iii. Multiple myelomaiv. Neoplastic diseases of bones.

Serum calcium is often increased following prolonged excessive intake of milk, alkali andantacids, the treatment for peptic ulcer.

90 Practical Biochemistry for Students 27Inorganic Phosphorus

(90)

Phosphorus is present in blood as:i. Inorganic phosphorus

ii. Ester phosphorus, i.e. organic phosphorusiii. Lipid phosphorus.

Of these inorganic phosphorus is determined most frequently.

Principle

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

Reactions

Inorganic phosphorus + Molybdate ______→ PhosphomolybdatePhosphomolybdate + Reducing agent ______→ Μolybdenum blue

Reagent

1. Trichloroacetic acid 10 gm%2. Ammonium molybdate 5 g%3. Perchloric acid4. 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, α-aminonaphthol sulphonic acid, 12 gm of sodium metabisulphite, 2.4 gm of sodium sulphitein distilled water and make the volume to 100 ml with distilled water. Store the reagent indark at cool place.

5. Standard phosphorus solution (1 mg/ml): Dissolve 438 mg of potassium dihydrogenphosphate in 100 ml of water, working standard (0.004 mg/ml). Prepared by diluting 1ml 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.

Inorganic Phosphorus 91

Test S1 S2 S3 S4 S5 Blank

Filtrate 5 ml — — — — — —Working phosphorus standard — 1 ml 2 ml 3 ml 4 ml 5 ml —Distilled water — 4 ml 3 ml 2 ml 1 ml — 5 ml

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

Precaution

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

Observation

Optical density of test =Optical density of standard S1= , conc. = 0.004 mg.Optical density of standard S2= , conc. = 0.008 mg.Optical density of standard S3= , conc. = 0.012 mg.Optical density of standard S4= , conc. = 0.016 mg.Optical density of standard S5= , conc. = 0.02 mg.

Calculation

The concentration of inorganic phosphorus is given by

Optical density of test Concentration of blood_____________________________________ × ____________________________________________ × 100 mgOptical 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 childrenwhose normal range is from 4 to 6 mg%.

Increased serum phosphorus levels may be found in:i. Hypervitaminosis D-↑Ca ↑P

ii. Hypoparathyroidism-↓Ca2+ ↑Piii. Renal failure-↓Ca2+ ↑P

Low serum phosphorous levels may be found in:‘i. Rickets-↓Ca2+ ↓P

ii. Osteomalacia-↓Ca2+ ↓Piii. Hyperparathyroidism-↑Ca2+ ↓Piv. Fanconi's syndrome

92 Practical Biochemistry for Students 28Serum Total Proteins andAlbumin: Globulin Ratio

(92)

Serum Proteins

It can be estimated by the number of methods such asa. Colorimetricallyb. Precipitation methodc. Microkjeldahl methodd. Electrophoretically.

The normal total serum proteins level is 5.5 to 7.5 gm%, albumin 3.5-5.5% and globulin 2-2.5g%.

Liver is the sole organ chiefly responsible for the formation of plasma albumin and at least80% of the plasma globulins, i.e. α-and β-globulins.

Principle

Proteins and peptide bonds react with alkaline copper sulphate (Biuret reagent) to give aviolet coloured coplex. The intensity of violet colour is measured colorimetrically and isproportional to the concentration of the total proteins in the sample.

Similarly globulin is precipitated by addition of sodium sulphate. The albumin left in thesupernatant is treated with Biuret reagent and violet colour produced is measuredcolorimetrically as above.

Reagents

1. Normal saline2. Biuret reagent3. 28% sodium sulphite4. Ether.

Procedure

For total proteins: Serum for total protein estimation is 1:10 diluted, i.e. 0.2 ml serum and 1.8ml normal saline.

Test S1 S2 S3 S4 S5 Blank

Serum (1:10 diluted) 0.5 ml — — — — — —Bovine albumin standard — 0.2 ml 0.4 0.6 0.8 1.0 —Normal saline 2.5 2.8 2.6 2.4 2.2 2.0 3

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= , conc, = 1.6 mg.Optical density of standard S2= , conc, = 3.2 mg.Optical density of standard S3= , conc, = 4.8 mg.Optical density of standard S4= , conc, = 6.4 mg.Optical density of standard S5= , conc, = 8 mg.

Calculation

The concentration of serum total proteins is given byOptical density of test Concentration of standard

_______________________________________ × __________________________________________ × 100Optical 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) assuch. 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 30minutes.

Centrifuge for 10 minutes. Carefully detach the globulin button formed at the interphaseof water: either layer and using a capillary dropper, take out the supernatant and transfer itinto another tube marked albumin.

Test Standard Blank

Supernatant 3 ml As in total protein —Normal saline — 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 Concentration of standard________________________________________ × _____________________________________________ × 100

Optical density of standard Effective volume of serum used


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 ini. Nephrotic syndrome

ii. Malnutritioniii. Kwashiorkoriv. Cirrhosis of liverv. Liver disorders

High protein values are observed in:i. Multiple myeloma

ii. Conditions associated with high globulin concentrationThe 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 liverii. Nephrotic syndrome

iii. Malnutritioniv. 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 9529Serum Bilirubin

(95)

Determination of Total and Conjugated Bilirubin in Serum

Any increase of bilirubin in the blood is an indication of jaundice (derived from French wordjaune 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 isexcreted 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 in-

soluble in water. It is soluble in alcohol that is why van den Bergh called the form ofbilirubin which reacted in the presence of alcohol is indirect (Unconjugated).

Total bilirubin is measured by adding methanol which permits reaction of direct and indirectbilirubin 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 10-30seconds 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 : 0.2 to 0.6 mg/100 ml• Total bilirubin in infants may go upto: 1.2 mg/100 ml• Conjugated bilirubin : 0.0 to 0.2 mg/100 ml

Serum Collection

Collect blood in a clean container without any anticoagulant and allowed to clot. Separateserum from the clot as soon as possible taking care to avoid hemolysis. Protect specimen fromlight.

If the test cannot be carried out immediately the sample should be frozen.


Principles

Direct Conjugated bilirubin couples with diazo reagent (diazotised sulphuric acid formingazobilirubin, a red-purple coloured product in acidic medium.

Indirect Unconjugated bilirubin is diazotised only in the presence of methanol. Thus the red-purple coloured azobilirubin produced in presence of methanol originates from both directand indirect fractions and thus total bilirubin concentration can be found out. The differenceof total and direct bilirubin gives indirect (conjugated) bilirubin result.

The intensity of red colour so developed above is measured colorimetrically and isproportional to the concentration of the appropriate fraction of bilirubin. The reaction can berepresented as.

ReactionH+

Bilirubin + Diazotised sulphanilic acid _________________→ Azobilirubin

H+

Red purple colour

Reagents

1. Diazo A: It contains sulphalinic acid dissolved in concentrated hydrochloric acid. DiazoA is prepared by dissolving 1 gm of sulphanilic acid in 15 ml of concentrated hydrochloricacid 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 preparedfreshly.

3. Diazo reagent: This reagent is prepared freshly before use by adding 10 ml of diazo A to0.3 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 mlwith chloroform.

7. Working bilirubin standard (2 mg%). Dilute the above bilirubin standard solution 1:5with 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.

Blank Standard Test Test

(For direct bilirubin) (For folic bilirubin)

Serum — — 0.2 ml 0.2 mlWorking bilirubin standard — 0.2 ml — —distilled water 2 ml 1.8 ml 1.8 ml 1.8 mlDiazo reagent — 0.5 ml 0.5 ml 0.5 mlHydrochloric acid 0.5 ml — — —Methanol 2.5 ml 2.5 ml 2.5 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) Concentration of standard ___________________________________________ × ___________________________________________ × 100

Optical density of standard Volume of serum used

Same formula can be used determining the concentration of conjugated bilirubin forconjugated 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 isless 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 thehepatocyte 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.

98 Practical Biochemistry for Students 30Prothrombin Time

(98)

Prothrombin time is the time required for clotting of citrated plasma to which optimum amountsof 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 3rdstages 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 3rdstages of extrinsic clotting system (Factors II (Prothrombin), V,VII, X or fibrinogen) will prolongthe 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 distilledwater.

Collection of blood

Blood is collected in 3.8% sodium citrate solution as anticoagulant. The ratio of anticoagulantto 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 forcontrol.

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 0.2ml 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 fibrinclot 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 asprothrombin index and prothrombin activity.

Prothrombin time of control plasmaProthrombin index (%) = ___________________________________________________

Prothrombin time of test plasma

Prothrombin time of control plasmaProthrombin ratio = ________________________________________________________

Prothrombin time of test plasma

100 Practical Biochemistry for Students 31Liver Function Tests

(100)

The functions of liver are many, and the term liver function tests is something of a misnomersince there is no single test of liver function. The best test of liver function depends on whatfunction of the liver is being tested. Several liver function tests should be performed as asingle 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 commonlyemployed have got significance for assessing the conditions of patients.

The first group of tests regarding the secretory, excretory and enzymatic functions areserum bilirubin estimation, bilirubin and urobilinogen in urine, BSP excretion test, serumalkaline phosphatase estimation and SGPT.

The second group meant for assessing the protein synthetic function are total proteinsestimation, A/G ratio and prothrombin time.

The final and third group that are meant for liver metabolic functions are; estimation ofserum 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. Inthis process X-amino acid is converted to its corresponding X-ketoacid.

α-keto- ALT or PyruvateAlanine + gluteric _________→ +

acid (SGPT) glutamate

α-keto- AST or OxaloacetateAspartic + gluteric _________→ +acid acid (SGOT) glutamate

Transamination reactions are important in intermediary metabolism because of their func-tion in synthesis and degradation of amino acid.

SGPT is found in hepatocytes in highest concentration. It is considered to be more liverspecific. SGOT is found in cardiac tissues, skeletal muscles, pancreas and kidney.

Normal values1. SGPT—3-15 IU/L at 25°C2. SGOT—4-17 IU/L at 25°C

Liver Function Tests 101

Diagnostic Significance

If the level is raised in serum, it indicates damage to tissues, so clinically its use is in evaluationof damage to tissues conditions in which SGPT is raised are hepatic disorders (like viralhepatitis). 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 isolatethem, so, a convenient way to quantitate them is to measure their catalytic activities and relatethis 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 colourand intensity of this colour is measured at 550 nm. The intensity of colour is directly propor-tional to activity of SGPT, NaOH is added to stop the reaction.

Calculation

ODT—ODCPlot 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 S1 S2 S3 S4

Standard pyurvate 0.5 ml 1.0 ml 1.5 ml 2.0 mlDistilled water 2.0 ml 1.5 ml 1.0 ml 0.5 ml –DNPH 0.5 ml 0.5 ml 0.5 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 activity1. End point2. Kinetic

— multipoint— continuous monitoring

1. End PointReactants are combined and reaction proceeds for designed time (SGPT—30 min.) and thenreaction is stopped.

2. Kineticsa. 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 willconvert 1μ mole of reactant into product per minute per litre.


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

Reagents

In Kindlking method following reagents are used.1. Substrate : Disodium phenyl phosphate2. Buffer : NaHCO3 + Na2CO3 (pH = 10)3. Phenol standard : 0.01 mg/ml4. NaOH stops the reaction5. 4-amino antipyrene6. Potassium ferricyanide oxidising agents

Test Control Std. Buffer

1. Substrate 1.0 ml 1.0 ml – –2. Buffer 1.0 ml 1.0 ml 1.1 ml 1.1 ml3. Phenol std – – 1.0 ml –4. Distilled water – – – 1.0 ml5. Serum 0.1 ml – – –

Incubate at 37°C for exactly 15 min6. NaOH 0.8 ml 0.8 ml. 0.8 ml. 0.8 ml7. Serum – 0.1 ml – –8. 4-amino antipyrene 1.0 ml 0.8 ml 1.0 ml 1.0 ml9. Potassium ferricyanide 1.0 ml 0.8 ml 1.0 ml 1.0 ml

Take OD at 540 nm after 10 min

Chemistry

Disodium pH = 10phenyl ___________________→ PhenolPhosphate AlkPhos |

↓ 4-amino antipyrene

Red colouredquinone.

Calculations

= ODT – ODC 100

× 0.01 ×ODS – ODB 0.1

= ODT – ODC × 0.10ODS – ODB

= ODT – ODC × 60 KA

KAU = Liberation of 1 mg of phenol in 15 min for 100 ml serumKAU × 7.1 = IV

= 3-13 KAUNormal value is 3-13 KAU is 2.5 times more in children.

Liver Function Tests 103

Causes of increased activitya. Physiological

1. Infants2. Pregnancy

b. Pathological1. Bone disorders

e.g. osteomalacia ricketsPaget's disease

2. Liver diseasesObstructive jaundiceIntrahepatic cholestasisLiver cancer

3. Certain drugse.g. steroids antidioxics

Causes of decreased activity1. Anaemia2. Kwashiorkor3. Scurvy

104 Practical Biochemistry for Students 32Demonstrations

(104)

ELECTROMETRIC DETERMINATION OF pH

The pH of solutions can be determined more accurately by potential measurements of certainelectrodes than by the use of indicators. They give rapid and accurate results. Common electricalmethods 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 intosolution in the forms of positive ions. When a strip of silver, copper or silver is placed in wateror aqueous solutions, an electrical double layer is thus set up at its surface. After a time therate at which Zn++ ions from the solution combine with electrons on the surface is equal to therate 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 solutionhas a fixed value. The magnitude of the electrical attraction between the two, determines thepotential of the metal in the solution and the tendency for zinc to zinc ions in solution isknown as solution tension of the metal. Potential developed depends upto the metal and alsoupon the concentration of metal ions in solution.

Electrode potential magnitude is in the order of their arrangement in the electrochemicalseries, 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 coatedwith patinum black absorb H2 gas. A Pt wire welded to the electrode makes contact with theouter circuit, the Pt strips is surrounding by glass tube with inlets and outlets for H2, which isadmitted at 1 atmosphere. Presume while the electrode is in a dilute solution of an acid.

H2_______→ 2H _______→ 2e– + 2H+

(On Pt surface) (Remain on Pt) (Pass into solution from electrode)

Since H2 is admitted at constant pressure, the solution tension of hydrogen atoms has aconstant value. Thus the reaction to the right is fixed at a constant rate and the equilibriumwill shift accordingly to the H+ ions concentration of the surrounding medium so that it can beused to measure pH.

If electrode 1 is maintained constant by immersion 1 N H+ solutions is called normalhydrogen electrode and is arbitrarily assigned a potential, Eno, of zero under all conditionsand is used as a standard reference for other electrodes.

emf = En + Eno

Demonstrations 105

If potential difference between the normal H2 electrode and the electrode in the unknownsolution is known for a given temperature, pH of the solution can be calculated from theformula.

emfpH = _____________________ , 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 solutionbut for a given temperature and KCl concentration, the potential of the calomel electrodeagainst the normal hydrogen electrode is constant, e.g. at 25°C potential of saturated calomelelectrode is +0.2458 volts. Now if a hydrogen electrode is placed in the solution of unknownpH and connected with a saturated calomel electrode, the potential difference registered willbe 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 hydrogenelectrode.

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 itcannot be used in oxidising solution.

Glass Electrode

The method determining pH is rapidly replacing the H2 electrode procedure. It is not effectedby oxidising or reducing agents. It is based on the principle that when a glass membraneseparates two different solution differing in pH, a potential difference is found to exist betweenthe two surfaces of the glass.

It consists of a thin walled glass bulb made out of a special type of low melting point. It isfilled with normal HCl solution in contact with Ag/AgCl electrodes. The platinum wire dippingin the electrolyte passes out of the glass tube at the bulb is placed in the solution of which pHis 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 dippinga Ag/AgCl electrode. The bulb is fragile as it is made up of a thin layer of glass. The glasselectrode and calomel electrode both dip in a solution of unknown pH. The electrodes areconnected by potentiometer.

Switch on the instruments and let it stabilise for 10 minutes. Take double distilled waterand read the pH. It should be zero or else adjust to zero. Then adjust the pH of the standardbuffers 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 solutionwhere pH is to be found out. Take the reading as the instrument is directly calibrated in pHunits. 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 convertedinto ammonium sulphate. From the ammonium sulphate so formed, the ammonia is distilledoff 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 backcalculation the nitrogen content is found out.

Reaction

DigestionNitrogen containing substance ______________________________→ (NH4)2SO450% H2SO4

Distillation40% NaOH

Boric acidTashiro’s indicator

Boric acid: NH3 complex ←__________________________ NH3(Green colour) (Violet colour)

N/70 H2SO4 Titrate

Violet colour

Procedure

The whole procedure for nitrogen estimation is divided into three parts:a. Digestionb. Distillationc. Estimation

a. Digestion

A known weight or the volume of the organic compound is taken in a small microKjeldahlflask, followed by 2 ml of concentrated sulphuric acid. To this a pinch of CuSO4 (which acts asa catalyst) and K2SO4 (it raises the boiling point and prevents bumping) are added. The mixturenow looks black. The flask is placed on a microburners and is heated slowly over a smallflame.

In the first stage, a low flame is used and later on a strong flame is used till the material iscompletely 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 ammoniumsulphate.

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

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

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Boric acidindicatorBurner

Water jacketDistillation

jacket

→Inlet

→Outlet

Vacuum

Trap

Steamgeneration

Clamp


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 yellowishorange colour formed by methyl red combine with methylene blue and a green colour resultsand 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 complexgreen colour

Indicator

AlkalineViolet Green

AcidicCalculations

Proteins contain 16% nitrogen.i.e. 16 gm of nitrogen is present in 100 gm of proteini.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-NH31 litre of 1 N-H2SO4 ≡ 14 gm of nitrogen1 ml of 1 N-H2SO4 ≡ 14 mg of nitrogen1 ml of N/79 H2SO4 ≡ 0.2 mg of nitrogenLet 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 × x1 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. Only process by which nitrogen in a mixture can be estimated.2. Percentage purity of nitrogen containing substance can be determined.3. Useful for the estimation of nitrogen in fertiliser, drugs and foodstuffs.4. 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 consistencyof pH has to maintained within very narrow limits in our body. Even the slightest change canlead to fatal conditions. This is done by electrolytes. The important anions are chloride andbicarbonates 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 accuratelyby flame photometry.

Principle

A stream of the sample passes over the flame as it burns, a colour of particular wavelength isformed which after passing through a set of filters, falls on a selenium cell and is changed intoelectricity. It is measured by potentiometer. The value is directly proportional to theconcentration 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 fineparticles. 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 aretwo other openings, one for gas supply and the other for compressed air supply. The burnerhas 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 10-20lbs/sq. inch. The mixture of gas and dry air burns in a nonluminous flame. The burner isarranged 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 photocell, 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 volumesof stock sodium and potassium standard which are given below. The volume is made upto100 ml with double distilled water and preserved in polythene bottle.

Stock sodium (ml) Stock patassium (ml) Sodium content (mEq/L) Patassium content (mEq/L)

0.55 0.20 1.1 0.020.60 0.30 1.2 0.030.65 0.40 1.3 0.040.70 0.50 1.4 0.050.75 0.60 1.5 0.060.80 0.70 1.6 0.070.85 0.80 1.7 0.08

The compressed air is started and its pressure is adjusted. Burner is lighted and the flameis made nonluminous. First the beaker (sampling up) is filled in the distilled water. It is suckedin and burn. The reading in the galvanometer is adjusted to zero with the help of a knob. Nowthe beaker with maximum strength of standard is placed under the suction needle. Thesensitivity knob is adjusted so that the deflection of galvanometer (maximum) is within thescale. Scale is graduated from 0-100. Now the different solutions are placed one by one underthe suction needle and the defections of galvanometer noted. The same procedure is repeatedwith the test sample. If this gives the same defection as one of the standards, then its Na+ andK+ concentration is that of the standard.

If Na+ and K+ content of the serum is to be estimated, the serum is diluted 1: 100 withdouble 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 PROTEINSElectrophoresis is defined as the migration of charged particles in the solution under theinfluence of electric field.

Demonstrations 111

This rate of migration is directly proportional to the number of charges present on acomponent. Proteins are colloidal particles and have charge either positive or negative whichdepend on the pH of the solution. In acidic medium, it acts as cation and in alkaline mediumas anion.

If uncharged particles are changed to charged ones, then they can be separated. If a potentialdifference is applied across them, current will flow and cations move towards cathode andanions towards anode.

Proteins are amphoteric in nature and have a particular pH (isoelectric pH) of which netcharge is zero and no migration takes place. At this stage, the ion is called zwitterion.

H H H| | |

R — C — COOH ______→ R — C — COO _____→ R — C — COO| _______ | ←_____ |

+NH3 +NH3 NH2

At pH = 1 Isoelectric pH At pH = 8Net charge= + 1 Charge = zero Charge = – 1

Separation is best at pH = 8.6. Since all plasma proteins have different isolectric pH, theirrate of migration is different.

Migration depends upon:1. pH of buffer2. Net charge of amphoteric particle3. Temperature4. Voltage and current.

This distance α time taken α potential gradient.∴ x α T α Σ

xor u = _______ sq. m/volts/sec.

T.Σ

Where u = Electrophoretic mobility (a constant)Potential gradient depends upon,

1. 1Current E α ______

1

2. r = Conductance of the solution3. q = Cross-sectional area.

r.qSo E = _______

1x

∴ μ = ________ sq.cm/volts/sec.Trq

1


Apparatus

Two main types of apparatus are used for separation of proteins.They are:

i. Free or moving boundry electrophoresisii. 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. Proteinshas 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 towardsanode 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 travelledconsiderably ahead of other proteins in above order. The light rays are passed through theseproteins. By appropriate optical apparatus, the degree of refraction can be photographed inthe form of so-called electrophoretic pattern. The proportion of each tube of protein isproportional to the area under its refractive peak in the pattern. It is called moving or freeboundry because proteins move freely in solution.2. Zone electrophoresis It is more important. Proteins are supported on various types of inert-

porous media.

Media

1. Paper2. Agar-gel3. Starch4. Cellulose acetate5. 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 whichit is dissolved.

Apparatus

Apparatus consists of two compartments filled with buffer solutions. Each is divided into alateral and a medical compartments connected holes plugged with cotton. In the two medicalcompartments filter paper is dipped and it rests on a purpose bridge. Whole apparatus iscovered by perspex cover.

In the actual experiment, we use constant voltage then the potential difference is usually3-4 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 paperno 1. Wet the paper in the buffer. Soak it in buffer and then fix on perspex bridge. Buffer usedis diethyl barbituric acid and sodium diethyl barbitol. Now cover the compartment and switchcurrent at voltage 110 for 5 minutes to saturate paper with current. After 5 minutes, take outcover and put a drop of serum sample on the paper. Cover it again and switch on the currentfor 16-18 hours. Afterwards take it out, remove strip and next step is fixation of proteins onthis 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 onsupporting 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 doneby bromophenol blue dye. Proteins gets boundry to the dye. Binding capacity of dye isproportional to the normal of protein present.

To remove unbound dye: Strip is rinsed with 0.5% acetic acid. Concentration of differentfractions or quantitation is done by

i. Densitometerii. Elution.

1. Densitometer: Stand strip in denstiometer and take optical density. By this we can calculateconcentration 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. Takedifferent optical density of different fractions. Wavelength is 520 nm.

Uses: Changes in electrophoretic pattern of serum proteins are indicated in many chemicalcondidtion.

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-globulinCirrhosis of liver : Increase in r-globulinNephrosis : Increase in α2fractionMultiple myeloma : Increase in r-fraction.


CHROMATOGRAPHY

For separation of constituents of a mixture one of the properties of the constituents of themixture 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 phaseis fixed and stationary and the moving phase flows past it. The mixture to be separated is putin the stationary phase and the moving phase carries along with it the different constituents ofthe mixture. As it does so, certain resistance is offered by stationary phase. Resistance offeredis different by different substances. It is therefore the resistance offered by the differentconstituents of the stationary phase which helps in the separation of the constituents bychromatography.

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 stationaryphase is solid like silicon, it can be packed in the phase, then the mixture to be separated isplaced on top and on appropriate moving phase which may be a liquid is allowed to runthrough it. Separation is effected by the resistance offered by stationary phase and differentconstituents of mixture reach the bottom of tube after different intervals of time. The movingphase is allowed to flow down the tube very very slowly, so that 10-15 drops fall down thecolumn per minute. As the rate is slow, separation is more effective. The moving phase whichcomes down can be collected in a series of test tube (i test tube for about 5 minutes). Substancewhich moves fastest reaches the test tube first and one which moves slowest reaches last theprocess is called elution.

If we do not want to separate the mixture, we can allow the moving phase to run till theend of the column and close the tap. This can then be detected by fluoroscopy. It is used for avariety 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 aninert gas is used, e.g. argon or sometimes nitrogen gas. The gas should be such that it shouldnot 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 andB and two solvents C and D and supposing C dissolves 90% of A and 10% of B and D dissolves90% of B and 10% of A. The two liquids A and B (mixture) should be immiscible. A and Bdissolved 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. Dcontains 90% of B of 10% of A. Then C is added to the funnel. If separated again 9% of it comesto B of 10% of A. So B is 99% pure. The process can be repeated many times. Every time, thepurity increases but separation is never 100%. The method depends upon the difference insolubility of solvents.

4. Paper Chromatography

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

Descending Chromatography

The paper hangs from the moving phase and the tip of the paper dips into the solvent at thetop. 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 tothe previous one. More spots are obtained in this direction, the constituents which could notbe 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 physicalproperties, e.g.

Isotopes are of two types-A. 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 allthe three particles while greatest ionisation power, because they have heavy particles, theycannot penetrate much.

β-rays—are fast moving electrons having the ionisation power less than α-particles andless than r-rays.

r-rays—are electromagnetic waves with maximum penetration power but least ionisationpower 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 displacingelectrons from them. Each positive ion so formed is paired either with the free detached electronor with a negative ion formed by the union of the electron with another neutral atom or moleculeα 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 intheir 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 orr-rays collide with the gas atoms or molecules. The cations and anions move to cathodeand anode respectively and produce electric impulse which is proportional to the activityof 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 areallowed 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 photomultiplier 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 notproduce 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 gain83.8 ergs of energy.

Application of Isotopes

A. In biochemistry Isotopes are used in working out metabolic pathways, e.g. cholesterolsynthesis from acetic acid, purine synthesis from glycine (4th, 5th and 7th carbon of purineskeleton is derived from glycine).

Some commonly used stable isotopes are used in the determination of turnover ofdifferent 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 beenstudied outside the body. When the albumin has been distributed evenly, the activity ofthe element is studied in a measured volume taken.

For measuring the volume of blood or emiscible pool or any other metabolic, the followingformula is used:

coV = v _______ = 1c

V = Volume of bloodV = Volume of labelled substance administered

co = Specific activity in vc = Specific activity in V.

Determination of average life of RBC Glycine labelled is used which gets incorporated inthe RBC. By determining the rate of decrease of activity over a certain period of time, averagelife can be calculated.

In iodine metabolism We can study iodine metabolism in the body. Iodine is essential forthe synthesis of thyroxine. Its distribution can be studied by the administration of radioactiveiodine I131. About 1/3 is taken up by thyroid and 2/3 is excreted by the kidneys. Inhyperthyroidism, there is increased uptake of I131 while in hypothyroidism there is decreaseduptake.

Hazards of Radioactivity

Radioactivity is responsible for genetic mutations. When an individual receives a massivedose 500 rad then that individual is susceptible to damage of germinal epithelium and damageto 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 r-rays.

2. Gloves must be used.3. Must get their blood count done every three weeks.

( )


4. Must carry plates which record radiations received by the body.5. Isotopes must not be disposed off in running water supply.6. Eating and drinking should be avoided.7. Radioactive solutions must not be sucked by mouth.8. 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 suspendedmatter 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 thebottom of the centrifuge tube.

For the purpose of comparison, centrifugal force is referred to as RCF, that is relativecentrifugal force. The RCF is taken as a guide to the separating capacity of a centrifuge. TheRCF 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 ofcentrifuge tube. ‘rpm’ is the number of revolution per minute of the centrifuge rotor. Smallmodels 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 tothe centrifugal which are fixed to the rotor.

The earlier centrifuge were operated by hands but modern laboratories are equipped withelectrically 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-outrotor is safer than angle rotor.

Operation

Select two centrifuge tubes of identical size. Place liquid to the centrifugal, water in other towithin 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 perminute is reached. When the tubes have been centrifugated sufficiently, switch off the motorand 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 electricalmethod depending upon the use of hydrogen electrode or glass electrode. The glass electrodemethod is most commonly used.

The principle of the method is that a small potential is set up between the glass and thesolution which is directly proportional to the hydrogen ion concentration of the solution. Thisvery small potential difference is indicated by a sensitive measuring devices.

This method is particularly useful when great accuracy is required. Glass electrode pHmeter covers the pH range 0-14 units.

It is ideal for use with biological fluids. The combination electrode system consists of areference electrode and a measuring electrode.

How to Operate

Keep function switch at pH positions. Set the temperature component control to read thesolution 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 respectivesockets. Connect the instrument to the main supply. Put on the main switch, clean the electrodeswith distilled water and dip them in buffer solution (prepared from the supplied buffer standardpH value of which is known). Being the READ switch at READ position. Adjust the zero andSTD control till the unit reads the correct pH value of supplied buffer solution. Bring the readswitch back in position. Take out the electrode from the buffer solution. Wash with distilledwater (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 (atleast within 10°C). In standardization a pH meter, it is advisable to use the supplied buffersolution. The pH value of which approx most closely to the test solution. A check with asecond buffer solution is suggested to ensure proper functioning of the equipment.

Electrode should be revised with distilled water and wiped with tissue paper gently beforeimmersing 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 becausetheir catalytic property also the detection of extremely small quantity of labelled immunereactant.

Principle

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


Glucose-6-phosphate dehydrogenase choice of enzyme depends upon1. Specificity2. Availability3. Satisfactory

Nowadays. ELISA is replacing RIA (Radio Immuno Assay) because it is cheaper, longershelf 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. Indirect2. Direct sandwich type3. Competitive.

1. Indirect ELISA It quantitates antibody. Sample containing primary antibody added toantigen coated microtitre wells, which reacts with antigen found to solid phase. Freeantibody washed away. Presence of antibody bound to antigen is detected by adding anenzyme conjugated second antibody which ends to primary antibody. Washing is doneto remove free antibody. Then substrate is added to develop colour which is measuredspectrophotometrically, e.g. detection of HIV antibodies.

2. Sandwich ELISA (Double antibody type) It quantitates antigen specific antibody and isadsorbed on solid phase sample with antigen added. Incubate and washing is done. Secondenzyme linked antibody is directed against a different epidope added. Complex is formed.Then enzyme substrate is added and is incubated. The colour developed is measuredunder specific wavelength. Uses are:1. All hormones can be detected and assayed2. Detection of viruses, pregnancy, hepatitis B

3. Competitive ELISA It is inhibition type assay for quantitating antigens. Competetive bindingof enzyme labelled antigen and unlabelled antigens to a high affinity antibody. The labelledantigen mixed with antibody at a concentration that just saturates the binding sites ofantibody molecule. Then increasing amount of unlabelled antizen of unknown concen-tration is added.

With increasing concentration of unlabelled antigen more and more labelled antigens willbe displaced.

By measuring the amount of labelled antigen we can determine that of labelled. Concentra-tion of antigens is inversely proportional to colour produced, e.g. HBs antigen, AustralianAntigen.

Appendix 121

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QUESTIONS ANALYSIS1. 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.

Appendix


A known case of homogentisic oxidase deficiency gave greenish—brown colourwith 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 broughtto the paediatrician. Biochemical investigations revealed hyperphenylalaninemia:

39. The urine of this patient gave ferric chloride test positive. Name the compound (s) responsiblefor 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 andother parameters were within normal range. Similar findings were present in hisparents:

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 thefunctional impairment is compatible with life. Sometimes they show beneficialeffect, 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 complaintof fever with chills from past 3 days. Suspecting a case of malaria he wasprescribed primaquine tablets but he developed haematuria after 4 days. Labo-ratory investigations revealed fall in Hb from 11 gm% to 7.4 gm% and rise in totalbilirubin 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 anddeep breathing. Biochemical investigation revealed:

Blood pH : 7.1HCO3

– : 5 mEq/LGlucose : 350 mg/dLUrinary 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 charac-terized 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) 60-80 mg%Blood glucose (folin) 80-100 mg%Blood urea 20-40 mg%Blood cholesterol (total) 150-250 mg%Blood cholesterol (ester form) 2/3rd of total

cholesterolBlood 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

Serum total proteins 5.5-7.5 gm%Albumin fraction 3.5-5.0 gm%Globulin fraction 2-2.5 g%A/G ratio 1.2 : 1Serum calcium (total) 9-11 mg%

or 4.5-5.5 mEq/LSerum bilirubin 0.2-0.6 μg%Serum iron 65-175 mg%Serum iron binding capacity 250-410 μg%Serum iodine (BEI) 3-6.5 μg%Serum protein bound iodine 4-8 μ%Serum sodium 130-145 mEq/L

or 310-340 mg%Serum potassium 3.5-5.0 mEq/L

14-20 mg%Serum chloride 100-126 mEq/L

or 350-375 mg%

Serum alkaline phosphatase 3-13 King-Armstrongunits (KAU)

Serum acid phosphatase 0-3 King-Armstrongunits (KAU)

SGOT 10-40 IU/LSGPT 9-32 IU/LSerum amylase 80-180 Somogyi

unit %Carbon dioxide combining power 55-65 vol %Prothrombin time 14-17 secondsCoagulation time 5-10 secondsBleeding time 1-3 seconds

CSF

Glucose 45-80 mg%Protein 15-80 mg%Chlorides 700-750 mg%

Urine

Reducing sugars 100 mg/dayUrea 30 gm/dayUric acid 0.8 mg/dayCreatinine 0.4-1.8 mg/dayChlorides as NaCl 10-15 gm/dayUrobilinogen 0.4 gm/dayKetone bodies 1 mg/dayPhosphates 10 gm/dayTitrable acidity 200-400 ml N/10

acid/day

Conversion of mg% to mEq/L

mg% ×10mEq/L = _________________________

Equivalent weight

mg% × 10millimole/litre = _____________________________

(mmole/L) molecular weight

Appendix 125

Molecular weight of some commonly usedsubstances in medical practice.

Substance Molecular weight

Alcohol 46Bilirubin 585Calcium 40Cholesterol 387Creatinine 113Glucose 180Iron 56Phosphorus 31Triglyceride (Triolein) 885Urea 60Uric acid 168

Tables of prefixes denoting decimal factorFactor Prefix Symbol

10-3 milli m10-6 Micro μ10-9 nano n10-12 pico p

Relation of biochemical tests with disease

Blood sugar Diabetes mellitusBlood urea Kidney diseaseSerum creatinineSerum uric acid GoutSerum cholesterol Heart disease, high blood pressureSerum bilirubin JaundiceSerum proteins Malnutrition, liver disease

Chronic infectionsSerum alkaline phosphatase Rickets, liver and bone diseaseSGOT Heart attackSGPT Liver disease

Serum Electrolytes Normal values

Sodium 130-145 mEq/L) Dehydration or loss of fluidsPotassium 2.5-5 mEq/L)Chloride 100-106 mEq/L) Acid-base balanceCalcium 9-11 mg% Vitamin D deficiencyPhosphorus 3-4.5 mg% Recurrent renal stones

List of recommended units

Molar concentration Mass concentration

mol/litre gm/litre (g/l)m mol/litre mg/lμmol/litre μg/ln mol/litre ng/l

Conversion factors of old units to SI units

Substance Old unitConversion

New unitfactor

Albumin g/100 ml 10 gm/lBilirubin mg/100 ml 104/585 μmol/lCalcium mg/100 ml 1/4 mmol/lCholesterol mg/100 ml 10/387 mmol/lCreatinine mg/100 ml 10/113 mmol/lGlucose mg/100 ml 1/18 mmol/lIron μg/100 ml 10/56 μmol/lPhosphorus mg/100 ml 10/31 mmol/lTriglyceride mg/100 ml 10/885 mmol/lUrea mg/100 ml 1/6 mmol/lUric acid mg/100 ml 10/168 mmol/l

Index

AAAAA

A/G ratio 93, 94

Achlorhydria 38

Achromic point 20Achylia gastrica 38

Adsorption 7

Albumin 26Alkaline phosphatase 101

Anticoagulants

ethylene diaminetetraacetate 51

heparin 52

oxalate 51sodium citrate 52

sodium fluoride-potassium

oxalate 52Anuria 41

Arterial blood 53

Ascorbic acid in urine 66vitamin C saturation test

66

BBBBB

Beer’s law 70

Bence Jones proteins 45

Bilirubinconjugated or direct 95

reactions 95

reagents 96unconjugated or indirect

95

Blood collection 52, 53Blood specimen 51

Blood sugar estimation

enzymatic method 74reduction method 74

Blood urea 80

postrenal 82prerenal 82

renal 82

CCCCC

Calcium 87Capillary puncture 53

Caraway method 68

Centrifugal force 118

Centrifuge 118

Chlorides in urine 60

Cholesterolstandard 85

Chromatography

column 114descending 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

DDDDD

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 (phenyl-

hydrazine test) 15

EEEEE

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 120sandwich 120

Eluters 7Enzymatic hydrolysis of starch 20Estimation of nitrogen 106

proceduredigestion 106distillation 107estimation 108

FFFFF

Flame photometry 109Food analysis

carbohydrate test 48fats (cholesterol) test 49mineral test 50test for proteins 49

Formal titration 35Formality 3

GGGGG

Gastric juice 37Globulin 26Glucose tolerance curve 78

flat GTT 79lag type 79

Glycosuria 43

HHHHH

Histamine test 37Hydrolysis

acid 17enzymatic 17

Hyperchlorhydria 38Hyperchlorhydria (gastric

hyperacidity) 39

IIIII

Identification of carbohydrate 18Inorganic phosphorus

reactions 90Iodine number 33Isoelectric pH 111Isoelectric point 27Isotopes 115


JJJJJ

Jaundice 95

LLLLL

Lambert’s (Bouguer) law 70

Liver function tests 100

MMMMM

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

NNNNN

Needle 52

Normal values 124

blood 124

CSF 124

serum 124

serum electrolytes 125

urine 124

Normality 2

OOOOO

Oliguria 41

Optical density 71

Osmosis 6

Osmotic pressure 6

PPPPP

Pentosuria 59

Percent 1

pH

electrometric determina-

nation 104

pH determination

selection of indicator 4

Gillespie’s drop method 5

indicator paper 5

universal indicator 5pH 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

QQQQQ

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

RRRRR

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

SSSSS

Saponification number 32Serum 53Serum aminotransferases

(transaminases) 100Serum calcium 87

calculation 88collection of sample 87

Serum proteins 92Serum urate 68Serum uric acid 68Solution tension 104Sucrose test 16Surface tension 7Syringes 52

TTTTT

Test for proteins 45Tests for lipids 29

acrolein test 30emulsification 29saponification 30solubility 29test for cholesterol 31unsaturation 31

Tourniquet 52Transmission 71

UUUUU

Urea 80Urea clearance 83Uric acid 68Urine analysis 40, 55

colour 41pH 42specific gravity 42total solids 43

bile pigments 46bile salts 47blood 47ketone bodies 46proteins 44urobilinogen 47reducing sugars usually

glucose 43volume 41

Urine specimenscollection 54preservatives 54

hibitane 55hydrochloric acid 54toluene 55

VVVVV

Vein puncture 52

Index 129

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