QUALITATIVE ANALYSIS—CARBOHYDRATESChapter One

I Trioses

3

Glyceraldehyde

Dihydroxy acetone

II. Tetroses

4

Erythrose

Erythrulose

III. Pentoses

5

Ribose

Ribulose

IV. Hexoses

6

Glucose

Fructose

V. Heptoses

7

Sedoheptose

Sedoheptulose

MALTOSE

→ Glucose + Glucose

SUCROSE

→ Glucose + Fructose

LACTOSE

→ Glucose + Galactose

Green (+)

0.1–0.5

Yellow (++)

0.5–1.0

Orange (+++)

1.0–2.0

Red (++++)

Above 2%

1.

Single Reagent - Kept in one bottle

Three different reagents A, B and C kept in 3 bottles. Has to be tested for autoreduction.

2.

Does not deteriorate on storage and so no need for autoreduction.

3.

Weak alkali (sodium carbonate) does not reduce other normal constituent like uric acid and creatinine.

Strong alkali (Sodium hydroxide) present here reduces other normal constituents also.

Needle shaped

10 minutes

Needle shaped

7 minutes (Sheaves of corn)

Sunflower shaped

30 minutes - formed after cooling.

Cotton ball or badminton ball shaped.

20 minutes - formed after cooling.

1. N/50 lodine solution.

2. Glacial acetic acid.

(i) Solid ammonium sulphate

(ii) Saturated solution of ammonium sulphate in water.

1.

Molisch's test

– To confirm the presence of carbohydrate.

2.

Iodine test

– To confirm the presence of polysaccharides.

3.

Benedict's test

– To confirm the presence of reducing sugars.

Fehling's test

(Both mono & disaccharides)

4.

Barfoed's test

– To confirm the presence of reducing monosaccharides only.

5.

Seliwanoff's test

– To confirm the presence of Fructose.

Foulger's test

6.

Bial's test

– To confirm the presence of pentose.

7.

Sucrose Hydrolysis test

– To confirm the presence of non-reducing disaccharide sucrose.

8.

Osazone test

– To confirm the type of reducing mono or disaccharide.

Take 2 ml of the given sample in a clean dry test tube. Add 3 drops of Molisch's reagent. Mix and incline the test tube slightly and add 2 ml of conc. sulphuric acid along the sides of the test tube. Then erect the tube slowly.

A reddish violet or purple ring is seen at the junction.

It is a general test for carbohydrates. Glucose is dehydrated by conc. sulphuric acid to form furfural derivatives which on condensation with Molisch's reagent forms a violet ring.

Take 2 ml of the given sample and add 5 drops of glacial acetic acid, add 3 drops of N/50 iodine solution drop by drop.

No characteristic color is seen. Only yellow color of lodine is seen.

This is a specific test for polysaccharide. Glucose being a Monosaccharide does not answer this.

Take 5 ml of Benedicts reagent in a test tube. Add 8 drops of glucose solution. Mix and boil for 2 minutes and cool.

A precipitate which may be green, yellow or red depending on the amount of sugar present in the solution is obtained.

It is a sensitive reduction test indicating the presence of reducing sugars. The precipitate formed is red cuprous oxide.

Take 2 ml of Barfoed's reagent, add 1 ml of glucose solution. Mix and boil for 30–60 secs. Allow the test tube to cool. Avoid prolonged heating.

A red precipitate is formed at the bottom of the test tube.

It is a specific reduction test for monosaccharide. Prolonged heating converts disaccharide to monosaccharides and gives false positive test. This test is used to distinguish monosaccharides from disaccharides.

Take 2 ml of Seliwanoff's reagent in a test tube. Add 3 drops of glucose solution. Boil for 30 seconds. Allow it to cool in the rack. Avoid prolonged heating.

No cherry red complex is formed.

This test is only for ketohexose. Glucose being an aldohexose does not answer this test. Prolonged heating may give false positive test for glucose and sucrose.

Take 3 ml of Foulger's reagent in a test tube. Add 5 drops of glucose solution. Boil for 1 minute vigorously. Leave the test tube to cool.

No blue color is seen.

This test is only for ketohexose. Glucose being an aldohexose does not answer this test.

Set up a boiling water bath. Take 5 ml of the sample in a test tube. Add 1/2 spatula of phenyl hydrazine hydrochloride and 1 spatula full of sodium acetate and 0.5 ml of glacial acetic acid. Shake well. Filter the contents. Keep it in a boiling water bath. Transfer few crystals onto a glass slide, cover it with a cover slip and observe under low power microscope.

Needle shaped yellow crystals of glucosazone appears within 10 minutes.

It is a test to identify glucose in the given sample.

Result : Reactions of the reducing monosaccharide-glucose are thus studied.

Take 2 ml of the given sample and add few drops of Molisch's reagent. Incline the test tube and add few drops of Conc.sulphuric acid.

A purple or violet ring is seen at the junction of two layers.

It shows the presence of carbohydrates. It is dehydrated by conc. Sulphuric acid to form furfural derivatives which on condensation with Molisch reagent gives violet ring.

Take 2 ml of the given sample and add 5 drops of glacial acetic acid and add 3 drops of N/50 iodine solution drop by drop.

No characteristic color is seen. Only yellow color of iodine is seen.

This is a specific test for polysaccharide. Fructose being a Monosaccharide doesnt answer this test.

Take 5 ml of Benedicts reagent in a test tube. Add 8 drops of the given sample. Mix and boil for 2 minutes and leave aside to cool.

A precipitate which may be green, yellow or red depending on the amount of sugar in solution is obtained.

It is a sensitive reduction test, indicating the presence of reducing sugar.

Take 2 ml of Barfoed's reagent, add 1ml of the given sample. Mix and boil for 30–60 seconds. Allow the test tube to cool.

A red precipitate is formed at the bottom of the test tube.

It is a specific test for monosaccharide. Prolonged heating converts disaccharide to monosaccharides and gives false positive test. This test is used to distinguish monosaccharides from disaccharides.

Take 2 ml of Seliwanoff's reagent in a test tube. Add 3 drops of the given sample to it and boil for 30 seconds. Leave it aside to cool.

Note the color of the sample. The sample turns into a cherry red colored complex.

This test is positive for ketohexoses only. Fructose being a ketohexose answers this test. Ketohexoses are dehydrated by Conc. HCl to form furfural derivatives which on condensation with resorcinol forms the red complex.

Take 3 ml of Foulger's reagent in a test tube. Add 5 drops of the given sample and boil it for 1 minute and allow it to cool.

A blue color is seen.

This test is positive for ketohexoses. Fructose being a ketohexose answers this test. This test is based on the formation of furfural which condenses with urea in the presence of stannous chloride to give a blue color.

Set up a boiling water bath. Take 5 ml of the sample in a test tube. Add 1/2 spatula of phenylhydrazine hydrochloride and 1 spatula full of sodium acetate and 0.5 ml of glacial acetic acid. Shake well. Filter the contents. Keep it in a boiling water bath. Transfer few crystals onto a glass slide. Cover it with a coverslip and observe under low power microscope.

Needle shaped yellow crystals of Fructosazone will be formed within seven minutes.

This confirms the presence of reducing sugar fructose.

Result : Reactions of Fructose are studied, which is a reducing ketohexose.

Take 2 ml of the pentose solution in a dry test tube. Add 3 drops of Molisch's reagent (1% of alphanaphthol in alcohol). Add 2 ml of conc. sulphuric acid along the sides of the test tube to form a layer at the bottom for a height of half an inch.

A purple ring or reddish violet ring is formed at the junction of the two layers.

This is a general test for carbohydrates. Pentoses are dehydrated by conc. sulphuric acid to form furfural derivatives which on condensation with Molisch's reagent forms a violet ring.

Take 2 ml of pentose solution and add 5 drops of glacial acetic acid and 3 drops of N/50 iodine solution drop by drop.

No characteristic color is seen. Only yellow color of lodine is seen.

This is a specific test for polysaccharides Pentoses do not answer this as they are monosaccharides.

Take 5 ml of Benedicts reagent in a test tube. Add 8 drops of the pentose solution. Mix. Boil for 2 minutes.

As per the amount of pentose present, green, yellow or red precipitate is obtained.

It is a sensitive reduction test, indicating the presence of reducing sugar. Precipitate formed is red cuprous oxide.

Take 2 ml of Barfoed's reagent, add 1 ml of Ribose solution. Mix and boil for 30–60 seconds and allow it to cool.

A red precipitate is formed at the bottom of the test tube.

It is specific reduction test for reducing monosaccharides. It helps to differentiate reducing monosaccharides from reducing disaccharides.

Take 2 ml of Seliwanoff's reagent in a test tube. Add 3 drops of pentose solution. Boil for 30 seconds and allow it to cool. Avoid prolonged heating.

No cherry red complex is formed.

This test answers only for ketohexoses. The given solution is an aldopentose and so it does not answer this test.

Take 3 ml of Foulger's reagent in a test tube. Add 5 drops of pentose solution. Boil vigorously for 1 minute and allow it to cool.

No blue color is obtained.

This test is also for ketohexoses. Aldopentoses will not answer this test.

Take 5 ml of Bials reagent (Dilute solution of Orcinol in 30% HCl) and add 0.5 ml of pentose solution and heat until the solution boils.

Green colored solution is obtained.

This is a confirmatory test for pentoses. Conc. HCl converts pentoses into furfural derivatives which then condenses with Orcinol to give green colored complex.

Result : Reactions of Pentoses are thus studied.

Take 2 ml of the given sample and add few drops of Molisch's reagent. Incline the test tube and add few drops of conc. sulphuric acid.

A reddish purple or reddish violet ring is seen at the junction of two liquids.

It is a general test for carbohydrate whether free or bound to substances such as proteins or lipids. Maltose being a disaccharide answers this test.

Take 2 ml of the given sample and add 5 drops of glacial acetic acid. Add 3 drops of N/50 iodine solution drop by drop.

No characteristic color is seen. Only yellow color of iodine is retained.

It is a specific test for polysaccharide. Maltose being a disaccharide does not answer this.

Take 5 ml of Benedicts reagent in a test tube. Add 8 drops of the given sample. Mix and boil for 2 minutes and leave aside to cool.

A precipitate which may be green, yellow or red is obtained depending on the amount of sugar.

It is a reduction test indicating the presence of reducing sugar. Maltose being a reducing sugar will answer this test.

Take 2 ml of Barfoed's reagent, add 1 ml of the given sample. Mix and boil for 30–60 seconds. Allow the test tube to cool.

No red precipitate is obtained.

It is a specific reduction test for monosaccharides. The given sample being a disaccharide does not answer this test.

Take 2 ml of Seliwanoff's reagent in a test tube. Add 3 drops of the given sample to it and boil for 30 seconds. Leave it aside to cool.

No cherry red colored complex is btained.

This is a positive test for keto hexoses only. Maltose will not answer this test.

Take 3 ml of Foulger's reagent in a test tube. Add 5 drops of the given sample and boil it for 1 minute and allow it to cool.

No blue color is obtained.

The sample being a disaccharide does not answer this test since it has no keto group.

Set up a boiling water bath. Take 5 ml of the sample in a test tube. Add 1/2 spatula of phenyl hydrazine hydrochloride and spatula full of sodium acetate and 0.5 ml of glacial acetic acid. Shake well. Filter the contents. Keep it in a boiling water bath. Transfer few crystals, onto a glass slide, cover it with a coverslip and observe under low power microscope.

Sunflower shaped maltosazone crystals appear in 30 minutes.

This confirms the presence of reducing sugar maltose.

Result: Reactions of maltose are thus studiedreducing disaccharide.

Take 2 ml of the given sample and add few drops of Molisch's reagent. Incline the test tube and add few drops of conc. sulphuric acid.

A reddish purple or reddish violet ring is seen at the junction of the two layers.

It is a general test for carbohydrate whether free or bound to substances such as proteins or lipids. Lactose being a disaccharide gives a positive reaction.

Take 2 ml of the given sample and add 5 drops of glacial acetic acid and add 3 drops of N/50 iodine solution drop by drop.

No characteristic color is seen. Only yellow color of iodine is retained.

It is a specific test for polysaccharide. Lactose being a disaccharide does not answer this test.

Take 5 ml of Benedicts reagent in a test tube. Add 8 drops of the given sample. Mix and boil for 2 minutes and leave aside to cool.

A precipitate which may be green, yellow or red is obtained depending on the amount of sugar.

It is a reduction test indicating the presence of reducing sugar. Lactose being a reducing sugar answers this test.

Take 2 ml of Barfoed's reagent, add 1ml of the given sample. Mix and boil for 30–60 seconds. Allow the test tube to cool.

No red precipitate is obtained.

It is a specific reduction test for monosaccharide. The given sample being a disaccharide doesnt answer this test.

Take 2 ml of Seliwanoff's reagent in a test tube. Add 3 drops of the given sample to it and boil for 30 seconds. Leave it aside to cool.

No cherry red colored complex is seen.

This test is positive for ketohexoses only. Lactose doesnt answer this test.

Take 3 ml of Foulger's reagent in a test tube. Add 5 drops of the given sample and boil it for 1 minute and allow it to cool.

No blue color is formed.

The sample being a disaccharide does not have a keto group, and so it shows negative reaction.

Set up a boiling water bath. Take 5 ml of the sample in a test tube. Add 1/2 spatula of phenylhydrazine hydrochloride and 1 spatula full of sodium acetate, 0.5 ml of glacial acetic acid. Shake well. Filter the contents. Keep it in a boiling water bath. Transfer few crystals into a glass slide, cover it with a coverslip and observe under low power microscope.

Cotton ball shaped or hedgehog shaped lactosazone crystals appear within 20 minutes.

It indicates the presence of reducing disaccharide namely lactose.

Result: Reactions of maltose are thus studied—reducing disaccharide.

Take 2 ml of the given sample and few drops of Molisch's reagent. Incline the test tube and add few drops of conc. sulphuric acid.

A reddish purple or reddish violet ring is seen at the junction of two layers.

It is a general test for carbohydrate whether free or bound to substances such as proteins or lipids. Sucrose being a disaccharide gives positive reaction.

Take 2 ml of the given sample and add 5 drops of glacial acetic acid and add few drops of N/50 iodine solution drop by drop.

No characteristic color is seen. Only yellow color of iodine is retained.

It is a specific test for polysaccharide. Sucrose being a disaccharide does not answer this test.

Take 5 ml of Benedicts reagent. Add 8 drops of the given sample. Boil for 2 minutes. Leave aside to cool.

No characteristic color is seen.

It is a reduction test indicating the presence of reducing sugar. Sucrose being a non-reducing sugar does not answer this test.

Take 2.5 ml of sucrose solution in a test tube. Add 3 drops of Conc. HCl and boil it for 2 minutes. Cool it under tap water. Add 20% of sodium carbonate drop-wise till there is no reducing done to neutralize the confirmatory Benedicts reagent. Boil for 2 minutes. Leave it to cool.

The precipitate which may be green, yellow or red color depending on the amount of sugar is obtained.

Sucrose is a non-reducing sugar but after hydrolysis by Conc. HCl it is converted to glucose and fructose which are reducing sugars. This is a confirmatory test for sucrose.

Take 2 ml of Barfoed's reagent, add 1ml of the given sample. Mix and boil for 30–60 seconds. Allow the test tube to cool.

No red precipitate is formed.

It is a specific reduction test for monosaccharides. The given sample being a disaccharide does not answer this test.

Result: The reactions of Sucrose (Non-reducing disaccharide) are thus studied.

The most commonly available polysaccharide is starch, which is a mixture of amylose and amylopectin. The individual glucose unit in amylose are linked by 1,4 glycosidic linkages. Amylopectins have branching points contributed by alpha 1,6 glycosidic linkages. Starch is insoluble in cold water, but forms a colloidal solution in hot water. Starch has no detectable reducing property.

Take 2 ml of the given sample. Add few drops of Molisch's reagent. Incline the test tube. Add few drops of Conc. sulphuric acid.

A reddish purple or violet ring is seen at the junction of two liquids.

It is a general test for carbohydrate. Starch being a polysaccharide answers this test.

Take 5 ml of the given sample and add 2 ml of glacial acetic acid and add few drops of N/50 iodine solution.

A blue color is seen.

It is a specific test for polysaccharide. Starch being a polysaccharide answers this test. Iodine forms colored complexes with polysaccharides.

Take 5 ml of the Benedicts reagent and add 8 drops of sample. Boil and then cool.

No characteristic reaction or colored precipitate is seen.

It is a reduction test for reducing sugar. Starch being a nonreducing sugar does not answer this test.

Take 5 ml of starch solution in a test tube. Add 5 ml of saturated ammonium sulphate. Shake vigorously. Allow to stand for 5 minutes in the rack. A white precipitate is formed. Filter in a test tube. To the filtrate add 2 drops of glacial acetic acid and 3 drops of N/50 iodine solution.

White precipitate is formed and no blue color is seen in the filtrate.

A white precipitate is obtained which indicates that starch is precipitated by half saturation with ammonium sulphate. The filtrate does not give blue color with iodine as it does not contain starch.

Result: The reactions of starch—a polysaccharide are thus studied.

They are formed as intermediate products during the course of hydrolysis of starch by dilute mineral acid and also by the reaction of amylases. These heterogenous compounds are grouped into amylodextrin, erythrodextrin and achrodextrin based on the color produced by N/50 iodine. These three groups which are in the order of decreasing molecular weight gives violet, red and no color in iodine reaction. Dextrins are partially soluble in water. They have more reducing activity.

Take 2 ml of dextrin solution in a test tube. Add 3 drops of Molisch's reagent. Incline the test tube slightly and add 2 ml of Conc. Sulphuric acid along the sides of the test tube.

Reddish purple ring is seen at the junction of the two liquids.

It is a general test for carbohydrate. Dextrin being a carbohydrate, answers this test.

Take 5 ml of the given sample and add 5 drops of glacial acetic acid and add few drops of N/50 iodine solution.

Reddish purple color is seen.

It is a specific test for polysaccharide. Dextrin being a polysaccharide answers this test as iodine forms colored complexes with polysaccharides.

Take 5 ml of Benedicts reagent and add 8 drops of the given solution. Boil and then cool.

A precipitate which may be green or yellow or red color is obtained.

Dextrin answers this test because it has free aldehyde group at the end of the chain of its molecule.

Take 5 ml of dextrin solution in a test tube. Add 5 ml of saturated ammonium sulphate. Shake vigorously. Allow it to stand for 5 minutes in the rack. Filter in a test tube. To the filtrate add 2 drops of glacial acetic acid and 3 drops of N/50 iodine solution.

No precipitate is formed.

Reddish purple color is seen in the filtrate.

Dextrin is not precipitated by half saturation with ammonium sulphate. So filtrate gives reddish purple color on addition of Iodine as it contains dextrin.

Take 5 ml of dextrin solution in a test tube. Add solid ammonium sulphate with constant shaking until it begins to settle at the bottom of the test tube. Filter it in another test tube. Add 2 drops of glacial acetic acid and 3 drops of N/50 iodine solution

A reddish purple color is seen in the filtrate.

Dextrins are not precipitated even by full saturation with ammonium sulphate. So the filtrate gives reddish purple color on addition of iodine as it contains dextrin.

Result: Thus the reactions of dextrin which is a polysaccharide are studied.

1. Molisch's test:

a. Purple ring is formed at the junction of 2 liquids.

Presence of carbohydrates

b. No purple ring.

Absence of carbohydrates

2. Iodine test:

a. Blue color is formed.

Presence of polysaccharide

b. No blue color

Absence of polysaccharide

3. Half saturation test:

a. White precipitate is formed. No blue color in filtrate

Presence of starch.

b. No white precipitate filtrate gives reddish purple color.

Presence of Dextrin.

4. Benedicts test:

a. Red precipitate is formed.

Presence of reducing sugar.

b. No red precipitate

Presence of non-reducing sugar.

5. Sucrose hydrolysis test:

a. Red precipitate

Presence of sucrose.

6. Barfoed's test:

a. Red precipitate

Presence of monosaccharides.

b. No red precipitate

Presence of reducing disaccharides.

7. Seliwanoff's test:

a. Cherry red color is formed.

Presence of fructose.

b. No cherry color

Absence of fructose.

8. Foulger's test:

a. Blue color is formed

Presence of fructose.

b. No blue color

Absence of fructose.

9. Bials test:

a. Green color is obtained.

Presence of Pentose.

b. No green color

Absence of Pentose.

10. Osazone test:

a. Needle shaped crystals

Glucose/fructose.

b. Sunflower shaped crystals.

Presence of Maltosazone.

c. Cotton ball shaped crystals.

Presence of Lactosazone.

Result: The unknown solution contains……………………. which is …………………….

1½

1

Blue

6

Blue

No Reduction

Solube starch

5 min

2

Violet

7

Green

Slight

Reduction (+)

Amylodextrin +

Maltose

8 min

3

Reddish violet

8

Red

More Reduction

(++)

Maltose

Erythrodextrin

+ More

12 min

4

No characteristic change (Yellow color of Iodine)

9

Red

Still more

Reduction (+++)

Achrodextrin + still more

Maltose

20 min

5

No characteristic change (yellow color of iodine)

10

Red

Maximum

Reduction

(++++)

Maltose +

Glucose

Note: Hydrolysis of starch by acid results in Glucose as end product. Hydrolysis of starch by enzyme results in Maltose as end product.

Result: The products yielded after the acid hydrolysis of starch are studied.

• Solid ammonium sulphate.

• Saturated solution of ammonium sulphate.

1. Take 2 ml of albumin solution in a test tube. Add 5–10 drops of 10% lead acetate solution. Mix well.

White precipitate is formed

This test indicates the presence of proteins. The metal cation combines with negatively charged group of proteins to cause precipitations of albumin.

2. Take 2 ml of albumin solution in a test tube. Add 5–10 drops of mercuric chloride prepared in 10% Sulphuric acid drop by drop.

White precipitate is formed

This test indicates the presence of proteins. The metal cation combines with negatively charged group of proteins to cause precipitations of albumin.

1. Take 2 ml of albumin solution in a test tube. Add 2 to 3 drops of sulpho salicylic acid. Mix well (This test is used to detect albumin in urine)

White precipitate is formed

Sulphosalicylic acid has negatively charged ions which neutralise the positive charged ions of albumin and cause denaturation resulting in precipitate formation.

2. Take 2 ml of albumin solution. Add 2 drops by freshly prepared tannic acid.

Light brown precipitate is formed.

Tannic acid has negatively charged ions which neutralize the positively charged ions of albumin and cause denaturation resulting in precipitate formation.

3. Take 2 ml of albumin solution. Add 2 to 3 drops of picric acid.

Yellow precipitate is formed.

Picric acid has negatively charged ions which neutralize the positive charge of albumin causing denaturation resulting in precipitation.

Take 3 ml of albumin solution in a test tube. Add equal volume of saturated Ammonium sulphate solution. Mix well. To this add 3 ml of 40% sodium hydroxide and 2–3 drops of copper sulphate.

Filtrate gives violet color

Albumin is not precipitated by half saturation with ammonium sulphate. So it gives violet color with Biuret reagent.

Take 3 ml of albumin solution in a test tube. Add solid ammonium sulphate until little salt settles at the bottom. Wait for 5 minutes. A white precipitate is formed. Filter it. To the filtrate add 3ml of 40% Sodium Hydroxide and 2–3 drops of 1% copper sulphate solution (Biuret reagent).

Filtrate gives blue color of copper sulphate.

Albumin is completely precipitated by full saturation with ammonium sulphate. Hence filtrate gives no color due to the precipitation of albumin.

Fill 3/4th of test tube with albumin solution. Hold the test tube at its bottom. Incline it slightly and heat the upper portion of the solution over the flame. Add 3 drops of 1% acetic acid.

A white precipitate is formed.

It indicates the presence of the coagulable protein albumin. Appearance of coagulation indicates denaturation of albumin and precipitation by acetic acid. Acetic acid gives the suitable pH to get maximum precipitate.

(This test is used to confirm the presence of albumin in urine)

Take 2 ml of albumin solution in a test tube and 2 ml of 5% sodium hydroxide. Mix and add 2 drops of Copper sulphate solution.

A violet colored complex is formed.

The color produced is due to denaturation and the formation of co-ordination complex between cupric ions and the nitrogen atoms of the peptide bonds.

(Precaution : If copper sulphate is added in excess violet color will not be formed)

Take 2 ml of albumin solution in test tube. Add 2–3 drops Ninhydrin and boil. Leave it aside and allow it to cool.

A bluish purple colored complex is formed.

Albumin contains free amino acids which react with Ninhydrin to give a blue color. This test is not answered by proline and hydroxy proline.

Take 3 ml of albumin solution in a test tube. Add 1ml of concentrated Nitric acid and mix. Heat it for a minute and cool it under tap water andadd 2 ml of 40% sodium hydroxide to make it alkaline. Mix

First a white precipitate is formed due to denaturation. It turns yellow on heating. On adding sodium hydroxide an orange yellow precipitate is formed

This test is specific for the benzene ring of the aromatic amino acids such as tyrosine and Tryptophan. Based on the nitration of the benzene ring it forms an yellow derivative which turns orange color. Phenylalanine may give a weak positive or negative reaction - due to the presence of unsubstituted phenyl group.

Take 1ml of albumin solution ml of 10% mercuric sulphate prepared in 10% Sulphuric acid. Boil for 30 seconds. Cool it under tapwater. Add 3 drops of sodium nitrite solution. Mix.

An yellow precipitate is formed which becomes red after boiling and adding sodium nitrite solution.

This test is specific for hydroxyl group of tyrosine. The red color is due to the mercury complex of nitrophenol derivative. It shows the presence of tyrosine in albumin.

Take 3 ml of albumin solution in a test tube. Add 2 drops of 1/500 Formalin and 1 drop of 10% mercuric sulphate in sulphuric acid. Add 3 ml of conc. H₂SO₄ along the sides of the test tube.

A purple or violet colored ring is formed at the junction of the two liquids.

This test is specific for tryptophan. The indole ring of tryptophan acts with formaldehyde in the presence of oxidized agents to form a violet colored ring. This shows the presence of tryptophan in albumin.

Take 3 ml of albumin solution in a test tube. Add 5 drops of 20% sodium hydroxide and 4 drops of Molisch's reagent (1% alpha naphthol). Mix and add 10 drops of bromine water.

A bright red color is formed.

This test is specific for guanidine group of arginine. In alkaline medium alpha naphthol combines with the guanidine group of arginine to form a complex which is oxidized by bromine water to produce red color. It shows the presence of arginine in albumin.

Take 2 ml of albumin solution in a test tube. Add 2 ml of 40% NaOH. Boil for one minute. Then add 5 drops of lead acetate.

A black or brownish precipitate is formed.

This test indicates the presence of sulphur containing amino acids such as cysteine and cystine. On boiling with NaOH, the sulphur present in the amino acid is converted to inorganic form of sodium sulphide which gives black precipitate. Methionine does not answer the test as the sulphur group is not released by alkaline digestion. It shows the presence of cysteine and cystine in albumin.

Take 1 ml of 0.5% sulphanilic acid in a test tube. Add 1 ml of sodium nitrite solution. Mix. Add 2 ml of albumin solution. Add 1 ml of 10% sodium carbonate to make it alkaline.

An orange red colored complex is formed.

This reaction is specific for imidazole group of histidine. Diazotized sulphanilic acid reacts with imidazole group of histidine to give a cherry red color under alkaline condition. It shows the presence of histidine in albumin.

Take 2 ml of albumin solution and add 2 drops of Molisch's reagent. Then add 2–3 ml of conc. H₂SO₄ along the sides of the tube.

No reddish purple or violet ring is seen at the junction of the two liquids.

This shows the absence of carbohydrate in albumin.

Result : The reactions of albumin are thus studied.

1. Albumin is simple heat coagulable protein.

2. It is precipitated only by full saturation with ammonium sulphate.

3. It has all amino acids.

1. Take 2 ml of gelatin solution in a test tube. Add 5–10 drops of Lead acetate solution. Mix well.

No precipitate is formed.

Gelatin is not precipitated by the positive metal ions.

2. Take 2 ml of gelatin solution in a test tube and add a few drops mercuric sulphate solution.

No precipitate is formed.

Gelatin is not precipitated by the positive metal ions.

1. Take 2 ml of gelatin solution. Add 2–3 drops of sulpho salicylic acid.

Sulphosalicylic acid has negatively charged ions which neutralize the positive charge of gelatin to cause denaturation to produce precipitation of protein salicylate.

2. Take 2 ml of gelatin solution. Add a few drops of freshly prepared tannic acid

3. Take 2 ml of gelatin solution. Add 1ml of picric acid.

Take 3 ml of gelatin solution in a test tube and add equal volume of saturated ammonium sulphate solution. Mix well and wait for 5 min. To the filtrate add 3 ml of 40% sodium hydroxideand 2–3 drops of 1% copper sulphate solution.

Filtrate does not give violet color.

Gelatin gets precipitated by half saturation method and so the filtrate does not give violet color with biuret reagent as it does not contain gelatin.

Fill 3/4 of the test tube with gelatin solution. Hold the test tube at the bottom. Incline it slightly and heat the upper part of the solution over a flame. Add 3–5 drops of 1% acetic acid.

No precipitate is formed.

It indicates that gelatin is not a heat coagulable protein.

Take 2 ml of gelatin solution in a test tube and add 2 ml of 5% sodium hydroxide. Mix and add 2 drops of 1% copper sulphate solution.

A violet colored complex is formed.

The color produced is due to denaturation and the formation of co-ordination complex between cupric ions and the nitrogen atoms of the peptide bonds.

Take 2 ml of gelatin solution in a test tube. Add 2–3 drops of ninhydrin and boil. Leave it aside and allow it to cool.

A bluish purple colored complex is formed.

Gelatin contains free amino acids which react with Ninhydrin to give a blue color. This test is not answered by proline and hydroxyproline.

Take 3 ml of gelatin solution in a test tube and add 1 ml of conc. nitric acid and mix. Heat it for a minute. Cool it under tap water and add 2 ml of 40% NaOH.

A faint yellow color is obtained.

Presence of trace amount of aromatic amino acids.

Take 1 ml of gelatin solution in test tube. Add 1 ml of 10% sulphate. Boil for 30 seconds. Cool under tap water. Add 3 drops of sodium nitrite solution. Mix.

A faint red color is obtained.

Shows the presence of a trace amount of mercuric tyrosine in gelatin.

Take 2 ml of gelatin solution in a test tube. Add 2 drops of 1/500 formalin and 1 drop of 10% mercuric sulphate in 10% H₂SO₄. Add 3 ml of conc. sulphuric acid along the sides of the test tube.

No purple or violet colored ring is formed at the junction of the two liquids.

This shows the absence of tryptophan in gelatin as this test is specific for the indole group of tryptophan.

Take 3 ml of gelatin solution in a test tube. Add 5 drops of 20% NaOH and 4 drops of Molisch's reagent (1% alpha naphthol). Mix. Add 10 drops of bromine water.

A bright red color is formed.

This test is specific for guanidine group of arginine. In alkaline medium alpha naphthol combines with the guanidine group of arginine to form a complex which is oxidized by sodium hypobromite to produce red color. It shows the presence of arginine in gelatin.

Take 3 ml of gelatin solution in a test tube. Add 2 ml of 40% NaOH. Boil for a minute. Then add 5 drops of lead acetate.

No black or brown precipitate is formed.

This indicates the absence of sulphur containing amino acidscysteine and cystine in gelatin.

Take 1 ml of 0.5% of sulphanilic acid in a test tube. Add 1 ml of sodium nitrite solution. Mix. Add 2 ml of gelatin solution. And then add 1 ml of 10% sodium carbonate to make it alkaline.

An orange red colored complex is formed.

This test is specific for imidazole group of histidine. Diazotized sulphanilic acid reacts with imidazole group of histidine to give a cherry red color under alkaline conditions. It shows the presence of histidine in gelatin.

Take 2 ml of gelatin solution and 2 drops of Molisch's reagent. Then add 2–3 ml of conc. sulphuric acid, along the sides of the tube.

No reddish purple or violet ring is seen at the junction of the two liquids.

This shows the absence of carbohydrates in gelatin.

Result: The reactions of gelatin are thus studied.

1. Gelatin is not heat coagulable.

2. It is precipitated by half saturation with ammonium sulphate.

3. Tryptophan and sulphur containing amino acids are not present in gelatin.

4. It has no carbohydrate.

1. Take 2 ml of peptone solution in a test tube. Add 5–10 drops of lead acetate solution. Mix well.

A white precipitate is formed.

This test indicates the presence of proteins. The metal cation combines with negatively charged group on peptones causing precipitations.

2. Take 2 ml of peptone solution in a test tube. Add 2 drops of mercuric sulphate solution drop by drop.

A white precipitate is formed.

This test indicates the presence of proteins. The metal cation combines with negatively charged group on peptones to cause precipitations.

1. Take 2 ml of peptone solution Add 2 drops of freshly prepared tannic acid

Brown precipitate is formed.

Tannic acid has negatively charged ions which neutralise the positive charges on peptones causing denaturation resulting in precipitation.

2. Take 2 ml of peptone solution. Add 2 to 3 drops of picric acid.

Yellow precipitate is formed.

Picric acid has negatively charged ions which neutralize the positive charge of peptones causing denaturation resulting in precipitation.

3. Take 2 ml of peptone solution. Add 3 drops of glacial acetic acid and add 2 drops of potassium ferricyanide solution.

Yellow precipitate is obtained.

Glacial acetic acid has negatively charged ions which neutralize the positive charge on peptones and cause denaturation resulting in precipitation.

Take 3 ml of peptone solution in a test tube. Add equal volume of saturated ammonium sulphate solution. Mix well. Wait for 5 minutes. Then add 3 ml of 40% sodium hydroxide and 2–3 drops of copper sulphate.

It gives rosy red color.

Peptone is not precipitated by half saturation with ammonium sulphate. So it gives rosy red color with biuret reagent as it contains peptones.

Take 3 ml of peptone solution in a test tube. Add solid ammonium sulphate, until little salt settles at the bottom. Wait for 5 minutes. To this add 3 ml of 40% sodium hydroxide and 2–3 drops of 1% copper sulphate solution. Mix.

Rosy red color is formed.

Peptones are not precipitated by full saturation with ammonium sulphate. Hence, it gives rosy red color as it contains peptones.

Fill ¾th of the test tube with peptone solution. Hold the test tube at its bottom. Incline it slightly and heat the upper portion of the solution over the flame. Add 3 drops of 1% acetic acid.

No white precipitate is formed.

Peptones are not heat coagulable proteins

Take 2 ml of peptone solution in a test tube. Add 2 ml of 5% sodium hydroxide. Mix. Then add 2–3 drops of 1% copper sulphate solution and mix.

A rosy red or pink colored complex is formed.

It is a general test for protein. This test is answered by compounds which contain 2 or more peptide bonds. The color produced is due to the formation of coordination complex of cupric ions and nitrogen atoms of the peptide bonds present in peptones.

Take 1 ml of peptone solution in a test tube. Add 2 to 3 drops of ninhydrin reagent. Mix and boil and allow it to cool.

A bluish purple colored precipitate is formed.

Peptones give positive reactions. This test is answered by all amino acids containing free alpha amino group.

Take 3 ml of peptone solution. Add 1 ml of conc. nitric acid. A white precipitate is formed. Boil it for 1 minute. An yellow colored precipitate is formed. Cool it under tap water. Add 2 ml of 40% sodium hydroxide to make the solution alkaline.

A faint orange colored precipitate is formed.

Peptones show faint reaction. This test is specific for benzene ring present in the aromatic amino acids such as tyrosine, and tryptophan.

Take 1 ml of the peptone solution Add 1ml of 10% mercuric sulphate in 10% sulphuric acid solution. Boil for 30 seconds. Cool under tap water. Add 3 drops of sodium nitrite solution Mix and warm.

A red precipitate is formed.

This test is specific for hydroxy phenyl group of tyrosine. The red precipitate is due to the mercury complex of nitrophenol derivative. It shows the presence of tyrosine in peptone.

Take 3ml of the peptone solution. Add 2 drops of 1/500 formalin and 1 drop of 10% mercuric sulphate in 10% H₂SO₄. Mix. Incline the test tube and add 3 ml of concentrated sulphuric acid drop by drop along the sides of the test tube.

A violet or purple colored ring is seen at the junction of 2 layers.

Peptones show positive reaction. This test is specific for indole group of tryptophan in peptones. The indole group combines with formaldehyde and oxidizing agent to form the violet ring. The presence of tryptophan in peptone is confirmed.

Take 3 ml of the peptone solution in a test tube. Add 5 drops of 20% sodium hydroxide, 4 drops of Molisch's reagent and 10 drops of bromine water.

A bright red color is seen

This test is specific for guanidine group of arginine in peptones. In alkaline medium alpha naphthol combines with guanidine group of arginine to form a complex which is oxidized by sodium hypobromite to produce a red color. It shows the presence of arginine in peptones.

Take 2 ml of peptone solution Add 2 ml of 40% sodium hydroxide. Mix. Boil for 1 minute. Add 5 drops of lead acetate solution.

A faint black or brown precipitate is formed.

Peptones show faint positive reaction. It shows the presence of cysteine and cystine and not methionine. The sulphur present is liberated as sodium sulphide which react with lead acetate to form a black precipitate of lead sulphide. It shows that small amount of cysteine and cystine are present in peptones.

Take 1 ml of 0.5% of sulphanilic acid in a test tube. Add 1ml of sodium nitrite solution. Mix and add 2 ml of peptone solution. Add 1 ml of 10% sodium carbonate to make it alkaline.

Orange red colored complex is seen.

Peptones show positive reaction. This test is specific for imidazole group of histidine. Diazotized sulphanilic acid reacts with imidazole group to form a cherry red colored complex under alkaline conditions. It shows the presence of histidine in peptones.

Take 2 ml of peptone solution and add 3 drops of Molisch's reagent. Mix. Incline the test tube and add 2 ml of concentrated sulphuric acid along the sides of the test tube.

No reddish purple ring is seen at the junction of the two layers.

Peptones show negative reaction due to the absence of carbohydrate.

Result: The reactions of peptones are thus studied.

1. Peptones are not heat coagulable.

2. They are not precipitated either by half or full saturation with ammonium sulphate.

3. Peptones contain all amino acids. Cystine and cysteine are present in small amount.

4. Peptones are not glycoproteins.

Take 3 ml of casein solution in a test tube. Equal volume of saturated ammonium sulphate solution is added and mixed. Wait for 5 min. A white precipitate is formed. Filter it. To the filtrate add 3 ml of 40% sodium hydroxide and 1drop of 1% copper sulphate.

Filtrate does not give violet color.

Casein in the solution is precipitated by half saturation with ammonium sulphate. The filtrate does not give violet color with biuret reagent as it does not contain casein.

Take 3 ml of casein solution in a test tube. Add 3 drops of bromocresol green indicator. Mix. A blue color is seen. Now add 2% acetic acid till the color changes to light green (pH 4.6).

A curdy white precipitate is formed.

It indicates that casein is precipitated at isoelectric pH of 4.6.

Take 15 ml of casein solution in a conical flask. Add N/10 HCl drop by drop with constant shaking until a maximum precipitate is obtained. Filter. Collect and dry the precipitate between the folds of the filter paper. Take a small amount of precipitate. Add 6 drops of conc. sulphuric acid and 2 drops of conc. nitric acid. Heat gently over the flame. Yellow fumes of nitrous oxide will be evolved first. Stop heating when white fumes of sulphuric acid are evolved. The solution becomes clear and colorless. If the solution is dark, add a drop of conc. nitric acid and heat it again till it becomes clear. Allow it to cool. Add 3 ml of distilled water and 1 drop of methyl red indicator. Add conc. ammonia drop by drop till the solution becomes yellow. Now add 2 ml of ammonium molybdate solution. Heat gently for 2 minutes.

A canary yellow precipitate is formed.

Indicates the presence of phosphorus in casein. On heating with sulphuric acid and nitric acid, casein is digested and phosphorus is liberated. It reacts with ammonium molybdate to form a canary yellow precipitate of ammonium phospho molybdate.

To 5 ml of casein solution, 0.5 ml of 20% NaOH is added, heated and cooled. 0.5 ml of conc. nitric acid is added and mixed. It is then filtered and to the filtrate, a pinch of solid ammonium molybdate is added and warmed.

A canary yellow precipitate is formed.

On boiling with NaOH, Organic phosphate is converted to inorganic phosphorus. On reacting with ammonium molybdate it is converted to ammonium phosphomolybdate.

Take 2 ml of casein solution in a test tube. Add 2 ml of 5% sodium hydroxide. Mix. Then add 2–3 drops of 1% copper sulphate.

A violet or purple colored complex is formed.

It is a general test for protein. This test is sodium answered by compounds which contain 2 or more peptide bonds. The color produced is due to the formation of coordination complex of cupric ions and nitrogen atoms of the peptide bonds present in casein.

Take 1 ml of casein solution in a test tube. Add 2 to 3 drops of ninhydrin reagent. Mix and boil and allow it to cool.

A bluish purple colored precipitate is formed.

Casein gives positive reactions. This test is answered by all amino acids containing free alpha amino group.

Take 3 ml of casein solution. Add 1ml of concentrated nitric acid. A white precipitate is formed. Boil it for 1 minute. An yellow colored precipitate is formed. Cool it under water. Add 2 ml of 40% sodium hydroxide to make the solution alkaline.

An orange colored precipitate is formed.

This test is specific for benzene ring present in the aromatic amino acids such as tyrosine and tryptophan.

Take 1 ml of the casein solution. Add 1 ml of 10% mercuric sulphate in 10% H₂SO₄. Boil for 30 seconds. Cool under tap water. Add 3 drops of sodium nitrite solution. Mix and again warm.

A red precipitate is formed.

This test is specific for hydroxy phenyl group of tyrosine. The red precipitate is due to the mercury complex of nitrophenol derivative.

Take 3 ml of the casein solution. Add 2 drops of 1/500 formalin and 1 drop of 10% mercuric sulphate in 10% H₂SO₄. Mix. Incline the test tube and add 3 ml of concentrated sulphuric acid drop by drop along the sides of the test tube.

A violet or purple colored ring is seen at the junction of 2 liquids.

Casein shows positive reaction. This test is specific for indole group of tryptophan in casein. The indole group combines with formaldehyde in the presence of oxidizing agents to form a violet colored ring. The presence of tryptophan in casein is confirmed.

Take 3 ml of casein solution in a test tube. Add 5 drops of 20% sodium hydroxide, 4 drops of Molisch's reagent and 10 drops of bromine water.

A bright red color is seen.

This test is specific for guanidine group of arginine. In alkaline medium alpha naphthol combines with guanidine group of arginine to form a complex which is oxidized by sodium hypobromite to produce a red color. It shows the presence of arginine.

Take 2 ml of casein solution. Add 2 ml of 40% sodium hydroxide Mix. Boil for 1 minute. Add 5 drops of lead acetate solution.

No black or brown precipitate is formed.

Casein shows negative reaction. It shows the absence of cysteine and cystine in casein.

Take 1ml of 0.5% of sulphanilic acid in a test tube. Add 1 ml of sodium nitrite solution. Mix and add 2 ml of casein solution. Add 1 ml of 10% sodium carbonate to make it alkaline.

An orange red colored complex is seen.

Casein shows positive reaction. This test is specific for imidazole group of histidine. Diazotized sulphanilic acid reacts with imidazole group to form a cherry red colored complex under alkaline conditions. It shows the presence of histidine in casein.

Take 2 ml of casein solution and add 3 drops of Molisch's reagent. Mix. Incline the test tube and add 2 ml of concentrated sulphuric acid along the sides.

No reddish purple ring is seen at the junction of the two layers.

Casein shows negative reaction due to the absence of carbohydrate.

Result: The reactions of casein are thus studied.

1. Casein is not heat coagulable.

2. It is precipitated by half saturation with ammonium sulphate.

3. It contains all amino acids except cysteine and cystine.

4. It is not a glycoprotein. It is a phosphoprotein.

1. Biuret test

Violet color is produced.

Presence of peptide bonds in protein.

2. Precipitation at isoelectric point

Curdy white precipitate No precipitate

Presence of casein. Absence of casein.

3. Heat coagulation test

Formation of coagulum. No coagulation.

Presence of albumin. Absence of albumin.

4. Half saturation test

Formation of violet color

No violet color

Presence of albumin or peptone.

Presence of gelatin or casein.

5. Full saturation test

Formation of violet color

No violet color

Presence of peptone

Presence of albumin

1. Biuret test

Violet colored complex.

Albumin, gelatin, peptone and casein answer this.

2. Ninhydrin test

Bluish purple colored

All give positive result.

3. Xanthoproteic test

Precipitate orange colored

All give positive reaction.

Gelatin gives faint positive.

4. Cole's mercuric nitrite test

Red precipitate

All give positive reaction.

Gelatin gives faint positive.

5. Aldehyde test

Purple ring seen at the junction of 2 liquids.

No ring formation Gelatin

Albumin/Casein /Peptones

6. Sulphur test

Black or brownish black precipitate

No black precipitate

Albumin or peptone

Gelatin or casein

7. Sakaguchi test

Bright red color

Albumin / gelatin/ peptone/ casein.

8. Pauly's test

Orange red color

–do-

9. Molisch's test

Reddish purple ring at the junction of 2 liquids

No ring formation

Albumin

Peptone/gelatin/casein

Result : The given protein is………………….

(a)

1% urea solution.

(b)

Alkaline hypobromite solution-100 ml of 40% sodium hydroxide with 10 ml of bromine water.

Normal value in blood:

Male

–

3–7 mg/100 ml

Female

–

2.5 – 6.5 mg/100 ml

Normal level in urine

250–750 mg/day.

1. Concentrated nitric acid.

2. Dilute ammonia (10%)

1. Saturated picric acid solution.

2. 10% sodium hydroxide solution.

Take 3 ml of urea solution in one test tube and 3 ml of distilled water in another test tube. Add 5 drops of freshly prepared alkaline hypobromite solution in both the test tubes and mix.

A brisk effervescence is observed in the tube containing urea

Alkaline hypobromite decomposes urea to N2, CO2 and H2O. Brisk effervescence is due to liberation of N2, CO2 gas. This principle is used in the quantitative estimation of urea in urine.

To 3 ml of urea solution add half spatula of horse gram powder and add a drop of phenolphthalein indicator. Warm the tube for few seconds with the hands and leave aside for 5–10 minutes

A pink color develops.

Horse gram is the source of the enzyme urease which acts on urea to form ammonium carbonate which is identified by the indicator.

Take 3 ml of uric acid solution add 1 ml of Benedict's uric acid reagent (Phospho tungstic acid) and 1 ml of 20% sodium carbonate solution and mix.

A deep blue color is seen.

Uric acid in alkaline condition reduces phosphotungstic acid to tungsten blue.

Moisten a piece of filter paper with few drops of ammoniacal silver nitrate solution. Add 2–3 drops of uric acid solution on the same paper.

A black color develops.

Uric acid reduces salts of silver nitrate to metallic silver.

Take a little quantity of uric acid solution in a china dish and add few drops of nitric acid and warm gently over a flame. A reddish yellow residue is got. Wait for 3 minutes for cooling and add 1–2 drops of dilute ammonia solution.

A purplish red color develops on the addition of ammonia.

Uric acid is oxidized by nitric acid to give purpuric acid which is reddish yellow in color. This combines with ammonia to form ammonium purpurate or murexide to form red color.

Take 3 ml of creatinine solution in test tube and 3 ml of water in another test tube. Add 1ml of saturated picric acid and 10 drops of 10% sodium hydroxide to each tube, mix and wait for 5 minutes.

A deep reddish orange color develops in the tube containing creatinine and yellow color is seen in the control tube.

Creatinine reacts with picric acid in the presence of alkali to form creatinine pictrate which is reddish orange in color.

Result : The reactions of NPN are studied.

a.

Simple lipids

– Fats, waxes

b.

Complex lipids

– Phospho lipids, Glycolipids

c.

Precursor and derived lipids

– Fatty acids, glycerol, sterol, alcohols etc.

a. To 1 ml of palmitic acid in alcohol, bromine water is added drop by drop.

Pale yellow color of Bromine is retained

Indicates the presence of saturated fatty acid.

b. To 1 ml of oleic acid (olive oil) in alcohol, bromine water is added drop by drop.

Yellow color of Bromine, disappeared.

Indicates the presence of unsaturated fatty acids. Bromine forms a dibromine with the double bonds of unsaturated fatty acid, and gets decolorised.

To 2 ml of cholesterol solution in chloroform taken in a clean dry test tube, 10 drops of acetic anhydride and 3 drops of conc. sulphuric acid are added and mixed.

Solution turns purplish red, then blue and finally emerald green in color.

Indicates the presence of cholesterol. The change in color is due to dehydration by sulphuric acid and acetic anhydride

To 2 ml of chloroform solution of cholesterol, taken in a clean, dry test tube, 2 ml of conc. sulphuric acid is added along the sides of the test tube.

A reddish violet color is seen in the upper chloroform layer and the lower acid layer becomes yellowish in color with green fluorescence.

It indicates the presence of cholesterol.

Result: The reactions of Lipids are thus studied.

1. How do you classify lipids?

2. How do you classify fatty acids?

3. What is the fat of our body called?

4. Name the essential fatty acids.

5. What are the functions of essential fatty acids?

6. How will you detect essential fatty acids?

7. How will you detect cholesterol?

8. What is the principle of this test?

9. What is the ring of cholesterol structure called as?

10. How many carbon atoms are there in cholesterol?

11. What are the derivatives of cholesterol?

12. What is the normal cholesterol level in blood?

Proteins

1.2

4.0

Fat

3.6

3.5

Lactose

6.8

4.6

Minerals

0.2

0.75

100 ml of milk is taken in a glass cylinder and the lactometer is allowed to float on it without touching the wall of the glass cylinder. The mark on the stem of the lactometer is noted.

The mark on the stem coinciding the surface of the milk is……

This is the normal specific gravity of milk which is between 1.028 to 1.034.

A litmus paper or indicator paper is dipped in milk and the color is compared to the standard chart.

pH of milk is ……

pH of fresh milk is 6.4 to 6.8

a. 20 ml of milk is taken in a beaker and 20 ml of water is added to it. Then 20 ml of 1% acetic acid is added drop by drop with slow stirring with a glass rod.

A white precipitate is obtained.

Casein with fat is precipitated

b. The precipitate is filtered using double filter paper. Then it is dried. The following experiments will be performed separately with (A) Precipitate and (B) Filtrate

A small amount of precipitate is taken in a test tube and 1ml of 5% sodium hydroxide is added to dissolve the precipitate. 5 ml of water is added and mixed and color reactions of casein are performed. (Refer Page 65 to 68)

All color reactions are positive except sulphur test

Casein does not contain sulphur containing amino acids such as cystine and cysteine.

A small amount of precipitate is taken in a test tube and Neumann's test is performed. (Refer Page 65)

Canary yellow precipitate is obtained

Indicates the presence of casein which is a phosphoprotein.

A small portion of the precipitate is taken in a test tube and 3 ml of ether is added to dissolve it. The supernatant is decanted to a clean dry china dish and allowed to evaporate at room temperature.

Oily spots are seen in the china dish.

The fat precipitated along with casein are extracted with ether. This shows the presence of fat in milk.

5 ml of filtrate is taken in a test tube and the top portion is heated over the flame. Few drops of 1% acetic acid is added.

Coagulum is seen in the heated portion.

Indicates the presence of the heat coagulable proteins—Lactalbumin and lactoglobulins

a. 2 ml of filtrate is taken in a test tube and 2 ml of 2% sodium carbonate is added and mixed. 3 ml of Benedict's qualitative reagent is added and boiled for 2 minutes.

Green or yellow or red precipitate is formed.

Indicates the presence of reducing sugar in milk.

b. Osazone test is performed with the filtrate (Refer Page 20)

Cottonball shaped lactosazone crystals appear within 20 minutes.

Presence of lactose confirmed.

5 ml of filtrate is taken in a test tube and 5 drops of strong ammonia solution is added and mixed well to get a precipitate which is filtered and filtrate is discarded. Precipitate is dissolved in 5 ml of dilute acetic acid and divided into 2 parts.

a. To one part, 2 ml of 2% potassium oxalate solution is added.

A white precipitate of calcium oxalate is formed.

Shows the presence of calcium in milk.

b. To the other portion, 1 ml of Conc. nitric acid and and 3 ml of ammonium molybdate reagent are added and warmed.

A canary yellow precipitate of phospho- molybdate is formed.

Indicates the presence of phosphorus in milk.

Result: Constituents of milk are thus analysed.

1. What are the composition of milk?

2. How does human milk differ from cow's milk?

3. Name the vitamins and minerals present in milk

4. Name the sugar of milk

5. Name the milk proteins

6. What type of protein is casein?

a. Hydrochloric acid

– Secreted by parietal cells.

b. Enzymes

– Pepsin - Secreted by chief cells.

c. Mucin

– secreted by mucous cells.

2 ml of gastric juice is taken in a test tube and 2 drops of Topfer's indicator is added and mixed.

Red color is seen.

Indicates the presence of hydrochloric acid.

To 2 ml of gastric juice, add few drops of Iodine.

Blue color is formed.

Indicates the presence of starch due to the stasis of food in stomach as in duodenal ulcer or due to delayed emptying of stomach.

To a pinch of benzidine powder, 1ml of glacial acetic acid is added and dissolved. Then 1ml of Hydrogen peroxide is added to it and then 1ml of gastric juice is added.

Blue or green color develops which turns to brownish black in color within few minutes.

This shows the presence of blood in gastric juice which may be due to gastric ulcer or carcinoma of stomach or due to injury.

Two test tubes are taken. In the first tube 2 ml of gastric juice and in the second 2 ml of water are taken. A pinch of sulphur powder is sprinkled over the surface of the liquid in each tube.

Sulphur powder sinks to the bottom of the tube containing gastric juice and floats in water

Indicates the presence of bile salts in gastric juice. Bile could enter the stomach by regurgitation.

To 5 ml of gastric juice few crystals of magnesium sulphate are added and dissolved by shaking. Then 2 ml of 10% barium chloride is added and mixed to get a precipitate. This is filtered in a filter paper and dried. Then 1–2 drops of Fouchet's reagent is added on the dry precipitate

A green color is developed.

Shows the presence of bile pigments. Bile enters the stomach by regurgitation.

To 2 ml of gastric juice, 2 ml of Ufflemann's reagent is added and mixed. (Ufflemann's reagent – 1% Phenol + 10% FeCl₃)

Violet color is changed to yellow color.

Indicates the presence lactic acid.

2 ml of gastric juice is taken in a test tube and a few drops of Maclean's reagent (Containing mercuric chloride and ferric chloride) is added. A control with 2 ml of distilled water and few drops of Maclean's reagent is performed.

Yellow color is developed in gastric juice. No color change in distilled water.

Indicates the presence of lactic acid which is formed due to the fermentation of carbohydrates due to low HCl as found in cancer of stomach.

Results: The gastric juice contains………………………

1. What are the normal constituents of gastric juice?

2. Name the types of gastric cells and the various secretions secreted by them.

3. What is the pH of gastric juice?

4. Which cells secrete HCl?

5. What is the function of HCl?

6. Name the enzymes present in the gastric juice?

7. What is the daily output of gastric juice?

8. What is the function of pepsin?

9. What is the precursor form of pepsin?

10. What are the possible abnormal constituents of gastric juice?

11. In which diseases, blood will be present in gastric juice?

12. If lactic acid is present in gastric juice, what pathology do you think of?

13. In which conditions bile will be present in gastric juice?

14. Which indicator is used to detect the presence of Hcl in gastric juice?

1. pH: Test the pH with red litmus paper.

Litmus paper turns blue

pH of bile is alkaline

2 test tubes are taken. In one test tube 3ml of diluted bile is taken and in the other test tube 3ml of water is taken as ‘control’. Little quantity of sulphur powder is sprinkled over the surface of the liquid in each tube. Should not mix the contents.

Sulphur powder sinks to the bottom of the tube in bile and floats in water.

Indicates the presence of bile salts. Surface tension of bile is lowered by bile salts and so sulphur powder sinks. (This is useful in detecting bile salts in urine)

3 ml of diluted bile is taken in a clean test tube and a little quantity of sucrose (table sugar is added and dissolved. 3 ml of conc. sulphuric acid is added along the sides of the test tube.

A reddish purple ring is seen at the junction of 2 liquids.

Indicates the presence of bile salts. Sucrose is hydrolysed and dehydrated by conc. H₂SO₄ to form furfural derivative which condenses with bile salt to form the reddish purple ring.

Note: (This test is not used to detect bile salts in urine due to presence of interfering substances in urine)

3 ml of diluted bile solution is taken in a test tube. Few crystals of magnesium sulphate is added and dissolved. Then 2 ml of 10% barium chloride is added and mixed. A white precipitate is formed which is then filtered by a filter paper kept in a funnel. The filter paper is then unfolded and dried by means of blotting by another filter paper. 1–2 drops of Fouchet's reagent is added on the dry precipitate.

Green color develops in the precipitate.

Shows the presence of bile pigments. The precipitate formed is barium sulphate which absorbs bile pigments. Ferric chloride present in the Fouchet's reagent oxidises the bile pigments to give green colored pigment biliverdin.

5 ml of conc. nitric acid is taken in a test tube. Bile is taken in a 10 ml pipette and the pipette is inserted into the tube containing nitric acid and then bile is added to the nitric acid drop by drop.

Green, blue and brown colored rings are obtained.

Shows the presence of bile pigments. Bilirubin is oxidised by nitric acid to biliverdin (green), bilicyanin (blue), and bilifuscin (brown).

To 3 ml of diluted bile, few drops of glacial acetic acid is added. Little more acetic acid is then added.

Greenish white precipitate is formed which is not soluble in excess of acetic acid.

Shows the presence of nucleoproteins in bile. Excess of acid makes the precipitate insoluble by bile salts.

Result: The reactions of bile are thus studied.

1 in 200 dilution of blood is prepared, and taken in a clean test tube and examined with spectroscope.

2 bands are seen in the green region. The band closer to D line is narrow while other is broader. The reading of 2 bands are 576 and 540 nm respectively (α and β band).

Shows the presence of oxyhemoglobin.

To 1 : 100 diluted blood a pinch of reducing agent sodium hydrosulphite is added and mixed. The solution is viewed through spectroscope.

Now shake the tube vigourously.

The contents turn purple. A single band in the green region is seen at 565 nm.

The purple color is changed to red.

Shows the presence of reduced Hb.

Due to the formation of Oxy Hb.

To 5 ml of 1 in 50 diluted blood taken in a test tube, a small amount of potassium ferricyanide is added and mixed well.

Red color changes to brown.

Due to the formation of met Hb

a. Examine through a spectro scope.

A band is seen in the red region with its center at 630 nm. 2 faint bands are seen in green in the same place as Oxy Hb.

b. A pinch of sodium hydrosulphite is added and mixed and viewed through spectroscope.

The solution becomes purple. A single band of reduced Hb is seen.

A drop of blood is spread on a glass slide to form a thin film and it is dried over a low flame. 2 drops of Nippe's fluid is added and a cover glass is placed in position. Then the slide is heated gently until gas bubbles are formed. 2 drops of Nippe's reagent is run underneath the glass and cooled. The glass slide is viewed under the microscope.

Brown rhomboid crystals of hemin are seen.

Upon heating with acid, Hb is denatured and haem is oxidised to hematin which is then converted to hematin chloride which is called hemin.

Result : The reactions of hemoglobin are thus studied.

Normal Value of total Inorganic sulphates

– 0.7 to 1gm/day

Increased in

– High protein diet

Decreased in

– Renal dysfunction.

Urinary sulphates

–

Increased in:

–

Inherited disorders - Cystinuria, Homocystinuria.

–

Cyanide poisoning - thiocyanate

Normal Urinary Urea

– 15–30 gms/24 hours.

Increased in:

– (Increased protein catabolism)

– High protein diet.

– Fever

– Diabetes mellitus.

Decreased in:

– Low protein diet

– Liver diseases

– Nephritis

– Acidosis.

Normal value

– 0.5 to 1 gm/day.

Increased in

– leukemias especially during cytotoxic drug therapy.

– Wilson's disease.

– Administration of cortisone/ACTH.

Increased urobilinogen

– Hemolytic jaundice, liver diseases.

Absence of urobilinogen

– Obstructive jaundice.

1. Color

–

Straw colored

2. Turbidity

–

Clear

3. pH

–

6.0

4. Specific gravity

–

1.015 – 1.025

5. Volume/day

–

1.5 litres

Solids

–

60 gm/L

Water

–

1200 ml

Sodium chloride

–

10 – 12 gm / day

Calcium

–

0.1–0.3 gm / day

Phosphates

–

0.8 – 1.3 gm / day

Sulphates

–

0.7– 1.0 gm / day

Potassium

–

3.0 gm / day

Urea

–

15–30 gm / day

Uric acid

–

0.5 to 1.0 gm / day

Ammonia

–

0.6– 0.7gm / day

Creatinine

–

1–2 gm / day

Hippuric acid

–

0.6 gm

Indican

–

0.01 gm

1. Color

Straw colored.

Due to the presence of urochrome.

2. Appearence

Clear.

Normal urine is clear and transparent.

3. Reaction (pH) Test the pH by using pH paper or Litmus paper

pH of the given urine is (Acidic or alkaline)

pH of normal urine is 4.6 – 8.0 (average 6.0)

4. Specific gravity Test: Fill 3/4 of glass cylinder with urine. Float the urinometer without touching the sides. Note the mark on the stem of the urinometer coinciding to the surface of urine. That gives the specific gravity of urine.

Specific gravity of the given urine is

Normal specific gravity of urine is 1.015 – 1.025.

To 3 ml of urine taken in a test tube, 0.5 ml of conc. nitric acid is added and mixed. 1 ml of 3% silver nitrate solution is then added.

A white precipitate of silver chloride is formed.

Shows the presence of chlorides.

To 10 ml of urine, 5–10 drops of strong ammonia is added and boiled, and cooled. After 5 minutes, a white precipitate is formed. Filter it and discard the filtrate. Add 5 ml of dilute acetic acid through the sides of the filter paper. Pierce the filter paper with a glass rod. Collect this in a test tube and divide it into 2 parts.

A white precipitate is obtained (Calcium oxalate)

It shows the presence of calcium in urine.

A canary yellow precipitate is obtained. (Ammonium phosphomolybdate)

Indicates the presence of phosphorus in urine.

To 2 ml of urine add a drop of phenolphthalein indicator. Mix and add 2% sodium carbonate drop by drop with constant mixing till a pink color is got. Boil this solution. While boiling, keep a glass rod dipped in phenolphthalein indicator near the mouth of the tube.

Phenolphthalein indicator in the glass rod changes to pink color.

The appearance of pink color is due to evolution of ammonia from decomposition of ammonia salts in urine.

Take 5 ml of urine in a test tube Add 1 ml of conc. hydrochloric acid. Mix well and add 5 ml of 10% barium chloride.

A white precipitate of barium sulphate is formed.

This indicates the presence of inorganic sulphates in urine.

Filter off the precipitate got in above experiment. Boil the filtrate.

A white precipitate of barium sulphate is obtained.

The filtrate from the above experiment already contains excess BaCl₂ and HCl. Free sulphate is formed from organic sulphate by heating with HCl which reacts with barium chloride to form a white precipitate of barium sulphate.

To 3 ml of urine, 5 drops of freshly prepared alkaline hypobromite solution is added and mixed.

Brisk effervescence is produced.

Shows the presence of urea in urine.

To 3 ml of urine, 3 drops of phenolphthalein indicator is added and mixed. Then a spatulaful of horsegram powder containing urease is added and mixed and the tube is rolled between the palms.

A pink color develops.

Presence of urea is confirmed.