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KENYATTA UNIVERSITY
SCHOOL OF MEDICINE.
DEPARTMENT OF MEDICAL BIOCHEMESTRY
NAME: LANDO ELVIS OTIENO
REG NO: P29S/16344/2015.
COURSE: MBCHB
LECTURER: Dr OKUN PHD.
HANDING DATE: 22 /10/2015
SUBJECT: PRACTICAL REPPORT ON TEST FOR
CARBOHYDRATES.
HANDING DATE: 22/10/2015
Introduction
Carbohydrates are polymers of sugar molecules that serve as storage molecules or as
structural molecules. Starch and glycogen are examples of storage molecules. Hydrolysis of
either starch or glycogen releases the monosaccharide glucose, which may then serve as fuel
for
cellular work or as a carbon source for synthesis of other organic molecules. Cellulose, which
makes plant cell walls tough, and chitin, the main ingredient of an insect’s exoskeleton, are
examples of structural molecules.
Carbohydrates classified as simple or complex. Simple carbohydrates, often called
monosaccharides or simple sugars, cannot be broken down into smaller carbohydrate
molecules.
Complex carbohydrates can be broken down into smaller carbohydrate units through a process
known as hydrolysis.
Carbohydrates as the name indicate are made from carbon (C) and hydrate (H2O) and
was originally applied to the compounds that gave the (CH2O)n formula (Applicable for
monosaccharides and for example C6H12O6) and often referred to as CHO. Disaccharides are
formed through the condensation (Polymreraization) reaction to remove one H2O molecule
and the formation of glycosidic bond between the two monosaccharides, disaccharides usually
have C2nH2n-2On-1 formula and for example C12H22O11. Polysaccharides form through the
addition of monosaccharides and the removal of one H2O moleculkes per each addition of the
monosaccharides. Polysaccharides usually have (CnH2n-2On-1)y formula, this formula
represent a
long monosaccharides residues minus H2O.
Chemically, carbohydrates are either aldehydes or ketones derivatives of polyhydroxylic
compound that known as aldoses and ketoses. Carbonyl group give aldoses and ketoses the
reducing properties. Sucrose is the most common non-reducing sugar as the reducing carbonyl
groups of glucose (C1) and fructose (C2) are involved in the glycosidic bond.
A:Monosaccharides:
Arabinose, Fructose, Glucose, Galactose, Mannose ,Ribose, Xylose
Disaccharides:
Sucrose
Lactose Maltose
Polysaccharides:
Starch
Cellulose
1 Molisch’s test
Principle:Concentrated sulphuric acid hydrolyses glycosidic bonds to give the monosaccharides,
which are then dehydrated to furfural and its derivatives. These products then combine with
sulphonated naphthol to give a purple complex.
This reaction is a general one for the presence of all carbohydrates (Monosaccharides
give a rapid positive test. Disaccharides and polysaccharides react slower) and other organic
compounds that give furfural with concentrated sulphuric acid.
Materials and reagents:
1. Carbohydrate solutions (1 g/l and 10 g/l).
2. Sulphuric acid (concentrated).
3. Molisch’s reagent (α-naphthol 50 g/l in ethanol, prepare fresh).
Method:
1.pipette 2ml of the sugar solution into 2 separate test tubes.
2.Add 2drops of the molisch reagent.
3. Mix thoroughly, incline the the tube, add 1 – 2mls of coc.H2SO4 down the slide, slowly and
carefully to form two layers.
4. The formation of a purple ring at the interface indicate presence of a carbohydrate. A green
color is not a positive reaction.In case of doubt, the test should be repeated in a more dilute
solution.
Observations and results.
Carbohydrates Color formed after adding
2mls of H2SO4 in the solution
fructose Yellowish-purple layer
arabinose Purple layer
maltose Slight purple leyer
glucose Purple layer
sucrose purple layer
Discussion Conclusion
-All the sugar solutions mixed with molisch reagent formed a purple ring, producing
heat(exothermic)as a result of reaction, when reacted with 2mls H2SO4.Hence are
carbohydrates.
This reaction is a general one for the presence of all carbohydrates (Monosaccharides
give a rapid positive test. Disaccharides and polysaccharides react slower) and other organic
compounds that give furfural with concentrated sulphuric acid.
Chemically, carbohydrates are either aldehydes or ketones derivatives of polyhydroxylic
compound that known as aldoses and ketoses. Carbonyl group give aldoses and ketoses the
reducing properties. Sucrose is the most common non-reducing sugar as the reducing carbonyl
groups of glucose (C1) and fructose (C2) are involved in the glycosidic bond. Therefore the acid
was used to break the bond in sucrose to its constituent monosaccharide’s , thereby forming a
purple furfural layer.
B:Reaction of reducing sugar
Reducing sugars are those sugars capable of reducing copper ions in a solution to cuprous
oxide. The reducing sugars include:
(1) All of the simple sugars (monosaccharide’s).
(2) A few of the disaccharides (including lactose and maltose, but not sucrose).
(3) None of the polysaccharides.
Background: If a suspension of copper hydroxide in alkaline solution is heated, then
black cupric oxide is formed.
Cu(OH)2 - CuO + H2O
However, if a reducing substance is present, then rust-brown cuprous oxide is precipitated.
2Cu(OH)2 - Cu2O + H2O + O2
In practice, an alkaline solution of a copper salt and an organic compound containing
alcoholic –OH is used rather than the above suspension. Under these conditions, the copper
forms a soluble complex and the reagent is stable.
Carbohydrates with a free or potentially free aldehyde or ketone groups have reducing
properties in alkaline solution. In addition, monosaccharides act as reducing agents in weakly
acidic solution.
2 Benedict’s test
Principle:
Reducing sugars are oxidized by the copper ions in the Benedict’s solution to form a
carboxylic acid and a reddish precipitate of cuprous oxide within three minutes in alkaline
medium. Benedict is a more sensitive alternative modified reagent to the original Fehling's test
to produce a single solution which is more convenient for tests, as well as being more stable
than
Fehling`s reagent.
Aldehyde (Carbonyl gp.) Benedicts solution Carboxylic acid Reddish precipitate
Materials and reagents:
1. Carbohydrate solutions (1 g/l and 10 g/l).
2. Benedicts reagent. (To prepare one litre of Benedicts reagent: mix 173 g of tri-sodium citrate
and 100 g of sodium carbonate with 800 ml of distilled water. Warm to dissolve, then cool and
filter, add more distilled water to make the volume to 850 ml. Dissolve 17.3 g of copper
sulphate
in 100 ml distilled water and stir slowly into the first solution. Bring the volume to 1 litre with
distilled water).
Method:
1- Add five drops of the test solution to 2 ml of Benedicts reagent and place in a boiling water
bath for 5 min.
2- Examine the sensitivity of benedicts reagent using increasing dilutions of glucose.
3- The formation of a orange-brown precipitate within 3-5 minutes indicates positive result.
Biochemistry laboratory notes / Carbohydrates by Amjad I.A. Hussein page 5 of 11
Negative result (left) Positive result (right)
Observations and results.
Carbohydrates Color after heating the
solution with benedicts
reagent for 5min
Color after
cooling
Fructose red brown
Arabinose red brown
Maltose red brown
Glucose red brown
Sucrose blue blue
Discusion/conclusion
-All the carbohydrates except sucrose were reduced by benedicts solution to form redish-
orange solution after heating for 5minutes, therefore are reducing sugars, and sucrose retained
the blue color solution after heating and therefore is a non-reducing sugar.
C: Tests for individual carbohydrates
3- Bial's (Orcinol) test
Principle:
This test is specific for pentoses. The concentrated HCl in Bial's reagent dehydrates pentoses to
form furfural. Furfural further reacts with orcinol and the iron ion to produce a bluish
(Blue-green) product extractable into amyl alcohol. The test is used for the determination of
RNA and the detection of pentosuria. The reaction is not absolutely specific for pentoses since
prolonged heating of some hexoses yields hydroxy-methyl furfural which also reacts with
orcinol
to give colored complexes.
Materials andreagents:
1. Carbohydrate solutions (1 g/l and 10 g/l).
2. Bials oricinol reagent (Dissolve 1.5 g of oricinol in 500 ml of conc. HCl and add 20 drops of
100 g/l solution of FeCl3).
3. Amyl alcohol.
Method:
1- Add 2 ml of the test solution to 3 ml of Bials reagent.
2- Heat until boiling commences.
Negative results (left & middle) Positive result (right)
2- The formation of a blue-green color indicates a positive result for pentoses.
3- Note that Hexoses generally react to form green, red, or brown products.
Carbohydrates Color in Bial´s reagent after heating for one
minute.
fructose brownish
arabinose greenish
maltose yellowish
glucose yellowish
sucrose brownish
Discusion
--Fructose and sucrose turned into blue color in Bials reagent after heating for 1 minute hence
are Hexoses, Arabinose turned to light green(positive) hence is a pentose, glucose and maltose
did not reactand retained the original yellow color. Sucrose is a disaccharide made of fructose
and glucose.
4- Seliwanoff’s test
Seliwanoff’s test distinguishes between aldose and ketose sugars. The test is based on
the fact that, when heated, ketoses are more rapidly dehydrated than aldoses to give furfural
derivatives. The reagents consist of resorcinol and hydrochloric acid; 1) the acid hydrolysis
polysaccharides and oligosaccharides yields simpler sugars; 2) Then the dehydrated ketose
reacts
with the resorcinol to produce a deep cherry red color. Aldoses may react slightly and more
slowly to produce a faint pink color. Fructose and sucrose (A disaccharide consisting of fructose
and glucose) are two common sugars that give a positive test.
Materials and reagents:
1. Carbohydrate solutions (1 g/l and 10 g/l).
2. Seliwanoff's reagent (Dissolve 0.5 g of resorcinol in 1000 ml of 3M HCl).
Method:
1- Add 2 drops of the test solution to 5 ml of Seliwanoff's reagent.
Resorcinol
Biochemistry laboratory notes / Carbohydrates by Amjad I.A. Hussein page 8 of 11
2- Warm in boiling water for 1-2 min.
Negative result (left) Positive result (right)
3- The formation of a red color within 2 min indicates a positive result for ketoses (Fructose and
Sucrose).
4- Note that prolonged heating with aldoses generally react to form faint pink color product.
Observations/results.
Carbohydrates Color in seliwanoff´s reagent
and heated for 60seconds
Color in seliwanoff´s reagent
and heated for 5minutes.
Fructose Orange/pick red
Arabinose colorless colorless
maltose colorless colorless
Glucose colorless colorless
Sucrose Orange/pick red
Distilled water colorless colorless
discusion
-Fructose and sucrose gradually turned into pinkish color at 1minute then to red color after
heating for 5minutes therefore are Ketose carbohydrates, the other carbohydrates( maltose,
arabinose and glucose) remained colorless because they are aldoses.
REFFERENCES:
1.https://www.scribd.com/doc/26652097/Carbohydrates-Tests-Practical
2.https://www.wattpad.com/100096-carbohydrate-biochemistry-tests
3.https://en.wikipedia.org/wiki/Molisch%27s_test
4.http://web.knust.edu.gh/oer/pages/index.php?siteid=knustoer&page=find_materials&cou
=37
5.Kenyatta university-laboratory-practical-manual- for-MBCHB-Year 1-semester1.

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Kenyatta university biochem carbohydrates test

  • 1. KENYATTA UNIVERSITY SCHOOL OF MEDICINE. DEPARTMENT OF MEDICAL BIOCHEMESTRY NAME: LANDO ELVIS OTIENO REG NO: P29S/16344/2015. COURSE: MBCHB LECTURER: Dr OKUN PHD. HANDING DATE: 22 /10/2015 SUBJECT: PRACTICAL REPPORT ON TEST FOR CARBOHYDRATES. HANDING DATE: 22/10/2015
  • 2. Introduction Carbohydrates are polymers of sugar molecules that serve as storage molecules or as structural molecules. Starch and glycogen are examples of storage molecules. Hydrolysis of either starch or glycogen releases the monosaccharide glucose, which may then serve as fuel for cellular work or as a carbon source for synthesis of other organic molecules. Cellulose, which makes plant cell walls tough, and chitin, the main ingredient of an insect’s exoskeleton, are examples of structural molecules. Carbohydrates classified as simple or complex. Simple carbohydrates, often called monosaccharides or simple sugars, cannot be broken down into smaller carbohydrate molecules. Complex carbohydrates can be broken down into smaller carbohydrate units through a process known as hydrolysis. Carbohydrates as the name indicate are made from carbon (C) and hydrate (H2O) and was originally applied to the compounds that gave the (CH2O)n formula (Applicable for monosaccharides and for example C6H12O6) and often referred to as CHO. Disaccharides are formed through the condensation (Polymreraization) reaction to remove one H2O molecule and the formation of glycosidic bond between the two monosaccharides, disaccharides usually have C2nH2n-2On-1 formula and for example C12H22O11. Polysaccharides form through the addition of monosaccharides and the removal of one H2O moleculkes per each addition of the monosaccharides. Polysaccharides usually have (CnH2n-2On-1)y formula, this formula represent a long monosaccharides residues minus H2O. Chemically, carbohydrates are either aldehydes or ketones derivatives of polyhydroxylic compound that known as aldoses and ketoses. Carbonyl group give aldoses and ketoses the
  • 3. reducing properties. Sucrose is the most common non-reducing sugar as the reducing carbonyl groups of glucose (C1) and fructose (C2) are involved in the glycosidic bond. A:Monosaccharides: Arabinose, Fructose, Glucose, Galactose, Mannose ,Ribose, Xylose Disaccharides: Sucrose Lactose Maltose Polysaccharides: Starch Cellulose 1 Molisch’s test Principle:Concentrated sulphuric acid hydrolyses glycosidic bonds to give the monosaccharides, which are then dehydrated to furfural and its derivatives. These products then combine with sulphonated naphthol to give a purple complex. This reaction is a general one for the presence of all carbohydrates (Monosaccharides give a rapid positive test. Disaccharides and polysaccharides react slower) and other organic compounds that give furfural with concentrated sulphuric acid. Materials and reagents: 1. Carbohydrate solutions (1 g/l and 10 g/l). 2. Sulphuric acid (concentrated). 3. Molisch’s reagent (α-naphthol 50 g/l in ethanol, prepare fresh). Method: 1.pipette 2ml of the sugar solution into 2 separate test tubes. 2.Add 2drops of the molisch reagent.
  • 4. 3. Mix thoroughly, incline the the tube, add 1 – 2mls of coc.H2SO4 down the slide, slowly and carefully to form two layers. 4. The formation of a purple ring at the interface indicate presence of a carbohydrate. A green color is not a positive reaction.In case of doubt, the test should be repeated in a more dilute solution. Observations and results. Carbohydrates Color formed after adding 2mls of H2SO4 in the solution fructose Yellowish-purple layer arabinose Purple layer maltose Slight purple leyer glucose Purple layer sucrose purple layer Discussion Conclusion -All the sugar solutions mixed with molisch reagent formed a purple ring, producing heat(exothermic)as a result of reaction, when reacted with 2mls H2SO4.Hence are carbohydrates. This reaction is a general one for the presence of all carbohydrates (Monosaccharides give a rapid positive test. Disaccharides and polysaccharides react slower) and other organic compounds that give furfural with concentrated sulphuric acid. Chemically, carbohydrates are either aldehydes or ketones derivatives of polyhydroxylic compound that known as aldoses and ketoses. Carbonyl group give aldoses and ketoses the reducing properties. Sucrose is the most common non-reducing sugar as the reducing carbonyl groups of glucose (C1) and fructose (C2) are involved in the glycosidic bond. Therefore the acid was used to break the bond in sucrose to its constituent monosaccharide’s , thereby forming a purple furfural layer. B:Reaction of reducing sugar
  • 5. Reducing sugars are those sugars capable of reducing copper ions in a solution to cuprous oxide. The reducing sugars include: (1) All of the simple sugars (monosaccharide’s). (2) A few of the disaccharides (including lactose and maltose, but not sucrose). (3) None of the polysaccharides. Background: If a suspension of copper hydroxide in alkaline solution is heated, then black cupric oxide is formed. Cu(OH)2 - CuO + H2O However, if a reducing substance is present, then rust-brown cuprous oxide is precipitated. 2Cu(OH)2 - Cu2O + H2O + O2 In practice, an alkaline solution of a copper salt and an organic compound containing alcoholic –OH is used rather than the above suspension. Under these conditions, the copper forms a soluble complex and the reagent is stable. Carbohydrates with a free or potentially free aldehyde or ketone groups have reducing properties in alkaline solution. In addition, monosaccharides act as reducing agents in weakly acidic solution. 2 Benedict’s test Principle: Reducing sugars are oxidized by the copper ions in the Benedict’s solution to form a carboxylic acid and a reddish precipitate of cuprous oxide within three minutes in alkaline medium. Benedict is a more sensitive alternative modified reagent to the original Fehling's test to produce a single solution which is more convenient for tests, as well as being more stable than
  • 6. Fehling`s reagent. Aldehyde (Carbonyl gp.) Benedicts solution Carboxylic acid Reddish precipitate Materials and reagents: 1. Carbohydrate solutions (1 g/l and 10 g/l). 2. Benedicts reagent. (To prepare one litre of Benedicts reagent: mix 173 g of tri-sodium citrate and 100 g of sodium carbonate with 800 ml of distilled water. Warm to dissolve, then cool and filter, add more distilled water to make the volume to 850 ml. Dissolve 17.3 g of copper sulphate in 100 ml distilled water and stir slowly into the first solution. Bring the volume to 1 litre with distilled water). Method: 1- Add five drops of the test solution to 2 ml of Benedicts reagent and place in a boiling water bath for 5 min. 2- Examine the sensitivity of benedicts reagent using increasing dilutions of glucose. 3- The formation of a orange-brown precipitate within 3-5 minutes indicates positive result. Biochemistry laboratory notes / Carbohydrates by Amjad I.A. Hussein page 5 of 11 Negative result (left) Positive result (right) Observations and results. Carbohydrates Color after heating the solution with benedicts reagent for 5min Color after cooling Fructose red brown Arabinose red brown Maltose red brown Glucose red brown Sucrose blue blue
  • 7. Discusion/conclusion -All the carbohydrates except sucrose were reduced by benedicts solution to form redish- orange solution after heating for 5minutes, therefore are reducing sugars, and sucrose retained the blue color solution after heating and therefore is a non-reducing sugar. C: Tests for individual carbohydrates 3- Bial's (Orcinol) test Principle: This test is specific for pentoses. The concentrated HCl in Bial's reagent dehydrates pentoses to form furfural. Furfural further reacts with orcinol and the iron ion to produce a bluish (Blue-green) product extractable into amyl alcohol. The test is used for the determination of RNA and the detection of pentosuria. The reaction is not absolutely specific for pentoses since prolonged heating of some hexoses yields hydroxy-methyl furfural which also reacts with orcinol to give colored complexes. Materials andreagents: 1. Carbohydrate solutions (1 g/l and 10 g/l). 2. Bials oricinol reagent (Dissolve 1.5 g of oricinol in 500 ml of conc. HCl and add 20 drops of 100 g/l solution of FeCl3). 3. Amyl alcohol. Method: 1- Add 2 ml of the test solution to 3 ml of Bials reagent. 2- Heat until boiling commences. Negative results (left & middle) Positive result (right) 2- The formation of a blue-green color indicates a positive result for pentoses.
  • 8. 3- Note that Hexoses generally react to form green, red, or brown products. Carbohydrates Color in Bial´s reagent after heating for one minute. fructose brownish arabinose greenish maltose yellowish glucose yellowish sucrose brownish Discusion --Fructose and sucrose turned into blue color in Bials reagent after heating for 1 minute hence are Hexoses, Arabinose turned to light green(positive) hence is a pentose, glucose and maltose did not reactand retained the original yellow color. Sucrose is a disaccharide made of fructose and glucose. 4- Seliwanoff’s test Seliwanoff’s test distinguishes between aldose and ketose sugars. The test is based on the fact that, when heated, ketoses are more rapidly dehydrated than aldoses to give furfural derivatives. The reagents consist of resorcinol and hydrochloric acid; 1) the acid hydrolysis polysaccharides and oligosaccharides yields simpler sugars; 2) Then the dehydrated ketose reacts with the resorcinol to produce a deep cherry red color. Aldoses may react slightly and more slowly to produce a faint pink color. Fructose and sucrose (A disaccharide consisting of fructose and glucose) are two common sugars that give a positive test. Materials and reagents: 1. Carbohydrate solutions (1 g/l and 10 g/l). 2. Seliwanoff's reagent (Dissolve 0.5 g of resorcinol in 1000 ml of 3M HCl). Method: 1- Add 2 drops of the test solution to 5 ml of Seliwanoff's reagent.
  • 9. Resorcinol Biochemistry laboratory notes / Carbohydrates by Amjad I.A. Hussein page 8 of 11 2- Warm in boiling water for 1-2 min. Negative result (left) Positive result (right) 3- The formation of a red color within 2 min indicates a positive result for ketoses (Fructose and Sucrose). 4- Note that prolonged heating with aldoses generally react to form faint pink color product. Observations/results. Carbohydrates Color in seliwanoff´s reagent and heated for 60seconds Color in seliwanoff´s reagent and heated for 5minutes. Fructose Orange/pick red Arabinose colorless colorless maltose colorless colorless Glucose colorless colorless Sucrose Orange/pick red Distilled water colorless colorless discusion -Fructose and sucrose gradually turned into pinkish color at 1minute then to red color after heating for 5minutes therefore are Ketose carbohydrates, the other carbohydrates( maltose, arabinose and glucose) remained colorless because they are aldoses.