It is Experimenting Time….muwhahaha!!!!
Hello there again my Biochem peeps. Today I want to do a bit of review of our lab session and highlight the important details that may be helpful for our quiz next week. Therefore, I will try and breakdown each test performed and details the expected results and the reasoning behind said results. This will be done in three parts since our lab consisted of testing for Carbohydrates, Lipids and Amino acids/proteins. Below is the first portion of the Lab that contained most of the Carbohydrate tests. I hope that this information is a bit helpful.
- MOLISCH TEST:
Not specific to carbohydrates. However, a negative test result may rule out the presence of carbohydrates.
The reaction: Carbohydrates, specifically monosaccharides are dehydrated in the presence of concentrated sulphuric acid to form an aldehyde known as furfural (pentoses) or hydroxymethyl furfural (hexoses) derivatives. Polysaccharides and disaccharides are converted to monosaccharides via hydrolysis of the glycosidic bonds. The monosaccharide/furfural derivatives then react with 1-naphthol in Molisch reagent via a condensation reaction to form a purple coloured compound.
The result: The purple-coloured compound appears as a ring layer at the interface between the sulphuric acid and test solution. The sulphuric acid is denser than the test solution and therefore the reaction will occur at the junction where both substances meet.
Lab results: Glucose, maltose, arabinose and starch will all display the purple ring compound at the interface of the acid and solution.
-Not specific to carbohydrates
-Generalized test that cannot distinguish carbohydrates and further testing must be undertaken to identify the carbohydrate.
2. BENEDICT’S TEST:
Test for the presence of a reducing sugar. All monosaccharides are reducing sugars as well as some disaccharides and polysaccharides as long as they possess the free aldehyde or ketone group to be oxidized.
The reaction: Benedict’s Solution is composed of copper sulphate solution, which gives a blue colouration. This reacts with the electrons from the ketone or aldehyde group of the free reactive carbonyl group on the carbohydrate to form cuprous oxide (a red-brown precipitate). This precipitate is formed due to the reduction of the Cu2+ ions to Cu+ ions. In the process, the carbohydrate is oxidized. The colour of the precipitate is dependent on the volume of reducing sugar present. This means that the more free carbonyl groups present, the increase in the amount of precipitate formed.
The result: The red-brown precipitate is formed that may range from red to brick red in colour.
Lab result: Glucose is a reducing sugar that will form the red-brown precipitate. However, at different dilutions, there is a decrease in precipitate formation since there are less free reactive carbonyl groups available to react with the Benedict’s Solution. Precipitates may range from green colouration to red-brown/brick red precipitate. At very low concentrations of glucose, no precipitate may be observed and the solution remains blue.
This a general test to identify reducing sugars and there is low specificity in identification of the carbohydrate present in solution. Further testing must be performed to identify specific carbohydrates.
3. SELIWANOFF’S TEST:
This test is utilized to distinguish ketoses from aldoses in carbohydrates. It is specific to identifying ketoses since the reaction occurs faster with ketoses. Aldose sugars also react but at a much slower rate. Hence distinction between a ketose and aldose sugar may be performed.
The reaction: Seliwanoff’s Reagent contains a non-oxidizing acid such as HCl and resorcinol. The HCl reacts with the ketone group on the ketose sugar to form a furfural derivative due to the dehydration reaction. The furfural then reacts with the resorcinol in a condensation reaction to form a cherry-red complex.
The result: The cherry-red complex is formed in the reaction indicating the presence of the ketose sugar.
Lab result: Fructose is a ketose sugar that will react with the reagent to form the cherry-red complex after 1 minute. The Glucose solution will also form a pink to red complex but not after 1 minute. It will form the complex after a longer period.
Although this test is able to adequately distinguish a ketose sugar from an aldose sugar, it is not very specific since an aldose may also form the complex. The difference is the time taken to do so.
This test also is a generalized test that does not differentiate the specific ketose present but rather illustrates that a ketose sugar is present. Specific ketose sugar identification must be performed by further testing.
4. MODIFIED BARFOED’S TEST:
This is used to distinguish monosaccharides from disaccharides. This test requires that the carbohydrate be a reducing sugar since a reduction in the reagent occurs in order to identify the solution. Hence, it is able to distinguish monosaccharides from disaccharides. Monosaccharides are reducing sugars and therefore, will react faster with the reagents to form the colour change. Whereas, reducing disaccharides will react slower than the monosaccharides. Some disaccharides are non-reducing sugars and will therefore not react at all.
The Reaction: Barfoed’s reagent comprises of cupric acetate and acetic acid in solution. Monosaccharides readily react with the reagent to causes a reduction in the Cu2+ ions to Cu+ ions forming Cu2O. This is because monosaccharides are oxidized readily in weak acid solutions. Disaccharides also can reduce the cupric ions; however, this reaction is much slower. This will result in a red-brown colouration formed within the solution. The monosaccharides then reduce the phosphomolybdic acid in the phosphomolybdate colour reagent to form phosphomolybdenum blue. Disaccharides also cause this reduction, however, the reaction may be slower or none existent since it is dependent on the reducing nature of the sugar.
The Results: the Phosphomolybdenum blue formed causes a deep blue colouration in the solution in the monosaccharide solution.
Lab result: Glucose and fructose are both monosaccharides and reducing sugars. Therefore, they will form this deep blue colouration in this reaction. Maltose and sucrose are disaccharides. However, sucrose is not a reducing agent and therefore will not be able to reduce the Barfoed’s reagent or phosphomolybdate and therefore there is no observed colour change and the solution remains blue. The maltose solution may form the deep blue colourations since it is a reducing sugar and is able to be oxidized. However, maltose will react much slower and therefore the colour change may not be as evident.
The test is non-specific to the exact monosaccharide present
This reaction will occur in disaccharides that are reducing sugars and therefore are not totally specific to distinguishing the monosaccharides from the disaccharides in solution.
This test is used to distinguish pentose sugars.
The reaction: Bial reagent contains HCl, orcinol and ferric chloride. The pentose sugars are hydrolysed by the HCl to form a furfural derivative. This derivative then reacts with the orcinol to form a green-yellow complex in the presence of ferric ions via a condensation reaction. Polysaccharides made up of pentose units are hydrolysed to break the glycosidic bonds and then undergo the same reaction to form the complex. Hexoses are also hydrolysed and react with the orcinol but form a red to brown complex rather than a green-yellow colouration.
The results: The condensation reaction of pentoses with orcinol will form a green yellow colour change in the solution. Hexoses may also form a colour change, but this ranges from red to brown. Green colour change is an indication of presence of a pentose sugar.
Lab results: Arabinose is a pentose sugar that will yield the green yellow colour change. Gum Arabic is a polymer of arabinose, rhamnose and galactose and therefore when hydrolysed, will react similar to arabinose and give the green-yellow colour change. However, glucose is a hexose sugar and therefore will not react similar to arabinose and will therefore not yield a green colour change. Rather it may produce a red to brown colour change or no change at all.
It is not specific to which pentose is present and further tests must be conducted to identify specific pentose sugar.
6. IODINE TEST:
This is an indicator for the presence of starch in solution. It is specific to identifying starch.
The reaction: Starch is composed of a straight chain subunit known as amylose. Amylose is responsible for the reaction with iodine. The amylose subunit of starch is made up of a straight chain of alpha-glucose monomers that are connected by alpha 1à4 glycosidic bonds. In solution, the amylose exists as a helically coiled structure. When the iodine is added, the molecules of iodine become trapped within this helical structure and form a complex. Iodine is usually blue-black in colour and will therefore cause a blue-black colouration with starch once amylose is present.
The results: Starch will stain blue-black in colour since amylose is present.
Lab results: Starch will stain blue-black in colour. The glycogen however, consists of branched chains of glucose monomers, will therefore not form the helical structural arrangement like amylose in solution, and therefore will not be able to trap the iodine molecules to be stained blue-black. However, glycogen may stain brown to purple with iodine. Dextrin is a similar structure to starch in that they possess the same general formula, however, dextrin is a smaller and less complex structure to starch that is formed during the hydrolysis of starch. Dextrin usually yields a blue colouration in reaction with iodine.
7. ACID HYDROLYSIS OF SUCROSE:
This is performed on disaccharides and polysaccharides to breakdown to the monomer monosaccharide units. This test will enable the breakdown of sucrose into fructose and glucose.
The reaction: HCl breakdown the 1à2 glycosidic bond that links the glucose and fructose monomer together to form sucrose. This break causes the release of two monosaccharides. Sodium hydroxide is then added to neutralize the pH of the solution since the presence of acid in the solution will disrupt the alkalinity of the Seliwanoff’s and Benedict’s reagent that were added after. When hydrolysed, two reducing sugars are formed. This means that when Benedict’s test is performed, the solution will be able to reduce the reagent and form a red-brown precipitate. In Seliwanoff’s test, the presence of fructose will give a positive indication of the presence of a ketose sugar present in solution.
Lab results: The hydrolysis of the sucrose enabled the following results. In Benedict’s test, the solution reduced the Cu2+ ions to Cu+ ions resulting in the formation of the red-brown precipitate. This means that there was the presence of a reducing sugar now in the solution and since sucrose is not a reducing sugar; the hydrolysis had created a monosaccharide that contains a free reactive carbonyl group. In the Seliwanoff’s test, the presence of fructose facilitated the reaction with the reagent to form cherry red compound.
- BioSci: Intro Bios Labs: Benedict’s Test for Reducing Sugars
- Biochemistry Class notes: Urinalysis: Chemical Examination
- Laney University. Carbohydrates.
- Harper College: Seliwanoff’s Test
- Wikipedia: Seliwanoff’s Test.
- Practical Biochemistry for Medical Students by Srinivas B Rao
- Differences in Protein, Carbohydrate and Caffeine Content by Michael Chen
- Harper College: Bial’s Test.
- Biochemistry lab I: Identification of an Unknown Carbohydrate.
- Chemical land 21. Dextrin.
- Virtual Chembook. Elmhurst College. Starch-Iodine
- Demonstration Experiment on Video. Artificial Honey- Formation of a Sweet Imitation by Peter Keusch.