GENERAL TEST FOR CARBOHYDRATES: MOLISCH TEST AND IODINE TEST
Performed by: Group 3
SPECIFIC OBJECTIVE: To
test for the presence of carbohydrates.
Preparation of Sample (Chicken liver extract) a.
Get a thumb-sized portion of chicken liver and finely cut or shred it with scissors.
b.
Transfer the cut liver into a mortar and grind it with a pestle until it reaches a semi-liquid consistency.
c.
Transfer the liver into a test tube and add 10ml water and 1ml of 0.1% acetic acid.
d.
Heat the mixture for 30 mins in a boiling water bath to precipitate the proteins.
e.
While warm, the mixture should be filtered using filter paper. Collect the filtrate to be used in the Molisch and Iodine tests.
1. Molisch Test
a.
Prepare 3 test tubes and label them as follows: “glucose”, “carrageenan”, and “chicken liver”
b.
In the 1st tube, put 2ml of the 10% glucose solution.
c.
In the 2nd tube, put a pinch (mongo size) of gelatin and add 2ml of hot distilled water to dissolve. Cool to room temperature.
D.
In the 3rd tube., put 2ml of chicken liver extract.
e.
Add two drops of the Molisch reagent into each tube and mix thoroughly.
f.
Using a dropper, slowly add 1ml (15 drops) of concentrated sulfuric acid alongside each tube to create an interface with the mixture.
g.
Note what is formed in the
RESULTS
Glucose
Carageenan
Chicken Liver
RESULTS SAMPLES
OBSERVATIONS
Glucose
Violet/purple ring
Carrageenan
Violet ring
Chicken liver extract
Light violet ring
2. Iodine Test a.
Prepare 3 test tubes and label them as follows: “starch”, “carrageenan”, and “chicken liver”
b.
In the 1st tube, put 1ml of cooked potato starch mixture.
c.
In the 2nd tube, put a pinch (mongo size) of gelatin and add 2ml of hot distilled water to dissolve.
d. e.
In the 3rd tube, put 2ml of chicken liver extract.
Add a drop of iodine solution to each of the tubes and mix thoroughly.
f.
Note the color of the resulting mixture.
g.
Warm the 3 tubes in water bath (60Oc) until a change is observed in one of the tubes.
h.
Quickly remove all the tubes from the bath. Cool at room temperature.
i.
Note the color of each mixture and record the results.
RESULTS Iodine
was combined
Tube 1(starch): Black precipitate Tube 2(Carrageenan): Green color
Tube 3(Chicken Liver):Light
Yellow
Samples
were heated and cooled:
Tube 1(starch): Yellow with Black
precipitate
Tube 2(Carrageenan): Green color
Tube 3(Chicken Liver):Light yellow
RESULTS OBSERVATIONS SAMPLES
Before Heating
After Heating
On Cooling
Potato starch
black
Yellow w/ black precipitate
Yellow w/ black precipitate
Carrageenan
green
green
green
Chicken Liver Extract
Light yellow
Light yellow
Light yellow
MOLISCH TEST A.
What is the principle involved in the Molisch test? Concentrated sulphuric acid causes dehydration of all carbohydrates to give “furfural” compounds, that react with αnaphthol (Molisch’s reagent) giving a violet or purple colored complex.
B. What is the purpose of this test?
It is a general test for the presence of carbohydrates.
It is used widely as compared to other tests because before every specific test for different carbohydrates, a general test is important and is performed via Molisch test.
This test is positive for all types of carbohydrates whether free or in combined form like glycoproteins, glycolipids, etc.
C. Provide the type of equation used in the test.
D. What reagents were used in the Molisch’s test? State the purpose of each component.
The Molisch reagent contains alpha-naphthol dissolved in ethanol. 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.
E. Provide a possible explanation for: the positive results obtained with your samples/s - A positive result is indicated by the appearance of a purple ring in the interface between the acid and test layers. It indicates the presence of carbohydrates.
the negative results obtained with your sample/s - There is no formation of a purple ring. This means that there is no carbohydrate present in the sample.
IODINE TEST Ba. What is the principle involved in the iodine test?
Distinguish between monosaccharide, disaccharides and polysaccharide, you should perform iodine test. This test would be positive for polysaccharide and negative for mono and disaccharides.
When
iodine dissolved in potassium iodide solution react with starch or glycogen, it react with it and the color of solution is changes, indicating the presence of these polysaccharides.
Some
polysaccharides have have the property of adsorption for iodine. So, they adsorb iodine and give coloration.
B. Explain the color noted when: •
Iodine was combined with the samples; 1. First test tube : Black color
•
2. Second test tube : Green color
3.
Third test tube : Light Yellow
the mixture of iodine and each sample were heated; First test tube : Yellow with black
precipitate
2.
Second test tube : Green
3.
Third test tube : light yellow
the
heated mixture of iodine and each sample were cooled.
The same with the heated samples
C. What is the importance of the iodine test? What carbohydrates were identified in you samples?
Iodine test is done to test for the presence of starch. The basic principle of this test is that when an iodine solution (i.e., aqueous solution of potassium iodide) comes in contact with starch, the solution turns blue black in color. Starch, which is found in carbohydrate rich foods like potatoes, rice, corn, and barley can be separated into two fractions .
The Carbohydrates found in the sample were amylose and amylopectin. In the presence of iodine, amylose in starch forms a deep blue color. As starch is nothing but carbohydrates, the test for carbohydrates is same as that of starch.
D. Structurally, how are amylose and amylopectin different from each another? From glycogen? From cellulose? How do their structural characteristics affect their solubility in water?
Amylose is known to be continuous. On the other hand, Amylopectin is branching. It is also found only on plants.
Similarities: They are all POLYSACCHARIDES Formed in glycosidic bonds Primary functions are energy storage and food reserve.
Difference: the difference has to do with the branching.
Glucose molecules are linked in straight chains via the alpha 1,4 bond. Glycogen is more highly branched than amylopectin. Humans can make their own glycogen in the liver and muscles. But it can also be made by glycogenesis within the brain and stomach. Cellulose is the main component in cell walls which gives plant their structure. Cellulose is made up of oxygen, hydrogen and carbon as well as a derivation of glucose. Their structural characteristic is affected by their hydrogen bonding.
Their
structural characteristic is affected by their hydrogen bonding.
