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Carbohydrates Lab Report Essay

Introduction

Carbohydrates are essential in foods as an energy source (starch is the main source of human calories), a flavouring (simple sugars are usually sweet) and as a functional ingredient (sucrose allows ice cream to be soft in the freezer; xanthan gum thickens a low-fat salad dressing). Carbohydrates are a type of macronutrient found in many foods and beverages. Most carbohydrates are naturally occurring in plant-based foods, such as grains. Food manufacturers also add carbohydrates to processed foods in the form of starch or added sugar. As with all our approaches to food ingredients/constituents we will first examine the structure of carbohydrates and then try to elucidate how their structures allow them to function as they do. As their name suggests, carbohydrates basically made up from sugar and water, i.e. Cx(H2O)y, although this ratio is often not strictly true and occasionally other atoms may be present. The carbons are arranges in a chain (most often 5-6 atoms) functionalized with alcohol groups. The terminal carbon either carries either an aldehyde or a ketone functional group.

Carbohydrates are classified based on size of base carbon chain, number of sugar units, location of C=O and stereochemistry. Classifications of carbohydrate are monosaccharides, disaccharides, oligosaccharides, and polysaccharides. Monosaccharide is the smallest possible sugar unit. Examples include glucose, galactose or fructose. When we talk about blood sugar we are referring to glucose in the blood; glucose is a major source of energy for a cell. In human nutrition, galactose can be found most readily in milk and dairy products, while fructose is found mostly in vegetables and fruit. When monosaccharides merge together in linked groups they are known as polysaccharides. Disaccharide is two monosaccharide molecules bonded together. Polysaccharides are polymers. A simple compound is a monomer, while a complex compound is a polymer which is made of two or more monomers. Disaccharides are polysaccharides – “poly…” specifies any number higher than one, while “di…” specifies exactly two.

Examples of disaccharides include lactose, maltose, and sucrose. If you bond one glucose molecule with a fructose molecule you get a sucrose molecule. Sucrose is found in table sugar, and is often formed as a result of photosynthesis (sunlight absorbed by chlorophyll reacting with other compounds in plants). If you bond one glucose molecule with a galactose molecule you get lactose, which is commonly found in milk. Starch, glycogen, dextran and cellulose are polysaccharides. Polysaccharides differ not only in the natural of their component monosaccharides but also in the length of their chains and in the amount of chain branching that occurs. Polysaccharides function as storage materials, structural components, or protective substances.

Thus, starch ( which exists in two forms: amylose and amylopectin ), glycogen and other storage polysaccharides, as readily metabolizable food, provide energy reserves for cells. Chitin and cellulose provide strong support for the skeletons of arthropods and green plants, respectively. In this experiment those activity that had been carried out means to determine the carbohydrate class of an unknown by carrying out a series of chemical reactions with the unknown and known compounds in each class of carbohydrate such as the Molisch test (general CHO), Barfoed’s test (monosaccharides), Fehling’s test (reducing sugars), Benedict’s test (reducing sugars) and Iodine test (amylose).

ACTIVITY 3.1, MOLISCH TEST: A GENERAL TEST FOR CARBOHYDRATES

OBJECTIVE:

To test the carbohydrate solution

MATERIALS:

1 % of carbohydrate solutions( lactose, glucose, starch, sucrose, cellulose, fructose, apple and cabbage ), distilled water(as control tube), concentrated sulphuric acid, Molisch reagent.

APPARATUS:

Test tubes, test tube holder, dropper, 5ml pipette, glass rod, test tube rack, fume cupboard

CAUTION:

Molish reagent contains concentrated sulfuric acid , which is toxic and corrosive. It can cause severe burns. Prevent eye, skin clothing, and combustible material contact. Avoid ingesting the substance.If you spill any reagent or acid, immediately notify your laboratory instructor.

NOTE:

Do not place your thumb over the open end of a test tube when mixing its contents. Your laboratory instructor will suggest ways in which you can safely and thoroughly mix the contents of a test tube.

PROCEDURES:

1. 2 ml of each of the 1% carbohydrate solutions that have been prepared is added into one set of labelled test tubes. 2. 2 drops of Molisch reagent are added to each test tube and is mixed well with a clean glass stirring rod. 3. The test tube is inclined. Then 3ml of concentrated sulphuric acid is added slowly and carefully down the side of the tube to form a layer below the sugar solution.( This step is performed inside the fume cupboard ). 4. The resulting solution did not been shook or mixed.

5. The change of the solution is observed and recorded. ( A purple ring at the interface is indicative of a carbohydrate ). 6. The test solutions containing Molisch reagent is discarded into the container provided by laboratory instructor.

RESULT:

DISCUSSION:

Carbohydrates undergo dehydration reactions (loss of water) in the presence of concentrated sulfuric acid. Pentoses and hexoses form five member oxygen containing rings on dehydration. The five member ring, known as furfural, further reacts with Molisch reagent to form colored compounds. Pentoses are then dehydrated to furfural, while hexoses are dehydrated to 5-hydroxymethylfurfural. Either of these aldehydes, if present, will condense with two molecules of naphthol to form a purple-colored product. A positive reaction is indicated by appearance of a purple ring at the interface between the acid and test layers.

Monosaccharides give a rapid positive test. Glucose and fructose are monosaccharide. Disaccharides and polysaccharides react more slowly than monosaccharide. Sucrose and lactose are disaccharide which also gave purple color ring. Starch and cellulose gave slightly purple color because they are polysaccharides. Distilled water gave negative test because it is not carbohydrate. A large apple has around 28-31 grams of carbohydrate. Apple and cabbage contain carbohydrate so that they gave purple ring in this test.

CONCLUSION:

Glucose, lactose, fructose, sucrose, starch and cellulose all are carbohydrates which give positive test for Molisch test. A sample of distilled water is prepared and tested as the controlling sample.

ACTIVITY 3.2, BARFOED’S TEST:A GENERAL TEST TO DISTINGUISH BETWEEN MONOSACCHARIDE AND DISACCHARIDES

OBJECTIVE:

To distinguish the given carbohydrate solutions as monosaccharides or disaccharides.

MATERIALS:

1 % of carbohydrate solutions( lactose, glucose, starch, sucrose, cellulose, fructose), distilled water (as control tube), Barfoed’s reagent

APPARATUS:

Test tubes, test tube holder, 5 ml pipette, pipette filler, stop watch, water bath

CAUTION :

Barfoed’s reagent is corrosive and an irritant. If you spill any of the solution on yourself or on the bench, immediately notify your laboratory instructor.

PROCEDURES

1. 5 ml of each of the carbohydrate solutions is added into one set of the labelled test tubes. 2. 5 ml of Barfoed’s reagent is added to each test. 3. The contents of each tube are shook well. All the tubes are placed in an actively boiling water bath at the same time. 4. After the water starts boiling again, the solutions is heated for 3.5 min. ( Timing is important since a false positive test can be obtained for monosaccharides with disaccharide, if the disaccharides are heated for more than 3.5 min thereby breaking down ( hydrolyzing ) to monosaccharides ). 5. During this period, the tubes are observed closely and any change of clarity of the solutions is noted. ( A positive test for monosaccharides is the appearance of a red precipitate of Cu?O within 1 or 2 minutes, if no precipitate forms it indicates the presence of a disaccharide).

RESULT:

1% CARBOHYDRATE SOLUTION

FORMATION OF RED PRECIPITATE

Fructose
Yes
Glucose
Yes
Cellulose
No
Lactose
No
Sucrose
No
Starch
No
Distilled water(as control tube)
No

DISCUSSION:

Barfoed’s test distinguishes monosaccharides from disaccharides. Positive test for monosaccharides is the appearance of red precipitate (Cu2O) within 1-2 minutes. If no precipitate formed, indicates the presence of disaccharide. The red precipitate come from the reaction between the reduction of copper (II) acetate to copper(I) oxide (Cu2O). RCHO + 2Cu2+ + 2H2O > RCOOH + Cu2Ov + 4H+ The aldehyde group of the monosaccharide which normally forms a cyclic hemiacetal is oxidized to the carboxylate. Glucose and fructose which are monosaccharides show positive result in this test. Reducing disaccharides undergo the same reaction, but do so at a slower rate. So, the timing to heat the sample is set to 3.5 minutes. However, the samples are heated no more than 3.5 minutes to prevent the disaccharide breaking down to monosaccharide. Lactose, sucrose, cellulose, starch and distilled water showed negative result in this test.

CONCLUSION:

Only monosaccharide will give an immediate red precipitate in Barfoed’s test that is glucose and fructose the other remaining solutions which are cellulose, lactose, sucrose, starch and distilled water do not show any changes.

ACTIVITY 3.3 FEHLING TEST: FOR REDUCING

OBJECTIVE:

To distinguish the reducing sugars and non-reducing sugars

MATERIALS:

1% of carbohydrate solutions ( glucose, fructose, cellulose, lactose, sucrose, starch ), distilled water(as control tube), Fehling solution A ( 69.28 grams copper (II) sulfate pentahydrate dissolved in 1 litre of distilled water), Fehling solution B ( 346 grams Rochelle salt ( potassium sodium tartrate tetrahydrate) and 120 grams sodium hydroxide in 1 litre of distilled water) APPARATUS:

5 ml pipette, test tubes, test tube holder, test tube rack, pipette filler, stop watch

PROCEDURES:

1. 5 ml of carbohydrate solutions is added into one set of test tubes. 2. By using different glass pipettes, 5 ml of Fehling A and 5 ml of Fehling B are added into each test tubes. 3. The solution is heated in a boiling water bath for 5-10 minutes. 4. Red brick precipitate is formed for positive results.

5. Changes in test tubes are recorded.

RESULT:

Samples
Result
Lactose
Positive-red brick precipitate
Glucose
Positive-red brick precipitate
Fructose
Positive-red brick precipitate
Starch
Negative-no changes
Distilled water
Negative-no changes
Cellulose
Negative-no changes
Sucrose
Negative-no changes

DISCUSSION:

Fehling’s solution is used to test for the presence of a reducing sugar. Fehling’s solution was based on the aldehyde or ketone groups in the sugar structures. A sugar is classified as a reducing sugar only if it has an open-chain form with an aldehyde group or a free hemiacetal group. the presence of aldehydes but not ketones is detected by reduction of the deep blue solution of copper(II) to a red precipitate of insoluble copper oxide.

Fructose, glucose and lactose show positive result in this test. All monosaccharides are reducing sugars. Many disaccharides, like lactose, also have a reducing form, as one of the two units may have an open-chain form with an aldehyde group. However, sucrose, in which the anomeric carbons of the two units are linked together, are non-reducing disaccharides since neither of the rings is capable of opening. Polysaccharides (sugars with multiple chemical rings) are non-reducing sugars. Polysaccharides have
closed structures, which use free atoms to bond together their multiple rings, and take a much longer time to be broken down. So, starch and cellulose which are polysaccharides have negative result in Fehling’s test. Distilled water is not reducing sugar also shows negative result.

CONCLUSION:

Fehling test is the common test which is used to determine the presence of reducing sugar. Fructose, lactose and glucose are reducing sugars which give brick red precipitate after the solutions are heated.

ACTIVITY 3.4 BENEDICT’S TEST: FOR REDUCING

OBJECTIVE:

To test for reducing sugars

MATERIALS:

1% of carbohydrate solutions ( glucose, fructose, cellulose, lactose, sucrose, starch ), 3M hydrochloric acid (HCl), Benedict’s reagent, distilled water

APPARATUS:

Test tubes, test tube holder, test tube rack, 5 ml pipette, pipette filler, dropper, stop watch, water bath

PROCEDURES

1. 5 ml of Benedict’s reagent and 2 ml of carbohydrate are added to a test tube and each tube is shook thoroughly. 2. All the tubes are placed in a boiling water bath at the same time. The solutions are heated for 5-6 min. 3. Any changes in color, in the transparencies and in the formation and color of any precipitate are observed and recorded. 4. Later, 4 drops of 3M HCl are added to 5 ml of 1 % sucrose solution and is heated in the boiling water bath for 5 min. 5. 1 % starch solution is treated in the same way but the heating period was extended to 25-30 min. 6. 1-2 ml of each of solution is applied with Benedict’s test in the same manner as before. 7. The results are compared with those obtained without acid treatment.

RESULT:
Sugar solution

Result of colour of the solution
Starch
Light blue
Lactose
Brick red precipitate are formed
Fructose
Brick red precipitate are formed
Sucrose
Light blue
Cellulose
Light blue + white precipitate
Glucose
Brick red precipitate are formed
Distilled water
Light blue
Sucrose + HCI
Brick red precipitate are formed
Starch + HCI
Light

DISCUSSION:

The Benedict’s test is used to detect the presence of reducing sugars (sugars with a free aldehyde or ketone group) such as glucose, fructose and lactose. All monosaccharides are reducing sugars; they all have a free reactive carbonyl group. Some disaccharides have exposed carbonyl groups and are also reducing sugars. Lactose which is disaccharides also called reducing sugar as it has the exposed carbonyl groups. Other disaccharides such as sucrose and starch are non-reducing sugars and will not react with Benedict’s solution. Benedict’s reagent is a mild oxidant with CuSO4, Cu (II) sulfate, as one of the reagents. In the presence of a reducing sugar, the blue solution of Cu (II) or Cu+2, is changed to a brick red/brown precipitate of Copper (I) or Cu+1 oxide,Cu2O. If there a small or large amount of the reducing sugar present, the color would range from green to brick red respectively. RCHO + 2Cu2+ + 4OH- > RCOOH + Cu2O + 2H2O Sucrose indirectly produces a positive result with Benedict’s reagent if heated with dilute hydrochloric acid prior to the test, although after this treatment it is no longer sucrose. The addition of HCl hydrolysed the non-reducing sugar, as it split it up into its component monomers.

The monomers are reducing sugars which gave the positive result on the second reducing sugar test. The acidic conditions and heat break the glycosidic bond in sucrose through hydrolysis. The products of sucrose decomposition are glucose and fructose, both of which can be detected by Benedict’s reagent, as described above. This same goes for starch. But since starch has larger component compare to sucrose so it took a longer time to hydrolyse. That the purpose of heat it in longer time compare to sucrose. Without the addition of acid to sucrose solution, starch solution, the test given is negative. The solutions remain clear blue after the addition of Benedict’s reagent and heating. Tap water is used only to show the example of negative result of Benedict’s test. Thus it will not show any changes compare to the carbohydrates.

CONCLUSION:

Benedict’s test is the common test which is used to determine the existence of reducing sugar. Fructose, lactose, and glucose are reducing sugars which give positive test. Starch and sucrose are non reducing sugars which give positive results after adding hydrochloric acid.

ACTIVITY 3.6, IODINE TEST: FOR POLYSACCHARIDES

OBJECTIVE:
To test for polysaccharides
MATERIALS:
0.01M iodine, 0.12M KI , 1% carbohydrate solutions (cellulose and starch) , distilled water

APPARATUS:
Test tubes, test tube rack , dropper.
PROCEDURE:
1. Few drop of 0.01M iodine in 0.12M KI added to 1% starch and cellulose solutions. 2. Any changes to the colour are observed.
RESULT:
1% CARBOHYDRATE SOLUTION
COLOUR OBSERVED
Starch
Vivid blue
Cellulose
Yellowish brown

DISCUSSION:

Starch gives positive result in Iodine test as the color of solution change from yellow to dark blue. The immediate formation of a vivid blue color indicates amylose. Vivid blue coloration forms due to the polyiodide complex formed. Cellulose is derived from D-glucose units, which condensed through beta(1->4)-glycosidic bond. This give cellulose to be a straight polymer therefore, it can’t coil around iodine to produce blue colour as starch does. Only starch gives the color of vivid blue, this is because it contains amylase. The iodine molecules slip inside of the amylase coil. The amylose, or straight chain portion of starch, forms helices where iodine molecules assemble, forming a dark blue color.

CONCLUSION:

The Iodine test is used to test for the presence of starch. Starch is a type of polysaccharide carbohydrate which is made up of amylose and amylopectin. It is one of the main sources of carbohydrate and present naturally in plant. Amylose in starch form dark blue complex with iodine.

ACTIVITY 2.2 SOLUBILITY AND DIGESTIBILITY TEST
SOLUBILITY TEST
OBJECTIVE:
To test the solubility in hot water and digestion by amylase.

MATERIALS:
5g of starch, 5g of cellulose, distilled water

APPARATUS:
Test tubes, test tube holder, glass rod, test tube rack, fume cupboard, 2 centrifuge tubes, analytical balance, cylinder, graduated pipette, pipette filler, 2 evaporator dishes.

PROCEDURE:
1 5 g of starch is measured and put into a centrifuge tube; 2 40 ml of distilled of water is measured and poured into the same centrifuge tube; 3 Step 1 to 2 is repeated by replacing the starch with cellulose; 4 Both of the tubes are heated: the tube containing starch is heated for about 2-3 minutes whereas the tube containing cellulose is heated for about 10 minutes; 5 After heating, both of the content of the tubes are allowed to cool down slightly; 6 The tubes are put into a centrifuge with 3500 rpm for 10 minutes; 7 Empty weight for both of the evaporator dishes is measured; 8 5 mL of the supernatant from both of the tubes is pipetted and poured into two separate evaporator dishes; 9 The evaporator dishes are left in the oven overnight

10 The weight of the evaporator dishes is measured again.
11 The solubility results are recorded and tabulated.

RESULT:
Solubility (%) = Weight of dried supernatant (g)
Weight of the dried carbohydrates (g)

For starch, solubility (%) = 0.0093g
50.0023g
= 0.01860 %
For cellulose, solubility (%) = 0.0010g
50.0027g
= 0.002000 %
Carbohydrates
Weight of dry carbohydrates (gram)
Weight of dried supernatant (gram)
Solubility (%)
Starch
50.0023g
24.8768g- 24.8675g= 0.0093g
0.01860
Cellulose
50.0027g
21.2150g- 21.2140g= 0.0010g
0.002000

DISCUSSION:

In this activity, the solubility is defined as the percentage ratio of the weight of dried supernatant to the weight of the dry starch. Solubility can be interpreted as the amount of the dissolved compound that is present in the test solution. From the calculations done, we can see that starch, with a percentage of solubility at 0.01860 %, whereas cellulose have 0.002000 %. Starch and cellulose are two very similar polymers. In fact, they are both made from the same monomer, glucose, and have the same glucose-based repeat units. Since the sugar molecules contain the hydroxyl group or –OH, Thus it can form hydrogen bonds with water molecules, which makes it soluble in water, but only to a limited extent.

However, the glucose units in starch are connected by alpha linkages while the glucose units in cellulose are connected by beta linkages. In starch, all the glucose repeat units are oriented in the same direction. But in cellulose, each succesive glucose unit is rotated 180 degrees around the axis of the polymer backbone chain, relative to the last repeat unit. Although cellulose contains hydroxyl groups too, but most of them are hydrogen-bonded to each other when the microfibrils stack together, which accounts for the strength of cellulose fibers. There’s less free hydroxyl groups that can hydrogen bond with water molecules, other than those hydroxyl groups that’s present at the end of each cellulose chain, which causes the cellulose to be less soluble in water when compared to starch.

CONCLUSION:

Although both starch and cellulose are complex carbohydrates, which have large molecular weight size, significantly reducing their affinity for water, but the hydroxyl groups that exist in the monomers itself actually contributes to their insignificant solubility. However, the solubility will increase when these complex carbohydrates are broken down into its monomers where the hydroxyl groups can form hydrogen bonds with other water molecules easily due to the reduced molecular weight and size that affects the affinity for water.

DIGESTIBILITY TEST

OBJECTIVE:

To determine the digestibility of complex carbohydrates

MATERIALS:

Starch powder, cellulose powder, enzyme amylase, benedict’s solution, distilled water

APPARATUS:
2 centrifuge tubes, measuring cylinder, analytical balance, pipette fillers, graduated pipettes, 2 droppers, 5 test tubes.

PROCEDURE:
1 5 g of starch is measured and put into a centrifuge tube; 2 40 ml of distilled of water is measured and poured into the same centrifuge tube; 3 Step 1 to 2 is repeated by replacing the starch with cellulose; 4 Both of the tubes are heated: the tube containing starch is heated for about 2-3 minutes whereas the tube containing cellulose is heated for about 10 minutes; 5 After heating, both of the tubes are allowed to cool down slightly; 6 5 mL of starch is pipetted into a test tube; 7 Step 6 is repeated using a different test tube but a drop of amylase is dropped into it; 8 5 mL of cellulose is pipetted into a test tube; 9 Step 8 is repeated using a different test tube but a drop of amylase is dropped into it; 10 5 mL of distilled water is pipetted into the last test tube, and a drop of amylase is dropped into it; 11 20 drops of benedict’s solution is dropped into five of the test tubes; 12 Any changes occurred is recorded and tabulated.

Result
Samples
Colours of the solutions
Benedict’ s test
5 g of starch
blue
Negative
5 g of starch with amylase
Brick red precipitate is formed
Positive
5 g of cellulose
blue
Negative
5 g of cellulose with amylase
blue
Negative
Distilled water with amylase
Blue
Negative

DISCUSSION:

Amylase is one of the many members of a class of enzyme, hydrolases, that catalyze the hydrolysis of starch into smaller carbohydrate molecules such as maltose (a molecule composed of two glucose molecules). Two categories of amylases, denoted alpha and beta, differ in the way they attack the bonds of the starch molecules. Alpha-amylase is widespread among living organisms. In the digestive systems of humans and many other mammals, an alpha-amylase called ptyalin is produced by the salivary glands, whereas pancreatic amylase is secreted by the pancreas into the small intestine. In the experiment, the test tube that contains only distilled water served as a control for this experiment.

As for the test tubes that contain starch and cellulose without the amylase, they give a negative result for Benedict’s test, because for starch and cellulose, since both of them are complex carbohydrates, thus they have very few carbonyl groups which contribute to the compound’s reducing properties. Starch is a non-reducing sugar which shows negative result in the Benedict’s test. As for the test tube that contains starch and cellulose with the addition of a drop of enzyme, amylase, the test tube with starch gives a positive result, but not the test tube with cellulose. As we all know, enzyme amylase can only catalyzes the breakdown of starch into simpler sugars, but not cellulose. Cellulose only digested by cellulase enzyme. It is impossible for human digestive enzymes to break the glycosidic bond. Therefore, only the test tube containing starch treated with amylase gives a positive result for the Benedict’s test.

CONCLUSION:

Enzyme is a highly specific catalyst which can only converts a specific set of reactants into specific products. Amylase only hydrolyze the starch but not cellulose. From here, we can say that the human digestive system would not be able to digest the cellulose, because our digestive system only contains amylase, and not cellulose. Therefore in the perspective of a human, we can conclude that the digestibility of starch is higher than cellulose, provided that the enzyme amylase is present.


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