The Chemical Constituents of Living Matter

The principal chemical constituents of living matter are: water, mineral salts, organic compounds such as carbohydrates, proteins, lipids, nucleic acids. In this exercise we shall concentrate on carbohydrates, lipids, proteins, and we shall estimate the concentration of Vitamin C, an organic compound, in a solution, by the iodometric technique.

• Carbohydrates include simple sugars, disaccharides and polysaccharides. They are the most important source of energy for most organisms. Polysaccharides change color in the presence of iodine solution: Glycogen gives a red-brown color and starch a dark blue-violet color. While simple sugars, having an aldehyde group, act as reducing agents in the presence of Benedict's reagent producing a range of colors from green to brown depending on the degree of reduction they exhibit.
• • Lipids such as fats and oils are important in cell membranes and also as an energy reserve. They produce translucent spots on paper. In addition they dissolve a nonpolar Sudan III /IV dye.
• • Proteins have a vital role in the growth and repair of tissues also have a role to play in cellular metabolism since most enzymes are proteins. The presence of two or more peptide bonds gives a violet color with Biuret's reagent (NaOH & CuSO4).
• • Vitamins are organic compounds required in small doses; most of them act as coenzymes.

Test for carbohydrates

A. Iodine test

Materials required

• Starch paste (cornstarch)
• Glucose solution
• Sucrose solution
• Distilled water
• 4 test tubes
• marking pen
• Iodine solution
• test tube rack

Procedure

• Fill 4 test tubes with 2 ml of the following: distilled water (tube 1), glucose solution (tube 2), sucrose solution (tube 3), starch solution (tube 4).
• Add 3 drops of iodine solution to each of the above test tubes and swirl to mix.
• Record any changes in color

Data collection

Qualitative data

Table1: colors detected in test tubes 1 through 4

Tube #

New color detected

1

Yellow

2

Yellow

3

Yellow

4

Dark blue

Illustration

Conclusion and evaluation

Since distilled water contains only H2O molecules, the iodine didn't react with the solution but became dilute, thereby giving a yellow color.

Tubes 2 and 3 contain carbohydrates; however sucrose and glucose are disaccharides, hence the iodine does not react with the solutions in the two tubes since this indicator only changes color in the presence of a polysaccharide. As a result tube 2 and 3 present the only change in color anticipated yellow, since the iodine added is merely being diluted because no reaction is taking place.

In tube 4 we can observe a change in color from colorless to dark blue. This occurs since starch is a polysaccharide it reacts with the iodine giving a dark blue color.

Glucose and sucrose are not polysaccharides since they do not react with iodine.

There can be almost no sources of error in this experiment since the results we are anticipating are qualitative, hence if more or less of each reactant was added, the results would be the same. There are no sources of error in this experiment since its results are qualitative.

B. Benedict's test

Material required:

• Glucose solution
• Dilute hydrochloric acid
• Galactose solution
• Benedict's reagent
• Sucrose solution
• Distilled water
• Starch suspension
• Boiling water bath
• Test tube holders
• Test tube rack
• Marking pen
• 6 test tubes

Procedure:

• Mark 6 test tubes and fill each test tubes with 2m1 of the following: distilled water (tube I), glucose sol (tube 2), Galactose sol-(tube 3), starch sol (tube 4), sucrose sol (tube 5). Fill tube 6 with 2m1 of sucrose sol and add few drops of hydrochloric acid.
• Add 2m1 of Benedict's reagent to each of the 6 test tubes.
• Place the test tubes either in a boiling water bath or in a beaker of boiling water and wait for few minutes.
• Record the change in color in the test tubes

Data collection:

Qualitative data

Table 2: colors detected in test tubes 1 through 6

Test tube #

Color observed during heating

Color observed after heating

1

Dilute blue

Dilute blue

2

Brick red

Red precipitation

3

Brown brick red

Red precipitation

4

Bright green

Change from green to red

5

Bright blue

Bright blue

6

Bright blue

Bright blue

Illustration

Conclusion and evaluation

Chemical equation illustrating how Benedict's reagent reacts with simple sugars (monosaccharides)

Most reducing sugars react with Benedict's reagent and contain either an aldehyde group or a ketone group which are responsible for reducing the reagent when heated to give either a green to red brick precipitate depending on the amount of the sugar present.

Test tube 1 contains only distilled water, i.e. no type of simple sugar is present, so no reaction occurs and the Benedict's reagent is merely diluted, which explains the bright blue color. Test tube 2 contains glucose, a reducing sugar. Hence when the glucose was heated it reacted with the reagent to give a red brick color which proves that glucose is a reducing sugar. Test tube 5 contains sucrose, a disaccharide. This disaccharide is not a simple sugar and hence is not reducing agent, so it is expected that no change in color (solution remains blue) i.e. no reaction occurs. Sucrose which is made of fructose and glucose can only react with Benedict's reagent if it is hydrolyzed and the bonds between the fructose and glucose molecules are broken, as a result there would be free glucose molecules to react with the reagent. Tube 3 contains Galactose, a monosaccharide which possesses aldehyde groups in its structure, so it is expected that the solution changes in color producing a red precipitate as a result of the reaction between the two substances (Benedict's reagent + Galactose). In tube 4, there is a starch solution. Starch is a polysaccharide of α glucose monomers bonded together by glycosidic bonds. Before heating, no change in color is observed, however during heating, the solution acquires a green color which progressively darkens giving a red color to the solution. After a while a red precipitate can be observed in suspension the solution which slowly sediments. During the heating the glucose subunits on the extremities of the polysaccharide are slowly breaking away from the macromolecule giving free glucose monomers in the solution. The heat is responsible for this, since it allows the glycosidic bonds to be broken and hydrolyzed in the presence of water. As it is stated before, Benedict's reagent gives a green color when simple sugars are present in small quantities and a red color when the concentration increases. As the test tube was subjected for a longer time to the heat, progressively the color darkened towards a red color. This occurs due to the fact that the glycosidic bonds are being progressively broken down, slowly but effectively. Test tube 6 contains sucrose and HCl. The acid is not concentrated enough to actually break the bonds of the polysaccharide, hence no reducing sugar is produced and no change in color occurs.

Test for lipids

A. Spot test

Material required:

• Brown wrapping paper
• Corn oil
• Distilled water
• Whole milk
• Skimmed milk
• Medicine droppers

Procedure

• With a medicine dropper add a drop of corn oil, of water, of whole milk, and of skimmed milk on different locations of a sheet of brown wrapping paper. You can mark the different sites beforehand.
• Let the paper dry, and then examine each spot by holding the paper up to the light.
• Record your observations.

Data collection:

Qualitative data

Table 3: observable changes in paper under source of light

Substance admitted on paper

Observable change in paper

Corn oil

Paper becomes translucent

Water

No change in translucency of paper

Whole milk

Slightly translucent

Skimmed milk

No change in translucency of paper

Conclusion and evaluation

When lipids are present in a solution that is put on a piece of unglazed paper (brown paper), it causes the space which it covers on the paper to become translucent. This is because this type of paper can readily absorb lipids and allow light to pass it more efficiently. Corn oil made the paper very translucent, so it contains lipid substances. Whole milk also made the paper translucent but not as intense as the corn oil did hence the concentration of lipids in whole milk is not as high as the concentration in corn oil. water and skimmed milk do not contain lipids (not in quantities detectable by this experiment that is).

From these two experiments, spot test and Sudan dye test, we determined that whole milk and corn oil contain lipids.

B. Sudan III/IV test

Material required:

• Distilled water
• Corn oil
• Sudan Ill/TV dye
• Test tube rack
• 'Whole milk
• Skimmed milk
• 4 test tubes
• marking pen

Procedure

• Fill 4 test tubes with 2 ml of the following: distilled water (tube 1), corn oil (tube 2), whole milk (tube 3), and skimmed milk (tube 4).
• Add to each of the 4 test tubes few grains of the Sudan III! IV dye.
• Record your observations in table 3.

Data collection:

Qualitative data

Table 4: behavior of Sudan dye in different solutions

Test tube #

Color observed during heating

1

No dissolving

2

Complete dissolving , red color appears

3

Partial dissolving, no red color appears

4

No dissolving

Conclusion and evaluation

There are three forms of lipids - triglycerides, sterols, and phospholipids - all of which contain primarily carbon, hydrogen, and oxygen. Lipids absorb pigments in fat-soluble dyes such as Sudan IV. A red color occurs at the interface of a lipid and Sudan IV dye. Since in corn oil with the dye, the solution became red, we can say that tube 2 contains lipids. Since no red color is observed in the other tubes, they do not contain any lipids.

Adding too much dye can have an effect and change our results, making the changes in the solutions more difficult to observe. This might be a source of error.

TEST FOR PROTEINS

Biuret's test

Material required:

• Albumin solution
• 10% Sodium hydroxide
• 0.5% Copper sulfate
• Distilled water
• Whole milk
• 3 test tubes
• test tube rack
• marking pen

Procedure

• Fill 3 test tubes with 2 ml of the following: distilled water (tube 1), albumin solution (tube 2), and whole milk (tube 3).
• Add an equal volume of 10% NaOH solution.
• Add 0.5% CuSO4 drop by drop and watch for a change in color, then stop adding.
• Record any change in color in each of the 3 test tubes in table 4.

Data collection:

Qualitative data

Table 5: Change in color with the addition of Biuret's reagent

Test tube #

Color observed

1

light Blue

2

Violet

3

Slightly violet

Conclusion and evaluation

Biuret's reagent consists of sodium hydroxide & copper sulfate. This solution reacts with the peptide bonds producing a color change. A deep violet or blue color indicates the presence of proteins. The violet or lavender color obtained is due to a complex ion formation with Cu as the central atom and the ligands will be N-donor ones of the protein fragments. In tubes 2 and 3 a change in color occurs which indicate that proteins are present in both test tubes. In test tube 2 the color is darker, meaning that the concentration of proteins in high. However in test tube 3 the color is lighter, meaning that the concentration of proteins in relatively lower. In tube 1, no change occurs meaning that no polypeptides are present. No sources of errors can be found since this experiment mainly demands adding Biuret's reagent to a solution and observing a change in color.

TEST FOR VITAMINS

Material required:

• 1 ml pipettes
• test tubes
• 0.O1M iodine solution
• 0.05% Vitamin C solution
• 2% Starch suspension
• teat pipettes

Procedure

• Use a pipette to measure 1 .Oml of vitamin C solution obtained by crushing 2x250 mg tablets in 1L distilled water.
• Using a teat pipette add 6 drops of starch solution to the vitamin solution.
• Using another clean teat pipette, carefully add the iodine solution 1 drop at a time. mixing after each drop until the natural colour of the solution is replaced by a blue black colour.
• Record the number of drops in table 5.
• Repeat steps 1 and 2 three times and record your results in table 5.

Data collection:

Quantitative data

• 1st try: 8 drops of iodine solution
• 2nd try: 9 drops of iodine solution
• 3rd try: 11 drops of iodine solution

Data processing and presentation:

Mean= (1st try+2nd try+3rd try)/3

= 9.33

Average: 9.33 drops of iodine solution

Knowing the relative mass of Ascorbic acid (176), using the solutions and technique suggested, and assuming 15 drops of 12 per mL, you can calculate the approximate concentration of Ascorbic acid (mg/L) in solution by multiplying the number of drops of iodine taken by 120.

Concentration of ascorbic acid= average # of drops of I2 X 120

= 9.33 X 120

=1119.6 mg/L

Conclusion and evaluation

The concentration of ascorbic acid is 1119.6 mg/L

The iodometric technique was used to estimate the concentration of Vit. C. as iodine is added to the solution it combines rapidly with ascorbic acid to form the iodocompound C6H8O6I2. the chemical equation of this reaction is :

C6H8O6+I2 C6H8O6I2

The starch in this solution is used as an indicator to show when the addition of I2 should stop. When there is no more vitamin C to react with the I2, the excess I2 reacts with the starch giving it a violet color.

A source of error may be that some of the vitamin C solution is on the walls of the test tube, and hence the amount of I2 reacted would not be exact since it is reacting only with part of the solution. One might correct this error by adding the drops of I2 to the walls of the tube to make sure every drop containing vitamin C reacted.

Resources

http://www.umanitoba.ca/Biology/lab2/biolab2_2.html#

http://www.geocities.com/CapeCanaveral/Hall/1410/lab-B-contents.html

http://www.olemiss.edu/courses/chem222/CH.24-Notes/CH.%2024-CHEM%20222.html

http://faculty.clintoncc.suny.edu/faculty/Michael.Gregory/files/Bio%20101/Bio%20101%20Laboratory/Chemical%20Composition%20of%20Cells/Chemical%20Composition%20of%20Cells.htm