Testing for Macromolecules

Essay by binky890High School, 11th gradeA, February 2004

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To determine which chemical indicators that will indicate the presence of glucose, starch, lipid, and protein in various samples.

Methods and Materials:

For materials, please refer to p.18 in Biology 11 (replace Sudan IV with filter paper)

For procedures, please refer to p.18-19 (except Part C)

For Part C in Inv. 1A, please read the following procedures:

1.Label sample 1-6 on the filter paper

2.Add one drop of each sample corresponding to their number indicated on the filter paper (for the number of each sample, please refer to step 4 in Part A on p.18)

3.Leave the filter paper to dry for approximately 5-10 minutes

4.Record observations


Table 1: Reactions of sample substances to indicators

SAMPLEBenedict's solution + heatIodine solutionFilter paperBiuret reagent

1.protein solutionPurpleNo changeDried upPurple

2.vegetable oilNo changeNo changePaper remained translucentNo change

3.glucose solutionOrangeNo changeDried upNo change

4.sucrose solutionNo changeNo changeDried upNo change

5.starch solutionNo changeDark blue-blackDried upNo change

6.distilled waterNo change (Clear blue)No change (Clear yellow)Dried upNo change

Investigation 1A: Testing for Macromolecules


By observing the table above, we could see that Benedict's test was for reducing sugars, iodine test was for the presence of starch, filter paper was for the presence of fatty acids, and the Biuret test was for amino groups present in proteins. Benedict's solution was used to test for the presence of simple sugars, such as glucose (monosaccharide). When heated, the solution mixed with monosaccharides produced a reddish-orange colour. This was because Benedict's solution is composed of sodium citrate, sodium carbonate, and cupric sulfate pentahydrate. When solution is heated, an oxidation-reduction reaction occurs: cupric ion (Cu+2) oxidizes into a cuprous ion (C+) and precipitates into cuprous oxide (Cu2O) because Benedict's solution loses an oxygen (Cu+2 ---> Cu+ forming Cu2O). Basically, it is a reaction between the aldehyde...