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Biological Molecules - Chemical tests for nutrients

Grade 12IGCSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The test for starch uses Iodine solution (iodine in potassium iodide). A positive result changes the color from orange-brown to blue-black due to the formation of a polyiodide complex within the starch helix.

Reducing sugars (like glucose, C6H12O6C_6H_{12}O_6) are detected using Benedict's reagent. The mixture must be heated in a water bath at approximately 80C80^\circ C to 100C100^\circ C. A color change from blue to green, yellow, orange, or brick-red indicates the presence of reducing sugars.

Non-reducing sugars (e.g., sucrose, C12H22O11C_{12}H_{22}O_{11}) do not react directly with Benedict's reagent. They must first be hydrolyzed by boiling with dilute hydrochloric acid (HClHCl), neutralized with sodium hydrogencarbonate (NaHCO3NaHCO_3), and then tested with Benedict's reagent.

The Biuret test is used to detect proteins. It reacts with the peptide bonds (CONH-CONH-) in the polypeptide chain. A positive result is a color change from light blue to purple or violet.

Lipids are tested using the Ethanol Emulsion test. Lipids are dissolved in ethanol (C2H5OHC_2H_5OH) and then added to water. Because lipids are insoluble in water, they precipitate out to form a cloudy white emulsion.

Vitamin C (ascorbic acid, C6H8O6C_6H_8O_6) is tested using DCPIPDCPIP solution. The blue DCPIPDCPIP is decolorized (becomes colorless) as it is reduced by the ascorbic acid.

📐Formulae

RCHO+2Cu2++5OHRCOO+Cu2O+3H2ORCHO + 2Cu^{2+} + 5OH^- \rightarrow RCOO^- + Cu_2O + 3H_2O

C12H22O11+H2OH+/heatC6H12O6+C6H12O6C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+/heat} C_6H_{12}O_6 + C_6H_{12}O_6

C6H8O6+DCPIPblueC6H6O6+DCPIPcolorlessC_6H_8O_6 + DCPIP_{blue} \rightarrow C_6H_6O_6 + DCPIP_{colorless}

💡Examples

Problem 1:

A student is given an unknown colorless solution. After adding Benedict's reagent and heating, the solution remains blue. However, after boiling a fresh sample with HClHCl, neutralizing with NaHCO3NaHCO_3, and re-testing with Benedict's, a brick-red precipitate forms. Identify the nutrient.

Solution:

The nutrient is a non-reducing sugar, most likely sucrose (C12H22O11C_{12}H_{22}O_{11}).

Explanation:

The negative initial Benedict's test rules out reducing sugars. The positive result after acid hydrolysis indicates that a disaccharide was broken down into its constituent reducing monosaccharides (C6H12O6C_6H_{12}O_6).

Problem 2:

Explain the observation when a piece of crushed peanut is shaken with ethanol, and the resulting liquid is poured into a test tube containing distilled H2OH_2O.

Solution:

A cloudy white emulsion forms in the H2OH_2O.

Explanation:

Peanuts contain lipids. Lipids are soluble in organic solvents like ethanol (C2H5OHC_2H_5OH) but insoluble in water. When the lipid-ethanol mixture is added to water, the lipid molecules aggregate into tiny droplets that scatter light, creating the 'emulsion' effect.

Problem 3:

A juice sample requires 5 drops5\text{ drops} to decolorize 2 cm32\text{ cm}^3 of 0.1%0.1\% DCPIPDCPIP, while a standard solution of 0.1%0.1\% Vitamin C requires 10 drops10\text{ drops}. Compare the concentration of Vitamin C.

Solution:

The juice sample is twice as concentrated (2×2 \times) as the standard Vitamin C solution.

Explanation:

The fewer the drops required to decolorize the DCPIPDCPIP, the higher the concentration of ascorbic acid (C6H8O6C_6H_8O_6). Since it took half as many drops (55 vs 1010), the juice has double the concentration.

Chemical tests for nutrients - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology