Triglycerides and evidence from food tests

What you’ll learn

• Describe triglyceride formation and distinguish saturated from unsaturated fatty acids.
• Interpret reducing-sugar and starch tests against colour standards.
• Explain why a semi-quantitative test gives an estimate.

Understand the idea

A triglyceride consists of glycerol joined to three fatty acids by three ester bonds. Each bond forms through condensation between a hydroxyl group of glycerol and the carboxyl group of a fatty acid. Hydrolysis reverses the process. Triglycerides are hydrophobic energy stores; they can also contribute to thermal insulation and cushioning. They are not polymers of repeating identical monomers.

A saturated fatty acid contains no carbon–carbon double bonds. An unsaturated fatty acid has at least one; cis double bonds produce bends that interfere with close packing. This often lowers the melting temperature of a mixture, although chain length and composition also matter. The words saturated and unsaturated describe fatty-acid structure, rather than whether a food contains any cholesterol.

In core practical 1, reducing-sugar and starch concentrations are estimated by comparing samples with known standards. A reducing sugar changes heated Benedict’s reagent from blue towards coloured precipitates; iodine solution turns blue-black with starch. Match sample and standard volumes, treatment time and temperature, and account for the sample’s own colour. A colour category brackets a concentration; it does not justify an exact value. Sucrose is non-reducing in the direct Benedict’s test, so a negative result does not establish absence of all sugars.

Key formulae

$$\text{glycerol}+3\text{ fatty acids}\rightleftharpoons\text{triglyceride}+3\mathrm{H_2O}$$

$$c_{\mathrm{original}}=c_{\mathrm{diluted}}\times\text{dilution factor}$$

Worked example

A food extract is diluted by making 2.0 cm³ up to 10.0 cm³. Its Benedict’s colour falls between standards of 0.4 and 0.6 g per 100 cm³. Estimate the original concentration range, assuming the sample matrix does not alter the test.

1. The dilution factor is 10.0 ÷ 2.0 = 5.
2. Multiply both limits by five: 0.4 × 5 = 2.0 and 0.6 × 5 = 3.0.
3. Report a range because the colour comparison only brackets the diluted concentration.

Answer: Approximately 2.0–3.0 g per 100 cm³ in the original extract.