Cambridge International AS & A Level Biology 9700: Testing for biological molecules – Study Guide

This study guide teaches sub-topic 2.1, Testing for biological molecules, from topic 2 (Biological molecules) of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content, so it is tested on Paper 1 (multiple choice) and Paper 2 (AS Level structured questions), and Paper 4 can draw on it because A Level questions require AS knowledge. The tests are also practical skills: Paper 3 expects you to identify the contents of unknown solutions and to estimate concentrations from qualitative results, and Paper 5 expects you to plan dilutions and controls.

Use it with the 9700 course hub and the printable 9700 checklist. Then condense it with the testing for biological molecules revision notes, test yourself with the testing for biological molecules practice questions, and take the free 10-minute 9700 AS diagnostic. For the whole of topic 2 (carbohydrates, lipids, proteins and water), see the biological molecules study guide.

What this sub-topic covers

9700 outcome	What you must be able to do
2.1.1	Describe and carry out the Benedict’s test for reducing sugars, the iodine test for starch, the emulsion test for lipids and the biuret test for proteins
2.1.2	Describe and carry out a semi-quantitative Benedict’s test: standardise the test and use the time to first colour change, or comparison with colour standards, to estimate the concentration of a reducing sugar
2.1.3	Describe and carry out a test for non-reducing sugars, using acid hydrolysis and Benedict’s solution

“Describe and carry out” means you must do each test at the bench and, in a written paper, give the method in a logical order, with the reagent, any heating, and both results as named colours.

2.1.1 The four standard tests

Benedict’s test for reducing sugars

Reducing sugars include glucose, fructose and maltose (outcome 2.2.4). Sucrose is not a reducing sugar.

1. Put about 2 cm³ of the sample solution in a test-tube. Grind solid food with a little water first.
2. Add an equal volume of Benedict’s solution, which is blue.
3. Heat in a water bath close to boiling for a few minutes.
4. Observe the colour.

Result. If a reducing sugar is present, the blue copper(II) ions are reduced and a precipitate forms. As the amount of reducing sugar rises, the colour goes blue → green → yellow → orange → brick-red. If no reducing sugar is present, the solution stays blue.

Iodine test for starch

1. Add a few drops of iodine in potassium iodide solution to the sample (a solution, a suspension, or the cut surface of a food).
2. Observe. No heating is needed.

Result. Starch present: the orange-brown solution turns blue-black. Starch absent: it stays orange-brown.

Emulsion test for lipids

1. Add about 2 cm³ of ethanol to the sample and shake well, so that any lipid dissolves in the ethanol.
2. Pour the ethanol into a clean test-tube containing about 2 cm³ of water (or add water to the ethanol). Do not shake hard.
3. Observe.

Result. Lipid present: a cloudy white emulsion forms. Lipid absent: the mixture stays clear.

Why it works: lipids dissolve in ethanol but are insoluble in water. When water is added, the lipid comes out of solution as tiny droplets dispersed through the liquid, and these make it look milky. The order matters. If you add water first, the lipid never dissolves, so there is nothing to come out of solution.

Biuret test for proteins

1. Add an equal volume of biuret reagent to about 2 cm³ of the sample. Alternatively, add potassium hydroxide solution, then a few drops of dilute copper sulfate solution.
2. Mix gently. No heating is needed.

Result. Protein present: the blue solution turns purple (lilac or mauve). Protein absent: it stays blue. The reagent reacts with peptide bonds, which is why it detects polypeptides and proteins.

Controls and safety

A negative control is the same test with distilled water in place of the sample. It shows what the reagent looks like when no target molecule is present, so you can judge a faint change. A positive control uses a known sample (for example glucose solution for Benedict’s) to show the reagent is working.

The syllabus materials list gives hazard codes for these reagents. Benedict’s solution is a moderate hazard; biuret reagent and potassium hydroxide are corrosive; ethanol is flammable. Wear eye protection, keep ethanol away from naked flames (heat with a water bath, not a Bunsen flame), and point heated tubes away from people.

Worked example 1: identifying unknown solutions

Four unlabelled tubes, P to S, each contain one of glucose, sucrose, starch or albumen (egg white protein). Results:

Tube	Benedict’s (heated)	Iodine	Biuret
P	stays blue	blue-black	stays blue
Q	brick-red	orange-brown	stays blue
R	stays blue	orange-brown	purple
S	stays blue	orange-brown	stays blue

• P gives blue-black with iodine, so P is starch.
• Q is the only positive Benedict’s test, so Q is glucose (a reducing sugar).
• R turns purple with biuret, so R is albumen.
• S is negative for all three. Of the four, only sucrose fits, so S is sucrose. To confirm it, you would carry out the non-reducing sugar test (2.1.3) on S.

S is identified by elimination, then confirmed.

2.1.2 The semi-quantitative Benedict’s test

Because the colour depends on how much reducing sugar is present, the Benedict’s test can estimate concentration. It is semi-quantitative: you compare with known solutions rather than measure directly.

Standardising the test

To compare tubes fairly, everything except the sugar concentration must be kept the same:

• the volume of sample (for example 2 cm³, measured with a syringe or graduated pipette);
• the volume of Benedict’s solution, enough to be in excess so that the colour depends on the sugar, not on running out of reagent;
• the temperature of the water bath (use a thermostatically controlled bath and a thermometer);
• the heating time (when you use colour standards);
• the way you judge colour, for example against a white tile, under the same lighting.

Making the known concentrations

You need a range of known concentrations of the same reducing sugar, usually glucose. There are two ways to make them.

Proportional dilution mixes different volumes of a stock solution with water to the same final volume.

Worked example 2: proportional dilution

Make 10 cm³ each of 0.10, 0.20, 0.30, 0.40 and 0.50 % glucose from a 0.50 % stock.

volume of stock = (required concentration / stock concentration) × final volume
0.10 %: (0.10 / 0.50) × 10 = 2.0 cm³ stock + 8.0 cm³ water
0.20 %: 4.0 cm³ stock + 6.0 cm³ water
0.30 %: 6.0 cm³ stock + 4.0 cm³ water
0.40 %: 8.0 cm³ stock + 2.0 cm³ water
0.50 %: 10.0 cm³ stock + 0.0 cm³ water

Serial dilution dilutes each solution by the same factor to make the next.

Worked example 3: serial dilution

Starting with 1.0 % glucose, make a series by halving the concentration each time. Transfer 5.0 cm³ of the 1.0 % solution into 5.0 cm³ of water and mix: this is 0.50 %. Then transfer 5.0 cm³ of that into 5.0 cm³ of water, and so on.

The series is 1.0, 0.50, 0.25, 0.125 and 0.0625 %.

Concentration units

Paper 5 may give concentrations as % (w/v) or in mol dm⁻³. A 1 % (w/v) solution contains 1 g of solute in 100 cm³ of solution. To convert for glucose, use its relative molecular mass, 180 (C₆H₁₂O₆).

Worked example 4: converting units

Express 0.20 mol dm⁻³ glucose as % (w/v).

mass per dm³ = 0.20 × 180 = 36 g dm⁻³
mass per 100 cm³ = 36 / 10 = 3.6 g
concentration = 3.6 % (w/v)

Method A: colour standards

Heat each known concentration with Benedict’s solution for the same time, in the same water bath. Line the tubes up as a set of colour standards. Heat the unknown in exactly the same way, then compare it with the standards. Its concentration lies between the two standards it matches most closely, or equals one it matches exactly.

The colours are ordinal data, so your estimate is a range, not a precise value.

Method B: time to first colour change

Put each tube into the water bath and start a timer. Record the time taken for the first visible change from blue (the first trace of green). The higher the reducing sugar concentration, the shorter the time. Repeat each concentration at least three times and calculate a mean.

You can also process times into a rate, 1/t, which increases with concentration.

Worked example 5: estimating an unknown from times

Glucose concentration / %	Mean time to first colour change / s	Rate (1/t) / s⁻¹
0.20	150	0.0067
0.40	95	0.0105
0.60	70	0.0143
0.80	55	0.0182
1.00	46	0.0217

An unknown solution, tested in the same way, took a mean of 80 s.

• 80 s lies between 95 s (0.40 %) and 70 s (0.60 %), so the unknown is between 0.40 % and 0.60 %.
• Plot time (y-axis) against concentration (x-axis), draw a smooth curve, and read across from 80 s. Joining the two nearest points with a straight line gives about 0.52 %, so a sensible estimate is about 0.5 %.

The estimate is only valid inside the range of your standards. Colour judgement is subjective, so the main errors are deciding when the first change happens and which standard matches.

2.1.3 The test for non-reducing sugars

Sucrose is a non-reducing sugar, so it gives a negative Benedict’s test. The glycosidic bond that joins its glucose and fructose units can be broken by acid hydrolysis, releasing these two reducing sugars.

1. Do a Benedict’s test on a sample of the solution. Record the colour. A non-reducing sugar on its own leaves it blue.
2. Put a fresh sample in another tube. Add about the same volume of dilute hydrochloric acid.
3. Heat in a water bath close to boiling for a few minutes. This hydrolyses glycosidic bonds.
4. Cool the tube, then neutralise by adding sodium hydrogencarbonate a little at a time until it stops fizzing. Check with pH paper or red litmus. Benedict’s test only works in alkaline conditions, so without this step a non-reducing sugar can be missed.
5. Carry out a Benedict’s test on the neutralised solution.

Result. If the first test stayed blue and the second gives green, yellow, orange or brick-red, a non-reducing sugar is present. Both tests staying blue means no sugar of either type.

Worked example 6: a solution with both types of sugar

A fruit extract gives green with Benedict’s before hydrolysis and orange after hydrolysis and neutralisation, with the same volumes and heating time in both tests.

• Green before hydrolysis: a small amount of reducing sugar is present.
• Orange after hydrolysis: more reducing sugar is present than before.
• The extra reducing sugar must have come from hydrolysis, so a non-reducing sugar is also present.

When the first test is already positive, you detect a non-reducing sugar only by comparing the two results, so standardise volumes and heating in both.

Common errors

• Writing “Benedict’s turns red” with no heating step, or heating the iodine and biuret tests.
• Describing a result as “it changes colour” instead of naming both colours.
• Calling sucrose a reducing sugar, or listing maltose as non-reducing.
• Leaving out neutralisation after acid hydrolysis, or neutralising with more acid.
• In the emulsion test, adding water before ethanol, or describing the positive result as “a white precipitate”.
• Saying the biuret test turns “blue-black” (that is iodine).
• Claiming the semi-quantitative test gives an exact concentration.
• Forgetting that a shorter time means a higher concentration.
• Estimating an unknown outside the range of the standards.

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for 2025, 2026 and 2027 (Version 1), Cambridge International (Cambridge University Press & Assessment). Topic 2, Biological molecules, sub-topic 2.1 Testing for biological molecules.