Cambridge IGCSE Biology 0610: Biological molecules – Study Guide

This study guide teaches topic 4, Biological molecules (sub-topic 4.1), of the Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028. Learning outcomes 1 to 3 are Core, so every candidate needs them. Outcome 4, the structure of DNA, is Supplement content, so it is Extended only. Core content is tested on Papers 1 and 3, Core and Supplement content on Papers 2 and 4, and food tests are one of the listed experimental contexts for the practical papers (Paper 5 or Paper 6).

Use it with the course hub and the printable 0610 checklist. When you have worked through it, condense it with the biological molecules revision notes and test yourself with the biological molecules practice questions.

What this topic covers

0610 outcome	What you must be able to do	Tier
4.1.1	List the chemical elements in carbohydrates, fats and proteins	Core
4.1.2	State that large molecules are made from smaller ones: starch, glycogen and cellulose from glucose; proteins from amino acids; fats and oils from fatty acids and glycerol	Core
4.1.3	Describe the iodine, Benedict’s, biuret, ethanol emulsion and DCPIP tests	Core
4.1.4	Describe the structure of DNA: double helix, bases, bonds between base pairs, A with T and C with G	Extended only

The topic is short, but it feeds almost everything that follows. Digestion in topic 7 breaks these large molecules back down, and enzymes (topic 5) are proteins. The enzymes study guide and the human nutrition study guide build directly on this page.

4.1.1 The elements in biological molecules

Three groups of nutrient molecules appear in this outcome. You need to list their elements by symbol or by name.

Molecule	Elements
Carbohydrates	carbon, hydrogen, oxygen (C, H, O)
Fats and oils	carbon, hydrogen, oxygen (C, H, O)
Proteins	carbon, hydrogen, oxygen and nitrogen (C, H, O, N); some also contain sulfur (S)

The difference that earns marks is nitrogen. Carbohydrates and fats contain only C, H and O. Proteins always contain N as well. This is why plants need nitrate ions to make amino acids, which you meet again in plant nutrition.

4.1.2 Large molecules from smaller molecules

Biological molecules are often large. Each is built by joining many smaller molecules together. The syllabus limits this outcome to three cases.

Large molecule	Made from
Starch	glucose
Glycogen	glucose
Cellulose	glucose
Proteins	amino acids
Fats and oils	fatty acids and glycerol

Three points to fix in your memory:

• Starch, glycogen and cellulose are all made from the same smaller molecule: glucose. Starch is found in plants, glycogen in animals and fungi, and cellulose in plant cell walls.
• Proteins are made from amino acids. Different proteins contain amino acids in different orders.
• Fats and oils are made from two kinds of smaller molecule: fatty acids and glycerol. Writing “fatty acids” alone loses the mark.

The reverse of these reactions is chemical digestion. Amylase breaks starch down to simple reducing sugars, proteases break proteins down to amino acids, and lipase breaks fats and oils down to fatty acids and glycerol.

4.1.3 The five food tests

These tests show whether a molecule is present. Solid foods are usually ground up with a little distilled water first, so that the test solution can mix with the food.

Iodine solution test for starch

1. Add a few drops of iodine solution to the food (solid or solution).
2. Starch present: the colour changes from orange-brown to blue-black.
3. Starch absent: the iodine stays orange-brown.

Benedict’s test for reducing sugars

1. Put about 2 cm³ of the food solution in a test-tube and add an equal volume of Benedict’s solution.
2. Heat the tube in a hot water bath (about 80 °C) for a few minutes. No heat, no result.
3. Reducing sugar present: the blue solution turns green, yellow, orange or brick-red. The further along that sequence it goes, the more reducing sugar there is.
4. Reducing sugar absent: the solution stays blue.

Glucose is a reducing sugar, so a glucose solution gives a positive result.

Biuret test for proteins

1. Add biuret solution to the food solution and mix. No heating is needed.
2. Protein present: the colour changes from blue to purple (violet or lilac).
3. Protein absent: it stays blue.

Ethanol emulsion test for fats and oils

1. Shake the food with about 2 cm³ of ethanol. Any fat dissolves in the ethanol.
2. Pour the ethanol into a test-tube containing water.
3. Fat or oil present: a cloudy white (milky) emulsion forms.
4. Fat absent: the mixture stays clear.

The order matters: ethanol first, then water. Fat dissolves in ethanol but not in water, so it comes out as tiny droplets when the ethanol meets the water.

DCPIP test for vitamin C

1. Put a measured volume of DCPIP solution, for example 1 cm³, in a test-tube.
2. Add the food solution drop by drop, shaking after each drop.
3. Vitamin C present: the DCPIP changes from blue to colourless.
4. The smaller the volume of food solution needed to decolourise the DCPIP, the more vitamin C the food contains.

Food-test summary

Test	Detects	Negative colour	Positive result	Heat?
Iodine solution	starch	orange-brown	blue-black	no
Benedict’s	reducing sugars	blue	green / yellow / orange / brick-red	yes
Biuret	protein	blue	purple	no
Ethanol emulsion	fats and oils	clear	cloudy white emulsion	no
DCPIP	vitamin C	stays blue	blue to colourless	no

Food tests as practical work

The practical papers expect more than the colour change. Be ready to:

• Run a control. Test distilled water in the same way. It shows the reagent alone does not give the positive colour.
• Keep variables constant when comparing foods: the same volume of food solution, the same volume of reagent, and for Benedict’s the same heating temperature and time.
• Record qualitative results in a table with clear headings: food, test, colour at the start, colour at the end, conclusion.
• Work safely. Ethanol is flammable, so keep it away from flames. Heat Benedict’s tests in a water bath, not over a direct flame. Wear eye protection, because biuret solution and other reagents can harm eyes.

Worked example 1: reading a results table

A student tests a cereal extract and records these results.

Test	Final colour
Iodine solution	blue-black
Benedict’s	green
Biuret	blue
Ethanol emulsion	clear

What does the cereal contain?

• Iodine turned blue-black, so starch is present.
• Benedict’s turned green, so a small amount of reducing sugar is present. Green is only just past blue.
• Biuret stayed blue, so no protein was detected.
• The ethanol mixture stayed clear, so no fat was detected.

Notice the wording “not detected”. A test can miss a very small amount, so “no protein detected” is safer than “the cereal has no protein”.

Worked example 2: comparing vitamin C using DCPIP

1 cm³ of DCPIP was decolourised by 1.6 cm³ of fresh orange juice and by 4.0 cm³ of the same juice after boiling.

• Fresh juice needed less, so it contains more vitamin C.
• How many times more? Vitamin C concentration is inversely related to the volume needed: 4.0 ÷ 1.6 = 2.5 times as concentrated.
• A sensible conclusion: boiling reduced the vitamin C content of the juice.

4.1.4 The structure of DNA – Extended only

DNA carries genetic information. You need four structural points.

1. DNA has two strands coiled together to form a double helix.
2. Each strand contains chemicals called bases.
3. Bonds between pairs of bases hold the two strands together.
4. The bases always pair up in the same way: A with T, and C with G.

The syllabus states that full names of the bases are not required, so the letters are enough. You do not need the detailed structure of the strands.

Using the pairing rule

Because A always pairs with T, and C with G, you can work out one strand from the other.

Strand 1:  A T G C C G T A
Strand 2:  T A C G G C A T

The same rule gives you base percentages. In a DNA molecule, the amount of A equals the amount of T, and the amount of C equals the amount of G.

Worked example 3: base percentages

In a sample of DNA, 34% of the bases are C. Calculate the percentage of each other base.

C pairs with G, so G = 34%
C + G = 34 + 34 = 68%
A + T = 100 − 68 = 32%
A pairs with T, so A = T = 32 ÷ 2 = 16%

Answer: G = 34%, A = 16%, T = 16%.

A quick check: 34 + 34 + 16 + 16 = 100.

This idea returns in inheritance (topic 17), where the sequence of bases in a gene decides the sequence of amino acids in a protein.

Common errors

• Giving proteins the elements C, H and O only. Nitrogen is the whole point of the question.
• Writing “fats are made of fatty acids”. The syllabus pairs fatty acids and glycerol.
• Saying glycogen is made of amino acids, or cellulose of fatty acids. All three polysaccharides here are made from glucose.
• Forgetting to heat in the Benedict’s test, or heating the biuret test, which does not need it.
• Giving only the final colour. “Turns blue-black” is weaker than “from orange-brown to blue-black”.
• Writing “black” for the iodine result, or “red” for any positive Benedict’s result. The Benedict’s colour depends on how much reducing sugar there is.
• Adding water before ethanol in the emulsion test.
• Reading DCPIP results the wrong way round: a smaller volume of juice means more vitamin C.
• (Extended) Pairing A with G, or C with T. Or saying the strands are held by bonds “between strands” without saying the bonds are between bases.

Next steps

• Condense this page with the revision notes.
• Try the practice questions with worked answers.
• Check your whole-course gaps with the free 0610 diagnostic: Core or Extended.
• Move on to the enzymes study guide.

Official syllabus

Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028 (Version 3), Cambridge International. Topic 4, Biological molecules, sub-topic 4.1.