Biological Molecules
Elements in carbohydrates, fats and proteins; how large biological molecules are built from monomers; the five food tests with expected results and procedures; and (Extended) the double-helix structure of DNA with complementary base pairing.
Biological Molecules
CORE EXTENDEDAll living organisms are built from the same small set of biological molecules. Understanding their elemental composition, how large molecules are assembled, and how to test for them in the lab underpins almost every later topic — from nutrition (Topic 7) to respiration (Topic 12) to biotechnology (Topic 21).
Elements in Biological Molecules
| Molecule | Elements present | Memory hint |
|---|---|---|
| Carbohydrates sugars, starch, glycogen, cellulose | Carbon (C), Hydrogen (H), Oxygen (O) | CHO — “CHO-colate” |
| Fats and oils triglycerides, lipids | Carbon (C), Hydrogen (H), Oxygen (O) | Same 3 elements as carbs, but very different H:O ratio (far more H relative to O) |
| Proteins enzymes, structural proteins, antibodies | Carbon (C), Hydrogen (H), Oxygen (O), Nitrogen (N) | CHON — proteins are the only group with nitrogen as a core element (some also contain sulfur) |
Carbohydrates and fats both contain only C, H, and O. Proteins are unique in always containing nitrogen (N). This is why nitrogen is an essential element in fertilisers — plants need it to make amino acids and proteins. Some proteins also contain sulfur (S), but the syllabus does not require you to state this.
Large Molecules from Small Monomers
The body builds large molecules (polymers) by joining many smaller molecules (monomers) together. The syllabus specifies three polymer-monomer relationships to know:
| Large molecule (polymer) | Made from (monomer) | Biological role |
|---|---|---|
| Starch | Glucose | Energy storage in plants (grains, potato tubers) |
| Glycogen | Energy storage in animals (liver, muscles) | |
| Cellulose | Structural — makes plant cell walls | |
| Proteins | Amino acids (20 different kinds) | Enzymes, structural support, antibodies, hormones |
| Fats and oils | Fatty acids + glycerol (1 glycerol + 3 fatty acids) | Long-term energy storage, insulation, cell membrane components |
The same glucose monomer, joined in different ways, produces three very different polymers: starch (energy store, coiled, easily digested), glycogen (compact, highly branched animal store), and cellulose (straight chains, rigid, indigestible to most animals). Their differences lie in how glucose units are bonded together — not in the monomer itself.
The Five Food Tests — Full Reference
These tests are core Paper 5/6 practical content and appear in theory papers. For each test you must know: the reagent, the procedure, the positive result, and the negative result.
| Test | Substance detected | Reagent (start colour) | Procedure | Positive result | Negative result |
|---|---|---|---|---|---|
| Iodine solution | Starch | Yellow-brown | Add a few drops of iodine solution to the sample (no heating required) | Blue-black colour | Stays yellow-brown |
| Benedict’s solution | Reducing sugars (glucose, maltose, fructose, lactose) |
Clear blue solution | Add Benedict’s solution; heat in a water bath at ~80°C for 5 minutes | Brick-red / orange precipitate (colour varies with sugar concentration: green → yellow → orange → red) | Stays blue |
| Biuret test | Proteins | Blue (NaOH + dilute CuSO₄) | Add sodium hydroxide solution, then add a few drops of dilute copper sulfate solution | Purple / violet colour | Stays blue |
| Ethanol emulsion test | Fats and oils | Colourless ethanol | Dissolve sample in ethanol; pour the ethanol solution into a tube of water | Milky-white emulsion (cloudy white) | Solution remains clear |
| DCPIP test | Vitamin C (ascorbic acid) | Blue DCPIP solution | Add sample (e.g. fruit juice) drop by drop to DCPIP solution | DCPIP decolorises (turns colourless / transparent) | DCPIP stays blue |
Procedure marks: Name the specific reagent, not just “a chemical”. State whether heating is required. Benedict’s requires a water bath — not a Bunsen flame directly (which causes bumping and uneven heating).
Result marks: Give the exact colour change, not just “it changes colour”. State both initial colour → final colour. For Benedict’s: “blue → brick red”. For iodine: “yellow-brown → blue-black”.
Control: In an investigation, a control tube with distilled water (no food) should show the negative result for each test, confirming the reagent is working correctly.
DCPIP titration: A more concentrated vitamin C solution decolorises DCPIP with fewer drops. To compare vitamin C content, record the volume of juice needed to decolorise a fixed volume of DCPIP — less volume = more concentrated.
A student tests a food sample with Benedict’s solution and heats it in a water bath. The solution turns brick-red. What does this result indicate?
- A. Starch is present in the sample
- B. A reducing sugar is present in the sample
- C. Protein is present in the sample
- D. Fat is present in the sample
A student tests a sample of milk for protein using the biuret test.
(a) Describe the procedure for the biuret test. [2 marks]
(b) State the expected result if protein is present. [1 mark]
(c) The student also wants to test whether the milk contains a reducing sugar. State the reagent needed and describe the expected positive result. [2 marks]
- (a) Add sodium hydroxide solution to the sample [1 mark]; then add a few drops of dilute copper sulfate solution [1 mark]
- (b) The solution turns purple / violet [1 mark]
- (c) Reagent: Benedict’s solution [1 mark]; positive result: solution turns brick-red / orange (accept: colour change from blue to brick-red/orange precipitate) [1 mark]
DNA Structure — Extended
DNA (deoxyribonucleic acid) carries the genetic information of all living organisms. The syllabus requires knowledge of four structural features:
1. Double helix: Two strands are coiled together to form a twisted ladder shape (the double helix).
2. Bases on each strand: Each strand is made up of units called nucleotides, each containing a chemical called a base. There are four bases: Adenine (A), Thymine (T), Cytosine (C), Guanine (G). Full names are not required — just the letters.
3. Base pairing holds strands together: Bonds between pairs of bases on opposite strands hold the two strands together. These are the “rungs” of the twisted ladder.
4. Complementary base pairing: Bases always pair in the same way — A pairs with T, and C pairs with G. This is called complementary base pairing.
Think of DNA as a ladder that has been twisted into a helix. The two “side rails” are the sugar-phosphate backbones of each strand. The “rungs” are the base pairs (A-T and C-G) held together by bonds.
If one strand reads: A-T-C-G-A
The complementary strand reads: T-A-G-C-T
A always pairs with T; C always pairs with G. This rule allows you to work out the complementary sequence for any section of DNA.
Complementary base pairing allows DNA to be copied accurately — each strand acts as a template, and the new complementary strand is assembled by matching bases (A→T, C→G). This is the basis of DNA replication before cell division.
A section of one strand of a DNA molecule has the base sequence: A-G-C-T-A-C. What is the complementary base sequence on the other strand?
- A. A-G-C-T-A-C
- B. T-C-G-A-T-C
- C. T-C-G-A-T-G
- D. A-U-C-G-A-G
Apply complementary base pairing to each base: A→T, G→C, C→G, T→A, A→T, C→G
Original: A – G – C – T – A – C
Complement: T – C – G – A – T – G
Option D is wrong because it contains U (uracil), which is found in RNA, not DNA. Option B has one error at the last base (C should pair with G, not with C).
Describe the structure of a DNA molecule. [4 marks]
- DNA consists of two strands coiled together [1 mark]
- … to form a double helix [1 mark]
- Each strand contains chemicals called bases [1 mark]
- Bonds between pairs of bases hold the two strands together [1 mark]
- Bases always pair in the same way: A with T, and C with G (complementary base pairing) [1 mark]
Examiner note: If the question says “describe”, you must give factual details. Simply writing “it is a double helix” without explaining base pairing or strand structure would score only 1 of 4 marks.
Comprehensive Practice Questions
Mixed questions in the style of Cambridge IGCSE 0610 Papers 1–4.
Which of the following correctly states the elements present in a protein molecule?
- A. Carbon, hydrogen and oxygen only
- B. Carbon and hydrogen only
- C. Carbon, hydrogen, oxygen and nitrogen
- D. Carbon, hydrogen, nitrogen and phosphorus
A student performs the DCPIP test on two fruit juices. Juice X decolorises 2 cm³ of DCPIP solution. Juice Y decolorises only 0.5 cm³ of the same DCPIP solution. What can be concluded?
- A. Juice X contains more vitamin C than Juice Y
- B. Juice Y contains a higher concentration of vitamin C than Juice X
- C. Both juices contain equal amounts of vitamin C
- D. Juice X contains no vitamin C
Starch, glycogen, and cellulose are all large carbohydrate molecules.
(a) State the monomer from which all three are made. [1 mark]
(b) State one difference in the biological role of starch (in plants) and glycogen (in animals). [1 mark]
(c) Explain why humans cannot digest cellulose, even though it is made from the same monomer as starch. [2 marks]
- (a) Glucose [1 mark]
- (b) Starch is the energy storage molecule in plants; glycogen is the energy storage molecule in animals (e.g. stored in the liver and muscles) [1 mark for any valid distinction]
- (c) Although the monomer is the same, the glucose units are bonded together differently in cellulose compared to starch [1 mark]; humans lack the enzyme (cellulase) needed to break these bonds / to hydrolyse cellulose [1 mark]
A student tests four food samples (A, B, C, D) using the iodine, Benedict’s and biuret tests. The results are shown in the table below:
Sample A: iodine = blue-black, Benedict’s = blue, biuret = blue
Sample B: iodine = yellow-brown, Benedict’s = brick-red, biuret = blue
Sample C: iodine = yellow-brown, Benedict’s = blue, biuret = purple
Sample D: iodine = blue-black, Benedict’s = brick-red, biuret = purple
(a) Which sample contains only starch and no protein or reducing sugar? [1 mark]
(b) Which sample contains both a reducing sugar and protein, but no starch? [1 mark]
(c) The student heats Sample B with Benedict’s solution in a beaker of boiling water rather than a water bath. Suggest one problem with this method. [1 mark]
- (a) Sample A — blue-black iodine (starch present), blue Benedict’s (no reducing sugar), blue biuret (no protein) [1 mark]
- (b) Sample C — yellow-brown iodine (no starch), brick-red Benedict’s (reducing sugar present), purple biuret (protein present) [1 mark]
- (c) Boiling water in an open beaker may cause the sample to boil and spit/bump, creating a safety hazard; OR uneven heating may occur; OR the test tube may break [1 mark for any valid safety or procedural problem]
A section of double-stranded DNA contains 120 base pairs. Analysis shows that 30 of the bases on one strand are adenine (A).
(a) How many thymine (T) bases are on the complementary strand in this section? [1 mark]
(b) The section contains 36 cytosine (C) bases in total across both strands. How many guanine (G) bases does this section contain in total? [2 marks]
(c) State how the two strands of DNA are held together. [1 mark]
- (a) 30 T bases [1 mark] — because A always pairs with T, so every A on one strand is opposite a T on the other; 30 A bases → 30 complementary T bases
- (b) 36 G bases [2 marks] — because C always pairs with G (complementary base pairing), so the total number of G bases must equal the total number of C bases [1 mark]; if there are 36 C across both strands, there are 36 G [1 mark]
- (c) The two strands are held together by bonds between complementary base pairs (A-T and C-G) [1 mark]
High-Frequency Mistakes — Topic 4 Overall
- ⚖Forgetting nitrogen in proteinsProteins contain C, H, O, and N. Carbohydrates and fats contain only C, H, O. Nitrogen is the one element that distinguishes proteins. Write “CHON” to remember.
- 🧲Iodine goes “blue” not “blue-black”The exact wording matters. The positive result for iodine is blue-black. Just saying “blue” may not receive the mark. Similarly, the negative result must be described as “stays yellow-brown”, not “stays brown”.
- 🍵Benedict’s tests for “sugar” in generalBenedict’s tests for reducing sugars specifically. Sucrose is a non-reducing sugar — it gives a negative Benedict’s result even though it is a sugar. If sucrose is hydrolysed (broken down) first, glucose is released and gives a positive result.
- 🧠DCPIP turns “yellow” or “pink”The positive DCPIP result is decolorisation — the blue solution turns colourless / transparent. It does not turn yellow or pink. Only full decolorisation counts as a positive result.
- 💧Ethanol emulsion gives a “white precipitate”More precisely, the result is a milky-white emulsion — a cloudy suspension, not a solid precipitate. The wording is important on mark schemes.
- 📄Saying starch is made from “sugars” or “carbohydrates”The monomer of starch, glycogen, and cellulose is specifically glucose. “Sugars” is too vague; mark schemes require “glucose”.
- 🍳Fats are made from “fatty acids and amino acids”Fats are made from fatty acids and glycerol. Amino acids are the monomers of proteins. Easy confusion — double-check: fats → fatty acids + glycerol; proteins → amino acids.
- 🔫Ext: Writing U (uracil) instead of T (thymine) in DNAUracil (U) is found in RNA, not DNA. In DNA, adenine always pairs with thymine (T). Writing A-U pairing in a DNA question is a common error and will lose marks.
- 🔗Ext: Mixing up base pairs (A-G, C-T, etc.)The only correct pairings are A with T, and C with G. No other combinations are used. A memory trick: A and T are both thin/straight letters; C and G look like they could fit together as curves. Or simply: “AT CG” — two letters per pair.
Food tests appear in almost every Paper 3/4 and are a guaranteed component of Paper 5/6. Memorise all five tests as a complete set: reagent (start colour) → procedure (heating?) → positive result (exact colour/appearance) → negative result. The most frequently lost marks are from vague colour descriptions and forgetting whether heating is required. For Extended, DNA base pairing calculations are straightforward if you remember A-T and C-G — practise a few complementary strand questions until it’s automatic.