AQA A Level Biology: Biological Molecules — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Biological Molecules revision notes

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Section A

1. Name the bond formed between two monosaccharides and the type of reaction. [2]

2. Explain why lipids are not classed as polymers. [2]

3. Name the four levels of protein structure and the bonds maintaining each. [4]

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Section B

4. Cellulose and starch are both polysaccharides of glucose.

(a) State which isomer of glucose forms each. [2]

(b) Explain, in terms of structure, why cellulose has high tensile strength. [3]

(c) Explain why starch is a good storage molecule. [3]

5. An enzyme-catalysed reaction was investigated at different substrate concentrations. The rate increased and then plateaued.

(a) Explain the shape of the curve. [3]

(b) Describe how the curve would differ with a competitive inhibitor present, and explain why. [3]

(c) Describe how the curve would differ with a non-competitive inhibitor, and explain why. [3]

6. A student is given an unknown solution and asked to test for a non-reducing sugar.

(a) Describe the full procedure. [4]

(b) Explain why a negative Benedict’s test must be carried out first. [2]

7. A polypeptide is 250 amino acids long. Calculate how many water molecules are released during its synthesis and how many are used during its complete hydrolysis. [2]

8. State the reagent and positive result for testing a food sample for lipid. [2]

9. Glycogen and starch are both storage polysaccharides of alpha glucose. State one structural difference and explain its functional significance. [2]

10. Explain why disulfide bridges are described as part of tertiary, not secondary, structure. [2]

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Answers

1. A glycosidic bond [1], formed by a condensation reaction (releasing water) [1].

2. A triglyceride consists of glycerol and three fatty acids — there is no repeating monomer unit [1], and it is not formed from many identical subunits joined in a chain [1].

3. Primary — peptide bonds [1]. Secondary — hydrogen bonds [1]. Tertiary — hydrogen, ionic, disulfide and hydrophobic interactions [1]. Quaternary — the same bonds, between separate polypeptides [1].

4. (a) Cellulose: β-glucose [1]. Starch: α-glucose [1].

(b) Alternate β-glucose molecules are inverted (rotated 180°) to form the glycosidic bond [1], giving straight, unbranched chains [1]. Many hydrogen bonds form between adjacent chains, holding them together as microfibrils [1].

(c) It is insoluble, so it does not affect water potential or cause osmotic problems [1]; it is compact (helical/branched), storing much glucose in little space [1]; amylopectin’s branching gives many ends for rapid hydrolysis when glucose is needed [1].

5. (a) At low substrate concentration, rate increases because more enzyme–substrate complexes form [1]. The rate plateaus because all active sites are occupied [1], so enzyme concentration becomes the limiting factor [1].

(b) The curve rises more slowly but eventually reaches the same maximum rate [1]. The inhibitor is a similar shape to the substrate and binds to the active site [1], but at high substrate concentration the substrate outcompetes it [1].

(c) The curve reaches a lower maximum rate [1]. The inhibitor binds elsewhere on the enzyme, changing the shape of the active site [1], so it cannot be overcome by adding more substrate [1].

6. (a) Boil the sample with dilute hydrochloric acid to hydrolyse it [1]; neutralise with sodium hydrogencarbonate [1]; add Benedict’s solution and heat [1]; a brick-red precipitate indicates a non-reducing sugar was present [1].

(b) To confirm no reducing sugar was present initially [1], since otherwise a positive result after hydrolysis could be due to that rather than a non-reducing sugar [1].

7. Bonds in a chain of n monomers = n − 1 = 249 [1]. Condensation releases 249 water molecules; hydrolysis consumes 249 water molecules [1]. The number of bonds, not the number of monomers, is what the calculation turns on.

8. Add ethanol, then add the mixture to water [1]; a white emulsion indicates lipid is present [1].

9. Glycogen is more highly branched than starch [1], giving more free ends for rapid hydrolysis — appropriate for animals with higher metabolic rates than plants [1].

10. Secondary structure is held only by hydrogen bonds between nearby amino acids in a regular repeating pattern (alpha helices/beta sheets) [1]; disulfide bridges form between R groups that may be far apart in the sequence, contributing to the overall 3D shape — a feature of tertiary structure [1].

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Where marks are usually lost

• Calling lipids polymers.
• Explaining cellulose strength without mentioning inverted alternate molecules.
• Saying more substrate overcomes a non-competitive inhibitor.
• Forgetting to neutralise after acid hydrolysis.
• Saying the plateau occurs because “the enzyme is used up” — enzymes are not consumed.
• Counting the number of monomers instead of the number of bonds (n − 1) when finding how many water molecules are released or used.
• Forgetting to add the ethanol/lipid mixture to water — a positive lipid test needs both steps, not ethanol alone.
• Describing glycogen as “the same as starch” without mentioning its greater branching and why that matters for a higher metabolic rate.
• Placing disulfide bridges at secondary rather than tertiary structure — secondary structure is hydrogen bonds only, in a regular repeating pattern.