Cambridge International AS & A Level Biology 9700: Carbohydrates and lipids – 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 from your board or school.

These questions cover section 2.2 Carbohydrates and lipids (learning outcomes 1 to 11) of Topic 2, Biological molecules, in the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content, examined on Papers 1 and 2, and Paper 4 can draw on it. Calculators are allowed in all 9700 papers.

Learn it first: Carbohydrates and lipids study guide · Recall: Carbohydrates and lipids revision notes · Whole-topic questions: Biological molecules practice questions · Course hub: Cambridge A Level Biology · Checklist: 9700 checklist

Questions

1. Glucose exists in two ring forms, α-glucose and β-glucose.

(a) State the number of carbon atoms in the ring of a glucose molecule. [1] (b) Describe how the structure of β-glucose differs from that of α-glucose. [2]

2.

(a) Define the term monomer. [1] (b) Define the term polysaccharide. [1] (c) Explain why a triglyceride is not described as a polymer. [1]

3. State which of these sugars are reducing sugars and which are non-reducing: glucose, sucrose, maltose, fructose. [2]

4. Maltose is formed from two molecules of α-glucose.

(a) Describe how the two α-glucose molecules are joined. [3] (b) Glucose has Mr 180 and water has Mr 18. Calculate the Mr of maltose. [1]

5. A sports drink contains sucrose as its only sugar. Describe how you would show that the drink contains a non-reducing sugar. Give the result you would expect at each stage. [5]

6. In a model, an amylopectin molecule and a glycogen molecule each contain 50 000 α-glucose units. The amylopectin molecule has one branch point for every 25 glucose units. The glycogen molecule has one branch point for every 10 glucose units.

(a) Calculate the number of α-1,6 glycosidic bonds in each molecule. [2] (b) Explain how the difference you calculated suits glycogen to its role in animals. [3]

7. Describe the molecular structure of cellulose and explain how the arrangement of cellulose molecules makes plant cell walls strong. [5]

8. Tristearin is a triglyceride made from glycerol and three molecules of stearic acid.

(a) Name the bond that forms between glycerol and each stearic acid molecule, and the type of reaction that forms it. [2] (b) State how many molecules of water are released when one molecule of tristearin forms. [1] (c) Glycerol has Mr 92, stearic acid has Mr 284 and water has Mr 18. Calculate the Mr of tristearin. [2]

9. The formulae of three fatty acids, each with 18 carbon atoms, are: A C₁₈H₃₆O₂; B C₁₈H₃₄O₂; C C₁₈H₃₂O₂.

(a) Identify the saturated fatty acid. [1] (b) State the number of C=C double bonds in B and in C. [1] (c) Describe how a C=C double bond affects the shape of a fatty acid chain. [1] (d) State the term used for a fatty acid such as C. [1]

10. A small migratory bird stores 30 g of triglyceride before its journey. Use these values: triglyceride releases 37 kJ g⁻¹ and carbohydrate releases 17 kJ g⁻¹ when respired.

(a) Calculate the energy stored in the bird’s triglyceride. [1] (b) Calculate the mass of glycogen that would store the same energy. Give your answer to 3 significant figures. [2] (c) With reference to molecular structure, explain why triglyceride is a better energy store than glycogen for a bird that flies long distances. [3] (d) State one other function of triglycerides in animals. [1]

11.

(a) Describe the structure of a phospholipid molecule, with reference to the polarity of its parts. [3] (b) A few drops of phospholipid are added to the surface of water in a dish. Predict how the molecules arrange themselves at the surface, and explain your prediction. [2]

12. A germinating seed hydrolyses its food stores of amylose, amylopectin and triglyceride.

(a) Name the product of complete hydrolysis of amylose, and the products of complete hydrolysis of a triglyceride. [2] (b) Calculate the number of water molecules needed to completely hydrolyse: (i) one unbranched amylose chain of 600 glucose units; (ii) one triglyceride molecule. [2] (c) Explain why amylopectin can be hydrolysed to glucose faster than the same mass of amylose. [2] (d) Suggest one advantage to a small, wind-dispersed seed of storing its energy as triglyceride rather than starch. [2]

Answers

1. (a) Five (the sixth ring atom is oxygen) [1] (b) The two forms differ only at carbon 1 [1]. In β-glucose the OH group on carbon 1 is above the ring (H below), while in α-glucose it is below the ring [1]. Examiner insight: “The OH is in a different position” is too vague for the second mark; you must say above/below the ring and name carbon 1.

2. (a) A relatively small molecule that can join with other similar molecules to form a polymer [1] (b) A polymer of many monosaccharides joined by glycosidic bonds [1] (c) It is made of one glycerol and three fatty acids, not a long chain of repeating monomer units [1] Examiner insight: For (b), “many sugars” without “joined by glycosidic bonds” or “monosaccharides” is usually not enough for a definition mark.

3. Reducing: glucose, maltose, fructose; non-reducing: sucrose. All four correct [2] (two or three correct scores 1 mark). Examiner insight: Maltose is a disaccharide but still reducing; listing every disaccharide as non-reducing loses the mark.

4. (a) The OH on carbon 1 of one α-glucose reacts with the OH on carbon 4 of the other [1]. A molecule of water is removed in a condensation reaction [1]. An α-1,4 glycosidic bond forms [1]. (b) (2 × 180) − 18 = 342 [1] Examiner insight: “Water is produced” earns the condensation mark only if the reaction is named or clearly described; “hydrolysis” here scores zero for that point.

5. Do a Benedict’s test on a sample and heat: it stays blue, so no reducing sugar is present [1]. Heat a fresh sample with dilute hydrochloric acid in a boiling water bath [1]. This hydrolyses the glycosidic bond in sucrose, giving glucose and fructose [1]. Cool and neutralise with sodium hydrogencarbonate [1]. Add Benedict’s solution and heat: an orange-red / brick-red precipitate forms, showing a non-reducing sugar was present [1]. Examiner insight: A colour change must name the colour; “it changes colour” or “turns positive” does not earn the result marks.

6. (a) Amylopectin: 50 000 ÷ 25 = 2000 [1]. Glycogen: 50 000 ÷ 10 = 5000 [1]. (b) Glycogen has more branches, so it has more chain ends [1]. Enzymes hydrolyse glucose from the ends, so many glucose molecules can be released at the same time / quickly [1]. Animals are active and can have a high, sudden demand for glucose for respiration [1]. Examiner insight: “More branches so it is more compact” is true but does not answer the question; the marks are for ends, enzyme access and rapid release.

7. Cellulose is made of β-glucose joined by β-1,4 glycosidic bonds [1]. Every other glucose is rotated 180°, so the chain is straight and unbranched [1]. Neighbouring chains lie parallel and are held together by many hydrogen bonds [1]. Groups of chains form microfibrils, which group into fibres laid in layers [1]. This gives high tensile strength, so the wall resists the pressure of water entering by osmosis and the cell does not burst [1]. Examiner insight: Reject “glycosidic bonds between chains”; the bonds between neighbouring chains are hydrogen bonds, and mixing them up costs the mark.

8. (a) Ester bond [1]; formed by condensation [1]. (b) Three [1] (c) 92 + (3 × 284) = 944 [1]; 944 − (3 × 18) = 890 [1] Examiner insight: If (b) is wrong, the method mark in (c) can still be earned for subtracting water correctly from your own count (error carried forward).

9. (a) A [1] (b) B: one; C: two (both needed) [1] (c) It puts a kink / bend in the hydrocarbon chain [1] (d) Polyunsaturated [1] Examiner insight: Answers in (b) are marked together; one correct value alone does not score.

10. (a) 30 × 37 = 1110 kJ [1] (b) 1110 ÷ 17 [1] = 65.3 g [1] (c) Triglycerides have many C–H bonds and little oxygen [1], so they release more energy per gram when respired [1]. The bird carries a smaller mass for the same energy, which reduces the energy cost of flight [1]. (d) Any one: thermal insulation; buoyancy; source of metabolic water; protection of organs [1] Examiner insight: In (b) the answer needs the unit and 3 significant figures; 65.29 or 65 loses the accuracy mark.

11. (a) Glycerol joined by ester bonds to two fatty acids [1]. A phosphate group is attached to the third carbon of glycerol [1]. The phosphate head is polar / hydrophilic and the fatty acid tails are non-polar / hydrophobic [1]. (b) A single layer forms, with phosphate heads in the water and tails pointing up into the air [1]. The heads are hydrophilic and attracted to water; the tails are hydrophobic and repelled by water [1]. Examiner insight: A bilayer is the wrong prediction for a surface; credit goes to a reasoned arrangement that matches the conditions given.

12. (a) Amylose: α-glucose [1]. Triglyceride: glycerol and fatty acids [1]. (b) (i) 600 − 1 = 599 [1]; (ii) 3 [1] (c) Amylopectin is branched, so it has more ends than unbranched amylose [1]. Enzymes act at the ends, so more glucose is released at once [1]. (d) Triglyceride releases more energy per gram [1], so the seed can be lighter for the same energy store and be carried further by wind [1]. Examiner insight: In (b)(i), answering 600 shows the n − 1 rule was missed; one water is used per glycosidic bond, not per glucose unit.

Where marks are usually lost

• Describing α/β glucose without naming carbon 1 or saying above/below the ring.
• Calling maltose non-reducing, or sucrose reducing.
• Mixing up condensation and hydrolysis, or which one releases water.
• Leaving out neutralisation, or not stating colours, in the non-reducing sugar test.
• Writing “glucose” where the question needs α-glucose or β-glucose.
• Saying cellulose chains are joined by glycosidic bonds to each other.
• Using n instead of n − 1 when counting bonds or water molecules in a chain.
• Forgetting to subtract the water when calculating the Mr of a condensation product.
• Linking fat to energy storage without the reason: more C–H bonds, more energy per gram.
• Calling a phospholipid tail hydrophilic, or a triglyceride polar.

Next steps

• Carbohydrates and lipids revision notes
• Carbohydrates and lipids study guide
• Biological molecules practice questions
• Cambridge A Level Biology hub
• 9700 printable checklist
• Free 9700 AS diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment – Topic 2, Biological molecules (section 2.2 Carbohydrates and lipids).