Biology HL

1. 1.

   An unbranched carbohydrate is made by joining 80 glucose molecules into one chain using condensation. Calculate how many water molecules are released, explain why hydrolysis needs water, and explain one advantage of storing glucose as a large relatively insoluble polymer.

   [4 marks] · no calculator

   Worked explanation

   1. The first monomer needs no link; each of the remaining 79 adds one link and releases one water.
   2. Reverse the chemistry using water to cleave bonds, not just heat or mechanical breaking.
   3. Compare polymer storage with the same number of free soluble monomers, not with an empty cell.

   Common mistake: Eighty monomers in a single linear chain form 79 links, not 80, because there is no link before the first monomer.

   Marking points

   • A linear chain of 80 monomers has 79 links, releasing 79 water molecules.
   • Hydrolysis consumes water to provide H and OH at the broken link.
   • Many glucose residues can be retained in a compact storage molecule.
   • Relative insolubility reduces osmotic effects and loss by diffusion compared with many free glucose molecules.

   Examiner tip: Eighty monomers in a single linear chain form 79 links, not 80, because there is no link before the first monomer.

2. 2.

   Compare cellulose with starch as glucose polymers. Explain how monomer orientation and interactions between chains suit cellulose to plant cell walls, whereas starch is suited to storage.

   [4 marks] · no calculator

   Worked explanation

   1. Different glucose orientation leads to different polymer geometry despite a shared monomer type.
   2. Parallel straight cellulose chains form strong hydrogen-bonded bundles for resisting tension.
   3. A compact storage polymer can be hydrolysed when glucose is needed; it is not a structural substitute for cellulose.

   Common mistake: Hydrogen bonds cross-link cellulose chains; they are not the covalent glycosidic links joining glucose within a chain.

   Marking points

   • Cellulose uses beta-glucose with alternating orientation; starch uses alpha-glucose.
   • Cellulose forms straight chains that align in bundles.
   • Hydrogen bonds between cellulose chains strengthen microfibrils and provide tensile strength.
   • Coiling and/or branching makes starch compact and its glucose can be mobilized by hydrolysis for energy use.

   Examiner tip: Hydrogen bonds cross-link cellulose chains; they are not the covalent glycosidic links joining glucose within a chain.

3. 3.

   Contrast a triglyceride with a phospholipid in composition and behaviour in water. Explain why phospholipids assemble into bilayers but triglycerides are better described as hydrophobic energy stores.

   [4 marks] · no calculator

   Worked explanation

   1. Identify the compositional difference before explaining the water interaction.
   2. Both sides of a cellular bilayer contact aqueous solution, so heads face outward and inward towards those solutions while tails meet centrally.
   3. A triglyceride droplet is not a bilayer with two aqueous-facing head layers.

   Common mistake: The tails point away from water into the bilayer interior, not out towards the cytoplasm and extracellular fluid.

   Marking points

   • A triglyceride contains glycerol linked to three fatty acids.
   • A typical phospholipid contains glycerol, two fatty acids and a phosphate-containing polar head.
   • Hydrophilic heads face water while hydrophobic tails are shielded inside a bilayer.
   • Triglycerides lack the comparable hydrophilic head and are largely water-insoluble energy stores rather than bilayer-forming amphipaths.

   Examiner tip: The tails point away from water into the bilayer interior, not out towards the cytoplasm and extracellular fluid.

4. 4.

   Compare a triglyceride with a phospholipid in terms of fatty-acid attachment and interaction with water. Explain why phospholipids can form bilayers whereas triglycerides are more suited to bulk energy storage than to forming the main membrane bilayer.

   [4 marks] · no calculator

   Worked explanation

   1. Identify the typical structures before connecting them to their different roles.
   2. In water on both sides of a membrane, phospholipid heads can face each aqueous compartment while tails face inward.
   3. The contrast concerns molecular architecture and water interactions, not a claim that triglycerides have no biological function.

   Common mistake: A phospholipid is not a triglyceride with three fatty-acid tails plus a phosphate group.

   Marking points

   • A triglyceride has glycerol ester-linked to three fatty acids.
   • A typical glycerol-based phospholipid has two fatty-acid tails and a phosphate-containing polar head.
   • Phospholipids are amphipathic, with heads interacting with water and tails sheltered inside a bilayer.
   • Triglycerides lack the comparable polar head arrangement and form hydrophobic stores rather than the main bilayer framework.

   Examiner tip: A phospholipid is not a triglyceride with three fatty-acid tails plus a phosphate group.

5. 5.

   A linear chain is assembled from 25 glucose molecules by condensation, with one glycosidic linkage between each successive pair. Calculate the number of glycosidic linkages and water molecules released. Explain what hydrolysis would require to separate the chain completely back into glucose molecules.

   [3 marks] · no calculator

   Worked explanation

   1. Build the counting rule from two monomers joined by one linkage: each additional monomer adds one linkage.
   2. For 25 monomers the rule is n - 1, not n, because the chain is linear and not cyclic.
   3. Hydrolysis reverses condensation at each linkage, with water consumed rather than released.

   Common mistake: For a linear chain the number of links is one less than the number of monomers; do not confuse hydrolysis with condensation.

   Marking points

   • A linear chain of 25 monomers has 24 glycosidic linkages.
   • Formation of those linkages releases 24 water molecules under the stated condensation model.
   • Complete hydrolysis breaks all 24 linkages, using 24 water molecules overall.

   Examiner tip: For a linear chain the number of links is one less than the number of monomers; do not confuse hydrolysis with condensation.

6. 6.

   Two fatty acids each contain 18 carbon atoms. X has no carbon-carbon double bonds; Y has one cis carbon-carbon double bond. Classify X and Y as saturated or unsaturated. Explain why the cis bond can make Y-containing lipid chains pack less closely than otherwise comparable X-containing chains, and predict which of two pure triglycerides would usually have the lower melting temperature if one contains only X chains and the other only Y chains. All other structural features are matched; do not generalize this comparison to every natural mixture of fats.

   [4 marks] · no calculator

   Syllabus placement evidence: Biology guide, first assessment 2025, updated March 2024: B1.1.10, printed p.50 (PDF p.56); Tool 1, physical and digital molecular modelling, p.30. B1.1.10 prescribes fatty-acid saturation classes and the effect of carbon-carbon double bonds on melting point at both SL and HL. The matched cis-chain comparison applies molecular geometry to that structure-function link. Placement in an HL topic does not make this content AHL-only; no claim of complete lipid coverage is made.

   Worked explanation

   1. Count carbon-carbon double bonds, not oxygen-containing bonds in the carboxyl group. Equal carbon counts isolate the supplied saturation difference rather than a chain-length difference.
   2. Connect cis geometry to a bend, reduced packing and weaker collective intermolecular attraction. Melting disrupts intermolecular organization; it does not require breaking the carbon-carbon double bond or the ester bonds.

   Common mistake: An unsaturated lipid need not melt at a higher temperature merely because a carbon-carbon double bond is strong: melting does not break that covalent bond.

   Marking points

   • X is saturated; Y is monounsaturated because it has one carbon-carbon double bond.
   • The cis arrangement introduces a bend/kink that prevents the chain adopting the same extended shape as X.
   • The bent chains pack less closely, reducing effective intermolecular attractions between otherwise comparable chains.
   • The triglyceride containing only Y chains would usually melt at a lower temperature because less thermal energy is needed to disrupt the less tightly packed assembly.

   Examiner tip: An unsaturated lipid need not melt at a higher temperature merely because a carbon-carbon double bond is strong: melting does not break that covalent bond.

7. 7.

   A hypothetical pure sample contains 0.20 mmol of triglyceride molecules. Each has one glycerol joined to three fatty acids by three ester bonds. Assume complete hydrolysis without side reactions. Name the products, calculate the amounts of glycerol and fatty acids formed and the amount of water consumed, and explain why melting the same sample would not produce those chemical products.

   [5 marks]

   Syllabus placement evidence: Biology guide, first assessment 2025, updated March 2024: B1.1.9, printed p.50; C1.1.3, p.65; Tool 3 Mathematics, p.30; official 2025 specimen HL Paper 1A Q4, printed p.3. B1.1.9 supplies the triglyceride composition/condensation basis and C1.1.3 the catabolic-hydrolysis context. The official specimen independently assesses triglyceride hydrolysis products. The 1:3 product and water calculation is an application to the explicitly supplied three ester bonds, not an extra memorized equation. The underlying lipid content is shared SL/HL, despite this HL-topic placement.

   Worked explanation

   1. Use triglyceride:glycerol:fatty acids:water = 1:1:3:3: one triglyceride yields one glycerol and three fatty acids, consuming three water molecules.
   2. Keep the supplied units in mmol throughout. Contrast covalent-bond hydrolysis with the physical-state change addressed in the earlier packing task.

   Common mistake: Hydrolysis consumes water; it does not release three waters as the reverse condensation model would.

   Marking points

   • Complete hydrolysis produces glycerol and fatty acids by breaking ester bonds.
   • Glycerol formed = 0.20 mmol.
   • Fatty acids formed = 3 x 0.20 = 0.60 mmol.
   • Water consumed = 3 x 0.20 = 0.60 mmol because each ester hydrolysis consumes one water molecule.
   • Melting changes intermolecular organization/physical state without cleaving the ester bonds; it does not convert the triglyceride into glycerol and fatty acids.

   Examiner tip: Hydrolysis consumes water; it does not release three waters as the reverse condensation model would.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.