OCR A Level Biology A: Water, carbohydrates and lipids (H420) – 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 learning outcomes 2.1.2(a) to 2.1.2(j) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025. Learn the content first in the study guide or the revision notes. Course pages: OCR A Level Biology hub and checklist. More mixed practice is in Foundations practice. Where a question gives data, use those values.

Questions

1. Explain how hydrogen bonding between water molecules makes water suitable as a coolant and as a solvent in living organisms. [3]

2. (a) State the elements found in a triglyceride, and the elements found in a protein. [2] (b) Explain what is meant by a polymer, and why condensation and hydrolysis are both important for polymers in cells. [2]

3. (a) State the one structural feature that distinguishes the α form of glucose from the β form. [1] (b) Ribose is a pentose and glucose is a hexose. State what this means and describe the ring in ribose. [3]

4. A solution contains 3.60 g of maltose, which is completely hydrolysed. The Mr of maltose is 342 and the Mr of glucose is 180. (a) Name the bond that is broken. [1] (b) Calculate the mass of glucose formed. [3]

5. A glycogen molecule contains 1 200 glucose residues joined by glycosidic bonds. Assume the Mr of glucose is 180 and of water 18. (a) State how many water molecules were released when the chain formed. [1] (b) Calculate the Mr of the molecule, giving your answer in standard form to 3 significant figures. [3]

6. A starch molecule and a glycogen molecule each contain 10 000 glucose residues. Branches occur every 25 residues in the amylopectin of the starch and every 10 residues in the glycogen. Assume each branch creates one extra free end and that the molecule begins with one free end. (a) Calculate the number of free ends in each molecule. [2] (b) Calculate how many times more free ends glycogen has. [1] (c) Explain why this difference suits the role of glycogen in muscle. [2]

7. Compare the structures of amylose and cellulose, and explain how each is suited to its function. [6]

8. Palmitic acid has the formula C16H32O2 and glycerol C3H8O3. (C = 12, H = 1, O = 16.) (a) Name the type of bond that joins a fatty acid to glycerol. [1] (b) Calculate the Mr of a triglyceride formed from glycerol and three palmitic acid molecules. [3] (c) State how many water molecules are used to hydrolyse one molecule of this triglyceride. [1]

9. A migrating bird stores 12 g of lipid. Assume lipid releases 38 kJ g⁻¹ and carbohydrate 17 kJ g⁻¹. (a) Calculate the energy in the stored lipid. [1] (b) Calculate the mass of carbohydrate that would store the same energy. [2] (c) Explain why lipid is a better energy store than carbohydrate for a flying animal. [2]

10. A student prepared artificial lipid vesicles containing a dye and measured the rate at which dye leaked out at 37 °C. The vesicles contained different percentages of cholesterol.

Cholesterol in membrane (%)	Leakage rate (arbitrary units per minute)
0	8.0
10	5.6
20	3.5
30	2.4

(a) Calculate the percentage decrease in leakage rate between 0% and 20% cholesterol. [2] (b) Describe the trend and suggest how cholesterol produces it. [3] (c) Explain why this result applies to eukaryote cell-surface membranes but not usually to prokaryotes. [1]

11. The table shows the density of water at different temperatures. The density of ice is 0.917 g cm⁻³.

Temperature (°C)	0	2	4	6	10
Density of water (g cm⁻³)	0.99984	0.99994	0.99997	0.99994	0.99970

(a) State the temperature at which water is densest. [1] (b) Calculate the percentage of the volume of floating ice that lies above the surface of water at 0 °C. [2] (c) Explain, using hydrogen bonding, why ice floats, and why this helps organisms in a pond in winter. [3]

12. Describe the structure of a phospholipid and explain how its properties lead to a bilayer in water and to the function of membranes in a bacterium and in an animal cell. [8]

Answers

1. Coolant: hydrogen bonds between molecules must be broken for water to evaporate [1], so evaporation (sweating or transpiration) removes a large amount of energy because of its high latent heat of vaporisation [1]. Solvent: water is polar, so its δ+ and δ− regions attract and surround ions and polar molecules, so reactants dissolve in cytoplasm [1]. [3] Examiner insight: A property named without its cause, or a cause without the biological role, usually earns only part of the credit; link all three.

2. (a) Triglyceride: carbon, hydrogen and oxygen [1]. Protein: carbon, hydrogen, oxygen, nitrogen and sulfur [1]. (b) A polymer is a long chain made of many monomers [1]. Condensation builds the chain and releases water; hydrolysis breaks it using water, for digestion or recycling of monomers [1]. [4] Examiner insight: Giving the element symbols and the names together avoids ambiguity; say which reaction uses water and which releases it.

3. (a) The –OH group on carbon 1 is below the ring in α-glucose and above it in β-glucose [1]. (b) A pentose has five carbon atoms and a hexose has six [1]. The ring of ribose contains four carbons and one oxygen [1], with the fifth carbon outside the ring as CH2OH [1]. [4] Examiner insight: Name the carbon (carbon 1) and the position (above or below the plane) so the comparison cannot be read two ways.

4. (a) Glycosidic bond [1]. (b) Moles of maltose = 3.60 ÷ 342 = 0.01053 mol [1]. One maltose gives two glucose, so 0.02105 mol [1]. Mass = 0.02105 × 180 = 3.79 g [1]. [4] Examiner insight: Show the moles and the two-for-one ratio so each step can be followed. Check that the answer is above 3.60 g, because water is added.

5. (a) 1 199 [1]. (b) Mass of residues = 1 200 × 180 = 216 000 [1]. Water lost = 1 199 × 18 = 21 582 [1]. Mr = 216 000 − 21 582 = 194 418 = 1.94 × 10⁵ [1]. [4] Examiner insight: Writing 1 200 water molecules gives a different final figure; the n − 1 step is the point of the question.

6. (a) Glycogen: 10 000 ÷ 10 = 1 000 branches, so 1 001 ends [1]. Starch: 10 000 ÷ 25 = 400 branches, so 401 ends [1]. (b) 1 001 ÷ 401 = 2.5 times [1]. (c) More free ends mean enzymes can hydrolyse many α-1,4 bonds at once [1], so glucose is released rapidly for respiration during muscle contraction [1]. [5] Examiner insight: Linking “more ends” to “enzymes act at many points at once” gives the explanation; “more surface” on its own is vague.

7. Amylose: α-glucose joined by α-1,4 bonds [1]; unbranched chain coils into a helix, which is compact [1]; insoluble, so it stores energy without affecting water potential [1]. Cellulose: β-glucose joined by β-1,4 bonds [1]; alternate residues are inverted so chains are straight [1]; hydrogen bonds between parallel chains form strong microfibrils, so the cell wall resists bursting [1]. [6] Examiner insight: A comparison needs both molecules addressed point by point; two separate descriptions without a stated similarity or difference can lose credit.

8. (a) Ester bond [1]. (b) Palmitic acid Mr = 192 + 32 + 32 = 256; glycerol Mr = 36 + 8 + 48 = 92 [1]. Mr = 92 + 3 × 256 [1] − 3 × 18 = 806 [1]. (c) Three [1]. [5] Examiner insight: Subtracting only one water, as in a two-molecule condensation, is a common slip; count the ester bonds first.

9. (a) 12 × 38 = 456 kJ [1]. (b) 456 ÷ 17 [1] = 26.8 g [1]. (c) Lipid stores more energy per gram, so the bird carries less mass [1]; carbohydrate is stored with more water attached, adding further mass, while lipid is hydrophobic [1]. [5] Examiner insight: Say “per gram” or “per unit mass” explicitly; “more energy” alone does not state the comparison that matters for flight.

10. (a) (8.0 − 3.5) ÷ 8.0 × 100 [1] = 56% (2 s.f.) [1]. (b) Leakage rate decreases as cholesterol percentage increases [1]. Cholesterol fits between phospholipids, filling gaps and restricting movement of the fatty acid tails [1], so the membrane is less permeable to the dye [1]. (c) Most prokaryotes do not make cholesterol; it is a component of eukaryote, especially animal, membranes [1]. [6] Examiner insight: Support a described trend with values from the data; “gets lower” alone is weaker than “falls from 8.0 to 2.4”.

11. (a) 4 °C [1]. (b) Fraction below the surface = 0.917 ÷ 0.99984 = 0.917 [1]. Above the surface = 1 − 0.917 = 0.083 = 8.3% [1]. (c) In ice, hydrogen bonds hold molecules in a fixed, open lattice, so they are further apart than in liquid water, giving a lower density [1]. Ice floats on the surface and forms an insulating layer [1], so the water below stays liquid and organisms survive [1]. [6] Examiner insight: Reference to hydrogen bonds is expected when the stem says “using hydrogen bonding”; “ice is lighter” without the lattice is not an explanation.

12. A phospholipid has a glycerol backbone [1], two fatty acids joined by ester bonds, which are hydrophobic tails [1], and a phosphate group, which is a charged hydrophilic head [1]. In water the heads point towards the surrounding water while the tails turn inwards, so two layers form a bilayer [1]. The hydrophobic core stops ions and polar molecules diffusing through [1], so the cell-surface membrane keeps the contents separate from the surroundings in both a bacterium and an animal cell [1]. In the animal cell, membranes also surround organelles, creating separate compartments [1], and cholesterol between phospholipids regulates fluidity [1]. [8] Examiner insight: Longer questions reward a clear sequence from structure to property to function; each region of the molecule should be named with its behaviour.

Where marks are usually lost

• Not linking a water property to its cause in hydrogen bonds and then to the organism.
• Quoting only human examples when the question asks for prokaryote or eukaryote contexts.
• Using n instead of n − 1 for water molecules and bonds in a polymer.
• Mixing up which carbon-1 position is α and which is β.
• Writing “starch” when the question needs amylose or amylopectin.
• Not giving the working for a calculation with supplied data.
• Describing cholesterol as part of every membrane.
• Explaining the bilayer without saying which region faces water.
• Giving a percentage change without stating the starting value used.

Next steps

• Go back to the revision notes or the study guide.
• Visit the OCR A Level Biology hub and the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025: section 2.1.2, learning outcomes (a) to (j).