Paper 4 style practice paper: Structured questions (Extended) — mark scheme
IGCSE · Cambridge IGCSE Biology (Extended)
Mark scheme
Structured questions (Extended)
Paper 4 style practice paper · Maximum mark: 80
Notes for markers
- Award one mark for each point marked [1], up to the marks shown for the part.
- Accept answers worded differently that show the same understanding.
- Units are required where the answer line shows one. ecf: allow an error carried forward.
Section A: Structured questions[40 marks]
| Question | Answer and marking points | Marks |
|---|---|---|
| 1(a) | Chloroplast; absorbs light energy for photosynthesis.
| 2 |
| 1(b) | Any two of: cell wall, cell membrane, cytoplasm, ribosomes.
| 2 |
| 1(c) | Actual size = image size ÷ magnification = 12 ÷ 600 = 0.02 mm.
| 2 |
| 1(d) | A tissue.
| 2 |
| Total for Question 1 | 8 | |
| 2(a) | Changed (independent variable): pH. Controlled: temperature and enzyme concentration (also: starch concentration, volume of solutions, time).
| 3 |
| 2(b) | Samples of the mixture are taken at intervals and mixed with iodine on a spotting tile; iodine turns blue-black while starch remains; the time until samples stop changing colour shows when all the starch has been digested — a shorter time means faster amylase activity.
| 3 |
| 2(c) | Activity at pH 2 is much lower or zero. Amylase has an optimum pH near neutral; at pH 2 the ion concentrations interfere with the bonds holding the enzyme’s shape, so the active site changes shape (denatures) and the substrate no longer fits.
| 4 |
| Total for Question 2 | 10 | |
| 3(a) | Water lost by transpiration from the leaves is replaced by water drawn up the xylem, so uptake approximately equals loss — though some water is retained in the tissues, so it is only an estimate.
| 2 |
| 3(b) | Water uptake falls. Higher humidity raises the water potential of the air around the leaves, reducing the water-potential gradient between leaf and air, so evaporation from the mesophyll and loss through the stomata slow down.
| 3 |
| 3(c) | Any two of: temperature, air movement (wind), light intensity, the shoot used (surface area/species).
| 2 |
| 3(d) | Volume = 45 × 2.0 = 90 mm³; rate = 90 ÷ 5 = 18 mm³ per minute.
| 3 |
| Total for Question 3 | 10 | |
| 4(a) | The left ventricle pumps oxygenated blood at high pressure around the whole body; the right ventricle only pumps blood a short distance to the lungs at lower pressure, so it needs less muscle.
| 3 |
| 4(b) | Vena cava → right atrium → right ventricle → pulmonary artery → lungs → pulmonary vein → left atrium → left ventricle → aorta.
| 4 |
| 4(c) | Platelets are cell fragments involved in clotting. When a vessel is damaged, platelets and damaged tissue trigger a cascade that converts soluble fibrinogen into insoluble fibrin; fibrin fibres form a mesh across the wound, trapping red cells and platelets; the mesh dries into a scab that prevents blood loss and blocks the entry of pathogens.
| 5 |
| Total for Question 4 | 12 |
Section B: Structured questions[40 marks]
| Question | Answer and marking points | Marks |
|---|---|---|
| 5(a) | Gametes B and b from each parent; offspring BB, Bb, Bb, bb.
| 2 |
| 5(b) | Genotypes BB : Bb : bb = 1 : 2 : 1; phenotypes black : brown = 3 : 1.
| 2 |
| 5(c) | The black parent was heterozygous (Bb). Brown offspring are bb, so they must have received b from both parents — including a b from the black parent, which therefore carried both B and b.
| 3 |
| 5(d) | Females are XX and males XY; the egg always carries X, while sperm carry X or Y; fertilisation gives XX or XY in equal proportions — a 1:1 ratio.
| 3 |
| 5(e) | Black is dominant, so both BB and Bb mice look black — the dominant phenotype hides the recessive allele; only genetic crosses (or testing) reveal the hidden allele.
| 2 |
| Total for Question 5 | 12 | |
| 6(a) | Any two of: very large total surface area; walls one cell thick (short diffusion distance); good blood supply maintaining gradients; good ventilation.
| 2 |
| 6(b) | Expired air has less oxygen (about 16% vs 21%), more carbon dioxide (about 4% vs 0.04%) and more water vapour. Limewater turns milky with expired air and stays clear with inspired air.
| 3 |
| 6(c) | During the sprint her muscles respired anaerobically, building up lactic acid and creating an oxygen debt; her heart rate stays fast, transporting lactic acid in the blood from the muscles to the liver; her breathing stays deep and fast, supplying the oxygen needed; the liver respires the lactic acid aerobically, removing it; once the debt is repaid, breathing and pulse return to resting rates.
| 5 |
| Total for Question 6 | 10 | |
| 7(a) | Population: a group of organisms of one species living in the same area at the same time. Community: all of the populations of different species in an ecosystem.
| 2 |
| 7(b) | e.g. algae → water flea → small fish → heron. Algae: producer (level 1); water flea: primary consumer (level 2); small fish: secondary consumer (level 3); heron: tertiary consumer (level 4).
| 4 |
| 7(c) | Fertiliser increases nitrate and other ions in the water; algae grow rapidly (algal bloom); the algae die and decomposers multiply; the decomposers respire aerobically, reducing dissolved oxygen; organisms needing dissolved oxygen, including fish, die.
| 4 |
| Total for Question 7 | 10 | |
| 8(a) | Any two of: rapid reproduction rate; ability to make complex molecules; few ethical concerns; presence of plasmids.
| 2 |
| 8(b) | Restriction enzymes cut the human insulin gene, leaving sticky ends; the same restriction enzymes cut the bacterial plasmid, forming complementary sticky ends; DNA ligase joins the human DNA into the plasmid, forming a recombinant plasmid.
| 4 |
| 8(c) | The bacteria are multiplied in fermenters; they express the human insulin gene, producing insulin that is harvested and purified.
| 2 |
| Total for Question 8 | 8 |
— End of mark scheme —