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Paper 4 style practice paper: Structured questions (Extended) — mark scheme

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QQ/IGCSE-BIO/P4S

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]

QuestionAnswer and marking pointsMarks
1(a)

Chloroplast; absorbs light energy for photosynthesis.

  • chloroplast [1]
  • absorbs light for photosynthesis [1]
2
1(b)

Any two of: cell wall, cell membrane, cytoplasm, ribosomes.

  • two structures named [1]
  • [1]
2
1(c)

Actual size = image size ÷ magnification = 12 ÷ 600 = 0.02 mm.

  • 12 ÷ 600 seen [1]
  • 0.02 mm with unit [1]
2
1(d)

A tissue.

  • group of similar cells doing one function [1]
2
Total for Question 18
2(a)

Changed (independent variable): pH. Controlled: temperature and enzyme concentration (also: starch concentration, volume of solutions, time).

  • pH is the independent variable [1]
  • temperature controlled [1]
  • enzyme concentration controlled [1]
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.

  • iodine tests samples at intervals [1]
  • blue-black = starch present [1]
  • time to no colour change measures rate [1]
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.

  • activity lower/zero at pH 2 [1]
  • optimum pH concept [1]
  • active site shape changed [1]
  • denatured / substrate no longer fits [1]
4
Total for Question 210
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.

  • loss replaced by uptake through xylem [1]
  • approximately / not exact because some water is retained [1]
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.

  • uptake decreases [1]
  • water-potential gradient reduced [1]
  • less evaporation/stomatal loss [1]
3
3(c)

Any two of: temperature, air movement (wind), light intensity, the shoot used (surface area/species).

  • two variables [1]
  • [1]
2
3(d)

Volume = 45 × 2.0 = 90 mm³; rate = 90 ÷ 5 = 18 mm³ per minute.

  • 45 × 2.0 = 90 mm³ [1]
  • divide by 5 minutes [1]
  • 18 mm³ per minute [1]
3
Total for Question 310
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.

  • left pumps to whole body at high pressure [1]
  • right only to lungs / shorter distance [1]
  • more muscle needed for systemic pressure [1]
3
4(b)

Vena cava → right atrium → right ventricle → pulmonary artery → lungs → pulmonary vein → left atrium → left ventricle → aorta.

  • right atrium → right ventricle [1]
  • pulmonary artery → lungs → pulmonary vein [1]
  • left atrium → left ventricle [1]
  • aorta [1]
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.

  • platelets for clotting [1]
  • fibrinogen converted to fibrin [1]
  • fibrin mesh traps cells [1]
  • dries to scab [1]
  • prevents blood loss / pathogen entry [1]
5
Total for Question 412

Section B: Structured questions[40 marks]

QuestionAnswer and marking pointsMarks
5(a)

Gametes B and b from each parent; offspring BB, Bb, Bb, bb.

  • gametes identified [1]
  • BB, Bb, Bb, bb [1]
2
5(b)

Genotypes BB : Bb : bb = 1 : 2 : 1; phenotypes black : brown = 3 : 1.

  • 1:2:1 [1]
  • 3:1 [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.

  • black parent is Bb [1]
  • brown offspring are bb [1]
  • so black parent supplied b [1]
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.

  • XX female, XY male [1]
  • egg X, sperm X or Y [1]
  • 1:1 [1]
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.

  • dominant allele masks recessive [1]
  • BB and Bb share a phenotype [1]
2
Total for Question 512
6(a)

Any two of: very large total surface area; walls one cell thick (short diffusion distance); good blood supply maintaining gradients; good ventilation.

  • two features [1]
  • [1]
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.

  • less oxygen / more CO₂ [1]
  • more water vapour [1]
  • limewater milky with expired air [1]
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.

  • anaerobic respiration → lactic acid / oxygen debt [1]
  • fast heart transports lactic acid to liver [1]
  • deep fast breathing supplies oxygen [1]
  • liver respires lactic acid aerobically [1]
  • debt repaid → rates return to normal [1]
5
Total for Question 610
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.

  • population [1]
  • community [1]
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).

  • chain begins with producer [1]
  • arrows point eater-ward [1]
  • four organisms [1]
  • trophic levels named [1]
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.

  • nutrients increase → algal bloom [1]
  • algae die → decomposition rises [1]
  • aerobic respiration by decomposers → oxygen falls [1]
  • fish die from lack of dissolved oxygen [1]
4
Total for Question 710
8(a)

Any two of: rapid reproduction rate; ability to make complex molecules; few ethical concerns; presence of plasmids.

  • two reasons [1]
  • [1]
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.

  • restriction enzymes cut gene — sticky ends [1]
  • same enzyme cuts plasmid — complementary sticky ends [1]
  • DNA ligase joins them [1]
  • recombinant plasmid formed [1]
4
8(c)

The bacteria are multiplied in fermenters; they express the human insulin gene, producing insulin that is harvested and purified.

  • multiplied in fermenters [1]
  • express the gene → insulin harvested [1]
2
Total for Question 88

— End of mark scheme —