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Paper 6 style practice paper: Alternative to practical — mark scheme

QuizQuipPractice Series
QQ/IGCSE-BIO/P6S

IGCSE · Cambridge IGCSE Biology (Extended)

Mark scheme

Alternative to practical

Paper 6 style practice paper · Maximum mark: 40

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: Practical skills questions[20 marks]

QuestionAnswer and marking pointsMarks
1(a)

Iodine — starch — blue-black. Benedict’s — reducing sugars — brick-orange/red precipitate on heating. Biuret — protein — purple. Ethanol — fats (lipids) — white emulsion.

  • iodine: starch, blue-black [1]
  • Benedict’s: reducing sugar, orange/red when heated [1]
  • Biuret: protein, purple [1]
  • ethanol: fat, white emulsion [1]
4
1(b)

Use a water bath rather than a direct flame; wear safety goggles (and point the tube away from people).

  • water bath not direct flame [1]
  • goggles / point tube away [1]
2
Total for Question 16
2(a)

So that surface solution is not weighed — only the change in the potato’s own water content is measured.

  • remove surface liquid so only the strip is weighed [1]
1
2(b)

Percentage change = −(0.6 ÷ 5.0) × 100 = −12%, a 12% loss of mass.

  • 0.6 ÷ 5.0 [1]
  • −12% with loss direction stated [1]
2
2(c)

In dilute solutions the water potential outside the cells is higher than inside, so water enters the cells by osmosis and the strips gain mass; in concentrated solutions the outside water potential is lower, so water leaves the cells and the strips lose mass.

  • dilute: outside higher water potential → water in [1]
  • concentrated: outside lower → water out [1]
  • osmosis named [1]
3
2(d)

At the concentration equal to the water potential of the potato cell sap — water moves in and out at equal rates.

  • equal water potential / no net osmosis [1]
1
Total for Question 27
3(a)

Place equal masses of apple pulp in identical flasks; add equal volumes of pectinase solution pre-warmed to each test temperature; hold each flask at its temperature in a water bath for the same time; filter and measure the juice volume. Control: pH of the enzyme solution and the type/mass of apple pulp (also: time, volume of enzyme).

  • equal pulp and equal enzyme volume [1]
  • water baths at set temperatures [1]
  • same time then filter and measure juice [1]
  • two controlled variables [1]
4
3(b)

40 °C is close to pectinase’s optimum temperature, so enzyme and substrate collide frequently and pectin is digested fast; at 60 °C the enzyme is denatured — its active site changes shape — so little pectin is digested and little juice is released.

  • 40 °C ≈ optimum — fast collisions [1]
  • 60 °C denatures enzyme [1]
  • active site shape lost → little juice [1]
3
Total for Question 37

Section B: Practical skills questions[20 marks]

QuestionAnswer and marking pointsMarks
4(a)

Oxygen — a product of photosynthesis, released when the rate of photosynthesis exceeds the rate of respiration.

  • oxygen [1]
  • product of photosynthesis / exceeds respiration [1]
2
4(b)

The bubble rate increases — light intensity rises as the lamp gets closer, increasing the rate of photosynthesis (until another factor limits it).

  • bubble rate increases [1]
  • higher light intensity → faster photosynthesis [1]
2
4(c)

Any two of: temperature of the water, carbon dioxide concentration (same water/hydrogencarbonate), the piece of pondweed used.

  • two variables [1]
  • [1]
2
4(d)

In darkness the weed respires but cannot photosynthesise, so carbon dioxide accumulates, turning the indicator yellow.

  • respiration continues, photosynthesis stops → CO₂ rises [1]
1
Total for Question 47
5(a)

Jar 2: dry, room temperature, open — no germination; water missing. Jar 3: water, refrigerator, open — no germination; suitable temperature missing. Jar 4: water, room temperature, sealed with limited air — no germination; oxygen missing.

  • jar 2 — water [1]
  • jar 3 — temperature [1]
  • jar 4 — oxygen [1]
3
5(b)

Water is needed to activate enzymes in the seed and swell the stored food, restarting growth; oxygen is needed for aerobic respiration, which releases the energy the growing seedling needs.

  • water activates enzymes / swells seed [1]
  • oxygen for aerobic respiration [1]
  • releases energy for growth [1]
3
Total for Question 56
6(a)

Total = 46; mean = 46 ÷ 10 = 4.6 daisies per quadrat.

  • total 46 seen [1]
  • ÷ 10 = 4.6 [1]
2
6(b)

Area = 20 × 30 = 600 m²; estimate = 4.6 × 600 = 2760 daisies.

  • area 600 m² [1]
  • 4.6 × 600 = 2760 [1]
2
6(c)

Random placement avoids bias — students might unconsciously choose daisy-rich spots, overestimating the population. Errors include the small sample size, or daisies on the quadrat border being counted inconsistently.

  • random avoids bias [1]
  • one valid error (small sample / border counting) [1]
2
6(d)

Daisies need light for photosynthesis, so fewer would grow in shade — the distribution is uneven.

  • light affects daisy growth/distribution [1]
1
Total for Question 67

— End of mark scheme —