Section A — Structured Questions (Q1–5 · 50 marks)
Q1
Cell Biology & Transport · 10 marks
This question is about cell structure and transport across cell membranes.
(a) State three structures found in plant cells but NOT in animal cells. [3]
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- Cell wall (made of cellulose) [1]
- Chloroplast [1]
- Large permanent vacuole (filled with cell sap) [1]
Do NOT accept: cell membrane, mitochondria, or nucleus — these are present in both.
(b) A plant cell is placed in a very concentrated sugar solution. Explain what happens, using the term "water potential." [4]
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- The concentrated solution has a lower water potential than the cell cytoplasm [1]
- Water moves out of the cell by osmosis, down the water potential gradient [1]
- The vacuole and cytoplasm shrink; the cell membrane pulls away from the cell wall [1]
- The cell becomes plasmolysed [1]
Must use "water potential" for [1]. "Dehydrated" alone scores 0 for that point.
(c) Explain why root hair cells need active transport to absorb mineral ions from the soil. [3]
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- Mineral ion concentration in the root hair cell cytoplasm is already higher than in the dilute soil water [1]
- Ions must move against the concentration gradient (from low to high concentration) [1]
- Active transport uses ATP (from respiration) and carrier proteins to move ions against the gradient [1]
Q2
Enzymes · 10 marks
A student investigated the effect of temperature on amylase activity. Amylase digests starch to maltose. Samples were tested with iodine every 30 s; time for blue-black colour to disappear was recorded.
Results: 10°C → 480 s | 20°C → 240 s | 37°C → 80 s | 45°C → 340 s | 60°C → no change (starch remained)
(a) State the optimum temperature for amylase in this experiment and explain what "optimum" means. [2]
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- Optimum temperature: 37°C (shortest time = fastest rate) [1]
- Optimum = the temperature at which the enzyme works at its maximum rate [1]
(b) Explain why no reaction occurs at 60°C. Use the terms "active site" and "denatured." [3]
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- High temperature breaks the hydrogen bonds maintaining the enzyme's 3D structure [1]
- The active site changes shape permanently — the enzyme is denatured [1]
- Starch (substrate) can no longer fit the active site → no enzyme-substrate complexes → no digestion [1]
⚠ Trap: Do NOT say the enzyme is "killed" — enzymes are not alive. Say "denatured."
(c) Explain why activity is lower at 10°C than at 37°C, but the enzyme is NOT denatured. [3]
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- At 10°C enzyme and substrate molecules have less kinetic energy [1]
- Collisions between enzyme and substrate are less frequent [1]
- Fewer enzyme-substrate complexes form per second → slower rate. The active site shape is unchanged; warming up restores full activity [1]
(d) State one variable the student must keep constant and explain why. [2]
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- Any one: pH / concentration of amylase / concentration of starch / volume of solutions [1]
- So that only temperature varies; any change in results is due to temperature alone [1]
Q3
Transport in Humans · 10 marks
This question is about the human circulatory system and blood components.
(a) Give three structural differences between arteries and veins. [3]
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Mark scheme — any 3 (must state BOTH sides of each difference)
- Arteries: thick muscular/elastic wall; Veins: thin wall [1]
- Arteries: narrow lumen; Veins: wide lumen [1]
- Arteries: no valves; Veins: valves to prevent backflow [1]
- Arteries: high-pressure blood; Veins: low-pressure blood [1]
(b) Explain why the left ventricle wall is much thicker than the right ventricle wall. [3]
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- Left ventricle pumps blood to the entire body (systemic circulation) [1]
- This is a much longer circuit with greater resistance than the pulmonary circuit [1]
- Greater muscular force (and thus thicker wall) is needed to generate sufficient pressure [1]
(c) Describe how red blood cells are adapted for their function of carrying oxygen. Give three adaptations. [4]
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Mark scheme — 4 marks for 3 adaptations with explanations
- Biconcave disc shape → increased surface area:volume ratio for faster diffusion of O₂ [1]
- Contains haemoglobin → binds O₂ in lungs (forms oxyhaemoglobin) and releases it in respiring tissues [1]
- No nucleus → more space for haemoglobin, maximising oxygen-carrying capacity [1]
- Flexible / thin membrane → can squeeze through narrow capillaries to reach all tissues [1]
Q4
Respiration & Gas Exchange · 10 marks
This question is about respiration and gas exchange in the human body.
(a) Write the balanced chemical equation for aerobic respiration. [2]
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- C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (ATP) [2] — award [1] for correct but unbalanced, or correct word equation
(b) Explain how alveoli are adapted for efficient gas exchange. Give four adaptations. [4]
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Mark scheme — any 4 (1 each)
- Large total surface area (300 million alveoli, ~70 m²) — maximises diffusion [1]
- Walls one cell thick + thin capillary wall → very short diffusion distance [1]
- Moist lining — gases dissolve in moisture for diffusion [1]
- Rich network of capillaries — maintains steep concentration gradient by continuously removing O₂ and supplying CO₂ [1]
- Good ventilation (breathing) — maintains steep concentration gradient on the air side [1]
(c) A marathon runner's muscles respire anaerobically near the end of the race. Explain what happens to lactic acid after the race and why breathing rate stays elevated. [4]
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- Lactic acid accumulates in muscles causing fatigue; it is transported in blood to the liver [1]
- In the liver, lactic acid is oxidised to CO₂ and water (or converted to glycogen) using extra oxygen [1]
- The extra oxygen needed to remove lactic acid is the "oxygen debt" [1]
- Breathing rate and depth remain elevated after exercise to repay this oxygen debt [1]
Q5
Homeostasis & Coordination · 10 marks
This question is about blood glucose regulation and the nervous system.
(a) Describe how blood glucose is regulated after a carbohydrate-rich meal. Name the organ, hormone, and target tissue. [5]
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- Blood glucose rises after a carbohydrate meal [1]
- The pancreas (beta cells of islets of Langerhans) detects the rise and secretes insulin [1]
- Insulin travels in blood to liver and muscle cells (target tissues) [1]
- Insulin causes cells to take up glucose and convert it to glycogen (glycogenesis) for storage [1]
- Blood glucose falls back to normal — this is negative feedback [1]
(b) Explain why a person with Type 1 diabetes must inject insulin rather than swallow it as a tablet. [2]
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- Insulin is a protein [1]
- It would be digested/broken down into amino acids by protease enzymes in the stomach/gut and would lose its function [1]
(c) Trace the path of a nerve impulse in a reflex arc when you touch a hot surface, from stimulus to response. [3]
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- Receptor (pain/heat receptor in skin) detects stimulus → impulse travels along sensory neurone to spinal cord [1]
- Relay neurone in spinal cord passes impulse to motor neurone [1]
- Motor neurone carries impulse to effector (muscle) → muscle contracts to withdraw hand. Brain is informed but does not initiate the reflex [1]
Section B — Extended Questions (Q6–8 · 30 marks)
Q6
Genetics & Inheritance · 10 marks
In peas, tall (T) is dominant over dwarf (t). A heterozygous tall plant is crossed with a dwarf plant.
(a) Construct a full genetic diagram showing all possible genotypes and phenotypes of the offspring and their expected ratio. [5]
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- Parent genotypes correctly stated: Tt (tall) × tt (dwarf) [1]
- Gametes correctly listed: T or t (from Tt parent), t only (from tt parent) [1]
- Punnett square completed: Tt, Tt, tt, tt [1]
- Offspring phenotypes: 2 tall (Tt) : 2 dwarf (tt) = 1 tall : 1 dwarf [1]
- Probability of each: 50% tall, 50% dwarf [1]
All 5 stages (parental genotypes, gametes, cross, offspring genotypes, phenotype ratio) must be shown for full marks.
(b) Define the terms: (i) dominant allele, (ii) recessive allele, (iii) homozygous. [3]
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- Dominant: an allele that is expressed/shown in the phenotype even when only one copy is present (in the heterozygous state) [1]
- Recessive: an allele only expressed in the phenotype when two copies are present (homozygous recessive) [1]
- Homozygous: having two identical alleles for a gene (e.g. TT or tt) [1]
(c) Red-green colour blindness is X-linked recessive. Explain why colour blindness is much more common in males than females. [2]
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- Males (XY) have only one X chromosome; if that X carries the recessive allele, there is no second X to mask it — the male is colour blind [1]
- Females (XX) need two copies of the recessive allele (one on each X) to be colour blind; one copy makes her only a carrier with normal vision [1]
Q7
Ecology & Environment · 10 marks
The food web below shows feeding relationships in a freshwater pond.
Algae → Water fleas → Small fish → Large fish
Algae → Water fleas → Dragonfly larvae → Large fish
Algae → Tadpoles → Large fish
Algae → Water fleas → Small fish → Large fish
Algae → Water fleas → Dragonfly larvae → Large fish
Algae → Tadpoles → Large fish
(a) Write one complete food chain with four trophic levels. Identify the producer and the secondary consumer. [3]
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- Correct 4-link chain, e.g. Algae → Water fleas → Small fish → Large fish [1]
- Producer: Algae [1]
- Secondary consumer: Small fish (or Dragonfly larvae, depending on chain written) [1]
(b) A pesticide kills most water fleas. Predict and explain two effects on other organisms in the food web. [4]
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Mark scheme — 2 effects × [prediction + reason]
- Algae population increases — fewer water fleas consuming algae [1+1]
- Small fish decrease — water fleas (their main food) are reduced [1+1]
- Large fish decrease — small fish (prey) have decreased, reducing food supply [1+1]
- Dragonfly larvae decrease — water fleas (their food) are reduced [1+1]
Award [1] for prediction (increase/decrease) and [1] for correct linked explanation each.
(c) Explain why energy transfer between trophic levels is inefficient, and give two reasons why energy is lost. [3]
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- Only ~10% of energy passes to the next trophic level; most is lost [1]
- Lost as heat during respiration (metabolic processes) [1]
- Lost in egestion (undigested food in faeces) / urine / parts of organism not eaten (bones, fur, roots) [1]
Q8
Plant Nutrition & Transport · 10 marks
This question is about photosynthesis and water transport in plants.
(a) A student uses a variegated leaf (with green and white areas) to investigate whether chlorophyll is needed for photosynthesis. Predict the result of the iodine test and explain your answer. [4]
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- Green areas: iodine turns blue-black — starch is present [1]
- White areas: iodine remains orange-brown — no starch present [1]
- Green areas contain chlorophyll (in chloroplasts) which absorbs light for photosynthesis, producing glucose converted to starch [1]
- White areas have no chlorophyll/chloroplasts, so cannot photosynthesise; no glucose/starch produced [1]
The leaf must be destarched before the experiment (kept in darkness 24–48 h), but do not penalise if student doesn't mention this unless asked.
(b) Describe how water travels from the soil to the leaf mesophyll. Include the process at root hair cells, the vessel used in the stem, and the driving force. [4]
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- Water enters root hair cells by osmosis — soil water has a higher water potential than the root cell cytoplasm [1]
- Water passes through root cells and into xylem vessels (dead, hollow, lignified tubes) [1]
- Water evaporates from leaf cells and exits through stomata (transpiration) [1]
- Transpiration creates a tension/suction that pulls water up the xylem — the transpiration stream / cohesion-tension mechanism [1]
(c) State two environmental conditions that increase the rate of transpiration and explain the effect of each. [2]
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Mark scheme — any 2 (1 each)
- Higher temperature → more energy for evaporation of water from leaf surfaces → faster transpiration [1]
- Lower humidity → greater water potential gradient between leaf air spaces and outside air → faster diffusion of water vapour out [1]
- Wind → carries water vapour away from leaf surface, maintaining steep water potential gradient → faster transpiration [1]
- Brighter light → stomata open wider → more water vapour can escape [1]