Scenario 1 · Plant Nutrition
Effect of CO₂ Concentration on Rate of Photosynthesis
A scientist measured the rate of photosynthesis in a tomato leaf at two different light intensities (low and high) as CO₂ concentration increased from 0.01% to 0.10%. The rate was measured as cm³ of O₂ produced per hour. The temperature was kept constant at 25°C throughout.
Table 1 — Rate of O₂ production (cm³/hour)
| CO₂ concentration / % | Low light intensity | High light intensity |
|---|---|---|
| 0.01 | 2 | 3 |
| 0.02 | 5 | 10 |
| 0.04 (atmospheric) | 8 | 22 |
| 0.06 | 9 | 34 |
| 0.08 | 9 | 42 |
| 0.10 | 9 | 42 |
(a) Describe the effect of increasing CO₂ concentration on the rate of photosynthesis at HIGH light intensity. Quote data in your answer. [3]
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Mark scheme [3]
- As CO₂ concentration increases from 0.01% to 0.08%, the rate of photosynthesis increases (from 3 to 42 cm³/hour) [1]
- The rate increases rapidly at first (e.g. from 3 to 22 cm³/hr as CO₂ goes from 0.01% to 0.04%) [1]
- Above 0.08% CO₂, the rate plateaus at 42 cm³/hour — increasing CO₂ further has no effect [1]
Must quote at least two data values for full marks.
(b) At LOW light intensity, the rate stops increasing above 0.06% CO₂. Explain what is limiting the rate at this point. [2]
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Mark scheme [2]
- CO₂ is no longer the limiting factor — another factor is now limiting the rate [1]
- At low light intensity, light is the limiting factor — there is insufficient light energy for the light-dependent reactions, so increasing CO₂ cannot increase the rate further [1]
(c) At 0.04% CO₂, the rate at high light intensity (22 cm³/hr) is more than double the rate at low light intensity (8 cm³/hr). Explain why light intensity affects the rate of photosynthesis. [3]
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Mark scheme [3]
- Light provides the energy for the light-dependent stage of photosynthesis [1]
- Higher light intensity provides more energy to split water molecules (photolysis) and produce ATP and NADPH [1]
- More ATP and NADPH are available for the light-independent stage (Calvin cycle) to fix CO₂ into glucose → faster overall rate of photosynthesis [1]
(d) A greenhouse grower increases CO₂ to 0.10% and doubles the light intensity. Predict whether the rate of photosynthesis will increase further from 42 cm³/hr. Justify your answer. [2]
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Mark scheme [2]
- Yes, the rate is likely to increase further [1]
- At 0.10% CO₂ with high light intensity, the plateau suggests another factor (temperature or enzyme activity) is now limiting; doubling light intensity alone at higher CO₂ could increase rate if the experiment data was recorded at a light intensity that was still sub-maximal [1] — OR: the rate may not increase if temperature is now the limiting factor at 25°C, as enzyme activity in the Calvin cycle is limiting [1]
Accept either a reasoned "yes" or "no/maybe" answer as long as the correct limiting factor concept is applied.
Scenario 2 · Health & Disease
Malaria Incidence and Mosquito Control in Two Regions
Malaria is caused by the Plasmodium parasite, transmitted by female Anopheles mosquitoes. A health authority monitored malaria cases per 100,000 people in two regions (A and B) over 8 years. Region B introduced mosquito-control measures at the start of Year 3.
Table 2 — Malaria cases per 100,000 people
| Year | Region A (no intervention) | Region B (control from Yr 3) |
|---|---|---|
| 1 | 420 | 415 |
| 2 | 435 | 428 |
| 3 | 441 | 390 |
| 4 | 438 | 310 |
| 5 | 450 | 210 |
| 6 | 462 | 120 |
| 7 | 455 | 68 |
| 8 | 468 | 22 |
(a) Calculate the percentage decrease in malaria cases in Region B between Year 3 and Year 8. Show your working. [2]
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Mark scheme [2]
- % decrease = (390 − 22) / 390 × 100 [1]
- = 368 / 390 × 100 = 94.4% (accept 94% to 94.4%) [1]
(b) Describe and compare the trends in malaria cases in Region A and Region B over the 8 years. Quote data. [3]
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Mark scheme [3]
- Region A: cases remained roughly constant / showed a slight increase throughout all 8 years (from 420 to 468 per 100,000) [1]
- Region B: cases were similar to Region A in Years 1–2 (~415–428), but decreased sharply from Year 3 onwards after intervention, falling to just 22 by Year 8 [1]
- The difference between the two regions grew substantially over time; by Year 8 Region A had 468 cases vs 22 in Region B — a 21× difference [1]
(c) Suggest two mosquito-control measures that Region B may have introduced and explain how each reduces malaria transmission. [4]
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Mark scheme — 2 measures × [measure + explanation]
- Draining standing water / covering water containers — removes mosquito breeding sites; fewer mosquitoes → less transmission [1+1]
- Insecticide-treated bed nets — physical barrier prevents mosquitoes reaching sleeping humans; insecticide kills mosquitoes that land on the net [1+1]
- Spraying insecticides (e.g. DDT, malathion) indoors or in standing water — kills adult mosquitoes or larvae, reducing the mosquito population [1+1]
- Biological control (e.g. introducing mosquito-eating fish or bacteria like Bt) — reduces mosquito larval population without chemicals [1+1]
(d) Malaria is caused by Plasmodium, a protozoan parasite — not a bacterium or virus. Explain why antibiotics would be ineffective as a treatment for malaria. [2]
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Mark scheme [2]
- Antibiotics work by targeting specific structures or processes found in bacteria (e.g. cell wall synthesis, bacterial ribosomes) [1]
- Plasmodium is a eukaryotic parasite — it does not have these bacterial targets, so antibiotics have no effect on it [1]
Note: some antibiotics (e.g. doxycycline) are actually used for malaria prevention, but they work by a different mechanism than their antibacterial action — this is beyond O-Level scope.
Scenario 3 · Genetics
Inheritance of ABO Blood Groups in a Family
ABO blood groups are controlled by three alleles: Iᴬ (produces antigen A), Iᴮ (produces antigen B), and i (produces no antigen). Iᴬ and Iᴮ are both dominant over i, but co-dominant with each other. A couple (father blood group A, mother blood group B) have four children with the blood groups shown below.
Family blood groups
| Individual | Blood group |
|---|---|
| Father | A |
| Mother | B |
| Child 1 | O |
| Child 2 | A |
| Child 3 | B |
| Child 4 | AB |
(a) Child 1 has blood group O (genotype ii). Use this information to deduce the genotypes of the father and mother. Explain your reasoning. [3]
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Mark scheme [3]
- Child 1 has genotype ii — must have received one i allele from each parent [1]
- Father (blood group A) must be Iᴬi (heterozygous) — must carry i to pass it to Child 1; Iᴬ is expressed so he is blood group A [1]
- Mother (blood group B) must be Iᴮi (heterozygous) — must carry i to pass it to Child 1; Iᴮ is expressed so she is blood group B [1]
(b) Construct a genetic diagram (Punnett square) for the cross Iᴬi × Iᴮi. State the genotype and blood group of Child 4 (AB) and explain how co-dominance produces the AB phenotype. [4]
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Mark scheme [4]
- Punnett square: Iᴬi × Iᴮi → offspring: IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B), ii (O) [1]
- Expected ratio: 1 AB : 1 A : 1 B : 1 O [1]
- Child 4 (AB): genotype IᴬIᴮ [1]
- Co-dominance: both Iᴬ and Iᴮ alleles are fully expressed — both A and B antigens are produced on the red blood cell surface — so neither allele is masked; both phenotypes are expressed simultaneously [1]
(c) A person with blood group O can donate blood to any ABO blood group. Explain why a person with blood group A cannot receive blood group B. [3]
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Mark scheme [3]
- A person with blood group A has anti-B antibodies in their plasma [1]
- Blood group B red blood cells carry B antigens on their surface [1]
- The anti-B antibodies bind to B antigens → agglutination (clumping) of red blood cells → can block capillaries and cause serious complications/death [1]
Scenario 4 · Enzymes — Unfamiliar Context
The Effect of Heavy Metal Ions on Enzyme Activity
A researcher investigated the effect of mercury ions (Hg²⁺) on the activity of catalase (which breaks down H₂O₂ → H₂O + O₂). Different concentrations of Hg²⁺ were added to tubes containing the same amount of catalase and H₂O₂. The volume of O₂ produced in 5 minutes was recorded.
Table 3 — Effect of Hg²⁺ on catalase activity
| Hg²⁺ concentration / mmol dm⁻³ | Volume of O₂ produced in 5 min / cm³ | Rate relative to control / % |
|---|---|---|
| 0 (control) | 24.0 | 100 |
| 0.5 | 19.2 | 80 |
| 1.0 | 14.4 | 60 |
| 2.0 | 7.2 | 30 |
| 4.0 | 2.4 | 10 |
| 8.0 | 0.0 | 0 |
(a) Describe the relationship between Hg²⁺ concentration and the rate of catalase activity. Quote data. [2]
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Mark scheme [2]
- As Hg²⁺ concentration increases, the rate of catalase activity decreases [1]
- The rate drops from 100% at 0 mmol/dm³ to 0% at 8.0 mmol/dm³; the relationship is approximately inversely proportional [1]
(b) Suggest how Hg²⁺ ions inhibit catalase activity. You are not expected to have prior knowledge of heavy metals — apply your understanding of enzyme structure. [3]
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Mark scheme [3]
- Hg²⁺ ions bind to the catalase enzyme (e.g. at the active site or at another site on the enzyme) [1]
- This changes the shape of the active site [1]
- H₂O₂ (the substrate) can no longer fit the active site properly → fewer enzyme-substrate complexes form → reduced rate of O₂ production [1]
This is an example of enzyme inhibition. At O-Level, you are expected to explain inhibition in terms of active site shape change — you do not need to know the specific mechanism of heavy metal binding.
(c) At 8.0 mmol/dm³ Hg²⁺, no O₂ is produced. Suggest one further experiment to determine whether this is because the enzyme is irreversibly denatured or reversibly inhibited. [2]
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Mark scheme [2]
- Dialyse (or dilute and wash) the enzyme solution to remove the Hg²⁺ ions [1]
- Then add fresh H₂O₂ and measure O₂ production — if activity is restored, the inhibition was reversible; if no O₂ is produced, the enzyme was irreversibly denatured [1]
Accept any logical experiment that involves removing the Hg²⁺ and testing if activity returns.
Scenario 5 · Ecology
Predator-Prey Population Dynamics — Lynx and Snowshoe Hare
Historical records from Canada showed cyclical changes in the populations of snowshoe hares (prey) and Canadian lynx (predator) over 90 years. The data below shows a simplified pattern of population changes over one cycle of approximately 10 years.
Table 4 — Population data (thousands)
| Year | Hare population / thousands | Lynx population / thousands |
|---|---|---|
| 0 | 20 | 5 |
| 2 | 60 | 8 |
| 4 | 90 | 20 |
| 6 | 40 | 25 |
| 8 | 15 | 10 |
| 10 | 20 | 5 |
(a) Between Year 4 and Year 8, the hare population fell from 90,000 to 15,000. Calculate the percentage decrease. [2]
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Mark scheme [2]
- % decrease = (90 − 15) / 90 × 100 [1]
- = 75 / 90 × 100 = 83.3% (accept 83% to 83.3%) [1]
(b) The lynx population peaked at Year 6, two years after the hare population peaked at Year 4. Explain this time lag. [3]
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Mark scheme [3]
- When hares are abundant (Year 4), lynx have more food and reproduce successfully → lynx population grows [1]
- It takes time (a generation) for the increase in food supply to translate into increased lynx births and survival to breeding age [1]
- So the lynx peak comes after the hare peak; by the time lynx numbers peak (Year 6), they have eaten so many hares that the hare population has already declined [1]
(c) Explain why the hare population begins to recover after Year 8, even though lynx are still present. [3]
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Mark scheme [3]
- The lynx population has also fallen (from 25,000 to 10,000) due to reduced food supply (fewer hares) [1]
- With fewer predators, predation pressure on hares decreases — fewer hares are eaten per unit time [1]
- Hare birth rate now exceeds death rate → hare population increases; the cycle repeats [1]
(d) Suggest one abiotic factor (other than predation) that could also cause the hare population to decline from Year 4 onwards. Explain your suggestion. [2]
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Mark scheme [2]
- Abiotic factor: severe winter / low temperature / deep snowfall / food plant shortage [1]
- Explanation: e.g. severe winter reduces availability of the hares' food plants → food shortage → increased starvation and death rate → population decline [1]
Accept any sensible abiotic factor with a logical explanation. Abiotic = non-living environmental factor.