OCR GCSE Biology A: Food security, agricultural solutions and selective breeding (J247) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover outcomes B6.2a, B6.2b and B6.2c of the OCR GCSE (9-1) Gateway Science Biology A specification (J247), Version 4.0 (August 2026), for first assessment in 2018. No statement here is in bold in the specification, so none is labelled Higher Tier only. Learn the content from the study guide and the revision notes. Course links: OCR GCSE Biology hub, printable checklist and the free diagnostics.

Questions

1. State two biological factors that can reduce food security. [2]

2. A field without pesticide loses 18% of its potential yield to insect pests. The actual yield is 8.5 tonnes per hectare. Calculate the potential yield. Give your answer to 3 significant figures. [3]

3. A field produces crops containing 80 000 kJ of energy. One person needs 10 000 kJ per day (supplied value). Assume 10% of the crop energy is passed to chickens when they eat the crop.

(a) For how many person-days could the crop provide energy if people ate it directly? [1] (b) Calculate the energy available from the chickens, and the number of person-days it supplies. [2] (c) Suggest why rising meat consumption in wealthier populations reduces food security. [1]

4. Explain how fertiliser washed from a field into a lake can cause fish to die. [4]

5. A farmer trials phosphate-rich fertiliser on potatoes.

Fertiliser (kg per hectare)	0	40	80	120	160
Yield (tonnes per hectare)	21	30	36	38	38.5

(a) Calculate the percentage increase in yield from 0 to 80 kg per hectare. [2] (b) Suggest which rate the farmer should use. Give two reasons. [2]

6. Five hydroponic plants have masses of 212, 198, 225, 205 and 210 g. The mean mass of soil-grown plants is 171 g.

(a) Calculate the mean mass of the hydroponic plants. [1] (b) Calculate how much greater this mean is than the soil mean, as a percentage of the soil mean. [2] (c) State one disadvantage of hydroponics. [1] (d) Give one reason why the conclusion from this data should be treated with caution. [1]

7. Ladybirds were released onto some bean plants to control aphids. The table shows mean aphids per plant.

Week	0	2	4	6	8
No ladybirds	20	95	310	640	820
Ladybirds released	20	60	120	45	30

(a) Calculate the percentage fewer aphids at week 8 on plants with ladybirds. [2] (b) Explain the fall in aphids between weeks 4 and 8 on plants with ladybirds. [2] (c) Give one advantage of this method over spraying insecticide. [1] (d) Give one disadvantage of this method. [1]

8. Describe how a plant breeder could produce tomato plants with large fruits that also resist a fungal disease. [4]

9. Hens were selectively bred for six generations. Mean egg production rose from 190 to 310 eggs per hen per year.

(a) Calculate the percentage increase. [2] (b) Calculate the mean increase per generation. [1] (c) Suggest how the breeder could show that the increase was caused by the hens’ genes, not their feed. [1]

10. A farmer grows a single high-yield wheat variety bred from only a few parent plants. A new fungal disease reaches the area.

(a) Explain why the whole crop might be lost. [4] (b) Suggest two ways to reduce this risk in the future. [2]

11. A growing country wants more food from limited land. Evaluate the use of fertilisers, pesticides, hydroponics and biological control. [8]

Answers

1. Any two from: increasing population; new pests or pathogens; changing diets (more meat); environmental change such as drought; cost or sustainability of inputs. First factor [1]; second factor [1]. [2] Examiner insight: Two distinct factors are needed, so “more people” and “population growth” count as one idea, not two.

2. Actual yield is 100 - 18 = 82% of potential [1]. Potential = 8.5 / 0.82 [1] = 10.4 tonnes per hectare [1]. [3] Examiner insight: Divide by 0.82 to find the whole; multiplying 8.5 by 1.18 uses the wrong base, a common slip.

3. (a) 80 000 / 10 000 = 8 person-days [1]. (b) 80 000 × 0.10 = 8 000 kJ [1]; 8 000 / 10 000 = 0.8 person-days [1]. (c) Energy is lost between trophic levels, so animals need much more crop than people would, and more land is needed per person fed [1]. [4] Examiner insight: Answers on diet gain more credit when they link to energy loss between feeding levels, not only to “more land”.

4. Fertiliser (nitrates and phosphates) washes into the lake in run-off [1]. Algae grow rapidly, forming a bloom [1]. The bloom blocks light, so plants below die and decomposers respire aerobically, using up oxygen [1]. With too little oxygen, fish die [1]. [4] Examiner insight: Naming oxygen use by decomposers is what separates a full explanation from “the algae poison the fish”.

5. (a) (36 - 21) / 21 [1] × 100 = 71.4% [1]. (b) 80 kg per hectare [1]. Above 80 the extra yield is small (2 then 0.5 tonnes), so more fertiliser costs more than it earns and raises run-off [1]. [4] Examiner insight: Quote figures from the table when justifying a choice; “it gives the best yield” alone ignores cost.

6. (a) (212 + 198 + 225 + 205 + 210) / 5 = 210 g [1]. (b) (210 - 171) / 171 [1] × 100 = 22.8% [1]. (c) Any one: high set-up or energy cost; needs skilled management; disease spreads in the solution; power failure [1]. (d) Only five plants were measured, so the mean may not represent all plants [1]. [5] Examiner insight: Divide by the soil mean, as the question asks; dividing by 210 gives a different answer.

7. (a) (820 - 30) / 820 [1] × 100 = 96.3% [1]. (b) Ladybirds feed on aphids [1], and ladybird numbers had risen, so aphids were eaten faster than they reproduced [1]. (c) No chemical residues, or specific to the pest, or no resistance [1]. (d) Slow to act, or does not remove the pest completely, or needs suitable conditions [1]. [6] Examiner insight: In an explanation, give the cause (predation) and the effect on numbers, and do not just repeat that aphids fell.

8. Choose parent plants with large fruit and parents with resistance [1]. Breed them together [1]. Select offspring with large fruit and resistance, and breed those [1]. Repeat over many generations [1]. [4] Examiner insight: Describe repetition over generations; a single cross rarely produces the finished variety.

9. (a) (310 - 190) / 190 [1] × 100 = 63.2% [1]. (b) 120 / 6 = 20 eggs per generation [1]. (c) Keep all generations on the same diet and housing [1]. [4] Examiner insight: A control question needs one named variable held constant, not “do a fair test”.

10. (a) Breeding from few parents reduces genetic variation [1]. The plants are genetically very similar [1]. If none carries an allele for resistance, all are susceptible [1]. The disease spreads through the whole crop [1]. (b) Any two from: keep seed banks or wild relatives to retain alleles; grow several varieties; breed in resistance from other varieties. First way [1]; second way [1]. [6] Examiner insight: Link the loss directly to missing alleles, not just “weak plants”.

11. Fertilisers replace removed ions and raise yield [1]; run-off causes eutrophication [1]. Pesticides protect yield from pests, weeds and fungi [1]; they harm non-target species and resistance evolves [1]. Hydroponics gives exact control, saves land and water [1]; it needs high energy cost and skill [1]. Biological control leaves no residues and is specific [1], but is slow and incomplete. A judgement with a reason, such as using integrated methods suited to the farm’s money and land [1]. [8] Examiner insight: Evaluation answers tend to gain more when each method has both a benefit and a limit, followed by a reasoned conclusion.

Where marks are usually lost

• Saying fertiliser is “plant food” instead of a source of mineral ions.
• Missing the oxygen step in eutrophication.
• Using the wrong base in percentage change questions.
• Forgetting units such as kJ, grams or tonnes per hectare.
• Writing “expensive” without saying what costs.
• Saying pests become “immune” rather than describing survival of resistant individuals.
• Describing selective breeding as one cross instead of repeated selection.
• Reading only one line of a table or graph.

Next steps

• Review the revision notes and the study guide.
• Tick off outcomes on the printable checklist and see the OCR GCSE Biology hub.
• Take the free 10-minute diagnostics.
• Talk to a teacher: book a free trial class.

Official syllabus

OCR, GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, topic B6.2 Feeding the human race, outcomes B6.2a to B6.2c.