OxfordAQA International AS and A-level Biology: Energy transfer, productivity and farming (9610) – 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 sections 3.3.4.1 and 3.3.4.2 of the OxfordAQA International AS and A-level Biology specification, Version 5.1 (Unit 3, International A-level only). Data are supplied in each question. Use the study guide and revision notes if needed, plus the course hub and printable checklist.

Questions

1. (a) Define gross primary production. [1] (b) Define net primary production and state the equation that links it to gross primary production. [2]

2. In a lake, the standing biomass of phytoplankton is 3.0 g m⁻² and of zooplankton is 11 g m⁻². The annual production is 450 g m⁻² for phytoplankton and 54 g m⁻² for zooplankton.

(a) State the shape of the pyramid of standing biomass. [1] (b) Explain how a smaller stock of phytoplankton can support the larger stock of zooplankton. [2] (c) Calculate the zooplankton production as a percentage of phytoplankton production. [1] (d) Explain why a pyramid of energy for this lake cannot be inverted over a year. [1]

3. A meadow stores 30 000 kJ m⁻² year⁻¹ in grass. Grasshoppers store 2 400 kJ m⁻² year⁻¹ and shrews that eat them store 288 kJ m⁻² year⁻¹.

(a) Calculate the efficiency of transfer from grass to grasshoppers. [2] (b) Calculate the efficiency of transfer from grasshoppers to shrews. [1] (c) Calculate the overall efficiency from grass to shrews. [1] (d) Name one way in which energy is lost between the grasshoppers and the shrews. [1]

4. A wetland has a GPP of 14 200 kJ m⁻² year⁻¹ and an NPP of 8 236 kJ m⁻² year⁻¹. Calculate the energy lost in plant respiration and express it as a percentage of GPP. [3]

5. A sheep ingests food containing 18 400 kJ in a day. Its faeces contain 5 520 kJ, its urine 920 kJ and its net production is 1 840 kJ.

(a) Calculate the energy lost in respiration. [2] (b) Calculate net production as a percentage of intake. [1] (c) Calculate the percentage of intake lost in respiration. [1]

6. Calves in a feedlot ingest 48 000 kJ each per day. Faeces hold 9 600 kJ, urine 2 400 kJ and net production is 9 600 kJ. The farmer changes the housing so that respiration falls by 20%, with intake, faeces and urine unchanged.

(a) Calculate the original respiratory loss. [2] (b) Calculate the new net production and give it as a percentage of intake. [2] (c) Suggest two changes to housing that could reduce respiratory loss. [2]

7. The wasp Encarsia formosa is released into a glasshouse to reduce whitefly on tomato plants.

(a) Explain how this increases productivity of the tomato crop. [2] (b) Explain one risk of using an introduced biological agent. [2]

8. A grower compares two controls of whitefly in glasshouse tomatoes. Without control, the yield is 52 t ha⁻¹. Tomatoes sell at £340 per tonne.

Control	Yield (t ha⁻¹)	Cost (£ ha⁻¹)
Chemical insecticide	71	1 950
Biological agent	68	2 600

(a) Calculate the extra revenue from each control compared with no control. [2] (b) Calculate the gain after subtracting cost for each control. [2] (c) Evaluate which control the grower should use. Refer to economic and environmental points. [3]

9. A crop stores 18 000 kJ in its edible tissue. Cattle fed on this crop store 6% of that energy in their bodies. People eating the beef retain 10% of the energy in the beef.

(a) Calculate the energy stored in the cattle. [1] (b) Calculate the energy in people from eating the beef. [1] (c) Express your answer to (b) as a percentage of the energy in the crop. [1] (d) Use your answer to explain why shorter food chains can feed more people per hectare. [1]

10. Pigs ingest 30 000 kJ per day. Faeces hold 4 500 kJ and urine 1 500 kJ. Respiratory loss is 16 800 kJ for pigs kept in a heated building and 19 500 kJ for pigs kept outdoors.

(a) Calculate net production for each group. [2] (b) Calculate net production as a percentage of intake for each group. [2] (c) Evaluate the use of heated buildings with restricted movement, using the data and economic, environmental and ethical points. [4]

11. A wheat farmer uses an integrated system: crop rotation, a variety resistant to a fungal disease, parasitic wasps against aphids, and an insecticide spray only when aphid numbers pass a set threshold. Protected fields yield 6.8 t ha⁻¹ and unprotected fields 5.1 t ha⁻¹.

(a) Calculate the percentage yield loss in the unprotected fields compared with the protected fields. [1] (b) Explain how two parts of this system increase the efficiency of energy transfer to the crop. [4] (c) Suggest why the farmer sprays only above a threshold instead of regularly. [2]

Answers

1. (a) The chemical energy store in plant biomass in a given area or volume, in a given time. [1] (b) The chemical energy store in plant biomass after respiratory loss (R) has been taken into account. [1] NPP = GPP - R. [1] Examiner insight: include “in a given area (or volume), in a given time”; it turns a vague statement about energy into a definition of a rate.

2. (a) Inverted: zooplankton (11 g m⁻²) are greater than phytoplankton (3.0 g m⁻²). [1] (b) Phytoplankton are replaced rapidly as they are eaten. [1] Production is 450 g m⁻² per year, far more than the 3.0 g m⁻² present at one time, so the stock supports a large flow of energy. [1] (c) 54 ÷ 450 × 100 = 12.0%. [1] (d) Each level gets its energy from the level below and some is lost at each transfer, so there is less energy at each level. [1] Examiner insight: in (b), compare a stock (g m⁻²) with a rate (g m⁻² per year), not just “they grow fast”.

3. (a) 2 400 ÷ 30 000 [1] × 100 = 8.0%. [1] (b) 288 ÷ 2 400 × 100 = 12.0%. [1] (c) 288 ÷ 30 000 × 100 = 0.96%. [1] (d) Any one of: some parts of a grasshopper not eaten; energy lost in the shrews’ faeces; energy lost in urine; energy released in respiration (as heat). [1] Examiner insight: show which value is divided by which. A “loss” must name a route, not say energy “disappears”.

4. R = GPP - NPP [1] = 14 200 - 8 236 = 5 964 kJ m⁻² year⁻¹. [1] 5 964 ÷ 14 200 × 100 = 42.0% of GPP. [1] Examiner insight: rearrange NPP = GPP - R first; adding the two values shows a misunderstanding of NPP.

5. (a) R = I - F - U - N [1] = 18 400 - 5 520 - 920 - 1 840 = 10 120 kJ. [1] (b) 1 840 ÷ 18 400 × 100 = 10.0%. [1] (c) 10 120 ÷ 18 400 × 100 = 55.0%. [1] Examiner insight: give units on energy values, and check that F, U, R and N add to I.

6. (a) R = 48 000 - (9 600 + 2 400 + 9 600) [1] = 26 400 kJ. [1] (b) New R = 26 400 × 0.80 = 21 120 kJ, so N = 48 000 - (9 600 + 2 400 + 21 120) = 14 880 kJ. [1] 14 880 ÷ 48 000 × 100 = 31.0%. [1] (c) Any two of: restrict movement of the calves [1]; keep the building warm so less energy is used to maintain body temperature [1]; control feeding to reduce activity. Examiner insight: link each housing change to less respiration; “keep them inside” gains more when it says why R falls.

7. (a) The wasps parasitise (and kill) whitefly, so there are fewer whitefly feeding on the tomato plants. [1] More of the energy fixed by the tomato plants stays in the tomato biomass, so the yield rises. [1] (b) It may feed on or compete with non-target species. [1] It may become a pest itself by spreading beyond the glasshouse. [1] Examiner insight: for a risk, state the harm and why it happens, not just its name.

8. (a) Chemical: (71 - 52) × 340 = £6 460. [1] Biological: (68 - 52) × 340 = £5 440. [1] (b) Chemical: 6 460 - 1 950 = £4 510. [1] Biological: 5 440 - 2 600 = £2 840. [1] (c) The chemical insecticide gives the higher yield and the greater gain (£4 510 compared with £2 840 per hectare). [1] The chemical may harm non-target insects, leave residues and lead to resistance in whitefly. [1] The biological agent costs more and yields slightly less, but avoids residues and resistance. A judgement is needed, for example chemical for short-term profit, biological for long-term control. [1] Examiner insight: an evaluation needs a decision supported by figures and at least one environmental point, not just a list of facts.

9. (a) 18 000 × 0.06 = 1 080 kJ. [1] (b) 1 080 × 0.10 = 108 kJ. [1] (c) 108 ÷ 18 000 × 100 = 0.6%. [1] (d) Each extra trophic level loses most of the energy, so people eating the crop directly receive far more of it than people eating the beef. [1] Examiner insight: multiply successive efficiencies. Adding 6% and 10% is a common wrong step.

10. (a) Heated: 30 000 - (4 500 + 1 500 + 16 800) = 7 200 kJ. [1] Outdoor: 30 000 - (4 500 + 1 500 + 19 500) = 4 500 kJ. [1] (b) Heated: 7 200 ÷ 30 000 × 100 = 24.0%. [1] Outdoor: 4 500 ÷ 30 000 × 100 = 15.0%. [1] (c) Economic: respiratory loss is 2 700 kJ lower indoors, so more feed becomes meat, though heating costs money. [1] Environmental: heating uses energy and concentrated pigs produce concentrated waste. [1] Ethical: restricted movement limits natural behaviour. [1] Judgement: efficiency is better indoors, but the decision depends on how costs and welfare are weighed. [1] Examiner insight: use the figures from (a) and (b) in the evaluation. An answer on ethics alone does not evaluate the data.

11. (a) (6.8 - 5.1) ÷ 6.8 × 100 = 25.0%. [1] (b) Any two of these. Crop rotation breaks the life cycles of pests and disease, so fewer organisms feed on the crop [1] and more energy stays in the crop [1]. The resistant variety stops the fungus damaging leaves [1], so more energy goes into growth and yield [1]. Wasps kill aphids, so less sap is lost and more energy stays in the crop. Four marks maximum. (c) Spraying only when needed reduces cost and the harm to non-target species. [1] It lowers the chance of insecticide resistance evolving in the aphids. [1] Examiner insight: “explain” needs a cause and a consequence. Name how the part reduces energy loss, then say what happens to the yield.

Where marks are usually lost

• Defining GPP or NPP without “in a given area or volume, in a given time”.
• Putting the higher level under the fraction, or adding efficiencies instead of multiplying them.
• Leaving urine out of N = I - (F + U + R), or forgetting that all four terms are energy values.
• Missing units on energy values or a percentage sign.
• Giving a farming practice but not linking it to a reduced loss of energy.
• Evaluating with a single viewpoint when economic, environmental and ethical points are all asked for.
• Listing facts in an evaluation with no justified judgement.

Next steps

• Recap with the revision notes and the study guide.
• Course hub: OxfordAQA A-level Biology; printable checklist.
• Take all the free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

OxfordAQA International AS and A-level Biology, Version 5.1, for International A-level exams May/June 2018 onwards, sections 3.3.4.1 and 3.3.4.2.