AQA A-level Biology: Energy Transfer and Productivity in Ecosystems (7402) – 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 section 3.5.3, Energy and ecosystems (A-level only), in the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015. Learn the ideas first from the study guide and the revision notes. Course pages: AQA A-level Biology hub and the printable checklist. Use 4.18 J g⁻¹ °C⁻¹ for the specific heat capacity of water, and 1 ha = 10 000 m². Other values are given in the question. To check your starting level, try the free diagnostics.

Questions

1. Define gross primary production (GPP) and net primary production (NPP), and state the equation that links them. [3]

2. A student collects 340 g of fresh grass from a 2.0 m² area. After drying, the grass has a mass of 91.8 g.

(a) Describe how the student should dry the grass so that all the water is removed without burning the tissue. [2] (b) Calculate the dry mass per m². [1] (c) Give this value in kg ha⁻¹. [1]

3. A sample of dried plant tissue of mass 0.75 g is burned under 40.0 g of water. The water temperature rises from 19.5 °C to 47.5 °C.

(a) Calculate the energy transferred to the water, in J, to 3 significant figures. [2] (b) Calculate the energy per gram of tissue, in kJ g⁻¹. [1] (c) The true energy content of the tissue is higher than your answer. Suggest one reason. [1]

4. A lake has gross primary production of 14 600 kJ m⁻² year⁻¹. Respiratory loss by the producers is 5 110 kJ m⁻² year⁻¹.

(a) Calculate the net primary production. [1] (b) Calculate the respiratory loss as a percentage of GPP. [1] (c) Express the net primary production in kJ ha⁻¹ year⁻¹. [1] (d) State what happens to the net primary production in the lake. [1]

5. A herd of sheep ingests 6.2 × 10⁶ kJ ha⁻¹ year⁻¹ of food energy. 2.1 × 10⁶ kJ ha⁻¹ year⁻¹ is lost in faeces and urine. Their net production is 8.8 × 10⁵ kJ ha⁻¹ year⁻¹.

(a) Calculate the energy lost through respiration. [2] (b) Calculate the net production as a percentage of the energy ingested. [1] (c) State what is meant by secondary productivity. [1]

6. In a meadow, net primary production is 8.40 × 10⁶ kJ ha⁻¹ year⁻¹. Primary consumers have net production of 6.72 × 10⁵ kJ ha⁻¹ year⁻¹ and secondary consumers have 5.04 × 10⁴ kJ ha⁻¹ year⁻¹.

(a) Work out what percentage of the energy in the producers’ net production is stored in the primary consumers, and what percentage of that is then stored in the secondary consumers. [2] (b) Calculate the overall efficiency from producers to secondary consumers. [1] (c) Explain why the overall efficiency is lower than either single transfer. [1] (d) State two ways in which energy leaves a consumer’s food intake without becoming new consumer biomass. [1]

7. Calves are reared for meat in two barns. In barn A they are kept in small pens. In barn B they are kept in large pens with room to run. Data are energy per calf over the rearing period, in MJ.

Barn	Ingested (I)	Lost in faeces and urine (F)	Lost in respiration (R)
A	118	31.0	61.5
B	131	33.0	83.7

(a) Calculate the net production of a calf in each barn. [2] (b) Calculate the net production as a percentage of ingested energy in each barn. [2] (c) Explain the difference between the barns. [2]

8. Explain how farmers can increase the efficiency of energy transfer to humans by simplifying food webs and by reducing respiratory losses in livestock. Give one disadvantage of either approach. [6]

9. A wheat field receives 3.2 × 10¹⁰ kJ ha⁻¹ year⁻¹ of light energy. Its GPP is 4.1 × 10⁸ kJ ha⁻¹ year⁻¹ and its NPP is 2.9 × 10⁸ kJ ha⁻¹ year⁻¹. The energy in the harvested grain is 1.16 × 10⁸ kJ ha⁻¹ year⁻¹.

(a) Calculate GPP as a percentage of the light energy received. [1] (b) Calculate the respiratory loss. [1] (c) Calculate the grain as a percentage of NPP. [1] (d) Suggest what happens to the rest of the NPP. [2]

10. The table shows productivity in a mature forest and in a managed crop. Both are in kJ m⁻² year⁻¹.

Ecosystem	GPP	Respiratory loss
Forest	38 500	21 600
Crop	29 400	9 800

(a) Calculate the net primary production of each ecosystem. [2] (b) Calculate the respiratory loss as a percentage of GPP for each. [2] (c) The forest has the higher GPP but the lower NPP. Explain this, using the data. [3]

11. The dry biomass of a woodland is 12 400 kg ha⁻¹. Carbon makes up 46% of dry mass in this tissue.

(a) Calculate the mass of carbon in kg ha⁻¹. [1] (b) Calculate the mass of carbon, in tonnes, in a 3.0 ha wood (1 tonne = 1 000 kg). [1] (c) Explain why biomass is measured as dry mass, or as mass of carbon, and not as fresh mass. [2]

Answers

1. GPP is the chemical energy store in plant biomass, in a given area or volume [1]. NPP is the chemical energy store in plant biomass after respiratory losses to the environment have been taken into account [1]. NPP = GPP - R, where R is respiratory loss [1]. [3] Examiner insight: State “after respiratory losses”; “what is left” alone is vague.

2. (a) Heat in an oven at a temperature low enough not to burn the tissue [1]. Re-weigh at intervals until the mass stops changing (constant mass) [1]. [2] (b) 91.8 ÷ 2.0 = 45.9 g m⁻² [1]. [1] (c) 45.9 × 10 000 = 459 000 g ha⁻¹ = 459 kg ha⁻¹ [1]. [1] Examiner insight: “Dry it” alone says little; the repeat weighing to constant mass is the point that shows how you know it is dry.

3. (a) Rise = 47.5 - 19.5 = 28.0 °C [1]. Energy = 40.0 × 4.18 × 28.0 = 4 681.6 J, so 4 680 J [1]. [2] (b) 4 681.6 ÷ 0.75 = 6 242 J g⁻¹ = 6.24 kJ g⁻¹ [1]. [1] (c) Heat was lost to the surroundings or apparatus, or combustion was incomplete [1]. [1] Examiner insight: Show the J to kJ conversion; writing the unit at each step makes a slip of 1 000 easy to spot.

4. (a) NPP = 14 600 - 5 110 = 9 490 kJ m⁻² year⁻¹ [1]. [1] (b) 5 110 ÷ 14 600 × 100 = 35.0% [1]. [1] (c) 9 490 × 10 000 = 9.49 × 10⁷ kJ ha⁻¹ year⁻¹ [1]. [1] (d) It is available for plant growth and reproduction, and for herbivores and decomposers [1]. [1] Examiner insight: Show the subtraction and the unit so that your working can be followed.

5. (a) R = I - F - N = 6.2 × 10⁶ - 2.1 × 10⁶ [1] - 8.8 × 10⁵ = 3.22 × 10⁶ kJ ha⁻¹ year⁻¹ [1]. [2] (b) 8.8 × 10⁵ ÷ 6.2 × 10⁶ × 100 = 14.2% [1]. [1] (c) The rate of secondary production, that is biomass of consumers in a given area in a given time, for example kJ ha⁻¹ year⁻¹ [1]. [1] Examiner insight: When R is unknown, rearrange first, then substitute.

6. (a) 6.72 × 10⁵ ÷ 8.40 × 10⁶ × 100 = 8.0% [1]; 5.04 × 10⁴ ÷ 6.72 × 10⁵ × 100 = 7.5% [1]. [2] (b) 5.04 × 10⁴ ÷ 8.40 × 10⁶ × 100 = 0.60% [1]. [1] (c) Energy is lost at each transfer, so the losses compound [1]. [1] (d) Any two of: lost in faeces and urine; lost as heat in respiration [1]. [1] Examiner insight: Divide by the level below, not the level above. A result above 100% is a signal that the fraction is upside down.

7. (a) A: 118 - (31.0 + 61.5) = 25.5 MJ [1]. B: 131 - (33.0 + 83.7) = 14.3 MJ [1]. [2] (b) A: 25.5 ÷ 118 × 100 = 21.6% [1]. B: 14.3 ÷ 131 × 100 = 10.9% [1]. [2] (c) Calves in barn A move less, so respiratory loss is a smaller share of intake (52.1% against 63.9%) [1]. More of the food energy is stored in biomass, so net production is a higher percentage of intake [1]. [2] Examiner insight: Back the comparison with numbers from the data; “A is better” on its own gains little.

8. Simplifying food webs: herbicides kill weeds that compete with the crop [1]. Pesticides kill pests that eat the crop [1]. Less energy is lost to non-human food chains, so more NPP is stored in crops or livestock feed [1]. Reducing respiratory losses: restricting movement or keeping animals warm reduces respiration [1]. R is smaller, so N is larger and more energy goes into growth [1]. Disadvantages, for example: pesticides may harm non-target species or reduce biodiversity; confinement reduces animal welfare [1]. [6] Examiner insight: Name the energy loss that each practice reduces. “More efficient” with no loss identified is a common weakness.

9. (a) 4.1 × 10⁸ ÷ 3.2 × 10¹⁰ × 100 = 1.28% [1]. [1] (b) R = GPP - NPP = 4.1 × 10⁸ - 2.9 × 10⁸ = 1.2 × 10⁸ kJ ha⁻¹ year⁻¹ [1]. [1] (c) 1.16 × 10⁸ ÷ 2.9 × 10⁸ × 100 = 40.0% [1]. [1] (d) It remains in parts that are not harvested, such as stems, leaves and roots [1]. Some passes to decomposers or to pests [1]. [2] Examiner insight: Check what the denominator is: light, GPP and NPP give three different percentages.

10. (a) Forest: 38 500 - 21 600 = 16 900 kJ m⁻² year⁻¹ [1]. Crop: 29 400 - 9 800 = 19 600 kJ m⁻² year⁻¹ [1]. [2] (b) Forest: 21 600 ÷ 38 500 × 100 = 56.1% [1]. Crop: 9 800 ÷ 29 400 × 100 = 33.3% [1]. [2] (c) NPP = GPP - R [1]. The forest loses 56.1% of its GPP in respiration against 33.3% in the crop [1]. The forest’s R is more than twice the crop’s (21 600 against 9 800), which outweighs its GPP advantage of 9 100 [1]. [3] Examiner insight: Use figures and a comparison, not just “one is bigger”.

11. (a) 12 400 × 0.46 = 5 704 kg ha⁻¹ [1]. [1] (b) 5 704 × 3.0 = 17 112 kg = 17.1 tonnes [1]. [1] (c) Fresh mass includes water, whose amount varies with conditions [1]. Dry mass or carbon measures the biological molecules that make up the biomass [1]. [2] Examiner insight: Say why water is a problem: its amount varies. “Inaccurate” is too vague.

Where marks are usually lost

• Omitting R from N = I - (F + R), or treating F as the only loss.
• Mixing up GPP and NPP, or adding R when finding NPP.
• Giving a rate with no time unit.
• Mixing m² and ha in one calculation.
• Dividing by the wrong trophic level in an efficiency.
• Writing “energy is lost” without saying how.
• Describing farming practices with no link to the energy loss that they reduce.
• Not stating constant mass when describing how to dry a sample.

Next steps

• Revise: Energy transfer and productivity revision notes
• Relearn: Energy transfer and productivity study guide
• Course hub: AQA A-level Biology
• Printable checklist: AQA A-level Biology checklist
• Check your gaps: all free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

AQA, AS and A-level Biology specification (7401 and 7402), first teaching 2015, section 3.5.3 Energy and ecosystems (A-level only).