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A-Level · Biology

AB5 Energy Transfers in and between Organisms

AQA 7402 · Calculator allowed · about 115 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Photosynthesis begins with the light-dependent reactions, which take place on and across the thylakoid membranes of the chloroplast.

(a)State the two products of the light-dependent reactions that are required for the light-independent reactions (the Calvin cycle).(2)

(b)Describe how a proton (H+) gradient is generated across the thylakoid membrane during non-cyclic photophosphorylation.(3)

(c)Explain how ATP synthase uses the proton gradient described in part (b) to produce ATP.(3)

(Total for Question 1 is 8 marks)

Required practical. A student carried out the required practical method to investigate the effect of light intensity on the rate of photosynthesis using algal balls (immobilised Chlorella suspended in alginate gel beads) placed in hydrogencarbonate indicator solution in sealed test tubes. A lamp was positioned at different distances from a tube, and the time taken for the indicator to change colour from red to purple was recorded as a measure of the rate of photosynthesis.

(a)Give two variables, other than light intensity, that should be controlled in this investigation to ensure a valid comparison between tubes.(2)

(b)Explain why the hydrogencarbonate indicator changes colour from red to purple as the rate of photosynthesis increases.(2)

(c)The rate of photosynthesis is directly proportional to light intensity, and light intensity follows the inverse square law with distance from the lamp (intensity is proportional to 1/distance²). At a distance of 10 cm from the lamp, the indicator took 40 s to change from red to purple. Assuming light remains the limiting factor throughout, calculate the time you would predict for the colour change to occur at a distance of 20 cm.(3)

(d)Suggest one limitation of using the time taken for a full colour change as a measure of the rate of photosynthesis.(1)

(Total for Question 2 is 8 marks)

Aerobic respiration releases energy from glucose in four linked stages: glycolysis, the link reaction, the Krebs cycle, and oxidative phosphorylation.

(a)Name the stage of aerobic respiration that occurs in the cytoplasm.(1)

(b)State the net ATP yield from glycolysis per molecule of glucose, and explain why this yield is described as a 'net' gain.(2)

(c)Explain why the link reaction and the Krebs cycle must take place in the mitochondrial matrix.(2)

(d)Per molecule of glucose, aerobic respiration produces 10 reduced NAD (NADH) and 2 reduced FAD (FADH2) in total (from glycolysis, the link reaction and the Krebs cycle combined). Oxidative phosphorylation yields 2.5 ATP per NADH and 1.5 ATP per FADH2. In addition, 2 ATP are produced by substrate-level phosphorylation in glycolysis and 2 ATP by substrate-level phosphorylation in the Krebs cycle. Calculate the maximum theoretical total number of ATP molecules produced per molecule of glucose.(3)

(Total for Question 3 is 8 marks)

Required practical. A student used a respirometer to investigate the rate of oxygen consumption by blowfly larvae (maggots). The apparatus had two tubes connected to a manometer containing coloured fluid: one tube contained the maggots with soda lime, the other (the control tube) contained an equal volume of inert glass beads with soda lime. As the maggots respired, the coloured fluid in the manometer moved towards the tube containing the maggots.

(a)Explain the purpose of the soda lime in the apparatus.(2)

(b)Explain the purpose of the control tube containing inert glass beads.(1)

(c)The manometer capillary tube has an internal radius of 1.0 mm. During the experiment, the coloured fluid moved 30 mm along the tube in 5.0 minutes. Using volume = π x r² x length, calculate the volume of oxygen consumed, in cm^3. Give your answer to 3 significant figures.(2)

(d)The mass of maggots used was 2.0 g. Using your answer to part (c), calculate the rate of oxygen consumption per gram of maggots, in cm³ g^-1 min^-1. Give your answer to 2 significant figures.(3)

(e)Suggest one reason why the value calculated in part (d) might be an underestimate of the true rate of aerobic respiration of the maggots.(2)

(Total for Question 4 is 10 marks)

In a woodland ecosystem, the producers had a gross primary productivity (GPP) of 18500 kJ m^-2 year^-1. The producers used 11200 kJ m^-2 year^-1 in respiration (R).

(a)Define the term 'net primary productivity' (NPP).(1)

(b)Calculate the net primary productivity (NPP) of the producers. Show your working.(2)

(c)The primary consumers in the woodland had a net productivity of 730 kJ m^-2 year^-1. Calculate the percentage of the producers' NPP (from part b) that is transferred to primary consumers.(2)

(d)Suggest one reason why not all of the energy in the producers' net primary productivity is transferred to primary consumers.(1)

(Total for Question 5 is 6 marks)

Explain why the efficiency of energy transfer between successive trophic levels in an ecosystem is generally low, and explain why food chains rarely contain more than about five trophic levels.

(Total for Question 6 is 6 marks)

The nitrogen cycle involves several groups of soil bacteria that convert nitrogen between different chemical forms.

(a)Name the type of bacteria responsible for each of the following processes: (i) conversion of atmospheric nitrogen gas (N2) into ammonium compounds; (ii) conversion of ammonium ions into nitrite ions; (iii) conversion of nitrate ions into atmospheric nitrogen gas.(3)

(b)Explain the mutualistic relationship between Rhizobium bacteria and leguminous plants (e.g. peas, clover) growing in root nodules.(2)

(c)Farmers can increase nitrate availability in soil either by growing leguminous crops as part of a crop rotation, or by applying inorganic nitrogen fertiliser. Evaluate the environmental impact of using inorganic nitrogen fertiliser instead of crop rotation with legumes.(3)

(Total for Question 7 is 8 marks)

Saprobiotic microorganisms (decomposers) play a key role in the carbon cycle.

(a)Describe the role of saprobiotic microorganisms in the carbon cycle.(2)

(b)Explain why deforestation, followed by burning of the cleared vegetation, increases the concentration of carbon dioxide in the atmosphere.(2)

(c)State two ways in which carbon can be stored (sequestered) for long periods of time.(2)

(Total for Question 8 is 6 marks)

Identify the correct answer for each of the following.

(a)Which molecule acts as the final electron acceptor in the electron transport chain during aerobic respiration?(1)

A) NAD
B) Oxygen
C) FAD
D) Pyruvate

(b)Which process converts nitrite ions (NO2-) into nitrate ions (NO3-)?(1)

A) Nitrogen fixation
B) Ammonification
C) Nitrification
D) Denitrification

(c)Which enzyme catalyses the fixation of carbon dioxide by combining it with RuBP in the light-independent reactions?(1)

A) ATP synthase
B) RuBisCO
C) DNA polymerase
D) Catalase

(d)Which statement best describes gross primary productivity (GPP)?(1)

A) The total energy available to primary consumers after respiratory losses
B) The total chemical energy store in producer biomass at a single point in time
C) The total chemical energy fixed by photosynthesis per unit area per unit time
D) The total energy transferred to secondary consumers

(e)Which term describes the loss of nitrogen from soil as nitrogen gas, occurring under anaerobic (waterlogged) soil conditions?(1)

A) Nitrification
B) Denitrification
C) Ammonification
D) Nitrogen fixation

(Total for Question 9 is 5 marks)

The table shows energy values for the trophic levels of a grassland food chain: grass (producers) -> grasshoppers (primary consumers) -> foxes (secondary consumers), recorded over one year. Producers: NPP = 9000 kJ m^-2 year^-1. Grasshoppers: energy intake = 900 kJ m^-2 year^-1; production (biomass produced) = 90 kJ m^-2 year^-1. Foxes: energy intake = 9 kJ m^-2 year^-1; production = 0.18 kJ m^-2 year^-1.

(a)Calculate the percentage efficiency of energy transfer from the producers' NPP to the grasshoppers' energy intake.(2)

(b)Calculate the production efficiency of the grasshoppers (production as a percentage of energy intake).(2)

(c)Calculate the production efficiency of the foxes (production as a percentage of energy intake). Give your answer to an appropriate number of significant figures.(2)

(d)Foxes are mammals and maintain a constant, relatively high body temperature (endotherms), whereas grasshoppers are insects that do not regulate their body temperature (ectotherms). Use your answers to parts (b) and (c) to explain why the production efficiency of the foxes is lower than that of the grasshoppers.(2)

(Total for Question 10 is 8 marks)

A student calculated an energy budget for a population of voles. The voles had a food energy intake (I) of 15000 kJ m^-2 year^-1. Energy lost in faeces and urine (F+U) was 9000 kJ m^-2 year^-1, and energy used in respiration (R) was 5850 kJ m^-2 year^-1.

(a)Using the equation P = I - (F+U) - R, show that the net production (P) of the vole population is 150 kJ m^-2 year^-1.(2)

(b)State what happens to the energy in the net production (P) that is not consumed by the next trophic level.(1)

(c)Explain, in terms of energy flow, why only a small percentage of the Sun's energy incident on a habitat is ultimately fixed by photosynthesis.(3)

(Total for Question 11 is 6 marks)

A plant physiologist measured the net (apparent) rate of oxygen production by an aquatic plant at different light intensities, and separately measured the rate of oxygen consumption in the dark (respiration rate) as 2.0 arbitrary units per hour. At a light intensity of 8 units, the net rate of oxygen production was 6.0 units per hour.

(a)Calculate the true (gross) rate of photosynthesis at this light intensity.(2)

(b)Define the term 'light compensation point'.(2)

(c)Explain what would happen to the net production of biomass by the plant if light intensity were maintained below the compensation point for a prolonged period.(2)

(d)A commercial glasshouse grower raises the carbon dioxide concentration well above normal atmospheric levels while keeping light intensity and temperature constant. Explain why, above a certain CO2 concentration, further increases in CO2 produce no further increase in the rate of photosynthesis.(2)

(Total for Question 12 is 8 marks)

A farmer wants to increase the efficiency of energy transfer to a herd of beef cattle. He is considering two intensive farming methods: (1) keeping the cattle indoors in temperature-controlled barns with restricted space to move; (2) treating the cattle with growth-promoting hormones. Discuss how each method could increase the efficiency of energy transfer to the cattle, and evaluate the ethical and environmental issues raised by these farming practices.

(Total for Question 13 is 6 marks)

Mark scheme · AB5 Energy Transfers in and between Organisms

Question 1

(a) B1 ATP
(a) B1 reduced NADP (NADPH)
(a) Answer: ATP and reduced NADP
(b) B1 photoactivation of chlorophyll at photosystem II releases (excited) electrons
(b) B1 electrons pass along the electron transport chain, releasing energy that is used to actively pump H+ into the thylakoid space (lumen)
(b) B1 photolysis of water also releases H+ (and electrons and O2) directly into the thylakoid space, adding to the gradient
(b) Answer: Electron transport chain pumps H+ into the thylakoid lumen; photolysis of water also releases H+ into the lumen, both increasing H+ concentration compared with the stroma.
(c) B1 protons diffuse down their concentration (electrochemical) gradient through ATP synthase, from the thylakoid space into the stroma
(c) B1 this process is called chemiosmosis
(c) B1 energy released as protons flow through the enzyme is used to bond ADP and inorganic phosphate (Pi) together, forming ATP
(c) Answer: Chemiosmosis: H+ diffuse through ATP synthase down their gradient, and the energy released drives ADP + Pi -> ATP.

Question 2

(a) B1 temperature (any two from this list, one mark each)
(a) B1 volume of hydrogencarbonate indicator solution / number and size of algal balls / concentration of hydrogencarbonate indicator / time left before readings begin
(a) Answer: Any two of: temperature; volume of indicator solution; number/size of algal balls; concentration of indicator.
(b) B1 photosynthesis removes carbon dioxide from solution (as CO2 is used in the light-independent reactions)
(b) B1 less dissolved CO2 means a lower concentration of carbonic acid/H+ ions, so pH rises and the indicator turns purple (more alkaline)
(b) Answer: CO2 is used up by photosynthesis, so [H+] falls, pH rises, and the indicator turns purple.
(c) M1 recognises that doubling the distance (10 cm to 20 cm) reduces light intensity, and therefore rate, to 1/4 of its original value (inverse square law: (10/20)² = 1/4)
(c) M1 time taken is inversely proportional to rate, so time increases by a factor of 4
(c) A1 160 (s) cao
(c) Answer: 160 s
(d) B1 the colour change is gradual and the exact end point is subjective/difficult to judge precisely (or: assumes rate is constant throughout, which may not be true), so results may be imprecise
(d) Answer: Judging the exact colour end-point is subjective, reducing precision.

Question 3

(a) B1 glycolysis
(a) Answer: Glycolysis
(b) B1 net yield = 2 ATP
(b) B1 because 2 ATP are used (invested) early in glycolysis (phosphorylation of glucose/intermediates) while 4 ATP are produced later, giving a net gain of 2
(b) Answer: Net yield = 2 ATP; 4 ATP produced minus 2 ATP invested = 2 ATP net.
(c) B1 the enzymes that catalyse the reactions of the link reaction and Krebs cycle are located in the matrix
(c) B1 this places the reduced coenzymes (NAD, FAD) produced close to the electron transport chain on the inner mitochondrial membrane, allowing efficient transfer of electrons/hydrogen
(c) Answer: Matrix enzymes catalyse these reactions and their products (reduced NAD/FAD) are positioned next to the inner membrane electron transport chain.
(d) M1 ATP from NADH = 10 x 2.5 = 25
(d) M1 ATP from FADH2 = 2 x 1.5 = 3
(d) A1 total = 25 + 3 + 2 + 2 = 32 (ATP) cao
(d) Answer: 32 ATP

Question 4

(a) B1 soda lime absorbs the carbon dioxide released by the maggots during respiration
(a) B1 so any change in gas volume/pressure detected is due only to the uptake of oxygen, allowing oxygen consumption to be measured
(a) Answer: Soda lime absorbs CO2 so the fluid movement reflects O2 uptake only.
(b) B1 controls for (corrects for) fluid movement caused by changes in temperature or atmospheric pressure rather than by respiration, so the comparison is valid
(b) Answer: It corrects for fluid movement caused by temperature/pressure changes, not respiration.
(c) M1 correct unit conversion: r = 1.0 mm = 0.1 cm, length = 30 mm = 3.0 cm
(c) A1 volume = π x (0.1)² x 3.0 = 0.0942 cm³ (awrt 0.094 cm³)
(c) Answer: 0.0942 cm³
(d) M1 rate = volume / time = 0.0942 / 5.0 = 0.01884 cm³ min^-1 (ft from (c))
(d) M1 rate per gram = 0.01884 / 2.0 (ft from (c))
(d) A1 0.0094 cm³ g^-1 min^-1 (2 s.f.) awrt
(d) Answer: 0.0094 cm³ g^-1 min^-1
(e) B1 small leaks at the joints/seals of the apparatus allow gas exchange with the atmosphere
(e) B1 this reduces the pressure change detected, so less fluid movement occurs than would be expected from respiration alone, giving a lower calculated rate than the true rate
(e) Answer: Leaks in the apparatus reduce the recorded pressure change, so the calculated rate is lower than the true rate.

Question 5

(a) B1 the rate at which energy (or biomass) is stored/made available to primary consumers, i.e. GPP minus respiratory losses (R): NPP = GPP - R
(a) Answer: NPP = GPP - R: the energy remaining after the producers' own respiratory losses.
(b) M1 NPP = GPP - R = 18500 - 11200
(b) A1 7300 kJ m^-2 year^-1 cao (correct unit required)
(b) Answer: 7300 kJ m^-2 year^-1
(c) M1 (730 / 7300) x 100 (ft from (b))
(c) A1 10 (%) cao
(c) Answer: 10%
(d) B1 not all plant material is eaten (e.g. roots are inaccessible, or the plant is not found/eaten before it dies) and/or some ingested material is indigestible (e.g. cellulose/lignin) and egested as faeces rather than absorbed
(d) Answer: Some plant material is not eaten, or is egested undigested (e.g. cellulose), rather than absorbed by consumers.

Question 6

Level 3 (5-6): A detailed and coherent explanation is given of why energy transfer between trophic levels is inefficient, and of why the number of trophic levels is limited. Ideas are logically linked using accurate scientific terminology throughout.
Level 2 (3-4): An explanation is given covering several reasons for energy loss and/or the limitation on trophic levels, generally clearly expressed, with mostly accurate terminology but some links between ideas may be missing.
Level 1 (1-2): Basic, generalised statements about energy loss are made (e.g. 'energy is lost as heat'), with little detail and limited or no reference to why trophic levels are limited.
Level 0 (0): No relevant content, or the response does not address the question.
Indicative content:
- Not all biomass/organisms at one trophic level are consumed by the next trophic level; some organisms die without being eaten and are broken down by decomposers, so their energy passes to decomposers rather than to the next trophic level
- Some ingested material is indigestible (e.g. bone, cellulose, lignin) and is egested as faeces without being absorbed, so this energy is not available for growth
- Energy that is absorbed may be lost in excretory products (e.g. urea in urine) rather than being incorporated into biomass
- A large proportion of absorbed energy is used in respiration to generate ATP for movement, active transport and (in mammals and birds) maintaining a constant body temperature
- Energy used in respiration is ultimately released as heat to the surroundings and is not available to the next trophic level (in line with the second law of thermodynamics)
- As a result of these losses, typically only around 10% of the energy available at one trophic level is transferred to and incorporated into the biomass of the next
- Because energy availability falls sharply (geometrically) at each successive trophic level, after four or five transfers there is too little energy/biomass remaining to support a further viable population of consumers
- Top predators would need an impractically large feeding area/population of prey to obtain sufficient energy, which limits the number of trophic levels that can be sustained

Question 7

(a) B1 (i) nitrogen-fixing bacteria (e.g. Rhizobium or Azotobacter)
(a) B1 (ii) nitrifying bacteria (e.g. Nitrosomonas)
(a) B1 (iii) denitrifying bacteria
(a) Answer: (i) Nitrogen-fixing bacteria; (ii) Nitrifying bacteria (Nitrosomonas); (iii) Denitrifying bacteria
(b) B1 Rhizobium fixes atmospheric nitrogen into ammonium/nitrogen-containing compounds that the plant can absorb and use, e.g. to make amino acids
(b) B1 in return, the plant supplies the bacteria with carbohydrates (organic compounds from photosynthesis) as a source of energy/carbon
(b) Answer: Rhizobium provides fixed nitrogen compounds to the plant; the plant provides carbohydrates to the bacteria.
(c) B1 excess inorganic nitrate not taken up by crops can leach into waterways, causing eutrophication (algal bloom, death of aquatic plants, decomposer respiration depletes dissolved oxygen, death of aquatic animals)
(c) B1 crop rotation with legumes is more sustainable as it does not rely on manufactured fertiliser, whose production (Haber process) is energy-intensive and produces greenhouse gases
(c) B1 valid comparative judgement, e.g. inorganic fertiliser gives a faster/more controllable increase in nitrate but at a greater long-term environmental cost than legume rotation
(c) Answer: Inorganic fertiliser risks leaching/eutrophication and has a high energy/carbon cost of manufacture; legume rotation is more sustainable but slower acting.

Question 8

(a) B1 they secrete digestive enzymes onto dead organic matter/waste, breaking it down extracellularly (saprobiotic nutrition) and absorbing the soluble products
(a) B1 carbon dioxide is released back into the atmosphere as a result of the decomposers' own respiration
(a) Answer: Decomposers digest dead matter extracellularly and release CO2 through respiration, returning carbon to the atmosphere.
(b) B1 burning releases the carbon stored in the biomass directly into the atmosphere as CO2 (combustion)
(b) B1 with fewer trees/plants remaining, less CO2 is removed from the atmosphere by photosynthesis, so the net concentration of atmospheric CO2 rises
(b) Answer: Combustion releases stored carbon as CO2, and reduced photosynthetic uptake means less CO2 is removed, so atmospheric CO2 rises.
(c) B1 as fossil fuels (coal, oil, natural gas or peat)
(c) B1 as calcium carbonate in limestone or in the shells/skeletons of marine organisms, or as wood in long-lived trees/forests
(c) Answer: Any two of: fossil fuels; limestone/marine shells (calcium carbonate); wood in long-lived trees.

Question 9

(a) B1 B (oxygen)
(a) Answer: B) Oxygen
(b) B1 C (nitrification)
(b) Answer: C) Nitrification
(c) B1 B (RuBisCO)
(c) Answer: B) RuBisCO
(d) B1 C
(d) Answer: C
(e) B1 B (denitrification)
(e) Answer: B) Denitrification

Question 10

(a) M1 (900 / 9000) x 100
(a) A1 10 (%) cao
(a) Answer: 10%
(b) M1 (90 / 900) x 100
(b) A1 10 (%) cao
(b) Answer: 10%
(c) M1 (0.18 / 9) x 100
(c) A1 2.0 (%) cao
(c) Answer: 2.0%
(d) B1 foxes (endotherms) use a much greater proportion of their absorbed/assimilated energy in respiration to generate metabolic heat and maintain a constant body temperature, compared with grasshoppers (ectotherms), which do not need to do this
(d) B1 this means less of the energy intake is available to be converted into new biomass (growth/production) in foxes, giving a lower production efficiency (2.0%) than in grasshoppers (10%)
(d) Answer: Foxes lose far more energy as metabolic heat (thermoregulation) than grasshoppers, leaving less available for growth, so their production efficiency is lower.

Question 11

(a) M1 15000 - 9000 - 5850 (working shown)
(a) A1 = 150 kJ m^-2 year^-1 cso (answer given)
(a) Answer: 150 kJ m^-2 year^-1 (given)
(b) B1 it remains as dead organic matter and is broken down by decomposers/detritivores (returning carbon/nutrients to the abiotic environment, with further energy lost as heat via decomposer respiration)
(b) Answer: It is broken down by decomposers, returning nutrients to the environment.
(c) B1 only certain wavelengths of light are absorbed by photosynthetic pigments (chlorophyll); other wavelengths (e.g. green light) are reflected or transmitted and not used
(c) B1 much of the light reaching a habitat is not intercepted by chlorophyll/leaves at all (e.g. it reflects off non-photosynthetic surfaces, hits bare ground, or is absorbed by clouds/atmosphere before reaching plants)
(c) B1 some absorbed light energy is lost as heat rather than being converted into chemical energy during the light-dependent reactions
(c) Answer: Wrong wavelengths are reflected/transmitted, much light misses chlorophyll entirely, and some absorbed energy is lost as heat rather than fixed.

Question 12

(a) M1 gross photosynthesis = net (apparent) photosynthesis + respiration = 6.0 + 2.0
(a) A1 8.0 units per hour cao
(a) Answer: 8.0 units per hour
(b) B1 the light intensity at which the rate of gross (true) photosynthesis equals the rate of respiration
(b) B1 so there is no net exchange of gas (O2 or CO2) between the plant and its environment (net photosynthesis = 0)
(b) Answer: The light intensity at which gross photosynthesis equals respiration rate, so net gas exchange is zero.
(c) B1 the rate of respiration would exceed the rate of (gross) photosynthesis
(c) B1 so the plant would have a net loss of organic material/mass (using up stored carbohydrate reserves), and could eventually die if this continued
(c) Answer: Respiration would exceed photosynthesis, causing a net loss of biomass over time.
(d) B1 carbon dioxide concentration is no longer the limiting factor once photosynthesis rate plateaus
(d) B1 light intensity or temperature (whichever is now in shortest supply) has become the new limiting factor, so the rate cannot increase further until that factor is increased
(d) Answer: CO2 stops being limiting; another factor (light intensity or temperature) becomes limiting instead.

Question 13

Level 3 (5-6): A detailed, well-linked discussion explains how both methods increase energy-transfer efficiency and gives a balanced evaluation of ethical and environmental issues, using accurate scientific terminology throughout.
Level 2 (3-4): Explains how one or both methods increase efficiency, with some evaluation of ethical or environmental issues; mostly clear but links between ideas may be incomplete.
Level 1 (1-2): Makes basic, largely unlinked statements about efficiency or about ethical/environmental issues, with limited detail or scientific reasoning.
Level 0 (0): No relevant content, or the response does not address the question.
Indicative content:
- Keeping cattle indoors in a temperature-controlled barn reduces energy lost as heat to the environment (less energy needed for thermoregulation), so a greater proportion of the energy from food is available for growth/production of biomass
- Restricting movement reduces the energy used in respiration for muscle contraction and other activity, again increasing the proportion of ingested energy that can be converted into biomass (meat)
- Growth-promoting hormones can increase the rate of protein synthesis and growth rate, increasing biomass produced per unit of food consumed (improved feed conversion efficiency)
- Selective breeding for fast growth rate or good feed conversion could also be mentioned as a further method to increase efficiency
- Ethical issue: restricting movement and housing cattle in small pens may compromise animal welfare by preventing natural behaviour, potentially causing stress, injury or disease
- Ethical/health issue: use of hormones is controversial and banned in some regions (e.g. the EU) due to concerns over animal welfare and possible effects on human health from residues in meat
- Environmental issue: intensive indoor housing concentrates large volumes of waste (slurry), which can pollute watercourses (eutrophication) if not managed carefully
- Environmental issue: heating/maintaining barns and producing concentrated feed requires additional energy input and resources, which may increase the overall carbon footprint of production despite improved biological efficiency
- A balanced conclusion should weigh the gain in energy-transfer efficiency (and associated commercial benefit) against the animal welfare and environmental costs of intensive methods

Energy Transfers in and between Organisms - Worksheets, Questions and Revision