AQA A-level Biology · 3.5

Energy transfers in and between organisms

50 revision questions with answers for 3.5 Energy transfers in and between organisms in AQA A-level Biology (spec 7402). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official AQA material.

All 50 questions and answers

Question 1 · Spec 3.5.1 · Chloroplast organisation · Recall
Where do the light-dependent reaction and Calvin cycle occur, and how is the chloroplast adapted?
Answer:
The light-dependent reaction occurs on thylakoid membranes, which carry pigments, electron carriers and ATP synthase. The thylakoid lumen allows proton accumulation. The Calvin cycle occurs in the stroma, which contains enzymes including rubisco.
Question 2 · Spec 3.5.1 · Photoionisation · Recall
What happens when chlorophyll absorbs light?
Answer:
Light energy raises electrons to a higher energy level. They leave chlorophyll in photoionisation and pass to electron acceptors. Their energy contributes to ATP production and reduction of NADP. Electrons from photolysis replace those lost from chlorophyll.
Question 3 · Spec 3.5.1 · Photolysis · Recall
What are the products of photolysis and how are they used?
Answer:
Light-driven splitting of water produces electrons, protons and oxygen. Electrons replace those lost by chlorophyll. Protons contribute to the proton gradient and reduction of NADP. Oxygen is a by-product that may leave the leaf or be used in respiration.
Question 4 · Spec 3.5.1 · Photophosphoryl ation · Recall
How does electron transfer lead to ATP production in chloroplasts?
Answer:
Electrons pass along membrane carriers and release energy that drives proton accumulation in the thylakoid lumen. Protons then move down their electrochemical gradient through ATP synthase into the stroma. ATP synthase couples this flow to ATP formation from ADP and phosphate.
Question 5 · Spec 3.5.1 · Reduced NADP · Recall
What is the role of reduced NADP in photosynthesis?
Answer:
NADP accepts electrons and hydrogen to become reduced NADP during the light-dependent reaction. It provides reducing power for conversion of glycerate 3-phosphate to triose phosphate in the Calvin cycle. Oxidised NADP returns to accept more electrons and hydrogen.
Question 6 · Spec 3.5.1 · Carbon fixation · Recall
Describe carbon dioxide fixation in the Calvin cycle.
Answer:
Rubisco catalyses the reaction of carbon dioxide with the five-carbon compound ribulose bisphosphate, RuBP. An unstable six-carbon intermediate forms and immediately splits into two molecules of the three-carbon compound glycerate 3-phosphate, GP.
Question 7 · Spec 3.5.1 · GP reduction · Recall
How is GP converted to triose phosphate?
Answer:
ATP supplies energy and reduced NADP supplies reducing power to convert GP into triose phosphate, TP. ADP, inorganic phosphate and NADP return to the light-dependent reaction to be reused.
Question 8 · Spec 3.5.1 · Triose phosphate fate · Recall
What happens to TP produced in the Calvin cycle?
Answer:
Most TP is used to regenerate RuBP, requiring ATP, so carbon fixation can continue. Some TP leaves the cycle to form useful organic compounds, including sugars, starch, cellulose and precursors of lipids and amino acids.
Question 9 · Spec 3.5.1 · Limiting factors · Application
What is a limiting factor, and which factors can limit photosynthesis?
Answer:
A limiting factor is the factor currently restricting the rate. Light intensity, carbon dioxide concentration and temperature can limit photosynthesis. Increasing one factor increases rate only while it is limiting. At high temperatures, enzyme disruption or stomatal closure may reduce the rate.
Question 10 · Spec 3.5.1 · Carbon dioxide changes · Application
Predict the immediate effects of lowering carbon dioxide concentration on GP and RuBP in an illuminated leaf.
Answer:
Less carbon dioxide is fixed, so less GP is formed. RuBP initially accumulates because less is consumed while regeneration continues. State these as immediate changes, because longer-term adjustments can alter the concentrations of several intermediates.
Question 11 · Spec 3.5.1 · Light changes · Application
Predict immediate changes in GP and TP when a photosynthesising leaf is moved into darkness.
Answer:
ATP and reduced NADP production fall. Conversion of GP to TP slows, so GP initially accumulates and TP falls. RuBP regeneration also slows. The Calvin cycle depends indirectly on light because it needs products of the light-dependent reaction.
Question 12 · Spec 3.5.1, PS 2.1 · Measuring photosynthesis · Analysis
Why can oxygen measurements underestimate gross photosynthesis?
Answer:
Respiration consumes some of the oxygen produced, so measured oxygen release is a net rate. Keep temperature and other factors controlled. Gas volume is generally more reliable than bubble counts because bubble sizes vary. Account for background changes using suitable controls.
Question 13 · Spec 3.5.1 · Agricultural practice · Analysis
How should the use of lighting, heating and carbon dioxide enrichment in greenhouses be evaluated?
Answer:
Each can increase photosynthesis when it removes a limiting factor. Consider interactions, crop yield, energy and carbon dioxide costs, environmental effects and diminishing returns. An increase in photosynthetic rate does not by itself establish that a practice is economically worthwhile.
Question 14 · Spec 3.5.1, RP7 · Required practical 7 · Practical
How are leaf pigments separated by chromatography?
Answer:
Extract pigments with a suitable solvent. Place a concentrated spot on a pencil baseline and keep it above the solvent level. Let solvent rise through the paper or plate, then mark the solvent front promptly. Pigments move different distances according to their interactions with both phases.
Question 15 · Spec 3.5.1, RP7, MS 0.3 · Required practical 7 · Calculation
A pigment moves 36 mm and the solvent front moves 60 mm. Calculate its Rf value.
Answer:
Rf = distance moved by pigment/distance moved by solvent front = 36/60 = 0.60. It has no units. Measure both distances from the baseline and compare Rf values only under the same solvent and stationary-phase conditions.
Question 16 · Spec 3.5.1, RP7 · Required practical 7 · Analysis
How can chromatograms from shade and sun leaves be compared fairly?
Answer:
Use comparable tissue masses, extraction volumes and spotting procedures, with the same solvent and stationary phase. Compare band positions, colours and intensities cautiously. A missing visible band may indicate a low concentration rather than complete absence of that pigment.
Question 17 · Spec 3.5.1, RP8 · Required practical 8 · Practical
Why is DCPIP useful for investigating dehydrogenase activity in chloroplast extracts?
Answer:
DCPIP is an artificial electron acceptor. Its oxidised form is blue and its reduced form is colourless. Light-driven electron transfer in chloroplast extracts reduces it. A fall in absorbance can therefore estimate activity under controlled assay conditions.
Question 18 · Spec 3.5.1, RP8 · Required practical 8 · Practical
How could light intensity effects on chloroplast extracts be investigated?
Answer:
Use equal volumes and concentrations of fresh extract and DCPIP at several light intensities. Control temperature, pH and total volume. Include dark and boiled-extract controls, repeat treatments and measure absorbance over time. Use an appropriate blank for the extract's background colour.
Question 19 · Spec 3.5.1, RP8, MS 3.5 · Required practical 8 · Calculation
Absorbance falls from 0.80 to 0.32 in 120 s. Calculate the mean reduction rate and state a limitation.
Answer:
Rate of decrease = (0.80 - 0.32)/120 = 0.0040 absorbance units s⁻¹. This is an average over the interval. If the curve is not linear, a tangent near the start gives a better estimate of initial rate.
Question 20 · Spec 3.5.2 · Glycolysis · Recall
Where does glycolysis occur and what are its main stages?
Answer:
Glycolysis occurs in the cytoplasm and does not require oxygen. ATP phosphorylates glucose. The six-carbon compound splits into two triose phosphates. These are oxidised to pyruvate, producing reduced NAD and ATP by substrate-level phosphorylation.
Question 21 · Spec 3.5.2 · Glycolysis yields · Recall
What is the net yield of glycolysis per glucose molecule?
Answer:
Two pyruvate, two reduced NAD and a net gain of two ATP. Four ATP are produced but two were used in the initial phosphorylation steps. Reduced NAD carries electrons and hydrogen to later reactions or is reoxidised during fermentation.
Question 22 · Spec 3.5.2 · Anaerobic pathways · Application
Why must reduced NAD be reoxidised during anaerobic respiration?
Answer:
Oxidised NAD is needed to accept hydrogen during glycolysis. When oxidative phosphorylation cannot reoxidise enough NAD, transferring hydrogen from reduced NAD to pyruvate or its derivative regenerates NAD. This allows glycolysis and its limited ATP production to continue.
Question 23 · Spec 3.5.2 · Fermentation products · Recall
Compare the fates of pyruvate in lactate and ethanol fermentation.
Answer:
In lactate fermentation, pyruvate is reduced to lactate. In ethanol fermentation, pyruvate loses carbon dioxide and is then reduced to ethanol. Both regenerate oxidised NAD from reduced NAD and yield no additional ATP beyond glycolysis.
Question 24 · Spec 3.5.2 · Link reaction · Recall
Describe the link reaction in aerobic respiration.
Answer:
Pyruvate enters the mitochondrial matrix by active transport. It is decarboxylated and oxidised, producing carbon dioxide and reduced NAD. The resulting two-carbon acetate combines with coenzyme A to form acetylcoenzyme A.
Question 25 · Spec 3.5.2 · Krebs cycle · Recall
Describe the main events of the Krebs cycle.
Answer:
Acetylcoenzyme A combines with a four-carbon compound to form a six-carbon compound, releasing coenzyme A. Oxidation and decarboxylation reactions release carbon dioxide, reduce NAD and FAD, and generate ATP by substrate-level phosphorylation. The four-carbon compound is regenerated.
Question 26 · Spec 3.5.2, MS 0.3 · Per-glucose accounting · Calculation
Why must link-reaction and Krebs-cycle yields be doubled when accounting for one glucose molecule?
Answer:
Glycolysis produces two pyruvate per glucose, so the link reaction occurs twice and the Krebs cycle turns twice. For example, the link reactions release two carbon dioxide molecules and the Krebs cycles release four, giving six overall.
Question 27 · Spec 3.5.2 · Oxidative phosphorylation · Recall
How is energy from reduced coenzymes used to make ATP?
Answer:
Reduced NAD and reduced FAD donate electrons to carriers on the inner mitochondrial membrane. Electron transfer releases energy used to move protons from the matrix into the intermembrane space. Proton movement back through ATP synthase drives ATP production.
Question 28 · Spec 3.5.2 · Oxygen in respiration · Application
Why is oxygen essential for continued aerobic respiration?
Answer:
Oxygen is the final electron acceptor in the mitochondrial electron transfer chain. It combines with electrons and protons to form water. Without it, electron transfer stops, reduced coenzymes cannot be reoxidised efficiently, and the link reaction and Krebs cycle cannot continue normally.
Question 29 · Spec 3.5.1, 3.5.2 · Chemiosmosis · Synoptic
What is shared by ATP production in mitochondria and chloroplasts?
Answer:
Both use membrane electron transfer to establish a proton electrochemical gradient. Protons diffuse down that gradient through ATP synthase, which catalyses ATP formation. The energy source and the compartments receiving pumped protons differ between the organelles.
Question 30 · Spec 3.5.2 · Other respiratory substrates · Application
How can lipids and proteins serve as respiratory substrates?
Answer:
Glycerol can enter glycolysis, while fatty acid breakdown produces acetylcoenzyme A for the Krebs cycle. Amino acids can be deaminated and their carbon skeletons enter respiratory pathways. These routes allow different stored or dietary molecules to contribute to ATP production.
Question 31 · Spec 3.5.2, 3.2.1.1 · Mitochondrial adaptations · Application
How does mitochondrial structure support high rates of aerobic respiration?
Answer:
Cristae provide a large area for electron carriers and ATP synthase. The inner membrane maintains the proton gradient. The matrix contains enzymes for the link reaction and Krebs cycle. Cells with high sustained ATP demand often contain many mitochondria.
Question 32 · Spec 3.5.2 · Energy transfer comparison · Recall
Compare substrate-level phosphorylation and oxidative phosphorylation.
Answer:
Substrate-level phosphorylation transfers a phosphate directly from an intermediate to ADP, producing ATP in glycolysis and the Krebs cycle. Oxidative phosphorylation uses energy from electron transfer to establish a proton gradient that drives ATP synthase. Most aerobic ATP is produced by the latter.
Question 33 · Spec 3.5.2, RP9 · Required practical 9 · Practical
How could temperature effects on yeast respiration be investigated?
Answer:
Use equal yeast concentrations, substrate concentrations and volumes at several temperatures. Allow equilibration, then measure carbon dioxide production over a fixed interval using suitable gas-collection equipment. Repeat and calculate mean rates. State whether conditions are intended to be aerobic or anaerobic.
Question 34 · Spec 3.5.2, RP9 · Required practical 9 · Analysis
What controls and limitations matter in a yeast respiration investigation?
Answer:
Control pH, substrate supply, culture age, yeast concentration, oxygen availability and measurement time. Check for leaks and allow equilibration. A no-yeast control tests non-biological gas changes. Carbon dioxide solubility changes with temperature, so gas release may not exactly equal production.
Question 35 · Spec 3.5.2, MS 0.5, MS 2.5 · Microbial growth · Calculation
A culture starts with 200 cells and doubles every 30 minutes. How many cells are expected after 3 hours?
Answer:
There are 180/30 = 6 doublings. Expected number = 200 x 2⁶ = 12,800 cells. This assumes a constant doubling time with no resource limitation or cell loss. Logarithmic axes can help display growth across several orders of magnitude.
Question 36 · Spec 3.5.3 · Primary production · Recall
Distinguish gross primary production, net primary production and productivity.
Answer:
GPP is the chemical energy incorporated into plant biomass over a stated interval. NPP is GPP minus respiratory losses, R. Productivity is production per unit time, usually expressed per unit area or volume. NPP supports plant growth, reproduction and other trophic levels.
Question 37 · Spec 3.5.3, MS 2.3 · Primary production calculation · Calculation
GPP is 24,000 kJ m⁻² year⁻¹ and plant respiratory losses are 9000 kJ m⁻² year⁻¹. Calculate NPP.
Answer:
NPP = GPP - R = 24,000 - 9000 = 15,000 kJ m⁻² year⁻¹. The percentage of GPP retained as NPP is (15,000/24,000) x 100 = 62.5%. Keep the area and time basis consistent.
Question 38 · Spec 3.5.3, MS 2.3 · Consumer production · Calculation
A consumer ingests 800 kJ, loses 250 kJ in faeces and urine, and loses 400 kJ in respiration. Calculate net production.
Answer:
N = I - (F + R) = 800 - (250 + 400) = 150 kJ. This energy is available for growth and reproduction and can contribute to the next trophic level if the consumer is eaten.
Question 39 · Spec 3.5.3, MS 0.3 · Transfer efficiency · Calculation
A trophic level produces 1200 kJ m⁻² year⁻¹ from 15,000 kJ m⁻² year⁻¹ at the preceding level. Calculate transfer efficiency.
Answer:
Efficiency = (production at the receiving level/production at the preceding level) x 100. Here, (1200/15,000) x 100 = 8.0%. Use equivalent areas and time periods and specify which energy quantities are being compared.
Question 40 · Spec 3.5.3 · Energy losses · Application
Why is less energy available to each successive trophic level?
Answer:
Some biomass is not eaten. Some ingested material is not digested and leaves in faeces, while energy-containing compounds can be lost in urine. Respiration transfers energy to the environment as heat. Only net production becomes new consumer biomass.
Question 41 · Spec 3.5.3 · Biomass and calorimetry · Practical
Why measure dry biomass, and how can its energy content be estimated?
Answer:
Dry mass removes variation caused by water content. A known dry mass can be burned to heat a measured water mass. Use E = mc deltaT, with an appropriate specific heat capacity, then divide by sample mass. Heat loss and incomplete combustion cause underestimation.
Question 42 · Spec 3.5.3 · Farming efficiency · Analysis
How can farming increase the proportion of production available to humans?
Answer:
Removing competing organisms or pests can simplify food webs. Keeping suitable animals warm or limiting unnecessary movement can reduce respiratory energy use. Evaluate gains alongside welfare, cost, disease risk and environmental effects rather than assuming maximum production is the only aim.
Question 43 · Spec 3.5.4 · Saprobionts · Recall
How do saprobionts contribute to nutrient recycling?
Answer:
They secrete enzymes onto dead organic material and waste. Extracellular digestion produces soluble molecules that they absorb. Their metabolism releases inorganic nutrients, making chemical elements available for uptake by other organisms and linking decomposition to nutrient cycles.
Question 44 · Spec 3.5.4 · Nitrogen fixation · Recall
What is nitrogen fixation and why is it important?
Answer:
Nitrogen-fixing prokaryotes convert atmospheric nitrogen into ammonia or ammonium compounds that can enter biological molecules. Some live freely and others associate with plant roots. Plants cannot directly use atmospheric nitrogen to make proteins or nucleic acids.
Question 45 · Spec 3.5.4 · Ammonification · Recall
How does nitrogen in dead material become available in soil?
Answer:
Saprobionts digest nitrogen-containing compounds such as proteins. Deamination and related processes release ammonia, which forms ammonium ions in soil. This ammonification supplies ammonium for plant uptake or nitrification.
Question 46 · Spec 3.5.4 · Nitrification and denitrification · Recall
Compare nitrification and denitrification.
Answer:
Nitrification is aerobic oxidation of ammonium ions to nitrite and then nitrate by bacteria. Denitrification reduces nitrate to nitrogen gas under oxygen-poor conditions. Nitrification increases nitrate availability, whereas denitrification returns nitrogen to the atmosphere.
Question 47 · Spec 3.5.4 · Phosphorus cycle · Recall
Outline the phosphorus cycle.
Answer:
Weathering releases phosphate ions from rocks. Plants absorb phosphate and incorporate it into biological molecules. Feeding transfers phosphorus between organisms. Decomposition releases phosphate from waste and dead material. Some phosphate is lost to sediments and can later form rock.
Question 48 · Spec 3.5.4 · Mycorrhizae · Application
How do mycorrhizal associations benefit plants and fungi?
Answer:
Fungal hyphae associated with roots increase the effective area and volume of soil explored, improving uptake of water and inorganic ions such as phosphate. The fungus receives organic compounds from the plant. Both partners can benefit from this association.
Question 49 · Spec 3.5.4 · Fertilisers · Application
Why are fertilisers needed in agriculture, and how do natural and artificial fertilisers differ?
Answer:
Harvesting and removing livestock export nutrients from an ecosystem. Natural fertilisers supply organic material that decomposers break down, while artificial fertilisers supply selected mineral ions more directly. Both can replace nitrate and phosphate, but excessive application can cause environmental damage.
Question 50 · Spec 3.5.4 · Eutrophication · Application
Explain how fertiliser leaching can reduce aquatic biodiversity.
Answer:
Rain carries soluble nutrients into water. Increased nutrients stimulate algal growth. Shading reduces photosynthesis of submerged plants, which die. Saprobionts decompose dead biomass and respire, consuming dissolved oxygen. Oxygen-sensitive organisms may die, reducing biodiversity.

← 3.4 Genetic information, variation and relationships between organisms 3.6 Organisms respond to changes in their internal and external environments →

Energy transfers in and between organisms

All 50 questions and answers

More AQA A-level Biology topics

Cookies on Knowledge Noodles

Cookie settings