AQA A-level Biology · 3.3

Organisms exchange substances with their environment

Revision questions with answers for 3.3 Organisms exchange substances with their environment 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.

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Question 1 · Spec 3.3.1 · Surface area to volume · Application
How does increasing size affect surface area to volume ratio and the need for transport systems?
Answer:
For similarly shaped organisms, volume increases faster than surface area. A smaller surface area to volume ratio and longer diffusion distances make diffusion alone less able to meet metabolic demand. Specialised exchange surfaces and mass transport systems maintain supply and removal.
Question 2 · Spec 3.3.1, MS 4.1 · Ratio calculation · Calculation
Calculate surface area to volume ratio for cubes of side 1 cm and 3 cm.
Answer:
For a cube, surface area = 6l² and volume = l³. The 1 cm cube has 6/1 = 6 cm⁻¹, or 6:1 using cm² and cm³. The 3 cm cube has 54/27 = 2 cm⁻¹, or 2:1.
Question 3 · Spec 3.3.1, 3.3.2 · Exchange surfaces · Recall
Which features enable rapid exchange, and how can a single-celled organism exchange gases?
Answer:
A large surface area, short diffusion distance and steep concentration gradient increase diffusion rate. A single-celled organism has a high surface area to volume ratio and a short pathway, allowing oxygen and carbon dioxide to diffuse across its cell surface.
Question 4 · Spec 3.3.2 · Insect gas exchange · Recall
How does the insect tracheal system deliver oxygen to cells?
Answer:
Air enters through spiracles and passes along tracheae into fine tracheoles close to respiring cells. Oxygen diffuses down its concentration gradient into tissues. Body movements can ventilate larger insects. Gases travel directly to tissues through this system.
Question 5 · Spec 3.3.2 · Insects and water loss · Application
How do terrestrial insects balance gas exchange with conserving water?
Answer:
Their relatively impermeable body covering limits evaporation. Spiracles can close to reduce water loss, although closure restricts gas exchange. Internal tracheal exchange surfaces reduce exposure to dry air. Opening patterns and ventilation vary with metabolic demand.
Question 6 · Spec 3.3.2 · Fish gills · Application
Explain how gill structure and counter-current flow maximise oxygen uptake.
Answer:
Filaments carry many lamellae, providing a large area with thin diffusion barriers and a blood supply. Water and blood flow in opposite directions. Water has a higher oxygen concentration than adjacent blood throughout the lamella, maintaining diffusion along its length.
Question 7 · Spec 3.3.2 · Leaf gas exchange · Application
How is a dicotyledonous leaf adapted for gas exchange?
Answer:
A thin leaf gives a short pathway. Stomata allow gases to move between the air and internal spaces. Air spaces connect to moist mesophyll cell surfaces with a large area. Photosynthesis and respiration help maintain concentration gradients.
Question 8 · Spec 3.3.2 · Xerophytes · Application
How do xerophytic features reduce water loss, and what trade-off can result?
Answer:
A thick cuticle, reduced leaf area, rolled leaves, hairs and sunken stomata can reduce evaporation or trap humid air. This reduces the water-vapour gradient. Features that restrict gas movement may also limit carbon dioxide entry and photosynthesis.
Question 9 · Spec 3.3.2 · Human gas exchange system · Recall
Trace air from the trachea to the alveoli.
Answer:
Air passes through the trachea, bronchi and progressively smaller bronchioles to alveoli. The lungs contain many alveoli, providing a large total exchange surface. Ventilation renews alveolar air while blood flow renews blood at the surface.
Question 10 · Spec 3.3.2 · Alveoli · Application
Why are alveoli effective exchange surfaces?
Answer:
Numerous alveoli provide a large area. Thin alveolar epithelium and capillary endothelium create a short diffusion pathway. Ventilation and a rich capillary supply maintain steep oxygen and carbon dioxide gradients. Their moist surface allows gases to dissolve.
Question 11 · Spec 3.3.2 · Inspiration · Recall
How do muscles produce inspiration?
Answer:
The diaphragm contracts and flattens. External intercostal muscles contract, moving ribs upwards and outwards. Thoracic and lung volume increase, lowering alveolar pressure below atmospheric pressure. Air moves into the lungs down the pressure gradient.
Question 12 · Spec 3.3.2 · Expiration · Recall
How do quiet and forced expiration occur?
Answer:
During quiet expiration, the diaphragm and external intercostal muscles relax and elastic recoil reduces lung volume. Pressure rises and air leaves. During forced expiration, internal intercostal and abdominal muscle contractions further reduce thoracic volume.
Question 13 · Spec 3.3.2, MS 2.2 · Pulmonary ventilation · Calculation
Tidal volume is 0.55 dm³ and breathing rate is 16 min⁻¹. Calculate pulmonary ventilation rate.
Answer:
Pulmonary ventilation rate = tidal volume x breathing rate = 0.55 x 16 = 8.8 dm³ min⁻¹. This is total air moved per minute. State the volume and time units and use consistent units throughout.
Question 14 · Spec 3.3.2 · Lung disease data · Analysis
How can lung disease reduce gas exchange, and how should risk-factor data be evaluated?
Answer:
A disease may reduce exchange area, increase diffusion distance or impair ventilation. Use the information given to explain the mechanism. When evaluating smoking or pollution data, examine sample size, confounding variables and study design. Correlation alone does not establish causation.
Question 15 · Spec 3.3.3 · Carbohydrate digestion · Recall
How do amylases contribute to carbohydrate digestion?
Answer:
Amylases hydrolyse glycosidic bonds in starch, producing maltose and other shorter carbohydrates. Digestion converts large molecules into smaller products that can ultimately cross the epithelial membranes of the small intestine.
Question 16 · Spec 3.3.3 · Membrane-boun d digestion · Recall
What is the role of membrane-bound disaccharidases?
Answer:
Enzymes on the ileum epithelial membrane hydrolyse disaccharides to monosaccharides. Maltase produces glucose from maltose, sucrase produces glucose and fructose from sucrose, and lactase produces glucose and galactose from lactose.
Question 17 · Spec 3.3.3 · Protein digestion · Recall
Compare endopeptidases, exopeptidases and dipeptidases.
Answer:
Endopeptidases hydrolyse peptide bonds within a polypeptide, creating shorter chains and more ends. Exopeptidases remove amino acids from chain ends. Membrane-bound dipeptidases hydrolyse dipeptides to amino acids that can be absorbed.
Question 18 · Spec 3.3.3 · Lipase and bile · Application
How do bile salts assist lipid digestion by lipase?
Answer:
Bile salts emulsify large lipid droplets into smaller droplets, increasing the surface area accessible to lipase. Lipase hydrolyses ester bonds, producing fatty acids and monoglycerides. Bile also helps provide suitable alkaline conditions in the small intestine.
Question 19 · Spec 3.3.3 · Micelles · Recall
How do micelles assist the absorption of lipid digestion products?
Answer:
Bile salts form micelles containing fatty acids and monoglycerides. Micelles deliver these products close to the epithelial surface, where they leave the micelles and diffuse across the cell membrane. The products are rebuilt into triglycerides within epithelial cells.
Question 20 · Spec 3.3.3, 3.2.3 · Ileum absorption · Application
How are monosaccharides and amino acids absorbed, and why are ileum cells well adapted?
Answer:
Sodium-linked co-transport uses a sodium gradient maintained by ATP-driven pumps. Products then pass towards the blood through suitable transport proteins. Villi and microvilli increase area, a thin epithelium shortens the pathway, and a blood supply maintains gradients.
Question 21 · Spec 3.3.4.1 · Haemoglobin · Recall
How does haemoglobin's structure enable oxygen transport?
Answer:
Haemoglobin has four polypeptide subunits, each associated with a haem group containing an iron ion. Each haem can bind one oxygen molecule reversibly. Red blood cells contain much haemoglobin, enabling substantially more oxygen transport than plasma alone.
Question 22 · Spec 3.3.4.1 · Cooperative binding · Application
Why is the oxyhaemoglobin dissociation curve sigmoid?
Answer:
The first oxygen binds relatively reluctantly. Its binding changes haemoglobin's shape, increasing affinity for further oxygen molecules. This cooperative binding gives a steep middle section, allowing substantial oxygen unloading over the partial-pressure range found in respiring tissues.
Question 23 · Spec 3.3.4.1 · Bohr effect · Application
How does increased carbon dioxide concentration affect oxygen unloading?
Answer:
Increased carbon dioxide lowers haemoglobin's affinity for oxygen, shifting the dissociation curve to the right. At a given oxygen partial pressure, saturation is lower. More oxygen is released in actively respiring tissues, where carbon dioxide production is high.
Question 24 · Spec 3.3.4.1 · Haemoglobin adaptations · Analysis
What does a left-shifted dissociation curve indicate, and why might it benefit an animal?
Answer:
It indicates higher oxygen affinity. The haemoglobin reaches a higher saturation at a given low oxygen partial pressure, supporting oxygen loading in an oxygen-poor environment. The benefit must be considered alongside its reduced tendency to unload oxygen.
Question 25 · Spec 3.3.4.1 · Mammalian circulation · Recall
Trace a red blood cell from the body through the heart and lungs and back to the body.
Answer:
Vena cava to right atrium, right ventricle, pulmonary artery, lungs, pulmonary vein, left atrium, left ventricle and aorta. Coronary arteries supply heart muscle. Renal arteries supply kidneys and renal veins return blood from them.
Question 26 · Spec 3.3.4.1 · Heart structure · Application
How do heart walls, the septum and valves support circulation?
Answer:
Atrial walls move blood into ventricles. Ventricular walls pump blood out of the heart, with the thicker left wall generating higher systemic pressure. The septum separates oxygenated and deoxygenated blood. Valves prevent backflow when pressure gradients reverse.
Question 27 · Spec 3.3.4.1 · Cardiac cycle · Recall
Describe pressure changes during atrial systole, ventricular systole and diastole.
Answer:
Atrial systole raises atrial pressure and completes ventricular filling. Ventricular systole raises ventricular pressure, closing atrioventricular valves and then opening semilunar valves. During ventricular diastole, pressure falls, semilunar valves close and filling resumes when atrioventricular valves open.
Question 28 · Spec 3.3.4.1 · Valve movements · Application
What causes heart valves to open and close?
Answer:
They respond passively to pressure differences. Atrioventricular valves open when atrial pressure exceeds ventricular pressure and close when the reverse occurs. Semilunar valves open when ventricular pressure exceeds arterial pressure and close when arterial pressure becomes greater.
Question 29 · Spec 3.3.4.1, MS 2.3 · Cardiac data · Calculation
How can stroke volume be obtained from a ventricular volume-time graph?
Answer:
Subtract end-systolic volume from end-diastolic volume. If maximum ventricular volume is 125 cm³ and minimum is 55 cm³, stroke volume is 70 cm³. Relate falling volume to ventricular ejection and rising volume to filling.
Question 30 · Spec 3.3.4.1, MS 2.2 · Cardiac output · Calculation
Heart rate is 72 min⁻¹ and stroke volume is 70 cm³. Calculate cardiac output.
Answer:
Cardiac output = heart rate x stroke volume = 72 x 70 = 5040 cm³ min⁻¹ = 5.04 dm³ min⁻¹. Rearrange the equation if heart rate or stroke volume is the unknown.
Question 31 · Spec 3.3.4.1 · Arteries · Application
How is an artery's structure suited to its function?
Answer:
A thick wall withstands high pressure. Elastic tissue stretches during systole and recoils during diastole, helping maintain pressure and smooth blood flow. Smooth muscle can alter lumen diameter. A smooth endothelial lining limits resistance to flow.
Question 32 · Spec 3.3.4.1 · Arterioles and veins · Recall
Compare the functions and structures of arterioles and veins.
Answer:
Arteriolar smooth muscle controls resistance and blood distribution to capillary beds. Veins return blood at lower pressure, with a relatively large lumen and thinner walls. Valves and surrounding skeletal muscle action help maintain venous return.
Question 33 · Spec 3.3.4.1 · Capillaries · Application
How are capillaries adapted for exchange with tissues?
Answer:
Their walls consist of a single endothelial layer, giving a short diffusion pathway. Numerous capillaries provide a large exchange area. Narrow lumens place blood close to the wall, and extensive capillary beds allow exchange with nearby tissue cells.
Question 34 · Spec 3.3.4.1 · Tissue fluid formation · Recall
How is tissue fluid formed at the arteriole end of a capillary bed?
Answer:
Relatively high capillary hydrostatic pressure drives water and small dissolved substances out through the capillary wall. Blood cells and most plasma proteins remain in the blood. Tissue fluid brings substances such as oxygen and glucose close to cells.
Question 35 · Spec 3.3.4.1 · Tissue fluid return · Recall
How does tissue fluid return to the circulatory system?
Answer:
Capillary hydrostatic pressure falls along the bed. Retained plasma proteins lower the blood's water potential, favouring water movement back into capillaries by osmosis in the conventional model. Excess tissue fluid enters lymph vessels and is returned to the blood.
Question 36 · Spec 3.3.4.1 · Cardiovascular evidence · Analysis
What makes evidence for a cardiovascular risk factor more persuasive?
Answer:
Look for a clear association in a large, representative sample, a dose-response pattern, control of confounding variables and a plausible biological mechanism. Compare absolute as well as relative risk and evaluate conflicting studies before drawing a causal conclusion.
Question 37 · Spec 3.3.4.1, RP5 · Required practical 5 · Practical
What could a heart dissection reveal, and how would you obtain reliable observations?
Answer:
Identify atria, ventricles, septum, major vessels and valves. Compare ventricular wall thickness at equivalent positions. Trace vessel connections carefully and make a labelled observational drawing. Use a dissecting board and direct sharp instruments away from hands.
Question 38 · Spec 3.3.2, RP5 · Required practical 5 · Practical
How could dissection of a fish head demonstrate adaptations of a gas exchange system?
Answer:
Expose the gills, identify arches and filaments, and examine a filament under magnification to observe lamellae. Relate numerous lamellae and thin surfaces to exchange. A dissection shows structure; counter-current flow requires physiological evidence as well.
Question 39 · Spec 3.3.4.2 · Xylem · Application
Which xylem features support water transport?
Answer:
Vessel elements form long, hollow tubes with little internal resistance. Lignified walls strengthen them against collapse under tension. Pits allow lateral water movement. Xylem carries water and dissolved inorganic ions through roots, stems and leaves.
Question 40 · Spec 3.3.4.2 · Cohesion-tensio n · Recall
Explain the cohesion-tension theory of xylem transport.
Answer:
Evaporation from mesophyll lowers leaf water potential. Water moves from xylem into leaf cells, creating tension in the xylem. Cohesion between water molecules transmits this pull down a continuous column, drawing water from roots. Transpiration supplies the driving force.
Question 41 · Spec 3.3.4.2 · Transpiration factors · Application
How do light, humidity, temperature and air movement affect transpiration?
Answer:
Light often opens stomata. Lower humidity steepens the water-vapour gradient. Higher temperature increases evaporation and diffusion. Air movement removes humid air near the leaf. Responses depend on stomatal behaviour and water availability, so controlled comparisons are needed.
Question 42 · Spec 3.3.4.2, AT b · Potometer method · Practical
How is a potometer used, and what are its main limitations?
Answer:
Cut and assemble the shoot under water, make joints airtight and track an air bubble in a capillary over time. Control other factors and allow acclimatisation. It measures water uptake, which estimates transpiration but also includes water retained or used by the plant.
Question 43 · Spec 3.3.4.2, MS 3.5 · Potometer calculation · Calculation
A bubble moves 40 mm in 120 s in a tube with cross-sectional area 0.50 mm². Calculate uptake rate.
Answer:
Volume taken up = area x distance = 0.50 x 40 = 20 mm³. Rate = 20/120 = 0.167 mm³ s⁻¹. If comparing different shoots, expressing uptake per unit leaf area improves the comparison.
Question 44 · Spec 3.3.4.2 · Phloem · Recall
What does phloem transport, and which cells are involved?
Answer:
Phloem translocates organic solutes, especially sucrose, from sources to sinks. Sieve tube elements form transport pathways with sieve plates. Companion cells support sieve tube elements and provide energy for processes such as solute loading.
Question 45 · Spec 3.3.4.2 · Source loading · Recall
How does sucrose loading create pressure at a phloem source?
Answer:
Sucrose is loaded into sieve tubes, often using active transport supported by companion cells. This lowers sieve-tube water potential. Water enters from nearby xylem by osmosis, increasing hydrostatic pressure at the source.
Question 46 · Spec 3.3.4.2 · Mass flow · Recall
Explain movement from a phloem source to a sink.
Answer:
Unloading sucrose at a sink raises sieve-tube water potential and allows water to leave, lowering hydrostatic pressure. The source-to-sink pressure gradient drives mass flow through sieve tubes. Different tubes can transport in different directions according to source-sink relationships.
Question 47 · Spec 3.3.4.2 · Tracer evidence · Analysis
How can a radioactive carbon tracer investigate translocation?
Answer:
Provide a leaf with labelled carbon dioxide so photosynthesis incorporates the label into organic compounds. Detect label at later times in phloem and sinks. Its distribution and movement support transport of photosynthetic products from sources towards sinks.
Question 48 · Spec 3.3.4.2 · Ringing evidence · Analysis
What does a ringing experiment show about phloem transport?
Answer:
Removing a ring of bark interrupts phloem while leaving xylem functioning initially. Sugars accumulate above the ring and roots may receive less. This supports phloem transport of organic solutes. Damage and effects on other tissues are limitations to consider.
Question 49 · Spec 3.3.4.2 · Testing mass flow · Analysis
Which observations support the mass-flow hypothesis, and what would need explaining?
Answer:
Pressure differences between sources and sinks and movement of labelled solutes support bulk flow. Active loading requires ATP. Different transport directions can occur in separate tubes. Evidence about speeds, pressures and sieve-plate resistance is needed to test whether the proposed gradient explains observed movement.
Question 50 · Spec 3.3.4.1, 3.3.4.2 · Synoptic link · Synoptic
How do mammalian and plant transport systems maintain diffusion or water-potential gradients?
Answer:
Blood flow replaces oxygen and removes carbon dioxide at tissues, sustaining diffusion. Xylem flow supplies water lost through transpiration. Phloem loading and unloading establish water-potential and pressure differences. Each system links bulk movement with exchange across membranes at its endpoints.

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