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A-Level · Biology

AB3 Organisms Exchange Substances with their Environment

AQA 7402 · Calculator allowed · about 190 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A biologist models organisms as simple cubes to investigate how size affects the surface area to volume (SA:V) ratio available for exchange with the environment. Cube P has a side length of 1 mm (representing a single-celled organism). Cube Q has a side length of 10 mm (representing a small multicellular organism).

(a)Calculate the surface area and the volume of cube P, and use these to give its surface area to volume ratio in its simplest form.(2)

(b)Calculate the surface area to volume ratio of cube Q, giving your answer as a ratio in its simplest form.(2)

(c)Using your answers to parts (a) and (b), explain why single-celled organisms such as cube P can rely on diffusion alone across their body surface, whereas an organism the size of cube Q generally needs a specialised gas exchange surface and a mass transport system.(3)

(d)State two features that a specialised gas exchange surface must have in order to maximise the rate of diffusion across it.(2)

(Total for Question 1 is 9 marks)

A student investigated the rate of diffusion using agar cubes containing phenolphthalein indicator, which is pink in alkaline conditions. A cube of side length 2 cm was placed in a beaker of dilute hydrochloric acid. Acid diffuses in from all six faces, and the agar loses its pink colour as it is neutralised. The cube took 320 seconds to become completely colourless throughout.

(a)Define the term diffusion.(1)

(b)Calculate the mean rate of diffusion of the acid into the cube, in mm per second. Show your working.(3)

(c)State three factors, other than distance, that affect the rate of diffusion across a surface.(3)

(d)The agar cube models diffusion into a living cell. Suggest one limitation of using an agar cube as a model for diffusion into a real cell.(2)

(Total for Question 2 is 9 marks)

Fish gills use a countercurrent system to maximise oxygen uptake from water. Water flows over the gill lamellae in the opposite direction to the flow of blood inside them. The table shows the percentage oxygen saturation of the water and of the blood at four equally spaced positions (1 to 4) along the length of a lamella, in the direction of water flow.
Position along lamella: 1, 2, 3, 4
Oxygen saturation of water (%): 90, 70, 50, 30
Oxygen saturation of blood (%): 80, 60, 40, 20

(a)Name the structures on a gill filament, referred to in the table above, across which gas exchange takes place.(1)

(b(i))Use the data in the table to explain how the countercurrent arrangement maintains a diffusion gradient for oxygen along the whole length of the lamella.(3)

(b(ii))Calculate the difference in oxygen saturation between the water and the blood at each of the four positions, and state what this shows about the diffusion gradient along the lamella.(2)

(c)In a hypothetical parallel-flow (concurrent) system, blood and water would flow in the same direction and their oxygen saturations would tend to equalise part-way along the lamella. Explain why the countercurrent system found in fish gills results in a greater overall uptake of oxygen by the blood than a parallel-flow system would.(3)

(Total for Question 3 is 9 marks)

Insects exchange gases using a tracheal system consisting of tracheae and tracheoles that carry air directly to respiring tissues, largely relying on diffusion rather than a mass transport system for gases.

(a)Name the pores in the exoskeleton through which air enters the tracheal system, and name the structures that control their opening and closing.(2)

(b)Suggest why the reliance on diffusion through the tracheal system, without a mass transport system to carry oxygen around the body, limits the maximum body size of insects.(3)

(c)During flight, an insect's flight muscles respire anaerobically for short periods, producing lactic acid. Explain how this can temporarily increase the rate of gas exchange at the ends of the tracheoles.(4)

(d)Some larger, more active insects use rhythmic abdominal pumping movements to ventilate the tracheal system. Suggest why this mass movement of air is necessary for these insects, in addition to diffusion.(2)

(Total for Question 4 is 11 marks)

Gas exchange in a leaf occurs mainly through pores in the lower epidermis. A student compared the leaves of a mesophyte (a plant adapted to average water availability) and a xerophyte (a plant adapted to a dry habitat) using a light microscope with a circular field of view of diameter 0.5 mm. The xerophyte leaf had a stomatal density of 50 stomata per mm², and the mesophyte leaf had a stomatal density of 200 stomata per mm^2.

(a)Name the pores described above, and name the specialised cells that control their opening and closing.(2)

(b)Explain, in terms of water potential and ion movement, how guard cells open a stoma in the light.(4)

(c)Calculate the number of stomata expected in one full field of view for the xerophyte leaf and for the mesophyte leaf. Use area of a circle = π x r², and π = 3.142. Give each answer to the nearest whole stoma.(4)

(d)State two structural adaptations, other than a lower stomatal density, that a xerophyte leaf may have to reduce water loss by transpiration.(2)

(e)Evaluate the extent to which a lower stomatal density is an effective adaptation for a xerophyte, considering both its benefit in a dry habitat and any disadvantage it may cause.(3)

(Total for Question 5 is 15 marks)

This question is based on the required practical involving the use of qualitative reagents to identify biological molecules produced by digestion, and relates to their subsequent absorption in the small intestine.

(a)(i) State the reagent(s) used, and the observation that indicates a positive result, in the Benedict's test for a reducing sugar. (ii) State the reagent used, and the observation that indicates a positive result, in the Biuret test for protein.(4)

(b)Suggest how the Benedict's test could be adapted, using a colorimeter, to estimate the concentration of reducing sugar in a food sample, rather than simply showing that reducing sugar is present.(3)

(c)Describe how glucose, a product of carbohydrate digestion, is absorbed across the epithelial cells lining the ileum into the blood.(4)

(d)Villi increase the surface area of the ileum for absorption by a factor of about 600 compared with a smooth-walled tube, and microvilli on each epithelial cell increase this further by a factor of about 20. Calculate the overall factor by which the total surface area for absorption is increased, compared with a smooth-walled tube.(2)

(Total for Question 6 is 13 marks)

The data below show how the percentage oxygen saturation of haemoglobin changes with the partial pressure of oxygen (pO2), for maternal haemoglobin, for fetal haemoglobin, and for maternal haemoglobin exposed to a raised carbon dioxide concentration (a Bohr shift), as found in actively respiring muscle tissue.
Resting curves (pO2 in kPa: 2, 4, 6, 8, 10, 12):
Maternal haemoglobin, % saturation: 20, 55, 80, 92, 96, 98
Fetal haemoglobin, % saturation: 35, 68, 88, 95, 97, 99
Bohr-shifted curve, representing blood near actively respiring muscle (pO2 in kPa: 2, 4, 6, 8, 10, 12):
Maternal haemoglobin, % saturation: 10, 40, 65, 80, 90, 95

(a)Define what is meant by the partial pressure of oxygen.(1)

(b)Use the resting maternal haemoglobin data to state the percentage saturation of maternal haemoglobin at a pO2 of 4 kPa.(1)

(c)Calculate the difference in percentage saturation between fetal haemoglobin and maternal haemoglobin (both resting curves) at a pO2 of 4 kPa.(1)

(d)Explain the significance of fetal haemoglobin having a higher affinity for oxygen than maternal haemoglobin, in terms of the transfer of oxygen across the placenta.(3)

(e)Explain why it is advantageous for maternal haemoglobin's affinity for oxygen to be reduced (a Bohr shift) in actively respiring muscle tissue during exercise, compared with resting muscle.(4)

(f)Using the resting maternal curve for blood leaving the lungs (pO2 = 12 kPa) and the Bohr-shifted curve for blood at the respiring muscle (pO2 = 4 kPa), calculate the percentage of oxygen unloaded to the muscle.(2)

(Total for Question 7 is 12 marks)

The heart pumps blood around the body in a rhythmic cardiac cycle.

(a)Name the blood vessel that carries oxygenated blood away from the left ventricle to the rest of the body.(1)

(b)Describe one structural difference between the wall of the left ventricle and the wall of the right ventricle, and explain the functional significance of this difference.(4)

(c)During ventricular systole, the atrioventricular (bicuspid and tricuspid) valves close. Explain the role of these valves at this point in the cardiac cycle.(3)

(d)A patient at rest has a heart rate of 72 beats per minute and a stroke volume of 70 cm^3. Calculate their cardiac output in dm³ per minute.(3)

(e)During exercise, the same patient's cardiac output increases to 20 dm³ per minute and their stroke volume increases to 120 cm^3. Calculate their heart rate during this exercise, to the nearest whole beat per minute.(3)

(Total for Question 8 is 14 marks)

Blood vessels are structurally adapted to their function.

(a)Name the blood vessel best matching each description. (i) Thick wall containing a relatively large amount of smooth muscle and elastic tissue; carries blood at high pressure away from the heart. (ii) Contains valves at intervals to prevent the backflow of blood; wide lumen and relatively thin wall; carries blood at low pressure back to the heart. (iii) Wall only one cell (endothelium) thick; narrow lumen, often only just wide enough for red blood cells to pass through in single file.(3)

(b)Explain why arteries have walls containing a thick layer of smooth muscle and elastic fibres.(3)

(c)Explain why the wall of a capillary is only one cell thick.(2)

(d)State one feature of capillaries, other than the thickness of their wall, that adapts them for efficient exchange of substances with tissue cells.(1)

(Total for Question 9 is 9 marks)

This question is based on the required practical investigating the effect of an environmental factor on the rate of transpiration using a potometer. A leafy shoot was set up in a potometer, and the shoot was left to acclimatise before an air bubble was introduced into the capillary tube.

(a)Describe how the potometer should be set up and used to ensure a valid measurement of water uptake, referring to the introduction of air bubbles and the sealing of the apparatus.(3)

(b)An air bubble took 4 minutes 10 seconds to travel 8.5 cm along the capillary tube, which had an internal diameter of 1 mm. Calculate the mean rate of water uptake, in mm³ per minute, giving your answer to 3 significant figures. Use volume of a cylinder = π x r² x length, with π = 3.142.(4)

(c)State two environmental variables, other than the one being investigated, that should be controlled during this investigation.(2)

(d)Explain why the rate of water uptake measured by a potometer is assumed to be approximately equal to the rate of transpiration, even though a potometer actually measures water uptake rather than water loss directly.(2)

(e)The investigation was repeated with a fan providing air movement across the shoot; under these conditions, the bubble travelled 15 cm in the same time (4 minutes 10 seconds). Calculate the percentage increase in the rate of water uptake caused by the air movement, compared with your answer to part (b).(3)

(Total for Question 10 is 14 marks)

Water moves from the roots to the leaves of a tall tree through xylem vessels, against the force of gravity, without the use of metabolic energy by the xylem itself.

(a)State two structural features of xylem vessels that adapt them for the transport of water.(2)

(b)Explain, using the cohesion-tension theory, how water is transported through the xylem from the roots to the top of a tall tree, without the use of metabolic energy.(6)

(c)Suggest why an air bubble (embolism) that becomes lodged in a xylem vessel would prevent water transport through that particular vessel.(2)

(Total for Question 11 is 10 marks)

Phloem tissue transports the products of photosynthesis (mainly sucrose) from sources to sinks. The mass flow hypothesis is the accepted explanation for translocation in phloem.

(a)Name the cells found alongside sieve tube elements that carry out the active loading of sucrose into the phloem, and name the type of transport process used to load sucrose into the sieve tubes at the source.(2)

(b)Explain, using the mass flow hypothesis, how a hydrostatic pressure gradient is generated between a source and a sink, resulting in the mass flow of phloem sap through the sieve tubes.(5)

(c)In a 'ringing experiment', a complete ring of bark (containing the phloem but not the xylem) is removed from around a tree trunk. Predict and explain what would happen, over the following weeks, to (i) the region of bark just above the ring and (ii) the tissue below the ring.(3)

(d)State one piece of experimental evidence that supports the mass flow hypothesis, and one observation that is more difficult for the mass flow hypothesis to fully explain.(4)

(Total for Question 12 is 14 marks)

This final question draws together ideas about exchange surfaces and transport systems from across different organisms.

(a)Compare the surface area to volume ratio of a small, single-celled organism such as an amoeba with that of a large multicellular organism such as a mammal, and explain how each is adapted to obtain sufficient oxygen for its metabolic needs, given the different SA:V ratio of each.(6)

(b)A student claims: 'All exchange surfaces need a large surface area, a thin diffusion distance, and a means of maintaining a diffusion gradient.' Evaluate this claim by giving one similarity and one difference between the gas exchange surface in fish gills, the gas exchange surface in insect tracheae, and the absorptive surface of the small intestine villi.(4)

(Total for Question 13 is 10 marks)

Mark scheme · AB3 Organisms Exchange Substances with their Environment

Question 1

(a) M1 surface area = 6 x (1 mm)² = 6 mm² and volume = (1 mm)³ = 1 mm³
(a) A1 SA:V ratio = 6:1 cao
(a) Answer: SA = 6 mm², V = 1 mm³, SA:V = 6:1.
(b) M1 surface area = 6 x (10 mm)² = 600 mm² and volume = (10 mm)³ = 1000 mm³
(b) A1 SA:V ratio = 0.6:1 (accept 3:5) cao
(b) Answer: SA = 600 mm², V = 1000 mm³, SA:V = 0.6:1 (3:5).
(c) M1 cube P has a much higher SA:V ratio (6:1) than cube Q (0.6:1), so cube P has a large surface area relative to its (metabolically active) volume oe
(c) M1 this means diffusion across the whole body surface of cube P can supply oxygen to, and remove waste from, every part of its volume quickly enough to meet its (relatively low) demand oe
(c) A1 cube Q has a much lower SA:V ratio, so its outer surface area is too small relative to its volume/metabolic demand for diffusion alone to supply the cells at its centre quickly enough, so a specialised exchange surface and mass transport system are needed to reduce diffusion distances oe
(c) Answer: Larger organisms have a much lower SA:V ratio, so diffusion across the body surface alone cannot meet metabolic demand; specialised surfaces and mass transport overcome this.
(d) B1 a large surface area (relative to the volume of the organism) oe
(d) B1 any one other valid feature: thin/short diffusion pathway (often one cell thick); permeable to the diffusing substance; good blood supply or ventilation to maintain a steep concentration gradient oe
(d) Answer: Large surface area; and one of thin/short diffusion pathway, permeable, or good blood supply/ventilation maintaining a concentration gradient.

Question 2

(a) B1 the net movement of particles (molecules or ions) from a region of higher concentration to a region of lower concentration, down a concentration gradient, resulting from the random motion of particles oe
(a) Answer: Net movement of particles from high to low concentration (down a concentration gradient) due to random motion.
(b) M1 acid diffuses inward from all 6 faces, so the greatest distance travelled to the centre of the cube = half the side length = 10 mm oe
(b) M1 rate = distance / time = 10 / 320
(b) A1 = 0.031 mm/s (2 s.f.) awrt 0.031
(b) Answer: 0.031 mm/s (2 s.f.).
(c) B1 temperature (higher temperature increases rate, as particles have more kinetic energy)
(c) B1 concentration gradient/difference in concentration (a steeper gradient increases rate)
(c) B1 surface area (a larger surface area increases rate); allow thickness of the membrane/barrier (thinner increases rate)
(c) Answer: Any three of: temperature, concentration gradient, surface area, thickness of membrane.
(d) M1 agar has a uniform composition throughout with no cell membrane/selectively permeable barrier, whereas a real cell membrane controls what enters the cell (e.g. via channel/carrier proteins) oe
(d) A1 so the agar model cannot show the effect of active transport, facilitated diffusion, or membrane selectivity on the movement of substances into a real cell oe
(d) Answer: Agar has no selectively permeable membrane, so it cannot model membrane-controlled processes such as facilitated diffusion or active transport.

Question 3

(a) B1 (gill) lamellae
(a) Answer: Lamellae.
(b(i)) M1 blood flows in the opposite direction to the water, so at every position along the lamella, the water has a higher oxygen saturation than the blood flowing alongside it oe
(b(i)) M1 for example, at position 4 the water is 30% saturated while the blood entering (before picking up oxygen) is only 20%; at position 1 the water is 90% saturated while the blood (about to leave, now highly oxygenated) is 80% oe
(b(i)) A1 because the oxygen saturation of the water always exceeds that of the blood next to it, a diffusion gradient for oxygen (from water to blood) is maintained along the entire length of the lamella, allowing oxygen to keep diffusing into the blood throughout its passage oe
(b(i)) Answer: Because blood and water flow in opposite directions, water is always more saturated than the adjacent blood, maintaining a gradient along the whole lamella.
(b(ii)) M1 differences = 90-80, 70-60, 50-40, 30-20 = 10, 10, 10, 10 (%)
(b(ii)) A1 the difference (gradient) is the same (10%) at every position, showing that a diffusion gradient for oxygen is maintained along the entire length of the lamella, rather than only at one end oe
(b(ii)) Answer: Difference = 10 percentage points at every position; shows a constant gradient is maintained along the whole lamella.
(c) M1 in a parallel-flow system, oxygen would diffuse from water to blood until their saturations became equal (an equilibrium), at which point no further net diffusion would occur, even though the blood may still be far from fully saturated oe
(c) M1 in the countercurrent system, blood and water saturations never equalise; the blood is always exposed to water with a higher saturation than itself as it passes along the lamella oe
(c) A1 so a diffusion gradient for oxygen is maintained along the entire length of the lamella in a countercurrent system, allowing the blood to become much more highly saturated with oxygen than would be possible with a parallel-flow arrangement oe
(c) Answer: Countercurrent flow avoids the equalisation that limits parallel flow, maintaining a gradient along the whole lamella so blood leaves much more oxygenated.

Question 4

(a) B1 spiracles
(a) B1 (valves formed by) sphincter/muscular valves
(a) Answer: Spiracles; controlled by sphincter muscles (valves).
(b) M1 the tracheal system relies on diffusion (and some mass flow of air) along the tracheae/tracheoles to deliver oxygen directly to tissues, without oxygen being carried by a circulatory fluid oe
(b) M1 diffusion is only efficient over short distances, so as body size (and the diffusion distance from spiracle to the furthest tissues) increases, diffusion alone becomes too slow to supply oxygen fast enough to meet respiratory demand oe
(b) A1 therefore beyond a certain body size, tissues at the centre of a large insect's body would not receive oxygen quickly enough to support aerobic respiration, limiting the maximum size insects can reach oe
(b) Answer: Diffusion is only rapid over short distances, so beyond a limited body size tissues far from spiracles would not receive oxygen fast enough, limiting insect size.
(c) B1 lactic acid builds up in the flight muscle cells, lowering their water potential oe
(c) B1 water moves out of the fluid-filled ends of the tracheoles and into the muscle cells by osmosis, down the water potential gradient oe
(c) B1 this draws air further along the tracheoles, into the space previously occupied by fluid oe
(c) B1 increasing the surface area of tracheole exposed directly to air (rather than fluid) and shortening the diffusion pathway for oxygen, increasing the rate of gas exchange to the muscle oe
(c) Answer: Lactic acid lowers water potential in muscle cells, drawing fluid out of tracheole endings by osmosis, so air is drawn further in, increasing the gas exchange surface/rate.
(d) M1 larger/more active insects have a higher oxygen demand and/or longer diffusion pathways than diffusion alone can supply quickly enough oe
(d) A1 abdominal pumping movements (changing body volume/pressure) cause mass flow of air in and out of the tracheal system, delivering air more quickly over the longer distances involved than diffusion alone could achieve oe
(d) Answer: Diffusion alone is too slow over the longer distances/higher demand of active insects; pumping movements produce mass flow of air, speeding up gas delivery.

Question 5

(a) B1 stomata (singular: stoma)
(a) B1 guard cells
(a) Answer: Stomata; guard cells.
(b) B1 in the light, guard cells actively transport (pump) potassium ions (K+) into themselves from surrounding epidermal cells, using ATP oe
(b) B1 this lowers the water potential inside the guard cells oe
(b) B1 water moves into the guard cells by osmosis, down the water potential gradient, and the guard cells become turgid oe
(b) B1 guard cells have unevenly thickened cellulose walls (thicker on the inner wall next to the pore), so as they become turgid they bend/change shape and pull apart, opening the stoma oe
(b) Answer: Active K+ uptake lowers water potential, water enters by osmosis, guard cells become turgid and their uneven wall thickening makes them bend apart, opening the stoma.
(c) M1 radius of field of view = 0.5/2 = 0.25 mm; area = 3.142 x 0.25² = 0.196 mm² (3 s.f.)
(c) A1 xerophyte: 0.196 x 50 = 9.8, so 10 stomata (nearest whole number)
(c) M1 mesophyte: 0.196 x 200
(c) A1 = 39.3, so 39 stomata (nearest whole number) awrt 39
(c) Answer: Xerophyte: about 10 stomata per field of view. Mesophyte: about 39 stomata per field of view.
(d) B1 a thick waxy cuticle (reduces evaporation from the leaf surface)
(d) B1 any one other valid adaptation: sunken stomata/stomata in pits (trap humid air, reducing the water potential gradient); rolled leaves or leaf hairs (trap a layer of still, humid air near the stomata); reduced leaf surface area/needle-shaped leaves oe
(d) Answer: Thick waxy cuticle; and one of sunken stomata, rolled leaves/hairs, or reduced leaf surface area.
(e) M1 a lower stomatal density reduces the total area available for water vapour to diffuse out of the leaf, reducing the rate of transpiration/water loss, which is beneficial for surviving in a dry habitat with limited water availability oe
(e) M1 however, fewer stomata also reduces the total area available for carbon dioxide to diffuse into the leaf for photosynthesis oe
(e) A1 so while lower stomatal density conserves water, it may limit the rate of photosynthesis/growth compared with a plant with a higher stomatal density, showing a trade-off between water conservation and gas exchange for photosynthesis oe
(e) Answer: Fewer stomata conserve water (beneficial in drought) but also limit CO2 uptake for photosynthesis, showing a trade-off.

Question 6

(a) B1 (i) add Benedict's reagent and heat in a boiling water bath oe
(a) B1 (i) a positive result is a colour change from blue to a brick-red/orange precipitate oe
(a) B1 (ii) add Biuret reagent (sodium hydroxide followed by dilute copper sulfate, or premixed Biuret reagent) oe
(a) B1 (ii) a positive result is a colour change from blue to purple/lilac/violet oe
(a) Answer: Benedict's: heat with Benedict's reagent, brick-red precipitate = positive. Biuret: add Biuret reagent, purple/lilac colour = positive.
(b) B1 prepare a range of glucose (reducing sugar) standard solutions of known concentration, carry out the Benedict's test on each and filter to remove the precipitate, then measure the absorbance/percentage transmission of light through each using a colorimeter oe
(b) B1 plot a calibration curve of absorbance (or % transmission) against known glucose concentration oe
(b) B1 carry out the same Benedict's test and colorimeter reading on the unknown food sample, and use the calibration curve to read off its glucose concentration oe
(b) Answer: Use colorimeter readings from a range of known-concentration standards to build a calibration curve, then read the unknown sample's absorbance against it.
(c) B1 sodium ions are actively transported out of the epithelial cells into the blood (by the sodium-potassium pump), maintaining a low sodium ion concentration inside the epithelial cells oe
(c) B1 sodium ions diffuse from the lumen of the ileum into the epithelial cells down their concentration gradient, through a cotransporter protein in the cell surface membrane, carrying glucose into the cell at the same time (against its own concentration gradient) oe
(c) B1 glucose then moves out of the epithelial cell into the blood by facilitated diffusion, through a separate carrier protein, down its concentration gradient oe
(c) B1 the large surface area of the villi and microvilli maximises the number of transport proteins/rate of absorption oe
(c) Answer: Sodium is pumped out of epithelial cells, creating a gradient that drives sodium-glucose cotransport into the cell from the lumen; glucose then leaves into the blood by facilitated diffusion.
(d) M1 overall factor = 600 x 20
(d) A1 = 12000 (times) cao
(d) Answer: 12000 times.

Question 7

(a) B1 the pressure exerted by oxygen alone within a mixture of gases (or dissolved in a liquid), which is a measure of the concentration of oxygen present oe
(a) Answer: The pressure contributed by oxygen alone in a gas mixture/dissolved in a liquid; a measure of oxygen concentration.
(b) B1 55% cao
(b) Answer: 55%.
(c) B1 13 (percentage points) cao (68 - 55)
(c) Answer: 13 percentage points.
(d) B1 at any given pO2 (e.g. in the placenta), fetal haemoglobin has a higher percentage saturation than maternal haemoglobin, i.e. a higher affinity for oxygen oe
(d) B1 this means that where maternal blood and fetal blood are close together in the placenta, oxygen will tend to dissociate from (be released by) maternal haemoglobin oe
(d) B1 and associate with (be picked up by) fetal haemoglobin, allowing oxygen to be transferred efficiently from mother to fetus across the placenta oe
(d) Answer: Fetal haemoglobin's higher oxygen affinity means oxygen released by maternal haemoglobin is preferentially picked up by fetal haemoglobin, enabling efficient placental oxygen transfer.
(e) M1 actively respiring muscle produces more carbon dioxide, which lowers the pH of the blood (forms carbonic acid/releases H+ ions) oe
(e) M1 the raised carbon dioxide concentration/lowered pH causes haemoglobin to change shape slightly, reducing its affinity for oxygen (shifting the dissociation curve to the right) oe
(e) M1 this means that at the same (relatively low) pO2 found in respiring muscle, haemoglobin releases/unloads more oxygen than it would at rest oe
(e) A1 supplying the actively respiring muscle cells with more oxygen for (aerobic) respiration, to meet their increased oxygen demand during exercise oe
(e) Answer: More CO2/lower pH in exercising muscle reduces haemoglobin's oxygen affinity, so more oxygen is unloaded at the muscle's pO2, meeting increased demand.
(f) M1 % saturation leaving lungs (resting curve, 12 kPa) = 98%; % saturation at muscle (Bohr-shifted curve, 4 kPa) = 40%
(f) A1 oxygen unloaded = 98 - 40 = 58 (percentage points) cao
(f) Answer: 58 percentage points of oxygen saturation are unloaded to the muscle.

Question 8

(a) B1 aorta
(a) Answer: Aorta.
(b) B1 the left ventricle wall is thicker (has more cardiac muscle) than the right ventricle wall oe
(b) B1 the left ventricle must pump blood at high pressure all the way around the body (the systemic circulation), overcoming greater resistance, so needs a thicker, more powerful muscular wall oe
(b) B1 the right ventricle wall is thinner (less muscular) than the left oe
(b) B1 the right ventricle only needs to pump blood a short distance to the lungs (the pulmonary circulation) at lower pressure, to avoid damaging the delicate capillaries in the lungs oe
(b) Answer: Left ventricle wall is thicker to generate high pressure for the systemic circulation; right ventricle wall is thinner as it only pumps blood the short distance to the lungs at lower pressure.
(c) M1 as the ventricles contract (ventricular systole), pressure in the ventricles rises above the pressure in the atria oe
(c) M1 this pressure difference pushes the atrioventricular valves shut oe
(c) A1 preventing the backflow of blood from the ventricles into the atria, ensuring blood is forced out through the semilunar valves into the aorta/pulmonary artery instead, maintaining efficient one-way flow oe
(c) Answer: Rising ventricular pressure during systole closes the AV valves, preventing backflow into the atria and ensuring blood is forced forward into the arteries.
(d) M1 cardiac output = heart rate x stroke volume = 72 x 70
(d) M1 = 5040 cm³/min
(d) A1 = 5.04 dm³/min cao
(d) Answer: 5.04 dm³/min.
(e) M1 heart rate = cardiac output / stroke volume = 20000 cm³/min / 120 cm³
(e) dM1 = 166.67 (dependent on correct rearrangement and unit conversion of cardiac output to cm³/min)
(e) A1 = 167 beats per minute (nearest whole beat) awrt 167
(e) Answer: 167 beats per minute.

Question 9

(a) B1 (i) artery
(a) B1 (ii) vein
(a) B1 (iii) capillary
(a) Answer: (i) artery (ii) vein (iii) capillary.
(b) M1 arteries carry blood at high pressure, produced by the pumping (contraction) of the ventricles oe
(b) M1 elastic fibres allow the artery wall to stretch and then recoil as the pressure pulse from each heartbeat passes, helping to maintain/even out the high pressure between heartbeats oe
(b) A1 smooth muscle allows the diameter of the artery to be adjusted (constrict/dilate) to help control blood flow/pressure to different organs oe
(b) Answer: Elastic fibres stretch and recoil to maintain high pressure between heartbeats; smooth muscle allows the vessel diameter to be regulated.
(c) M1 a capillary wall consisting of a single layer of flattened endothelial cells provides the shortest possible diffusion pathway/distance oe
(c) A1 allowing substances (e.g. oxygen, glucose, carbon dioxide, waste products) to diffuse rapidly/efficiently between the blood and surrounding tissue cells oe
(c) Answer: A one-cell-thick wall gives the shortest diffusion distance, allowing rapid exchange of substances between blood and tissues.
(d) B1 a very narrow lumen/diameter, so red blood cells are squeezed close against the capillary wall, reducing the diffusion distance to tissue cells oe (allow: a large branching network/large total surface area; allow: small gaps/pores between cells allowing plasma to leak out)
(d) Answer: Narrow lumen, so red blood cells pass close to the wall, reducing diffusion distance (or another valid feature such as a large branching network).

Question 10

(a) B1 the shoot should be cut under water (at a slant), and the whole apparatus assembled/the shoot inserted under water, to prevent air entering the xylem and blocking water uptake oe
(a) B1 all joints (e.g. where the shoot enters the apparatus) should be sealed, for example with petroleum jelly, to ensure the apparatus is airtight/watertight oe
(a) B1 a single air bubble is introduced by briefly removing the end of the capillary tube from the beaker of water (or opening a tap), then returning the tube end to the water/closing the tap, and timing the position of the bubble as it moves along a scale oe
(a) Answer: Cut and assemble the shoot under water, seal joints (e.g. with petroleum jelly), then introduce a single air bubble and time its movement along the scale.
(b) M1 time = 4 + (10/60) = 4.167 minutes
(b) M1 radius = 1/2 = 0.5 mm; length = 8.5 cm = 85 mm; volume = 3.142 x 0.5² x 85
(b) M1 volume = 66.77 mm³ (3 s.f.)
(b) A1 rate = 66.77 / 4.167 = 16.0 mm³/min (3 s.f.) awrt 16.0
(b) Answer: 16.0 mm³ per minute (3 s.f.).
(c) B1 light intensity
(c) B1 any one other valid variable: humidity; air movement/wind speed; temperature (whichever is not the independent variable)
(c) Answer: Light intensity; and one of humidity, air movement, or temperature.
(d) M1 almost all of the water taken up by the shoot is eventually lost as water vapour by transpiration from the leaves, since only a very small proportion is used in processes such as photosynthesis or retained for cell turgor/growth oe
(d) A1 so the rate of water uptake by the shoot is assumed to be approximately equal to (a reasonable estimate of) the rate of water loss by transpiration oe
(d) Answer: Almost all water taken up is eventually transpired (only a tiny fraction is used elsewhere), so uptake rate closely approximates transpiration rate.
(e) M1 new volume = 3.142 x 0.5² x 150 = 117.8 mm³; new rate = 117.8 / 4.167 = 28.3 mm³/min (3 s.f.)
(e) dM1 percentage increase = ((28.3 - 16.0) / 16.0) x 100 (dependent on a correct rate from part (b) and the new rate calculated above)
(e) A1 = 76.9% (accept awrt 76 to 77%) ft from candidate's own value in part (b)
(e) Answer: About 77% increase (awrt 76 to 77%).

Question 11

(a) B1 the vessels are made of dead cells with no cytoplasm/organelles, forming a continuous, hollow tube with little resistance to flow oe
(a) B1 any one other valid feature: lignified walls (provide strength, preventing the vessel collapsing under tension/negative pressure); no end walls between cells (forming an unbroken, continuous column); pits in the walls (allow lateral movement of water between vessels) oe
(a) Answer: Hollow, dead cells forming a continuous tube; and one of lignified walls, absent end walls, or pits.
(b) L3 (5-6): A detailed, logically sequenced account linking transpiration, cohesion, tension and adhesion to explain the continuous movement of water from roots to leaves, using accurate scientific terminology throughout (e.g. water potential, hydrogen bonding, negative pressure/tension).
(b) L2 (3-4): A generally accurate account of the main stages of the cohesion-tension mechanism, but the explanation may be incomplete, may omit a stage (e.g. root uptake or adhesion), or the links between transpiration, tension and cohesion may not be fully explained. Terminology mostly appropriate.
(b) L1 (1-2): Basic or isolated statements are made about water movement in xylem (e.g. transpiration occurs, or water evaporates from leaves), with little reference to cohesion, tension or the pulling of a continuous water column. Limited use of scientific terminology.
(b) L0 (0): No relevant content is provided.
(b) Answer: Full marks require the transpiration-driven water potential gradient, cohesion pulling up a continuous unbroken column under tension, and adhesion to vessel walls, to be linked in a coherent account, judged holistically against the level descriptors.
(c) M1 an air bubble breaks the continuity of the water column in that vessel, since air, unlike water, cannot experience/transmit cohesive forces (hydrogen bonding) between water molecules oe
(c) A1 so the tension created by transpiration higher up the vessel cannot be transmitted down through the air bubble to pull water up from the roots, and water transport through that vessel stops (although water can often be diverted sideways into a neighbouring vessel through pits) oe
(c) Answer: An air bubble breaks the continuous water column, so cohesive tension cannot be transmitted through it, halting water transport in that vessel.

Question 12

(a) B1 companion cells
(a) B1 active transport, using a proton (H+) cotransport mechanism
(a) Answer: Companion cells; active transport (via a proton/H+ cotransport mechanism).
(b) B1 at the source (e.g. a photosynthesising leaf), sucrose is actively loaded into the companion cells and sieve tube elements, lowering their water potential oe
(b) B1 water moves into the sieve tube elements from the xylem/surrounding cells by osmosis, down the water potential gradient, increasing the hydrostatic (turgor) pressure inside the sieve tube at the source oe
(b) B1 at the sink (e.g. a growing root or storage organ), sucrose is unloaded/used up (e.g. converted to starch or used in respiration), raising the water potential of the sieve tube elements there oe
(b) B1 water therefore leaves the sieve tube elements by osmosis at the sink, lowering the hydrostatic pressure there oe
(b) B1 this difference in hydrostatic pressure (high at the source, low at the sink) causes phloem sap to flow (mass flow) along the sieve tubes from source to sink oe
(b) Answer: Active loading of sucrose at the source lowers water potential there, drawing water in and raising pressure; unloading at the sink has the opposite effect, so sap flows from the high-pressure source to the low-pressure sink.
(c) M1 (i) the bark just above the ring would swell, because sucrose (and water drawn in by osmosis) continues to arrive from the leaves/source but can no longer move past the ring, since the phloem has been removed, so it accumulates above the ring oe
(c) M1 (ii) the tissue and roots below the ring would gradually be starved of sucrose, although xylem, and therefore water/mineral transport, is unaffected since it is left intact oe
(c) A1 so over time the tissue below the ring would be unable to respire sufficiently and would eventually die, while the tree may survive short-term only on stored reserves below the ring oe
(c) Answer: Bark above the ring swells as sucrose accumulates; tissue below the ring is deprived of sucrose (though not water) and eventually dies, since only the phloem, not the xylem, has been removed.
(d) B1 supporting evidence: ringing experiments show that sugars accumulate above (and are depleted below) a ring where phloem has been removed, consistent with mass flow occurring only through intact phloem oe
(d) B1 this shows that sucrose moves through the phloem specifically, from source to sink, as the mass flow hypothesis predicts oe
(d) B1 difficult observation: sieve plates (with their pores partially blocked by phloem protein/cytoplasmic strands) would be expected to create considerable resistance to a purely passive mass flow of sap, yet measured flow rates are faster than a simple pressure gradient would be predicted to allow through such a resistant pathway oe
(d) B1 similarly, not all solutes appear to move at the same rate as would be predicted by a single, purely passive mass flow of the whole solution oe
(d) Answer: Supporting: sugar accumulation above a ring/depletion below shows source-to-sink movement via phloem. Against: sieve plates should slow mass flow more than observed, so a purely passive mechanism does not fully account for the measured flow rate.

Question 13

(a) L3 (5-6): A detailed, well-linked account that compares the SA:V ratios of the two organisms and explains, with specific reference to named adaptations (e.g. a specialised gas exchange surface and a mass transport system in the mammal), why each organism's strategy is sufficient for its own metabolic demand. Scientific terminology used accurately throughout.
(a) L2 (3-4): A generally correct comparison of SA:V ratio and its consequences, but the explanation of adaptations may be less detailed, may focus on only one organism, or the link between SA:V ratio and the need for adaptations may not be fully developed. Terminology mostly appropriate.
(a) L1 (1-2): Basic or isolated statements about size, surface area or diffusion are made, with little comparison between the two organisms and little or no reference to specific adaptations. Limited use of scientific terminology.
(a) L0 (0): No relevant content is provided.
(a) Answer: Full marks require an explicit comparison of SA:V ratio between the two organisms plus a reasoned link to why the mammal specifically needs both a specialised exchange surface and a mass transport system, judged holistically.
(b) B1 similarity: all three surfaces (gill lamellae, tracheole walls, and villi/microvilli) have a large surface area and a thin (often one cell thick) wall/layer to maximise the rate of diffusion, supporting the student's claim oe
(b) B1 this is because all rely, at least in part, on diffusion, whose rate depends on surface area and diffusion distance in the same way regardless of the organism/substance involved oe
(b) B1 difference: fish gills and villi both maintain a diffusion/concentration gradient using a mass flow of blood (countercurrent blood flow in gills; blood flow beneath the villi), whereas the tracheal system of insects delivers air (and maintains the gradient at the tracheole tips) mainly by diffusion and mass flow of air itself, without any blood/circulatory fluid carrying the respiratory gas oe
(b) B1 so while the student's claim holds well for surface area and thinness, the specific method of maintaining a gradient varies (mass transport of a circulatory fluid versus direct mass flow/diffusion of air), showing the claim is only partly complete without this detail oe
(b) Answer: Similarity: all three have a large surface area and thin diffusion pathway. Difference: gills and villi rely on blood flow to maintain the gradient, whereas insect tracheae rely on the direct mass flow/diffusion of air itself, without a circulatory fluid.

Organisms Exchange Substances with their Environment - Worksheets, Questions and Revision