Revision Library

revisionlibrary.co.uk

HIGHER

B2 Organisation

AQA 8461 · Calculator allowed · about 110 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A biology student is describing the structures found in a typical human cheek cell (an animal cell) as seen under a light microscope.

(a)State the function of the nucleus.(1)

(b)State the function of the mitochondria.(1)

(c)State the function of the cell membrane.(1)

(d)State the function of the ribosomes.(1)

(Total for Question 1 is 4 marks)

Cells become specialised (differentiated) to carry out a particular function.

(a)Explain two ways in which a sperm cell is adapted to its function of fertilising an egg cell.(2)

(b)Explain how a root hair cell is adapted to absorb water and mineral ions efficiently.(1)

(c)Explain how a red blood cell is adapted to carry oxygen efficiently.(1)

(d)Define the term 'differentiation'.(1)

(Total for Question 2 is 5 marks)

A student used a light microscope as part of the required practical to observe onion epidermal cells. She measured the width of one cell on the eyepiece grid as 45 mm. The actual width of the cell is 0.03 mm.

(a)Calculate the magnification used to view the cell. Show your working.(3)

(b)State one technique used to prepare a temporary mount of onion cells for viewing.(1)

(Total for Question 3 is 4 marks)

Mitosis is a form of cell division that is part of the cell cycle.

(a)Give two reasons why mitosis is important in a multicellular organism.(2)

(b)State how the number of chromosomes in each of the two daughter cells produced by mitosis compares with the number in the parent cell.(1)

(c)A skin cell contains 46 chromosomes before it undergoes mitosis. State the number of chromosomes in each daughter cell produced.(1)

(d)Explain why cancer can result from changes to cells that control the cell cycle.(1)

(Total for Question 4 is 5 marks)

Stem cells are undifferentiated cells that can divide to produce more cells like themselves, some of which can then become specialised.

(a)Define the term 'stem cell'.(1)

(b)Name one source of stem cells found in an adult human.(1)

(c)Name the structure from which embryonic stem cells are obtained.(1)

(d)Give one medical use of stem cells.(1)

(e)Discuss one ethical concern associated with using embryonic stem cells in medical research.(2)

(Total for Question 5 is 6 marks)

Enzymes are biological catalysts.

(a)Define the term 'catalyst'.(1)

(b)Using the lock and key model, explain why a particular enzyme will only catalyse a reaction with one specific substrate.(2)

(c)Name the type of large biological molecule from which enzymes are made.(1)

(Total for Question 6 is 4 marks)

A student carried out the required practical investigation into the effect of pH on the rate of reaction between amylase and starch. Method: (1) place a drop of iodine solution into each well of a spotting tile; (2) mix starch solution, buffer solution and amylase solution in a test tube and start a stop clock; (3) every 30 seconds, use a dropping pipette to transfer a sample of the mixture into a well of iodine solution and observe the colour; (4) repeat using buffer solutions of pH 5, 6, 7 and 8, keeping the volumes and concentrations of starch and amylase, and the temperature, the same each time. The time taken for the iodine to stay orange-brown (showing all the starch had been broken down) was: pH 5 = 210 s, pH 6 = 120 s, pH 7 = 90 s, pH 8 = 180 s.

(a)Identify the independent variable in this investigation.(1)

(b)Give two variables, other than pH, that should be controlled in this investigation to make it a fair test.(2)

(c)Calculate the rate of reaction at pH 7. Use the equation: rate = 1000 / time. Give your answer to 3 significant figures.(2)

(d)The rate of reaction was fastest at pH 7 but slower at pH 8. Explain, in terms of enzyme structure, why the rate decreased at pH 8.(2)

(Total for Question 7 is 7 marks)

In an investigation into an enzyme-controlled reaction, the rate of reaction was measured at temperatures between 0 degC and 70 degC. The enzyme's optimum temperature is 37 degC. Between 0 degC and 37 degC the rate of reaction increased steadily. Above 40 degC the rate fell rapidly, reaching zero by around 60 degC.

(a)Explain, in terms of particles and collisions, why the rate of reaction increased steadily between 0 degC and 37 degC.(3)

(b)Explain, in terms of enzyme structure, why the rate of reaction fell rapidly above 40 degC.(3)

(Total for Question 8 is 6 marks)

The human digestive system contains several organs, each with a specific role.

(a)Name the organ, other than the salivary glands, that produces amylase to digest starch.(1)

(b)Name the organ where most water is absorbed from undigested food.(1)

(c)Name the organ that produces hydrochloric acid, which kills bacteria and provides the optimum pH for protease enzymes to work.(1)

(d)State the role of the liver in the digestive system.(1)

(e)Name the organ where most digested food molecules are absorbed into the blood.(1)

(Total for Question 9 is 5 marks)

A student carried out the required practical to test an unknown food sample for the presence of different nutrients.

(a)Describe the test, and the positive result, for the presence of reducing sugars.(2)

(b)Describe the test, and the positive result, for the presence of starch.(2)

(c)Describe the test, and the positive result, for the presence of protein.(1)

(d)The student added ethanol, then water, to a sample of the food. A white/milky emulsion formed. State what this result indicates about the food sample.(1)

(Total for Question 10 is 6 marks)

Bile is produced by the liver and stored in the gall bladder before being released into the small intestine.

(a)State two functions of bile in digestion.(2)

(b)Explain why emulsifying fat increases the rate at which lipase digests it.(2)

(Total for Question 11 is 4 marks)

The wall of the small intestine is lined with villi. Explain how the small intestine is adapted for the efficient absorption of digested food molecules into the blood.

(Total for Question 12 is 6 marks)

The heart is a muscular organ that pumps blood around the body through a double circulatory system.

(a)Name the blood vessel that carries deoxygenated blood from the body into the right atrium.(1)

(b)Name the chamber of the heart that pumps oxygenated blood into the aorta.(1)

(c)Explain why the wall of the left ventricle is more muscular than the wall of the right ventricle.(2)

(d)Name the valves that prevent the backflow of blood from the ventricles into the atria.(1)

(e)Evaluate the advantage of a double circulatory system, compared with a single circulatory system, in supplying the body with oxygen.(1)

(Total for Question 13 is 6 marks)

Arteries, veins and capillaries are the three types of blood vessel in the human circulatory system.

(a)State one structural feature of arteries that relates to the high pressure of the blood flowing through them.(1)

(b)State one structural feature of veins that helps blood, which is at low pressure, return to the heart.(1)

(c)Explain why capillary walls are only one cell thick.(2)

(d)State one other feature of capillaries that adapts them for the efficient exchange of substances.(1)

(Total for Question 14 is 5 marks)

Blood is made up of plasma, red blood cells, white blood cells and platelets.

(a)State the function of platelets.(1)

(b)State one way in which red blood cells are adapted to carry oxygen.(1)

(c)State the function of white blood cells.(1)

(d)A person has 5 000 000 red blood cells per mm³ of blood, and a total blood volume of 5 litres. Calculate the total number of red blood cells in their body. (1 litre = 1 000 000 mm³)(2)

(Total for Question 15 is 5 marks)

Coronary heart disease occurs when the coronary arteries become narrowed by a build-up of fatty material within their walls.

(a)Explain why narrowed coronary arteries can lead to a heart attack.(2)

(b)Compare how a stent and a statin each reduce the risk of a heart attack in a patient with coronary heart disease.(3)

(Total for Question 16 is 5 marks)

Scientists studied 10 000 adults and found a positive correlation between the amount of alcohol consumed per week and the risk of developing liver disease.

(a)Define the term 'risk factor'.(1)

(b)Explain the difference between a correlation and a cause, using the alcohol and liver disease example.(2)

(c)Give one other lifestyle risk factor, other than alcohol consumption, and name a non-communicable disease it is linked to.(2)

(Total for Question 17 is 5 marks)

Cancer is caused by changes in cells that lead to uncontrolled growth and division.

(a)State two differences between a benign tumour and a malignant tumour.(2)

(b)Give two risk factors that increase a person's chance of developing cancer.(2)

(Total for Question 18 is 4 marks)

Plants have two types of transport tissue: xylem and phloem.

(a)State the direction in which water and mineral ions move through xylem tissue.(1)

(b)State the direction in which phloem transports dissolved sugars, and name this process.(1)

(c)Describe one structural adaptation of xylem cells that suits their function.(2)

(d)Name the type of specialised cell found alongside phloem sieve tube cells, and state its function.(1)

(Total for Question 19 is 5 marks)

A student used a potometer as part of the required practical to investigate the effect of moving air (from an electric fan) on the rate of transpiration from a leafy shoot. She measured the distance an air bubble moved along the capillary tube in a fixed time of 10 minutes. In still air, the bubble moved 24 mm. With the fan switched on, the bubble moved 60 mm.

(a)State why an air bubble is introduced into the capillary tube of a potometer.(1)

(b)Give two variables that should be controlled to make this a valid comparison.(2)

(c)Calculate the rate of water uptake, in mm per minute, in still air and with the fan switched on.(2)

(d)Using your answers to part c, explain the effect of moving air on the rate of transpiration in terms of water vapour molecules at the leaf surface.(1)

(Total for Question 20 is 6 marks)

A root hair cell can be modelled as a cuboid measuring 1 mm long, 0.01 mm wide and 0.01 mm high.

(a)Calculate the surface area to volume ratio of the root hair cell. Give your answer as a ratio in its simplest whole-number form. Show your working.(3)

(b)Explain why a large surface area to volume ratio is important for water uptake by root hair cells, and explain why mineral ions often need to be absorbed by active transport rather than diffusion.(2)

(Total for Question 21 is 5 marks)

Mark scheme · B2 Organisation

Question 1

(a) B1 controls the activity of the cell / contains genetic material (DNA) organised as chromosomes (oe)
(a) Answer: Controls the cell's activities and contains the genetic material.
(b) B1 site of (aerobic) respiration, releasing energy for the cell (oe)
(b) Answer: Site of aerobic respiration, releasing energy.
(c) B1 controls what substances enter and leave the cell (oe)
(c) Answer: Controls the movement of substances into and out of the cell.
(d) B1 site of protein synthesis (oe)
(d) Answer: Where proteins are made (protein synthesis).

Question 2

(a) B1 long tail (flagellum) so the cell can swim/move towards the egg cell (oe)
(a) B1 many mitochondria in the mid-section to release energy (from respiration) for swimming (oe)
(a) Answer: Long flagellum for swimming; many mitochondria to provide energy for movement.
(b) B1 long hair-like extension gives a large surface area (to volume ratio) for absorption (oe)
(b) Answer: It has a long extension giving a large surface area for absorption.
(c) B1 biconcave disc shape gives a large surface area for gas exchange OR no nucleus, leaving more room for haemoglobin (oe, any one)
(c) Answer: Biconcave shape (large surface area) or lack of nucleus (more room for haemoglobin).
(d) B1 the process by which a cell changes/becomes specialised to carry out a particular function (oe)
(d) Answer: The process in which a cell changes to become specialised for its function.

Question 3

(a) M1 correctly states or uses magnification = image size / actual size (oe)
(a) M1 correct substitution: 45 / 0.03
(a) A1 x1500 (cao, no units for magnification)
(a) Answer: x1500
(b) B1 e.g. adding a stain such as iodine to increase contrast, OR lowering the coverslip at an angle to avoid trapping air bubbles (oe, any one valid technique)
(b) Answer: Adding iodine stain to increase contrast (or lowering the coverslip at an angle to avoid air bubbles).

Question 4

(a) B1 growth (production of new cells to increase body size) (oe)
(a) B1 repair of damaged tissue (replacing damaged/dead cells) (oe)
(a) Answer: Growth and repair of tissues.
(b) B1 the same (number)/identical to the parent cell (oe)
(b) Answer: The same number of chromosomes as the parent cell.
(c) B1 46 (cao)
(c) Answer: 46 chromosomes.
(d) B1 cells divide (by mitosis) in an uncontrolled way, forming a mass of cells called a tumour (oe)
(d) Answer: Cells divide uncontrollably, forming a tumour.

Question 5

(a) B1 an undifferentiated cell that can divide (by mitosis) to produce more cells of the same type, which can then differentiate into different cell types (oe)
(a) Answer: An undifferentiated cell that can divide and become specialised.
(b) B1 bone marrow (oe, e.g. certain adult tissues)
(b) Answer: Bone marrow.
(c) B1 (early) embryo / blastocyst
(c) Answer: The embryo (blastocyst).
(d) B1 e.g. treating diabetes, replacing damaged tissue after paralysis/spinal injury, treating some blood disorders/leukaemia via bone marrow transplant (oe, any one valid use)
(d) Answer: For example, treating some forms of paralysis by replacing damaged nerve cells.
(e) B1 identifies a valid concern, e.g. the embryo is destroyed in the process, which some people believe is destroying a potential life (oe)
(e) B1 links the concern to a balanced consideration, e.g. this must be weighed against the potential medical benefits to patients, so views on its acceptability differ (oe)
(e) Answer: Using embryonic stem cells destroys the embryo, which some consider morally wrong; this must be weighed against the medical benefits.

Question 6

(a) B1 a substance that increases the rate of a reaction, without being used up/changed itself (oe)
(a) Answer: A substance that speeds up a reaction without being used up.
(b) B1 the enzyme has an active site with a shape that is complementary to (fits) the substrate (oe)
(b) B1 only a substrate of the correct/matching shape can bind to the active site (to form an enzyme-substrate complex), so other substrates will not fit/react (oe)
(b) Answer: Only a substrate whose shape fits the enzyme's active site can bind and react; other substrates do not fit.
(c) B1 protein
(c) Answer: Protein.

Question 7

(a) B1 pH (of the buffer solution)
(a) Answer: pH of the buffer solution.
(b) B1 volume/concentration of starch solution (oe)
(b) B1 volume/concentration of amylase solution, OR temperature, OR volume of iodine used (any one different valid variable, oe)
(b) Answer: For example: volume/concentration of starch solution; volume/concentration of amylase solution; temperature.
(c) M1 correct substitution: 1000 / 90
(c) A1 awrt 11.1 (units: s^-1 or 1/s, cao to 3 s.f.)
(c) Answer: 11.1 s^-1 (3 s.f.)
(d) B1 pH 7 is the optimum pH for amylase; at pH 8 the enzyme's active site starts to change shape (denature) (oe)
(d) B1 the substrate (starch) can no longer bind to the active site as well/at all, so fewer successful enzyme-substrate complexes form and the rate of reaction decreases (oe)
(d) Answer: At pH 8 the enzyme's active site begins to change shape (denature), so the substrate fits less well and the rate decreases.

Question 8

(a) B1 as temperature increases, enzyme and substrate particles gain (more) kinetic energy and move faster (oe)
(a) B1 particles collide more frequently (oe)
(a) B1 more successful collisions between enzyme active site and substrate (per unit time), forming more enzyme-substrate complexes, so the rate of reaction increases (oe)
(a) Answer: Increased kinetic energy causes more frequent, successful collisions between enzyme and substrate, increasing the rate of enzyme-substrate complex formation.
(b) B1 above the optimum temperature, the enzyme gains too much energy and its structure/active site starts to change shape (oe)
(b) B1 the enzyme is described as denatured; the active site is no longer complementary to the substrate (oe)
(b) B1 the substrate can no longer bind to the active site, so enzyme-substrate complexes cannot form and the reaction rate falls to zero (oe)
(b) Answer: Above the optimum temperature the enzyme denatures: its active site changes shape so the substrate no longer fits, and the rate falls to zero.

Question 9

(a) B1 pancreas
(a) Answer: The pancreas.
(b) B1 large intestine
(b) Answer: The large intestine.
(c) B1 stomach
(c) Answer: The stomach.
(d) B1 produces bile
(d) Answer: It produces bile.
(e) B1 small intestine
(e) Answer: The small intestine.

Question 10

(a) B1 add Benedict's solution and heat in a water bath (oe)
(a) B1 positive result: colour changes from blue to (brick-red/orange) precipitate (oe)
(a) Answer: Add Benedict's solution and heat; a positive test turns the solution from blue to brick-red.
(b) B1 add iodine solution (oe)
(b) B1 positive result: colour changes from orange-brown to blue-black (oe)
(b) Answer: Add iodine solution; a positive test turns the solution from orange-brown to blue-black.
(c) B1 biuret test: add sodium hydroxide solution then dilute copper sulfate solution; positive result = colour changes from blue to purple/lilac (oe)
(c) Answer: Biuret test: add sodium hydroxide then copper sulfate solution; positive test turns purple/lilac.
(d) B1 the food contains lipid/fat
(d) Answer: The food contains lipid (fat).

Question 11

(a) B1 neutralises stomach acid, making conditions in the small intestine alkaline (the optimum pH for protease enzymes) (oe)
(a) B1 emulsifies fats, breaking large fat droplets into smaller droplets (oe)
(a) Answer: Neutralises stomach acid; emulsifies fats.
(b) B1 emulsification breaks fat into smaller droplets, increasing the (total) surface area of fat (oe)
(b) B1 this allows lipase to come into contact with (bind to/react with) more fat molecules at once, increasing the rate of digestion (oe)
(b) Answer: Smaller droplets give a larger surface area, so lipase can act on more fat molecules at once, increasing the rate of reaction.

Question 12

Level 1 (1-2): Simple statement(s) about villi, e.g. that they increase surface area, with little or no linkage to the mechanism of absorption. Answer may be a list with limited scientific detail.
Level 2 (3-4): Some relevant structural adaptations of villi are described (e.g. large surface area, thin walls, good blood supply) with a partial explanation of how at least one adaptation increases the rate of absorption/diffusion.
Level 3 (5-6): A detailed and coherent explanation linking several structural adaptations of villi (large surface area, single layer of thin epithelial cells, dense capillary network maintaining a steep concentration gradient, and/or microvilli) to the mechanisms of absorption (diffusion and, where relevant, active transport), with clear logical reasoning throughout.
Indicative content:
- Villi greatly increase the surface area of the small intestine wall available for absorption
- The wall of each villus is only one cell thick (thin epithelium), reducing the diffusion distance between the intestine and the blood
- Villi have a dense network of capillaries, which maintains a steep concentration gradient (removing absorbed molecules quickly), increasing the rate of diffusion
- Villi have a lacteal (part of the lymphatic system) for the absorption of fatty acids and glycerol
- Microvilli on the epithelial cells further increase the surface area for absorption
- Some nutrients (e.g. some glucose and amino acids) are absorbed by active transport against a concentration gradient, using energy from respiration, so absorption can continue even when concentration inside the blood is higher than in the gut

Question 13

(a) B1 vena cava
(a) Answer: The vena cava.
(b) B1 left ventricle
(b) Answer: The left ventricle.
(c) B1 the left ventricle pumps blood around the whole body (systemic circulation), a much longer distance/at higher pressure, than the right ventricle, which only pumps blood to the (nearby) lungs (oe)
(c) B1 more muscle is needed to generate the greater force/pressure required (oe)
(c) Answer: The left ventricle pumps blood around the whole body at higher pressure, so it needs more muscle than the right ventricle, which only pumps blood to the lungs.
(d) B1 atrioventricular valves (bicuspid and tricuspid valves) (oe)
(d) Answer: The atrioventricular valves (bicuspid and tricuspid).
(e) B1 blood passes through the heart twice per full circuit, so its pressure/speed is boosted again after picking up oxygen at the lungs, meaning oxygenated blood reaches body cells faster/more efficiently than in a single circulatory system (oe)
(e) Answer: Blood is re-pressurised by the heart after leaving the lungs, so oxygenated blood reaches the body faster than in a single circulatory system.

Question 14

(a) B1 thick, muscular and/or elastic walls (oe)
(a) Answer: Thick, muscular, elastic walls.
(b) B1 valves (to prevent backflow of blood) (oe, wide lumen also accepted)
(b) Answer: Valves to prevent the backflow of blood.
(c) B1 reduces the diffusion distance between the blood and body cells/tissues (oe)
(c) B1 increases the rate of diffusion of substances (e.g. oxygen, glucose, carbon dioxide, waste products) between the blood and cells (oe)
(c) Answer: A thin wall reduces the diffusion distance, increasing the rate of exchange of substances between blood and cells.
(d) B1 e.g. very narrow (lumen), permeable walls, or a large total number/surface area of capillaries (oe, any one)
(d) Answer: Very narrow lumen, or permeable walls, or a large total surface area.

Question 15

(a) B1 help blood clot at a wound (oe)
(a) Answer: They help blood to clot at the site of a wound.
(b) B1 contain haemoglobin (which binds oxygen), OR biconcave disc shape (large surface area), OR no nucleus (more room for haemoglobin) (oe, any one)
(b) Answer: They contain haemoglobin, which binds to oxygen.
(c) B1 defend the body against disease/pathogens, e.g. engulfing pathogens (phagocytes) or producing antibodies (lymphocytes) (oe)
(c) Answer: Defend the body against disease, e.g. by engulfing pathogens or producing antibodies.
(d) M1 converts blood volume to mm³: 5 x 1 000 000 = 5 000 000 mm³, then multiplies by cell concentration (ft candidate's own correct method even if conversion figure differs)
(d) A1 2.5 x 10¹³ (accept 25 000 000 000 000 or 25 x 10¹², cao)
(d) Answer: 2.5 x 10¹³ red blood cells.

Question 16

(a) B1 narrowed arteries reduce blood flow (and the supply of oxygen and glucose) to the heart muscle (myocardium) (oe)
(a) B1 heart muscle cells become oxygen-starved and can die, so part of the heart may stop working properly (oe)
(a) Answer: Narrowing reduces oxygen/glucose supply to heart muscle, which can die, causing a heart attack.
(b) B1 a stent is a mesh tube inserted into the narrowed artery and expanded, physically holding the artery open and restoring blood flow immediately (oe)
(b) B1 a statin is a drug taken regularly that reduces the amount of LDL ('bad') cholesterol in the blood, slowing the rate at which fatty material builds up in the arteries (oe)
(b) B1 valid comparison point, e.g. a stent gives an immediate mechanical effect but involves surgery/risk, whereas a statin acts more slowly over time, requires long-term use and can have side effects (oe)
(b) Answer: A stent mechanically holds the artery open, giving immediate relief but with surgical risk; a statin lowers blood cholesterol over time, reducing further build-up but requiring ongoing use.

Question 17

(a) B1 something linked to an increased likelihood/chance that a person will develop a particular disease (oe)
(a) Answer: Something linked to an increased chance of developing a disease.
(b) B1 a correlation is a pattern/link seen between two variables (e.g. as alcohol consumption increases, so does the incidence of liver disease) but does not by itself prove one causes the other (oe)
(b) B1 a cause means it has been shown, through further investigation of the mechanism, that a change in one variable (alcohol intake) directly produces a change in the other (liver damage), rather than the link being coincidental or due to a third factor (oe)
(b) Answer: A correlation is just a pattern between two variables; a causal link means it has been proven that one variable directly causes the change seen in the other.
(c) B1 names a valid lifestyle risk factor, e.g. smoking, obesity/poor diet, lack of exercise, high salt intake
(c) B1 correctly links it to a disease it increases the risk of, e.g. smoking - lung disease/cardiovascular disease; obesity - type 2 diabetes/cardiovascular disease; high salt intake - high blood pressure (ft candidate's own risk factor)
(c) Answer: For example: smoking is linked to an increased risk of cardiovascular disease and lung disease.

Question 18

(a) B1 a benign tumour grows in one place/is contained (often within a membrane) and does not spread to other parts of the body (oe)
(a) B1 a malignant tumour invades neighbouring (healthy) tissue and can spread via the blood or lymph to other parts of the body, forming secondary tumours (metastasis) (oe)
(a) Answer: Benign tumours stay in one place and do not spread; malignant tumours invade other tissue and can spread around the body.
(b) B1 e.g. smoking/tobacco use, OR exposure to UV/ionising radiation, OR obesity, OR exposure to carcinogens
(b) B1 a second, different valid risk factor from the list above, or inherited (genetic) factors
(b) Answer: For example: smoking and prolonged exposure to UV radiation.

Question 19

(a) B1 one direction only, upwards from the roots to the leaves (as part of the transpiration stream) (oe)
(a) Answer: One-way, upwards from roots to leaves.
(b) B1 both directions, from sources (e.g. leaves) to sinks (e.g. roots, storage organs, growing parts); process = translocation
(b) Answer: Both directions, from sources (e.g. leaves) to sinks (e.g. roots); the process is called translocation.
(c) B1 xylem cells are dead and have no cytoplasm/end walls, forming continuous hollow tubes (oe)
(c) B1 cell walls are strengthened/thickened with lignin, providing support and preventing the tube from collapsing under the tension created as water is drawn upwards (oe)
(c) Answer: Xylem cells are dead, hollow tubes strengthened with lignin, which supports the tissue and prevents collapse under tension.
(d) B1 companion cells; they provide energy (e.g. ATP from respiration) to support the active transport/loading of sugars into the sieve tubes (oe)
(d) Answer: Companion cells; they provide the energy needed to load sugars into the sieve tubes.

Question 20

(a) B1 to act as a marker so the distance moved (and hence the rate of water uptake, used to estimate the rate of transpiration) can be measured (oe)
(a) Answer: It acts as a marker so the rate of water uptake can be measured.
(b) B1 temperature (oe)
(b) B1 light intensity, OR humidity, OR using the same shoot/leaf area, OR starting the bubble from the same position each time (any one different valid control, oe)
(b) Answer: For example: temperature and light intensity kept the same in both readings.
(c) M1 correct method shown: distance / time for at least one condition
(c) A1 still air = 2.4 mm/min AND with fan = 6.0 mm/min (both required, cao)
(c) Answer: Still air: 2.4 mm/min. With fan: 6.0 mm/min.
(d) B1 moving air removes water vapour from around the leaf/stomata more quickly, maintaining a steeper concentration gradient of water vapour between the leaf air spaces and the air outside, which increases the rate of diffusion of water vapour out of the leaf (and so increases the rate of water uptake), consistent with the higher rate (6.0 vs 2.4 mm/min) measured (oe)
(d) Answer: Moving air removes water vapour from the leaf surface faster, maintaining a steeper concentration gradient and increasing the rate of diffusion (transpiration), which matches the higher rate of water uptake measured with the fan.

Question 21

(a) M1 correct surface area method: 2 x [(1 x 0.01) + (1 x 0.01) + (0.01 x 0.01)] = 0.0402 mm² (ft to final ratio if only arithmetic slip made)
(a) M1 correct volume method: 1 x 0.01 x 0.01 = 0.0001 mm³
(a) A1 402 : 1 (cao)
(a) Answer: 402 : 1
(b) B1 a large surface area to volume ratio maximises the area of membrane in contact with the soil water, increasing the rate of diffusion/osmosis of water into the root hair cell (oe)
(b) B1 the concentration of mineral ions is often lower in the soil than inside the root hair cell, so ions cannot move in down a concentration gradient by diffusion; they must be absorbed by active transport, which requires energy released by respiration (in mitochondria) (oe)
(b) Answer: A large surface area to volume ratio increases the rate of water uptake by diffusion/osmosis. Mineral ions are often more concentrated inside the root hair cell than in the soil, so they must be moved in against the gradient by active transport, using energy from respiration.

Organisation - Worksheets, Questions and Revision