References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Membrane structure & cell signalling · Core · Paper 1 style · 1 mark
(a)Which components of the cell surface membrane are mainly involved in cell recognition?[1]
Show answer
(a) B glycoproteins and glycolipids
The carbohydrate chains of glycoproteins and glycolipids (the glycocalyx) act as recognition sites / antigens. Cholesterol (C) only regulates fluidity and stability; channel proteins (D) let ions and polar molecules through.
Q2Membrane structure & cell signalling · Standard · Paper 2 style · 6 marks
(a)Explain why the structure of the cell surface membrane is described as a 'fluid mosaic'.[2]
(b)Fish living in very cold water under polar ice have cell membranes with a different composition from fish in tropical rivers such as the Indus. State the role of cholesterol in a cell membrane.[2]
(c)Explain why sodium ions cannot cross the phospholipid bilayer by simple diffusion.[2]
Show answer
(a) Fluid: phospholipids (and many proteins) can move about / diffuse sideways within the bilayer. Mosaic: proteins are scattered among the phospholipids in a pattern, like tiles of a mosaic.
1 mark: fluid – molecules move laterally within their layer. 1 mark: mosaic – proteins (of different kinds) scattered/embedded through the phospholipid bilayer.
(b) Cholesterol regulates membrane fluidity: it stops phospholipids packing too closely at low temperature (keeps the membrane fluid) and restricts their movement at high temperature (keeps it stable); it also reduces permeability to ions/polar molecules.
1 mark each for any two of: regulates/buffers fluidity; prevents tight packing / keeps membrane fluid at low temperature; reduces fluidity / adds stability at high temperature; reduces permeability to water-soluble ions/polar molecules; mechanical stability.
(c) Sodium ions are charged (and surrounded by water molecules); the interior of the bilayer is made of hydrophobic fatty acid tails, which repel/do not let charged particles through, so they must use channel or carrier proteins.
1 mark: ions are charged / hydrated / water-soluble. 1 mark: core of bilayer is hydrophobic (fatty acid tails), so ions need channel/carrier proteins.
Q3Membrane structure & cell signalling · Standard · Paper 1 style · 1 mark
(a)Proteins in the membrane of a human cell were labelled with a green fluorescent dye, and proteins of a mouse cell with a red dye. The two cells were fused. After 40 minutes at 37 °C the red and green labels were evenly mixed over the whole hybrid cell. At 5 °C the labels stayed in separate halves.What do these results show?[1]
Show answer
(a) B Membrane proteins can move sideways in the bilayer, and this movement is faster at higher temperature.
Mixing shows lateral movement of proteins (the 'fluid' in fluid mosaic); it needs kinetic energy so it is slowed at 5 °C. C is wrong because the labels were on existing proteins; D is about phospholipids, which were not labelled.
Q4Membrane structure & cell signalling · Challenge · Paper 2 style · 7 marks
(a)Outline the stages by which a water-soluble hormone such as glucagon, arriving in the blood, causes a specific response inside a liver cell.[4]
(b)Steroid hormones, such as oestrogen, have receptors inside the cell rather than on the cell surface membrane. Suggest why.[2]
(c)Suggest why only some cells in the body respond to glucagon even though all cells are exposed to it in the blood.[1]
Show answer
(a) Glucagon binds to a specific receptor on the cell surface membrane (complementary shape); binding changes the receptor's shape; this activates a G protein / transduction inside the cell; an enzyme makes a second messenger (cyclic AMP); the second messenger activates a chain of enzymes (amplification); the final enzyme causes the response (glycogen broken down to glucose).
1 mark each for any four of: hormone (ligand) binds to specific receptor on cell surface membrane; receptor has complementary shape / only target cells have the receptor; receptor changes shape / signal transduction; G protein activated; second messenger (cAMP) produced; cascade/amplification by enzymes; specific response, e.g. glycogen breakdown.
(b) Steroids are lipid-soluble (non-polar), so they can diffuse through the hydrophobic phospholipid bilayer and reach receptors in the cytoplasm or nucleus.
1 mark: steroids are lipid-soluble / non-polar / hydrophobic. 1 mark: so they pass directly through the phospholipid bilayer (by simple diffusion).
(c) Only target cells (e.g. liver cells) have receptors with a shape complementary to glucagon.
1 mark: only target cells have the specific receptor.
Q5Membrane structure & cell signalling · Challenge · Paper 2 style · 6 marks
(a)Describe the structure of a phospholipid molecule and explain why phospholipids form a bilayer when mixed with water.[4]
(b)A species of bacterium lives in hot springs at 80 °C. Its membrane phospholipids contain a higher proportion of saturated fatty acids than those of bacteria living at 25 °C. Suggest how this helps the bacterium to survive.[2]
Show answer
(a) Glycerol joined to two fatty acids and a phosphate group; the phosphate head is hydrophilic (polar) and the fatty acid tails are hydrophobic (non-polar); in water the heads face the water on both sides and the tails point inwards, away from water, forming a bilayer.
1 mark: glycerol + two fatty acids + phosphate (group). 1 mark: phosphate head hydrophilic/polar. 1 mark: fatty acid tails hydrophobic/non-polar. 1 mark: heads face aqueous surroundings, tails face inwards / away from water, so two layers form.
(b) Saturated fatty acid tails are straight (no kinks), so they pack closely together; this reduces fluidity / keeps the membrane stable and not too leaky at high temperature.
1 mark: saturated tails straight / no C=C kinks, pack tightly. 1 mark: membrane less fluid / more stable / less permeable at high temperature (does not become too leaky).
Q6Diffusion, osmosis & water potential · Core · Paper 1 style · 1 mark
(a)Two neighbouring plant cells, X and Y, have water potentials of −450 kPa and −620 kPa respectively.Which statement is correct?[1]
Show answer
(a) B Water moves from X to Y because X has the higher water potential.
Water moves by osmosis from a higher (less negative) water potential to a lower (more negative) one: −450 kPa is higher than −620 kPa. A reverses the direction; osmosis is passive, so D is wrong.
Q7Diffusion, osmosis & water potential · Standard · Paper 2 style · 6 marks
(a)A plant epidermal cell is placed in a sucrose solution that has a lower water potential than the cell. Describe and explain what happens to the cell.[3]
(b)A red blood cell and a plant cell are both placed in distilled water. Explain why the red blood cell bursts but the plant cell does not.[3]
Show answer
(a) Water leaves the cell by osmosis, down the water potential gradient, through the partially permeable cell surface membrane (and tonoplast); the vacuole and protoplast shrink; the membrane pulls away from the cell wall (plasmolysis).
1 mark: water moves out by osmosis. 1 mark: from higher to lower water potential / down water potential gradient through partially permeable membrane. 1 mark: protoplast/vacuole shrinks; membrane pulls away from wall = plasmolysis (cell becomes flaccid).
(b) In both cells water enters by osmosis because distilled water has the higher water potential. The red blood cell has no cell wall, so it swells until the membrane bursts (haemolysis). The plant cell has a strong cellulose cell wall that resists expansion, so the cell becomes turgid but does not burst.
1 mark: water enters both cells by osmosis (water has higher water potential). 1 mark: animal cell has no cell wall, membrane bursts / haemolysis. 1 mark: plant cell wall (cellulose) resists expansion / exerts pressure, cell becomes turgid.
Q8Diffusion, osmosis & water potential · Challenge · Paper 2 style · 6 marks
The rate of glucose uptake by human red blood cells was measured at different external glucose concentrations.External glucose / mmol dm⁻³: 0, 2, 4, 8, 12, 16, 20Rate of uptake / arbitrary units: 0, 7.5, 12.0, 16.2, 18.1, 18.9, 19.0
(a)Describe how the rate of glucose uptake changes as the external glucose concentration increases.[2]
(b)Glucose enters red blood cells by facilitated diffusion through carrier proteins. Explain the shape of the curve at concentrations above 16 mmol dm⁻³.[2]
(c)Suggest how the curve would differ for a substance that crosses the membrane only by simple diffusion through the phospholipid bilayer. Explain your answer.[2]
Show answer
(a) The rate increases steeply at first (roughly in proportion) up to about 4 mmol dm⁻³, then increases more slowly and levels off at about 19 arbitrary units from about 16 mmol dm⁻³.
1 mark: rate increases as concentration increases (steeply at low concentration). 1 mark: levels off / plateaus at about 19 a.u. (from about 16 mmol dm⁻³) – data quote needed.
(b) All the carrier proteins are occupied (saturated), so the number of carrier proteins is now the limiting factor and increasing concentration cannot increase the rate further.
1 mark: carrier proteins all in use / saturated. 1 mark: number of carriers limits the rate (not the concentration gradient).
(c) The rate would keep rising in proportion to concentration (a straight line, no plateau), because no carrier proteins are involved, so there is nothing to become saturated; rate depends only on the concentration gradient.
1 mark: straight line / rate continues to increase / no plateau. 1 mark: no proteins to saturate; rate proportional to concentration gradient.
Q9Diffusion, osmosis & water potential · Challenge · Paper 1 style · 1 mark
(a)The rate of diffusion across an exchange surface is proportional to (surface area × difference in concentration) ÷ thickness of the surface.In one tissue the surface area doubles and the thickness of the surface halves, while the concentration difference stays the same. What happens to the rate of diffusion?[1]
Show answer
(a) C It increases four times.
Rate ∝ (2 × area) ÷ (0.5 × thickness) = 4 × original rate. B forgets the effect of thickness; A wrongly assumes the two changes cancel out.
Q10Active transport, endo- & exocytosis · Core · Paper 2 style · 5 marks
(a)Define active transport.[2]
(b)Cells in the pancreas release the enzyme amylase, a protein, into a duct. Name the process by which amylase leaves the cell and describe how it happens.[3]
Show answer
(a) The movement of molecules or ions across a membrane against their concentration gradient (from lower to higher concentration), using carrier proteins and energy from ATP (respiration).
1 mark: against a concentration gradient / low to high concentration. 1 mark: uses carrier proteins and ATP / energy from respiration.
(b) Exocytosis: amylase is packaged by the Golgi body into secretory vesicles; vesicles move to the cell surface membrane (along microtubules); the vesicle membrane fuses with the cell surface membrane and the contents are released outside; this needs ATP.
1 mark: exocytosis. 1 mark each for any two of: protein packaged into (secretory) vesicles by Golgi; vesicles move to cell surface membrane; vesicle fuses with membrane, contents released; requires ATP.
Q11Active transport, endo- & exocytosis · Standard · Paper 2 style · 6 marks
Root tissue from a wheat seedling was placed in a solution containing 0.5 mmol dm⁻³ nitrate ions. The concentration of nitrate inside the root cells was already 12 mmol dm⁻³. The rate of nitrate uptake was 4.8 µmol g⁻¹ h⁻¹. Cyanide, which inhibits aerobic respiration, was then added, and the rate fell to 0.6 µmol g⁻¹ h⁻¹.
(a)Calculate the percentage decrease in the rate of nitrate uptake caused by adding cyanide. Give your answer to 1 decimal place.[2]
(b)Explain how these results show that most nitrate ions are taken up by active transport.[3]
(c)Suggest why some nitrate uptake still occurs in the presence of cyanide.[1]
Show answer
(a) 87.5 %
M1: (4.8 − 0.6) ÷ 4.8 × 100. A1: 87.5 %.
(b) Cyanide stops aerobic respiration, so little ATP is made; uptake falls greatly, so uptake needs ATP; nitrate is taken up although its concentration is higher inside the cells, i.e. against the gradient – both features of active transport.
1 mark: cyanide inhibits respiration / ATP production. 1 mark: uptake depends on ATP (falls without it). 1 mark: ions moved against concentration gradient (higher inside cells) with data quote.
(c) Some nitrate enters by (facilitated) diffusion, or some ATP is still made by anaerobic respiration / from stored ATP.
1 mark for either: some passive uptake (facilitated diffusion) or some ATP still available (anaerobic respiration / reserves).
Q12Active transport, endo- & exocytosis · Standard · Paper 1 style · 1 mark
(a)Which pair of processes both require ATP from respiration?[1]
Show answer
(a) A exocytosis and active transport
Exocytosis and active transport both use ATP (vesicle movement/fusion; carrier protein shape change). Osmosis, facilitated diffusion and simple diffusion are passive, so B, C and D each contain a passive process.
Q13Active transport, endo- & exocytosis · Challenge · Paper 2 style · 6 marks
(a)A nerve cell membrane contains sodium–potassium pumps. Each pump moves 3 sodium ions out of the cell and 2 potassium ions into the cell for each molecule of ATP hydrolysed. The pumps in one cell hydrolyse 1.5 × 10⁸ molecules of ATP per second. Calculate the number of sodium ions moved out of this cell in one minute. Give your answer in standard form.[2]
(b)Explain why the sodium–potassium pump is an example of active transport rather than facilitated diffusion.[2]
(c)Suggest why cells that carry out a lot of active transport, such as those lining kidney tubules, have many mitochondria.[2]
Show answer
(a) 2.7 × 10¹⁰ sodium ions
M1: 3 × 1.5 × 10⁸ × 60. A1: 2.7 × 10¹⁰.
(b) The pump moves both ions against their concentration gradients (Na⁺ out to where it is more concentrated, K⁺ in to where it is more concentrated) and uses energy from ATP hydrolysis to change the shape of the carrier protein.
1 mark: against concentration gradient. 1 mark: uses ATP / energy to change shape of carrier protein (facilitated diffusion is passive, down the gradient).
(c) Mitochondria are the site of aerobic respiration, which produces large amounts of ATP; active transport requires a continuous supply of ATP.
1 mark: mitochondria carry out aerobic respiration / make ATP. 1 mark: active transport needs much ATP.
Q14Surface area : volume · Core · Paper 1 style · 1 mark
(a)What is the surface area : volume ratio of an agar cube with sides of 3 cm?[1]
Show answer
(a) B 2 : 1
Surface area = 6 × 3² = 54 cm²; volume = 3³ = 27 cm³; 54 : 27 = 2 : 1. D (6 : 1) is the ratio for a 1 cm cube.
Q15Surface area : volume · Standard · Paper 2 style · 6 marks
Cubes of agar containing an alkali and pink phenolphthalein indicator were placed in dilute hydrochloric acid. The indicator becomes colourless as acid diffuses in. The time for each cube to become completely colourless was recorded.Cube side / cm: 0.5, 1.0, 1.5, 2.0Time / s: 70, 260, 590, 1050
(a)Calculate the surface area : volume ratio of the 0.5 cm cube and of the 1.5 cm cube.[2]
(b)Describe the relationship between surface area : volume ratio and the time taken for the cube to become colourless.[1]
(c)Use the results to explain why large multicellular organisms such as mammals need specialised gas exchange surfaces, while a single-celled organism such as Amoeba does not.[3]
1 mark each. 0.5 cm: SA = 6 × 0.25 = 1.5, V = 0.125, ratio 12 : 1. 1.5 cm: SA = 13.5, V = 3.375, ratio 4 : 1.
(b) As the surface area : volume ratio decreases (cubes get larger), the time to become colourless increases.
1 mark: smaller ratio → longer time (or larger ratio → shorter time).
(c) Large organisms have a small surface area : volume ratio, so the body surface is too small to supply all cells; diffusion distances to the centre are too long, so diffusion would be too slow (as in the large cube). Specialised surfaces (alveoli) give a large surface area and short distance. Amoeba has a large ratio and short distances, so diffusion across its surface is enough.
1 mark: large organism has small SA:V / surface too small for its volume. 1 mark: long diffusion distance, diffusion too slow to meet needs (link to cube times). 1 mark: Amoeba has large SA:V / short distance so diffusion across surface is sufficient.
Q16Skill: planning a method · Standard · Paper 5 style · 6 marks
(a)Plan an investigation to find the effect of temperature on the permeability of the cell membranes of beetroot tissue, using a colorimeter. Beetroot cells contain a red pigment in their vacuoles that leaks out when membranes are damaged.[6]
Show answer
(a) Cut beetroot cylinders with a cork borer and trim to the same length (e.g. 1 cm); rinse in distilled water until no more pigment leaks; place equal numbers of pieces in the same volume of distilled water in tubes in thermostatically controlled water baths at e.g. 20, 30, 40, 50, 60 and 70 °C; leave for the same time (e.g. 30 min); remove tissue, shake; measure absorbance of each solution with a colorimeter (blue-green filter), zeroed with distilled water; repeat three times at each temperature and calculate means; higher absorbance = more pigment = more permeable.
1 mark each for any six of: IV temperature with at least 5 values over a suitable range (e.g. 20–70 °C); water bath/thermostatic control, checked with thermometer; same size pieces (cork borer, same length) – same surface area; rinse to remove pigment from cut cells; same volume of distilled water; equilibrate water to temperature before adding tissue; same time in water; DV absorbance (or transmission) measured by colorimeter; blue/green filter and zeroed with water blank; at least 3 replicates and mean; beetroot from same root/same age.
Q17Skill: planning a method · Challenge · Paper 5 style · 7 marks
(a)A student has a 1.0 mol dm⁻³ sucrose solution and distilled water. Copy and complete the volumes needed to make 10 cm³ of each of these concentrations: 0.2, 0.4, 0.6 and 0.8 mol dm⁻³.[2]
(b)Outline how the student could use these solutions to estimate the water potential of potato tissue. A table converting sucrose concentration into water potential is available.[4]
(c)Explain why the student should calculate percentage change in mass rather than change in mass.[1]
Volume of stock = required concentration × 10 ÷ 1.0. 2 marks all four correct; 1 mark for two or three correct.
(b) Cut potato cylinders of the same size (cork borer, same length), blot and weigh each; place one set in each sucrose solution (and water) for the same time (e.g. 24 h) at constant temperature; remove, blot dry in the same way and reweigh; calculate percentage change in mass; plot percentage change against concentration and read off the concentration where the line crosses zero (no net water movement); use the table to convert this concentration to water potential = water potential of the tissue.
1 mark each for any four of: same size cylinders/same surface area; blot and weigh before and after; same time and temperature; calculate % change in mass; graph % change against concentration; read concentration at zero change (no net osmosis); convert using table to water potential; replicates and means.
(c) The cylinders may start with different masses, so percentage change allows a fair comparison.
Q18Skill: graphs & charts · Standard · Paper 3 style · 6 marks
A student measured the percentage change in mass of potato cylinders after 2 hours in sucrose solutions.Sucrose concentration / mol dm⁻³: 0.0, 0.2, 0.4, 0.6, 0.8, 1.0Percentage change in mass / %: +18.2, +9.6, +1.8, −6.0, −12.4, −16.1
(a)State which variable should be plotted on the x-axis and explain why.[1]
(b)Describe how the student should plot and draw the graph so that the point where the line crosses zero can be read accurately.[3]
(c)Estimate, by linear interpolation between the readings at 0.4 and 0.6 mol dm⁻³, the sucrose concentration at which there would be no change in mass. Give your answer to 2 decimal places.[2]
Show answer
(a) Sucrose concentration, because it is the independent variable (the one the student changed).
1 mark: concentration on x-axis as it is the independent variable.
(b) Label axes with quantities and units (concentration / mol dm⁻³; percentage change in mass / %); choose linear scales that use over half the grid and include negative values with a clear zero line; plot points accurately as small crosses; join points with ruled straight lines (or a smooth best-fit line) without extrapolating beyond the data.
1 mark each for any three of: full axis labels with units; linear scale using most of grid, including negative values; points plotted exactly (small × or encircled dot); ruled lines point to point / best-fit line, not extended beyond data.
Q19Skill: graphs & charts · Challenge · Paper 5 style · 6 marks
Beetroot discs were placed in water or in solutions of three household products for 20 minutes. The absorbance of the surrounding liquid was measured (higher absorbance = more pigment released).Treatment: water | detergent P | detergent Q | ethanol solution RMean absorbance: 0.20 | 1.14 | 0.79 | 0.3495 % confidence interval (±): 0.05 | 0.10 | 0.08 | 0.07
(a)State the type of graph that should be used to display these results and give a reason.[2]
(b)Describe how the 95 % confidence intervals should be shown on the graph.[1]
(c)A student concluded that all three solutions damaged the membranes more than water did, and that detergent P damaged them most. Use the confidence intervals to evaluate this conclusion.[3]
Show answer
(a) A bar chart, because the independent variable (type of treatment) is categorical / discontinuous.
1 mark: bar chart (bars of equal width, not touching). 1 mark: independent variable is categorical / not continuous.
(b) As error bars drawn above and below the top of each bar, from (mean − 95 % CI value) to (mean + 95 % CI value).
1 mark: vertical error bars centred on each mean extending ± the CI value.
(c) The intervals are water 0.15–0.25, P 1.04–1.24, Q 0.71–0.87 and R 0.27–0.41. None of P, Q or R overlaps with water (R's lower limit 0.27 is just above water's upper limit 0.25), so all three released significantly more pigment than water, though R only just. P's interval does not overlap with Q's or R's, so P is significantly the most damaging. The conclusion is supported, but only for these concentrations and this time.
1 mark: calculates/compares intervals – water 0.15–0.25 does not overlap with P, Q or R, so all significantly greater than water. 1 mark: P's interval (1.04–1.24) does not overlap Q (0.71–0.87), so P significantly the most damaging. 1 mark: R only just separated from water / limited to these conditions / small sample, so treat with caution.
Q20Skill: calculations & data processing · Core · Paper 3 style · 3 marks
(a)A potato cylinder had a mass of 2.45 g before and 2.71 g after being left in a dilute sucrose solution. Calculate the percentage change in mass. Give your answer to 1 decimal place.[2]
(b)State what this result shows about the water potential of the potato tissue compared with the solution.[1]
(b) The tissue had a lower water potential than the solution, so it gained water by osmosis.
1 mark: tissue water potential lower than solution / water moved into tissue.
Q21Skill: calculations & data processing · Challenge · Paper 5 style · 5 marks
A length of Visking (dialysis) tubing was filled with 1.0 mol dm⁻³ sucrose solution, attached to a vertical capillary tube of internal radius 0.5 mm and placed in a beaker of distilled water. In 15 minutes the liquid in the capillary tube rose by 36 mm.
(a)Calculate the volume of solution that entered the tubing in 15 minutes. Use π r² l and give your answer in mm³ to 3 significant figures.[2]
(b)Calculate the mean rate of water uptake in mm³ min⁻¹. Give your answer to 3 significant figures.[1]
(c)Explain why the liquid in the capillary tube rises.[2]
(c) The sucrose solution in the tubing has a lower water potential than the distilled water outside; water moves in by osmosis through the partially permeable tubing (sucrose molecules are too large to leave), increasing the volume inside.
1 mark: lower water potential inside / water potential gradient. 1 mark: water enters by osmosis through partially permeable membrane; sucrose cannot pass out.
Q22Skill: statistics (SD, SE, CI, t-test, Spearman, chi-squared) · Challenge · Paper 5 style · 6 marks
A student compared two potato varieties grown in Okara. Six cylinders of each were left in 0.3 mol dm⁻³ sucrose solution and the percentage change in mass was found.Variety P: mean −4.2 %, standard deviation 1.1, n = 6Variety Q: mean −6.9 %, standard deviation 1.4, n = 6
(a)State a null hypothesis for this investigation.[1]
(b)Calculate the value of t, using t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂). Give your answer to 2 decimal places.[3]
(c)The critical value of t at p = 0.05 for 10 degrees of freedom is 2.23. Explain what your value of t shows.[2]
Show answer
(a) There is no significant difference between the mean percentage change in mass of the two varieties (any difference is due to chance).
1 mark: no (significant) difference between means of P and Q.
(c) 3.71 is greater than 2.23, so the null hypothesis is rejected: the difference between the varieties is significant (probability of it being due to chance is less than 5 %); variety Q loses more water, so its tissue has a higher water potential.
1 mark: t > critical value so reject null hypothesis. 1 mark: difference significant / not due to chance (p < 0.05).
Q23Skill: recording results in a table · Core · Paper 3 style · 5 marks
A student cut four potato strips, each 50.0 mm long, and placed one in each of four sucrose solutions. After 1 hour she wrote in her notebook: '0.00 mol dm⁻³ strip now 53.5 mm; 0.25 mol dm⁻³ strip 51.0 mm; 0.50 strip 48.5 mm; 0.75 strip 47.0 mm.'
(a)Describe how the student should set out a table to record these results, including the headings.[3]
(b)Calculate the percentage change in length for each concentration.[2]
Show answer
(a) Columns (or rows) for: sucrose concentration / mol dm⁻³; initial length / mm; final length / mm; percentage change in length / %. Units only in the headings, not with each value; independent variable in the first column; all lengths to the same number of decimal places (1 d.p.).
1 mark: table with ruled columns and IV (concentration) in first column. 1 mark: headings include initial length, final length and % change. 1 mark: units in headings only (mol dm⁻³, mm, %) and consistent decimal places.
M1: (final − 50.0) ÷ 50.0 × 100. A1: all four correct with signs.
Q24Skill: explaining results with biology · Standard · Paper 2 style · 6 marks
Samples of sheep red blood cells were placed in sodium chloride solutions of different concentrations. After 30 minutes the percentage of cells that had burst (haemolysed) was found and the cells were examined with a microscope.NaCl / %: 0.30, 0.45, 0.60, 0.85, 1.20Cells burst / %: 100, 50, 5, 0, 0At 1.20 % the cells appeared shrunken with a crinkled outline.
(a)Explain the results for the 0.30 % and 0.45 % solutions.[3]
(b)Explain why cells in the 1.20 % solution appeared shrunken with a crinkled outline.[2]
(c)Suggest why a 0.85 % sodium chloride solution is used when giving fluid to a dehydrated patient through a drip.[1]
Show answer
(a) These solutions have a higher water potential than the cytoplasm of the red blood cells; water enters the cells by osmosis; the cells swell and, with no cell wall, the membrane bursts (haemolysis); at 0.45 % the gradient is smaller, so only the cells with the weakest membranes burst (50 %).
1 mark: solution has higher water potential than cell. 1 mark: water enters by osmosis, cells swell and burst (no cell wall). 1 mark: smaller gradient at 0.45 % so fewer burst / cells vary in resistance.
(b) The 1.20 % solution has a lower water potential than the cells, so water leaves the cells by osmosis and they shrink (crenation).
1 mark: lower water potential outside. 1 mark: water lost by osmosis, cells shrink / crenated.
(c) It has the same water potential as blood (is isotonic), so blood cells neither gain nor lose water.
1 mark: same water potential as plasma/cells, no net osmosis, no haemolysis or crenation.
Q25Skill: errors, reliability & improvements · Standard · Paper 3 style · 6 marks
A student cut pieces of beetroot of about the same size with a kitchen knife and placed one piece straight into a test-tube of water at each of five temperatures, heating the water over a Bunsen burner in a beaker. After about 15 minutes she judged the depth of red colour in each tube by eye, on a scale from 0 (none) to 5 (very dark).
(a)Identify three sources of error in this method and suggest an improvement for each.[6]
Show answer
(a) Pieces cut by knife differ in size/surface area – use a cork borer and ruler to cut equal cylinders; pigment from cells damaged by cutting not washed off – rinse pieces in water until the water stays clear; water bath temperature not controlled – use a thermostatically controlled water bath and check with thermometer; colour judged by eye is subjective – use a colorimeter to measure absorbance; only one tube per temperature – do three replicates and calculate means; timing not the same – use a stopwatch and fixed time.
Award 1 mark for each error and 1 mark for its matching improvement, for three pairs (max 6): unequal size → cork borer/scalpel and ruler; unwashed cut cells → rinse until no pigment; uncontrolled temperature → thermostatic water bath; subjective colour → colorimeter; no replicates → at least 3 repeats + mean; time varied → stopwatch, same time.
Q26Skill: conclusions & interpreting data · Challenge · Paper 5 style · 5 marks
Pieces of small intestine from a rat were placed in a solution containing 1.0 mmol dm⁻³ of amino acid X. The concentration of X inside the cells was measured after 40 minutes. At the start, the cells contained no X.Condition: oxygen bubbled through | nitrogen bubbled through (no oxygen) | oxygen + respiratory inhibitorConcentration of X inside cells / mmol dm⁻³: 4.6 | 0.9 | 0.8
(a)State two conclusions that can be drawn from these results.[2]
(b)A student concluded that amino acid X enters the cells only by active transport. Evaluate this conclusion.[3]
Show answer
(a) Amino acid X accumulates inside the cells to a concentration higher than outside (against the gradient) only when oxygen is present; without oxygen (or with the inhibitor) uptake is much slower and the inside concentration does not rise above the outside concentration, so most uptake depends on respiration / ATP.
1 mark each for any two of: X is taken up against its concentration gradient in oxygen (data: 4.6 inside vs 1.0 outside); uptake needs oxygen/aerobic respiration / ATP; without oxygen/with inhibitor uptake is much less and reaches only about the outside concentration (0.9, 0.8).
(b) Partly supported: uptake against the gradient and dependence on oxygen/respiration show active transport. But some uptake still occurs without oxygen and with the inhibitor (concentration rises from 0 to 0.8–0.9 mmol dm⁻³), which suggests some entry by (facilitated) diffusion down the gradient; so 'only' is not justified; also only one amino acid / one tissue / no repeats or variability shown.
1 mark: supporting evidence – accumulation above outside concentration only with oxygen. 1 mark: against – some uptake without O₂ / with inhibitor up to about the outside concentration = diffusion. 1 mark: limitation – no replicates/SD, single concentration, so conclusion tentative.
Q27Skill: variables & controls · Core · Paper 5 style · 8 marks
A student plans to investigate the effect of ethanol concentration (0 %, 10 %, 20 %, 30 %, 40 % and 50 %) on the permeability of beetroot cell membranes. She will measure the absorbance of the solution around the beetroot after 20 minutes with a colorimeter.
(a)Identify the independent variable and the dependent variable.[2]
(b)State three variables that should be kept constant and describe how to keep one of them constant.[4]
(c)Describe a suitable control for this investigation and explain its purpose.[2]
Show answer
(a) Independent: concentration of ethanol. Dependent: absorbance of the solution (amount of pigment released).
1 mark each.
(b) Any three: temperature; volume of solution; size/surface area of beetroot pieces; time in solution; source/age of beetroot; washing procedure. E.g. temperature – keep all tubes in a thermostatically controlled water bath at 25 °C.
1 mark each for three suitable variables (max 3). 1 mark for a valid method of control for one of them.
(c) Beetroot pieces in distilled water (0 % ethanol) treated in the same way; shows how much pigment leaks without ethanol (e.g. from cut cells), so any extra can be attributed to ethanol.
1 mark: same pieces in water, 0 % ethanol. 1 mark: baseline – shows effect is due to ethanol.
Q28Skill: measurement, graticule & magnification · Standard · Paper 3 style · 5 marks
A student calibrated an eyepiece graticule with a stage micrometer using the ×40 objective lens. 40 eyepiece divisions lined up exactly with 10 stage micrometer divisions; each stage division is 10 µm. She then measured red blood cells. In an isotonic solution a cell measured 3.0 eyepiece divisions across. In a slightly dilute solution a cell measured 3.6 divisions.
(a)Calculate the length of one eyepiece graticule division in µm.[1]
(b)Calculate the actual diameter of a red blood cell in the isotonic solution and in the dilute solution.[2]
(c)Calculate the percentage increase in diameter.[1]
(d)Explain why the eyepiece graticule must be recalibrated if a different objective lens is used.[1]
(d) The graticule scale stays the same size in the eyepiece but the specimen is magnified by a different amount, so each division represents a different actual length.
1 mark: magnification changes so value of each eyepiece division changes.
Q29Skill: hypotheses & predictions · Standard · Paper 5 style · 5 marks
A student plans to use Visking tubing containing 0.5 mol dm⁻³ sucrose solution, attached to a capillary tube, to investigate the effect of temperature (10, 20, 30 and 40 °C) on the rate of osmosis.
(a)State a testable hypothesis for this investigation.[1]
(b)State the null hypothesis.[1]
(c)Predict the results and explain your prediction using biological knowledge.[3]
Show answer
(a) The rate of water uptake into the tubing increases as temperature increases (from 10 °C to 40 °C).
1 mark: clear direction of relationship between temperature and rate of osmosis.
(b) There is no relationship between temperature and the rate of water uptake.
1 mark.
(c) Rate of rise will increase as temperature increases, because water molecules gain kinetic energy and move faster, so more cross the partially permeable membrane per unit time; the water potential gradient is the same at each temperature, so the difference is due to the speed of molecules.
1 mark: rate increases with temperature. 1 mark: molecules have more kinetic energy / move faster. 1 mark: more water molecules cross membrane per unit time by osmosis (diffusion of water).
Q30Skill: risk assessment · Core · Paper 5 style · 5 marks
A student will cut cubes of agar containing phenolphthalein and dilute sodium hydroxide with a scalpel, then place them in 0.1 mol dm⁻³ hydrochloric acid in beakers and time how long each takes to lose its colour.
(a)Identify two hazards in this investigation and state a precaution for each.[4]
(b)State the overall level of risk of this investigation and give a reason.[1]
Show answer
(a) Scalpel used to cut agar – sharp blade, cut away from fingers on a tile; hydrochloric acid – irritant to eyes and skin, wear eye protection (goggles) and gloves, wipe spills; phenolphthalein dissolved in ethanol – flammable/harmful, keep away from flames; glassware may break – handle carefully.
1 mark each for two hazards (blade, acid, glass, flammable indicator) and 1 mark each for a matching precaution.
(b) Low risk, because the acid is dilute and the hazards are easily controlled with simple precautions.
1 mark: low with reason (dilute acid / small volumes / precautions easily taken).
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
