An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 (Multiple Choice - Extended)
Answer all 40 multiple-choice questions. For each question, choose the one correct option from A, B, C, or D.
40 Question · 40 marks
Question 1 · multipleChoice
1 marks
Which statement about anaerobic respiration in human muscle cells and in yeast is correct?
A.Yeast cells produce carbon dioxide, whereas human muscle cells do not.
B.Both yeast and human muscle cells produce lactic acid.
C.Both yeast and human muscle cells produce ethanol.
D.Human muscle cells produce carbon dioxide, whereas yeast cells do not.
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Worked solution
During anaerobic respiration in yeast, glucose is converted into ethanol and carbon dioxide. In contrast, anaerobic respiration in human muscle cells produces lactic acid only, with no carbon dioxide released.
Marking scheme
1 mark for option A. - Option B is incorrect as yeast produces ethanol, not lactic acid. - Option C is incorrect as yeast produces ethanol, not muscle cells. - Option D is incorrect as it is the reverse of the correct statement.
Question 2 · multipleChoice
1 marks
Two different organisms are classified in the same class but in different orders. What can be correctly deduced about these two organisms?
A.They belong to the same genus.
B.They belong to the same phylum.
C.They share more features than two organisms in the same family.
D.They have more similar DNA base sequences than two organisms in the same genus.
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Worked solution
The taxonomic hierarchy is: Kingdom, Phylum, Class, Order, Family, Genus, Species. If two organisms are in the same class, they must also belong to the same broader taxonomic groups above it, which are Phylum and Kingdom.
Marking scheme
1 mark for option B. - Option A is incorrect because if they are in different orders, they cannot belong to the same genus. - Option C is incorrect because organisms in the same family share more features than those in the same class but different orders. - Option D is incorrect because organisms in the same genus have more similar DNA sequences.
Question 3 · multipleChoice
1 marks
An enzyme-controlled reaction is carried out at \(50^\circ\text{C}\), which is well above its optimum temperature. The rate of reaction is found to be zero. Which statement explains this observation?
A.The kinetic energy of the enzyme and substrate is too low for successful collisions.
B.The shape of the active site has changed, so the substrate no longer fits.
C.The substrate molecules have changed shape, so they no longer fit the active site.
D.The bonds between enzyme and substrate have become too strong to be broken.
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Worked solution
At high temperatures above the optimum, the active site of the enzyme denatures. This permanent change in shape means the substrate can no longer fit into the active site, stopping the reaction.
Marking scheme
1 mark for option B. - Option A is incorrect because kinetic energy is high at high temperatures. - Option C is incorrect as high temperature does not reduce activation energy to zero. - Option D is incorrect as it is the enzyme (protein) that denatures, not the substrate.
Question 4 · multipleChoice
1 marks
Which row correctly matches a cell structure to its function?
| | cell structure | function | |---|---|---| | A | chloroplast | site of aerobic respiration | | B | ribosome | site of protein synthesis | | C | mitochondrion | site of photosynthesis | | D | vacuole | controls the entry and exit of substances |
A.A
B.B
C.C
D.D
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Worked solution
Ribosomes are the site of protein synthesis. Chloroplasts carry out photosynthesis, mitochondria carry out aerobic respiration, and the cell membrane (not the vacuole) controls the entry and exit of substances.
Marking scheme
1 mark for option B. - Option A is incorrect as the chloroplast is for photosynthesis. - Option C is incorrect as the mitochondrion is for respiration. - Option D is incorrect as the cell membrane controls entry and exit.
Question 5 · multipleChoice
1 marks
A population of yeast is cultured in a closed flask with a fixed volume of nutrient solution. Which statement explains why the population enters the stationary phase of the sigmoid growth curve?
A.The rate of cell reproduction is equal to the rate of cell death.
B.The rate of cell reproduction is greater than the rate of cell death.
C.The rate of cell reproduction is less than the rate of cell death.
D.No cell reproduction or cell death occurs.
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Worked solution
In the stationary phase of a sigmoid population growth curve, limiting factors (such as nutrient depletion and accumulation of toxic waste products) cause the rate of cell reproduction to equal the rate of cell death, leading to a constant population size.
Marking scheme
1 mark for option A. - Option B describes the log (exponential) phase. - Option C describes the death phase. - Option D is incorrect because cell reproduction and death still occur at equal rates.
Question 6 · multipleChoice
1 marks
Which statement correctly describes active immunity?
A.It is short-term because antibodies are injected directly into the blood.
B.It is long-term because the body produces its own antibodies and memory cells.
C.It is short-term because no memory cells are produced in response to a vaccine.
D.It is long-term because antibodies are transferred from mother to fetus across the placenta.
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Worked solution
Active immunity is long-term defense against a pathogen. It is achieved when the body produces its own antibodies and memory cells after being exposed to a pathogen or vaccine.
Marking scheme
1 mark for option B. - Option A describes passive immunity. - Option C is incorrect as active immunity is long-term and produces memory cells. - Option D describes passive immunity.
Question 7 · multipleChoice
1 marks
In early spring, a deciduous tree transports stored food to its developing buds so they can grow into new leaves. What are the source and the sink for translocation in this tree?
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Worked solution
In early spring, before leaves develop, food stored as starch in the roots is converted to sucrose and transported upwards to growing regions. The roots are the source, and the developing buds are the sink.
Marking scheme
1 mark for option C. - Option A has source and sink reversed. - Option B refers to the summer state where leaves are the source. - Option D has source and sink incorrect for early spring.
Question 8 · multipleChoice
1 marks
Why is the muscular wall of the left ventricle much thicker than the muscular wall of the right ventricle?
A.The left ventricle must pump blood under higher pressure to the lungs.
B.The left ventricle must pump blood under higher pressure to the rest of the body.
C.The left ventricle must pump a larger volume of blood than the right ventricle.
D.The left ventricle receives blood directly from the vena cava under very high pressure.
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Worked solution
The left ventricle pumps blood to the systemic circulation (all body organs except the lungs), which is a much larger system with higher resistance. Thus, it needs to generate a much higher pressure than the right ventricle, which only pumps blood to the nearby lungs.
Marking scheme
1 mark for option B. - Option A is incorrect as the left ventricle pumps to the body. - Option C is incorrect as both ventricles pump equal volumes of blood. - Option D is incorrect as the left ventricle receives blood from the pulmonary vein under low pressure.
Question 9 · multipleChoice
1 marks
The pathways of anaerobic respiration in different organisms are described. Path 1: glucose is converted to compound X, carbon dioxide and energy. Path 2: glucose is converted to compound Y and energy. Which row correctly identifies compound X, compound Y, and an organism that performs Path 1?
A.Compound X = lactic acid, Compound Y = ethanol, Organism = human muscle
B.Compound X = ethanol, Compound Y = lactic acid, Organism = yeast
C.Compound X = lactic acid, Compound Y = ethanol, Organism = yeast
D.Compound X = ethanol, Compound Y = lactic acid, Organism = human muscle
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Worked solution
Path 1 describes anaerobic respiration in yeast, which yields ethanol (Compound X), carbon dioxide, and energy. Path 2 describes anaerobic respiration in mammalian muscles, which yields lactic acid (Compound Y) and energy.
Marking scheme
1 mark for the correct option B.
Question 10 · multipleChoice
1 marks
The table shows the percentage similarity of DNA sequences in a shared gene among four species of insects: W, X, Y, and Z. W vs X is 85%; W vs Y is 45%; W vs Z is 70%; X vs Y is 40%; X vs Z is 72%; Y vs Z is 50%. Which statement is supported by this genetic evidence?
A.Species W and Y share the most recent common ancestor.
B.Species W and X are the most closely related.
C.Species Y and Z are more closely related than species W and Z.
D.Species X and Y diverged most recently.
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Worked solution
The highest percentage of DNA sequence similarity is 85% between species W and X, indicating they share the most recent common ancestor and are the most closely related of the group.
Marking scheme
1 mark for the correct option B.
Question 11 · multipleChoice
1 marks
An enzyme-catalysed reaction was carried out at 30 °C. The experiment was then repeated at 40 °C (which is below the optimum temperature for this enzyme) while keeping all other variables constant. Which statement correctly describes the changes at 40 °C compared to 30 °C?
A.The kinetic energy of the substrate molecules decreases, resulting in fewer collisions per second.
B.The enzyme molecules denature, decreasing the rate of reaction.
C.The frequency of successful collisions between enzyme and substrate molecules increases.
D.The activation energy of the reaction is lowered further by the increase in temperature.
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Worked solution
An increase in temperature from 30 °C to 40 °C (below the optimum) increases the kinetic energy of both the enzyme and substrate molecules. This leads to faster movement, more frequent collisions, and a higher frequency of successful collisions, increasing the reaction rate.
Marking scheme
1 mark for the correct option C.
Question 12 · multipleChoice
1 marks
An unknown organism was examined under an electron microscope, revealing a cell wall, 70S ribosomes, circular DNA free in the cytoplasm, and an absence of mitochondria. From which group of organisms is this sample most likely to be?
A.animals
B.bacteria
C.fungi
D.plants
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Worked solution
Prokaryotes (bacteria) possess a cell wall, 70S ribosomes, and a circular DNA molecule (nucleoid) free in the cytoplasm. They lack membrane-bound organelles such as mitochondria.
Marking scheme
1 mark for the correct option B.
Question 13 · multipleChoice
1 marks
The population growth of a bacterial culture in a closed environment follows a sigmoid growth curve. During which phase of this growth curve is the birth (or division) rate of the cells equal to the death rate?
A.lag phase
B.log (exponential) phase
C.stationary phase
D.deceleration phase
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Worked solution
In the stationary phase of a sigmoid population growth curve, the rate of cell division equals the rate of cell death due to limiting factors like food depletion and accumulation of toxic waste, resulting in a constant population size.
Marking scheme
1 mark for the correct option C.
Question 14 · multipleChoice
1 marks
A person recovers from an infection and develops active immunity against the pathogen. Which statement correctly explains this form of defense?
A.Phagocytes undergo rapid mutations to recognize the pathogen's antigens.
B.Memory cells survive in the body and can rapidly produce antibodies upon reinfection.
C.Antibodies are directly transferred across the placenta to provide immediate defense.
D.The physical barrier of the skin becomes permanently impermeable to the pathogen.
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Worked solution
Active immunity occurs after exposure to antigens (either via infection or vaccination), leading to the production of memory cells. These memory cells persist in the body and allow a much faster, larger production of antibodies if the same pathogen is encountered again.
Marking scheme
1 mark for the correct option B.
Question 15 · multipleChoice
1 marks
Which statement explains how mechanical digestion in the mouth facilitates chemical digestion of proteins in the stomach?
A.Chewing mixes the food with saliva, which contains protease enzymes.
B.Chewing breaks the food into smaller pieces, increasing the surface area for stomach acid and pepsin to act on.
C.Chewing activates the alkaline secretion of the stomach wall.
D.Chewing emulsifies protein molecules, making them soluble in water.
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Worked solution
Mechanical digestion (chewing) reduces the size of food particles without changing their chemical nature. This significantly increases the total surface area of the food exposed to digestive enzymes (like pepsin) in the stomach, thereby accelerating chemical digestion.
Marking scheme
1 mark for the correct option B.
Question 16 · multipleChoice
1 marks
Xylem vessels are highly adapted to transport water under high tension. Which feature prevents these vessels from collapsing inward when water is pulled up the stem?
A.the presence of end walls forming sieve plates
B.thin, elastic membranes made of cellulose
C.spiral or annular rings of lignin reinforcing the cell walls
D.active transport proteins situated in the cell membrane
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Worked solution
Lignin is a strong, woody substance deposited in the walls of xylem vessels. It provides structural support, preventing the vessels from collapsing inward under the extreme negative pressure (tension) generated by transpiration pull.
Marking scheme
1 mark for the correct option C.
Question 17 · multipleChoice
1 marks
The percentage similarity in DNA base sequences between four species of primates is: species W and X = 94%; species W and Y = 88%; species W and Z = 76%; species X and Y = 85%; species X and Z = 74%; species Y and Z = 81%. Which statement is correct?
A.Species W and Y are more closely related than species W and X.
B.Species Z shared a more recent common ancestor with Y than with X.
C.Species X is more closely related to Z than to Y.
D.Species W is the ancestor of all the other species.
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Worked solution
Species Z and Y share 81% similarity, which is higher than the 74% similarity between Z and X. This means species Z shared a more recent common ancestor with Y than with X. The other options are incorrect: W and X are more closely related (94%) than W and Y (88%); X is more closely related to Y (85%) than to Z (74%); and percentage similarity does not prove W is the common ancestor of all others.
Marking scheme
Award 1 mark for selecting B.
Question 18 · multipleChoice
1 marks
Which statement describes the effect of increasing the temperature from 20 °C to 30 °C on an enzyme-controlled reaction?
A.The activation energy of the reaction is lowered.
B.The frequency of successful collisions between enzyme and substrate increases.
C.The shape of the active site changes to fit the substrate more tightly.
D.The enzyme molecules gain kinetic energy and denature.
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Worked solution
Increasing the temperature increases the kinetic energy of both the enzyme and substrate molecules. This increases the frequency of successful collisions, thereby increasing the rate of reaction. Activation energy is not altered by temperature; the shape of the active site does not change to fit the substrate more tightly; and denaturation typically happens at much higher temperatures.
Marking scheme
Award 1 mark for selecting B.
Question 19 · multipleChoice
1 marks
During vigorous exercise, muscle cells respire anaerobically. Which statement about anaerobic respiration in human muscle cells is correct?
A.It produces carbon dioxide, water, and a small amount of energy.
B.It produces lactic acid and releases less energy per glucose molecule than aerobic respiration.
C.It occurs in the mitochondria and requires a small volume of oxygen.
D.It produces ethanol and carbon dioxide, which causes muscle fatigue.
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Worked solution
Anaerobic respiration in human muscles converts glucose into lactic acid without using oxygen. Because glucose is only partially broken down, it releases significantly less energy per molecule of glucose compared to aerobic respiration. Carbon dioxide and water are products of aerobic respiration, while ethanol is produced during anaerobic respiration in yeast, not humans.
Marking scheme
Award 1 mark for selecting B.
Question 20 · multipleChoice
1 marks
Which row correctly identifies the roles of the rough endoplasmic reticulum and ribosomes in protein synthesis?
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Worked solution
Ribosomes are the actual site where amino acids are assembled into proteins (protein synthesis). The rough endoplasmic reticulum has ribosomes attached to its surface and is responsible for transporting these synthesised proteins throughout the cell.
Marking scheme
Award 1 mark for selecting C.
Question 21 · multipleChoice
1 marks
A population of bacteria is grown in a nutrient broth in a closed fermenter. After reaching the stationary phase, what is the primary reason why the population size stops increasing?
A.The temperature of the fermenter rises above the optimum.
B.The rate of reproduction equals the rate of death due to limiting nutrients and toxin accumulation.
C.The bacteria run out of oxygen for anaerobic respiration.
D.Mutations occur that make the bacteria reproduce more slowly.
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Worked solution
In a closed system, the stationary phase is reached when the growth rate equals the death rate. This is primarily caused by limiting factors such as the depletion of nutrients and the accumulation of toxic waste products.
Marking scheme
Award 1 mark for selecting B.
Question 22 · multipleChoice
1 marks
Which statement describes the role of memory cells in active immunity?
A.They produce large quantities of antibodies immediately during the primary immune response.
B.They remain in the blood and can rapidly divide to produce plasma cells upon re-infection with the same pathogen.
C.They engulf and digest pathogens through phagocytosis to prevent them from multiplying.
D.They are injected into the body during vaccination to provide immediate, short-term protection.
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Worked solution
Memory cells persist in the lymphatic system and blood stream after a primary response or vaccination. If the same pathogen enters the body again, memory cells quickly recognize it, divide rapidly, and differentiate into plasma cells that produce massive quantities of antibodies, preventing the disease.
Marking scheme
Award 1 mark for selecting B.
Question 23 · multipleChoice
1 marks
A plant is supplied with carbon dioxide containing radioactive carbon-14. This radioactive carbon is incorporated into sucrose during photosynthesis. In which tissue will the radioactive sucrose be transported, and in which direction?
A.phloem, from sink to source
B.phloem, from source to sink
C.xylem, from source to sink
D.xylem, from sink to source
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Worked solution
Sucrose is manufactured in photosynthesising leaves (the source) and must be transported to areas of growth or storage (the sinks) via the phloem. Xylem transports water and mineral ions from the roots to the leaves.
Marking scheme
Award 1 mark for selecting B.
Question 24 · multipleChoice
1 marks
Root hair cells take up mineral ions from the soil. Which conditions would lead to the lowest rate of active transport of mineral ions into the root hair cells?
A.high oxygen concentration in soil, high temperature
B.low oxygen concentration in soil, low temperature
C.high carbon dioxide concentration in soil, high temperature
D.low carbon dioxide concentration in soil, high temperature
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Worked solution
Active transport is an active process that requires energy in the form of ATP, which is produced during aerobic respiration. Low oxygen concentrations and low temperatures reduce the rate of respiration, thereby reducing the energy available for active transport.
Marking scheme
Award 1 mark for selecting B.
Question 25 · multipleChoice
1 marks
During vigorous exercise, human muscle cells respire anaerobically. How does this process compare to anaerobic respiration in yeast?
A.Human muscle cells produce carbon dioxide, whereas yeast cells do not.
B.Yeast cells produce ethanol, whereas human muscle cells produce lactic acid.
C.Human muscle cells release far more energy per glucose molecule than yeast cells.
D.Yeast cells require oxygen to break down their products, whereas human muscle cells do not.
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Worked solution
Anaerobic respiration in yeast produces ethanol and carbon dioxide, whereas in human muscles it produces only lactic acid. Both pathways release a very small, identical amount of energy (2 ATP) per glucose molecule. Lactic acid must be broken down using oxygen in humans, whereas yeast does not break down ethanol in this manner.
Marking scheme
Award 1 mark for identifying that yeast produces ethanol and muscles produce lactic acid.
Question 26 · multipleChoice
1 marks
Two different species of beetles are classified in the same genus. Which statement about these two species must be correct?
A.They can interbreed to produce fertile offspring.
B.They share a more recent common ancestor with each other than with a species in a different genus.
C.They have identical base sequences in their DNA.
D.They occupy identical ecological niches in their habitat.
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Worked solution
Organisms in the same genus are more closely related evolutionary and share a more recent common ancestor than organisms in different genera. Since they are different species, they cannot interbreed to produce fertile offspring, nor do they have identical DNA or identical niches.
Marking scheme
Award 1 mark for the correct explanation of taxonomic relationships within the same genus.
Question 27 · multipleChoice
1 marks
An enzyme-controlled reaction was carried out at \(20\ ^\circ\text{C}\). The temperature was then increased to \(30\ ^\circ\text{C}\). Which statement explains the increase in the rate of reaction?
A.The activation energy of the reaction is lowered.
B.The enzymes and substrate molecules have more kinetic energy, leading to more frequent collisions.
C.The shape of the active site changes to fit the substrate molecule more tightly.
D.The concentration of the enzyme-substrate complexes decreases.
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Worked solution
Increasing temperature increases the kinetic energy of both enzyme and substrate molecules. This causes them to move faster, increasing the frequency of successful collisions and thus increasing the rate of reaction.
Marking scheme
Award 1 mark for explaining the role of kinetic energy and collision frequency in temperature effects on enzymes.
Question 28 · multipleChoice
1 marks
Which cell structures are present in both a human liver cell and a leaf palisade cell?
A.cell membrane, nucleus and ribosomes
B.cell wall, cytoplasm and mitochondria
C.cytoplasm, chloroplasts and cell membrane
D.nucleus, permanent vacuole and mitochondria
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Worked solution
Both animal cells (liver) and plant cells (palisade) contain a cell membrane, nucleus, ribosomes, cytoplasm, and mitochondria. Cell walls, chloroplasts, and large permanent vacuoles are only found in plant cells.
Marking scheme
Award 1 mark for identifying the organelles common to both plant and animal cells.
Question 29 · multipleChoice
1 marks
During which phase of a sigmoid population growth curve is the birth rate of the population equal to the death rate?
A.lag phase
B.log (exponential) phase
C.stationary phase
D.deceleration phase
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Worked solution
In the stationary phase of a sigmoid population growth curve, the population size remains constant (reaches carrying capacity) because the birth rate and death rate are equal.
Marking scheme
Award 1 mark for correctly identifying the stationary phase.
Question 30 · multipleChoice
1 marks
Which statement correctly describes passive immunity?
A.It is achieved only after a person has been infected by a pathogen.
B.It involves the production of antibodies by the body’s own lymphocytes.
C.It provides long-term protection through the creation of memory cells.
D.It provides immediate, temporary protection by injecting antibodies from another organism.
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Worked solution
Passive immunity is the short-term defense against a pathogen by antibodies acquired from another individual or organism. Because the body does not produce its own antibodies or memory cells, the protection is temporary but immediate.
Marking scheme
Award 1 mark for defining passive immunity as immediate, temporary protection via injected antibodies.
Question 31 · multipleChoice
1 marks
Which process is an example of physical digestion?
A.The breakdown of starch to maltose by amylase in the mouth.
B.The emulsification of fats into smaller droplets by bile salts in the duodenum.
C.The absorption of amino acids through the microvilli of the ileum.
D.The breakdown of proteins to peptides by pepsin in the stomach.
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Worked solution
Physical digestion involves breaking food down into smaller pieces without changing its chemical structure. Bile emulsifies large fat globules into tiny droplets to increase their surface area. Amylase and pepsin breakdown are chemical digestion, and absorption is not digestion.
Marking scheme
Award 1 mark for identifying fat emulsification by bile as physical digestion.
Question 32 · multipleChoice
1 marks
How do the structures of xylem vessels and phloem sieve tubes compare?
A.Xylem vessels have cytoplasm, whereas phloem sieve tubes are empty dead cells.
B.Xylem vessels have walls reinforced with lignin, whereas phloem sieve tubes do not.
C.Xylem vessels have sieve plates, whereas phloem sieve tubes have open ends.
D.Xylem vessels transport sucrose, whereas phloem sieve tubes transport mineral ions.
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Worked solution
Xylem vessels are dead, hollow cells with cell walls reinforced with waterproof lignin. Phloem sieve tubes are living cells with cytoplasm (but no nucleus or ribosomes), contain sieve plates, are not lignified, and transport sucrose and amino acids.
Marking scheme
Award 1 mark for correctly distinguishing between xylem and phloem structural features.
Question 33 · multipleChoice
1 marks
Which row correctly describes the products of anaerobic respiration and the relative energy released per glucose molecule compared to aerobic respiration?
A.Yeast products: carbon dioxide and ethanol; Muscle products: lactic acid; Energy released: much less
B.Yeast products: carbon dioxide and water; Muscle products: lactic acid; Energy released: much more
C.Yeast products: ethanol only; Muscle products: lactic acid and carbon dioxide; Energy released: much less
D.Yeast products: lactic acid; Muscle products: ethanol; Energy released: same amount
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Worked solution
Anaerobic respiration in yeast produces ethanol and carbon dioxide. In human muscles, it produces lactic acid (lactate). Both processes release substantially less energy per glucose molecule (approximately 2 ATP) compared to aerobic respiration (approximately 36-38 ATP).
Marking scheme
1 mark for the correct option (A).
Question 34 · multipleChoice
1 marks
Two different species of beetles look extremely similar physically but occupy different ecological niches. Which method would provide the most accurate evidence to determine how closely related these two species are?
A.Comparing the anatomical structures of their jointed limbs.
B.Comparing the base sequences of their DNA.
C.Comparing the habitats in which they feed and reproduce.
D.Comparing the overall size and mass of adult individuals.
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Worked solution
Comparing DNA base sequences (or amino acid sequences in proteins) provides the most direct and accurate evidence for evolutionary relationships and classification, as physical features can sometimes be misleading due to convergent evolution.
Marking scheme
1 mark for the correct option (B).
Question 35 · multipleChoice
1 marks
An enzyme-controlled reaction is carried out at pH 2, pH 7, and pH 12. The enzyme has an optimum pH of 7. Which statement about the enzyme molecules at pH 12 is correct?
A.They have gained kinetic energy, causing more frequent successful collisions with the substrate.
B.Their active sites have permanently changed shape, preventing substrate binding.
C.They are fully active because high pH prevents them from denaturing.
D.They have reverted to their original amino acid chain structure due to hydrolysis.
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Worked solution
Extreme pH levels (such as pH 12 for an enzyme with an optimum of pH 7) cause the enzyme to denature. Denaturation involves the permanent alteration of the active site's three-dimensional shape, meaning the substrate can no longer fit.
Marking scheme
1 mark for the correct option (B).
Question 36 · multipleChoice
1 marks
A single-celled organism is found to possess a cell wall, ribosomes, and a circular loop of DNA, but completely lacks mitochondria and a nucleus. To which group of organisms does this cell belong?
A.animals
B.bacteria
C.fungi
D.plants
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Worked solution
Bacterial cells are prokaryotic; they contain a cell wall, ribosomes, and a circular loop of DNA (plasmid/nucleoid), but they lack membrane-bound organelles such as mitochondria and a defined nucleus.
Marking scheme
1 mark for the correct option (B).
Question 37 · multipleChoice
1 marks
During which phase of a sigmoid population growth curve is the rate of reproduction equal to the rate of death?
A.lag phase
B.log (exponential) phase
C.stationary phase
D.death phase
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Worked solution
During the stationary phase of a sigmoid growth curve, the population size remains constant because the rate of cell division/reproduction is equal to the rate of death due to limiting factors like food supply and waste accumulation.
Marking scheme
1 mark for the correct option (C).
Question 38 · multipleChoice
1 marks
A child is bitten by a venomous snake and immediately given an injection of antivenom containing specific antibodies. Two years later, the child is bitten by the same species of snake but has no protection and suffers severe symptoms. Which type of immunity was provided by the antivenom injection?
A.active artificial immunity
B.active natural immunity
C.passive artificial immunity
D.passive natural immunity
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Worked solution
The antivenom injection provides passive artificial immunity because pre-made antibodies are injected directly into the body. This provides immediate but temporary protection, as no memory cells are created by the child's own immune system.
Marking scheme
1 mark for the correct option (C).
Question 39 · multipleChoice
1 marks
Which statement correctly describes a process of physical digestion in the human alimentary canal?
A.Hydrochloric acid in the stomach breaks large protein molecules into soluble peptides.
B.Amylase in saliva breaks down insoluble starch into maltose molecules.
C.Bile salts emulsify large fat droplets into smaller droplets, increasing their surface area.
D.Proteases in the small intestine alter the chemical bonds of polypeptides.
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Worked solution
Bile salts emulsifying fats is a mechanical/physical process because it breaks large lipid droplets into smaller lipid droplets, increasing the overall surface area for lipase to act upon. It does not alter the chemical structure of the lipid molecules themselves.
Marking scheme
1 mark for the correct option (C).
Question 40 · multipleChoice
1 marks
In early spring, a deciduous tree transports stored nutrients from its roots to growing buds before new leaves have developed. Which row correctly identifies the source, the sink, and the form of carbohydrate transported?
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Worked solution
In early spring, before leaves are present to photosynthesise, stored nutrients in the roots (source) are transported upwards to the growing buds (sink). Carbohydrates are transported through the phloem sieve tubes as sucrose.
Marking scheme
1 mark for the correct option (A).
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6 Question · 80 marks
Question 1 · structured
13 marks
A biology student discovered a small marine animal on a sandy shore. The animal has a segmented body, a hard exoskeleton, and jointed appendages.
(a) (i) State the phylum to which this marine animal belongs. [1] (ii) State two characteristics of this phylum that are visible without dissection. [2]
(b) Explain how modern scientists use DNA base sequencing to classify organisms and establish evolutionary relationships more accurately than by observing physical features alone. [4]
(c) Table 1.1 shows the percentage of genetic similarity in a specific protein-coding gene between a reference species (Species X) and four other related marine species (A, B, C, and D).
| Species | Percentage similarity of gene to Species X (%) | | :--- | :---: | | A | 94.2 | | B | 78.5 | | C | 89.1 | | D | 62.4 |
(i) Identify which species is most closely related to Species X, and explain your choice using the data in Table 1.1. [3] (ii) State what is meant by a dichotomous key, and describe how it is used to identify unfamiliar organisms. [3]
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Worked solution
1(a)(i) The presence of a hard exoskeleton, segmented body, and jointed limbs are characteristic diagnostic features of the phylum Arthropoda. 1(a)(ii) Visible features include jointed limbs, segmented body, and the exoskeleton. 1(b) DNA sequencing determines the exact order of nucleotides in a gene. Organisms with a more recent common ancestor have had less time to accumulate mutations, so their DNA sequences are more similar. Physical traits can sometimes be misleading due to convergent evolution (analogous structures), making molecular classification far more reliable. 1(c)(i) Species A shows the highest percentage genetic similarity (94.2%) to Species X. This indicates they share a more recent common ancestor compared to the other species listed. 1(c)(ii) A dichotomous key is an identification tool consisting of a series of choices, each with two options (di-). By choosing the option that matches the specimen's features, the user is guided systematically to the correct taxonomic name.
Marking scheme
1(a)(i) Arthropoda / arthropod; [1] 1(a)(ii) Any two from: segmented body; exoskeleton; jointed legs / limbs / appendages; [2] 1(b) Any four from: base sequence of DNA varies between species; more closely related organisms have more similar base sequences; fewer differences in sequences indicates a more recent common ancestor; physical features can be misleading due to convergent evolution; biochemical analysis is objective / quantitative; AVP (e.g., can compare non-coding regions / mitochondrial DNA); [4] 1(c)(i) Species A; [1] It has the highest genetic similarity (94.2%) to Species X; [1] This indicates they shared a common ancestor most recently / have had less time for mutations to accumulate; [1] 1(c)(ii) A system of identifying organisms using a series of paired contrasting choices (binary steps); [1] By choosing the statement that matches the characteristics of the specimen; [1] And following the directions at each step until the organism's identity / name is reached; [1]
Question 2 · structured
13 marks
Fig. 2.1 is an incomplete diagram of a palisade mesophyll cell from a leaf.
(a) (i) List two cell organelles or structures that are present in both a palisade mesophyll cell and a root hair cell. [2] (ii) List two cell structures that are found in a palisade mesophyll cell but are absent from a root hair cell. [2]
(b) Explain how the internal structure of a palisade mesophyll cell is highly adapted to maximize the efficiency of photosynthesis. [5]
(c) Most leaf epidermal cells do not contain chloroplasts. Suggest why this is an advantage for the leaf as a photosynthesizing organ. [4]
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Worked solution
2(a)(i) Both plant cell types share standard eukaryotic features like the cell wall, cell membrane, cytoplasm, nucleus, and a large central vacuole. 2(a)(ii) Root hair cells are located underground and do not perform photosynthesis, thus they lack chloroplasts and associated structures like starch grains. 2(b) Palisade mesophyll cells are situated directly below the upper epidermis. They contain many chloroplasts, which are pushed to the periphery of the cytoplasm by a large central vacuole to shorten the diffusion distance for carbon dioxide and capture maximum light. Their elongated columnar shape allows many cells to pack tightly together under the leaf surface. 2(c) Epidermal cells form the outer boundary. Lacking chloroplasts means they are transparent, permitting light to travel unimpeded to the light-harvesting palisade mesophyll beneath.
Marking scheme
2(a)(i) Any two from: cell wall; cell membrane; cytoplasm; nucleus; (large permanent) vacuole; mitochondria; ribosomes; [2] 2(a)(ii) Chloroplasts; starch grains (reject chlorophyll as it is a pigment, not a structure); [2] 2(b) Any five from: many chloroplasts; to absorb maximum light; chloroplasts can move within cytoplasm to receive optimum light; large permanent vacuole pushes chloroplasts to edge of cell; shorter diffusion distance for gases; tall / column-shaped cell / packed tightly vertically; allows many cells to fit in the upper layer; thin cell wall for rapid diffusion of carbon dioxide / water; [5] 2(c) Any four from: allows light to pass through the epidermal layer; light reaches the palisade layer below; where most photosynthesis occurs; protects the inner layers without blocking light; prevents energy from being wasted on chloroplast production in cells that receive less optimal gas exchange; [4]
Question 3 · structured
13 marks
A student investigated the effect of pH on the activity of pepsin, a protease enzyme found in the mammalian stomach.
(a) Define the term *optimum pH* and explain how a pH far below or above this value can decrease the rate of an enzyme-controlled reaction. [4]
(b) Table 3.1 shows the rate of reaction of pepsin at different pH values.
(i) Describe the effect of pH on the rate of reaction of pepsin shown in Table 3.1. [3] (ii) Calculate the percentage decrease in pepsin activity when the pH increases from pH 2.0 to pH 3.0. Show your working. [3]
(c) Pepsin acts on proteins in the stomach. Describe what happens to the protein molecules when they are digested by pepsin. [3]
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Worked solution
3(a) Optimum pH is the narrow range where the enzyme's three-dimensional shape is perfectly maintained, leading to the highest rate of reaction. Drastic pH shifts alter the charge of amino acids, breaking hydrogen and ionic bonds. This denatures the active site, making it impossible for the substrate to bind. 3(b)(i) From pH 1 to 2, the rate of reaction increases significantly from 12 to 48 a.u. Above pH 2, the rate of reaction drops back down to 32 a.u. (pH 3), 8 a.u. (pH 4), and reaches complete inactivity (0 a.u.) by pH 5. 3(b)(ii) Change in activity: \( 48 - 32 = 16 \) a.u. Initial activity at pH 2.0: 48 a.u. Percentage decrease: \( \frac{16}{48} \times 100 = 33.33 \% \). 3(c) Protease enzymes target the peptide bonds connecting amino acids. Digestion cleaves these bonds, hydrolyzing large, insoluble proteins into smaller, soluble polypeptides and peptides.
Marking scheme
3(a) Optimum pH is the pH at which the enzyme works at its maximum rate / most active; [1] Extreme pH alters the ionic charges of amino acids; [1] This breaks ionic/hydrogen bonds holding the tertiary structure together; [1] This denatures the enzyme / changes the shape of the active site so the substrate can no longer fit / form enzyme-substrate complexes; [1] 3(b)(i) Rate of reaction increases as pH increases from pH 1.0 to pH 2.0; [1] Reaches peak / optimum at pH 2.0 (rate is 48 a.u.); [1] Rate of reaction decreases above pH 2.0 until it reaches 0 at pH 5.0; [1] 3(b)(ii) Working: difference = 48 - 32 = 16; [1] Percentage decrease = \( \frac{16}{48} \times 100 \); [1] Answer = 33% / 33.3% (accept 33.33%); [1] 3(c) Proteins are large, insoluble molecules; [1] Pepsin breaks the peptide bonds; [1] Converting them into smaller, soluble peptides / polypeptides / amino acids; [1]
Question 4 · structured
14 marks
Vigorous physical exercise requires a rapid supply of energy to contracting muscles.
(a) (i) Write the balanced chemical equation for anaerobic respiration in human muscle cells. [2] (ii) Contrast this by writing the word equation for anaerobic respiration in yeast cells. [2]
(b) During a 100-meter sprint, an athlete breathes heavily but cannot supply enough oxygen to the muscles for aerobic respiration alone. Explain why lactic acid builds up in the muscles during this sprint, and explain how the body processes this lactic acid after the sprint has ended. [6]
(c) Suggest two advantages to the athlete of having a well-developed anaerobic respiration pathway, despite its low yield of ATP per glucose molecule. [4]
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Worked solution
4(a)(i) Under anaerobic conditions, glucose is converted to lactic acid: \( C_6H_{12}O_6 \rightarrow 2C_3H_6O_3 \). 4(a)(ii) In yeast, the pathway produces alcohol and gas: glucose \( \rightarrow \) ethanol + carbon dioxide. 4(b) During high-intensity bursts, aerobic pathways are limited by oxygen delivery. Cells shift to anaerobic glycolysis, generating lactic acid. Post-exercise, the lactic acid is cleared via blood circulation to the liver, where it requires oxygen (oxygen debt) to be metabolized back into glucose or fully oxidized. 4(c) Even though anaerobic respiration is inefficient (2 ATP per glucose compared to 36-38 ATP aerobically), it produces ATP at a much faster rate. This instantaneous availability of energy is crucial for short-duration power outputs.
Marking scheme
4(a)(i) \( C_6H_{12}O_6 \rightarrow 2C_3H_6O_3 \); [2] (1 mark for correct formulas, 1 mark for correct balancing) 4(a)(ii) glucose \( \rightarrow \) ethanol + carbon dioxide; [2] (1 mark for glucose, 1 mark for products) 4(b) During intense exercise, oxygen demand exceeds supply; [1] Muscle cells respire anaerobically to meet energy demands; [1] Glucose is partially broken down into lactic acid; [1] After exercise, heart rate and breathing rate remain high to pay back the oxygen debt; [1] Lactic acid is transported by the blood from muscles to the liver; [1] Where it is oxidized (aerobically broken down) to carbon dioxide and water / converted back to glucose; [1] 4(c) Any two from: Allows energy to be produced very rapidly (faster than aerobic respiration); [2] Does not require oxygen, allowing muscles to contract when oxygen levels are depleted; [2] Enables short bursts of high-intensity performance (e.g., sprinting / escaping danger); [2]
Question 5 · structured
14 marks
Immunity plays a vital role in protecting the body against transmissible diseases.
(a) State what is meant by the term *pathogen*. [1]
(b) Distinguish between *active immunity* and *passive immunity*, making reference to the source of antibodies and the duration of protection. [4]
(c) Explain how vaccination protects a child from future infection by a specific pathogen, such as the measles virus. [5]
(d) Some vaccines require a 'booster' dose several years after the initial vaccination. Suggest the immunological reason for this booster dose. [4]
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Worked solution
5(a) A pathogen is defined simply as any biological agent (organism) that causes disease. 5(b) Active immunity occurs when the host's immune system manufactures its own antibodies in response to antigen exposure, producing memory cells for long-term defense. Passive immunity occurs when pre-formed antibodies are introduced from the outside, giving immediate but temporary protection without forming memory cells. 5(c) Vaccines contain dead or inactive antigens. These stimulate the primary immune response where lymphocytes produce complementary antibodies and differentiate into memory cells. Upon actual pathogen exposure, memory cells launch a highly rapid, robust secondary response. 5(d) Over time, circulating antibody titers and the population of specific memory cells may dwindle. A booster dose mimics a secondary infection, stimulating memory cells to rapidly multiply and elevate antibody titers back to protective thresholds.
Marking scheme
5(a) A disease-causing organism; [1] 5(b) Active immunity involves the production of antibodies by the individual's own body, whereas passive immunity involves receiving antibodies from an external source; [2] Active immunity provides long-term protection (due to memory cells), whereas passive immunity provides only short-term protection (no memory cells, antibodies are broken down); [2] 5(c) Any five from: vaccine contains weakened / dead / harmless version of the pathogen / antigen; antigens trigger an immune response; lymphocytes produce specific antibodies; lymphocytes clone themselves; some lymphocytes become memory cells; memory cells remain in the blood for a long time; if the real pathogen enters later, memory cells recognize it and produce antibodies much faster / in larger quantities; [5] 5(d) Any four from: concentration of antibodies / memory cells declines over time; booster dose re-introduces the antigen; stimulating memory cells to divide / produce more memory cells; causing a secondary immune response; which is faster / produces higher levels of antibodies; ensuring long-term / life-long protection remains active; [4]
Question 6 · structured
13 marks
An ecological study monitored the population of a herbivorous rodent species on an isolated reserve over a 30-year period.
(a) Describe the three main phases of a typical sigmoid population growth curve that occur before a population reaches its carrying capacity. [4]
(b) State three abiotic or biotic factors that can act as limiting factors to prevent a herbivore population from increasing indefinitely. [3]
(c) At Year 15, a carnivorous predator was introduced to the reserve to control the rodent population. Explain the likely effects of this introduction on the food web of the reserve, and describe why careful population monitoring is essential in conservation management. [6]
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Worked solution
6(a) Sigmoid growth starts with the Lag phase (organisms adapt, numbers are low). This is followed by the Log/Exponential phase (excess resources, low competition, rapid doubling). As resources begin to deplete, the Deceleration phase is reached (growth rate slows down, mortality rates rise toward birth rates). 6(b) Primary limiting factors for land herbivores are food resources, density-dependent transmissible diseases, competition for nesting spaces, and water availability. 6(c) Introducing a predator creates top-down control. By preying on the rodents, their population drops, reducing the grazing pressure on vegetation. This cascade can allow native flora to recover, benefiting other primary consumers. However, if not monitored, the predator could cause local extinctions, destabilizing other trophic interactions.
Marking scheme
6(a) Lag phase: slow growth because population is small / adapting to environment; [1] Log / exponential phase: rapid growth because birth rate exceeds death rate / resources are abundant; [1] Deceleration phase: growth slows down as limiting factors begin to affect the population; [1] Stationary phase: population size stabilizes / birth rate equals death rate; [1] (Max 4 marks total) 6(b) Any three from: availability of food; disease / spread of pathogens; availability of water; shelter / nesting sites; accumulation of toxic waste; temperature / light (abiotic); [3] 6(c) Any six from: predator will feed on the rodent, causing its population to decline; this reduces overgrazing / consumption of primary producers (plants); plant biomass / biodiversity increases; other herbivores might have less competition for food; introducing a new predator could disrupt existing food chains (e.g., predator might eat other non-target native species); monitoring ensures the predator does not drive the prey to extinction; helps ecologists detect imbalance early and adjust management strategies; [6]
Paper 62 (Alternative to Practical)
Answer all questions. Show your experimental design, drawing, and calculations where appropriate.
3 Question · 40 marks
Question 1 · practical
13 marks
A student investigated the rate of anaerobic respiration in yeast when supplied with different carbohydrate substrates: glucose, maltose, and starch.
They set up three identical boiling tubes, each containing a yeast suspension and one of the carbohydrates. A gas syringe was connected to each tube to collect the carbon dioxide gas produced. The volume of gas collected in the syringe after 10 minutes was recorded.
(a) (i) Prepare a table to record these results. [3] (a) (ii) State a conclusion for these results. [1] (a) (iii) Suggest why the rate of respiration was lowest with starch as the substrate. [2]
(b) Describe a test that could be used to show that the gas produced was carbon dioxide, and state the positive result of this test. [2]
(c) Plan an investigation to determine the effect of temperature on the rate of anaerobic respiration in yeast. [5]
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Worked solution
(a) (i) Table with two columns: 'Carbohydrate substrate' and 'Volume of gas collected / cm³'. The three substrates (glucose, maltose, starch) should be listed with their corresponding values. (a) (ii) Yeast respires glucose the fastest/most rapidly, followed by maltose, and starch is respired the slowest/least. (a) (iii) Starch is a large, insoluble polymer/polysaccharide that cannot be directly absorbed or used by yeast cells; it must first be broken down/hydrolysed into simple sugars (like glucose) by enzymes, which takes time.
(b) Bubble/pass the gas through limewater; the limewater turns cloudy/milky.
(c) Plan: 1. Independent variable: Use at least 5 different temperatures (e.g., \(10\,^{\circ}\text{C}\), \(20\,^{\circ}\text{C}\), \(30\,^{\circ}\text{C}\), \(40\,^{\circ}\text{C}\), \(50\,^{\circ}\text{C}\)) controlled using thermostatically controlled water-baths. 2. Dependent variable: Measure the volume of gas produced in a set time (e.g., 5 minutes) using a gas syringe or count the number of bubbles produced per minute. 3. Constant variables: Keep the concentration and volume of yeast suspension constant; keep the concentration and volume of glucose solution constant. 4. Repeats: Repeat the experiment at least three times at each temperature and calculate a mean. 5. Safety: Wear eye protection / handle hot water-baths with care to prevent scalds.
Marking scheme
1(a)(i) [Total: 3 marks] - Table drawn with clear boundary lines and a header row [1] - Column headings with appropriate units: 'Carbohydrate substrate' and 'Volume of gas collected / cm^3' (accept 'Volume of gas / cm3') [1] - All data correctly entered [1]
1(a)(ii) [Total: 1 mark] - Correct conclusion: glucose is respired the fastest / rate of respiration is highest with glucose / starch is respired the slowest [1]
1(a)(iii) [Total: 2 marks] - Starch is a large molecule/polymetric/polysaccharide [1] - Starch must be digested/broken down/hydrolysed into glucose before respiration can occur [1]
1(b) [Total: 2 marks] - Pass gas into limewater [1] - Limewater turns cloudy / milky / chalky [1] (Reject: white precipitate alone without stating limewater starts clear)
1(c) [Total: 5 marks] - Use of thermostatically controlled water-baths to achieve at least 5 different temperatures [1] - Measurement of volume of gas in a fixed time / count bubbles per unit time [1] - Standardise yeast concentration/volume AND sugar concentration/volume [1] - Repeat trials (at least 3 times) and calculate mean [1] - Relevant safety precaution: safety goggles / care with hot water [1]
Question 2 · practical
13 marks
Sweet potato tissues contain the enzyme catalase, which breaks down hydrogen peroxide into water and oxygen gas. A student investigated the effect of pH on catalase activity.
Sweet potato cylinders of equal length (\(30\text{ mm}\)) were placed in boiling tubes containing hydrogen peroxide solutions adjusted to different pH values. The volume of oxygen gas produced in 3 minutes was measured.
(a) (i) Plot a line graph on a grid of the volume of oxygen produced against pH. [4] (a) (ii) Determine the optimum pH for catalase in sweet potato from your graph. [1] (a) (iii) State the independent and dependent variables in this investigation. [2]
(b) (i) Describe how the sweet potato cylinders should be prepared to ensure this investigation is a fair test. [2] (b) (ii) Suggest why the volume of oxygen produced decreased at pH 11. [2]
(c) Identify one potential source of error in the measurement of the gas produced, and suggest an improvement to overcome this error. [2]
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Worked solution
(a) (i) The graph should have pH on the x-axis (scale 0 to 12, linear) and Volume of oxygen on the y-axis (scale 0 to 16, linear). Both axes must be fully labelled with units where appropriate. Points plotted accurately with a small 'x' or dot, and connected with a smooth curve or neat straight lines. (a) (ii) \(7.0\) (or based on the peak of the student's plotted graph). (a) (iii) Independent variable: pH. Dependent variable: Volume of oxygen produced in 3 minutes.
(b) (i) Cut cylinders using the same cork borer to ensure they have the same diameter/surface area; use a scalpel and ruler to ensure they are exactly \(30\text{ mm}\) long; remove any skin from the potato. (b) (ii) The highly alkaline pH (pH 11) denatured the catalase enzyme; the shape of the active site changed, so it was no longer complementary to the hydrogen peroxide substrate.
(c) Error: Gas may escape when inserting the bung/cork into the boiling tube at the start. Improvement: Use a divided flask (reaction flask with a partition) so sweet potato and hydrogen peroxide are mixed only after the bung is fully secured.
Marking scheme
2(a)(i) [Total: 4 marks] - Axes labelled with units: 'pH' on x-axis and 'Volume of oxygen produced / cm3' on y-axis [1] - Suitable linear scale, where plotted points occupy at least half of the grid in both directions [1] - All 5 points plotted accurately to within half a small square [1] - Points connected with a smooth curve or straight lines from point to point with no extrapolation [1]
2(a)(ii) [Total: 1 mark] - Optimum pH: 7.0 (allow value corresponding to peak of graph) [1]
2(a)(iii) [Total: 2 marks] - Independent variable: pH [1] - Dependent variable: Volume of oxygen / gas produced [1] (Reject: rate of reaction without reference to volume or time)
2(b)(i) [Total: 2 marks] - Use same cork borer / ensure same diameter [1] - Measure accurately with ruler to ensure same length (30 mm) / peel skin to ensure uniform tissue exposure [1]
2(b)(ii) [Total: 2 marks] - Enzyme/catalase is denatured [1] - Active site changes shape so substrate/hydrogen peroxide can no longer bind [1]
2(c) [Total: 2 marks] - Source of error: Gas escapes when inserting stopper/bung OR temperature fluctuations [1] - Improvement: Use a delivery tube assembly with a partition flask to mix reactants after sealing OR perform in a water-bath to maintain constant temperature [1]
Question 3 · practical
14 marks
A student investigated osmosis in potato tissue using different concentrations of sucrose solution.
Potato cylinders were weighed and placed into sucrose solutions of concentrations \(0.0\), \(0.2\), \(0.4\), \(0.6\), and \(0.8\text{ mol/dm}^3\) for 2 hours. After 2 hours, the cylinders were removed, blotted dry, and reweighed.
The results for the \(0.8\text{ mol/dm}^3\) sucrose solution were: - Initial mass of potato cylinder = \(4.10\text{ g}\) - Final mass of potato cylinder = \(3.48\text{ g}\)
(a) (i) Calculate the percentage change in mass for the potato cylinder in the \(0.8\text{ mol/dm}^3\) sucrose solution. Show your working and round your answer to two significant figures. [3] (a) (ii) Explain why it is better to calculate the percentage change in mass rather than the absolute change in mass. [1]
(b) (i) Describe the net movement of water that caused the mass of the potato cylinder to increase in the \(0.0\text{ mol/dm}^3\) solution (pure water). [3] (b) (ii) State two variables that must be controlled in this investigation. [2]
(c) The student observed one of the potato cells under a microscope after it had been in the \(0.8\text{ mol/dm}^3\) sucrose solution.
(i) Describe the expected appearance of this cell. [2] (ii) State the term used to describe a plant cell in this condition. [1] (iii) Suggest how the student could restore this cell to its normal, turgid state. [2]
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Worked solution
(a) (i) Working: Change in mass = \(3.48\text{ g} - 4.10\text{ g} = -0.62\text{ g}\) Percentage change = \(\left(\frac{-0.62}{4.10}\right) \times 100 = -15.12\%\) Rounded to two significant figures = \(-15\%\) (or a loss of \(15\%\)).
(a) (ii) The initial masses of the potato cylinders were not exactly the same. Calculating percentage change allows for a fair comparison between them.
(b) (i) Water moved into the potato cells by osmosis; down a water potential gradient (from a higher water potential outside to a lower water potential inside the cells); across a partially permeable membrane.
(b) (ii) Any two from: - Temperature of the solutions - Volume of the sucrose solution - Time the cylinders are left in the solution (2 hours) - Source/variety of the potato
(c) (i) The cell membrane has pulled away from the cell wall; the vacuole is shrunken/smaller. (c) (ii) Plasmolysed (accept plasmolysis). (c) (iii) Place the cell in pure water / distilled water; water will enter the cell by osmosis, expanding the vacuole and pushing the cytoplasm back against the cell wall.
Marking scheme
3(a)(i) [Total: 3 marks] - Correct calculation of mass change (-0.62 g) [1] - Correct calculation of percentage change (-15.12%) [1] - Correct rounding to two significant figures: -15% or 15% decrease [1]
3(a)(ii) [Total: 1 mark] - To allow comparison because the starting/initial masses of potato cylinders were different [1]
3(b)(i) [Total: 3 marks] - Water moves into the potato cells by osmosis [1] - From a high water potential to low water potential / down a water potential gradient [1] - Across a partially permeable membrane [1]
3(b)(ii) [Total: 2 marks] - Any two controlled variables: temperature, volume of solution, duration of immersion, surface area/dimensions of cylinders, potato variety [2]
3(c)(i) [Total: 2 marks] - Cytoplasm/cell membrane pulled away from the cell wall [1] - Vacuole is shrunken / smaller volume [1]
3(c)(ii) [Total: 1 mark] - Plasmolysed [1]
3(c)(iii) [Total: 2 marks] - Place the potato cell/tissue in distilled water / pure water / solution of higher water potential [1] - Water enters by osmosis to make it turgid again [1]
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