2025 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Antibodies transferred from mother to baby via breast milk represent passive immunity. Because the baby's own immune system does not undergo an immune response to produce these antibodies, no memory cells are formed.

1 mark for identifying the correct combination: passive immunity and no memory cells produced.

Goblet cells secrete mucus to trap dust particles and pathogens. Ciliated cells have tiny hair-like structures called cilia that beat in a coordinated way to move the mucus up and away from the lungs towards the throat, where it can be swallowed.

1 mark for identifying the correct functions of both goblet cells and ciliated cells.

Yeast respires anaerobically (fermentation) in the dough, producing carbon dioxide and ethanol. The carbon dioxide gas is trapped in the dough, forming bubbles that expand and cause the dough to rise. The ethanol evaporates during baking.

1 mark for identifying anaerobic respiration and carbon dioxide as the key process and product.

At high temperatures above the optimum, the increased thermal energy causes the bonds maintaining the enzyme's three-dimensional shape to break. This changes the shape of the active site permanently (denaturation), so the substrate can no longer bind to it.

1 mark for identifying that the active site changes shape permanently, preventing substrate binding.

Concentrated sucrose solution has a lower water potential than the cell cytoplasm. Water moves out of the cells by osmosis down the water potential gradient, causing the vacuole and cytoplasm to shrink and pull away from the cell wall, leaving the cells plasmolysed.

1 mark for identifying water leaving the cells and the cells becoming plasmolysed.

Xerophytes have adaptations to reduce transpiration in dry environments. Stomata sunk in pits or grooves trap moist air near the leaf surface, reducing the water potential gradient between the inside of the leaf and the atmosphere, thereby slowing down transpiration.

1 mark for identifying sunken stomata as a xerophytic adaptation that reduces transpiration.

In the nitrogen cycle, nitrification is the process where nitrifying bacteria convert ammonium ions first into nitrites and then into nitrates, which can be absorbed by plants.

1 mark for identifying nitrification as the correct process.

Crossing two heterozygous plants (Rr x Rr) produces offspring with genotypes RR, Rr, and rr in a \(1:2:1\) ratio. Since R is dominant, RR and Rr genotypes result in red flowers, while rr results in white flowers, giving a phenotypic ratio of \(3:1\) (red to white).

1 mark for the correct phenotypic ratio of \(3:1\) (red to white).

Option C is correct because recovering from an infection like measles stimulates the body's own immune system to produce its own antibodies and memory cells, which is natural active immunity. Option A is natural passive immunity. Option B is artificial passive immunity. Option D is artificial active immunity.

1 mark: Correct option chosen (C).

Inspiration is an active process where the external intercostal muscles contract while the internal intercostal muscles relax, pulling the ribcage upwards and outwards. At the same time, the diaphragm contracts and flattens, increasing the volume of the thorax and decreasing the pressure inside the lungs below atmospheric pressure, drawing air into the lungs.

1 mark: Correct option chosen (A).

At extreme pH values such as pH 3, the hydrogen bonds holding the three-dimensional shape of the enzyme amylase are disrupted. This denatures the enzyme, permanently changing the shape of its active site. As a result, the substrate (starch) can no longer fit into the active site, and no reaction occurs. Starch is a polysaccharide and does not denature in the way protein enzymes do, and pH does not directly reduce kinetic energy (which is affected by temperature).

1 mark: Correct option chosen (A).

The \(0.0\text{ mol/dm}^3\) solution (pure water) has a higher water potential than the cell sap of the potato cells. Water enters the potato cells by osmosis, causing them to expand and show the greatest percentage increase in mass. The \(0.8\text{ mol/dm}^3\) sucrose solution has a much lower water potential than the cell sap. Water leaves the potato cells by osmosis, causing the cell membrane to pull away from the cell wall, which is plasmolysis.

1 mark: Correct option chosen (A).

Temperature is kept constant in a fermenter using a water jacket because the respiration of the fungus releases heat energy, which could otherwise raise the temperature and denature the enzymes needed for growth and penicillin synthesis. Penicillium requires aerobic conditions, so oxygen is supplied (not carbon dioxide). pH is controlled using acids/alkalis, not nutrients. Stirring paddles are for mixing and heat distribution; Penicillium is a fungus and cannot photosynthesise.

1 mark: Correct option chosen (A).

Xerophytes are adapted to conserve water. Sunken stomata in pits trap moist air, reducing the diffusion gradient for water vapour out of the leaf. A thick, waxy cuticle reduces evaporation from the leaf surface. A small leaf surface area to volume ratio (like spines or needle-like leaves) reduces the total area available for transpiration.

1 mark: Correct option chosen (A).

The heterozygous black rabbit has the genotype \(Bb\). The brown rabbit, having the recessive phenotype, must be homozygous recessive with the genotype \(bb\). A cross between \(Bb\) and \(bb\) results in gametes \(B\) and \(b\) from the heterozygous parent, and gamete \(b\) from the homozygous parent. The offspring genotypes will be \(Bb\) (black fur) and \(bb\) (brown fur) in a 1:1 ratio. Therefore, the expected phenotypic ratio is 1 black : 1 brown.

1 mark: Correct option chosen (B).

During the log (exponential) phase, growth is rapid and the reproduction rate far exceeds the death rate because there is an abundance of food, water, and space, and waste products have not yet reached toxic levels. In the lag phase, bacteria are adapting to the environment and not dying rapidly. In the stationary phase, reproduction rate equals death rate because resources are limiting. In the death phase, death rate exceeds reproduction rate because nutrients are depleted and toxic waste has accumulated.

1 mark: Correct option chosen (B).

Passive immunity involves the transfer of antibodies from another source (e.g., across the placenta or via breast milk, or through an injection). Because the host's immune system is not stimulated to make its own antibodies, no memory cells are produced. Consequently, this immunity is short-term as the foreign antibodies are eventually broken down by the body.

1 mark for B. Correctly identifies that passive immunity does not produce memory cells, involves antibodies from another organism, and provides short-term protection.

During expiration, the external intercostal muscles relax and the diaphragm relaxes (moving upwards into its dome shape). This decreases the volume of the thorax, which increases the pressure inside the lungs relative to the atmosphere, forcing air out. During forced expiration, the internal intercostal muscles contract.

1 mark for B. Correctly identifies the muscle movements and pressure changes that drive expiration.

Pectinase digests pectin, a polysaccharide found in plant cell walls. By breaking down the cell walls, it helps release more juice from the plant cells (increasing yield) and clarifies the juice by removing suspended pectin particles.

1 mark for A. Identifies that pectinase digests plant cell walls to increase juice yield and clarity.

As temperature increases up to the optimum (typically around \(37^\circ\text{C}\) to \(40^\circ\text{C}\) for many animal enzymes), the kinetic energy of both enzyme and substrate molecules increases. This causes them to move faster, leading to a higher frequency of successful collisions between substrate molecules and the active site of the enzymes, thus raising the rate of reaction.

1 mark for A. Connects temperature increase with kinetic energy and the frequency of successful collisions.

The concentrated sucrose solution has a lower water potential than the cell sap. Water moves out of the cell by osmosis. The cytoplasm and vacuole shrink, pulling the cell membrane away from the cell wall (plasmolysis). The cell wall is semi-rigid and maintains its structural shape.

1 mark for B. Correctly identifies water loss, plasmolysis of the cytoplasm, and the stability of the cell wall.

Hydrophytes live in or on water. Floating hydrophytes have a thin cuticle since water conservation is not an issue. Stomata are found only on the upper surface of the leaves so they can exchange gases with the air rather than being submerged. They contain large air spaces (aerenchyma) in their tissues to provide buoyancy and allow gas diffusion.

1 mark for B. Recognises thin cuticle, upper surface stomata, and abundant air spaces as adaptations of a floating hydrophyte.

Denitrification is the conversion of nitrate ions (\(\text{NO}_3^-\)) into nitrogen gas (\(\text{N}_2\)). This process is carried out by denitrifying bacteria under anaerobic conditions, which are typically found in waterlogged soils.

1 mark for C. Correctly identifies denitrification as the nitrate-reducing pathway occurring in waterlogged soil.

The cross is between a pink plant (\(RW\)) and a white plant (\(WW\)). The gametes produced by the pink parent are \(R\) and \(W\), while the white parent produces only \(W\) gametes. The offspring genotypes will be \(1\,RW\) (pink) to \(1\,WW\) (white). Thus, the phenotypic ratio is 1 pink : 1 white.

1 mark for A. Correctly applies co-dominance rules to derive the 1:1 ratio of pink to white phenotypes.

Injecting ready-made antibodies provides passive immunity because the body did not manufacture the antibodies itself. This type of immunity is temporary and does not result in the production of memory cells, as the recipient's immune system is not activated to produce its own response.

Award 1 mark for the correct answer D. No marks are given for incorrect choices.

Expired air contains more carbon dioxide (about 4% compared to 0.04% in inspired air) because it is a waste product of respiration. It contains less oxygen (about 16% compared to 21% in inspired air) because oxygen is absorbed for aerobic respiration. It also contains more water vapor (it is saturated) compared to inspired air, which has variable water vapor content.

Award 1 mark for the correct answer A. No marks are given for incorrect choices.

Sterile air is bubbled through the fermenter to provide a continuous supply of oxygen, which is essential for the aerobic respiration of Penicillium. The bubbling action also helps to stir and mix the liquid, keeping the fungus in suspension so it remains in contact with nutrients.

Award 1 mark for the correct answer B. No marks are given for incorrect choices.

Increasing the temperature above the optimum breaks bonds within the enzyme molecule. This changes the tertiary structure and permanently alters the shape of the active site, meaning the substrate can no longer fit. This process is called denaturation.

Award 1 mark for the correct answer B. No marks are given for incorrect choices.

A concentrated salt solution has a lower water potential than the plant cells' cytoplasm. Water therefore moves out of the cells by osmosis down a water potential gradient. This causes the cell membrane to pull away from the cell wall, making the cells plasmolysed.

Award 1 mark for the correct answer C. No marks are given for incorrect choices.

Xerophytic plants live in dry environments and have features to reduce the rate of transpiration. These include sunken stomata (which trap moist air and reduce the concentration gradient), a thick waxy cuticle (which acts as a barrier to evaporation), and rolled leaves (which keep stomata sheltered in a humid microclimate).

Award 1 mark for the correct answer B. No marks are given for incorrect choices.

Nitrification is the biological process where nitrifying bacteria in the soil oxidize ammonium ions into nitrite ions, and then into nitrate ions, which can be absorbed by plant roots.

Award 1 mark for the correct answer B. No marks are given for incorrect choices.

Since the parents have normal pigmentation but produced an albino child (genotype \(aa\)), both parents must be heterozygous carriers (genotype \(Aa\)). A cross of two heterozygous individuals (\(Aa \times Aa\)) results in a offspring ratio of 1 \(AA\) : 2 \(Aa\) : 1 \(aa\). The probability of an offspring inheriting the homozygous recessive genotype (\(aa\)) is 1 in 4, or 25%.

Award 1 mark for the correct answer B. No marks are given for incorrect choices.

The injection of antivenom containing antibodies provides passive immunity. In passive immunity, antibodies are introduced from an external source, which offers immediate but temporary protection. No memory cells are produced by the patient's immune system, and the injected antibodies are eventually broken down and removed from the blood. Therefore, upon a second bite three years later, the patient has no immunological memory of the venom, and their immune system will respond with the same slow primary response as before.

Award 1 mark for the correct option C. Reject A because passive immunity does not produce memory cells. Reject B because antibodies from passive immunity are temporary and do not persist for years. Reject D because the first injection does not suppress the immune system's future responses.

Carbon dioxide produced in respiring cells diffuses into the capillaries. Deoxygenated blood travels through veins to the vena cava, enters the right side of the heart, and is pumped via the pulmonary artery to the lungs. In the lungs, carbon dioxide diffuses from the capillaries into the alveoli to be exhaled.

Award 1 mark for the correct option B. Option A is incorrect because the pulmonary vein carries oxygenated blood from lungs to heart. Options C and D have incorrect pathways of systemic and pulmonary circulation.

Egg white is composed primarily of proteins (such as albumin). Protease is an enzyme that specifically catalyses the breakdown of proteins into soluble peptides and amino acids, which are easily washed away. Butter contains lipids, grass contains chlorophyll, and orange juice contains sugars and organic acids, none of which are substrates for protease.

Award 1 mark for the correct option B. Reject A because lipid stains require lipase. Reject C and D because chlorophyll and sugars are not digested by protease.

The data shows that the enzyme is highly active at pH 1.5 (relative activity 95) and completely inactive at pH 7.0 and 9.0. In the human digestive system, the stomach is highly acidic due to the secretion of hydrochloric acid, typically maintaining a pH of around 1.5 to 2.0. This is the optimum pH for stomach proteases (like pepsin). The mouth, duodenum, and ileum are neutral to slightly alkaline, where this enzyme would be inactive.

Award 1 mark for the correct option B. Reject A, C, and D because these regions of the digestive system do not have the highly acidic pH (around 1.5) required for this enzyme's activity.

Solution P (0.1 mol/dm³) has a higher water potential than the potato cells (0.3 mol/dm³), so water enters the cells by osmosis, causing an increase in mass. Solution Q (0.3 mol/dm³) is isotonic, so there is no net movement of water and the mass remains constant. Solution R (0.6 mol/dm³) has a lower water potential than the potato cells, so water leaves the cells by osmosis, causing a decrease in mass. This matches Option B.

Award 1 mark for selecting the correct option B. Reject A, C, and D because they misidentify the direction of water movement via osmosis.

Xerophytes are adapted to survive in dry environments by minimizing transpiration. Rolled leaves trap moist air, reducing the diffusion gradient. A thick waxy cuticle acts as a physical barrier to water loss. Sunken stomata trap moist air in pits, reducing the rate of diffusion. Broad leaves, thin cuticles, and lack of cuticles are features of hydrophytes or mesophytes.

Award 1 mark for the correct option B. Reject A and C because these are adaptations of hydrophytes (water plants). Reject D because thin cuticles and large flat leaves increase transpiration rates.

Processes 2 (respiration of decomposers breaking down dead organic matter) and 3 (respiration of plants) release carbon dioxide gas into the atmosphere. Process 1 (photosynthesis) and Process 4 (shell formation) remove carbon dioxide from the atmosphere or water.

Award 1 mark for the correct option C. Reject options containing 1 or 4 because these processes absorb rather than release carbon dioxide.

The man has brown eyes, which is a dominant trait, meaning he has at least one allele \(B\) (genotype \(B\_\)). His father had blue eyes, which is a recessive trait, meaning his father's genotype was \(bb\). Since his father could only pass on the allele \(b\), the man must be heterozygous (\(Bb\)). The woman has blue eyes, so her genotype is \(bb\). Crossing a heterozygous male (\(Bb\)) with a homozygous recessive female (\(bb\)) yields offspring with genotypes \(Bb\) (brown eyes) and \(bb\) (blue eyes) in a 1:1 ratio. Thus, there is a \(50\%\) chance that their child will have blue eyes.

Award 1 mark for the correct option C. Reject A because the father is a carrier of the blue eye allele. Reject B and D because the cross is a test cross (Bb x bb), giving a 1:1 phenotypic ratio (50%).

(a) Active immunity is the defense against a pathogen by antibody production in the body. It is gained after an infection by a pathogen or by vaccination.

(b) Vaccination involves introducing weakened, harmless, or dead pathogens (or their antigens) into the body. These antigens trigger an immune response where lymphocytes detect them. Lymphocytes produce specific antibodies that bind to the antigens and destroy the pathogens. At the same time, memory cells are produced. These memory cells persist in the blood, and if the real pathogen enters the body in the future, they can rapidly produce large quantities of antibodies to destroy it before symptoms occur.

(c) Three differences:
1. Active immunity involves the production of antibodies by the person's own body, whereas passive immunity involves receiving antibodies made by another organism.
2. Active immunity provides long-term protection (due to memory cells), while passive immunity provides short-term protection.
3. Active immunity takes time to develop, whereas passive immunity provides immediate protection.

(d) Mechanical barriers physically block pathogens from entering (e.g., skin or hairs in the nose). Chemical barriers use chemical substances to destroy or inhibit pathogens (e.g., hydrochloric acid in the stomach or mucus that traps pathogens).

(a) [Max 2 marks]
- defense against a pathogen by antibody production [1]
- in the body / by the person's own immune system [1]
- gained after an infection or by vaccination [1]

(b) [Max 5 marks]
- harmless/weakened/dead pathogen or antigens introduced [1]
- triggers immune response / lymphocytes activated [1]
- lymphocytes produce antibodies [1]
- antibodies are specific to the antigens [1]
- memory cells are produced [1]
- memory cells persist/remain in the blood [1]
- rapid/larger antibody production upon secondary exposure [1]

(c) [Max 3 marks]
- Active involves own antibody production vs Passive involves receiving ready-made antibodies [1]
- Active produces memory cells vs Passive does not produce memory cells [1]
- Active provides long-term protection vs Passive provides short-term protection [1]
- Active takes time to develop vs Passive is immediate [1]

(d) [Max 3 marks]
- mechanical barrier definition + example: physically blocks entry e.g. skin / hairs in nose [1]
- chemical barrier definition + example: chemicals that destroy/inhibit pathogens e.g. stomach acid / mucus / tears [1]
- correct identification of which is mechanical and which is chemical [1]

(a) Goblet cells produce and secrete sticky mucus. This mucus traps dust, dirt, and pathogens (such as bacteria and viruses) inhaled from the air. Ciliated cells have tiny hair-like structures called cilia on their surface. These cilia beat continuously in a coordinated rhythm to sweep the mucus up and away from the lungs towards the throat, where it can be swallowed and digested by stomach acid.

(b) During physical exercise, the rate of aerobic respiration in muscle cells increases to produce more ATP for muscle contraction. This increased respiration produces carbon dioxide (\(CO_2\)) as a waste product, which diffuses into the blood. The concentration of \(CO_2\) in the blood increases, lowering the blood pH (making it more acidic). The brain (specifically the medulla) detects this decrease in pH of the blood. The brain sends electrical impulses along nerves to the diaphragm and intercostal muscles, stimulating them to contract more rapidly and with greater force, which increases the rate and depth of breathing. This helps to excrete the excess \(CO_2\) and bring in more oxygen.

(c) Three features of alveoli:
1. Large surface area for faster diffusion of oxygen and carbon dioxide.
2. Very thin walls (one cell thick) which provide a short diffusion distance.
3. Surrounded by an extensive network of blood capillaries, maintaining a steep concentration gradient for gas exchange.

(a) [Max 4 marks]
- goblet cells produce/release mucus [1]
- mucus traps dust / pathogens / bacteria [1]
- ciliated cells have cilia / hair-like structures [1]
- cilia beat/sweep mucus upwards / away from lungs / to the back of the throat [1]

(b) [Max 6 marks]
- exercise increases respiration rate in muscle cells [1]
- more carbon dioxide (\(CO_2\)) is produced [1]
- \(CO_2\) concentration in blood increases / blood pH decreases [1]
- brain / medulla detects this change in blood pH / \(CO_2\) [1]
- brain sends impulses/signals to breathing muscles (diaphragm and intercostal muscles) [1]
- muscles contract harder and faster [1]
- causes increased rate and depth of breathing [1]

(c) [Max 3 marks]
- large surface area [1]
- thin walls / one cell thick / short diffusion distance [1]
- dense/good capillary network / good blood supply [1]
- moist surface (allows gases to dissolve) [1]

(a) First, the human insulin gene is identified and isolated using restriction enzymes, which cut DNA leaving 'sticky ends'. A plasmid is extracted from a bacterium and cut open using the same restriction enzymes to create complementary sticky ends. The human insulin gene and the plasmid are joined together using the enzyme DNA ligase to form a recombinant plasmid. This recombinant plasmid is then inserted back into a bacterial cell. The bacteria are then selected and cloned, reproducing rapidly to produce the human insulin.

(b) (i) Temperature: Bacteria contain enzymes that catalyze metabolic reactions. If the temperature is too low, the rate of reaction is slow due to low kinetic energy. If the temperature is too high, the enzymes denature, stopping bacterial growth and insulin production. Thus, the temperature must be kept at the optimum level for enzyme activity.
(ii) Supply of oxygen: Oxygen is required for aerobic respiration by the bacteria. Aerobic respiration provides energy for growth, reproduction, and synthesis of insulin. If oxygen is limited, bacteria may undergo anaerobic respiration or die, leading to low yield.

(c) Three reasons:
1. They have rapid reproduction rates / fast generation times.
2. They possess plasmids (which can be easily extracted and manipulated).
3. They share the same genetic code as humans (universal genetic code), allowing human genes to be translated.
4. There are fewer ethical concerns compared to using animals.

(a) [Max 5 marks]
- isolate/cut out human insulin gene using restriction enzymes [1]
- cut plasmid using the same restriction enzyme [1]
- complementary sticky ends are produced [1]
- join gene and plasmid using DNA ligase [1]
- forms a recombinant plasmid [1]
- insert plasmid into bacterial cell [1]

(b) (i) [Max 3 marks]
- temperature affects enzyme-controlled reactions [1]
- too low: low kinetic energy / slow rate of reaction [1]
- too high: enzymes denature [1]
- kept at optimum temperature for maximum growth / yield [1]

(ii) [Max 2 marks]
- oxygen needed for aerobic respiration [1]
- provides energy for growth / reproduction / protein synthesis [1]
- prevents anaerobic respiration (which produces toxic by-products or lower energy) [1]

(c) [Max 3 marks]
- rapid rate of reproduction [1]
- presence of plasmids [1]
- universal genetic code [1]
- lack of ethical concerns [1]
- ability to make complex molecules [1]

(a) Enzyme-substrate specificity means that an enzyme can only catalyze a reaction for one specific type of substrate molecule. According to the lock and key hypothesis, the enzyme's active site has a specific three-dimensional shape (the lock) that is complementary to the shape of the substrate (the key). The substrate fits precisely into the active site to form an enzyme-substrate complex, allowing the reaction to occur. Other molecules with different shapes cannot fit into the active site.

(b) (i) At the optimum pH, amylase activity is at its maximum. This is because the shape of the active site is perfect for the starch substrate to fit in, resulting in the highest frequency of successful collisions between amylase and starch, and the fastest formation of enzyme-substrate complexes.
(ii) At pH 2, amylase will have little to no activity. The extreme acidity (high concentration of hydrogen ions) disrupts the ionic and hydrogen bonds maintaining the three-dimensional structure of the enzyme. This permanently alters the shape of the active site, denaturing the enzyme. Consequently, starch can no longer fit into the active site, and no enzyme-substrate complexes can form.

(c) Two factors:
1. Temperature
2. Enzyme concentration (or Substrate concentration)

(a) [Max 4 marks]
- active site has a specific 3D shape [1]
- active site shape is complementary to the substrate shape [1]
- substrate fits into the active site / forms enzyme-substrate complex [1]
- idea of 'lock and key' model (enzyme = lock, substrate = key) [1]
- only one specific substrate fits a specific enzyme [1]

(b) (i) [Max 3 marks]
- maximum/highest activity of amylase [1]
- active site has the optimal shape / is not altered [1]
- highest rate of successful collisions / enzyme-substrate complexes forming [1]

(ii) [Max 4 marks]
- amylase is denatured [1]
- extreme pH alters / disrupts bonds in the protein structure [1]
- shape of the active site is changed [1]
- substrate / starch can no longer fit [1]
- no enzyme-substrate complexes can form / reaction stops [1]

(c) [Max 2 marks]
- temperature [1]
- enzyme concentration / substrate concentration [1]
- inhibitor concentration [1]

(a) Osmosis is the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution), down a water potential gradient, through a partially permeable membrane.

(b) The \(0.0\text{ mol/dm}^3\) sucrose solution (pure water) has a higher water potential than the cytoplasm of the potato cells. Water enters the potato cells by osmosis, passing down the water potential gradient across the partially permeable cell membranes. As water enters, the vacuole expands, pushing the cytoplasm against the cell wall. This creates turgor pressure inside the cell, which prevents the cell from bursting. The entry of water increases the mass of the cells, causing the potato cylinder to gain mass and become turgid.

(c) In the \(0.8\text{ mol/dm}^3\) sucrose solution, the solution has a lower water potential than the cytoplasm of the potato cells. Water leaves the potato cells by osmosis down the water potential gradient. As water is lost, the vacuole shrinks, and the cytoplasm pulls away from the cell wall, a process called plasmolysis. The cells lose turgidity, become flaccid, and this leads to a decrease in the overall mass of the potato cylinder.

(d) The student should plot percentage change in mass (y-axis) against sucrose concentration (x-axis) and draw a line of best fit. The concentration of sucrose inside the potato cells is the point where the line crosses the x-axis, where the percentage change in mass is 0% (indicating no net movement of water).

(a) [Max 3 marks]
- net movement of water molecules [1]
- from higher water potential to lower water potential / down a water potential gradient [1]
- through a partially permeable membrane [1]

(b) [Max 4 marks]
- pure water / 0.0 \(\text{mol/dm}^3\) solution has higher water potential than inside the potato cell [1]
- water enters cells by osmosis [1]
- increases volume/pressure inside cell / pushes against cell wall [1]
- develops turgor pressure / cells become turgid [1]
- reference to cell wall preventing bursting [1]

(c) [Max 4 marks]
- 0.8 \(\text{mol/dm}^3\) solution has lower water potential than inside cell [1]
- water leaves cell by osmosis [1]
- cell membrane pulls away from cell wall / plasmolysis occurs [1]
- cells become flaccid / lose turgidity [1]
- mass decreases [1]

(d) [Max 2 marks]
- plot graph and find the point where there is 0% change in mass / line crosses x-axis [1]
- at this point, water potential inside and outside the cells is equal / no net water movement [1]

(a) (i) Sunken stomata trap moist air in pits on the leaf surface. This reduces the water potential gradient between the inside of the leaf and the atmosphere immediately outside the stomata, which significantly reduces the rate of transpiration.
(ii) A thick waxy cuticle acts as a waterproof barrier on the surface of the leaf. This reduces the evaporation of water directly from the epidermal cells, conserving water.
(iii) Rolled leaves trap a layer of humid air inside the rolled space, covering the stomata. This increases humidity around the stomata and reduces the water potential gradient, thereby reducing transpiration.

(b) Three adaptive features of hydrophytes:
1. Large air spaces (aerenchyma) in the stems and leaves: This provides buoyancy to help the leaves float on the water surface to access light, and allows oxygen to diffuse down to submerged parts for respiration.
2. Stomata located only on the upper epidermis of floating leaves: This allows gas exchange to occur directly with the air rather than being blocked by water.
3. Absent or highly reduced root systems and xylem vessels: Since water is abundant around the plant, extensive roots are not needed for absorption, and reduced xylem saves energy.
4. Flexible leaves and stems: This prevents damage by water currents.

(c) An adaptive feature is an inherited feature that helps an organism to survive and reproduce in its environment.

(a) (i) [Max 2 marks]
- traps moist/humid air near the stomata [1]
- reduces the water potential gradient (between inside and outside) [1]
- reduces transpiration / water loss [1]

(ii) [Max 2 marks]
- waterproof barrier [1]
- reduces water loss by evaporation from the leaf surface/epidermis [1]

(iii) [Max 2 marks]
- traps humid air inside the roll [1]
- protects stomata from wind/air movement [1]
- reduces water potential gradient / reduces transpiration [1]

(b) [Max 6 marks (2 marks for each well-explained feature, max 3 features)]
- Feature 1: Air spaces / aerenchyma [1]
- Explanation 1: provides buoyancy to float / allows transport of oxygen to roots [1]
- Feature 2: Stomata on upper epidermis only [1]
- Explanation 2: allows gas exchange with the air / prevents blockage of stomata by water [1]
- Feature 3: Reduced root system / reduced xylem [1]
- Explanation 3: water is easily absorbed directly through leaves/stems / no need for structural support / saves energy [1]
- Feature 4: Flexible stems / divided leaves [1]
- Explanation 4: prevents damage from water currents [1]

(c) [1 mark]
- inherited feature that helps an organism to survive and reproduce in its environment [1]

(a) Independent variable: pH.
Dependent variable: time taken for starch to be completely broken down / rate of amylase activity.

(b) Table:
| pH | Time taken / s | Rate of reaction / arbitrary units |
|----|----------------|------------------------------------|
| 4 | 180 | 5.6 |
| 5 | 110 | 9.1 |
| 6 | 40 | 25.0 |
| 7 | 20 | 50.0 |
| 8 | 120 | 8.3 |

(c) Temperature affects enzyme activity because higher temperatures increase kinetic energy and the rate of collisions between amylase and starch, while extreme heat denatures the enzyme. Keeping temperature constant ensures that only the effect of pH is being measured. This can be achieved by placing all test-tubes in a thermostatically-controlled water bath.

(d) Source of error 1: The testing interval of 10 seconds is too large, meaning the exact endpoint might have occurred between measurements.
Improvement 1: Decrease the testing interval to every 5 seconds.
Source of error 2: Determining the color change endpoint is subjective and difficult to judge consistently.
Improvement 2: Use a colorimeter to measure absorbance or compare the mixture against a standard reference card.

(a) [Max 2 marks]
- 1 mark for identifying pH as the independent variable.
- 1 mark for identifying time taken or rate of reaction as the dependent variable.

(b) [Max 4 marks]
- 1 mark for a complete table structure with appropriate column headings and units (e.g., pH, Time / s, Rate / arbitrary units).
- 1 mark for entering all raw time values correctly.
- 2 marks for all calculations of rate correct and rounded to one decimal place (1 mark if only 3 or 4 are correct or if rounding is incorrect).

(c) [Max 3 marks]
- 1 mark for explaining that temperature affects enzyme activity / rate of reaction (e.g., kinetic energy / collisions / denaturation).
- 1 mark for stating that keeping it constant ensures a fair test / pH is the only independent variable.
- 1 mark for suggesting a thermostatically-controlled water bath.

(d) [Max 4.3 marks]
- 1 mark for identifying error 1 (e.g., sampling interval too long) and 1 mark for corresponding improvement (e.g., sample more frequently).
- 1 mark for identifying error 2 (e.g., subjective color endpoint) and 1 mark for corresponding improvement (e.g., use a colorimeter / color standard).
- 0.3 marks for logical, clear presentation of both pairs of errors and improvements.

(a) Change in mass = \(3.57\text{ g} - 4.20\text{ g} = -0.63\text{ g}\).
Percentage change = \(\frac{-0.63}{4.20} \times 100 = -15.0\%\).
This is a loss of 15.0% (or percentage change of -15.0%).

(b) Three procedures:
1. Use a cork borer to cut all cylinders to ensure they have the exact same diameter / surface area.
2. Remove all outer potato peel/skin from the cylinders because skin acts as a barrier to osmosis.
3. Take all potato cylinders from the same potato tuber to ensure the initial water potential of the cells is identical.

(c) The paper towel removes excess solution adhering to the outside of the potato cylinder. If left on, this extra liquid would contribute to the final mass measurement, leading to an inaccurate and overstated final mass.

(d) Plot a line graph with the concentration of NaCl on the x-axis and the percentage change in mass on the y-axis. Draw a line of best fit through the plotted points. Locate the point where the line of best fit crosses the x-axis (where the percentage change in mass is 0%). The concentration at this intersection point is the isotonic concentration because there is no net movement of water.

(a) [Max 3 marks]
- 1 mark for correct working shown: \(\frac{3.57 - 4.20}{4.20} \times 100\).
- 1 mark for correct value: 15% / 15.0% / -15%.
- 1 mark for identifying it as a 'loss' or showing a negative sign.

(b) [Max 3 marks]
- 1 mark for using a cork borer to keep diameter / surface area constant.
- 1 mark for cutting cylinders to the same length using a ruler.
- 1 mark for removing all skin / peel.
- 1 mark for using potato cylinders from the same tuber.
- 1 mark for repeating the trials at each concentration to identify anomalies.
(Accept any three valid points)

(c) [Max 2 marks]
- 1 mark for stating that it removes surface solution / excess water on the outside.
- 1 mark for explaining that this liquid would add extra mass and decrease the accuracy of the final weight.

(d) [Max 5.3 marks]
- 1 mark for plotting concentration on the x-axis and percentage change in mass on the y-axis.
- 1 mark for plotting a line of best fit.
- 2 marks for stating that the isotonic concentration is found where the line crosses the x-axis / where there is 0% change in mass.
- 1.3 marks for explaining the biological reasoning: 0% change means water potentials are equal / no net movement of water.

(a) The student breathes through a central mouthpiece connected to a T-tube. When they inhale (inspiration), air is drawn from the room into Flask A, bubbling through its submerged tube into the limewater. The connection to Flask B is blocked by a valve (or by liquid level) so air cannot be pulled from Flask B. When they exhale (expiration), the expired air is pushed through Flask B, bubbling through its limewater, while a valve prevents air from entering Flask A.

(b) Expected observations:
- Flask A (inspired air): The limewater remains clear (or only turns very slightly cloudy after a long time).
- Flask B (expired air): The limewater turns cloudy / milky quickly.
Explanation: Inspired air has a low concentration of carbon dioxide (about 0.04%), which is not enough to turn the limewater cloudy quickly. Expired air contains a high concentration of carbon dioxide (about 4%) produced by aerobic respiration in body cells, which reacts with the calcium hydroxide in limewater to form an insoluble precipitate of calcium carbonate.

(c) Safety precaution: Wear safety goggles to protect the eyes from limewater, because limewater is alkaline and corrosive. Alternatively, inhale gently to prevent liquid from splashing into the mouth.

(d) Plan:
1. Measure the time taken in seconds for the limewater in Flask B to turn from clear to a standard level of cloudiness while the student is at rest.
2. Empty Flask B, rinse it, and refill it with the same volume of fresh limewater.
3. Have the student perform 2 minutes of vigorous exercise (e.g., step-ups).
4. Immediately after exercise, have the student breathe into the mouthpiece at the same breathing rate/intensity and measure the time taken for Flask B's limewater to reach the same level of cloudiness.
5. Repeat the entire procedure three times to calculate a mean time for both at rest and after exercise. A shorter time to turn cloudy indicates a faster rate of carbon dioxide production.

(a) [Max 4 marks]
- 1 mark for mentioning the mouthpiece / T-tube connection.
- 1 mark for stating that inspiration pulls atmospheric air down a tube submerged in Flask A's limewater.
- 1 mark for stating that expiration pushes air down a tube submerged in Flask B's limewater.
- 1 mark for mentioning valves or tube arrangements that prevent backflow / ensure air travels in one direction only.

(b) [Max 4 marks]
- 1 mark for Flask A remaining clear / slow to turn cloudy AND Flask B turning cloudy rapidly.
- 1 mark for stating inspired air has a low concentration of CO2 (~0.04%).
- 1 mark for stating expired air has a high concentration of CO2 (~4%).
- 1 mark for linking the cloudiness to the reaction of CO2 with limewater to form a precipitate.

(c) [Max 2 marks]
- 1 mark for safety hazard (limewater is alkaline / corrosive / toxic if swallowed).
- 1 mark for prevention (wear eye protection / goggles OR instruct the student to inhale gently to avoid sucking liquid into the mouth).

(d) [Max 3.3 marks]
- 1 mark for stating the dependent variable is the *time taken* for the limewater to turn cloudy.
- 1 mark for standardizing variables (same volume/concentration of limewater, same duration of exercise - 2 minutes).
- 1 mark for mentioning replication (repeating the trials to find a mean time).
- 0.3 marks for explaining how the results will show a difference (e.g., shorter time to turn cloudy after exercise = higher rate of CO2 production).