2026 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Water moves by osmosis down a water potential gradient from a region of higher water potential (inside the potato cells) to a region of lower water potential (the concentrated sucrose solution). As water leaves the cells, the cytoplasm pulls away from the cell wall, and the cells become plasmolysed.

B is correct [1]; A is incorrect because water leaves down a water potential gradient; C is incorrect because osmosis is passive; D is incorrect because the loss of water causes plasmolysis/flaccidity, not turgidity.

Transpiration is fastest when the water potential and concentration gradients for water vapour between the inside of the leaf and the atmosphere are steepest, and when kinetic energy of water molecules is high. High temperature increases evaporation rate, low humidity steepens the diffusion gradient, and high wind speed continuously removes water vapour from the leaf surface to maintain a steep concentration gradient.

C is correct [1]; A and B are incorrect because high humidity lowers the concentration gradient for transpiration; D is incorrect because low wind speed allows a boundary layer of humid air to accumulate, reducing the transpiration rate.

When core body temperature falls, negative feedback mechanisms act to reduce heat loss. Arterioles supplying the surface capillary loops constrict (vasoconstriction), redirecting blood through shunt vessels deeper in the dermis, and sweat production by sweat glands decreases to minimise evaporative cooling.

A is correct [1]; B is incorrect because sweat secretion decreases; C is incorrect because arterioles constrict, not dilate; D is incorrect because arterioles constrict to divert blood away from surface capillaries.

The correct sequence of eutrophication is: fertiliser leaches into water (4) -> rapid algal growth at the surface (2) -> algae block light causing submerged plants to die (5) -> decomposers multiply and respire aerobically breaking down dead plants (1) -> oxygen depletion causes aquatic fauna to die (3).

D is correct [1]; other sequences incorrectly order the causal events leading to deoxygenation.

During ventricular systole, the muscular walls of the ventricles contract. The pressure in the ventricles rises above that in the atria, causing the atrioventricular (bicuspid and tricuspid) valves to close to prevent backflow into the atria. The pressure forces the semilunar valves open, pumping blood into the aorta and the pulmonary artery.

C is correct [1]; A and B have incorrect atrioventricular valve states; D has incorrect semilunar valve state and destination.

Restriction enzymes cut both the bacterial plasmid DNA and the human DNA at specific base sequences, leaving complementary sticky ends. DNA ligase is used to join the sticky ends of the human gene and the cut plasmid together, forming the recombinant plasmid.

B is correct [1]; A reverses the roles; C and D attribute wrong enzymatic functions.

Actual size \(= \text{Image size} / \text{Magnification}\).
\(\text{Actual size} = 16\text{ mm} / 2000 = 0.008\text{ mm}\).
To convert \(\text{mm}\) to \(\mu\text{m}\), multiply by \(1000\):
\(0.008\text{ mm} \times 1000 = 8.0\ \mu\text{m}\).

C is correct [1]; A results from dividing by \(100\); B is off by a factor of 10; D is off by a factor of 10 from incorrect unit conversion.

Mitosis produces 2 daughter cells that are diploid (containing 46 chromosomes in humans) and genetically identical to each other and the parent cell. Meiosis produces 4 daughter cells that are haploid (containing 23 chromosomes in humans) and genetically non-identical.

A is correct [1]; B incorrectly states mitosis yields genetically different cells; C incorrectly states meiosis yields diploid cells; D gives incorrect chromosome ploidies for both processes.

Eutrophication begins when leached fertiliser enters water, stimulating rapid algal growth (algal bloom). This algal layer blocks sunlight from reaching submerged aquatic plants, causing them to die. Decomposers (aerobic bacteria) feed on the dead plant matter and multiply rapidly, consuming dissolved oxygen through aerobic respiration, which results in a marked decrease in dissolved oxygen.

A is correct [1 mark]. B, C, and D present incorrect chronological ordering of events in eutrophication.

Transpiration rate increases with higher temperature (increases kinetic energy of water molecules, speeding up evaporation), lower humidity (maintains a steep water potential/concentration gradient between the inside of the leaf and the outside air), and higher air movement (blows away humid air from around the stomata, maintaining the steep gradient).

B is correct [1 mark]. High temperature, low humidity, and high air movement all independently maximise the rate of water vapour diffusion out of stomata.

When body temperature rises, negative feedback mechanisms are triggered: arterioles supplying skin capillaries dilate (vasodilation) and shunt vessels constrict, directing more blood through surface capillaries so heat is lost by radiation. Sweat glands secrete more sweat, which cools the body as water evaporates from the skin surface.

B is correct [1 mark]. Arterioles dilate (vasodilation), shunt vessels constrict to redirect blood to surface capillaries, and sweat secretion increases.

The 0.7 mol / dm\(^3\) sucrose solution has the lowest water potential. Therefore, it creates the steepest water potential gradient between the cytoplasm/vacuole of the potato cells and the external solution, resulting in the greatest net movement of water out of the cells by osmosis and thus the greatest percentage loss in mass.

D is correct [1 mark]. The highest external solute concentration (0.7 mol / dm\(^3\)) produces the greatest water potential gradient driving water out of the tissue by osmosis.

Luteinising hormone (LH) is produced and secreted by the pituitary gland. A surge in LH triggers ovulation (release of an egg from an ovarian follicle). FSH is produced by the pituitary gland (not ovaries), oestrogen is produced by the ovaries (not pituitary), and progesterone maintains the uterus lining rather than causing shedding.

B is correct [1 mark]. LH is secreted by the pituitary gland and causes ovulation.

At high temperatures, the excessive kinetic energy breaks hydrogen and other bonds holding the 3D protein structure together. This causes the enzyme to denature, altering the shape of its active site so that the substrate molecule no longer fits (is no longer complementary), preventing the formation of enzyme-substrate complexes.

B is correct [1 mark]. High temperatures cause denaturation of the enzyme's active site due to disruption of bonds maintaining tertiary structure.

During ventricular systole, the ventricles contract. The pressure inside the ventricles rises above that in the atria, causing the atrioventricular valves to close (preventing backflow into atria). As ventricular pressure exceeds the pressure in the aorta and pulmonary artery, the semilunar valves are forced open, ejecting blood into the arteries.

A is correct [1 mark]. Ventricular contraction forces atrioventricular valves closed and opens semilunar valves as ventricular pressure exceeds arterial pressure.

Using the formula \(\text{actual size} = \frac{\text{image size}}{\text{magnification}}\):
First convert \(28\text{ mm}\) to micrometres: \(28 \times 1000 = 28\,000\,\mu\text{m}\).
Then calculate actual size: \(\frac{28\,000\,\mu\text{m}}{4000} = 7.0\,\mu\text{m}\).

C is correct [1 mark]. \(\text{Actual size} = \frac{28\text{ mm}}{4000} = 0.007\text{ mm} = 7.0\,\mu\text{m}\).

Eutrophication begins when mineral ions from fertiliser cause rapid growth of algae (algal bloom). The algae block sunlight, preventing submerged plants from photosynthesising and causing them to die. Aerobic decomposers (bacteria) feed on the dead plant matter and multiply rapidly, consuming dissolved oxygen in aerobic respiration. This causes dissolved oxygen levels to drop, leading to the death of fish and other aquatic animals.

A is correct [1].
B, C, and D describe incorrect chronological orders or inverted effects.

Transpiration is fastest when the temperature is high (increases kinetic energy of water molecules and rate of evaporation), humidity is low (maintains a steep water potential/concentration gradient between leaf air spaces and external air), and wind speed is high (moves humid air away from the leaf surface, maintaining the gradient).

B is correct [1].
A, C, and D contain conditions that reduce the rate of transpiration (e.g. high humidity or low temperature/wind speed).

In response to a decrease in external temperature, arterioles in the skin constrict (vasoconstriction) and shunt vessels dilate, directing blood flow away from the superficial capillary beds near the skin surface. Sweat production by sweat glands decreases to reduce heat loss by evaporation.

A is correct [1].
B describes vasodilation which increases heat loss. C and D state incorrect combinations of shunt vessel action and sweat production.

Water enters plant cells by osmosis down a water potential gradient (from a higher/less negative water potential to a lower/more negative water potential). Solution P caused the largest gain in mass (from \(5.0\text{ g}\) to \(5.8\text{ g}\)), showing that it had the highest water potential, creating the steepest water potential gradient into the potato cells.

A is correct [1].
B had a lower water potential than P. C and D had lower water potentials than the potato tissue as mass was lost.

As temperature increases, enzyme and substrate molecules gain kinetic energy and move faster. This increases the frequency of collisions between substrate molecules and enzyme active sites, as well as the proportion of successful collisions that form enzyme-substrate complexes.

C is correct [1].
A is incorrect because active sites do not change specificity. B is incorrect because enzymes do not alter substrate activation energy by changing structure this way. D is incorrect because denaturation decreases reaction rate.

The left ventricle has thick muscular walls and contracts to pump oxygenated blood under high pressure into the aorta to be distributed around the body. The right ventricle pumps deoxygenated blood into the pulmonary artery.

B is correct [1].
A pairs the left ventricle with the pulmonary artery (incorrect vessel and oxygenation). C and D pair the right ventricle with incorrect vessels and oxygenation levels.

Magnification is calculated using the formula:
\[\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{24\text{ mm}}{0.8\text{ mm}} = \times 30\]

C is correct [1].
A is calculated as actual/image (inverted). B is \(24 \times 0.8\). D is an arithmetic error by a factor of 10.

In genetic modification, restriction enzymes are used to cut both the human DNA and the bacterial plasmid at specific base sequences, creating complementary sticky ends. DNA ligase is then used to join the sticky ends of the human gene and the cut plasmid together, forming a recombinant plasmid.

D is correct [1].
A describes the role of restriction enzymes. B describes the role of ribosomes/translation. C describes plasmid uptake/transformation.

Transpiration rate increases when humidity is low (which maintains a steep concentration gradient of water vapour between the leaf interior and surrounding air), temperature is high (which increases the kinetic energy of water molecules), and wind speed is high (which rapidly blows away humid air from around the stomata).

[1] C is correct; low humidity, high temperature, and high wind speed all act together to maximise the rate of water vapour evaporation and diffusion from leaves.

When blood glucose concentration rises above the set point, the pancreas detects this change and secretes the hormone insulin. Insulin stimulates liver and muscle cells to absorb glucose from the blood and convert it into insoluble glycogen for storage, lowering blood glucose levels back to normal.

[1] C is correct; insulin is released by the pancreas and triggers the conversion of soluble glucose to insoluble glycogen in the liver.

Water moves by osmosis down a water potential gradient from a region of higher water potential (inside the cell) to a region of lower water potential (the concentrated sucrose solution). As water leaves the vacuole and cytoplasm, the volume decreases and the cell membrane pulls away from the cell wall, making the cells plasmolysed.

[1] C is correct; net movement of water is out of the cells by osmosis, causing plasmolysis.

At high temperatures beyond the optimum, increased molecular vibrations break bonds maintaining the specific tertiary shape of the enzyme. This alters the shape of the active site (denaturation), preventing substrate molecules from fitting and forming enzyme-substrate complexes.

[1] B is correct; denaturation occurs as the active site changes shape, losing its complementary fit to the substrate.

Restriction enzymes are used to cut both the human DNA and the bacterial plasmid at specific base sequences to create sticky ends. DNA ligase is then used to join the sticky ends of the human insulin gene to the plasmid, forming a recombinant plasmid.

[1] B is correct; DNA ligase joins the sticky ends of the human gene and the cut plasmid together.

The muscular wall of the left ventricle is significantly thicker than that of the right ventricle because it must generate a much higher blood pressure to pump oxygenated blood through the aorta to the rest of the body (systemic circulation), whereas the right ventricle only pumps blood a short distance to the lungs.

[1] B is correct; the left ventricle has a thicker wall, pumps blood to the systemic circulation (rest of body), and generates higher pressure.

Meiosis is a reduction division that halves the chromosome number from diploid (
2n
) to haploid (
n
), producing four genetically non-identical daughter cells (gametes).

[1] D is correct; meiosis produces 4 genetically different haploid daughter nuclei.

Eutrophication begins with the runoff of mineral ions (fertiliser) into water bodies, causing rapid growth of algae (algal bloom). This layer blocks sunlight, causing submerged plants to die. Decomposers (bacteria) increase in population as they feed on dead plant tissue, consuming dissolved oxygen through aerobic respiration, which leads to the death of fish and other aquatic animals due to lack of oxygen.

[1] A is correct; sequence runs from fertiliser runoff to algal bloom, death of submerged plants, aerobic decomposition by bacteria, and suffocation of aquatic organisms.

Water moves from a region of higher water potential (-150 kPa) to a region of lower water potential (-400 kPa) down a water potential gradient by osmosis. As water enters the plant cell, the vacuole expands and pushes against the cell wall, generating turgor pressure. The plant cell becomes turgid. Plant cells do not burst due to the presence of the rigid cellulose cell wall.

A is correct [1 mark]. B, C and D are incorrect because water enters down the water potential gradient and the cell wall prevents bursting.

Eutrophication begins with increased availability of nitrate and other ions, leading to rapid algal growth (algal bloom). The algae block sunlight from reaching bottom-dwelling aquatic plants, which then die. Decomposers (bacteria) feed on dead plant matter and multiply rapidly, consuming dissolved oxygen through aerobic respiration. This causes dissolved oxygen levels to drop, leading to the death of fish and other aquatic organisms.

B is correct [1 mark]. A, C and D describe incorrect causal orders or physiological effects.

Transpiration rate increases with higher temperature (more kinetic energy for evaporation), lower humidity (steeper concentration gradient of water vapour between leaf air spaces and external air), and higher wind speed (removes water vapour boundary layer from the leaf surface, maintaining a steep diffusion gradient).

A is correct [1 mark]. B, C and D all include conditions that decrease the rate of transpiration.

Digestion and absorption of carbohydrates cause blood glucose concentration to rise. The pancreas detects this increase and secretes insulin. Insulin stimulates liver cells to absorb glucose from the blood and convert it into the insoluble storage polysaccharide glycogen.

C is correct [1 mark]. A and B have the wrong hormone; D shows insulin causing the opposite process.

When the muscular walls of the ventricles contract, intraventricular pressure rises sharply above atrial and arterial pressure. This causes the atrioventricular (bicuspid and tricuspid) valves to close to prevent backflow into the atria, while the semilunar valves are forced open, allowing blood to flow into the aorta and pulmonary artery.

D is correct [1 mark]. A, B and C state incorrect valve positions or incorrect vessel flow during ventricular systole.

In double-stranded DNA, complementary base pairing dictates that adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). Therefore: Percentage of T = Percentage of A = 31%. The combined percentage of A and T = 31% + 31% = 62%. The remaining percentage consists of G and C: 100% - 62% = 38%. Since Percentage of C = Percentage of G: Percentage of C = 38% / 2 = 19%.

A is correct [1 mark]. B (31%) is the percentage of thymine; C (38%) is the combined percentage of G + C; D (62%) is the combined percentage of A + T.

Meiosis is a reduction division in which the chromosome number is halved from diploid (2n = 46) to haploid (n = 23). It results in four genetically non-identical daughter cells (gametes) due to independent assortment and crossing over.

B is correct [1 mark]. A describes mitosis; C and D confuse cell counts and ploidy levels.

In genetic modification, restriction enzymes cut DNA at specific base sequences to isolate the target gene and open the plasmid vector, leaving sticky ends. DNA ligase is then used to join the complementary sticky ends of the human gene and the bacterial plasmid together, forming recombinant DNA.

C is correct [1 mark]. A is the role of restriction enzymes; B is the role of ribosomes/RNA polymerase; D is the role of DNA polymerase.

Phloem is the vascular tissue that translocates sucrose and amino acids from sources to sinks in flowering plants.

phloem ;

At high temperatures, the bonds maintaining the three-dimensional tertiary structure of the enzyme are disrupted, causing the active site to lose its complementary shape. The enzyme is denatured.

denatured / denaturation ; [R 'killed' / 'dead']

Glucagon is released by the alpha cells of the pancreas in response to low blood glucose levels and stimulates liver cells to convert stored glycogen into glucose.

glucagon ;

Lacteals are specialised lymphatic capillaries located in the core of each villus that absorb fatty acids and glycerol following lipid digestion.

lacteal / lymph vessel / lymphatic vessel ;

DNA ligase catalyses the formation of phosphodiester bonds between adjacent nucleotides, sealing the gene insert into the plasmid vector to form recombinant DNA.

(DNA) ligase ;

Stem cells are undifferentiated cells capable of self-renewal via mitosis and differentiation into various specialised cellular lineages.

stem cell(s) ;

The pulmonary vein returns newly oxygenated blood from the alveolar capillary beds of the lungs into the left atrium.

pulmonary vein ;

Lymphocytes (specifically B-lymphocytes) produce specific protein antibodies complementary to antigens on pathogens.

lymphocyte(s) / B-cell(s) / B-lymphocyte(s) ; [R phagocyte]

Deamination is the removal of the nitrogen-containing part of amino acids to form urea. This biochemical process occurs exclusively in the liver.

liver;

Translocation of soluble organic substances such as sucrose and amino acids is carried out by the phloem tissue.

phloem / phloem sieve tubes;

DNA ligase is the enzyme that catalyses the formation of phosphodiester bonds to join together fragments of DNA with complementary sticky ends.

DNA ligase / ligase;

The fovea (or fovea centralis) is the central area of the retina where cone photoreceptor cells are packed at the highest concentration, providing detailed colour vision.

fovea / fovea centralis / yellow spot; [R] blind spot

(i) Starch is the substrate broken down by amylase into maltose.

(ii) Pepsin is the specific protease enzyme secreted by gastric glands in the stomach.

(iii) The pancreas is the organ that secretes trypsin into the duodenum.

(iv) Lipids (or fats/oils) are broken down by lipase into fatty acids and glycerol.

(v) Glucose is the final product formed when maltase hydrolyses maltose on the epithelium of the small intestine.

(i) starch ; [1]
(ii) pepsin ; [1] (I gastric protease; R trypsin / protease unqualified)
(iii) pancreas ; [1]
(iv) lipid(s) / fat(s) / oil(s) ; [1]
(v) glucose ; [1] (I simple sugars / monosaccharides)

(i) Arterioles supplying skin surface capillaries undergo vasoconstriction (constrict/narrow) to reduce blood flow near the skin surface.

(ii) Shunt vessels dilate (widen), allowing blood to bypass the superficial capillary beds.

(iii) Sweat glands decrease or cease secretion of sweat.

(iv) Less evaporation of sweat occurs from the skin surface, preventing latent heat loss.

(v) Erect hairs trap a layer of still air next to the skin, which acts as a thermal insulator and reduces heat loss by convection.

(i) vasoconstrict / constrict / narrow ; [1] (R capillaries constrict / capillaries move)
(ii) dilate / widen / open ; [1]
(iii) decrease / stop / reduce sweat production / secretion ; [1]
(iv) less / no evaporation (of water / sweat) / less heat lost by evaporation ; [1]
(v) traps a layer of air AND air acts as an insulator / reduces heat loss by convection / conduction ; [1]

1. Water evaporates from the wet surfaces of mesophyll cells into the leaf air spaces.
2. Water vapour diffuses out of the leaf through open stomata into the atmosphere.
3. The loss of water reduces the water potential of mesophyll cells, so they draw water out of xylem vessels by osmosis.
4. This creates tension (a suction force or negative pressure) at the top of the xylem column.
5. Cohesive forces between polar water molecules hold the water molecules together in a continuous column, pulling water upwards from the roots to the leaves.

Any four from:
1. Evaporation of water from the surface of mesophyll cells (into air spaces) ;
2. Loss of water vapour / diffusion of water vapour through stomata (into atmosphere) ;
3. Lowers water potential in mesophyll cells / water moves from xylem to mesophyll cells (by osmosis) ;
4. Creates tension / negative pressure / suction force (at the top of xylem) ;
5. Cohesion / attraction between water molecules (due to hydrogen bonds) ;
6. Maintains a continuous / unbroken column of water (pulled up the xylem) ;

1. Temperature receptors in the skin or hypothalamus detect the lower temperature.
2. Vasoconstriction occurs: the muscle in the walls of arterioles supplying skin surface capillaries contracts, narrowing the lumen.
3. Less warm blood flows through surface capillaries, reducing heat loss by radiation, conduction, and convection.
4. Sweat glands reduce or stop the secretion of sweat, which minimises latent heat loss from evaporation.
5. Hair erector muscles contract, pulling hair shafts upright to trap a layer of warm air that acts as an insulator.

Any four from:
1. (Drop in temperature) detected by thermoreceptors / temperature receptors (in skin / hypothalamus) ;
2. Vasoconstriction / arterioles supplying surface capillaries constrict / narrow ;
3. Less blood flows through surface capillaries / shunt vessels dilate ;
4. Reduces heat loss by radiation / conduction / convection ;
5. Sweat glands stop / reduce sweat secretion, reducing heat loss by evaporation ;
6. Hair erector muscles contract, raising hairs to trap an insulating layer of air ;

1. The 1.0 mol/dm³ sucrose solution has a lower water potential (more negative) than the inside of the potato cells.
2. Water leaves the potato cells by osmosis through the partially permeable cell membranes down the water potential gradient.
3. The volume of the cytoplasm and vacuole decreases.
4. The cell membrane detaches and pulls away from the cell wall, meaning cells become plasmolysed.
5. The loss of internal turgor pressure leaves the cells flaccid, making the potato tissue soft and bendable.

Any four from:
1. Water potential of sucrose solution is lower / more negative than inside potato cells (ora) ;
2. Water moves out of cells by osmosis / across a partially permeable membrane ;
3. Down a water potential gradient / from high to low water potential ;
4. Cytoplasm / vacuole decreases in volume / shrinks ;
5. Cell membrane pulls away from cell wall / cells become plasmolysed ;
6. Loss of turgor pressure / cells become flaccid / tissue becomes soft / limp / flexible ;

1. Large surface area provided by many folding chorionic villi (and microvilli) to maximise the rate of substance exchange.
2. Very thin barrier separating maternal and fetal bloodstreams, minimising the diffusion distance.
3. Dense capillary network within the villi combined with maternal blood pools ensures continuous movement of blood, maintaining steep concentration gradients for oxygen, glucose, carbon dioxide, and urea.
4. Specialized transport proteins in cell membranes enable active transport of ions and nutrients.

Any four from:
1. (Chorionic) villi / microvilli provide a large surface area (for diffusion / exchange) ;
2. Thin wall / thin membrane / one-cell thick barrier gives a short diffusion pathway ;
3. Dense / extensive network of fetal capillaries in villi ;
4. Continuous blood flow (maternal and fetal) maintains a steep concentration gradient ;
5. Named substance correctly matched to direction of transport (e.g. oxygen / glucose / amino acids to fetus OR urea / carbon dioxide to mother) ;
6. Presence of transport / carrier proteins for active transport (of glucose / amino acids / ions) ;

1. High temperatures supply excessive kinetic energy, causing the atoms in the enzyme protein to vibrate violently.
2. Chemical bonds (such as hydrogen bonds) holding the specific tertiary 3D structure together are disrupted and broken.
3. The shape of the active site changes / is distorted.
4. Substrate molecules can no longer fit into the active site (loss of complementarity).
5. Fewer or no enzyme-substrate complexes can form, which means the enzyme is denatured and the reaction rate drops sharply.

Any four from:
1. Enzyme molecules gain excessive kinetic energy / vibrate more strongly ;
2. Bonds holding the 3D / tertiary protein structure break ;
3. Active site changes shape / conformation changes ;
4. Substrate is no longer complementary / can no longer fit into the active site ;
5. Fewer / no enzyme-substrate complexes / ESCs formed ;
6. Enzyme is denatured (permanent loss of catalytic activity) ;

1. The left ventricle pumps blood around the entire systemic circulation (to all body tissues), whereas the right ventricle pumps blood only to the lungs (pulmonary circulation).
2. The systemic circulation is much longer and has higher vascular resistance, requiring higher force.
3. The thicker myocardium of the left ventricle generates the higher blood pressure necessary to overcome this resistance and deliver blood to the extremities.
4. The lower pressure generated by the right ventricle protects the delicate, thin-walled capillaries in the alveoli of the lungs from bursting.

Any four from:
1. Left ventricle pumps blood to the whole body / systemic circulation, whereas right ventricle pumps blood only to the lungs / pulmonary circulation ;
2. Systemic circulation is longer / has higher resistance than pulmonary circulation ;
3. Thicker muscular wall / myocardium contracts with greater force / generates higher blood pressure ;
4. High pressure ensures blood reaches all body organs / extremities ;
5. Right ventricle generates lower pressure to prevent damage / bursting of delicate capillaries in the lungs / prevent pulmonary oedema ;

1. The pathogen Vibrio cholerae secretes a protein toxin into the small intestine.
2. The toxin causes the epithelial cells lining the intestine to actively transport chloride ions (\( \text{Cl}^- \)) into the gut lumen.
3. The accumulation of chloride ions lowers the water potential of the fluid inside the intestinal lumen.
4. Water is drawn out of epithelial cells and blood plasma into the lumen by osmosis down a water potential gradient.
5. The excess fluid cannot be reabsorbed, producing voluminous watery diarrhoea that rapidly depletes the body of water and essential mineral salts.

Any four from:
1. Bacteria produce / secrete cholera toxin ;
2. Toxin causes chloride ions (\( \text{Cl}^- \)) to be secreted into the lumen of the small intestine ;
3. Lowers the water potential in the intestinal lumen ;
4. Water moves out of intestinal cells / blood into the lumen by osmosis ;
5. Down a water potential gradient / from high to low water potential ;
6. Results in watery diarrhoea / large loss of fluid and essential mineral ions / dehydration ;

1. Restriction enzymes cut DNA at specific base sequences to isolate the human insulin gene and cut the bacterial plasmid.
2. Restriction enzymes generate complementary single-stranded overhanging ends known as sticky ends.
3. DNA ligase is an enzyme that joins the sticky ends of the human gene and plasmid together by forming phosphodiester bonds, yielding a recombinant plasmid.
4. The plasmid functions as a vector to introduce the foreign gene into the bacterial host cell so it can express the insulin protein.

Any four from:
1. Restriction enzymes cut human DNA (to isolate insulin gene) AND cut plasmid DNA ;
2. Restriction enzymes produce complementary sticky ends / unpaired base overhangs ;
3. DNA ligase joins / seals the insulin gene into the plasmid (by forming phosphodiester bonds) ;
4. Produces a recombinant plasmid ;
5. Plasmid acts as a vector to transfer the gene into host bacteria ;
6. Bacteria take up plasmid and are cultured to express / produce human insulin ;

1. Water evaporates from the cell walls of spongy and palisade mesophyll cells into the sub-stomatal air spaces.
2. Water vapour diffuses out of the leaves through open stomata down a water potential/concentration gradient.
3. The loss of water reduces the water potential in the mesophyll, creating tension or a suction force (transpiration pull) at the top of the xylem column.
4. Cohesion (hydrogen bonding/attraction between adjacent water molecules) holds the water molecules together, allowing a continuous unbroken column of water to be pulled up from roots to leaves.

1. evaporation of water from the surface/walls of mesophyll cells (into air spaces);
2. diffusion of water vapour out of the leaf through stomata (down a concentration gradient);
3. creates tension / transpiration pull / negative pressure (in xylem vessels);
4. cohesion / attraction between water molecules (maintains a continuous / unbroken column of water);
5. adhesion of water molecules to the (lignified) walls of xylem vessels;
[max 4]

When describing the differences between the two profiles:
1. Initial baseline difference: Person B has a higher initial / fasting blood glucose concentration (\(8.5\text{ mmol dm}^{-3}\)) than Person A (\(5.0\text{ mmol dm}^{-3}\)).
2. Peak and return difference: Person A reaches a maximum earlier (at \(45\text{ min}\)) and returns to baseline (\(5.0\text{ mmol dm}^{-3}\)) by \(120\text{ min}\), whereas Person B peaks later (at \(120\text{ min}\)) and remains elevated without returning to the initial level.
3. Magnitude of change / data comparison: The peak concentration reached by Person B (\(16.2\text{ mmol dm}^{-3}\)) is much higher than that of Person A (\(7.8\text{ mmol dm}^{-3}\)).

Any three from:
1. Person B has a higher initial / starting blood glucose concentration than Person A / ora ;
2. Blood glucose in Person A peaks earlier / at \(45\text{ min}\) compared to \(120\text{ min}\) in Person B / peaks later in Person B ;
3. Person A returns to the starting concentration / baseline (by \(120\text{ min}\)), whereas Person B remains high / does not return to baseline (at \(180\text{ min}\)) / ora ;
4. Person B reaches a much higher peak / maximum concentration than Person A / has a larger overall increase / ora ;
5. Comparative data quote with values, time and units for both individuals (e.g. peak of \(7.8\text{ mmol dm}^{-3}\) for A vs \(16.2\text{ mmol dm}^{-3}\) for B / at \(180\text{ min}\) A is \(5.0\text{ mmol dm}^{-3}\) vs B is \(15.5\text{ mmol dm}^{-3}\)) ;

To answer a 3-mark comparative trend question:
1. State a similarity in the overall pattern: In both species, increasing light intensity causes an increase in the rate of water loss.
2. State a key difference: Species B shows a much greater / steeper increase in water loss and maintains a higher rate than Species A at every light intensity tested.
3. Identify a plateau or specific difference in the trend limit: The rate of water loss in Species A levels off / plateaus at 40 arbitrary units (at 12 mg min⁻¹), whereas Species B continues to increase up to 50 arbitrary units (reaching 48 mg min⁻¹).
4. Provide a comparative data quote with units (e.g. at 30 arbitrary units, Species B loses water at 36 mg min⁻¹ while Species A loses water at 10 mg min⁻¹).

Award 1 mark for each valid point up to a maximum of 3 marks:

1. (Similarity) both increase as light intensity increases (from 0 to 40 / 50 a.u.) ;
2. (Difference in magnitude/rate) Species B always has a higher rate of water loss than Species A / rate in Species B increases faster / more steeply than Species A (ora) ;
3. (Difference in plateau) Species A plateaus / reaches maximum at 40 a.u. (at 12 mg min⁻¹), whereas Species B continues to increase to 50 a.u. / reaches a much higher maximum (48 mg min⁻¹) ;
4. (Data manipulation / comparative data quote) paired comparative data with units for both species at a specified light intensity (e.g. at 30 a.u., Species B is 36 mg min⁻¹ and Species A is 10 mg min⁻¹ / difference of 26 mg min⁻¹) ;

1. Use the dilution formula \(C_1 V_1 = C_2 V_2\):
\[V_1 = \frac{C_2 \times V_2}{C_1} = \frac{0.4\text{ mol/dm}^3 \times 50.0\text{ cm}^3}{2.0\text{ mol/dm}^3} = 10.0\text{ cm}^3\]

2. Calculate the volume of distilled water needed to make the total volume up to \(50.0\text{ cm}^3\):
\[V_{\text{water}} = 50.0\text{ cm}^3 - 10.0\text{ cm}^3 = 40.0\text{ cm}^3\]

volume of \(2.0\text{ mol/dm}^3\) stock sucrose solution = \(10.0\text{ cm}^3\) / \(10\text{ cm}^3\) [1];
volume of distilled water = \(40.0\text{ cm}^3\) / \(40\text{ cm}^3\) [1];

[Note: Correct units (\(\text{cm}^3\)) must be given at least once if not printed. Allow ecf for volume of water = \(50.0 - \text{calculated stock volume}\).]

(i) Total volume of gas produced = Final reading - Initial reading
\[\text{Volume} = 19.5\text{ cm}^3 - 1.5\text{ cm}^3 = 18.0\text{ cm}^3\]

(ii) Mean rate of gas production = \(\frac{\text{Volume of gas}}{\text{Time}}\)
\[\text{Rate} = \frac{18.0\text{ cm}^3}{5\text{ min}} = 3.6\text{ cm}^3/\text{min}\]

(i) \(18.0\text{ cm}^3\) / \(18\text{ cm}^3\) [1];
(ii) \(3.6\text{ cm}^3/\text{min}\) [1];

[Allow ecf for (ii) from candidate's answer to (i) divided by 5.]

1. For Tube \(\mathbf{P}\) (\(0.1\%\)):
\[V_{\text{stock}} = \frac{0.1}{1.0} \times 20.0\text{ cm}^3 = 2.0\text{ cm}^3\]

2. For Tube \(\mathbf{Q}\) (\(0.5\%\)):
\[V_{\text{stock}} = \frac{0.5}{1.0} \times 20.0\text{ cm}^3 = 10.0\text{ cm}^3\]

volume of stock for \(\mathbf{P}\) = \(2.0\text{ cm}^3\) / \(2\text{ cm}^3\) [1];
volume of stock for \(\mathbf{Q}\) = \(10.0\text{ cm}^3\) / \(10\text{ cm}^3\) [1];

(i) Volume of lemon juice added = Initial reading - Final reading
\[\text{Volume} = 4.85\text{ cm}^3 - 2.20\text{ cm}^3 = 2.65\text{ cm}^3\]

(ii) To ensure accurate measurement when reading a graduated syringe or measuring cylinder:
- View the scale at eye level (perpendicular to the scale) to avoid parallax error.
- Read consistently from the edge of the plunger seal / bottom of the meniscus.

(i) \(2.65\text{ cm}^3\) [1];
(ii) view scale at eye level / avoid parallax error / read from the same position on the plunger seal each time / read at right angles to the scale [1];

[Ignore: repeat the experiment / use a more accurate syringe]

To plot the graph correctly:

1. Axes (A):
- Horizontal axis (x-axis): labelled as `Hydrogen peroxide concentration / %`.
- Vertical axis (y-axis): labelled as `Volume of oxygen gas collected in 2 minutes / cm³` (or `Volume of oxygen collected / cm³`).

2. Scale (S):
- Select sensible, linear scales with evenly spaced major intervals (e.g., x-axis: 1.0% per 2 or 4 major divisions; y-axis: 10 cm³ per 2 or 4 major divisions).
- Ensure the plotted points cover more than half the grid in both the horizontal and vertical directions.

3. Plotting (P):
- Plot all 6 coordinate pairs accurately within ±0.5 of a small grid square: (0.5, 6.0), (1.0, 11.5), (2.0, 22.0), (3.0, 31.5), (4.0, 39.0), and (5.0, 44.0).
- Mark coordinates using sharp, small crosses (×) or encircled dots (⊙).

4. Line (L):
- Draw a single, clean, thin unbroken smooth curve of best fit or neat point-to-point ruled straight lines connecting the plotted coordinates.
- Do not make the line thick, feathery, or double, and do not extrapolate beyond 5.0% unless starting accurately at the origin (0, 0).

A (Axes): Both axes labelled with variable name AND correct units matching the table (`Hydrogen peroxide concentration / %` on x-axis and `Volume of oxygen (gas) collected (in 2 minutes) / cm³` on y-axis) [1];
S (Scale): Linear, sensible scale where the plotted points occupy more than half of the grid along both axes [1];
P (Plotting): All 6 points plotted accurately to within half a small square using small neat crosses or encircled dots [1];
L (Line): Single, thin, unbroken smooth curve of best fit OR points joined with neat ruled straight lines from point to point; no feathery lines and no inappropriate extrapolation [1].

(a) Formula for magnification:
\[\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{I}{A}\]

(b) Calculation:
Measured image size \(I = 48\text{ mm}\)
Actual size \(A = 1.2\text{ mm}\)
\[\text{Magnification} = \frac{48\text{ mm}}{1.2\text{ mm}} = \times 40\]

(c) Valid biological drawing rules include:
- Use single, continuous, unbroken lines.
- No shading or cross-hatching.
- Use a sharp HB pencil.
- Ensure accurate proportions.
- Occupy at least half the allocated drawing space.

(a) \(\text{magnification} = \frac{\text{image size}}{\text{actual size}}\) / \(M = \frac{I}{A}\); [0.5]

(b) substitution: \(\frac{48}{1.2}\); [1]
\(\times 40\) / \(40\); [1]
(ecf applies if incorrect measurement value is substituted correctly)

(c) any one from:
- clear, continuous / unbroken outlines / single line (no feathery / sketch lines); [1]
- no shading / no cross-hatching / no stippling; [1]
- larger than original / occupies more than half the available box / space; [1]
- drawn with correct proportions / shape; [1]
- (if labels required) label lines drawn with a ruler and touch the structure without crossing; [1]
[Total: 3.5]

(a) Unit conversion from micrometres to millimetres:
\[15\ \mu\text{m} = \frac{15}{1000}\text{ mm} = 0.015\text{ mm}\]

(b) Calculation of magnification:
\[\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\]
Convert both to the same units:
Image size \(= 45\text{ mm}\)
Actual size \(= 0.015\text{ mm}\)
\[\text{Magnification} = \frac{45}{0.015} = \times 3000\]
(Alternatively: \(\frac{45000\ \mu\text{m}}{15\ \mu\text{m}} = \times 3000\))

(c) Common biological drawing errors to avoid:
- Shading or hatching.
- Feathery, sketchy, or broken outlines.
- Drawing too small (less than 50% of the allocated space).

(a) \(0.015\) (\(\text{mm}\)); [0.5]

(b) correct unit conversion / formula substitution: \(\frac{45}{0.015}\) OR \(\frac{45\,000}{15}\); [1]
\(\times 3000\) / \(3000\); [1]
(allow ecf from part (a))

(c) any one from:
- shading / hatching / stippling; [1]
- sketchy / feathery / discontinuous / broken lines; [1]
- drawing too small / occupying less than half the available space; [1]
- overlapping / intersecting label lines (if labels drawn); [1]
- inaccurate proportions / wrong relative thickness of inner guard cell wall; [1]
[Total: 3.5]

A complete and logical investigation plan must address the following key points:

1. Independent Variable:
- Vary wind speed by using at least 3 different fan settings (e.g. speeds 1, 2, 3, and fan switched off for still air) OR by placing the fan at a minimum of 3 set distances from the shoot (e.g. 20 cm, 40 cm, 60 cm, 80 cm, 100 cm measured with a metre ruler).

2. Dependent Variable:
- Measure the distance moved by the air bubble (or water meniscus) in the capillary tube (in mm or cm) using the scale/ruler over a fixed period of time (e.g. 10 minutes timed using a stopwatch).
- Calculate the rate of water uptake as distance moved divided by time (e.g. mm/min).

3. Controlled Variables (at least two):
- Temperature: conduct the experiment in a temperature-controlled room or monitor room temperature using a thermometer.
- Light intensity: keep distance from an artificial light source constant / keep in constant ambient room light.
- Humidity: perform all tests in the same room away from open windows/draughts.
- Plant material: use shoots with the same number of leaves / similar total surface area and from the same plant species.

4. Method / Experimental Details:
- Cut the stem of the leafy shoot underwater (to prevent air entering xylem vessels).
- Assemble the potometer underwater and ensure all connections are airtight using petroleum jelly.
- Allow the shoot to equilibrate/acclimatise to each wind condition for 5 minutes before taking initial readings.
- Introduce an air bubble into the capillary tube and record its starting position.

5. Reliability:
- Repeat the procedure at least twice more for each wind speed (minimum of 3 trials per setting) and calculate a mean rate of water uptake (identifying and excluding any anomalies).

6. Safety Precaution:
- Take care when cutting the shoot with a scalpel/blade by cutting downwards onto a cutting board/tile away from fingers; OR keep the electric fan and plugs dry and away from water spills to prevent electrical hazards.

Award 1 mark for each of the following marking points (maximum 6 marks total):

- MP1 (Independent variable): at least three different fan speeds / wind speeds OR fan placed at at least three named distances (e.g. 20, 40, 60 cm) / includes a still air control (fan off) ;
- MP2 (Dependent variable): measure distance moved by air bubble / water meniscus in capillary tube in a stated / fixed time (e.g. in 5–15 minutes) OR measure time taken for bubble to travel a set distance ;
- MP3 & MP4 (Controlled variables - max 2):
- constant temperature (e.g. thermostatically controlled room / monitor with thermometer) ;
- constant light intensity (e.g. fixed distance from lamp / constant ambient light) ;
- constant humidity (e.g. enclosed chamber / same room) ;
- shoot characteristics kept constant (e.g. same species / same number of leaves / same surface area) ;
- MP5 (Experimental procedure detail): cut stem underwater / assemble underwater / apply petroleum jelly to joints to make airtight / allow shoot time to equilibrate before timing ;
- MP6 (Reliability): repeat at each wind speed / distance at least twice (minimum 3 trials per condition) AND calculate a mean ;
- MP7 (Safety): suitable precaution for using sharp blade (e.g. cut away from hands / on a board) OR electrical safety with fan near water (e.g. keep electrical equipment dry / wipe up spills immediately) ;

[Total: 6 marks]