2025 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Solution S caused the greatest decrease in length of the potato cylinder (from 50 mm to 46 mm, a decrease of 4 mm). This indicates that the net movement of water out of the potato cells by osmosis was the greatest in this solution. Water moves from a region of higher water potential to a region of lower water potential. Therefore, Solution S must have the lowest water potential relative to the potato cells.

Award 1 mark for identifying the correct solution (S) corresponding to Option D. Reject other options because they represent solutions with higher water potentials (less water loss or net water gain).

The purple color in the Biuret test indicates the presence of protein. The formation of a milky-white emulsion in the ethanol emulsion test indicates the presence of fat (lipids). The Benedict's test remaining blue means that reducing sugars are absent. Therefore, the food sample contains protein and fat.

Award 1 mark for Option B. Reject Option A because starch is not tested/indicated. Reject Options C and D because reducing sugar is absent (Benedict's remained blue).

Glucose is a small molecule that is filtered out of the blood in the glomerulus into the kidney tubule. In a healthy individual, 100% of the filtered glucose is actively reabsorbed back into the blood in the proximal convoluted tubule, so none is excreted in the urine. Proteins are too large to be filtered. Urea is an excretory product and is not completely reabsorbed. Water is reabsorbed, but not completely, as some is needed to form urine.

Award 1 mark for Option A (Glucose). Reject Option D because proteins do not pass through the filtration barrier. Reject Options B and C as they are not completely reabsorbed.

In bright light, the pupil reflex occurs to constrict the pupil and reduce the amount of light entering the eye. This is achieved by the contraction of the circular muscles and the relaxation of the radial muscles in the iris.

Award 1 mark for Option A. Reject Options B, C, and D because they describe incorrect combinations of muscle action or result in pupil dilation, which would allow too much light into the eye.

Typically, about 10% of the energy at one trophic level is transferred to the next. The krill are at the second trophic level (24,000 kJ). The energy transferred to the penguin population (third level) is approximately 10% of 24,000 kJ = 2,400 kJ. The energy transferred to the leopard seal population (fourth level) is approximately 10% of 2,400 kJ = 240 kJ.

Award 1 mark for Option B. Reject other options as they represent incorrect trophic level calculations or incorrect percentage transfer rates.

Protein synthesis begins in the nucleus where a DNA template is transcribed to form mRNA. The mRNA then leaves the nucleus through a nuclear pore and enters the cytoplasm, where it binds to a ribosome. Finally, tRNA molecules bring specific amino acids to the ribosome, matching their anticodons with the codons on the mRNA to assemble the protein.

Award 1 mark for Option A. Reject Options B, C, and D because they describe biochemically incorrect flows of genetic information or incorrect roles of DNA, mRNA, and tRNA.

Auxin is a plant hormone that is redistributed away from unilateral light. If light comes from the left, auxin moves to and concentrates on the shaded (right) side of the shoot tip. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, cells on the right side elongate more rapidly than those on the left, causing the shoot to bend towards the light source on the left.

Award 1 mark for Option C. Reject Options A and B because auxin concentrates on the shaded side (right), not the lit side (left). Reject Option D because auxin stimulates (rather than decreases) cell elongation in shoots.

During the stationary phase of a sigmoid growth curve, resources (like food and space) become limited and toxic waste products accumulate. As a result, the rate of bacterial cell division slows down and equals the rate of cell death, keeping the overall population size constant.

Award 1 mark for Option C. Reject Option A (lag phase, where cells are adapting and growth is slow but division exceeds death). Reject Option B (log phase, where division greatly exceeds death). Reject Option D (death phase, where death exceeds division).

At the concentration where there is no net movement of water into or out of the potato cells, the percentage change in mass would be 0%. Looking at the data, 0% change lies between +4% (at 0.2 mol/dm³) and -3% (at 0.4 mol/dm³). The concentration closest to this is 0.3 mol/dm³.

1 mark: Identify that 0% mass change indicates isotonic conditions, and interpolate between 0.2 and 0.4 mol/dm³ to find the correct value.

The Biuret test remaining blue indicates that protein is absent. The ethanol emulsion test turning cloudy-white indicates that fat (lipid) is present. Iodine solution turning blue-black indicates that starch is present. Benedict's test remaining blue indicates that reducing sugar is absent. Therefore, fat and starch are present.

1 mark: Correctly interpret each test result and select the option indicating the presence of fat and starch.

Glucose is small enough to be filtered out of the glomerulus into the Bowman's capsule, but is completely reabsorbed back into the blood in the proximal convoluted tubule so that none is lost in the urine of a healthy person. Urea is a waste product and is excreted. Water is only partially reabsorbed. Proteins are too large to be filtered in the first place.

1 mark: Identify glucose as the substance that is filtered and completely reabsorbed.

In bright light, the pupil constricts to protect the retina from damage. This is achieved by the contraction of the circular muscles of the iris and the relaxation of the radial muscles.

1 mark: Correctly identify the antagonistic action of circular and radial muscles and the resulting pupil constriction in bright light.

Energy is lost at each trophic level (about 90% loss) due to processes like respiration (where energy is lost as heat), excretion, and parts of organisms that are not eaten. Thus, the energy available decreases significantly at each subsequent level.

1 mark: Identify respiration heat loss as the main reason for energy loss between trophic levels.

tRNA anticodons are complementary to the mRNA codons. Adenine (A) pairs with Uracil (U), and Cytosine (C) pairs with Guanine (G). tRNA contains Uracil (U) instead of Thymine (T). Therefore, AUG pairs with UAC, GUG with CAC, CAC with GUG, and CUG with GAC.

1 mark: Correctly determine the complementary RNA base pairs for the given mRNA sequence.

In a horizontal shoot, gravity causes auxin to accumulate on the lower side. In shoots, a higher concentration of auxin stimulates cell elongation. Therefore, the cells on the lower side elongate more than those on the upper side, causing the shoot to curve upwards.

1 mark: Identify that auxin accumulates on the lower side of the shoot and stimulates elongation there, causing upward bending.

During the stationary phase, the population size remains constant because the rate of cell reproduction is balanced by the rate of cell death.

1 mark: Identify the stationary phase as the stage where the rate of reproduction equals the rate of death.

A solution with a higher water potential than the cell sap of the potato will cause water to enter the potato cells by osmosis, resulting in an increase in mass. The solution in which the potato gained the most mass (Solution 1, +5.0%) must have the highest water potential. Conversely, a solution with a lower water potential causes water to leave the cells, resulting in a loss of mass. The solution in which the potato lost the most mass (Solution 4, -8.0%) must have the lowest water potential.

Award 1 mark for identifying that the most positive percentage change in mass corresponds to the highest water potential of the external solution, and the most negative percentage change corresponds to the lowest water potential.

Since red blood cells are animal cells, they lack a cell wall. When placed in a solution with a higher water potential (hypotonic solution), water moves down its water potential gradient into the cells by osmosis. Because there is no cell wall to resist the internal turgor pressure, the red blood cells swell and eventually burst (undergo haemolysis).

Award 1 mark for the correct explanation that water enters by osmosis and causes the cells to swell and burst due to the absence of a cell wall.

The iodine test is used to detect starch; since the solution remains yellow-brown, starch is absent. Benedict's test is used to detect reducing sugars and requires heating; a change from blue to brick-red indicates a high concentration of reducing sugar is present.

Award 1 mark for correctly matching the negative iodine test to the absence of starch and the positive Benedict's test to the presence of reducing sugar.

Proteins are large molecules (polymers) synthesized by joining together smaller basic units called amino acids. Glycogen and starch are polymers of glucose only, and fats are made of glycerol and fatty acids.

Award 1 mark for identifying that proteins are made from amino acids.

Urea is produced in the liver from excess amino acids through the process of deamination. It is then transported in the blood to the kidneys, which filter it out to be excreted in urine. The bladder only stores urine, but does not excrete it from the blood.

Award 1 mark for identifying the liver as the site of production and the kidneys as the main organ of excretion.

For near vision (accommodation), the ciliary muscles contract, which decreases the diameter of the ciliary ring. This reduces the tension on the suspensory ligaments, causing them to slacken (become loose). Consequently, the elastic lens bulges and becomes fatter (more convex), increasing its refractive power to focus near light rays onto the retina.

Award 1 mark for identifying that near accommodation requires contraction of ciliary muscles, slackening of suspensory ligaments, and a fatter lens.

Energy is lost from a trophic level through egestion (undigested food in faeces), excretion (metabolic waste products like urea), and respiration (released as heat). Energy used for growth and stored in tissues (assimilation) is the only energy transferred to the next trophic level.

Award 1 mark for identifying egestion, excretion, and respiration as the processes that result in energy loss rather than transfer.

Alleles are defined as alternative or different forms of the same gene. A gene (not a chromosome) is a length of DNA coding for a specific protein. DNA is made of nucleotides, and human diploid cells contain 46 chromosomes in total.

Award 1 mark for identifying that alleles are alternative forms of the same gene.

(a) Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane.
(b) A high concentration of sucrose solution has a lower water potential than the cell sap of the potato cells. Therefore, water moves out of the cells by osmosis, down a water potential gradient. The vacuole and cytoplasm shrink, leading to the cells becoming flaccid or plasmolysed, which results in a loss of overall mass.
(c) Using percentage change in mass accounts for differences in the initial starting masses of the potato cylinders, allowing for a fair and direct comparison between them.
(d) Plot percentage change in mass on the y-axis against sucrose concentration on the x-axis, and draw a line of best fit. Find the point where the line crosses the x-axis (0% change in mass). At this point, the water potential inside the potato cells is equal to the water potential of the external sucrose solution.

(a) 1 mark for: net movement of water molecules; 1 mark for: from a region of higher water potential to a region of lower water potential / dilute to concentrated; 1 mark for: through a partially permeable membrane. [Total: 3]

(b) Max 5 marks from: external sucrose solution has lower water potential than cell sap / cytoplasm [1]; water moves out of the cells [1]; by osmosis [1]; down a water potential gradient [1]; vacuole / cytoplasm shrinks or cells become flaccid / plasmolysed [1]; loss of water leads to decrease in mass [1]. [Total: 5]

(c) 1 mark for: allows comparison because starting masses were different / not identical. 1 mark for: shows relative change rather than absolute. [Total: 2]

(d) 1 mark for: plot percentage change in mass against concentration; 1 mark for: draw a line of best fit; 1 mark for: find the concentration where there is 0% change in mass / where line crosses x-axis; 0.33 marks for: stating that at this concentration, water potential inside and outside are equal / isotonic. [Total: 3.33]

(a) (i) Carbon, hydrogen, oxygen, and nitrogen. (ii) Amino acids are the simple chemical subunits that are joined together by peptide bonds to form long polypeptide chains, which fold into a specific three-dimensional protein shape.
(b) To test for protein, add Biuret reagent (or a mixture of copper sulfate and sodium hydroxide) to the milk sample. If protein is present, the solution will change color from blue to purple/violet.
(c) To test for lipids, perform the ethanol emulsion test. Add ethanol to the sample and shake to dissolve the lipids, then pour the mixture into water. A positive result is indicated by the formation of a cloudy white emulsion.
(d) Antibodies have specific shapes that bind to specific antigens on the surface of pathogens. This helps to clump the pathogens together, neutralizing them or making it easier for phagocytes to engulf and destroy them.

(a) (i) 2 marks for naming all four: carbon, hydrogen, oxygen, nitrogen (1 mark for any two or three). [Total: 2]
(ii) 1 mark for: amino acids are the subunits; 1 mark for: joined by peptide bonds / forming a polypeptide chain. [Total: 2]

(b) 1 mark for: add Biuret reagent / copper sulfate and sodium hydroxide; 1 mark for: adding it to the milk sample; 1 mark for: positive color change from blue to purple / violet / mauve. [Total: 3]

(c) 1 mark for: add ethanol to sample and shake/mix; 1 mark for: pour mixture into water; 1.33 marks for: observation of a white / cloudy emulsion. [Total: 3.33]

(d) Max 3 marks from: antibodies bind to antigens [1]; on pathogens / bacteria / viruses [1]; they have a complementary shape [1]; clumping / agglutination of pathogens [1]; helps phagocytes engulf them / neutralizes toxins [1]. [Total: 3]

(a) Excess amino acids undergo deamination in the liver. The nitrogen-containing amino group is removed from the amino acid, forming ammonia. Since ammonia is highly toxic, the liver rapidly converts it into a less toxic substance called urea, which is then released into the blood.
(b) In the glomerulus, the high pressure of the blood forces small molecules (such as water, glucose, salts, and urea) through the capillary walls and the basement membrane into the Bowman's capsule, forming the glomerular filtrate. Large molecules like plasma proteins and blood cells are too big to pass through and remain in the blood.
(c) Glucose is a small molecule that passes freely into the nephron during ultrafiltration. However, in a healthy person, all of the filtered glucose is completely reabsorbed back into the blood from the proximal convoluted tubule via active transport against its concentration gradient. Consequently, no glucose remains in the fluid to be excreted in urine.
(d) When a person is dehydrated, the brain detects low water potential in the blood, causing the pituitary gland to release more ADH. ADH increases the permeability of the collecting ducts in the kidney, allowing more water to be reabsorbed back into the blood. This results in the production of a small volume of highly concentrated, dark-colored urine.

(a) Max 4 marks from: excess amino acids are broken down in liver [1]; nitrogen-containing part / amino group is removed [1]; process is called deamination [1]; forms ammonia [1]; ammonia is toxic [1]; ammonia is converted to urea [1]. [Total: 4]

(b) Max 3 marks from: high blood pressure in glomerulus [1]; forces small molecules / water / glucose / salts / urea through capillary walls [1]; into Bowman's capsule / forming filtrate [1]; large molecules / proteins / red blood cells cannot pass [1]. [Total: 3]

(c) Max 3.33 marks from: glucose is small enough to be filtered [1]; all glucose is reabsorbed [1]; in the proximal convoluted tubule [1]; by active transport [1.33] / against concentration gradient. [Total: 3.33]

(d) Max 3 marks from: brain / hypothalamus detects low water potential / dehydration [1]; pituitary gland secretes more ADH [1]; collecting ducts become more permeable [1]; more water reabsorbed into blood [1]; urine volume decreases AND concentration increases [1]. [Total: 3]

(a) To focus on a near object, the ciliary muscles in the eye contract, which causes the suspensory ligaments to slacken (become loose). This reduces the pull on the lens, allowing the lens to become thicker, more rounded, and more convex. This increased curvature refracts (bends) light rays more strongly, focusing them sharply onto the fovea of the retina.
(b) When bright light enters the eye, it is detected by photoreceptors in the retina. Electrical impulses travel along the sensory neurones of the optic nerve to the brain. The brain coordinates a response, sending impulses along motor neurones to the muscles of the iris. In response, the circular muscles of the iris contract while the radial muscles relax. This antagonistic action causes the pupil to constrict, reducing the amount of light entering the eye to prevent damage to the retina.
(c) Rod cells are distributed across most of the retina but are absent at the fovea. They are highly sensitive to low light intensities and provide black-and-white vision in dim light. Cone cells are concentrated at the fovea, are sensitive to high light intensities, and provide color vision and high-acuity (sharp) vision.

(a) Max 5 marks from: ciliary muscles contract [1]; suspensory ligaments slacken / loosen [1]; lens becomes thicker / more convex / more rounded [1]; less tension on lens [1]; light rays are refracted / bent more [1]; image focused onto retina / fovea [1]. [Total: 5]

(b) Max 5.33 marks from: light detected by retina / photoreceptors [1]; impulses along sensory neurone / optic nerve [1]; to brain [1]; impulses along motor neurone [1]; to iris / effector [1]; circular muscles contract [1]; radial muscles relax [1]; pupil constricts / smaller [1]; protective role / prevents too much light entering [0.33]. [Total: 5.33]

(c) Max 3 marks from: rods found outside fovea / all over retina AND cones concentrated at fovea [1]; rods detect low light intensity / low-light vision AND cones detect high light intensity / color vision [1]; three types of cones (red, green, blue) [1]; rods give low acuity / cones give high visual acuity [1]. [Total: 3]

(a) The principal source of energy for most ecosystems is the Sun. Producers, such as green plants, capture this light energy using the green pigment chlorophyll inside their chloroplasts. Through the process of photosynthesis, they convert this light energy into chemical energy stored in organic molecules (like glucose), which can then be passed along the food chain when they are consumed.
(b) Energy is lost at each trophic level in several ways, meaning only a small fraction is passed on. First, not all parts of an organism are eaten (such as woody branches, roots, or bones). Second, some of the food consumed is indigestible and is lost as feces (egestion). Third, a significant amount of energy is released through respiration and used for active processes like movement, muscle contraction, and maintaining body temperature, with much of this energy being dissipated as heat to the environment. Finally, energy is lost in metabolic waste products like urea.
(c) To find the percentage efficiency of energy transfer, use the formula:
Percentage efficiency = (Energy in primary consumers / Energy in producers) * 100
Percentage efficiency = (2250 / 25000) * 100 = 9.0%.

(a) Max 4 marks from: Sun / sunlight is principal source of energy [1]; light energy captured by chlorophyll [1]; in chloroplasts / in producers [1]; converted to chemical energy [1]; during photosynthesis [1]. [Total: 4]

(b) Max 6 marks from: not all of the organism is consumed [1]; some parts are indigestible / lost in feces / egestion [1]; energy lost as heat [1]; from respiration [1]; energy used for movement / metabolic processes [1]; energy lost in excretory products / urine [1]. [Total: 6]

(c) 1 mark for correct formula: (energy at higher level / energy at lower level) * 100 [1]; 1 mark for correct numbers substituted: (2250 / 25000) * 100 [1]; 1.33 marks for correct final answer of 9.0% or 9% [1.33]. (Accept correct answer with no working for full marks). [Total: 3.33]

(a) (i) Gravitropism is a response in which a plant grows towards or away from gravity. (ii) Roots grow towards the direction of gravity, showing positive gravitropism, whereas shoots grow away from gravity, showing negative gravitropism.
(b) Auxin is produced in the actively growing shoot tip. Under unilateral light (light from one side), auxin molecules migrate or diffuse away from the light, accumulating on the shaded side of the shoot. In shoots, a high concentration of auxin stimulates cell elongation. Because the cells on the shaded side have more auxin, they elongate more rapidly than the cells on the illuminated side. This unequal rate of growth causes the shoot to bend towards the light source.
(c) Unlike in shoots, a high concentration of auxin in roots inhibits cell elongation. When a root is placed horizontally, gravity causes auxin to accumulate on the lower side of the root tip. Since high auxin levels slow down cell elongation in roots, the cells on the upper side of the root (which have less auxin) elongate faster than the cells on the lower side. This differential growth causes the root to bend downwards, towards gravity.

(a) (i) 1 mark for: growth response to stimulus [1]; 1 mark for: gravity [1]. (ii) 1 mark for: roots show positive gravitropism [1]; 1 mark for: shoots show negative gravitropism [1]. [Total: 4]

(b) Max 6 marks from: auxin made / produced in shoot tip [1]; auxin diffuses down the shoot [1]; auxin moves to the shaded side [1]; high concentration of auxin on shaded side [1]; auxin stimulates cell elongation in shoots [1]; cells on shaded side elongate more / faster [1]; shoot bends towards the light [1]. [Total: 6]

(c) Max 3.33 marks from: in roots, high auxin concentration inhibits cell elongation [1]; gravity causes auxin to accumulate on lower side of horizontal root [1]; lower side elongates less / upper side elongates more / faster [1]; root bends downwards [0.33]. [Total: 3.33]

(a) (i)
Table should have clear rows and columns with suitable headings including units:
- Sucrose concentration / mol dm\(^{-3}\)
- Initial mass / g
- Final mass / g
- Change in mass / g
- Percentage change in mass / %

Calculated values:
- 0.0 mol dm\(^{-3}\): Change in mass = +0.48 g; Percentage change = +12.0%
- 0.2 mol dm\(^{-3}\): Change in mass = +0.16 g; Percentage change = +3.9% (accept +3.90%)
- 0.4 mol dm\(^{-3}\): Change in mass = -0.12 g; Percentage change = -3.0% (accept -3.04%)
- 0.6 mol dm\(^{-3}\): Change in mass = -0.40 g; Percentage change = -9.9% (accept -9.88%)
- 0.8 mol dm\(^{-3}\): Change in mass = -0.60 g; Percentage change = -15.0%

(a) (ii) Any two from:
- Volume of sucrose solution used (e.g., 20 cm\(^3\))
- Dimensions/diameter/length of the sweet potato cylinders
- Duration of immersion (60 minutes)
- Temperature of the environment/solutions
- Source/variety of the sweet potato

(a) (iii) To remove excess/adhering surface liquid; which would add extra mass and lead to an inaccurate/overestimated final mass measurement.

(b) (i) Identify where the line of best fit crosses the x-axis / where the percentage change in mass is 0%; at this point, the water potential inside the sweet potato cells is equal to the external sucrose concentration (isotonic), resulting in no net water movement.

(b) (ii) Error: Only one cylinder was tested per concentration (no replicates).
Improvement: Perform 3 or more replicates at each sucrose concentration and calculate a mean percentage change.
(Accept other valid errors: e.g., evaporation of solution -> use a lid/bung; variations in cylinder surface area -> use a standardized cutter/template).

(c) Plan:
- Independent Variable: Use at least 5 different temperatures (e.g., 10 °C, 20 °C, 30 °C, 40 °C, 50 °C) using thermostatically controlled water baths.
- Dependent Variable: Measure initial and final mass of cylinders and calculate the percentage change in mass over a fixed time period (e.g., 30 minutes).
- Constant Variables: Use the same concentration of sucrose solution (e.g., 0.4 mol dm\(^{-3}\)) across all temperatures. Ensure cylinders are cut from the same sweet potato using the same cork borer to keep surface area and volume constant.
- Replicates: Use at least 3 sweet potato cylinders at each temperature to calculate a mean and identify anomalies.
- Safety: Wear heat-resistant gloves when handling hot water baths, and use a cutting board when using sharp scalpels or cork borers to avoid injury.

(a) (i) Max [6 marks]:
- [1] Table drawn with clear cells, columns, and rows.
- [1] Headings containing both parameter and unit (Concentration / mol dm\(^{-3}\), Masses / g, Changes / %).
- [1] All raw mass data correctly transcribed.
- [1] Change in mass calculated correctly with correct signs (+ and -).
- [1] Percentage change calculated correctly for all concentrations.
- [1] Signs (+ and -) explicitly included in both 'change in mass' and 'percentage change' columns.

(a) (ii) [2 marks]:
- [1] for each correct controlled variable (volume of solution, cylinder size, duration, temperature, or tissue source) up to max 2.

(a) (iii) [2 marks]:
- [1] for removing surface water/liquid.
- [1] for preventing overestimation of final mass / ensuring accurate mass reading.

(b) (i) [2 marks]:
- [1] Find concentration where line crosses the x-axis / where change in mass is zero.
- [1] Explain that at this point, water potential inside and outside the cells is equal / isotonic (no net movement of water).

(b) (ii) [2 marks]:
- [1] for identifying a valid error (e.g., no replicates, evaporation, cutting variation).
- [1] for a matching, scientifically valid improvement.

(c) Max [6 marks]:
- [1] IV: At least 5 temperatures using water baths.
- [1] DV: Measure mass change / calculate percentage mass change.
- [1] CV 1: Same sucrose concentration.
- [1] CV 2: Same cylinder dimensions / same sweet potato source.
- [1] Replicates: At least 3 repeats per temperature to calculate a mean.
- [1] Safety: Specific precaution related to temperature (heat-resistant gloves) or cutting tool (cutting board/away from fingers).

(a) (i)
Table should feature clear boundaries with three column headings:
- Juice sample
- Time taken for first color change / s (or seconds)
- Final color

Data rows:
- Sample X: 45 | Brick-red
- Sample Y: 180 | Greenish-yellow
- Sample Z: No change (or >300 / 300) | Blue

(a) (ii)
- Independent variable: The juice sample / type of fruit juice.
- Dependent variable: Time taken for the first color change (or final color of the mixture).

(a) (iii)
- Sample X.
- It had the shortest time to change color (45 s) and the final color was brick-red, indicating the highest concentration of reducing sugars.

(a) (iv) Any two from:
- Volume of Benedict's reagent (2 cm\(^3\))
- Volume of juice sample (2 cm\(^3\))
- Temperature of the water bath (80 °C)
- Maximum heating time (300 seconds)

(a) (v)
- Safety: To prevent liquid from boiling over/spitting violently out of the tube (bumping), which can cause chemical burns.
- Control: Water bath provides uniform/even heating across all tubes, ensuring temperature is kept constant.

(b) (i)
- Add Biuret reagent (or sodium hydroxide and copper sulfate solution) to the food sample.
- Mix / shake the tube gently.
- Positive result: Color changes from blue to lilac / purple / mauve.

(b) (ii)
- Step 1: Set up 5 test tubes containing equal volumes (e.g., 2 cm\(^3\)) of each known glucose concentration (1%, 2%, 3%, 4%, 5%).
- Step 2: Add equal volumes of Benedict's reagent (2 cm\(^3\)) to each tube.
- Step 3: Heat all tubes simultaneously in the same 80 °C water bath.
- Step 4: Record the exact time taken for the first color change to occur in each tube.
- Step 5: Plot a calibration curve of glucose concentration (x-axis) against time taken for color change (y-axis).
- Step 6: Test sample X under the same conditions, measure its time to change color, and read the corresponding concentration off the calibration curve.

(a) (i) [4 marks]:
- [1] Table drawn with clear columns and rows, including row for Z.
- [1] Headings with appropriate units (Juice sample, Time / s, Final color).
- [1] All times recorded accurately (45, 180, and >300 or 'no change').
- [1] All final colors correct (brick-red, greenish-yellow, blue).

(a) (ii) [2 marks]:
- [1] Independent variable: Juice sample / concentration of sugar.
- [1] Dependent variable: Time for color change / final color.

(a) (iii) [2 marks]:
- [1] Correctly identifies Sample X.
- [1] Justified by shortest time to change color / brick-red final color.

(a) (iv) [2 marks]:
- [1] for each correct variable (volume of reagent, volume of juice, temperature of water bath) up to max 2.

(a) (v) [2 marks]:
- [1] Safety: Benedict's contains alkaline chemicals that can spit/spill easily upon direct heating.
- [1] Control: Temperature is regulated evenly / direct flame creates hot spots.

(b) (i) [3 marks]:
- [1] Name reagent: Biuret reagent / Biuret solution.
- [1] Method: Add and mix/shake.
- [1] Result: Blue to lilac / purple / mauve (reject pink/red).

(b) (ii) Max [5 marks]:
- [1] Test all five known glucose concentrations under the same conditions (same volume, same temperature).
- [1] Record the time taken for the first color change for each known concentration.
- [1] Plot a calibration curve / graph of concentration vs time to first change.
- [1] Test sample X under the exact same conditions.
- [1] Use the measured time for sample X to read off the glucose concentration from the calibration curve.