2025 CCEA GCSE Biology 1010 Practice Paper with Answers

The nucleus is the control centre of the cell. It contains the cell's DNA, organised into chromosomes, which carries the genetic instructions that control the cell's activities, including directing cell division.

1 mark: contains genetic material (DNA/chromosomes) which controls the cell's activities; 1 mark: controls cell division. Accept any two distinct, correct functions of the nucleus.

Photosynthesis combines carbon dioxide and water to form glucose and oxygen, using light energy absorbed by the green pigment chlorophyll. Checking the equation balances: carbon, 6 = 6; hydrogen, 6×2 = 12 on the left and 12 in \( C_6H_{12}O_6 \) on the right; oxygen, (6×2)+(6×1) = 18 on the left and 6+(6×2) = 18 on the right. The equation \( 6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2 \) is therefore correctly balanced, and chlorophyll is the pigment that absorbs the light energy.

1 mark: correctly balanced symbol equation \( 6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2 \) (all formulae and balancing numbers correct); 1 mark: chlorophyll named as the light-absorbing pigment.

The rate of photosynthesis can be limited by light intensity, since insufficient light means too little energy is captured for the light-dependent reactions; by carbon dioxide concentration, since CO2 is a raw material needed to make glucose; and by temperature, since photosynthesis is enzyme-controlled. Any two of these factors are correct.

1 mark each for any two of: light intensity; carbon dioxide concentration; temperature. Max 2 marks.

To test for protein, sodium hydroxide solution is added to the food sample, followed by a few drops of dilute copper sulfate solution (Biuret test). If protein is present, the solution changes colour from blue to purple (lilac).

1 mark: sodium hydroxide + dilute copper sulfate solution added to the sample (Biuret test); 1 mark: positive result = colour change from blue to purple/lilac.

Iron deficiency in the diet leads to anaemia, since iron is an essential component of haemoglobin, the oxygen-carrying pigment found in red blood cells. Without sufficient iron, the body cannot make enough haemoglobin, reducing the blood's capacity to transport oxygen.

1 mark: iron; 1 mark: needed to make haemoglobin, which carries oxygen (in red blood cells).

The enzyme amylase, present in saliva, catalyses the breakdown of starch (the substrate) into the sugar maltose (the product), by breaking the bonds joining the glucose sub-units of starch.

1 mark: amylase named; 1 mark: substrate correctly identified as starch and product as maltose (both required).

Gas exchange takes place at the alveoli, tiny air sacs found in vast numbers (hundreds of millions) at the end of the bronchioles. Because there are so many of them, the total internal surface area of the lungs is very large, which maximises the rate at which oxygen and carbon dioxide can diffuse across the alveolar walls.

1 mark: alveoli; 1 mark: large numbers give a large (total) surface area for (rapid) diffusion of gases.

An impulse generated at a receptor first travels along a sensory neurone to the spinal cord. Within the spinal cord it passes to a relay neurone (connector neurone), which then passes the impulse on to a motor neurone. The motor neurone carries the impulse to an effector (muscle or gland).

1 mark: sensory neurone then relay (connector) neurone; 1 mark: then motor neurone, in this correct order. Comparison/order required; reject if order is incorrect.

A synapse is the tiny gap between the end of one neurone and the next. When an electrical impulse reaches the end of the first neurone, it triggers the release of a chemical called a neurotransmitter, which diffuses across the synaptic gap and binds to specific receptor molecules on the membrane of the next neurone, triggering a new electrical impulse in that neurone.

1 mark: synapse = the (small) gap between two neurones; 1 mark: neurotransmitter released, diffuses across the gap and binds to receptors on the next neurone, generating a new impulse.

Insulin and glucagon are both produced by the pancreas. When blood glucose concentration rises above normal, insulin is released, causing cells (particularly liver and muscle cells) to take up glucose from the blood and convert it to glycogen for storage, thereby lowering blood glucose concentration back towards normal.

1 mark: pancreas; 1 mark: insulin decreases/lowers blood glucose concentration.

Adrenaline is released by the adrenal glands, situated above the kidneys. It prepares the body for 'fight or flight' by, for example, increasing heart rate, which delivers oxygen and glucose to the muscles more quickly.

1 mark: adrenal gland(s); 1 mark: any one correct effect, e.g. increased heart rate, increased breathing rate, pupils dilate, blood diverted to muscles, blood glucose concentration increases.

The rabbit, which eats the producer (grass), occupies the primary consumer trophic level. Energy transfer between trophic levels is never 100% efficient because a large proportion of the energy taken in is lost as heat through respiration, and further energy is lost in materials that are not eaten (e.g. roots) or not digested and so passed out in faeces (egestion).

1 mark: primary consumer; 1 mark: energy lost as heat (via respiration) and/or in undigested/uneaten material (egestion), not passed on to the next trophic level.

(a) As distance from the lamp increases, the rate of bubble production (rate of photosynthesis) decreases. The decrease is not linear: it is rapid at first (48→26→12 between 10–30 cm) and levels off at greater distances (12→7→5 between 30–50 cm).
(b) Light intensity decreases as distance from the lamp increases. Since light provides the energy for the light-dependent reactions of photosynthesis, less light energy reaching the pondweed means a slower rate of photosynthesis, and so fewer oxygen bubbles are produced per minute.
(c) Water/solution temperature should be kept constant, since temperature also affects the rate of the enzyme-controlled reactions of photosynthesis (carbon dioxide concentration and the length/species of pondweed used are also acceptable).

(a) 1 mark: rate decreases as distance increases (inverse relationship); 1 mark: valid additional detail, e.g. decrease is not linear, rate falls rapidly at first then levels off. [2]
(b) 1 mark: light intensity decreases with distance from the lamp; 1 mark: less light energy available for the light-dependent reaction, so a reduced rate of photosynthesis (fewer bubbles). [2]
(c) 1 mark: any valid controlled variable, e.g. water/solution temperature, CO2/sodium hydrogencarbonate concentration, size/species of pondweed, time allowed before counting. [1]

(a) Sample Z is positive for iodine (starch) and Biuret (protein), and negative for the emulsion test, so Z contains starch and protein but not fat.
(b) In the emulsion test, the food sample is shaken thoroughly with ethanol, and the resulting ethanol solution is then poured into a test tube of water. A positive result is shown by a cloudy white (milky) emulsion forming in the water; a negative result leaves the water clear.
(c) Benedict's test (Benedict's solution added and the mixture heated in a water bath).

(a) 1 mark: starch and protein (both required; must not include fat). [1]
(b) 1 mark: sample shaken/mixed with ethanol; 1 mark: resulting solution poured/added into water; 1 mark: positive result = cloudy white/milky emulsion forms. [3]
(c) 1 mark: Benedict's test/Benedict's solution (heated). [1]

(a) pH 7 gives the shortest time (25 s), meaning the fastest rate of reaction, so pH 7 is the optimum pH for this enzyme.
(b) At pH 4, the enzyme is far from its optimum pH. The excess H+ ions disrupt the bonds holding the enzyme's tertiary structure in shape, changing the shape of the active site. Because the active site's shape is no longer complementary to the shape of the starch substrate, fewer enzyme-substrate complexes can form, so the enzyme works much more slowly and the reaction takes far longer to complete.
(c) Repeating the reading and calculating a mean reduces the effect of anomalous results/random error, improving the reliability of the result.

(a) 1 mark: pH 7. [1]
(b) 1 mark: pH 4 is far from the optimum; 1 mark: bonds maintaining the enzyme's (tertiary) structure/shape are disrupted, changing the shape of the active site; 1 mark: active site no longer complementary to the substrate, so fewer enzyme-substrate complexes form and the reaction rate is much slower. [3]
(c) 1 mark: to identify/reduce the effect of anomalous results and improve the reliability of the mean. [1]

(a) Increase = 28 − 14 = 14 breaths/min. Percentage increase = (14 ÷ 14) × 100 = 100%. Checking by a second route: 28 is exactly double 14, and a doubling corresponds to a 100% increase, confirming the answer.
(b) During exercise, muscles respire faster to release more energy, using more oxygen and producing more carbon dioxide. Breathing rate increases to increase the rate of gas exchange in the lungs, supplying oxygen to the blood faster and removing carbon dioxide faster.
(c) Oxygen debt.

(a) 1 mark: correct working shown, e.g. (28−14)/14 × 100; 1 mark: correct answer of 100%. Accept ECF from an incorrect subtraction provided the method is correct. [2]
(b) 1 mark: increased respiration in muscles requires more oxygen and produces more CO2; 1 mark: faster breathing increases the rate of gas exchange, supplying O2 and removing CO2 faster. [2]
(c) 1 mark: oxygen debt. [1]

(a) The reflex response is faster than the voluntary reaction. Difference = 280 − 40 = 240 ms.
(b) A reflex response follows a short, fixed neural pathway (reflex arc): receptor → sensory neurone → relay neurone in the spinal cord → motor neurone → effector, bypassing the brain, so the response happens automatically. A voluntary reaction involves impulses travelling all the way to the brain, where the sensory information must be processed and a conscious decision made before a response is initiated, adding extra synapses and processing time. The reflex pathway is therefore shorter, with fewer synapses, so the response is faster.

(a) 1 mark: reflex is faster; 1 mark: correct difference of 240 ms shown with working. [2]
(b) 1 mark: reflex pathway is short/fixed and does not involve the brain (spinal cord only); 1 mark: voluntary reaction involves impulses travelling to the brain for conscious processing/decision-making; 1 mark: reflex pathway has fewer synapses/shorter distance, so is faster. Comparison required. [3]

(a) Blood glucose concentration rises sharply from 4.5 to a peak of 7.8 mmol/dm3 at 0.5 hours after the meal, then falls steadily, returning to the original (normal) level of 4.5 mmol/dm3 by around 2.5–3 hours.
(b) The rise in blood glucose concentration above normal is detected, causing the pancreas to release more insulin. Insulin causes cells (especially liver and muscle cells) to take up glucose from the blood and convert it into glycogen for storage, lowering blood glucose concentration back towards the normal set point. Because this corrective response acts to reverse/cancel out the original change (the rise in glucose concentration), this is an example of negative feedback.

(a) 1 mark: rises to a peak (at 0.5 h); 1 mark: then falls/decreases back to the original (starting) level by about 2.5-3 hours. [2]
(b) 1 mark: rise in blood glucose detected, pancreas releases (more) insulin; 1 mark: insulin causes uptake of glucose by cells/conversion to glycogen (in liver/muscle), lowering blood glucose; 1 mark: correctly identifies this as negative feedback because the response counteracts/reverses the original change, returning the level to normal. [3]

(a) Percentage transferred = (energy at primary consumers ÷ energy at producers) × 100 = (1800 ÷ 18000) × 100 = 10%. Checking by a second route: 10% of 18000 = 1800, which matches the given primary consumer value, confirming the calculation.
(b) The pyramid of energy would have a wide base (producers, largest energy content), narrowing sharply at each successive level, always forming an upright pyramid shape, since energy content always decreases from one trophic level to the next.
(c) Energy is lost between trophic levels because: (1) a large proportion of energy is lost as heat through respiration; (2) some energy is lost in materials that are not eaten (e.g. roots, bark) by the next consumer; (3) some energy is lost in material that is eaten but not digested, and is egested as faeces rather than being converted into new biomass.

(a) 1 mark: correct method shown, e.g. (1800/18000) × 100; 1 mark: correct answer of 10%. [2]
(b) 1 mark: pyramid shape narrows/decreases at each level, always upright/never inverted, since energy always decreases up the chain. [1]
(c) 1 mark each for any three of: energy lost as heat via respiration; energy lost in uneaten parts/material; energy lost in undigested/egested material (faeces); energy used for movement/other life processes not converted to biomass. Max 3 marks. [3]

(a) Rate at intensity 1 = 5 bubbles/min; rate at intensity 4 = 25 bubbles/min. Increase = 25 − 5 = 20 bubbles/min. Percentage increase = (20 ÷ 5) × 100 = 400%. Checking by a second route: 25 is 5 × 5, so the rate has become five times as great, corresponding to an increase of (5−1) × 100% = 400%, confirming the answer.
(b) The rate levels off from light intensity 5 onwards (values at intensity 5 and 6 are both 30 bubbles/min, showing no further increase). Above this point, carbon dioxide concentration (or temperature) has become the limiting factor rather than light intensity.

(a) 1 mark: correct working shown, e.g. (25−5)/5 × 100 or equivalent; 1 mark: correct final answer, 400%. Accept ECF. [2]
(b) 1 mark: light intensity 5 (accept 'above 5'); 1 mark: carbon dioxide concentration or temperature identified as the new limiting factor. [2]

The hypothalamus, in the brain, contains thermoreceptors that continuously monitor the temperature of the blood flowing through it. When core body temperature rises above the normal set point (about 37°C), the hypothalamus detects this change and coordinates several responses to lower temperature back to normal.

Vasodilation: arterioles supplying blood to capillaries near the surface of the skin widen (dilate), increasing blood flow close to the skin surface. This increases the rate of heat loss from the blood to the environment by radiation, cooling the blood.

Sweating: sweat glands increase their production of sweat, released onto the skin surface. As the water in sweat evaporates, it takes heat energy from the skin and body, cooling the body (evaporative cooling).

Hair erector muscles relax, causing body hairs to lie flat, reducing the layer of insulating air trapped near the skin and allowing heat to be lost more easily.

These responses reduce body temperature back towards the normal set point of 37°C. Because the corrective responses act to reverse the original rise in temperature, restoring the body to its normal state, this is an example of negative feedback.

Level 0 [0]: No relevant content.
Level 1 [1-2]: Basic, partial description — e.g. names one relevant response (such as sweating) with little or no reference to the hypothalamus or to how the response lowers temperature. Little use of specialist terms; weak organisation.
Level 2 [3-4]: Satisfactory description of at least two correct responses (e.g. vasodilation and sweating) with some explanation of how each helps lower body temperature; hypothalamus mentioned as the coordinating/detecting centre; some use of specialist terminology; adequately organised.
Level 3 [5-6]: Comprehensive and accurate description of at least two responses (vasodilation, increased sweating, hair erector muscles relaxing) with a clear, correct explanation of how each lowers body temperature (via radiation/evaporation), correctly identifies the hypothalamus as the centre that detects the temperature rise, and identifies the mechanism as negative feedback; uses accurate specialist terminology throughout (e.g. hypothalamus, vasodilation, thermoreceptors, negative feedback); well organised with accurate grammar and spelling.
Award marks according to the level that best fits the response as a whole; use the full range within a level to reflect how fully the response meets the level descriptor.

An allele is one of two or more alternative versions of a gene, which may produce different forms of a characteristic. The genotype of an organism is the combination of alleles it possesses for a particular gene/characteristic (e.g. TT, Tt or tt).

1 mark: allele = an alternative version of a gene; 1 mark: genotype = the (combination of) alleles present in an organism for a characteristic.

A gene is a short section of DNA that carries the code for (codes for) a specific protein, which in turn controls a particular characteristic of an organism. Genes are found at specific positions along chromosomes, which are located in the nucleus of a cell.

1 mark: gene = a section of DNA that codes for a (specific) protein/characteristic; 1 mark: located on chromosomes, in the nucleus.

A normal human body cell contains 46 chromosomes, arranged as 23 homologous pairs. Gametes (sperm and egg cells) contain only 23 chromosomes, one from each pair, as a result of meiosis halving the chromosome number.

1 mark: 46 (body cell); 1 mark: 23 (gamete).

Gametes are produced by meiosis. This process creates genetic variation in the daughter cells in several ways, including crossing over (the exchange of sections of DNA between homologous chromosomes) and independent assortment (the random arrangement of homologous chromosome pairs before they are separated), both of which produce new combinations of alleles.

1 mark: meiosis; 1 mark: valid source of variation, e.g. crossing over (exchange of DNA between homologous chromosomes) or independent/random assortment of chromosomes.

Oestrogen is the hormone responsible for repairing and thickening the lining of the uterus (endometrium) following menstruation, in preparation for a possible pregnancy. It is produced by the ovaries.

1 mark: oestrogen; 1 mark: produced by the ovaries.

A hormonal method of contraception, such as the combined contraceptive pill, contains hormones that prevent pregnancy. A barrier method, such as a condom, works by physically preventing sperm from reaching an egg.

1 mark: any valid hormonal method, e.g. the (contraceptive) pill, implant, injection; 1 mark: any valid barrier method, e.g. condom, diaphragm/cap.

Fertilisation, the fusion of a sperm and egg nucleus, normally occurs in the oviduct (fallopian tube). The resulting fertilised egg (zygote) then travels to the uterus, where it implants into the thickened uterus lining (endometrium).

1 mark: oviduct/fallopian tube (fertilisation); 1 mark: uterus/uterus lining (implantation).

The human heart has four chambers: two upper chambers, the right atrium and left atrium, and two lower chambers, the right ventricle and left ventricle.

1 mark: right atrium and left atrium named; 1 mark: right ventricle and left ventricle named. All four required across the two marking points.

Arteries have thicker walls, containing more muscle and elastic tissue, than veins. This structural difference allows arteries to withstand the high pressure of blood as it is pumped away from the heart, and the elastic tissue allows the artery wall to stretch and recoil, helping to maintain blood pressure and flow.

1 mark: valid structural difference, e.g. arteries have thicker/more muscular/more elastic walls than veins (or veins have valves, arteries do not); 1 mark: difference correctly linked to function, e.g. thick/elastic walls withstand high pressure from the heart. Comparison required.

The pulmonary artery carries deoxygenated blood away from the heart to the lungs (it is the one artery in the body that carries deoxygenated blood). It originates from the right ventricle, which pumps blood into it.

1 mark: pulmonary artery; 1 mark: right ventricle.

Phagocytes are white blood cells that defend the body by phagocytosis: the phagocyte engulfs the pathogen, taking it inside the cell within a vesicle, and then digests/destroys it using digestive enzymes.

1 mark: phagocyte; 1 mark: engulfs the pathogen and digests it (using enzymes).

A vaccine contains a dead or inactive form of a pathogen (or an antigen from it) and is given before infection to stimulate the immune system to produce antibodies and memory cells, so preventing future disease. An antibiotic is a medicine given after infection has occurred, to kill or stop the growth of bacteria causing the illness; antibiotics are not effective against viral infections.

1 mark: vaccine given to prevent disease, by stimulating an immune response/antibody or memory cell production, before infection; 1 mark: antibiotic given to treat/cure an existing bacterial infection (and does not work against viruses).

Tuberculosis (TB) is caused by a bacterium (Mycobacterium tuberculosis). It is spread between people through airborne droplets, released into the air when an infected person coughs or sneezes and then inhaled by another person.

1 mark: bacterium/bacteria; 1 mark: spread via airborne droplets from coughing/sneezing (droplet infection).

Continuous variation describes a characteristic that can take any value across a range, with no clear-cut categories (e.g. human height or body mass), and is usually influenced by both genes and the environment. Discontinuous variation describes a characteristic that falls into a small number of distinct, separate categories with no intermediates (e.g. human blood group, A/B/AB/O), and is usually controlled by a single gene, largely unaffected by the environment.

1 mark: continuous variation = a full range/no distinct categories, with a valid example (e.g. height, mass); 1 mark: discontinuous variation = distinct/separate categories, with a valid example (e.g. blood group). Comparison required.

Osmosis is the diffusion (net movement) of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane.

1 mark: movement of water (molecules) from a higher to a lower water potential/from a dilute to a more concentrated solution, through a partially permeable membrane. The listing principle applies: all key terms required for the mark.

In a double circulatory system, blood passes through the heart twice for each complete circuit of the body: once through the pulmonary circuit (heart to lungs and back) and once through the systemic circuit (heart to body and back). Blood pressure and flow rate drop significantly as blood passes through the many narrow capillaries of the lungs. By returning to the heart before being sent to the rest of the body, the blood's pressure can be re-boosted by the left side of the heart. This maintains a faster rate of blood flow, and therefore a faster rate of oxygen delivery to body tissues, than would be possible with a single circulation.

1 mark: double circulation = blood passes through the heart twice for each complete circuit (pulmonary circuit to lungs, systemic circuit to body); 1 mark: blood pressure/speed drops significantly after passing through the (low-pressure) capillaries of the lungs; 1 mark: returning to the heart before travelling to the body allows the pressure to be re-boosted; 1 mark: this maintains a faster rate of blood flow/more efficient oxygen delivery to body tissues than a single circulation would allow.

(i) The biconcave disc shape increases the red blood cell's surface area relative to its volume, and keeps the cell thin, reducing the diffusion distance. Both effects allow oxygen to diffuse into and out of the cell more quickly.
(ii) Because red blood cells have no nucleus, there is more internal space available to be packed with haemoglobin, the pigment that binds and carries oxygen, so allowing each red blood cell to carry a greater quantity of oxygen.

1 mark (i): biconcave shape increases surface area (to volume ratio) and/or reduces diffusion distance; 1 mark (i): allowing faster/more efficient diffusion of oxygen into/out of the cell; 1 mark (ii): no nucleus leaves more internal space for haemoglobin; 1 mark (ii): allowing the cell to carry/transport more oxygen.

Both parents must be heterozygous carriers of the condition, with genotype Ff (one dominant, normal allele F and one recessive, cystic fibrosis allele f). Because the allele is recessive, a carrier with genotype Ff does not show the condition, since the dominant allele F is expressed and masks the effect of the recessive allele. However, if both parents happen to pass on their recessive f allele to a child, that child inherits the genotype ff. With no dominant allele present to mask it, the recessive allele's effect is expressed, and the child has cystic fibrosis.

1 mark: both parents must be heterozygous carriers (genotype Ff); 1 mark: carriers do not show the condition because the dominant (normal) allele masks/is expressed over the recessive allele; 1 mark: for the child to have cystic fibrosis, they must inherit the recessive allele from both parents (genotype ff); 1 mark: with no dominant allele present, the recessive allele's effect (cystic fibrosis) is expressed in the child.

Human females have the sex chromosome genotype XX, so every egg a mother produces carries an X chromosome. Human males have the genotype XY, so a father produces two types of sperm in equal numbers: those carrying an X chromosome and those carrying a Y chromosome. If an X-carrying sperm fertilises the egg, the resulting offspring is XX (female); if a Y-carrying sperm fertilises the egg, the offspring is XY (male). Since the mother's contribution (X) is always the same, it is the type of sperm (X or Y) contributed by the father that determines the sex of the offspring.

1 mark: mother always contributes an X chromosome (all her eggs carry X); 1 mark: father produces both X-carrying and Y-carrying sperm (in equal numbers); 1 mark: whichever type of sperm fertilises the egg determines the offspring's sex (XX = female, XY = male), so it is the father's gamete that determines sex.

The oestrogen and progesterone in the pill inhibit (suppress) the release of FSH (and LH) from the pituitary gland. Without sufficient FSH, no follicle matures in the ovary, so ovulation does not occur (no egg is released). Since fertilisation requires an egg to be present, without ovulation, fertilisation and therefore pregnancy cannot take place. (Progesterone in the pill also thickens the mucus at the entrance to the uterus, making it harder for any sperm to pass through.)

1 mark: hormones inhibit/suppress release of FSH (and LH) from the pituitary gland; 1 mark: without FSH, no follicle/egg matures in the ovary; 1 mark: therefore ovulation does not occur (no egg released); 1 mark: without an egg being released, fertilisation cannot take place (accept progesterone thickening cervical mucus, hindering sperm, as an alternative fourth marking point).

FSH (follicle stimulating hormone), released by the pituitary gland, stimulates a follicle in the ovary to mature/develop, and also stimulates the ovary to produce oestrogen. Rising oestrogen levels cause the uterus lining to thicken and repair, and, once oestrogen reaches a sufficiently high level, stimulate the pituitary gland to release a surge of LH (luteinising hormone). This surge of LH triggers ovulation, the release of the mature egg from the ovary.

1 mark: FSH released by the pituitary gland, stimulates a follicle to mature/develop in the ovary; 1 mark: FSH also stimulates the ovary to produce oestrogen; 1 mark: rising oestrogen causes the uterus lining to thicken/repair and stimulates release of LH from the pituitary; 1 mark: LH surge triggers ovulation (release of the egg).

A vaccine contains a dead or inactivated form of a pathogen, or a harmless fragment of it (an antigen). This stimulates white blood cells (lymphocytes) to produce antibodies specific to that pathogen's antigens. Importantly, memory cells are also produced, which remain in the body for a long time after vaccination. If the same pathogen infects the body again in future, these memory cells allow a much faster and stronger secondary immune response, producing large quantities of the correct antibody quickly enough to destroy the pathogen before it can cause the symptoms of disease.

1 mark: vaccine contains a dead/inactive pathogen (or antigen fragment), which stimulates white blood cells to produce specific antibodies; 1 mark: memory cells are also produced and remain in the body; 1 mark: on a second/future exposure to the same pathogen, memory cells enable a faster and stronger (secondary) immune response; 1 mark: pathogen destroyed before the person becomes ill/shows symptoms of disease.

Within a bacterial population there is natural variation (arising from mutation), so a small number of bacteria may, by chance, already carry an allele giving resistance to the antibiotic. When the antibiotic is used, the non-resistant bacteria are killed, but the resistant bacteria survive. These surviving resistant bacteria reproduce (rapidly, by binary fission), passing the resistance allele on to their offspring. Over time, and especially with repeated use of the antibiotic, the proportion of resistant bacteria in the population increases, until eventually the antibiotic may become ineffective against most of the population.

1 mark: variation exists within the bacterial population (due to mutation), so a few bacteria may already be resistant; 1 mark: antibiotic use kills non-resistant bacteria, but resistant bacteria survive; 1 mark: resistant bacteria reproduce, passing the resistance allele to their offspring; 1 mark: over time/generations, the proportion of resistant bacteria in the population increases.

Variation in colour (light or dark) already existed within the moth population, due to different alleles. As soot blackened the tree bark, dark moths became better camouflaged against the darkened bark, making them less visible to predatory birds, while light moths became more visible and more likely to be eaten. Because dark moths were more likely to survive to reproduce, they passed on the allele for dark colouration to more of their offspring than light moths did. Over many generations, the proportion of dark-coloured moths in the population increased.

1 mark: variation in colour exists within the population (light/dark alleles); 1 mark: dark moths were better camouflaged against sooty bark, so were less likely to be eaten by predators (light moths more visible/more likely to be eaten); 1 mark: dark moths more likely to survive and reproduce, passing on the allele for dark colour, so the proportion of dark moths increased over generations.

Parent Tt produces gametes T and t (in equal proportion). Parent tt produces only gamete t.

Punnett square:
t t
T Tt Tt
t tt tt

Offspring genotypes: 2 Tt : 2 tt, i.e. 1 Tt : 1 tt. Since T is dominant, Tt offspring are tall and tt offspring are short.
Final answer: expected ratio of tall to short offspring = 1 : 1.

1 mark: correct gametes identified for each parent (T and t from Tt; t only from tt); 1 mark: correctly completed Punnett square/genetic diagram showing all four offspring genotype combinations (Tt, Tt, tt, tt); 1 mark: correct genotype ratio, 1 Tt : 1 tt (or 2:2); 1 mark: correct phenotype ratio, 1 tall : 1 short.

Each Ee parent produces gametes E and e (in equal proportion).

Punnett square:
E e
E EE Ee
e Ee ee

Offspring genotypes: 1 EE : 2 Ee : 1 ee. Since E is dominant, EE and Ee offspring have unattached earlobes, and only ee offspring have attached earlobes. Out of 4 possible offspring, 1 is ee.
Percentage with attached earlobes = (1 ÷ 4) × 100 = 25%. Check by a second route: the phenotype ratio is 3 unattached : 1 attached, and 1 part out of 4 total parts is 25%, confirming the answer.

1 mark: correct gametes (E and e) identified from each Ee parent; 1 mark: correctly completed Punnett square showing EE, Ee, Ee, ee; 1 mark: correct genotype ratio 1 EE : 2 Ee : 1 ee (phenotype ratio 3 unattached : 1 attached); 1 mark: correct final answer, 25% (1 in 4) expected to have attached earlobes.

(a) Cardiac output = heart rate × stroke volume = 72 × 70 = 5040 cm3/min. Check by a second route: 72 × 70 = 72 × 7 × 10 = 504 × 10 = 5040, confirming the answer.
(b) Since 1 dm3 = 1000 cm3, 5040 cm3 ÷ 1000 = 5.04 dm3/min.

(a) 1 mark: correct method (72 × 70); 1 mark: correct answer, 5040, with correct unit (cm3/min). [2]
(b) 1 mark: correct method (÷ 1000); 1 mark: correct answer, 5.04 dm3/min, with correct unit. Accept error carried forward from (a). [2]

(a) Number of 20-minute divisions in 2 hours (120 minutes) = 120 ÷ 20 = 6 divisions. Population after n divisions = 2^n, so population = 2^6 = 64. Check by repeated doubling: 1 → 2 → 4 → 8 → 16 → 32 → 64 (6 doublings), confirming the answer of 64.
(b) Exponential growth.
(c) In a real, closed environment, growth eventually slows because nutrients/food become limited, toxic waste products build up, or there is increased competition for space or oxygen.

(a) 1 mark: correct method, e.g. 120 ÷ 20 = 6 divisions, population = 2^6; 1 mark: correct answer, 64. [2]
(b) 1 mark: exponential growth. [1]
(c) 1 mark: any valid limiting factor, e.g. nutrients/food become limited, waste products build up, competition for space/oxygen increases. [1]

(i) The skin acts as a physical barrier covering the body, preventing most pathogens from entering the underlying tissues. If the skin is cut, platelets help clot the blood at the wound site, resealing the barrier and reducing further pathogen entry.
(ii) The lining of the airways produces mucus, a sticky substance that traps pathogens (and dust) before they can reach the lungs. Cilia, tiny hair-like structures on the cells lining the airways, beat continuously to move the mucus, along with any trapped pathogens, up and out of the airways towards the throat, where it is swallowed or expelled, preventing pathogens from reaching the lungs.
(iii) White blood cells provide two main lines of defence. Phagocytes engulf and digest pathogens by phagocytosis, destroying them non-specifically, regardless of the type of pathogen. Lymphocytes recognise specific antigens on a pathogen's surface and produce antibodies that bind to and help destroy that particular pathogen; lymphocytes can also produce memory cells, providing longer-term immunity to that pathogen.

Level 0 [0]: No relevant content.
Level 1 [1-2]: Basic, partial description of one defence (e.g. skin as a barrier) with little explanation of mechanism. Little use of specialist terms; weak organisation.
Level 2 [3-4]: Satisfactory description of at least two of the three defences, with some explanation of how each works; some use of specialist terminology (e.g. mucus, cilia, phagocyte); adequately organised.
Level 3 [5-6]: Comprehensive, accurate description of all three defences — skin as a physical barrier, mucus/cilia trapping and removing pathogens from the airways, and white blood cells providing both non-specific (phagocytosis) and specific (antibody/lymphocyte) defence; uses accurate specialist terminology throughout (e.g. phagocytosis, antigen, antibody, non-specific/specific defence); well organised with accurate grammar and spelling.
Award marks according to the level that best fits the response as a whole; use the full range within a level to reflect how fully the response meets the level descriptor.

An accurate volume such as 5 cm3 of a liquid is measured out using a measuring cylinder of an appropriate size (e.g. 5 cm3 or 10 cm3), or alternatively a graduated pipette or syringe.

1 mark: measuring cylinder (or other valid accurate volume-measuring apparatus, e.g. graduated pipette, syringe, burette).

Iodine solution is orange/brown in its original form. When it is added to a sample containing starch, it forms a blue-black colour, which is the positive result for starch.

1 mark: colour change from orange/brown to blue-black. Both the starting and end colour required for the mark (listing principle applies).

A suitable table has a heading row identifying the quantity being measured and its unit, e.g. 'Time for starch to disappear (s)', with separate columns for Repeat 1, Repeat 2 and Repeat 3, and a further column for the calculated Mean. Units (s) should appear once, in the column heading, not repeated after every value in the table.

1 mark: table includes clearly labelled columns for the repeats, with the unit (s) given once in the heading, not repeated in every cell; 1 mark: table includes a column for the mean of the repeats.

Both axes should be scaled so the data occupies more than half of the grid: time (0-60 s) on the x-axis and volume of oxygen (0-25 cm3, or slightly higher to allow for the curve to be drawn comfortably) on the y-axis, each labelled with the correct quantity and unit. All seven points — (0,0), (10,9), (20,16), (30,21), (40,24), (50,25), (60,25) — should be plotted accurately. A single smooth curve of best fit should be drawn through the points (not joined point-to-point), rising steeply over the first 20-30 seconds and then levelling off to a horizontal plateau at 25 cm3 from around 50 seconds onward, since the reaction has finished (all the hydrogen peroxide has been used up). The graph should be given a suitable title, e.g. 'Graph to show the volume of oxygen produced over time in the reaction between catalase and hydrogen peroxide'.

1 mark: sensible scales chosen for both axes, using more than half of the available grid; 1 mark: both axes correctly labelled with quantity and unit (Time/s; Volume of O2 collected/cm3); 1 mark: at least 5 of the 7 points plotted accurately (within ±half a small square); 1 mark: all 7 points plotted accurately; 1 mark: a single smooth curve of best fit drawn through the points (not point-to-point), correctly showing the plateau; 1 mark: a suitable, relevant title given to the graph.

(a) Rate = volume of O2 produced ÷ time taken = (16 − 0) cm3 ÷ 20 s = 0.8 cm3/s. Check: 0.8 × 20 = 16, confirming the answer.
(b) Rate = (25 − 24) cm3 ÷ 20 s = 1 ÷ 20 = 0.05 cm3/s. Check: 0.05 × 20 = 1, confirming the answer. The rate has fallen from 0.8 cm3/s to 0.05 cm3/s (16 times slower) because the concentration of hydrogen peroxide (the substrate) decreases as the reaction proceeds, meaning there are fewer substrate molecules available to collide with and react at the catalase active sites in a given time; by 40-60 s almost all of the hydrogen peroxide has already reacted.

(a) 1 mark: correct method, (16-0)/20; 1 mark: correct answer, 0.8 cm3/s. [2]
(b) 1 mark: correct method and answer, (25-24)/20 = 0.05 cm3/s; 1 mark: valid explanation that the rate falls because substrate (H2O2) concentration decreases as it is used up, reducing the frequency of successful collisions with the enzyme's active sites. [2]

Variables that should be kept constant include the temperature of the reaction mixture and the volume (and concentration) of yeast suspension used, since both temperature and enzyme quantity affect the rate of an enzyme-catalysed reaction. If temperature were not controlled, any change observed in the rate of reaction could be caused by the change in temperature rather than (or as well as) the change in hydrogen peroxide concentration, making it impossible to draw a valid conclusion about the effect of the independent variable alone.

1 mark: valid controlled variable 1, e.g. temperature; 1 mark: valid controlled variable 2, e.g. volume/concentration of yeast suspension (accept other valid variables, e.g. total reaction volume); 1 mark: valid explanation of why controlling one named variable is important, referring to it otherwise affecting the rate and confounding the results.

Taking regular readings throughout the reaction, rather than only a single final reading, allows the student to see how the volume of oxygen produced (and therefore the rate of reaction) changes over the course of the experiment, and to plot a graph of volume against time. This makes it possible to calculate the rate of reaction at different stages (e.g. the initial rate, when it is fastest) and to identify the point at which the reaction has finished (where the graph plateaus), none of which could be determined from a single end-point reading alone.

1 mark: regular readings allow the rate to be tracked/calculated over time, not just the total volume; 1 mark: allows a graph of volume against time to be plotted; 1 mark: allows identification of how/when the rate changes and when the reaction has finished (the plateau), which a single final reading could not show.

Hydrogen peroxide solution is an irritant, and can cause harm if it comes into contact with the skin or, particularly, the eyes. To reduce this risk, the student should wear eye protection (safety goggles) throughout the practical, and should avoid direct skin contact with the solution, washing any spillage off the skin immediately with plenty of water.

1 mark: valid hazard, e.g. hydrogen peroxide is an irritant/can harm skin or eyes; 1 mark: valid, relevant precaution that reduces this specific risk, e.g. wearing eye protection/goggles, avoiding skin contact, washing off spillages immediately.

If there is a gas leak, some of the oxygen produced during the reaction escapes through the leak rather than being pushed into the gas syringe, so the recorded volume is lower than the true volume of oxygen actually produced by the reaction. To check for and minimise this source of error, the student should ensure the bung fits tightly into the flask, seal any joints (for example with a small amount of Vaseline), and check the apparatus is airtight before adding the hydrogen peroxide and yeast suspension.

1 mark: gas escapes through the leak rather than being collected, so the recorded volume is an underestimate of the true volume produced; 1 mark: valid method to check the apparatus is airtight before starting, e.g. ensuring a tight-fitting bung/sealing joints and testing for leaks.

(a) Repeating the experiment several times reduces the effect of random errors and one-off anomalous readings on the final result. Calculating a mean from several repeats gives a value that is more representative of the 'true' result than any single reading, making the data more reliable (more likely to give a similar result if the experiment were repeated again).
(b) An anomalous result should be identified, as it lies well outside the range of the other repeats and is likely due to a measurement or procedural error rather than a genuine change in the reaction. This result should be excluded (not included) when calculating the mean, and the mean should instead be calculated using only the two consistent (concordant) results; ideally, the student should also repeat that trial again to check whether the anomaly recurs.

(a) 1 mark: repeating reduces the effect of random error/one-off anomalies on the result; 1 mark: mean is more representative of the true value, improving reliability. [2]
(b) 1 mark: the anomalous result should be identified and excluded/not included when calculating the mean; 1 mark: mean recalculated using only the remaining (concordant) results, and/or the trial repeated to check. [2]

Reliability refers to how consistent (repeatable) a set of results is: if the experiment were repeated under the same conditions, reliable results would be very similar each time. Validity refers to whether the experiment actually measures/tests what it is intended to: a valid experiment for this investigation is one that correctly investigates only the effect of hydrogen peroxide concentration on rate of reaction, with every other variable (e.g. temperature, enzyme concentration) properly controlled, so that any change in rate observed can be confidently attributed to the change in hydrogen peroxide concentration alone. The validity of the experiment could be improved by ensuring hydrogen peroxide concentration is the only variable that is deliberately changed between trials, and that all other variables (temperature, volume and concentration of yeast suspension, total reaction volume) are kept identical across every trial.

1 mark: reliability = consistency/repeatability of results (similar results if repeated); 1 mark: validity = whether the experiment truly tests/measures what it is intended to, with all other variables controlled; 1 mark: valid suggestion to improve validity, e.g. ensuring hydrogen peroxide concentration is the only variable changed and all others are properly controlled; 1 mark: suggestion correctly justified/explained.

One improvement is to use a gas syringe with finer graduations (for example, one that can be read to the nearest 0.5 cm3 rather than 1 cm3); this reduces the uncertainty in each individual volume reading, improving the precision (and therefore accuracy) of the collected data. A second improvement is to keep the reaction flask in a water bath at a constant, controlled temperature throughout the experiment; this ensures that any change in the rate of oxygen production is due only to the change in hydrogen peroxide concentration being tested, and not to uncontrolled fluctuations in room temperature, which would otherwise introduce error into the results.

1 mark: valid improvement 1, e.g. use a gas syringe with finer graduations; 1 mark: correct explanation of how improvement 1 increases accuracy (more precise individual readings); 1 mark: valid improvement 2, e.g. use a water bath to control temperature; 1 mark: correct explanation of how improvement 2 increases accuracy (removes a confounding variable affecting rate).

(a) A suitable table has a column (or row) for each of the three tests — iodine test, Biuret test, Benedict's test — and a row (or column) for each of the three samples P, Q and R, with a clear heading and space to record a + or − result in every one of the nine sample/test combinations.
(b) Completed table:

Sample Iodine test Biuret test Benedict's test
P + − −
Q − + +
R − − −

(c) The food group not tested for by the iodine, Biuret or Benedict's tests is fat. Because none of these three tests can detect fat, a negative result in all three only shows that starch, protein and reducing sugar are absent; it does not rule out the sample containing fat, so sample R could still be (or contain) a fat, such as an oil.

(a) 1 mark: table has a column for each of the three tests, clearly labelled; 1 mark: table has a row for each of the three samples, clearly labelled; 1 mark: table logically organised with space to record a result in every sample/test combination. [3]
(b) 1 mark each for sample P, Q and R correctly completed to match the given results. [3]
(c) 1 mark: fat correctly named, with a valid explanation that none of the three tests used detects fat, so it cannot be ruled out. [1]

(a) Sorting each of the 20 values into the four class intervals:
10-19 mm: 12, 18, 19, 15, 17 → 5 leaves
20-29 mm: 24, 27, 22, 26, 29, 21, 28, 25 → 8 leaves
30-39 mm: 31, 35, 33, 37, 32 → 5 leaves
40-49 mm: 41, 44 → 2 leaves
Check: 5 + 8 + 5 + 2 = 20, matching the total number of leaves measured, confirming no value has been miscounted or omitted.

Frequency table:
Length (mm) Tally Frequency
10-19 IIII 5
20-29 IIII III 8
30-39 IIII 5
40-49 II 2

(b) The modal class interval is the one with the highest frequency, which is 20-29 mm (8 leaves).
(c) Grouping a large set of continuous raw values into class intervals makes the overall pattern/distribution of the data much easier to see and interpret at a glance (e.g. that most leaves are 20-29 mm long) than a long list of 20 individual values; the trade-off is that some precision is lost, since the exact length of each individual leaf within a class can no longer be read directly from the table.

(a) 1 mark: four class intervals of equal width (10-19, 20-29, 30-39, 40-49) correctly defined with no gaps or overlaps; 1 mark: correct tally/frequency for the 10-19 and 20-29 intervals (5 and 8); 1 mark: correct tally/frequency for the 30-39 and 40-49 intervals (5 and 2); 1 mark: total frequency of 20 shown/matches the number of leaves measured. [4]
(b) 1 mark: 20-29 mm. [1]
(c) 1 mark: grouping makes the overall pattern/distribution of the data easier to interpret; 1 mark: valid related point, e.g. this comes at the cost of losing precision about the exact value of each individual measurement. [2]

(a) A suitable table includes columns headed Concentration (M), Initial mass (g), Final mass (g), Change in mass (g) and Percentage change in mass (%), with one row for each of the five concentrations tested.
(b) Change in mass at 0.6 M = 4.95 − 5.25 = −0.30 g. Percentage change = (−0.30 ÷ 5.25) × 100 = −5.7% (to 1 decimal place). Check by a second route: 5.25 × 0.057 ≈ 0.30, confirming the size of the change; the sign is negative since the potato lost mass.
(c) At 0.4 M sucrose concentration, the potato chip's mass did not change (5.30 g → 5.30 g, a 0% change), meaning there was no net movement of water into or out of the potato tissue by osmosis. This suggests that the water potential of the 0.4 M sucrose solution is approximately equal to the water potential of the potato tissue itself, since osmosis only produces a net change in mass when there is a difference in water potential between the tissue and the surrounding solution.

(a) 1 mark: columns for initial mass, final mass and change in mass, correctly labelled with units (g); 1 mark: column for percentage change in mass, correctly labelled; 1 mark: five rows, one for each concentration, correctly organised. [3]
(b) 1 mark: correct method shown, (4.95-5.25)/5.25 × 100; 1 mark: correct answer, −5.7% (accept −5.71% or equivalent rounding), with correct sign. [2]
(c) 1 mark: no net mass change at 0.4 M means no net osmotic movement of water; 1 mark: therefore the water potential of the potato tissue is approximately equal to that of the 0.4 M solution. [2]

(a) Temperature rise = 63 − 21 = 42°C. Energy transferred = volume × 4.2 × temperature rise = 50 × 4.2 × 42 = 8820 J. Check by a second route: 50 × 42 = 2100, and 2100 × 4.2 = 8820, confirming the answer.
(b) Energy content per gram = total energy ÷ mass of sample = 8820 ÷ 1.2 = 7350 J/g. Check: 7350 × 1.2 = 8820, which matches the value from part (a), confirming the answer.
(c) Much of the heat released by the burning food is lost to the surrounding air and to the apparatus itself, rather than being transferred to the water, so the measured energy content is lower than the food's true energy content (incomplete combustion is also an acceptable reason).

(a) 1 mark: correct method, 50 × 4.2 × 42; 1 mark: correct answer, 8820 J. [2]
(b) 1 mark: correct method, 8820 ÷ 1.2; 1 mark: correct answer, 7350 J/g. Accept ECF from (a). [2]
(c) 1 mark: valid reason, e.g. heat lost to the surroundings/apparatus, or incomplete combustion of the sample. [1]

(a) Increase in rate = 21 − 8 = 13 bubbles/min. Percentage increase = (13 ÷ 8) × 100 = 162.5%. Check by a second route: 8 × 1.625 = 13, and 21 = 8 + 13, confirming the increase and the percentage.
(b) Between 6% (21 bubbles/min) and 8% (22 bubbles/min), the rate increases by only 1 bubble/min, a much smaller increase than seen at lower concentrations. This suggests that the rate of fermentation is levelling off above 6% glucose, so glucose concentration is no longer the main factor limiting the rate; some other factor, such as the amount of enzyme available in the yeast, has become limiting instead.

(a) 1 mark: correct increase calculated, 21 − 8 = 13; 1 mark: correct method for percentage, (13/8) × 100; 1 mark: correct final answer, 162.5%. [3]
(b) 1 mark: rate levels off/increases only slightly between 6% and 8%; 1 mark: suggests glucose is no longer the main limiting factor above 6%, with some other factor now limiting the rate. [2]

(a) Change in mass = 5.52 − 4.80 = 0.72 g. Percentage change = (0.72 ÷ 4.80) × 100 = 15%. Check by a second route: 4.80 × 0.15 = 0.72, confirming the answer.
(b) Distilled water has a very high water potential (the highest possible), higher than the water potential inside the potato cells, which contains dissolved solutes that lower it. Water therefore moved by osmosis from the region of higher water potential (the distilled water) to the region of lower water potential (inside the potato cells), across the partially permeable cell membranes, causing the potato to gain mass.

(a) 1 mark: correct method, (5.52-4.80)/4.80 × 100; 1 mark: correct answer, +15%. [2]
(b) 1 mark: distilled water has a higher water potential than the potato cells; 1 mark: water moves by osmosis from high to low water potential (into the cells) across the partially permeable membrane, increasing mass. [2]

(a) Volume per hour = 250 × 60 = 15000 cm3. Since 1 dm3 = 1000 cm3, this is 15000 ÷ 1000 = 15 dm3/hour. Check by a second route: 15 dm3 = 15000 cm3, and 15000 ÷ 60 = 250 cm3/min, matching the original rate, confirming the answer.
(b) Increase = 750 − 250 = 500 cm3/min. Percentage increase = (500 ÷ 250) × 100 = 200%. Check by a second route: 250 × 3 = 750, so the rate has tripled, which corresponds to an increase of (3−1) × 100% = 200%, confirming the answer.

(a) 1 mark: correct method, 250 × 60 (and ÷1000); 1 mark: correct answer, 15 dm3/hour, with correct unit. [2]
(b) 1 mark: correct method, (750-250)/250 × 100; 1 mark: correct answer, 200%. [2]

A suitable hypothesis is that as temperature increases (within a suitable range), the rate of respiration of the germinating peas will increase, because respiration is an enzyme-controlled process and enzyme activity generally increases with temperature up to an optimum. The independent variable (the one deliberately changed) is temperature. The dependent variable (the one measured) is the rate of respiration, for example the volume of oxygen consumed, or carbon dioxide produced, per unit time. A variable that should be controlled includes the mass or number of peas used, the volume of respirometer solution, or the concentration of the substance (e.g. soda lime) used to absorb carbon dioxide.

1 mark: valid hypothesis linking temperature to rate of respiration, with a reason (e.g. enzyme-controlled process); 1 mark: independent variable = temperature; 1 mark: dependent variable = rate of respiration, with a valid measure; 1 mark: one valid controlled variable, e.g. mass/number of peas, volume of solution, concentration of CO2-absorbing substance.

The control tube accounts for any change in gas volume or pressure within the apparatus that is caused by factors other than respiration, such as small fluctuations in atmospheric pressure or temperature during the experiment. Because the control tube contains glass beads, which do not respire, any movement recorded in the control is known to be due to these external physical factors rather than respiration. Comparing the reading from the tube containing peas with the reading from the control therefore allows the student to calculate a corrected value that reflects the effect of respiration alone.

1 mark: control accounts for changes in gas volume/pressure due to factors other than respiration (e.g. temperature/atmospheric pressure changes); 1 mark: any change seen in the control is not due to respiration, since glass beads do not respire; 1 mark: comparing the two tubes allows a corrected/true value for respiration to be calculated.

Germinating peas are actively metabolising and respiring, producing a measurable rate of gas exchange, whereas dry, non-germinating peas have a very low metabolic rate and would respire far too little to give measurable results. Peas of a similar size should be used in each tube so that a similar (comparable) total mass of respiring tissue is present in every tube; otherwise, a tube containing larger peas might show a greater rate of respiration simply because of the greater mass of tissue present, rather than because of the temperature or condition being tested, making the comparison unfair.

1 mark: germinating peas respire actively (produce a measurable rate), unlike dry/dormant peas; 1 mark: similar-sized peas ensure a comparable mass of respiring tissue is used in each tube; 1 mark: this makes the comparison between conditions/tubes a fair test.

(a) The diameter of the clear zone is measured using a (millimetre) ruler, held against or under the base of the plate.
(b) Sealing the plate with a small piece of adhesive tape, rather than fully sealing all the way round, prevents the lid from being accidentally removed and reduces the risk of contamination by unwanted microorganisms from the air, while still allowing a small amount of air exchange, avoiding a build-up of excessive condensation or anaerobic conditions inside the plate.
(c) Bacteria are unable to grow within the clear zone surrounding the disc, because the antiseptic has diffused outwards from the disc and killed or inhibited bacterial growth in that region. A larger diameter clear zone therefore means the antiseptic diffused effectively and was able to inhibit bacterial growth over a wider area, so a more effective antiseptic produces a larger clear zone diameter.

(a) 1 mark: ruler (millimetre ruler). [1]
(b) 1 mark: tape prevents contamination/lid coming off, while still allowing some air exchange. [1]
(c) 1 mark: bacteria cannot grow within the clear zone around the disc; 1 mark: larger diameter means the antiseptic inhibited/killed bacteria over a greater area, i.e. was more effective. [2]

A variable that should be controlled between plates includes the incubation temperature, the incubation time, the species/strain of bacteria used, the volume or concentration of bacterial culture spread on each plate, or the size of the paper disc. A relevant safety precaution when handling bacterial cultures is to incubate the plates below 37°C, since this reduces the risk of encouraging the growth of pathogens that are adapted to thrive at human body temperature and could pose a risk to humans; plates should also be sealed and disposed of safely (e.g. by autoclaving).

1 mark: valid controlled variable, e.g. incubation temperature/time, species of bacteria, volume of culture, disc size; 1 mark: valid safety precaution relevant to handling bacterial cultures; 1 mark: precaution correctly linked to reducing a specific risk, e.g. incubating below 37°C reduces the risk of culturing pathogens that could infect humans.

(a) The student's conclusion makes a general claim, using the word 'always', but the investigation only tested the effect of the antiseptic on one named species of bacteria, under one set of conditions (one type of agar, one temperature, one incubation time). The results obtained for this single species cannot necessarily be generalised to all species of bacteria, since different species may respond differently to the antiseptic (for example, some species may naturally be more resistant), so the conclusion goes beyond what the data can actually support.
(b) The investigation could be extended by repeating it using several different species of bacteria (and/or testing a wider range of antiseptic concentrations), before drawing a general conclusion about how antiseptic concentration affects bacterial growth in general.

(a) 1 mark: the conclusion is a general claim not justified by results from a single species/investigation; 1 mark: results may not apply to other species, which could respond differently or be resistant. [2]
(b) 1 mark: valid extension, e.g. repeat the investigation using several different bacterial species; 1 mark: developed/justified, e.g. and/or over a wider range of concentrations, to check the pattern holds generally. [2]

An anomalous result is one that does not fit the overall pattern or trend shown by the rest of the data — for example, if one plate at a high antiseptic concentration produced an unexpectedly small clear zone when a much larger zone would be expected based on the other results, this would be anomalous, and is usually caused by an error in procedure or measurement. In contrast, a result at the extreme of a genuine trend, such as the largest clear zone occurring at the highest concentration tested, is still consistent with (continues) the overall pattern shown by the rest of the data, so it is not anomalous, even though it is the largest value recorded.

1 mark: anomalous result = does not fit the overall pattern/trend shown by the rest of the data; 1 mark: valid example/explanation relevant to the antiseptic investigation; 1 mark: a result at the extreme of a genuine trend still follows the pattern, so is not anomalous, even if it is the largest/smallest value.

The value 22 mm is the anomalous result, since it lies far outside the range of the other four readings, which are consistently 14-15 mm. This value should be identified and excluded (not included) when calculating the mean, as it does not fit the pattern of the other repeats and is likely to be the result of an experimental or measurement error. The mean should instead be calculated using only the remaining four concordant results: (14 + 15 + 14 + 15) ÷ 4 = 58 ÷ 4 = 14.5 mm. Check by a second route: 14.5 × 4 = 58, which matches the sum of the four values, confirming the answer.

1 mark: 22 mm correctly identified as the anomalous result; 1 mark: excluded from further analysis (e.g. the calculation of the mean), as it does not fit the pattern of the other repeats/likely due to error; 1 mark: correct recalculated mean using the remaining four values, 14.5 mm, shown with working.

(i) Reliability: using three repeats at each concentration, and calculating a mean, improves the reliability of the results, since it reduces the effect of random error or a single anomalous reading on the final result for each concentration. However, three repeats is still a relatively small sample, so the calculated mean could still be noticeably affected by one unusual reading; reliability could be further improved by using more repeats (for example, five or more) at each concentration.
(ii) Source of error: measuring the diameter of each clear zone by eye with a ruler is subject to human measurement error (e.g. parallax error, or difficulty judging the exact edge of the zone if it is not perfectly sharp or circular). This source of error could be reduced by taking several diameter measurements across each zone, at different angles, and calculating a mean diameter for each individual plate, or by using image-analysis software to measure the zone more precisely and objectively.
(iii) Validity: for the investigation to validly test only the effect of antiseptic concentration, every other variable — the bacterial species and the density of the bacterial inoculum, incubation temperature and time, disc size, and the type and depth of agar used — must be kept identical across all plates. The validity of the overall conclusion could be improved by testing more than one species of bacteria, since a conclusion based on results from a single species cannot be confidently generalised to the effectiveness of the antiseptic against bacteria in general.

Level 0 [0]: No relevant content.
Level 1 [1-2]: Basic comment on one aspect (reliability, error, or validity) without development or clear reference to the specific investigation described.
Level 2 [3-4]: Satisfactory discussion of at least two of the three aspects, with some development and some reference to the specific method described.
Level 3 [5-6]: Comprehensive, well-developed discussion of all three aspects, each correctly explained and specifically linked to the antiseptic investigation described (e.g. correctly explaining why repeats improve reliability but that a small sample size still limits this; a specific, plausible source of measurement error together with a valid way to reduce it; a valid way to improve the generalisability/validity of the conclusion, such as testing further bacterial species); uses accurate specialist terminology (reliability, validity, anomalous, random error) throughout; well organised, with accurate grammar and spelling.
Award marks according to the level that best fits the response as a whole; use the full range within a level to reflect how fully the response meets the level descriptor.