2023 CCEA GCSE Biology 1010 Practice Paper with Answers

The mitochondrion is the site of the aerobic stages of respiration. Its inner membrane is highly folded into cristae, which greatly increases the surface area available for the enzymes and electron carriers involved in ATP production.

1 mark: mitochondrion (mitochondria); 1 mark: folded inner membrane/cristae increases surface area for respiratory enzymes.

Plant cells possess a rigid cellulose cell wall external to the cell membrane, which animal cells lack. Plant cells also commonly contain chloroplasts (for photosynthesis) and a large permanent vacuole filled with cell sap, structures not normally found in animal cells.

1 mark each for any two of: cell wall present in plant cells only; chloroplasts present in plant cells only; large permanent (sap) vacuole present in plant cells only. Max 2 marks.

During photosynthesis, carbon dioxide and water are combined, using light energy absorbed by chlorophyll, to produce glucose and oxygen.

1 mark: correct reactants (carbon dioxide and water) and products (glucose and oxygen) in the correct positions; 1 mark: light energy identified as the energy source.

Chlorophyll is the main photosynthetic pigment. It is embedded within the thylakoid membranes, which are stacked into structures called grana within the chloroplast.

1 mark: chlorophyll; 1 mark: thylakoid membrane/grana.

A few drops of iodine solution are added directly to the food sample. If starch is present, the iodine solution changes colour from its original orange-brown to a dark blue-black.

1 mark: iodine (solution) added directly to the sample; 1 mark: positive result = blue-black colour change (from orange/brown).

A prolonged lack of vitamin C in the diet causes scurvy, since vitamin C is required for the production of healthy connective tissue. Symptoms include bleeding or spongy gums, joint pain and slow healing of wounds.

1 mark: scurvy; 1 mark: any correct symptom, e.g. bleeding/spongy gums, poor wound healing, joint pain.

An enzyme is a protein molecule that acts as a biological catalyst, increasing the rate of a specific chemical reaction without itself being permanently changed or used up. The active site is a pocket on the enzyme's surface whose three-dimensional shape is complementary to that of its substrate, allowing the substrate to bind and form an enzyme-substrate complex.

1 mark: enzyme = a (protein) biological catalyst that speeds up a specific reaction without being used up; 1 mark: active site = region with a shape complementary to the substrate, to which the substrate binds.

At temperatures above the optimum, the additional kinetic energy breaks the bonds holding the enzyme's tertiary structure in shape. This causes the active site to change shape (denaturation), so it is no longer complementary to the substrate and an enzyme-substrate complex can no longer form.

1 mark: activity decreases/enzyme is denatured; 1 mark: (tertiary) structure of the active site changes shape so the substrate can no longer bind/enzyme-substrate complex cannot form.

Gas exchange occurs at the alveoli, tiny air sacs at the end of the bronchioles. Their walls are only one cell thick, which minimises the distance over which oxygen and carbon dioxide must diffuse, and they are surrounded by a dense network of capillaries.

1 mark: alveoli; 1 mark: any correct adaptation, e.g. thin (one-cell-thick) walls, large surface area, good blood supply, moist lining.

A sensory neurone carries nerve impulses from a receptor towards the central nervous system (brain and spinal cord). Where one neurone meets the next, a small gap called a synapse separates them, across which the impulse is transmitted chemically.

1 mark: sensory neurone; 1 mark: synapse.

A producer is an organism, typically a green plant or alga, that manufactures its own organic food molecules from simple inorganic substances, usually via photosynthesis. A primary consumer is an organism that obtains its energy by feeding directly on producers; primary consumers are herbivores.

1 mark: producer = organism that synthesises its own organic/food molecules (by photosynthesis); 1 mark: primary consumer = organism that feeds directly on producers/a herbivore.

Minute ventilation is calculated as tidal volume multiplied by breathing rate: \( 500 \text{ cm}^3 \times 15 = 7500 \text{ cm}^3 \text{ per minute} \), which is equivalent to 7.5 dm³ min⁻¹.

1 mark: correct method shown (tidal volume × breathing rate); 1 mark: correct calculation, 7500; 1 mark: correct unit, cm³ per minute (or dm³ min⁻¹), given.

\( \% \text{ decrease} = \dfrac{48-6}{48} \times 100 = \dfrac{42}{48} \times 100 = 87.5\% \)

1 mark: correct decrease calculated (48 − 6 = 42); 1 mark: correct method (÷ original value × 100); 1 mark: correct final answer, 87.5% (accept 87–88%).

\( \% \text{ efficiency} = \dfrac{1800}{20000} \times 100 = 9\% \)

1 mark: correct fraction set up (secondary ÷ primary consumer energy); 1 mark: correctly multiplied by 100; 1 mark: correct final answer, 9%.

\( (12 \times 17) + (5 \times 17) + (3 \times 37) = 204 + 85 + 111 = 400 \text{ kJ} \)

1 mark: each macronutrient mass correctly multiplied by its energy value (204, 85, 111 — allow ECF); 1 mark: values correctly summed; 1 mark: correct final total, 400 kJ.

\( \text{Magnification} = \dfrac{\text{image size}}{\text{actual size}} = \dfrac{45 \text{ mm}}{0.03 \text{ mm}} = 1500 \)

1 mark: correct formula stated/used (image size ÷ actual size); 1 mark: correct substitution with consistent units; 1 mark: correct final answer, ×1500.

\( \% \text{ increase} = \dfrac{4.2-3.6}{3.6} \times 100 = \dfrac{0.6}{3.6} \times 100 = 16.7\% \)

1 mark: correct increase calculated (0.6 dm³); 1 mark: correct method (÷ original value × 100); 1 mark: correct final answer, 16.7% (accept 16.6–16.7%).

The lungs contain millions of alveoli, giving a very large total surface area for gas exchange. Both the alveolar walls and the surrounding capillary walls are only one cell thick, minimising the diffusion distance for oxygen and carbon dioxide. A dense network of capillaries surrounds each alveolus, constantly bringing deoxygenated blood and removing oxygenated blood, which maintains a steep concentration gradient for diffusion. The alveolar surface is also kept moist, allowing gases to dissolve before diffusing across the membranes.

1 mark per correct point, up to 4: large surface area (many alveoli in the lungs); alveolar and capillary walls are one cell thick, minimising diffusion distance; dense capillary network maintains a concentration gradient by constantly removing oxygen/delivering carbon dioxide; moist alveolar lining allows gases to dissolve before diffusing.

During inhalation, the diaphragm contracts and flattens, while the external intercostal muscles also contract, pulling the rib cage upwards and outwards. Together these movements increase the volume of the thoracic cavity and lungs. As the same mass of air now occupies a larger volume, the air pressure inside the lungs falls below atmospheric pressure, so air moves into the lungs down this pressure gradient.

1 mark: diaphragm contracts and flattens; 1 mark: external intercostal muscles contract, ribs move up and out; 1 mark: volume of thorax/lungs increases; 1 mark: pressure inside the lungs falls below atmospheric pressure, so air moves in along the pressure gradient.

A receptor in the skin detects the painful stimulus and generates a nerve impulse. This impulse travels along a sensory neurone to the spinal cord. Within the spinal cord, a relay neurone forms synapses connecting the sensory neurone directly to a motor neurone, without first passing through the brain. The motor neurone then carries the impulse to an effector, such as a muscle, causing it to contract and pull the hand away. Because the pathway does not require conscious processing by the brain, the response is very rapid and involuntary.

1 mark: receptor detects the (painful) stimulus and generates a nerve impulse; 1 mark: impulse carried by a sensory neurone to the spinal cord/CNS; 1 mark: relay neurone connects sensory to motor neurone (via synapses) in the spinal cord; 1 mark: motor neurone carries the impulse to an effector (muscle), causing a response — bypassing the brain, making the response rapid/involuntary.

After a meal, blood glucose concentration rises above normal. This rise is detected by cells in the pancreas, which respond by secreting insulin into the blood. Insulin causes liver and muscle cells to take up more glucose from the blood, and stimulates the conversion of glucose into glycogen (glycogenesis) for storage in these cells. As a result, blood glucose concentration falls back towards its normal level.

1 mark: rise in blood glucose detected by the pancreas, which secretes insulin into the blood; 1 mark: insulin causes liver and muscle cells to take up more glucose from the blood; 1 mark: insulin stimulates the conversion of glucose to glycogen (glycogenesis) in the liver/muscle for storage; 1 mark: as a result, blood glucose concentration falls/returns to normal.

At each trophic level, a large proportion of the energy taken in by an organism is lost rather than passed on to the next level. Energy is lost as heat generated during respiration, and is used for movement and other life processes. Not all of an organism is eaten by the next consumer, and much of what is eaten is not digested and is lost in faeces (egestion) or lost through excretion. As a result, typically only around 10% of the energy present at one trophic level is transferred to the next. Because so little energy remains after several such transfers, there is usually insufficient energy to support a viable population beyond four or five trophic levels.

1 mark: energy lost as heat through respiration; 1 mark: energy lost through movement/other life processes; 1 mark: energy lost in egestion (undigested material in faeces) and excretion, and not all parts of an organism are eaten; 1 mark: because only around 10% of energy is transferred to the next level, insufficient energy remains to support further trophic levels beyond four or five.

According to the lock-and-key model, an enzyme's active site has a specific three-dimensional shape, much like a lock. Only a substrate whose shape is complementary to this active site, like a matching key, can bind to it to form an enzyme-substrate complex and be catalysed. A substrate with a different shape does not fit the active site, so no complex forms and no reaction is catalysed, which is why each enzyme is specific to one substrate (or a small group of very similar substrates).

1 mark: the active site has a specific three-dimensional shape that is complementary to (fits) only one substrate; 1 mark: only a substrate of matching shape can bind to form an enzyme-substrate complex; 1 mark: substrates of a different shape do not fit the active site, so the reaction is not catalysed (enzyme specificity).

The root hair cell has a long, thin extension of its cell membrane (the root hair) projecting into the soil, which greatly increases the surface area in contact with the soil water for absorption. The cell contains a large number of mitochondria, which provide the ATP needed for the active transport of mineral ions into the cell against their concentration gradient. The cell also has a large permanent vacuole, which helps maintain a low (more negative) water potential inside the cell relative to the soil water, so that water continues to move in by osmosis.

1 mark: long, thin root hair extension increases surface area for absorption; 1 mark: many mitochondria provide ATP for active transport of mineral ions against a concentration gradient; 1 mark: large (permanent) vacuole maintains a low water potential, aiding water uptake by osmosis; 1 mark: thin cell wall/membrane minimises the diffusion distance for water and ions entering the cell.

A strong answer explains that photosynthesis removes carbon dioxide from the atmosphere, fixing it into carbohydrates within producers; this carbon is then passed along food chains as organisms feed. Respiration by producers, consumers and decomposers releases carbon dioxide back into the atmosphere. Decomposers play a key role by breaking down dead organisms and waste material, respiring as they do so and returning the carbon they contain to the atmosphere or soil, which recycles nutrients back into the ecosystem. Human activities disrupt this natural balance: the combustion of fossil fuels releases carbon that has been locked away for millions of years, adding extra carbon dioxide to the atmosphere, while deforestation reduces the number of photosynthesising producers available to remove carbon dioxide. Together, these activities have raised atmospheric carbon dioxide concentration, enhancing the greenhouse effect.

Band A (5–6 marks): specialist terms (e.g. photosynthesis, respiration, decomposition, combustion) used accurately throughout; at least 5 indicative points addressed with a clear, logical structure; excellent spelling, punctuation and grammar. Band B (3–4 marks): some specialist terms used; at least 3 indicative points addressed; generally coherent structure; satisfactory SPG. Band C (1–2 marks): limited specialist vocabulary; at least 1 indicative point addressed; basic structure/SPG. Band D (0 marks): response not worthy of credit. Indicative content: photosynthesis removes CO2 from the atmosphere into producers; carbon passed along food chains by feeding; respiration by producers/consumers/decomposers releases CO2; decomposers break down dead organisms/waste and respire, returning carbon to the atmosphere/soil; combustion of fossil fuels releases long-stored carbon; deforestation reduces CO2 removal by producers; human activity has increased atmospheric CO2, enhancing the greenhouse effect.

White blood cells provide immunity in two main ways. Phagocytes engulf and digest pathogens directly by phagocytosis, while lymphocytes recognise antigens and produce antibodies against them.

1 mark: phagocyte; 1 mark: lymphocyte.

Active immunity results from the body's own immune system producing antibodies and memory cells in response to exposure to an antigen. It can be acquired naturally, for example, by being infected with and recovering from a disease.

1 mark: active immunity = body produces its own antibodies in response to exposure to an antigen; 1 mark: correct natural example, e.g. following infection/recovering from a disease.

Tuberculosis is caused by a bacterium (Mycobacterium tuberculosis).

1 mark: bacterium/bacteria (accept named genus, e.g. Mycobacterium).

Intact skin forms a continuous physical barrier that most pathogens cannot penetrate. In addition, glands in the skin secrete sebum, which is acidic and antimicrobial, inhibiting the growth of many microorganisms on the skin's surface.

1 mark each for any two of: intact skin acts as a physical barrier; sebum is antimicrobial/lowers pH, inhibiting growth of microorganisms; skin flora out-competes pathogens. Max 2 marks.

The two upper chambers of the heart are the left atrium and right atrium. Blood is returned to the atria by veins: the vena cava returns deoxygenated blood to the right atrium, and the pulmonary vein returns oxygenated blood to the left atrium.

1 mark: atria (left and right atrium); 1 mark: veins/vena cava/pulmonary vein.

The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs. Arteries in general have thick, muscular walls containing elastic tissue, which allows them to withstand and maintain the high pressure of blood pumped from the heart.

1 mark: pulmonary artery; 1 mark: thick, muscular (and elastic) walls able to withstand high pressure.

Oxygen is carried by red blood cells (erythrocytes), which contain the pigment haemoglobin that binds reversibly to oxygen.

1 mark: red blood cells (erythrocytes)/haemoglobin.

Osmosis is the diffusion of water molecules specifically, moving from an area of higher water potential to an area of lower water potential, across a partially permeable membrane.

1 mark: net movement of water (molecules) from high to low water potential; 1 mark: through a partially permeable membrane.

Because the surrounding solution has a higher water potential than the cell contents, water moves into the cell by osmosis. As water enters, the cell contents push against the cell wall, and the cell becomes turgid.

1 mark: water moves into the cell by osmosis; 1 mark: cell becomes turgid.

Fertilisation involves the fusion of a male gamete (sperm) with a female gamete (egg/ovum), and normally takes place in the oviduct (fallopian tube).

1 mark: sperm and egg/ovum; 1 mark: oviduct/fallopian tube.

A surge in luteinising hormone (LH), released from the pituitary gland, triggers the release of an egg from the ovary (ovulation).

1 mark: luteinising hormone (LH); 1 mark: pituitary gland.

Barrier methods of contraception physically prevent sperm from reaching an egg. Examples include the male or female condom, and the diaphragm or cap, which is usually used together with spermicide.

1 mark each for any two of: condom (male or female); diaphragm/cap; spermicide used with a barrier. Max 2 marks.

The genotype of an organism is the combination of alleles it carries for a particular gene or genes. The phenotype is the observable, physical expression of those alleles, which results from the interaction of the genotype with the environment.

1 mark: genotype = the genetic make-up/alleles present; 1 mark: phenotype = the observable/physical characteristics (resulting from genotype and environment).

A normal human body cell is diploid and contains 46 chromosomes, arranged as 23 homologous pairs. A gamete is haploid and contains only 23 chromosomes, one from each pair.

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

\( \% \text{ change} = \dfrac{4.68-5.20}{5.20} \times 100 = \dfrac{-0.52}{5.20} \times 100 = -10\% \). Since the mass decreased, this represents a net loss of water from the potato tissue by osmosis.

1 mark: correct change in mass calculated (−0.52 g); 1 mark: correct method (÷ initial mass × 100); 1 mark: correct answer, −10%, correctly identified as a net loss of water (allow ECF).

The percentage change in mass crosses zero between 0.2 mol dm⁻³ (+4%) and 0.4 mol dm⁻³ (−3%). Using linear interpolation: \( 0.2 + \dfrac{4}{4+3} \times 0.2 \approx 0.31 \text{ mol dm}^{-3} \), so the isotonic point is approximately 0.3–0.35 mol dm⁻³.

1 mark: recognises the zero point lies between the 0.2 and 0.4 mol dm⁻³ readings; 1 mark: valid interpolation method shown (or a graph plotted and value read off); 1 mark: final estimate in the range 0.3–0.35 mol dm⁻³.

\( \text{Cardiac output} = \text{heart rate} \times \text{stroke volume} = 72 \times 70 = 5040 \text{ cm}^3 \text{ min}^{-1} = 5.04 \text{ dm}^3 \text{ min}^{-1} \)

1 mark: correct formula used (heart rate × stroke volume); 1 mark: correct calculation, 5040 cm³ min⁻¹; 1 mark: correctly converted to dm³ min⁻¹, 5.04.

\( \% \text{ decrease} = \dfrac{132-118}{132} \times 100 = \dfrac{14}{132} \times 100 = 10.6\% \)

1 mark: correct decrease identified (14 mmHg, using systolic values only); 1 mark: correct method (÷ original × 100); 1 mark: correct final answer, 10.6% (accept 10.5–10.6%).

\( \% \text{ difference} = \dfrac{26-4}{4} \times 100 = \dfrac{22}{4} \times 100 = 550\% \)

1 mark: correct difference calculated (22 mm); 1 mark: correct method (÷ smaller value × 100); 1 mark: correct final answer, 550%.

\( \% \text{ success} = \dfrac{79}{340} \times 100 = 23.2\% \)

1 mark: correct fraction set up (successful cycles ÷ total cycles); 1 mark: correctly multiplied by 100; 1 mark: correct final answer to 1 d.p., 23.2%.

A Punnett square cross of Tt × Tt gives offspring genotypes TT : Tt : Tt : tt, a ratio of 1 : 2 : 1 for TT : Tt : tt. Only the tt genotype (1 in 4) is dwarf, so the probability is \( \dfrac{1}{4} \times 100 = 25\% \).

1 mark: correct genotype ratio identified from a Punnett square/cross (1 TT : 2 Tt : 1 tt); 1 mark: correct final answer, 25%.

A vaccine contains a dead, inactivated or weakened form of a pathogen, or its antigens, which is introduced into the body. This stimulates a primary immune response, in which lymphocytes produce specific antibodies, and some lymphocytes remain in the body as memory cells. If the person is later exposed to the live pathogen, these memory cells enable a much faster and stronger secondary immune response, destroying the pathogen before it can cause symptoms of disease.

1 mark: vaccine contains a dead/inactivated/weakened form of the pathogen or its antigens, which is introduced into the body; 1 mark: this stimulates a primary immune response, producing antibodies and memory cells; 1 mark: on future exposure to the live pathogen, memory cells enable a faster, stronger secondary immune response, destroying the pathogen before symptoms occur.

Antibiotics work by targeting structures or processes that are specific to bacterial cells, such as bacterial cell wall synthesis or bacterial ribosomes, which are absent or different in human cells. Viruses have no such independent cell structures or metabolism of their own; instead, they replicate by using the host cell's own machinery. Because there is no bacteria-specific target present in a virus, antibiotics have no effect on viral infections.

1 mark: antibiotics target structures/processes specific to bacterial cells (e.g. bacterial cell wall synthesis, bacterial ribosomes) that are absent in human cells; 1 mark: viruses have no such independent cell structures/metabolism of their own; 1 mark: viruses replicate using the host cell's own machinery, so there is no bacteria-specific target for antibiotics to act on.

Random mutation may occur in the bacterial population, giving a small number of bacteria a resistance allele that protects them from a particular antibiotic. When that antibiotic is used, non-resistant bacteria are killed, but resistant bacteria survive, since the antibiotic acts as a selection pressure. These surviving resistant bacteria then reproduce, passing on the resistance allele to their offspring, so the proportion of resistant bacteria in the population increases over successive generations.

1 mark: random mutation may occur, giving some bacteria a resistance allele; 1 mark: when the antibiotic is used, non-resistant bacteria are killed but resistant bacteria survive (selection pressure); 1 mark: surviving resistant bacteria reproduce and pass on the resistance allele, so the proportion of resistant bacteria in the population increases over successive generations.

The left ventricle pumps oxygenated blood around the whole body in the systemic circulation, a much greater distance and against greater resistance than the right ventricle's pulmonary circuit, which only supplies the nearby lungs. This means the left ventricle must generate a much higher pressure to push blood all the way around the body. A thicker, more muscular wall allows the left ventricle to contract with the greater force needed to produce this higher pressure.

1 mark: left ventricle pumps blood around the whole body (systemic circulation), a much greater distance than the right ventricle's lungs-only (pulmonary) circuit; 1 mark: greater pressure is needed to overcome the higher resistance of the systemic circulation; 1 mark: a thicker, more muscular wall allows the left ventricle to contract with greater force to generate this higher pressure.

Capillary walls are only one cell thick, which minimises the diffusion distance for substances moving between the blood and surrounding tissue cells. The extensive, dense network of capillaries throughout the body provides a very large total surface area for exchange. Capillaries also have a narrow lumen, which slows the rate of blood flow through them, allowing more time for exchange of substances such as oxygen, glucose and carbon dioxide to occur.

1 mark: walls are one cell thick, minimising the diffusion distance for exchanged substances; 1 mark: an extensive/dense network of capillaries provides a large surface area for exchange; 1 mark: narrow lumen slows the rate of blood flow, allowing more time for exchange (or: small pores in the walls allow plasma to leak out and form tissue fluid).

Regular aerobic exercise helps a person maintain a healthy body mass, which reduces the risk of associated high blood pressure and reduces the strain placed on the heart. Exercise also improves the balance of blood cholesterol, raising HDL ('good') cholesterol and lowering LDL ('bad') cholesterol, which reduces the build-up of fatty deposits (atheroma) in the walls of the coronary arteries and so lowers the risk of a blockage restricting blood flow to the heart muscle.

1 mark: point 1 identified, e.g. exercise helps maintain a healthy body mass/lowers blood pressure; 1 mark: an explained consequence of point 1, e.g. reducing strain on the heart; 1 mark: point 2 identified with a linked consequence, e.g. exercise improves the HDL:LDL cholesterol ratio, reducing atheroma build-up in the coronary arteries.

Distilled water has a higher (less negative) water potential than the contents of either cell, so water moves into both cells by osmosis. The red blood cell has no cell wall, so as water continues to enter, the cell swells until it eventually bursts (lyses). The plant cell, however, is surrounded by a rigid cellulose cell wall. As water enters and the cell contents press against this wall, the wall exerts an inward pressure that resists further net entry of water once the cell becomes turgid, preventing it from bursting.

1 mark: distilled water has a higher (less negative) water potential than the cell contents, so water enters both cells by osmosis; 1 mark: the red blood cell has no cell wall, so it continues to swell and eventually bursts (lyses); 1 mark: the plant cell's rigid cellulose cell wall exerts an inward (wall) pressure once turgid, preventing further net water entry and bursting.

Water enters root hair cells from the soil by osmosis, because the soil water has a higher water potential than the contents of the root hair cell. The water then passes across the root and enters the xylem vessels, up which it is pulled by transpiration pull, generated by cohesion between water molecules as water evaporates from the leaves above. On reaching the leaf, water evaporates from the surface of the mesophyll cells into the air spaces within the leaf, and then diffuses out of the leaf into the atmosphere through the stomata, a process called transpiration.

1 mark: water enters root hair cells from the soil by osmosis (soil has a higher water potential); 1 mark: water passes across the root and up the xylem vessels, pulled by transpiration pull/cohesion between water molecules; 1 mark: water evaporates from the surface of mesophyll cells in the leaf into the air spaces, then diffuses out through the stomata into the atmosphere (transpiration).

Hormonal contraceptives such as the combined pill contain synthetic oestrogen and progesterone. These synthetic hormones inhibit the release of FSH and LH from the pituitary gland. Without a sufficient rise in FSH and LH, no egg matures within the ovary and ovulation is prevented, so there is no egg available for fertilisation.

1 mark: contains synthetic hormones (oestrogen and/or progesterone); 1 mark: these inhibit the release of FSH and LH from the pituitary gland; 1 mark: without sufficient FSH/LH, no egg matures and ovulation is prevented (accept: also thickens cervical mucus to block sperm, or thins the uterus lining to prevent implantation).

Oestrogen, secreted by the developing follicle in the first half of the cycle, stimulates the repair and thickening (proliferation) of the uterus lining following menstruation. After ovulation, the corpus luteum secretes progesterone, which maintains and further thickens the uterus lining, keeping it well supplied with blood ready to receive a fertilised egg. If fertilisation does not occur, the corpus luteum breaks down, the level of progesterone falls, and the uterus lining breaks down, resulting in menstruation.

1 mark: oestrogen (from the developing follicle) stimulates the repair/thickening of the uterus lining after menstruation; 1 mark: progesterone (from the corpus luteum after ovulation) maintains and further thickens the uterus lining, keeping it ready for implantation; 1 mark: if fertilisation does not occur, progesterone level falls (as the corpus luteum breaks down) and the lining breaks down/menstruation occurs.

A strong answer explains that vaccination introduces a dead, inactivated or weakened pathogen, or its antigens, stimulating a primary immune response that produces antibodies and memory cells without causing disease, so the vaccinated individual gains active immunity and can mount a rapid secondary response on future exposure. It should then explain herd immunity: when a sufficiently high proportion of a population is vaccinated, the spread of a pathogen is limited because few susceptible hosts remain for it to infect, which indirectly protects unvaccinated or vulnerable individuals (e.g. those who cannot be vaccinated for medical reasons) by reducing their chance of exposure. Finally, it should discuss factors limiting success, such as some individuals being unable to receive the vaccine (e.g. due to allergies or a weakened immune system), vaccine uptake falling below the threshold needed for herd immunity (vaccine hesitancy), and the pathogen mutating (antigenic variation) so existing vaccines become less effective.

Band A (5–6 marks): specialist terms (e.g. antigen, antibody, memory cell, herd immunity) used accurately throughout; at least 5 indicative points addressed with a clear, logical structure; excellent spelling, punctuation and grammar. Band B (3–4 marks): some specialist terms used; at least 3 indicative points addressed; generally coherent structure; satisfactory SPG. Band C (1–2 marks): limited specialist vocabulary; at least 1 indicative point addressed; basic structure/SPG. Band D (0 marks): response not worthy of credit. Indicative content: vaccine contains a dead/inactivated/weakened pathogen or antigens; stimulates a primary immune response producing antibodies and memory cells; individual gains active immunity/rapid secondary response on re-exposure; herd immunity occurs when enough of the population is vaccinated, limiting spread and protecting the unvaccinated; not everyone can be vaccinated (e.g. immunocompromised individuals, allergies); vaccine hesitancy/uptake below the threshold needed reduces herd immunity; pathogens may mutate (antigenic variation), reducing vaccine effectiveness.

Vessel W, entering the right atrium and carrying deoxygenated blood from the body, is the vena cava. Vessel X, leaving the left ventricle and carrying oxygenated blood to the body, is the aorta.

1 mark: W = vena cava; 1 mark: X = aorta.

Structure Y, the organ in which a fetus develops, is the uterus (womb). Structure Z, the tube along which the egg travels and in which fertilisation normally takes place, is the oviduct (fallopian tube).

1 mark: Y = uterus/womb; 1 mark: Z = oviduct/fallopian tube.

Tissue M, positioned centrally and carrying water and mineral ions, is the xylem. Tissue N, positioned towards the outside of the vascular bundle and transporting dissolved sugars (translocation), is the phloem.

1 mark: M = xylem; 1 mark: N = phloem.

A well-constructed results table has clear column headings, with the unit given once in each heading rather than repeated in every cell, and is ruled into rows and columns. Here the independent variable (sucrose concentration) forms the first column, followed by initial mass and final mass, with one row per concentration tested (five rows), and values recorded to a consistent number of decimal places matching the precision of the balance used.

1 mark: correct column headings with units given in the heading only (sucrose concentration, initial mass, final mass); 1 mark: table correctly ruled/organised with one row per concentration tested (five rows) and values recorded to a consistent number of decimal places.

The table should have two ruled columns, headed 'Time (s)' and 'Volume of oxygen collected (cm³)', with the unit stated once in each heading. Since readings were taken every 20 seconds up to 120 seconds, the table needs seven rows (for 0, 20, 40, 60, 80, 100 and 120 seconds).

1 mark: correct column headings with units given in the heading only (time in seconds, volume of oxygen in cm³); 1 mark: table correctly ruled with one row for each 20-second reading from 0 to 120 seconds (7 rows).

Time is the independent variable and is plotted on the x-axis; volume of oxygen, the dependent variable, is plotted on the y-axis. Plotting the data gives a curve that rises steeply at first, since substrate concentration and enzyme activity are both high, then becomes progressively less steep as the hydrogen peroxide is used up and its concentration falls, until the curve becomes a horizontal plateau once all the substrate has been decomposed, at which point no further oxygen is produced. The final, constant volume read from the table is 32 cm³.

1 mark: time correctly identified for the x-axis and volume of O₂ for the y-axis; 1 mark: shape correctly described as a steep initial rise that gradually levels off to a plateau/curve of decreasing gradient; 1 mark: shape correctly explained in terms of decreasing substrate (hydrogen peroxide) concentration slowing the rate of reaction as it proceeds, until all substrate is used up; 1 mark: correct final volume read from the table, 32 cm³.

A correct biological drawing shows a single cell, large enough to see clearly, with the cell wall drawn as a smooth continuous outline. Inside this, the cell membrane and cytoplasm are drawn as a smaller, shrunken shape that has pulled away from the wall in places, with a visible gap between the two. Ruled label lines, which do not cross, point to and correctly name the cell wall, the cell membrane, and the plasmolysis gap between them.

1 mark: clear, large single-cell outline with a smooth continuous line for the cell wall, drawn with a sharp pencil and no shading/hatching; 1 mark: cell membrane and cytoplasm/contents shown pulled away from the cell wall, correctly proportioned as a shrunken shape within the wall; 1 mark: correct label lines (ruled, not crossing) to the cell wall, cell membrane and the plasmolysis gap, with all three structures correctly named.

A correct diagram shows a conical flask, drawn large enough to be clear, containing the hydrogen peroxide solution and a source of catalase (such as a piece of potato or liver). The flask is sealed with a bung through which a delivery tube passes, connecting the flask to a gas syringe, which is drawn with a scale and plunger to show it collects and measures the volume of oxygen gas produced as the reaction proceeds.

1 mark: conical flask correctly drawn and labelled, containing the hydrogen peroxide/catalase reaction mixture; 1 mark: bung and delivery tube correctly drawn connecting the flask to the gas syringe, with an airtight seal implied; 1 mark: gas syringe correctly drawn and labelled as collecting/measuring the volume of oxygen gas produced.

The percentage change in mass crosses zero between the 0.2 mol dm⁻³ reading (+2.0%) and the 0.4 mol dm⁻³ reading (−4.5%). By linear interpolation: \( 0.2 + \dfrac{2.0}{2.0+4.5} \times 0.2 \approx 0.26 \text{ mol dm}^{-3} \). At this concentration, the sucrose solution and the potato tissue have the same water potential, so there is no net movement of water in either direction (the solution is isotonic to the tissue).

1 mark: recognises the zero crossing point lies between 0.2 and 0.4 mol dm⁻³; 1 mark: valid interpolation calculation shown; 1 mark: final estimate within the range 0.25–0.30 mol dm⁻³ (or correctly read from a plotted graph); 1 mark: (b) correctly states that the external solution and the potato cells have the same/equal water potential at this concentration (isotonic).

\( \text{Rate} = \dfrac{\text{change in volume}}{\text{time}} = \dfrac{12-0}{20} = 0.6 \text{ cm}^3 \text{ s}^{-1} \)

1 mark: correct change in volume identified (12 cm³) over the correct time interval (20 s); 1 mark: correct formula used (change in volume ÷ time); 1 mark: calculation correctly carried out; 1 mark: correct final answer with unit, 0.6 cm³ per second.

Original mean rate \( = \dfrac{32}{120} = 0.27 \text{ cm}^3 \text{ s}^{-1} \). New mean rate \( = \dfrac{32}{60} = 0.53 \text{ cm}^3 \text{ s}^{-1} \). \( \dfrac{0.53}{0.27} \approx 2 \), so the new rate is approximately twice the original rate.

1 mark: original mean rate correctly calculated (32 ÷ 120 ≈ 0.27 cm³ s⁻¹, allow ECF); 1 mark: new mean rate correctly calculated (32 ÷ 60 ≈ 0.53 cm³ s⁻¹); 1 mark: correctly identifies the new rate as twice (2×) the original rate.

Surface moisture clinging to the outside of the cylinder is not water that has actually been taken up by, or lost from, the tissue by osmosis. If it were not removed, this surface water would be included in the recorded mass, making the calculated change in mass inaccurate.

1 mark: to remove excess surface water/solution from the cylinder; 1 mark: this would otherwise be included in the mass recorded, making the mass (and hence the % change) inaccurate.

If the cylinders differed in diameter or length, they would have different surface area to volume ratios and different starting masses, which would themselves affect the rate and extent of water movement independently of the sucrose concentration. Keeping the size of every cylinder the same ensures that sucrose concentration is the only variable that differs between cylinders, making the investigation a fair test.

1 mark: ensures the surface area to volume ratio/size of each cylinder is the same/comparable, so that mass change is not affected by cylinder size; 1 mark: this keeps the investigation a fair test, with sucrose concentration as the only variable affecting the results.

Enzyme activity, and therefore the rate of oxygen production, is affected by temperature. If the flask were left at room temperature, small fluctuations in temperature during the investigation could speed up or slow down the reaction independently of the variable actually being tested. Using a water bath keeps the temperature constant and controlled, ensuring that any difference in the rate of oxygen production is due only to the intended independent variable, making the investigation a fair test.

1 mark: keeps temperature constant, preventing changes in temperature affecting enzyme activity/rate of reaction; 1 mark: ensures a fair test, so any difference in the rate of oxygen production is due only to the independent variable being investigated.

The diameter of the clear zone around an antibiotic disc, where bacterial growth has been inhibited, is measured directly with a ruler or callipers and recorded in millimetres.

1 mark: ruler/callipers; 1 mark: millimetres (mm).

When working with microorganisms, aseptic technique (such as working near a Bunsen burner flame and sterilising equipment) should be used to reduce the risk of contamination. Once inoculated, Petri dishes should be sealed with tape and never opened, and should be incubated at a maximum of 25 °C rather than at body temperature, since this reduces the growth of any pathogens that could otherwise infect humans.

1 mark each for any two of: use aseptic technique (e.g. work near a Bunsen flame, sterilise equipment); seal the Petri dish with tape and do not open it after inoculation; incubate at a maximum of 25 °C rather than at body temperature, to reduce growth of pathogens that could infect humans. Max 2 marks.

Breathing rate is typically measured by counting the number of breaths taken over a fixed time period (e.g. one minute), timed with a stopwatch. To make comparisons between people fair, variables such as the level of exercise or activity undertaken immediately before measurement should be controlled.

1 mark: stopwatch (used to count breaths over a timed period); 1 mark: any valid controlled variable, e.g. level of exercise/activity immediately beforehand, time spent resting before measurement, age/fitness matched where relevant.

A spirometer is used to measure lung volumes. The volume of air breathed in during one normal, resting breath is called the tidal volume.

1 mark: spirometer; 1 mark: tidal volume.

Preparing a root tip squash requires a slide and cover slip to mount the tissue, a scalpel or mounted needle to dissect and spread the tissue, a stain such as orcein or toluidine blue to make the chromosomes visible, and a light microscope to view the finished slide.

1 mark each for any two of: microscope slide and cover slip; light microscope; mounted needle/scalpel for dissection; stain (e.g. orcein or toluidine blue) to make chromosomes visible. Max 2 marks.

An electron microscope has a much higher resolution than a light microscope, allowing it to reveal fine internal detail such as the folded internal membranes (cristae) of mitochondria or individual ribosomes, which are far too small to be resolved with a light microscope.

1 mark: electron microscope; 1 mark: any correct sub-cellular structure only visible with an electron microscope, e.g. internal structure of mitochondria/cristae, ribosomes, detailed membrane structure.

Gel electrophoresis separates DNA fragments by size, using an electric current to pull negatively charged DNA fragments through a gel, with smaller fragments travelling further. This is carried out in an electrophoresis tank.

1 mark: gel electrophoresis; 1 mark: electrophoresis tank/apparatus.

A heart rate monitor or pulse oximeter can be used to measure pulse rate electronically. Manually, a pulse can be felt at points where an artery passes close to the skin, such as the wrist (radial artery) or the neck (carotid artery).

1 mark: heart rate monitor/pulse oximeter; 1 mark: correct pulse point, e.g. wrist/radial artery or neck/carotid artery.

Since light intensity is inversely proportional to distance², \( \text{intensity} \propto \dfrac{1}{d^2} \). If intensity at 10 cm = 1 unit, then at 20 cm: \( \dfrac{1/20^2}{1/10^2} = \dfrac{10^2}{20^2} = \dfrac{100}{400} = 0.25 \) units.

1 mark: correctly applies the inverse square relationship (intensity ∝ 1/distance²); 1 mark: correct substitution/method shown; 1 mark: correct final answer, 0.25 units (¼ of the intensity at 10 cm).

\( \% \text{ decrease} = \dfrac{45-9}{45} \times 100 = \dfrac{36}{45} \times 100 = 80\% \)

1 mark: correct decrease calculated (36 bubbles); 1 mark: correct method (÷ original value × 100); 1 mark: correct final answer, 80%.

\( \% \text{ increase} = \dfrac{34-16}{16} \times 100 = \dfrac{18}{16} \times 100 = 112.5\% \)

1 mark: correct increase calculated (18 breaths per minute); 1 mark: correct method (÷ original value × 100); 1 mark: correct final answer, 112.5%.

\( \dfrac{240}{90} \approx 2.7 \)

1 mark: correct values identified for comparison (240 s and 90 s); 1 mark: correct method (240 ÷ 90); 1 mark: correct final answer, approximately 2.7 (times faster).

(a) The 640 ms reading is much higher than the other four, consistent readings, and is likely due to a momentary lapse in the student's concentration or a delay in releasing/catching the ruler. (b) Excluding this anomaly: \( \dfrac{210+195+200+205}{4} = \dfrac{810}{4} = 202.5 \text{ ms} \).

1 mark: (a) 640 ms correctly identified as anomalous, with a plausible reason (e.g. momentary lapse in concentration, delayed release of the ruler); 1 mark: (b) correct sum of the four valid results (810 ms) divided by 4; 1 mark: correct final mean, 202.5 ms.

\( \% \text{ reduction} = \dfrac{86-5}{86} \times 100 = \dfrac{81}{86} \times 100 = 94.2\% \)

1 mark: correct reduction calculated (81 colonies); 1 mark: correct method (÷ original value × 100); 1 mark: correct final answer to 1 d.p., 94.2%.

The independent variable, which the student deliberately changes, is temperature. The dependent variable, which is measured as a result, is the volume (or rate) of carbon dioxide gas produced. To make the investigation a fair test, other variables that could affect gas production, such as the concentration and volume of glucose solution used or the mass/concentration of yeast, must be kept constant across all temperatures tested.

1 mark: independent variable = temperature; 1 mark: dependent variable = volume/rate of carbon dioxide produced; 1 mark: any valid controlled variable, e.g. concentration/volume of glucose solution, mass/concentration of yeast used, reaction time.

To ensure the investigation gives valid results, the student should include a control (e.g. distilled water in place of the antiseptic) so that any effect of the mouthwash itself can be identified by comparison. Aseptic technique should be used throughout to reduce contamination and the health and safety risk of working with bacteria. Finally, the student should test a suitable, evenly spaced range of antiseptic concentrations, with each concentration repeated (e.g. three times) so that a mean can be calculated, improving the reliability of the results.

1 mark each for any three of: include a control with no antiseptic (e.g. distilled water) for comparison; use aseptic technique to reduce contamination and health/safety risk; use a suitable, evenly spaced range of concentrations; repeat each concentration (e.g. three times) and calculate a mean to improve reliability. Max 3 marks.

Widely varying repeat results suggest problems with technique. Reading the burette scale at eye level avoids parallax error when recording volumes. Using smaller, more precise volume increments near the expected end point, and judging the colour-change end point consistently between repeats (for example by comparing against a reference colour), improves consistency. Since vitamin C degrades over time once a sample is exposed to air, a freshly prepared sample of juice should be used for each repeat rather than reusing one that has been standing.

1 mark each for any three of: read the burette at eye level to avoid parallax error; use smaller volume increments/more precise glassware; judge the end point (colour change) consistently between repeats, e.g. by comparing against a reference; use a freshly prepared sample of juice for each repeat, since vitamin C degrades over time. Max 3 marks.

A single reading may be affected by random error or a measurement mistake, and so may not be truly representative of the underlying result. Taking several repeats allows any anomalous (outlying) result to be identified and excluded. Calculating a mean from the remaining, concordant repeats reduces the overall effect of random error, giving a more accurate and reliable estimate of the true value.

1 mark: a single reading may be affected by random error/measurement mistakes and may not be representative; 1 mark: repeating allows anomalous results to be identified and excluded; 1 mark: calculating a mean from repeats reduces the effect of random error, giving a more reliable/accurate result.

Accuracy describes how close a measured result is to the true, actual value. Precision describes how close repeated measurements are to one another, regardless of whether they are close to the true value. It is possible for results to be precise (closely clustered together) without being accurate, for example if a systematic error affects every reading equally in the same direction.

1 mark: accuracy = how close a measured result is to the true value; 1 mark: precision = how close repeated measurements are to each other; 1 mark: correctly notes results can be precise without being accurate (or vice versa), e.g. due to a systematic error affecting all readings equally.

Judging colour change by eye is subjective: different people, or even the same person at different times, may perceive and record slight colour differences inconsistently. A colorimeter instead provides an objective, numerical measurement of light absorbance or transmission, and is more sensitive to small colour changes than the human eye. Because it is not affected by individual bias, a colorimeter produces more precise and repeatable data, both between repeats and between different people carrying out the investigation.

1 mark: estimating colour by eye is subjective and open to bias/inconsistency between observers or readings; 1 mark: a colorimeter provides an objective, numerical measurement of light absorbance/transmission; 1 mark: a colorimeter is more sensitive to small colour changes than the human eye, giving more precise/repeatable data.

A control is treated in exactly the same way as the test sample, except that it lacks the antibiotic being investigated. It shows how the bacteria grow under normal conditions in the absence of any treatment. By comparing the treated sample against this control, the student can attribute any reduction in bacterial growth specifically to the effect of the antibiotic, rather than to other uncontrolled factors, such as temperature or nutrient availability, making the conclusion drawn from the investigation valid.

1 mark: a control lacks the antibiotic/treatment being tested but is otherwise treated identically; 1 mark: it shows the growth/result expected in the absence of the treatment; 1 mark: comparing the treated sample to the control allows any difference to be attributed specifically to the antibiotic, making the conclusion valid.

An anomalous result lies noticeably away from the trend shown by the rest of the data and is likely caused by a random error or mistake specific to that one reading. It should be clearly identified, for example by circling the point, and should not be used when drawing the line of best fit, since including it would distort the line away from the true underlying trend. It should similarly be excluded from any mean calculated from the data, and, where possible, that particular measurement should be repeated to obtain a more reliable replacement value.

1 mark: the point is identified as anomalous because it lies well away from the general trend of the other data; 1 mark: it should be excluded when drawing the line of best fit, as it would otherwise distort the line away from the true trend; 1 mark: it should also be excluded from any mean calculated, and ideally the reading should be repeated.

An investigation is valid if it genuinely measures the effect of the variable it is intended to test, with all other variables properly controlled. In a photosynthesis investigation, validity would be reduced, for example, if moving the lamp closer to the pondweed also raised the temperature of the water due to heat given off by the lamp. In that case, any increase in the rate of bubble production could be caused, at least partly, by the temperature rise rather than by the change in light intensity alone, meaning the investigation would no longer purely be testing the effect it claims to test.

1 mark: validity = whether the investigation actually tests/measures what it claims to test, with other variables controlled; 1 mark: a plausible way validity could be reduced is identified, e.g. an uncontrolled variable (such as heat from the lamp raising temperature) also changing alongside the intended variable; 1 mark: explains that this means any change in the dependent variable could be due to the uncontrolled variable rather than (or as well as) light intensity.

Carbon dioxide is one of the raw materials required for photosynthesis, so increasing its concentration provides more substrate for the light-independent reactions, increasing the rate of photosynthesis. However, once another factor, such as light intensity or temperature, becomes the factor in shortest supply relative to the plant's needs (the limiting factor), that factor restricts the rate regardless of how much extra carbon dioxide is available. Beyond this point, further increases in carbon dioxide concentration therefore produce no further increase in the rate of photosynthesis.

1 mark: carbon dioxide is a raw material for photosynthesis, so increasing it increases the rate while it remains in short supply; 1 mark: rate levels off once another factor (e.g. light intensity or temperature) becomes limiting; 1 mark: once that other factor is limiting, further increases in carbon dioxide concentration have no additional effect on the rate.

In a simple reaction time task, there is only one possible stimulus and one possible response, so the nervous system can respond almost automatically once the stimulus is detected. A choice reaction time task instead requires the brain to first identify which of two (or more) stimuli has occurred and then select the correct corresponding response from the available options. This additional decision-making involves extra neural processing, with impulses passing across more synapses in the brain, and this additional processing takes extra time, making choice reaction time slower than simple reaction time.

1 mark: in a simple reaction task there is only one stimulus and response, requiring minimal processing; 1 mark: a choice reaction task requires the brain to identify which stimulus occurred and select between possible responses; 1 mark: this additional decision-making/processing (across more synapses) takes extra time, making choice reaction time slower.

A strong answer explains that tar in cigarette smoke irritates the airways, stimulating excess mucus production, while also paralysing and destroying the cilia that normally sweep mucus (and trapped pathogens/particles) out of the airways, allowing mucus to accumulate and leading to conditions such as chronic bronchitis. Smoke also damages the walls of the alveoli, breaking them down (emphysema), which reduces the total surface area available for gas exchange and so reduces the efficiency of oxygen diffusion into the blood. Separately, carbon monoxide in cigarette smoke binds to haemoglobin more readily than oxygen does, forming carboxyhaemoglobin, which reduces the oxygen-carrying capacity of the blood. Together, these effects mean less oxygen is delivered to respiring tissues, including muscles, during exercise; as a result, more energy must be released by anaerobic respiration, lactic acid builds up more quickly, and the person experiences breathlessness and fatigue sooner, reducing their overall fitness and exercise capacity.

Band A (5–6 marks): specialist terms (e.g. cilia, alveoli, carboxyhaemoglobin, anaerobic respiration) used accurately throughout; at least 5 indicative points addressed with a clear, logical structure; excellent spelling, punctuation and grammar. Band B (3–4 marks): some specialist terms used; at least 3 indicative points addressed; generally coherent structure; satisfactory SPG. Band C (1–2 marks): limited specialist vocabulary; at least 1 indicative point addressed; basic structure/SPG. Band D (0 marks): response not worthy of credit. Indicative content: tar irritates airways and increases mucus production; tar paralyses/destroys cilia, so mucus and pathogens are not cleared (chronic bronchitis); smoke breaks down alveoli walls, reducing surface area for gas exchange (emphysema); carbon monoxide binds to haemoglobin (forming carboxyhaemoglobin) more readily than oxygen; this reduces the oxygen-carrying capacity of the blood; less oxygen reaches muscles during exercise, increasing reliance on anaerobic respiration and lactic acid build-up; overall exercise capacity/fitness is reduced, with breathlessness and fatigue occurring sooner.