2024 OCR GCSE Gateway Science - Biology A - J247 Practice Paper with Answers

At high temperatures, the thermal energy breaks the bonds holding the tertiary structure of the amylase enzyme together. This causes the active site to change shape (denature), meaning the starch substrate can no longer fit, and the rate of reaction drops rapidly.

1 mark for identifying that the active site changes shape so substrate can no longer bind (denaturation).

A reflex arc starts with a receptor detecting a stimulus, sending an impulse along a sensory neurone to a relay neurone in the central nervous system. This passes the impulse to a motor neurone and finally to an effector (like a muscle) to produce a response.

1 mark for the correct sequence of components from receptor to effector.

On a sunny winter day, light intensity is already high. However, the cold temperature and atmospheric level of carbon dioxide are likely to be limiting factors. Increasing the temperature and carbon dioxide concentration inside the greenhouse will overcome these limits and maximize photosynthesis.

1 mark for identifying the combination of increasing temperature and carbon dioxide concentration to overcome limiting factors.

For the child to have cystic fibrosis, they must inherit two recessive alleles (ff), one from each parent. Since neither parent shows symptoms of the condition, they must both carry one copy of the recessive allele alongside a dominant allele. Therefore, both parents are heterozygous (Ff).

1 mark for identifying that both parents must be carriers (Ff) to pass on the recessive allele without showing symptoms.

High carbohydrate meals increase blood glucose levels. The pancreas detects this rise and releases insulin into the blood. Insulin stimulates muscle and liver cells to take up glucose and convert it into glycogen for storage, reducing blood glucose concentration back to normal.

1 mark for identifying that insulin is released by the pancreas to trigger glucose uptake and glycogen storage.

First, a random mutation occurs in a bacterium, introducing resistance. When an antibiotic is introduced, it acts as a selective pressure, killing non-resistant bacteria. The resistant bacteria survive, reproduce, and pass on their resistance alleles, leading to an increased proportion of resistant bacteria over generations.

1 mark for identifying the sequence beginning with mutation before antibiotic exposure, leading to selective survival and reproduction.

Photosynthesis is the only biological process that absorbs carbon dioxide from the atmosphere to manufacture glucose. Aerobic respiration, combustion of fossil fuels, and decomposition all release carbon dioxide into the atmosphere.

1 mark for identifying photosynthesis as the sole carbon-removing process.

Xylem vessels are dead, hollow tubes that carry water and dissolved mineral ions from the roots upwards to the leaves in a continuous transpiration stream. Phloem vessels transport sucrose and amino acids (translocation) both up and down the plant.

1 mark for identifying that xylem transports water and mineral ions upwards from roots.

The temperature coefficient, Q10, is calculated using the formula: Q10 = (Rate at T + 10 °C) / (Rate at T °C). Substituting the given values: Q10 = 5.0 / 2.5 = 2.0. Therefore, the temperature coefficient is 2.0, which is option B.

1 mark for correct calculation of Q10 and selection of option B.

Using a Punnett square for the cross Ff x Ff, the possible genotypes are: FF (25%), Ff (50%), and ff (25%). Heterozygous carriers have the genotype Ff, which has a probability of 50%. This is option B.

1 mark for the correct probability of 50% (Option B).

Random mutations occur in the bacterial population first, producing genetic variation (some bacteria possess a mutation that makes them resistant). When the antibiotic is introduced, it acts as a selective pressure, killing non-resistant bacteria. The resistant bacteria survive and reproduce by binary fission, passing on the allele for resistance to their offspring. This increases the frequency of the resistance allele in the population. Therefore, option B is correct.

1 mark for selecting B.

The primary consumer is the Caterpillar (biomass = 2,000 kg) and the secondary consumer is the Blue tit (biomass = 160 kg). The percentage efficiency of biomass transfer is: (Biomass of secondary consumer / Biomass of primary consumer) * 100 = (160 / 2,000) * 100 = 8.0%. Therefore, the correct option is C.

1 mark for correct calculation and selection of option C.

When blood glucose levels rise, the pancreas secretes insulin. Insulin causes cells, especially in the liver and muscles, to take up glucose from the blood and convert it into the storage carbohydrate glycogen. This lowers blood glucose back to normal levels. Therefore, option A is correct.

1 mark for selecting A.

In a spinal reflex arc, the sensory neurone transmits impulses from the receptor to the central nervous system (spinal cord). In the spinal cord, it synapses with a relay neurone (X), which then passes the impulse to a motor neurone. Thus, X is the relay neurone located in the central nervous system. This corresponds to option B.

1 mark for selecting B.

The waist-to-hip ratio is calculated as: Waist circumference / Hip circumference = 90 cm / 120 cm = 0.75. A lower waist-to-hip ratio indicates less abdominal fat, which is associated with a lower risk of cardiovascular disease. Thus, a ratio of 0.75 indicates a lower risk than a higher ratio such as 0.95. This matches option B.

1 mark for the correct calculation of 0.75 and selecting option B.

At lower substrate concentrations, the substrate is the limiting factor because there are free active sites available. As the concentration of hydrogen peroxide (substrate) increases, more enzyme-substrate complexes form. Eventually, a point is reached where every active site of the catalase enzymes is constantly occupied. At this point of saturation, adding more substrate cannot increase the rate further, and enzyme concentration becomes the limiting factor.

1 mark: Identifies that at lower substrate concentrations, substrate is the limiting factor / enzyme active sites are free. 1 mark: Explains that as substrate concentration increases, more active sites become occupied. 1 mark: Identifies that at the plateau, all enzyme active sites are saturated / occupied. 0.5 mark: Identifies that enzyme concentration has become the limiting factor.

First, a mouse or other mammal is injected with the chosen antigen to stimulate an immune response. The mouse's B-lymphocytes (plasma cells) begin producing antibodies specific to this antigen. These B-lymphocytes are then harvested from the mouse's spleen. Since B-lymphocytes cannot divide indefinitely in culture, they are fused with myeloma (cancerous white blood) cells, which can divide rapidly. This fusion creates hybridoma cells that both produce the specific antibody and divide continuously.

1 mark: Mouse/animal is injected with the specific antigen to stimulate an immune response / lymphocyte production. 1 mark: B-lymphocytes are harvested / extracted from the spleen. 1 mark: B-lymphocytes are fused with myeloma (tumour/cancer) cells. 0.5 mark: To produce hybridoma cells (which can divide rapidly and produce the antibody).

Since neither parent shows symptoms but they have an affected child (genotype \(ff\)), both parents must be heterozygous carriers with the genotype \(Ff\). A genetic cross (Punnett square) between two heterozygous individuals (\(Ff \times Ff\)) yields: 1 \(FF\) (healthy, non-carrier), 2 \(Ff\) (healthy carriers), and 1 \(ff\) (affected). Out of 4 possible outcomes, 2 are carriers (\(Ff\)). Therefore, the probability of having a carrier child is \(2/4 = 1/2\) or 50%.

1 mark: Correctly identifies both parents as heterozygous carriers with genotype \(Ff\). 1 mark: Demonstrates correct gamete combinations using a diagram or Punnett square (\(FF\), \(Ff\), \(Ff\), \(ff\)). 1 mark: Correctly identifies that the probability of a carrier is 50% / 0.5 / \(2/4\). 0.5 mark: Explicitly states that carriers are healthy because they possess one dominant allele \(F\) which masks the recessive allele \(f\).

First, a random mutation occurs in the DNA of a bacterium within the population, creating an allele that provides resistance to the antibiotic. When the antibiotic is introduced, it acts as a strong selective pressure, killing the normal, non-resistant bacteria. The mutated, resistant bacteria survive the treatment. They then reproduce rapidly by binary fission, passing the resistant gene to their offspring. Over several generations, the proportion of resistant bacteria in the population increases significantly.

1 mark: Mentions that a random genetic mutation occurs to produce a resistant allele/gene. 1 mark: Identifies that the antibiotic acts as a selection pressure, killing non-resistant bacteria. 1 mark: Explains that the resistant bacteria survive and reproduce. 0.5 mark: States that they pass the resistant allele on to their offspring, increasing its frequency over generations.

Decomposers secrete extracellular enzymes to digest complex organic compounds in dead organisms and waste. They absorb the soluble products and use them for aerobic respiration, which releases carbon dioxide (\(\text{CO}_2\)) back into the atmosphere. Raising the temperature from \(10\,^{\circ}\text{C}\) to \(25\,^{\circ}\text{C}\) increases the kinetic energy of both enzymes and substrate molecules, leading to more frequent successful collisions. This accelerates the rate of decomposition and therefore increases the rate of carbon release.

1 mark: Explains that decomposers digest/break down dead organic matter and release carbon dioxide via respiration. 1 mark: Explains that increasing temperature to \(25\,^{\circ}\text{C}\) increases the kinetic energy of enzymes and substrates. 1 mark: Explains that this leads to more frequent successful collisions / faster enzyme-controlled reactions, increasing the decomposition rate. 0.5 mark: Identifies that if the temperature rose too high (above optimum), enzymes would denature, slowing the process (not seen at \(25\,^{\circ}\text{C}\)).

When blood glucose concentration falls below normal, receptor cells in the pancreas detect this change and secrete the hormone glucagon into the blood. Glucagon travels to the liver, where it triggers liver cells to convert stored glycogen back into glucose, which is then released into the bloodstream. This raises the blood glucose concentration. Once the blood glucose concentration returns to the normal set point, negative feedback signals the pancreas to reduce or stop secreting glucagon, keeping blood sugar stable.

1 mark: The pancreas detects the low blood glucose and secretes the hormone glucagon. 1 mark: Glucagon travels to the liver to stimulate the conversion of glycogen into glucose (releasing it into the blood). 1 mark: Blood glucose levels rise back to normal. 0.5 mark: Explains that negative feedback reduces glucagon secretion once blood glucose is restored to the set-point to maintain homeostasis.

When an electrical impulse reaches the end of the pre-synaptic neurone, it stimulates tiny sacs called vesicles to release chemical messenger molecules called neurotransmitters. These neurotransmitters diffuse across the synaptic cleft (the microscopic gap between the neurones) down a concentration gradient. They bind to complementary, specific receptor molecules located only on the membrane of the post-synaptic neurone. This binding triggers a new electrical impulse in the post-synaptic neurone. Since receptors are only on the post-synaptic membrane, the signal can only travel in one direction.

1 mark: Arrival of the impulse triggers the release of neurotransmitters from vesicles in the pre-synaptic neurone. 1 mark: Neurotransmitter chemicals diffuse across the synaptic cleft/gap. 1 mark: Neurotransmitters bind to specific receptor molecules on the post-synaptic neurone, triggering a new electrical impulse. 0.5 mark: Explains that transmission is one-way because receptors are only present on the post-synaptic membrane / vesicles are only on the pre-synaptic side.

According to the inverse square law, light intensity (\(I\)) is inversely proportional to the square of the distance (\(d\)). The distance increases by a factor of 3 (from \(15\,\text{cm}\) to \(45\,\text{cm}\), since \(45 / 15 = 3\)). Squaring this factor: \(3^2 = 9\). Therefore, the light intensity at \(45\,\text{cm}\) is \(1/9\) (approximately 11.1%) of the intensity at \(15\,\text{cm}\). This drastic reduction in light intensity decreases the rate of photosynthesis because there is less light energy available to drive the light-dependent stage of photosynthesis, meaning light is the limiting factor.

1 mark: Shows calculation of distance factor change (\(45 / 15 = 3\)) and squares it (\(3^2 = 9\)). 1 mark: States that light intensity decreases by a factor of 9 / is reduced to \(1/9\) (or equivalent decimals like 0.11 of the original). 1 mark: Explains that the rate of photosynthesis will decrease because there is less light energy available for the reaction. 0.5 mark: Identifies light intensity as the limiting factor at this range.

To calculate \(Q_{10}\):
\(Q_{10} = \frac{\text{Rate at }25\text{ }^\circ\text{C}}{\text{Rate at }15\text{ }^\circ\text{C}}\)
\(Q_{10} = \frac{4.8}{2.4} = 2.0\)

A \(Q_{10}\) value of 2.0 indicates that the rate of this enzyme-controlled reaction doubles for every \(10\text{ }^\circ\text{C}\) increase in temperature (within its non-denaturing range).

- 1 mark for correct working shown: \(4.8 / 2.4\)
- 0.5 marks for correct calculation of temperature coefficient value: 2 (or 2.0)
- 1 mark for stating that the rate of reaction doubles
- 1 mark for stating this doubling happens for every \(10\text{ }^\circ\text{C}\) increase in temperature

Using the inverse square law, doubling the distance from \(10\text{ cm}\) to \(20\text{ cm}\) (a factor of 2) reduces the light intensity by a factor of \(2^2 = 4\).
Expected light intensity = \(\frac{1000}{4} = 250\text{ lux}\).
Since temperature and carbon dioxide are constant, light intensity is the limiting factor. Decreasing the light intensity reduces the light energy absorbed by chlorophyll, which directly decreases the rate of photosynthesis.

- 1 mark for showing working using the inverse square law relation (e.g., \((10/20)^2 \times 1000\) or dividing the intensity by 4)
- 0.5 marks for the correct final value: 250 (lux)
- 1 mark for stating that the rate of photosynthesis will decrease
- 1 mark for explaining that light intensity is the limiting factor / less light energy is absorbed by chlorophyll

When an electrical impulse reaches the end of the presynaptic neurone, it triggers the release of neurotransmitters from vesicles. These chemical neurotransmitters diffuse across the synaptic cleft (the microscopic gap between neurones). They bind to specific receptor molecules located on the postsynaptic membrane. This binding causes ions to enter and generates a new electrical impulse in the postsynaptic neurone. This process is strictly unidirectional because neurotransmitter vesicles are only present in the presynaptic neurone terminal, and the specific receptors are only present on the postsynaptic membrane.

- 0.5 marks for stating that neurotransmitters diffuse across the cleft/gap
- 1 mark for stating that they bind to specific receptors on the postsynaptic membrane to trigger a new electrical impulse
- 1 mark for explaining that vesicles/neurotransmitters are only stored/released on the presynaptic side
- 1 mark for explaining that receptor molecules are only present on the postsynaptic side

A decrease in blood thyroxine levels below the normal setpoint is detected by the hypothalamus. The hypothalamus is stimulated to release Thyrotropin-Releasing Hormone (TRH). TRH acts on the pituitary gland, stimulating it to release Thyroid-Stimulating Hormone (TSH). TSH then travels through the bloodstream to the thyroid gland, stimulating it to secrete more thyroxine into the blood. As thyroxine levels return to normal, the increased level of thyroxine inhibits the release of both TRH from the hypothalamus and TSH from the pituitary gland, completing the negative feedback loop.

- 0.5 marks for identifying that low thyroxine levels stimulate the hypothalamus to release TRH
- 1 mark for describing that TRH causes the pituitary gland to release TSH
- 1 mark for describing that TSH stimulates the thyroid gland to release thyroxine
- 1 mark for explaining that the increase in thyroxine levels inhibits further release of TRH/TSH (negative feedback loop)

Transpiration is the passive movement of water and dissolved mineral ions through xylem vessels. It is a unidirectional process, moving substances upwards from the roots, through the stem, and out of the leaves via stomata. Translocation is an active process that transports organic solutes, primarily sucrose and amino acids, through phloem tissue (consisting of sieve tubes and companion cells). Translocation is bidirectional, moving sugars from where they are produced or stored (sources, like leaves) to where they are needed for growth or storage (sinks, like roots or growing fruits).

- 1.5 marks for describing transpiration: 0.5 for xylem, 0.5 for water/mineral ions, 0.5 for unidirectional/upward transport
- 2 marks for describing translocation: 0.5 for phloem, 0.5 for sucrose/amino acids/organic solutes, 1.0 for bidirectional transport (or movement from source to sink)

Since the parents do not have cystic fibrosis (which is recessive, requiring genotype \(ff\) to express) but have a child who has it, both parents must carry one dominant healthy allele (\(F\)) and one recessive cystic fibrosis allele (\(f\)). Therefore, their genotypes must be heterozygous (\(Ff\)). A child must inherit one recessive \(f\) allele from the mother and one recessive \(f\) allele from the father to be affected (\(ff\)). A monohybrid cross of \(Ff \times Ff\) yields the following offspring genotypes: \(FF\) (normal), \(Ff\) (carrier), \(Ff\) (carrier), and \(ff\) (affected). This results in a \(25\%\) (or \(0.25\) or \(1\text{ in }4\)) probability that any subsequent child will have the disorder.

- 1 mark for identifying both parents' genotypes as heterozygous/carriers (\(Ff\))
- 1.5 marks for the inheritance explanation: 0.5 for stating cystic fibrosis is recessive/requires two alleles (\(ff\)), and 1 mark for explaining each heterozygous parent must pass on their recessive allele (\(f\))
- 1 mark for stating the correct probability: 0.25 / 25% / 1 in 4

Antibiotic resistance develops because random genetic mutations occur in bacteria, creating alleles that offer protection against a specific antibiotic. When the population of bacteria is exposed to that antibiotic, it acts as a strong selection pressure. The normal, non-resistant bacteria are killed, while the mutated, resistant bacteria survive. These survivors reproduce rapidly (binary fission) and pass on the gene/allele for antibiotic resistance to their offspring, increasing its frequency in the population. Human activities, such as overprescribing antibiotics for viral infections or patients failing to finish their prescribed course, accelerate this because it repeatedly exposes bacterial populations to non-lethal doses, selecting for and leaving behind the most resistant strains.

- 1 mark for explaining that a random mutation creates variation/a resistant allele
- 1 mark for explaining that antibiotics act as a selection pressure, killing non-resistant bacteria but allowing resistant ones to survive and reproduce (passing on the resistance allele)
- 1.5 marks for human acceleration: 0.5 for identifying a human action (e.g. overprescribing, not finishing courses, use in livestock farming) and 1 mark for explaining that this creates unnecessary selection pressure / leaves partially resistant bacteria alive to multiply

Fertilizers containing nitrates and phosphates wash into the lake, causing rapid growth of algae on the surface (an algal bloom). This algal layer blocks sunlight from reaching submerged aquatic plants, preventing photosynthesis and causing these plants to die. Decomposers, such as bacteria, multiply rapidly as they feed on the dead plant matter. These bacteria respire aerobically, consuming vast amounts of dissolved oxygen from the water. The resulting oxygen depletion (anoxia) causes fish and other aquatic organisms to suffocate and die, reducing the lake's biodiversity. Species such as stonefly nymphs or mayfly nymphs require very high levels of dissolved oxygen and serve as indicator species for clean, unpolluted water.

- 1 mark for describing that algal blooms block light, preventing photosynthesis and killing submerged plants
- 1 mark for explaining that bacteria decompose dead plants and use up dissolved oxygen via aerobic respiration
- 0.5 marks for stating that oxygen depletion leads to the death of fish / aquatic organisms, reducing biodiversity
- 1 mark for identifying a valid indicator species for clean, oxygenated water (e.g., stonefly nymph / mayfly nymph)

To calculate the efficiency of biomass transfer, divide the biomass of the consumer (blue tits) by the biomass of the prey (caterpillars) and multiply by 100: (108 / 1440) * 100 = 7.5%.

1 mark for identifying the correct biomass values (108 kg and 1440 kg). 1.5 marks for the correct working: (108 / 1440) * 100. 1 mark for the correct final answer of 7.5%.

Adrenaline prepares the body for physical action. It increases heart rate and breathing rate. An increased heart rate pumps blood faster, delivering glucose and oxygen more rapidly to respiring muscle cells. This increases the rate of cellular respiration, releasing energy needed for muscle contractions.

1 mark for stating first correct effect (e.g. increased heart rate, increased breathing rate, pupil dilation, or diversion of blood to muscles). 1 mark for stating second correct effect. 1.5 marks for explaining how one effect prepares the body (e.g. explaining that increased heart rate delivers more oxygen/glucose to muscles to increase cellular respiration and release energy for muscle contraction).

To treat cardiovascular disease (CVD), doctors can use a combination of lifestyle changes, medications, and surgical interventions depending on the severity of the condition.

1. Lifestyle changes:
- Examples: Eating a balanced diet low in saturated fats to reduce blood cholesterol, regular cardiovascular exercise, quitting smoking, and reducing alcohol intake.
- Benefits: Non-invasive, free or cheap to implement, improves overall physical and mental health, and addresses the underlying behavioral causes of the disease.
- Risks/Limitations: Requires high levels of patient self-discipline over a long period. Results are not immediate and may not be sufficient on their own if the disease is already severe or has a strong genetic component.

2. Medicines:
- Examples: Statins (to lower LDL cholesterol levels), antihypertensive drugs (to lower blood pressure), and daily low-dose aspirin (an anticoagulant to reduce blood clotting risk).
- Benefits: Non-invasive, highly effective at reducing the risk of heart attacks and strokes, easily administered daily, and backed by extensive clinical research.
- Risks/Limitations: Must be taken daily for life, requiring high compliance. Can cause side effects (e.g., muscle pain or liver damage from statins, or increased risk of internal bleeding from aspirin), and does not physically remove existing advanced blockages.

3. Surgical interventions:
- Examples: Inserting stents (mesh tubes to keep coronary arteries open), coronary artery bypass grafts (CABG, where a healthy blood vessel is used to bypass a blocked artery), and heart transplants for end-stage heart failure.
- Benefits: Provides rapid, life-saving, and highly effective relief for severe, life-threatening blockages by immediately restoring blood flow to the heart muscle.
- Risks/Limitations: Highly invasive, carrying general surgical risks such as infection, blood clots, and complications from anaesthetics. Heart transplants are limited by donor shortages, and recipients must take lifelong immunosuppressant drugs, making them highly vulnerable to infectious diseases.

Level 3 (5-6 marks):
- Detailed discussion covering all three categories of treatment (lifestyle changes, medicines, and surgical interventions).
- Provides clear, correct examples for each category.
- Evaluates both benefits and risks/limitations for all three categories.
- The answer is structured logically, presents ideas clearly, and uses scientific terminology accurately.

Level 2 (3-4 marks):
- Discusses at least two categories of treatment in reasonable detail.
- Provides examples and some benefits or risks/limitations for these categories.
- The answer is generally structured logically, but some key points or evaluations may be omitted.

Level 1 (1-2 marks):
- Identifies at least one treatment option or lists some simple details about treatments.
- Shows limited evaluation of advantages or disadvantages.
- The response lacks scientific detail and structure.

0 marks:
- No relevant content.

Indicative content:
- Lifestyle: Diet, exercise, smoking cessation. Benefits: No side effects, low cost, addresses root cause. Risks: Relies on patient discipline, slow, not enough for severe blockages.
- Medicines: Statins, antihypertensives, anticoagulants. Benefits: Easy to take, preventative. Risks: Side effects, daily compliance required.
- Surgery: Stents, bypass, transplants. Benefits: Immediate blood flow restoration. Risks: Surgery risk, infection, transplant rejection / lifelong immunosuppressants.