We read the actual downloaded question papers and mark schemes for every AQA Biology Paper 1 Higher Tier sitting we have available, including the two 2020 and 2021 papers which use AQA's Jun20 and Jun21 series codes on the printed page even though one of them was sat in May 2020. Below is what each question type has actually asked, what the real experimental setups and data showed, and a complete worked answer written to the mark scheme's top level for the extended response questions. This is the closest you can get to seeing exactly what a full mark answer looks like without a real exam paper in front of you.
Questions © AQA, quoted for analysis. Diagrams, tables and experimental setups described in our own words, not reproduced. Mark scheme content translated into plain English, not copied. PrepWise is independent and not endorsed by AQA.
Every single sitting we analysed has at least one 6-mark extended response question built the same way: it gives you a starting fact (a blockage, damaged roots, a dying pond snail, damaged tissue) and asks you to trace the biological consequences through a logical chain, or it asks you to describe how a named structure's features suit its job. These are marked on a three level scheme (5 to 6, 3 to 4, 1 to 2) and the difference between the levels is almost never about knowing more facts, it is about linking the facts you already know into a clear, logical account.
It wants you to build the full chain from reduced blood flow, through reduced oxygen and glucose delivery to heart muscle cells, to the switch to anaerobic respiration, and finally to the physical symptoms this causes, such as breathlessness, tiredness and chest pain.
The question follows a data question showing how blood flow through a coronary artery falls as the percentage of the artery that is blocked increases, with data plotted on a graph, so the idea of reduced blood flow with increasing blockage has already been established before this extended response is asked.
A partly blocked coronary artery reduces the amount of blood that can flow to the heart muscle cells, because the narrowed space inside the artery restricts how much blood can pass through per minute. This means less oxygen and less glucose reach the heart muscle cells than normal.
Because there is less oxygen available, the heart muscle cells cannot respire aerobically at the normal rate, so more of their energy comes from anaerobic respiration instead. Anaerobic respiration releases much less energy per glucose molecule than aerobic respiration and produces lactic acid as a waste product, so as the blockage gets worse, lactic acid starts to build up in the heart muscle tissue.
The build up of lactic acid causes muscle fatigue and pain in the chest, because the heart muscle cannot keep contracting normally when energy release is reduced and a waste product is accumulating around the muscle fibres. At the same time, because less blood is reaching the rest of the body from a heart that cannot pump normally, less oxygen and glucose reach other tissues too, which is why a person with a partly blocked coronary artery often also feels tired and short of breath, since their body is forced to try to compensate by breathing faster to bring in more oxygen.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise organisation and gas exchange questionsIt wants you to trace what root damage does to water uptake, mineral ion uptake, and the plant's own transport tissues, and then follow each of those failures through to a reason the whole plant could die, such as reduced photosynthesis, reduced protein synthesis or reduced anchorage.
The question describes hornet moth larvae as living inside tree roots, feeding on the root tissue and damaging it, alongside an illustration of the larvae, before asking for the biological consequences of this root damage for the tree.
Damage to the roots reduces the surface area of root hair cells available to absorb water from the soil, so less water is taken up into the plant. Since water is needed as a raw material for photosynthesis, a lower rate of water uptake leads to a lower rate of photosynthesis, meaning less glucose is produced. Because glucose is used to make cellulose for new cell walls and provides the substrate for respiration to release energy, reduced photosynthesis limits both new growth and the energy available for normal cell processes, and with less water reaching the cells generally, cells also start to lose their turgidity, weakening the support the plant relies on to hold its stems and leaves up.
Damaged roots also absorb fewer mineral ions from the soil, particularly nitrate ions and magnesium ions, since active transport of these ions depends on healthy root hair cells with working transport proteins. Nitrate ions are needed to make amino acids, which are used to build proteins for growth, so less nitrate uptake means less protein is made and the plant cannot grow or repair damaged tissue properly. Magnesium ions are needed to make chlorophyll, so reduced magnesium uptake means less chlorophyll is produced, which further reduces the rate of photosynthesis on top of the water shortage already described, compounding the energy problem in a second, independent way.
Root damage does not only reduce how much water and minerals the roots absorb in the first place, it can also destroy the xylem and phloem tissue that carries those substances once they are inside the root. This is a completely separate failure from the absorption problem above, because even if the roots below the damaged section are still absorbing water and minerals normally, none of it can get past a break in the xylem, so every cell above that point in the stem and leaves is cut off regardless of how well the roots underneath are working. In the same way, damaged phloem stops sugars made by photosynthesis in the leaves from being carried down to root cells, so root cells below the break lose their supply of glucose for respiration even though they may be absorbing water and minerals perfectly well. Because roots also physically anchor the tree in the soil, extensive root damage further reduces anchorage, making the tree more likely to be blown over in strong wind, which is a mechanical rather than a biochemical reason the tree could be lost.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise organisation and gas exchange questionsIt wants you to trace each of the three named symptoms back to a shortage of one specific type of blood cell or cell fragment, explaining the biological role of each and how a shortage of it directly causes that particular symptom.
The question follows a diagram showing three tumours of the same type found in the same patient, one in the lung and two in the liver, establishing that the cancer has spread (metastasised), before asking about the effect of a low blood component count on the body.
Tiredness is caused by a fall in the number of red blood cells, since red blood cells contain haemoglobin, which binds to oxygen in the lungs and carries it around the body. Fewer red blood cells means less haemoglobin overall, so less oxygen is transported to body cells. With less oxygen available, cells cannot respire aerobically at their normal rate, so more anaerobic respiration takes place instead, which releases far less energy per glucose molecule than aerobic respiration and also produces lactic acid, causing the muscle fatigue that a person experiences as tiredness.
Frequent infections are caused by a fall in the number of white blood cells, since white blood cells, including phagocytes and lymphocytes, are responsible for defending the body against pathogens. Phagocytes normally engulf and digest bacteria and other pathogens, while lymphocytes produce antibodies specific to particular pathogens. With fewer white blood cells available, fewer pathogens are engulfed by phagocytosis and fewer antibodies are produced, so bacteria and viruses that enter the body are not destroyed as quickly or as effectively as normal, which is why infections become more frequent and can become more serious.
Bleeding that will not stop is caused by a fall in the number of platelets, since platelets are cell fragments responsible for helping blood clot at the site of a cut or wound. Normally, platelets clump together at a wound and trigger a series of reactions that form a clot, sealing the break in the skin and blood vessels. With fewer platelets in the blood, clots form more slowly or do not form properly at all, so a cut that would normally stop bleeding within a short time continues to bleed for much longer, which explains this third symptom as a direct consequence of platelet shortage rather than any problem with the blood vessels themselves.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise organisation and gas exchange questionsIt wants you to name the specific structural features of the alveoli and the surrounding capillaries that maximise the rate of diffusion, and explain what each feature actually does to increase how quickly oxygen and carbon dioxide can move across the exchange surface.
The question shows a diagram of the human breathing system with a magnified inset of a single capillary wrapped around an alveolus, before asking how the lungs are adapted for efficient gas exchange.
The lungs contain many millions of tiny alveoli, and this huge number gives the lungs an extremely large total surface area in contact with the surrounding capillaries. A larger surface area means more gas molecules can diffuse across the exchange surface at the same time, since diffusion happens across the whole membrane simultaneously rather than at just one point, so the total rate of gas exchange for the lungs as a whole is much higher than it would be with only a few larger exchange surfaces.
The walls of the alveoli and the walls of the capillaries surrounding them are each only one cell thick, so together they provide a very short diffusion path between the air inside the alveolus and the blood inside the capillary. Because the rate of diffusion is faster over a shorter distance, this thin, one cell thick barrier allows oxygen to move into the blood and carbon dioxide to move out of the blood much more quickly than if the walls were made of several layers of cells.
Breathing constantly moves air in and out of the lungs, bringing in air with a high oxygen concentration and removing air that has a lower oxygen concentration and a higher carbon dioxide concentration after gas exchange has taken place. At the same time, the dense network of capillaries around each alveolus constantly brings in deoxygenated blood and carries away oxygenated blood. Together, ventilation of the lungs and the constant blood flow through the capillaries maintain a steep concentration gradient for both oxygen and carbon dioxide across the alveolar wall, and since diffusion depends on a concentration gradient, maintaining this steep gradient keeps the rate of gas exchange high rather than letting it slow down as concentrations equalise.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise organisation and gas exchange questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
How many chambers does the human heart have?
This question type always asks you to trace a chain of cause and effect, not just list facts. Practise linking your points with 'so' and 'this means' until the whole answer reads as one continuous account.
Practise organisation and gas exchange questionsA second 6-mark question type turns up almost every sitting: instead of tracing a biological cause and effect chain, it gives you a table of data and asks you to judge or evaluate it, or it asks you to design or describe the steps of an investigation. These use a different mark scheme, one built around whether your answer would actually produce a valid outcome or a supported judgement, not just whether you stated correct facts.
It wants you to compare the two types of burger using the actual numbers in the table (protein, fibre, fat, carbohydrate, cholesterol) and use your own knowledge of what each nutrient does in the body to reach a supported judgement about which is better for a balanced diet, rather than a one sided answer.
A table giving the mass of protein, fibre, fat, carbohydrate and cholesterol per 100g for burgers made from meat compared with meat-free burgers. Meat burgers are shown with more protein, more fat and more cholesterol; meat-free burgers are shown with more fibre and a similar amount of carbohydrate.
| Mass per 100g of burger | Burgers made from meat | Meat-free burgers |
|---|---|---|
| Protein in g | 14 | 9 |
| Fibre in g | 0.9 | 5.5 |
| Fat in g | 16 | 5.2 |
| Carbohydrate in g | 15.5 | 15.1 |
| Cholesterol in mg | 120 | 0 |
Meat burgers contain more protein per 100g than meat-free burgers, and protein is needed for growth and repair of body tissues, so meat burgers are the stronger source of this nutrient in the table. However, meat burgers also contain more fat and much more cholesterol than meat-free burgers, and a diet high in saturated fat and cholesterol increases the risk of narrowing of the arteries (atherosclerosis), which raises the risk of coronary heart disease and heart attacks over time. This means that although meat burgers are useful for protein, eating them regularly carries a genuine long term cardiovascular health risk that the table figures directly support.
Meat-free burgers contain considerably more fibre than meat burgers, and fibre aids digestion and helps prevent constipation by adding bulk to food moving through the intestines, which is a genuine benefit the table supports. The two types of burger have a similar amount of carbohydrate, so this table does not show one type being clearly better than the other as an energy source. The table also gives no information about vitamins or minerals in either burger, so a full judgement about which is more balanced overall cannot be made from this data alone, only from the specific nutrients that were actually measured.
Overall, meat burgers provide more protein but carry a greater long term cardiovascular health risk because of their higher fat and cholesterol content, while meat-free burgers provide less protein but more fibre and a lower cardiovascular risk. Neither is automatically the healthier choice on its own, since a balanced diet needs both adequate protein for growth and repair and enough fibre for healthy digestion, so the better choice depends on what else a person eats and how often they eat each type of burger, rather than one burger type being straightforwardly the healthiest option.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise evaluation and practical design questionsIt wants a complete, logically sequenced method that would actually produce valid, comparable results: testing a proper sample of athletes, using at least two different exercise types, measuring heart rate before and after in a consistent way, and controlling the variables that could otherwise make the comparison unfair.
Select a group of at least 30 athletes to take part, since a larger sample size reduces the effect of individual differences on the results and makes any pattern found more reliable. Record the resting heart rate of each athlete before any exercise takes place, by counting their pulse for a fixed period of time (for example, counting beats for 15 seconds and multiplying by four to get beats per minute) while they are sitting still.
Choose at least two different types or intensities of exercise to compare, for example light exercise such as walking and vigorous exercise such as running, and have every athlete in the group complete each type of exercise for the same fixed length of time. Immediately after each type of exercise, record each athlete's heart rate again using the same counting method as before, and calculate the increase in heart rate for each athlete by subtracting their resting heart rate from their heart rate after exercise.
Calculate the mean increase in heart rate across all athletes for each type of exercise, then compare the mean increases between the different exercise types to see which type of exercise causes the largest rise in heart rate. To make this comparison fair, keep other variables constant across all athletes and both exercise sessions: use a mix of biological sexes so the sample is representative, make sure all athletes are a similar age and have a similar level of overall fitness or activity, ensure no athlete has taken caffeine or medication that could affect heart rate before testing, and allow each athlete's heart rate to return to their resting rate before starting the next type of exercise.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise evaluation and practical design questionsIt wants the standard staged structure of a clinical trial: starting with a very low dose given to healthy volunteers to test safety, then moving to patients with the disease to test dose and effectiveness, ideally in a double blind design with a control group, described as a clear sequence.
The question describes a monoclonal antibody treatment for a fungal infection (Candida albicans) that has already been shown effective in laboratory tests on infected tissue culture cells and infected live animals, and is now ready to move to human clinical trials.
The trial should first be given to a small group of healthy volunteers at a very low dose, to test that the treatment is safe and to check for any side effects before it is given to anyone who is actually unwell. Only once this stage shows the treatment is safe at low doses should the dose be gradually increased in further healthy volunteers, still checking for side effects at each stage.
Once safety has been established, the treatment should then be trialled on patients who actually have the Candida albicans infection, to test both the correct or optimum dose needed to treat the infection and to check for side effects in people who are unwell, since a safe dose in a healthy person is not necessarily the correct dose to actually treat a disease. This stage should also test whether the treatment is effective (its efficacy) at treating the infection, not just whether it is safe.
The trial with patients should be a double blind trial, where neither the patients nor the doctors treating them know who has been given the real mAbs treatment and who has been given a placebo or an alternative existing treatment, since this prevents the patients' or doctors' expectations from influencing the results. The trial should also involve a reasonably large number of patients and run for a sufficient duration to be confident the results are reliable, and appropriate control variables, such as similar severity of infection between groups, should be used so that the comparison between the treatment group and the control group is fair.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise evaluation and practical design questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which organ produces hydrochloric acid to kill bacteria in food?
Evaluation and method design questions are marked on whether your answer actually works, not just whether it contains true facts. Practise judging data and designing investigations that would genuinely produce a fair result.
Practise evaluation and practical design questionsA third, distinct 6-mark question type asks you to directly compare the structure of two cells or two tissues. Unlike the extended response and evaluation questions above, this type has only two levels in the mark scheme, not three, so a genuinely thorough comparison of both similarities and differences reaches the top of the range, and there is no separate 'top tier' language to reach for.
It wants a direct, paired comparison of specific structural features (nucleus, cell wall, shape, pigments, vacuole) between a red blood cell and a plant cell, stated clearly enough that a reader knows exactly which cell has which feature.
A red blood cell has no nucleus, since it loses its nucleus during development in order to make more room to carry oxygen, but a plant cell has a nucleus, which contains the genetic material controlling the cell's activities. A red blood cell also has no cell wall, since animal cells never have a cell wall, but a plant cell has a cell wall made of cellulose, which gives the cell a fixed, rigid shape and support.
A red blood cell is a biconcave disc shape, a shape that increases its surface area for absorbing oxygen, whereas plant cells come in many different regular shapes depending on their type and function, often described as rectangular or box shaped because of their rigid cell wall. A red blood cell contains haemoglobin, a red pigment that binds to oxygen, while a plant cell does not contain haemoglobin, though some plant cells do contain the green pigment chlorophyll for photosynthesis, which a red blood cell never contains.
A red blood cell also has no permanent vacuole, whereas many plant cells have a large, permanent vacuole filled with cell sap that helps keep the cell rigid through turgor pressure. Despite all of these differences, both cell types share some features in common: both have cytoplasm, where chemical reactions take place, and both have a cell membrane, which controls what substances move into and out of the cell.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise cell and tissue comparison questionsIt wants paired comparisons covering both what each tissue is physically made of (structure) and what each tissue actually does and how (function), since the question explicitly asks for both, and the mark scheme requires reference to both for the top level.
Xylem tissue is made of dead cells, which have no cytoplasm and are essentially hollow tubes, whereas phloem tissue is made of living cells, which retain their cytoplasm. Xylem cell walls are strengthened with a substance called lignin, giving xylem vessels a rigid structure, but phloem cells do not contain lignin. Phloem cells also have small pores in their end walls, forming structures called sieve plates that allow substances to pass between adjacent cells, while xylem cells do not have pores in their end walls, since they form continuous hollow tubes instead.
In terms of function, xylem transports water and dissolved mineral ions, moving them only in one direction, upwards from the roots to the stem and leaves, as part of the transpiration stream. Phloem, on the other hand, transports dissolved sugars made during photosynthesis, and it can move substances in both directions around the plant, upwards or downwards depending on where sugars are needed, in a process called translocation.
Despite these differences, xylem and phloem do share some features in common: both are made up of columns of cells forming continuous tubes running through the plant, and both are involved in transporting liquids and dissolved substances around the stem, roots and leaves. The key difference in what they transport and in which direction is why a plant needs both tissues rather than just one, since neither tissue on its own could deliver both water and minerals upwards and sugars to wherever they are needed.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise cell and tissue comparison questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which part of the cell contains DNA and controls the cell's activities?
These comparison questions only have two levels, so a full, clearly paired answer covering several genuine differences (and structure AND function where asked) reaches full marks. Practise stating exactly which cell or tissue has which feature.
Practise cell and tissue comparison questionsEvery sitting we analysed builds at least one full question around a required practical: microscopy and magnification calculations, the effect of pH or temperature on enzyme activity, aseptic technique when growing bacteria, or the effect of light intensity or wavelength on photosynthesis. These aren't extended response questions but they appear in nearly every paper and often carry several linked marks across sub parts, so knowing the standard vocabulary for each required practical is worth real marks every single sitting.
It wants two specific, named aseptic techniques used when growing bacteria on agar plates, not a vague answer about hygiene, since AQA's required practical on culturing microorganisms has a defined list of accepted techniques.
A diagram of a Petri dish with paper discs containing different antibiotics placed on agar with Salmonella bacteria growing, used to test antibiotic effectiveness.
The scientists should have passed the inoculating loop or any forceps through a Bunsen burner flame immediately before touching the paper discs or the bacteria, since the heat kills any unwanted microorganisms already on the equipment. They should also have only lifted the lid of the Petri dish a small amount, or tilted it at an angle, whenever discs were added, rather than removing it fully, since this cuts down how long the agar is exposed to microorganisms floating in the air.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise required practical questionsA diagram showing a student's set up for an antibiotic investigation, with an agar plate divided by three filter paper discs each containing a different antibiotic, and a fourth image showing the same plate after incubation with clear zones around some of the discs.
Before setting up the plate, the student should have sterilised the Petri dish and the culture medium so that no unwanted microorganisms were already present on the equipment or the agar. When placing the three filter paper discs onto the agar, the student should also have secured the lid of the Petri dish with adhesive tape afterwards, since this stops microorganisms from the air getting in once the plate is set up and stops the bacteria being cultured from escaping.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise required practical questionsIt wants you to explain that 37 degrees Celsius matches human body temperature and gives bacteria the best conditions to grow for scientific study, while 25 degrees Celsius is used in schools specifically to reduce the risk of growing pathogens that could infect students at their body temperature.
Scientists incubate bacteria at 37 degrees Celsius because this matches normal human body temperature, and bacteria such as Salmonella grow best at this temperature, so using 37 degrees Celsius gives scientists the fastest, most reliable growth for their investigation.
Students in school must use 25 degrees Celsius instead, because this lower temperature reduces or prevents the growth of bacteria that could be harmful to humans. Since many of the bacteria that cause disease in people grow best around body temperature, keeping the incubator below that range in a school lab lowers the risk to students handling the culture.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise required practical questionsIt wants the standard magnification triangle rearranged to find real size, with the image length converted to a consistent unit before dividing, and the final answer converted into micrometres as specifically instructed.
A microscope image of onion epidermis cells at times 400 magnification, with one labelled cell (cell Z) and a scale line showing the measured width of that cell on the image.
Magnification = size of image / size of real object, so rearranging this gives: size of real object = size of image / magnification.
Real length of cell Z = 4.8 / 400 = 0.012 cm. Converting this to micrometres: 0.012 cm = 0.12 mm = 120 micrometres.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise required practical questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
What is magnification?
Required practical questions reward precise vocabulary and every step of a calculation shown, not just a final number. Practise the standard method vocabulary for each required practical so you never lose an easy mark to a vague answer.
Practise required practical questionsA recurring shorter question type asks you to explain enzyme behaviour at two contrasting temperatures or pH values using data from a results table, one where the enzyme works too slowly and one where it does not work at all. The reasoning required is different at each extreme, and mixing them up is the single most common way marks are lost here.
It wants two entirely separate explanations: why the iodine test stayed positive (starch not broken down) at 5 degrees Celsius because there is too little kinetic energy for successful collisions, and why it also stayed positive at 80 degrees Celsius because the amylase has denatured and its active site no longer fits the starch.
A results table showing the time taken for an iodine test on a starch-amylase mixture to stop turning blue-black (indicating starch was fully broken down) at six different temperatures, with the results at 5 degrees Celsius and 80 degrees Celsius both showing the mixture never stopped turning blue-black within the time tested.
| Temperature in °C | Time taken until iodine solution stays yellow-brown in minutes |
|---|---|
| 5 | did not become yellow-brown |
| 20 | 5 |
| 35 | 2 |
| 50 | 7 |
| 65 | 14 |
| 80 | did not become yellow-brown |
At both 5 degrees Celsius and 80 degrees Celsius, the iodine solution never turned blue-black, showing that starch was still present because it had not been broken down by the amylase. At 5 degrees Celsius, the amylase and starch molecules have low kinetic energy, so there are fewer successful collisions between the enzyme and its substrate, meaning enzyme-substrate complexes form far more slowly than at a warmer temperature.
At 80 degrees Celsius, the amylase has been denatured, since the high temperature has broken the bonds holding the enzyme in its normal three dimensional shape. This changes the shape of the amylase's active site, so the starch molecule can no longer fit into it, meaning no new enzyme-substrate complexes can form at all at this temperature.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise enzyme activity questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
What are enzymes?
Low temperature and high temperature enzyme questions need two completely different mechanisms. Practise telling kinetic energy explanations and denaturing explanations apart until you never mix them up under time pressure.
Practise enzyme activity questionsAcross the 4 sittings we have full papers for, these are the topics with the most exam appearances and marks at stake on Paper 1.
Stem cells and cancer cell control as a standalone extended response topic (has appeared in shorter structured questions only) · Vaccination and herd immunity as a standalone extended response topic in these four papers · Plant diseases and defences as a standalone extended response topic (has appeared as shorter structured questions on plant defence responses)
These topics have not carried a full extended response question in the papers we analysed, but can still appear as shorter structured questions, so do not skip them entirely.
The context and data are described in our own words, not reproduced, and the worked answers are written entirely by us, aimed at the actual level descriptors of the real AQA mark schemes for each sitting. They are not copied from AQA's own exemplar materials, since that would breach copyright, but they are built to hit exactly what the real mark scheme rewarded that year. PrepWise is independent of AQA and not endorsed by them.
Sometimes the exact style of a question repeats, and topics like heart and circulation, plant transport, enzymes and microscopy calculations return in some form almost every sitting. But you cannot rely on repeats alone, since the exact numbers, contexts and diagrams change every time even when the question type is similar. Use this page to see which TOPICS and QUESTION TYPES keep returning and make sure you know the underlying biology cold, whatever the exact wording turns out to be.
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