An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 AQA GCSE Biology 8461 paper. Not affiliated with or reproduced from AQA.
Paper 1F
Answer all questions in the spaces provided. Show clear working for calculations.
7 PastPaper.question · 100.1 PastPaper.marks
PastPaper.question 1 · Structured
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Figure 1 shows a simplified diagram of a human cheek cell as seen under a light microscope.
[Figure 1: Diagram showing an irregular shape with outer boundary labeled A, a central dark circle labeled B, and the fluid-filled interior labeled C.]
**a)** Identify the parts of the cell labeled A, B, and C. [3 marks]
**b)** Plant cells contain structures that are not found in animal cells. (i) Name two structures found in a plant cell but not in an animal cell. [2 marks] (ii) Give the function of one of the structures you named in part (b)(i). [1 mark]
**c)** A student wants to observe some cheek cells. Describe how the student should prepare a microscope slide of cheek cells. [3 marks]
**d)** The actual length of a cheek cell is \(0.06 \text{ mm}\). An image of the cell is magnified \(800\) times. Calculate the length of the magnified image in millimeters (mm). Use the equation: $$\text{magnification} = \frac{\text{size of image}}{\text{real size of object}}$$ Show your working. [2 marks]
**e)** Convert your answer from part (d) into micrometers (\(\mu\text{m}\)). Remember: \(1 \text{ mm} = 1000 \mu\text{m}\). [1 mark]
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a) The cell membrane (A) controls the movement of substances in and out of the cell. The nucleus (B) contains DNA and controls cell activity. The cytoplasm (C) is where chemical reactions take place. b) (i) Plant cells contain a cellulose cell wall, chloroplasts, and a permanent vacuole filled with cell sap, which are absent in animal cells. (ii) The cell wall provides strength and structural support. c) The slide preparation involves obtaining cells (swab cheek), transferring them to the slide, staining to make organelles visible, and placing a coverslip gently to protect the specimen. d) Rearranging the magnification formula: \(\text{size of image} = \text{magnification} \times \text{real size of object} = 800 \times 0.06\text{ mm} = 48\text{ mm}\). e) To convert mm to micrometers (\(\mu\text{m}\)), multiply by 1000: \(48 \times 1000 = 48000 \mu\text{m}\).
b) (i) [2 marks] - Any two from: Cell wall, chloroplast, (permanent) vacuole. (2) (ii) [1 mark] - Correct function matching one of the named structures (1): - Cell wall: strengthens/supports cell - Chloroplast: absorbs light (for photosynthesis) - Vacuole: contains cell sap / keeps cell turgid
c) [3 marks] - Rub inside of cheek with swab/cotton bud (1) - Put swab on slide and add a stain / dye (1) - Lower coverslip slowly/gently to avoid air bubbles (1)
A student investigated the effect of different concentrations of sugar solution on the mass of potato chips. This is the method used: 1. Cut five potato chips of equal size. 2. Record the mass of each potato chip. 3. Place each chip into a different tube containing a different concentration of sugar solution. 4. Leave the chips for 2 hours. 5. Remove the chips, dry them with a paper towel, and record the final mass.
**a)** Why did the student dry the potato chips before weighing them? [1 mark]
**b)** Explain why the mass of a potato chip would increase when placed in a very dilute sugar solution. Use the term 'osmosis' in your answer. [3 marks]
**c)** In one test, a potato chip had a starting mass of \(2.40 \text{ g}\). After 2 hours in a concentrated sugar solution, its final mass was \(2.16 \text{ g}\). Calculate the percentage change in the mass of this potato chip. Use the equation: $$\text{percentage change in mass} = \frac{\text{change in mass}}{\text{starting mass}} \times 100$$ Show your working. [3 marks]
**d)** Root hair cells in plants absorb water and mineral ions from the soil. (i) State the process by which water is absorbed by root hair cells. [1 mark] (ii) State the process by which mineral ions are absorbed by root hair cells when their concentration in the soil is very low. [1 mark] (iii) Root hair cells contain many mitochondria. Explain how mitochondria help the cell absorb mineral ions. [3 marks]
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PastPaper.workedSolution
a) Excess liquid on the surface of the potato would add extra, uncontrolled weight to the final reading. Drying ensures only the mass of the potato tissue is measured. b) Osmosis is the movement of water from a dilute solution (high water concentration) to a concentrated solution (low water concentration) across a partially permeable membrane. Since the dilute sugar solution has a higher water concentration than the potato cell cytoplasm, water enters the cells, increasing their mass. c) First, calculate the change in mass: \(2.16\text{ g} - 2.40\text{ g} = -0.24\text{ g}\). Next, calculate percentage change: \((-0.24 / 2.40) \times 100 = -10\\%\). d) (i) Water moves into root cells down a water concentration gradient via osmosis. (ii) When mineral ion concentration in the soil is lower than inside the cell, active transport must be used. (iii) Active transport is an active process requiring energy from respiration. Mitochondria are the site of aerobic respiration, releasing the energy required.
PastPaper.markingScheme
a) [1 mark] - To remove excess/surface water/liquid (so it does not add to the measured mass) (1)
b) [3 marks] - Water enters the potato/cells (1) - By osmosis from a high water concentration to a lower water concentration / down a concentration gradient (1) - Through a partially permeable membrane (1)
d) (i) [1 mark] - Osmosis (1) (ii) [1 mark] - Active transport (1) (iii) [3 marks] - Mitochondria release energy (1) - From (aerobic) respiration (1) - Energy is needed for active transport / to move ions against a concentration gradient (1)
PastPaper.question 3 · Structured
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The human digestive system breaks down food so it can be absorbed into the blood.
**a)** Figure 2 shows parts of the human digestive system.
[Figure 2: A diagram of the human digestive system showing the esophagus, stomach (labeled X), pancreas (labeled Y), and small intestine (labeled Z).]
Identify organs X, Y, and Z. [3 marks]
**b)** Enzymes are chemical catalysts that speed up chemical reactions. (i) Name the enzyme that breaks down starch into simple sugars. [1 mark] (ii) Name the organ that produces bile. [1 mark] (iii) Explain two ways bile helps in the digestion of lipids (fats). [4 marks]
**c)** A student wants to test a food sample for the presence of protein. (i) State the chemical reagent used to test for protein. [1 mark] (ii) Describe the colour change seen if protein is present in the sample. [1 mark] (iii) State the reagent and the condition needed to test for reducing sugars. [2 marks]
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PastPaper.workedSolution
a) X is the stomach, which churns food and adds hydrochloric acid. Y is the pancreas, which produces digestive enzymes. Z is the small intestine, where digestion is completed and nutrients are absorbed. b) (i) Amylase is the carbohydrase that specifically breaks down starch into glucose/maltose. (ii) Bile is produced in the liver and stored in the gall bladder. (iii) Bile has two key roles: neutralising hydrochloric acid from the stomach to provide alkaline conditions for pancreatic enzymes in the small intestine, and emulsifying fats into tiny droplets to increase the surface area for lipase activity. c) (i) Biuret reagent is used to detect peptide bonds in proteins. (ii) A positive result changes from blue to lilac/purple. (iii) To test for reducing sugars (like glucose), Benedict's solution is added, and the tube must be heated (e.g., in a hot water bath at around \(80^{\circ}\text{C}\)).
PastPaper.markingScheme
a) [3 marks] - X: Stomach (1) - Y: Pancreas (1) - Z: Small intestine (1)
b) (i) [1 mark] - Amylase / carbohydrase (1) (ii) [1 mark] - Liver (1) (iii) [4 marks] - (Bile) neutralises acid (from stomach) (1) - To make conditions alkaline / provide optimum pH for lipase (1) - Emulsifies fats/lipids (1) - Breaking them into smaller droplets to increase the surface area (for lipase to work faster) (1)
c) (i) [1 mark] - Biuret (solution/reagent) (1) (ii) [1 mark] - Blue to purple / lilac / violet (1) (iii) [2 marks] - Benedict's (reagent) (1) - Heat / warm water bath (1)
PastPaper.question 4 · Structured
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The heart is a muscular organ that pumps blood around the body.
**a)** Describe the double circulatory system of the human body. Explain where blood is pumped by each side of the heart. [2 marks]
**b)** Figure 3 shows a cross-section of three types of blood vessel, P, Q, and R.
[Figure 3: P has a very thick muscular and elastic wall with a narrow lumen. Q has a thin wall with a large lumen and a valve inside. R is extremely tiny with a wall only one cell thick.]
(i) Identify blood vessels P, Q, and R. [3 marks] (ii) Explain how the structure of blood vessel P is adapted to its function. [2 marks] (iii) What is the function of the valves in vessel Q? [1 mark]
**c)** Coronary heart disease (CHD) occurs when fatty material builds up inside the coronary arteries, reducing blood flow. (i) Explain how CHD can lead to a heart attack. [3 marks] (ii) Name two medical treatments used to treat coronary heart disease. [2 marks] (iii) For one of the treatments named in part (c)(ii), state one advantage and one disadvantage. [2 marks]
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PastPaper.workedSolution
a) The human circulatory system is a double system: the pulmonary circulation (right side to lungs and back) and the systemic circulation (left side to body organs and back). b) (i) Arteries (P) carry blood away from the heart at high pressure. Veins (Q) carry blood back to the heart at low pressure and have valves. Capillaries (R) are microscopic vessels where exchange occurs. (ii) Arterial walls are thick and elastic to cope with high-pressure surges from heartbeats. (iii) Valves prevent low-pressure venous blood from flowing backward. c) (i) Reduced blood flow in coronary arteries restricts oxygen and glucose delivery to cardiac muscle tissue. This prevents aerobic respiration, leading to heart muscle damage/infarction. (ii) Treatments include stents (mechanical widening) and statins (cholesterol-lowering drugs). (iii) Stents keep arteries open but carry surgical risk. Statins are effective but require long-term compliance and have potential side effects.
PastPaper.markingScheme
a) [2 marks] - Right side pumps blood to the lungs (1) - Left side pumps blood to the rest of the body / organs (1)
b) (i) [3 marks] - P: Artery (1) - Q: Vein (1) - R: Capillary (1) (ii) [2 marks] - Thick wall / muscle / elastic fibres (1) - To withstand high blood pressure (1) (iii) [1 mark] - Prevent the backflow of blood (1)
c) (i) [3 marks] - Restricts / narrows the lumen of the coronary artery, reducing blood flow (1) - Less oxygen / glucose delivered to the heart muscle (1) - Heart muscle cells cannot respire / die (1) (ii) [2 marks] - Any two from: Stents, Statins, Coronary bypass surgery, Valve replacement (2) (iii) [2 marks] - One matching advantage (1) and one matching disadvantage (1): - Stent: (Adv) lowers risk of heart attack / immediately effective (1); (Disadv) risk of surgical complications / blood clot (1) - Statin: (Adv) lowers LDL cholesterol / cheap / no surgery (1); (Disadv) must be taken daily / potential side effects e.g. liver damage (1)
PastPaper.question 5 · Structured
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Pathogens are microorganisms that cause communicable (infectious) diseases.
**a)** Draw one line from each disease to the type of pathogen that causes it. [3 marks]
**Diseases:** - Measles - Rose black spot - Salmonella food poisoning
**b)** The human body has several non-specific defense systems to prevent pathogens from entering and causing infection. Explain how each of the following helps defend the body: (i) The skin [1 mark] (ii) The trachea and bronchi [2 marks] (iii) The stomach [1 mark]
**c)** If pathogens enter the body, white blood cells help to destroy them. (i) Describe the process of phagocytosis. [2 marks] (ii) Explain how antibodies help defend the body against pathogens. [3 marks] (iii) Explain how antitoxins help defend the body against pathogens. [2 marks]
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PastPaper.workedSolution
a) Measles is a viral disease. Rose black spot is a fungal disease of plants. Salmonella is caused by bacteria. b) (i) Skin is a continuous physical barrier. (ii) Trachea and bronchi secrete sticky mucus to trap inhaled pathogens; ciliated cells sweep this mucus upwards away from the lungs. (iii) Stomach acid has a very low pH (around 1.5-2.0), which destroys most ingested microbes. c) (i) Phagocytes detect foreign cells, change shape to engulf them (ingestion), and release digestive enzymes to destroy them. (ii) Lymphocytes produce specific proteins called antibodies. These attach to the specific antigens of pathogens, rendering them inactive or tagging them for destruction by phagocytes. (iii) Antitoxins are specialized proteins that specifically neutralize the toxins produced by bacterial cells, preventing cellular damage.
PastPaper.markingScheme
a) [3 marks] - Measles linked to Virus (1) - Rose black spot linked to Fungus (1) - Salmonella food poisoning linked to Bacterium (1)
b) (i) [1 mark] - Physical barrier / secretes antimicrobial substances (1) (ii) [2 marks] - Produce mucus to trap pathogens (1) - Cilia waft / sweep mucus up to throat (to be swallowed) (1) (iii) [1 mark] - Contains acid / hydrochloric acid that kills pathogens (1)
c) (i) [2 marks] - Engulfing / swallowing the pathogen (1) - Digesting / breaking down using enzymes (1) (ii) [3 marks] - White blood cells / lymphocytes produce antibodies (1) - Antibodies bind to specific antigens on pathogens (1) - Pathogens are clumped / inactivated / targeted for destruction (1) (iii) [2 marks] - Bacteria produce toxins (1) - Antitoxins neutralise / counteract the toxins (1)
PastPaper.question 6 · Structured
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Vaccination can protect populations against dangerous communicable diseases.
**a)** Describe how a vaccine works to protect an individual from a specific pathogen. [4 marks]
**b)** A student researched how antibody concentration changes in the blood after vaccination. Figure 4 shows the antibody concentration in a person's blood after their first injection (vaccine) and after a second injection (booster) of the same vaccine.
[Figure 4: A graph showing antibody levels over time. After the first injection, there is a small, slow rise in antibody levels, which then declines. After the second injection, there is a very rapid, much higher rise in antibody levels, which remains high for a long time.]
(i) Describe two differences between the antibody response after the first injection and the response after the second injection. [2 marks] (ii) Explain why the antibody concentration changes so rapidly and reaches a higher level after the second injection. [2 marks]
**c)** Antibiotics are drugs used to treat some diseases. (i) What type of pathogen can be killed by antibiotics? [1 mark] (ii) Why can antibiotics not be used to treat viral infections like flu? [2 marks] (iii) Explain why doctors are concerned about the overuse of antibiotics. [3 marks]
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PastPaper.workedSolution
a) Vaccines stimulate primary immunity by introducing safe antigens. White blood cells produce antibodies and establish immunological memory (memory cells). Upon real infection, secondary response is much faster, stronger, and longer-lasting. b) (i) Comparison: First response is slow and has a low peak. Second response is rapid and has a high peak. (ii) This is due to memory cells remaining in circulation after the first exposure, allowing near-instantaneous synthesis of specific antibodies. c) (i) Antibiotics only target prokaryotes (bacteria). (ii) Viruses are intracellular parasites, shielded inside host eukaryotic cells. (iii) Antibiotic resistance occurs when non-resistant bacteria are killed, leaving resistant mutants to reproduce without competition, creating 'superbugs' like MRSA.
PastPaper.markingScheme
a) [4 marks] - Inject inactive/dead/weakened pathogen (1) - White blood cells / lymphocytes detect antigens (1) - White blood cells produce specific antibodies (1) - Memory cells are formed / if real pathogen returns, antibodies are produced rapidly/in large amounts (1)
b) (i) [2 marks] - Second response is faster / shorter delay (1) - Second response produces a higher concentration of antibodies / lasts longer (1) (ii) [2 marks] - Memory cells (from first injection) are present (1) - They rapidly recognize the antigen and produce antibodies (1)
c) (i) [1 mark] - Bacteria (1) (ii) [2 marks] - Viruses live / replicate inside body cells (1) - Antibiotics cannot destroy them without damaging body cells / tissues (1) (iii) [3 marks] - Resistant bacteria survive (when antibiotics are used) (1) - They reproduce / pass on resistance gene (1) - Eventually the antibiotic will no longer be effective against infections (1)
PastPaper.question 7 · Structured
14.3 PastPaper.marks
Green plants carry out photosynthesis to produce glucose.
**a)** Complete the word equation for photosynthesis. [2 marks] $$\text{carbon dioxide} + \text{\quad(i)\quad} \xrightarrow{\text{light}} \text{\quad(ii)\quad} + \text{oxygen}$$
**b)** Photosynthesis takes place inside plant cells. (i) Name the subcellular structure where photosynthesis occurs. [1 mark] (ii) Name the green pigment inside this structure that absorbs light energy. [1 mark]
**c)** A student set up an experiment to investigate the effect of light intensity on the rate of photosynthesis in pondweed. Figure 5 shows the experimental setup.
[Figure 5: Pondweed placed in a beaker of water with a funnel over it and an inverted test tube to collect gas. A light source is placed at a measured distance from the beaker.]
(i) State how the student could measure the rate of photosynthesis in this experiment. [1 mark] (ii) Suggest one variable that the student must control to make this investigation a fair test. [1 mark] (iii) The student moved the light source to different distances from the pondweed. Explain how the rate of photosynthesis would change as the light source is moved further away from the pondweed. [2 marks]
**d)** Figure 6 shows a graph of the rate of photosynthesis against light intensity.
[Figure 6: A line graph showing rate of photosynthesis on the y-axis and light intensity on the x-axis. The line rises steadily from the origin and then plateaus (flattens out) at higher light intensities.]
(i) Describe the relationship between light intensity and the rate of photosynthesis shown in Figure 6. [2 marks] (ii) Explain why the rate of photosynthesis does not increase any further when the light intensity is very high. Use the term 'limiting factor'. [2 marks] (iii) Name two other factors that can limit the rate of photosynthesis. [2 marks]
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PastPaper.workedSolution
a) Photosynthesis combines carbon dioxide and water using light energy captured by chlorophyll to synthesize glucose and release oxygen. b) (i) Chloroplasts contain the photosynthetic machinery. (ii) Chlorophyll is the light-absorbing pigment. c) (i) Photosynthetic activity releases oxygen bubbles; counting these per unit time provides an estimate of rate. (ii) Control variables like temperature (using a water bath or LED bulb to avoid heating) or carbon dioxide levels (using excess sodium hydrogencarbonate) must remain constant. (iii) Light intensity is inversely proportional to the square of the distance (inverse square law). Increasing distance lowers light intensity, slowing down light-dependent reactions. d) (i) The graph shows a linear, proportional increase initially, which then curves and levels off. (ii) The plateau indicates that light is no longer the factor restricting the reaction rate; another factor (e.g., \(\text{CO}_2\) availability or temperature limits enzyme activity) has become limiting. (iii) Temperature and carbon dioxide are the primary non-light limiting factors.
PastPaper.markingScheme
a) [2 marks] - (i) water (1) - (ii) glucose (1)
b) (i) [1 mark] - Chloroplast (1) (ii) [1 mark] - Chlorophyll (1)
c) (i) [1 mark] - Count bubbles of gas / oxygen per minute (or measure volume of gas) (1) (ii) [1 mark] - Any one from: Temperature, concentration of carbon dioxide / sodium hydrogencarbonate, type of pondweed (1) (iii) [2 marks] - Rate of photosynthesis decreases (1) - Because light intensity decreases (as distance increases) (1)
d) (i) [2 marks] - As light intensity increases, rate of photosynthesis increases (1) - Then the rate plateaus / flattens / stops increasing (1) (ii) [2 marks] - Light intensity is no longer the limiting factor (1) - Another factor (such as temperature or \(\text{CO}_2\) concentration) is limiting the rate (1) (iii) [2 marks] - Carbon dioxide (concentration) (1) - Temperature (1)
Paper 2F
Answer all questions in the spaces provided. Show clear working for calculations.
This question is about hormones and the regulation of blood glucose.
1.1. Identify the organ in the body that monitors and controls blood glucose concentration. [1 mark]
1.2. Compare Type 1 and Type 2 diabetes. In your answer, you should state how each condition is caused and how they are typically treated. [4 marks]
1.3. A student measured their blood glucose level every 15 minutes after drinking a sugary drink. The table below shows the results. - Time (minutes): 0, 15, 30, 45, 60, 75, 90 - Blood glucose concentration (mmol/dm³): 4.5, 6.2, 7.8, 6.5, 5.0, 4.6, 4.5 Describe the pattern shown by the blood glucose data. [3 marks]
1.4. Explain how the hormone insulin works to lower blood glucose concentration when it rises too high. [3 marks]
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PastPaper.workedSolution
1.1. The pancreas monitors blood glucose concentration using sensory cells and secretes hormones to regulate it. 1.2. Type 1 diabetes is an autoimmune condition where the pancreas produces little or no insulin, requiring insulin therapy. Type 2 is lifestyle-linked, where body cells resist insulin, requiring diet, exercise, or oral medication. 1.3. The trend shows an initial steep rise from 0 to 30 minutes (from 4.5 to 7.8 mmol/dm³), followed by a gradual decline from 30 to 90 minutes (from 7.8 to 4.5 mmol/dm³). 1.4. High blood glucose levels trigger the release of insulin. Insulin facilitates the uptake of glucose by body cells (decreasing blood concentration) and stimulates the conversion of glucose to glycogen in liver and muscle cells.
PastPaper.markingScheme
1.1. [1 mark] - Pancreas
1.2. [4 marks] - Type 1 cause: Pancreas does not produce (enough) insulin [1 mark] - Type 1 treatment: Insulin injections [1 mark] - Type 2 cause: Cells are resistant / do not respond to insulin [1 mark] - Type 2 treatment: Low-carbohydrate diet / exercise / weight loss [1 mark]
1.3. [3 marks] - Rises from 4.5 to peak at 7.8 mmol/dm³ [1 mark] - Peak occurs at 30 minutes [1 mark] - Decreases back to original level / 4.5 mmol/dm³ by 90 minutes [1 mark]
1.4. [3 marks] - Insulin is released by the pancreas [1 mark] - Glucose moves from the blood into cells [1 mark] - Excess glucose is converted to glycogen (in liver/muscle) [1 mark]
PastPaper.question 2 · structured
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This question is about the human nervous system and reflex actions.
2.1. Explain why reflex actions are rapid and automatic, and how they protect the body. [2 marks]
2.2. A person accidentally touches a hot pan. Describe the pathway of the electrical impulse through the reflex arc. Name the three neurones involved in the correct order. [3 marks]
2.3. Explain what a synapse is and how an impulse is transmitted across it. [3 marks]
2.4. Two groups of students investigated the effect of caffeine on reaction time using the ruler-drop test. - Group A (no caffeine): mean drop distance of 18.5 cm. - Group B (caffeinated drink): mean drop distance of 14.2 cm. Explain what these results show about the effect of caffeine on reaction time. [3 marks]
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PastPaper.workedSolution
2.1. Reflexes are automatic because they do not involve conscious decision-making by the brain. This minimizes the time taken to respond to danger, preventing tissue damage. 2.2. The stimulus (heat) is detected by temperature receptors in the skin. Electrical impulses travel along the sensory neurone to the central nervous system (spinal cord), cross to a relay neurone, then travel down a motor neurone to the effector (muscle) to cause a contraction. 2.3. Electrical impulses cannot cross the physical gap of a synapse. The impulse causes chemical molecules to diffuse across the synaptic cleft, translating the electrical signal into a chemical one and back to electrical. 2.4. Drop distance is directly proportional to reaction time. A lower mean drop distance (14.2 cm vs 18.5 cm) indicates a faster reaction time, proving caffeine increases nervous system alertness.
PastPaper.markingScheme
2.1. [2 marks] - Does not involve conscious parts of the brain / is automatic [1 mark] - Prevents or minimizes damage/harm to the body [1 mark]
2.2. [3 marks] - Sensory neurone [1 mark] - Relay neurone [1 mark] - Motor neurone [1 mark] (Must be in correct order for full marks; allow receptor -> effector context)
2.3. [3 marks] - Synapse is a gap/junction between neurones [1 mark] - Chemical / neurotransmitter released [1 mark] - Diffuses across the gap to trigger electrical impulse in next neurone [1 mark]
2.4. [3 marks] - Group B caught the ruler quicker / shorter distance [1 mark] - Caffeine decreases/speeds up reaction time [1 mark] - Caffeine acts as a stimulant [1 mark]
PastPaper.question 3 · structured
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This question is about how organisms are adapted to survive in their environment.
3.1. Plants in a woodland ecosystem compete with each other for resources. State three resources that plants compete for. [3 marks]
3.2. Explain how having a deep and widespread root system helps a desert plant to survive. [2 marks]
3.3. Describe two structural adaptations of animals living in very cold environments, and explain how each adaptation helps them survive. [4 marks]
3.4. Some microorganisms can survive in extreme environments, such as deep-sea hydrothermal vents. State the term used to describe these organisms and give one example of an extreme environmental condition they survive in. [2 marks]
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PastPaper.workedSolution
3.1. Plants do not compete for food (they make their own). They compete for light (for photosynthesis), water (for photosynthesis/structure), minerals (for healthy growth), and space. 3.2. Desert soils are dry. Deep roots reach deep water sources; widespread roots maximize water collection from rare rainfall events before it evaporates. 3.3. Structural adaptations refer to physical body features. Thick fur/blubber insulates against cold climates. Small ears reduce surface area, reducing heat loss. 3.4. Extremophiles are adapted to live in conditions that would denature proteins in most organisms, such as temperatures over 80 °C or high salt concentrations.
PastPaper.markingScheme
3.1. [3 marks] - Any three from: Light, Water, Minerals / Ions / Nutrients, Space [3 marks] (Reject: Food)
3.2. [2 marks] - Deep roots reach water deep underground [1 mark] - Widespread roots absorb water over a large area / catch rain quickly [1 mark]
3.3. [4 marks] - Adaptation 1: Thick fur / layer of fat / blubber [1 mark] - Explanation 1: Insulates body / reduces heat loss [1 mark] - Adaptation 2: Low surface area to volume ratio / small ears / round body shape [1 mark] - Explanation 2: Minimizes heat loss to environment [1 mark]
3.4. [2 marks] - Extremophile(s) [1 mark] - High temperature / high pressure / high salinity / extreme pH [1 mark]
PastPaper.question 4 · structured
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This question is about decay and the recycling of materials in an ecosystem.
4.1. Name the process by which green plants remove carbon dioxide from the atmosphere. [1 mark]
4.2. State two ways in which carbon is returned to the atmosphere as carbon dioxide by living organisms. [2 marks]
4.3. Microorganisms are decomposers. Explain how microorganisms recycle carbon and mineral ions from dead plant material back into the environment. [4 marks]
4.4. Gardeners build compost heaps to decay organic waste. - Name three environmental conditions that are required for rapid decay in a compost heap. - Explain why one of these conditions is necessary for decomposers. [4 marks]
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PastPaper.workedSolution
4.1. Photosynthesis is the only biological process that absorbs CO2 from the atmosphere: \(6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2\). 4.2. Carbon is released as CO2 via aerobic respiration in plants, animals, and decomposers. Combustion of wood/fossil fuels also releases CO2, but the question asks for 'by living organisms'. 4.3. Decomposers break down large complex molecules (like cellulose) extracellularly. They absorb the glucose and respire it, producing CO2. Inorganic mineral ions like nitrates and phosphates are returned to the soil. 4.4. Decay is biological. High temperature (warmth) increases enzyme activity (but too hot will denature them). Water/moisture keeps cells hydrated. Oxygen allows aerobic respiration.
PastPaper.markingScheme
4.1. [1 mark] - Photosynthesis
4.2. [2 marks] - Respiration by plants / animals [1 mark] - Respiration by decomposers / microorganisms [1 mark]
4.3. [4 marks] - Microorganisms / decomposers digest/break down organic matter [1 mark] - They secrete enzymes [1 mark] - They respire and release carbon dioxide (into air) [1 mark] - They release mineral ions / nutrients into the soil [1 mark]
4.4. [4 marks] - Warmth, oxygen, moisture [3 marks, 1 mark each] - Explanation: Oxygen needed for aerobic respiration OR warmth needed for optimal enzyme activity OR moisture needed to dissolve substances / prevent drying out [1 mark]
PastPaper.question 5 · structured
11.1 PastPaper.marks
This question is about reproduction and cell division.
5.1. Name the male gamete and the female gamete in flowering plants. [2 marks]
5.2. Compare sexual and asexual reproduction. You must include references to: - the number of parents involved - whether fusion of gametes occurs - the genetic variation in the offspring. [4 marks]
5.3. Name the type of cell division that produces gametes in humans, and state where this occurs in males and in females. [3 marks]
5.4. Explain why all offspring produced by asexual reproduction are genetically identical to the parent. [2 marks]
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PastPaper.workedSolution
5.1. In flowering plants, pollen grains are the male gametes, and the egg cells (found inside ovules) are the female gametes. 5.2. Sexual reproduction uses meiosis to create haploid gametes that fuse to create genetic variation. Asexual reproduction uses mitosis to produce diploid offspring that are exact genetic clones of the single parent. 5.3. Meiosis halves the chromosome number from diploid (46) to haploid (23) to produce gametes. This happens in the gonads: testes in males and ovaries in females. 5.4. Because there is only one parent, there is no fertilization or fusion of gametes. Cell division is strictly mitotic, producing genetically identical clones.
5.4. [2 marks] - Produced by mitosis [1 mark] - No mixing of genetic material / only one parent's genes [1 mark]
PastPaper.question 6 · structured
11.1 PastPaper.marks
This question is about selective breeding and genetic inheritance.
6.1. Explain the steps a farmer would take to selectively breed cows that produce a high yield of milk. [4 marks]
6.2. Describe two disadvantages of selective breeding in animals. [2 marks]
6.3. Polydactyly is an inherited disorder that causes a person to have extra fingers or toes. It is caused by a dominant allele (D). - A male parent is heterozygous for polydactyly (Dd). - The female parent is homozygous recessive (dd) and does not have the disorder. Complete a genetic cross (Punnett square) to show the possible genotypes of their children. [3 marks] State the probability (as a percentage) that their children will have polydactyly. [2 marks]
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6.1. Selective breeding is an artificial process where humans select individuals with desired traits to breed over multiple generations to accentuate that characteristic. 6.2. Disadvantages include inbreeding, which narrows the gene pool. This increases the likelihood of inheriting recessive genetic disorders and makes the entire population vulnerable to being wiped out by a single disease outbreak. 6.3. The father has genotype Dd (gametes D and d), and the mother has genotype dd (gametes d and d). The genetic diagram is: | | d | d | |---|---|---| | **D** | Dd | Dd | | **d** | dd | dd | Offspring genotypes: 2 Dd : 2 dd. Since D is dominant, the Dd children will have polydactyly (2 out of 4), which is 50%.
PastPaper.markingScheme
6.1. [4 marks] - Select cows with highest milk yield and bulls from high-yielding mothers [1 mark] - Breed them together [1 mark] - Select offspring with the highest milk yield and breed them [1 mark] - Repeat over many generations [1 mark]
6.2. [2 marks] - Any two from: - Inbreeding / reduced gene pool [1 mark] - More susceptible to diseases [1 mark] - Increased risk of genetic defects [1 mark]
6.3. [5 marks] - Correct maternal gametes (d, d) and paternal gametes (D, d) [1 mark] - Correct offspring genotypes in Punnett square: Dd and dd [2 marks] - Probability = 50% / 1 in 2 / 0.5 [2 marks] (Allow 1 mark if correct ratio 1:1 or 2:2 is given but percentage is missing)
PastPaper.question 7 · structured
11.1 PastPaper.marks
This question is about food security and efficient food production.
7.1. Define the term 'food security'. [2 marks]
7.2. Many livestock animals, such as chickens, are kept in warm, indoor, crowded cages (intensive farming) rather than being free-range. - Explain, in terms of energy transfer, how keeping animals indoors in a warm environment increases the efficiency of food production. [4 marks] - State one ethical objection to this intensive farming method. [1 mark]
7.3. Explain why eating crops directly is more energy-efficient for human populations than feeding the crops to livestock and then eating the livestock. [2 marks]
7.4. The fungus Fusarium is grown in industrial fermenters to produce mycoprotein. - Name the food source added to the fermenter to allow Fusarium to grow. - State why air (oxygen) is bubbled through the fermenter. [2 marks]
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7.1. Food security means ensuring that all people have access to safe, nutritious, and sufficient food at all times. 7.2. Maintaining a high ambient temperature minimizes heat transfer from the animal to its surroundings, reducing the rate of respiration required for thermoregulation. Restricting space prevents energy loss through muscle movement. More energy is converted to meat/eggs/biomass. 7.3. Approximately 90% of energy is lost at each trophic level. By eating plants (producers) directly, humans capture maximum energy directly. Introducing a primary consumer (livestock) wastes most of the plant energy. 7.4. Fusarium is grown on glucose syrup under aerobic conditions inside industrial fermenters. Oxygen is bubbled through to keep the mixture aerobic, preventing anaerobic respiration which produces less energy and yields toxic by-products.
PastPaper.markingScheme
7.1. [2 marks] - Having enough food [1 mark] - To feed a population / for everyone [1 mark]
7.2. [5 marks] - Warm environment reduces heat loss / energy spent on maintaining body temperature [1 mark] - Restricted movement reduces energy used in respiration/muscle contraction [1 mark] - More energy goes towards growth / biomass [1 mark] - Increases the efficiency of food production [1 mark] - Ethical objection: Cruelty to animals / restricted movement / unnatural lifestyle / rapid disease spread [1 mark]
7.3. [2 marks] - Energy is lost at each trophic level (e.g., via respiration / faeces) [1 mark] - Direct consumption has fewer stages / shorter food chain, so less energy is lost overall [1 mark]
7.4. [2 marks] - Glucose (syrup) [1 mark] - (Oxygen is needed) for aerobic respiration [1 mark]
PastPaper.question 8 · structured
11.1 PastPaper.marks
This question is about human impact on biodiversity and the environment.
8.1. Deforestation has occurred on a large scale in tropical regions. Explain how deforestation increases the level of carbon dioxide in the atmosphere. [3 marks]
8.2. Peat bogs are being destroyed to make compost for gardeners. Describe how the destruction of peat bogs affects the atmosphere and biodiversity. [3 marks]
8.3. Global warming is a major threat to global biodiversity. Describe three consequences of global warming on ecosystems. [3 marks]
8.4. Give two positive actions humans can take to maintain or restore biodiversity in an area. [2 marks]
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8.1. Forests are massive carbon sinks. Removing trees reduces photosynthesis, which normally absorbs CO2. If trees are burned or left to rot, combustion and respiration by decomposers release stored carbon back into the atmosphere. 8.2. Peat is partially decayed organic matter. When bogs are drained and dried, oxygen enters, allowing aerobic decomposers to decay the peat and release CO2. Burning peat also releases CO2. Peat bogs are unique biodiverse environments; destroying them drives specialized species to extinction. 8.3. Global warming increases surface temperatures, melting glaciers, causing thermal expansion of oceans, flooding habitats. Warm temperatures force species to migrate to cooler areas, or go extinct if they cannot adapt. 8.4. Humans can implement conservation programs, reduce deforestation, recycle, establish national parks, and plant hedgerows on farms to create green corridors.
PastPaper.markingScheme
8.1. [3 marks] - Less photosynthesis to lock up / absorb carbon dioxide [1 mark] - Combustion / burning of trees releases carbon dioxide [1 mark] - Microorganisms / decomposers decay wood, releasing carbon dioxide during respiration [1 mark]
8.2. [3 marks] - Decay / burning of peat releases carbon dioxide into the atmosphere [1 mark] - (Carbon dioxide is a) greenhouse gas contributing to global warming [1 mark] - Reduces biodiversity of unique bog plants/animals [1 mark]
8.3. [3 marks] - Any three from: - Loss of habitats due to flooding / rising sea levels [1 mark] - Changes in migration patterns [1 mark] - Extinction of species / loss of biodiversity [1 mark] - Extreme weather / drought / desertification [1 mark] - Change in distribution of species [1 mark]
8.4. [2 marks] - Any two from: - Breeding programmes for endangered species [1 mark] - Protection / regeneration of rare habitats [1 mark] - Replanting hedgerows / field margins [1 mark] - Reducing deforestation / carbon emissions [1 mark]
PastPaper.question 9 · Structured
11 PastPaper.marks
A student investigated the effect of background noise on reaction time using the ruler drop test.
In the test, one student drops a vertical ruler and another student catches it between their thumb and forefinger as quickly as possible. The distance the ruler falls before being caught is measured.
**(a)** (i) Name the organ that contains the receptor cells that detect the falling ruler. [1 mark] (ii) Name the effector that is involved in catching the ruler. [1 mark]
**(b)** Identify two variables that the students must keep the same (control variables) to ensure the results are valid. [2 marks]
**(c)** Table 1 shows the distance the ruler fell before being caught for one participant when there was no background noise (quiet).
(i) Identify the anomalous result in Table 1. [1 mark] (ii) Calculate the mean distance the ruler fell, ignoring the anomalous result. Show your working. [2 marks]
**(d)** Describe the pathway of a nerve impulse from when the participant sees the ruler drop to when they catch it. In your answer, use the following terms: - sensory neurone - motor neurone - central nervous system (CNS) - receptor - effector [4 marks]
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**Part (a)** (i) The eye contains the photoreceptors (retina) that detect light and movement of the falling ruler. (ii) The effector is a muscle in the hand, finger, or arm that contracts to catch the ruler.
**Part (b)** Variables that must be kept constant include: - Using the same hand (e.g., dominant hand) for catching. - Positioning the thumb and finger at the same initial point on the ruler (e.g., the 0 cm mark). - Keeping the distance between the thumb and finger constant before dropping the ruler. - Using the same ruler throughout the experiment. - Ensuring the same person drops the ruler each time. - Dropping the ruler with no prior warning (no countdown).
**Part (c)** (i) The anomalous result is 29 cm (Trial 3), as it is much higher than the other results (which are clustered closely between 13 cm and 15 cm). (ii) To find the mean while ignoring the anomalous result: - Sum the remaining values: \(15 + 14 + 13 + 14 = 56\) - Divide by the number of valid trials (4): \(56 \div 4 = 14\) cm.
**Part (d)** - The receptor in the eye detects the stimulus (the movement of the falling ruler) and generates an electrical impulse. - This impulse travels along a sensory neurone to the central nervous system (CNS), specifically the brain. - The CNS processes this sensory information and sends an electrical impulse down a motor neurone. - The impulse reaches the effector, which is the muscle in the hand/arm, causing it to contract and catch the ruler.
PastPaper.markingScheme
**Part (a)** - **(i)** [1 mark] Eye / retina (Accept: photoreceptors in the eye) - **(ii)** [1 mark] Muscle (in the hand/arm/finger) (Accept: effector muscle)
**Part (b)** [2 marks] Any two control variables from the following (1 mark per correct variable): - Same hand used (e.g., dominant hand) - Same starting height of finger and thumb (e.g., at the 0 cm mark) - Same initial distance/width between thumb and finger - Same ruler used - Same person dropping / catching the ruler - No verbal or visual warning given before dropping the ruler *Do not accept 'same background noise' as this is the independent variable being changed.*
**Part (c)** - **(i)** [1 mark] 29 (cm) / Trial 3 - **(ii)** [2 marks] - **1 mark** for showing correct working: \(15 + 14 + 13 + 14 = 56\) OR \(56 \div 4\) - **1 mark** for the correct mean: 14 (cm) (or 14) *Allow 1 mark for the correct mean of 14 even if no working is shown.*
**Part (d)** [4 marks] Points to look for: - **Mark 1:** Receptor (in eye/retina) detects stimulus (movement) and generates an electrical impulse. - **Mark 2:** Electrical impulse travels along a sensory neurone to the central nervous system (CNS) / brain. - **Mark 3:** The CNS passes the impulse to a motor neurone. - **Mark 4:** Electrical impulse travels along the motor neurone to the effector / muscle, causing it to contract (and catch the ruler). *Note: To achieve all 4 marks, the process must be described in the correct logical sequence using all 5 required terms.*