An original Thinka practice paper modelled on the structure and difficulty of the Specimen 2018 Oxford AQA International GCSE Biology (9201) paper. Not affiliated with or reproduced from Oxford.
卷一 (Core Concepts, Physiology and Evolution)
Answer all questions in the spaces provided. You must have a ruler and a calculator.
9 題目 · 90 分
題目 1 · structured
10 分
Figure 1 shows a multicellular green alga (*Volvox*) cell and a human cheek cell.
**Figure 1**
*Volvox* cell: contains a cell wall, chloroplasts, mitochondria, a nucleus, and ribosomes.
Human cheek cell: contains a cell membrane, cytoplasm, a nucleus, mitochondria, and ribosomes.
**1.1** State two structures present in the *Volvox* cell that are not present in the human cheek cell. [2 marks]
**1.2** A student observes a cheek cell under a microscope. The real width of the cheek cell is \(60\ \mu\text{m}\). The image width measured with a ruler is \(24\ \text{mm}\).
Calculate the magnification used. Show your working. [3 marks]
**1.3** Human cheek cells take in glucose from the surrounding tissue fluid.
Explain how glucose can enter these cells when the concentration of glucose is higher inside the cell than outside. [3 marks]
**1.4** The lining of the human trachea contains ciliated epithelial cells.
Describe how these cells are adapted to their function. [2 marks]
**1.3** - Glucose enters by active transport (1 mark) - Which moves substances against a concentration gradient / from a region of low concentration to high concentration (1 mark) - This process requires energy released during respiration (1 mark)
**1.4** - They have microscopic hair-like projections called cilia (1 mark) - Which sweep mucus containing trapped dust/pathogens away from the lungs / up the trachea (1 mark)
評分準則
**1.1** [2 marks] - 1 mark for cell wall - 1 mark for chloroplast / vacuole
**1.2** [3 marks] - 1 mark for correct conversion of mm to \(\mu\text{m}\) - 1 mark for correct substitution into magnification formula - 1 mark for correct final answer of \(400\) or \(\times 400\) (Award 3 marks for correct final answer with no working shown)
**1.3** [3 marks] - 1 mark for identifying active transport - 1 mark for stating movement against the concentration gradient - 1 mark for stating that energy is required from respiration
**1.4** [2 marks] - 1 mark for stating they have cilia - 1 mark for describing the movement of mucus away from lungs
題目 2 · structured
10 分
A student investigated the rate of water loss from a leafy shoot using a potometer. The potometer measures the distance moved by an air bubble in a capillary tube over a period of 10 minutes.
**2.1** Name the tissue that transports water up the stem of a plant. [1 mark]
**2.2** In one trial, the air bubble moved a distance of \(45\ \text{mm}\) in 10 minutes. The internal cross-sectional area of the capillary tube was \(0.8\ \text{mm}^2\).
Calculate the rate of water uptake by the shoot in \(\text{mm}^3\) per minute.
Give your answer to 2 significant figures. Show your working. [3 marks]
**2.3** The student repeated the experiment under three different environmental conditions: - Condition A: Still air, \(20\ ^\circ\text{C}\) - Condition B: Moving air (using a fan), \(20\ ^\circ\text{C}\) - Condition C: Still air, \(30\ ^\circ\text{C}\)
Predict and explain how the rate of water loss in Condition B and Condition C would compare to Condition A. [4 marks]
**2.4** Explain how root hair cells are adapted for efficient water absorption. [2 marks]
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解題
**2.1** - Xylem
**2.2** - Volume of water taken up = \(\text{Distance} \times \text{Area} = 45 \times 0.8 = 36\ \text{mm}^3\) (1 mark) - Rate of water uptake = \(\frac{36}{10} = 3.6\ \text{mm}^3/\text{min}\) (1 mark) - Final answer to 2 sig figs: \(3.6\) (1 mark)
**2.3** - Condition B (moving air): Rate is higher than A because wind removes accumulated water vapour from around the leaf, maintaining a steep concentration/diffusion gradient (2 marks) - Condition C (higher temperature): Rate is higher than A because warmer water molecules have more kinetic energy, increasing the rate of evaporation from cell surfaces (2 marks)
**2.4** - They have long, hair-like extensions which provide a very large surface area (1 mark) - This increases the rate of water absorption by osmosis (1 mark)
評分準則
**2.1** [1 mark] - 1 mark for xylem
**2.2** [3 marks] - 1 mark for calculating volume of \(36\ \text{mm}^3\) - 1 mark for dividing volume by 10 minutes - 1 mark for correct rounding to 2 significant figures (Award 3 marks for correct final answer of 3.6 with no working shown)
**2.3** [4 marks] - 1 mark for predicting rate increases in B - 1 mark for explaining B in terms of wind removing water vapour / maintaining steep concentration gradient - 1 mark for predicting rate increases in C - 1 mark for explaining C in terms of higher temperature increasing kinetic energy / evaporation rate
**2.4** [2 marks] - 1 mark for referencing a large surface area - 1 mark for link to increased osmosis / water uptake
題目 3 · structured
10 分
Respiration is a cellular chemical process that releases energy. Muscle cells respire anaerobically during intense physical exercise.
**3.1** Write the word equation for anaerobic respiration in human muscle cells. [1 mark]
**3.2** After a period of intense exercise, an athlete continues to breathe deeply and rapidly for several minutes.
Explain why. [4 marks]
**3.3** Explain how the human circulatory system responds to exercise to ensure adequate delivery of oxygen to the working muscles. [5 marks]
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解題
**3.1** - \(\text{Glucose} \rightarrow \text{Lactic acid}\) (1 mark) (Ignore references to energy; reject if oxygen or carbon dioxide is included)
**3.2** - Anaerobic respiration produces lactic acid (1 mark) - Lactic acid is toxic and causes muscle fatigue/pain (1 mark) - Extra oxygen is needed to break down/oxidise lactic acid in the liver to carbon dioxide and water (1 mark) - This extra oxygen demand is called the oxygen debt (1 mark)
**3.3** - Heart rate increases to pump blood more quickly (1 mark) - Stroke volume increases / heart beats with more force so more blood is ejected per beat (1 mark) - Blood vessels supplying the working muscles dilate (vasodilation) to allow more blood to flow to them (1 mark) - Blood vessels supplying non-essential organs (like the digestive system) constrict (1 mark) - This overall response increases the delivery rate of oxygen and glucose to respiring muscle cells (1 mark)
評分準則
**3.1** [1 mark] - 1 mark for the correct word equation: Glucose \(\rightarrow\) Lactic acid (do not accept if oxygen or carbon dioxide are reactant/product)
**3.2** [4 marks] - 1 mark for identifying lactic acid is produced anaerobically - 1 mark for stating lactic acid causes fatigue/pain / needs removal - 1 mark for explaining oxygen is needed to break down lactic acid (into CO2 and H2O / in liver) - 1 mark for identifying this as the oxygen debt
**3.3** [5 marks] - 1 mark for increased heart rate - 1 mark for increased stroke volume - 1 mark for vasodilation/widening of arteries supplying muscles - 1 mark for vasoconstriction of arteries supplying non-active organs - 1 mark for linking these changes to increased rate of oxygen/glucose delivery to respiring muscles
題目 4 · structured
10 分
Coordination in the human body is achieved through both nervous and hormonal systems.
**4.1** Describe the pathway of a nerve impulse during a reflex action when a person accidentally touches a sharp object. Name the components of the reflex arc in your answer. [4 marks]
**4.2** A student investigated the effect of caffeine on human reaction time using the ruler drop test. In each trial, the student caught a falling ruler between their thumb and forefinger. The drop distance was recorded.
Table 1 shows the results for five trials before and after drinking a caffeinated beverage.
Identify any anomalous results in the 'before caffeine' data and calculate the mean drop distance for the remaining valid trials before caffeine. [2 marks]
**4.3** Explain the effect of caffeine on the student's reaction time based on the data in Table 1. [2 marks]
**4.4** State two differences between the nervous system and the endocrine (hormonal) system. [2 marks]
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解題
**4.1** - Receptor detects pain/stimulus from the sharp object (1 mark) - Electrical impulse travels along a sensory neurone to the central nervous system / spinal cord (1 mark) - Impulse crosses a synapse to a relay neurone, and then across another synapse to a motor neurone (1 mark) - Impulse travels along the motor neurone to the effector / muscle, causing it to contract and pull the hand away (1 mark)
**4.2** - Anomalous result is Trial 4 (24.5 cm) (1 mark) - Mean of remaining valid trials: \(\frac{18.5 + 19.2 + 17.8 + 18.0}{4} = 18.375\ \text{cm}\) (accept \(18.4\)) (1 mark)
**4.3** - Caffeine decreases the reaction time / makes response faster (1 mark) - Because the mean drop distance decreased from \(18.4\ \text{cm}\) to \(14.2\ \text{cm}\) after drinking caffeine, meaning the ruler fell less distance before being caught (1 mark)
**4.4** - Nervous responses are faster, while endocrine responses are slower (1 mark) - Nervous responses use electrical impulses (via neurones), while endocrine responses use chemical hormones (via the bloodstream) (1 mark) - Nervous effects are short-lived, while endocrine effects are longer-lasting (1 mark) - Nervous signals target a highly localized area, while hormones act more widely (1 mark) *(Any 2 differences for 2 marks)*
評分準則
**4.1** [4 marks] - 1 mark for receptor detecting stimulus - 1 mark for sensory neurone transmitting impulse to CNS/spinal cord - 1 mark for relay neurone and synapses - 1 mark for motor neurone and effector (muscle) contracting
**4.2** [2 marks] - 1 mark for identifying Trial 4 (24.5 cm) as the anomalous result - 1 mark for calculating the correct mean of the remaining trials (18.375 or 18.4)
**4.3** [2 marks] - 1 mark for stating that caffeine decreases reaction time / makes reactions faster - 1 mark for supporting this with the decrease in drop distance
**4.4** [2 marks] - 1 mark per valid difference up to 2 marks (comparison must be explicit, e.g., 'electrical impulses vs chemical messengers')
題目 5 · structured
10 分
The human body maintains a constant internal environment through homeostasis.
**5.1** Explain how the pancreas and the liver cooperate to lower blood glucose concentration when it becomes too high. [4 marks]
**5.2** Thermoregulation is the control of core body temperature. During a marathon on a hot day, an athlete's body temperature starts to rise.
Describe how the skin responds to transfer excess heat away from the body. [4 marks]
**5.3** Explain why shivering can help to raise body temperature if an athlete suddenly becomes too cold. [2 marks]
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解題
**5.1** - The increase in blood glucose is detected by the pancreas (1 mark) - The pancreas secretes the hormone insulin into the blood (1 mark) - Insulin causes glucose to move from the blood into liver (and muscle) cells (1 mark) - In the liver, glucose is converted into insoluble glycogen for storage, lowering the blood glucose concentration (1 mark)
**5.2** - Sweat glands produce more sweat which evaporates from the skin surface (1 mark) - Evaporation of sweat transfers heat energy away from the skin, cooling the body (1 mark) - Blood vessels (arterioles) supplying skin capillaries dilate / vasodilation occurs (1 mark) - This allows more blood to flow close to the skin surface, increasing heat loss by radiation (1 mark) (Reject 'capillaries move closer to skin')
**5.3** - Shivering causes rapid, involuntary contraction of muscles (1 mark) - Muscle contraction requires energy from aerobic respiration, which releases heat as a waste product to warm the body (1 mark)
評分準則
**5.1** [4 marks] - 1 mark for high glucose detected by pancreas - 1 mark for pancreas secreting insulin - 1 mark for insulin causing liver/muscle cells to absorb glucose - 1 mark for liver converting glucose to glycogen
**5.2** [4 marks] - 1 mark for sweat production - 1 mark for heat loss via sweat evaporation - 1 mark for vasodilation / widening of arterioles supplying skin capillaries - 1 mark for increased heat loss by radiation (Do not accept 'capillaries move/expand')
**5.3** [2 marks] - 1 mark for describing muscle contractions - 1 mark for linking muscle contraction to increased respiration releasing heat
題目 6 · structured
10 分
Pathogens are microorganisms that cause infectious diseases.
**6.1** Explain how vaccination leads to immunization against a specific pathogen without causing the disease. [4 marks]
**6.2** Antibiotics are used to treat bacterial infections. Explain why antibiotics are ineffective against viral diseases. [2 marks]
**6.3** Before a new drug is approved for public use, it must undergo clinical trials. Explain why clinical trials start with very low doses of the drug on healthy volunteers, rather than patients. [2 marks]
**6.4** What is a double-blind trial, and why is it used in clinical drug testing? [2 marks]
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解題
**6.1** - A vaccine contains a dead, inactive, or weakened form of the pathogen (1 mark) - This introduces the pathogen's antigens to the body, stimulating white blood cells (lymphocytes) (1 mark) - Lymphocytes produce specific antibodies to destroy the vaccine antigens (1 mark) - Some lymphocytes remain as memory cells; if the live pathogen enters the body in the future, memory cells produce antibodies much faster and in larger quantities to prevent illness (1 mark)
**6.2** - Viruses live and reproduce inside host cells (1 mark) - Antibiotics cannot target viruses without damaging the body's host cells / viruses lack the cell walls or metabolic machinery that antibiotics affect (1 mark)
**6.3** - Healthy volunteers are used to check for safety and find any side effects (1 mark) - Very low doses are used first to ensure the drug is not highly toxic before increasing the dosage (1 mark)
**6.4** - In a double-blind trial, neither the volunteers nor the doctors/researchers know who is receiving the active drug or the placebo (1 mark) - This is used to prevent bias when reporting or interpreting the results of the trial (1 mark)
評分準則
**6.1** [4 marks] - 1 mark for vaccine containing dead/inactive/weakened pathogens - 1 mark for antigens stimulating antibody production - 1 mark for white blood cells/lymphocytes producing antibodies - 1 mark for memory cells providing rapid future response
**6.2** [2 marks] - 1 mark for stating viruses live inside host cells - 1 mark for noting antibiotics cannot destroy viruses without damaging host cells / lack of target structures
**6.3** [2 marks] - 1 mark for testing safety / side effects on healthy individuals - 1 mark for stating that low doses protect against unexpected toxicity
**6.4** [2 marks] - 1 mark for stating neither doctor nor patient knows who got the placebo - 1 mark for linking this to the elimination of bias
題目 7 · structured
10 分
Decomposers play a vital role in recycling nutrients in ecosystems through decay.
**7.1** State two optimal environmental conditions required for rapid decay of organic matter in a compost bin. Explain your answer. [4 marks]
**7.2** Figure 2 shows a simplified carbon cycle.
**Figure 2**
- Process A: Atmospheric Carbon Dioxide \(\rightarrow\) Green Plants - Process B: Green Plants \(\rightarrow\) Atmospheric Carbon Dioxide - Process C: Animals \(\rightarrow\) Atmospheric Carbon Dioxide - Process D: Fossil Fuels \(\rightarrow\) Atmospheric Carbon Dioxide
Identify the biological/chemical processes A, B, C, and D. [2 marks]
**7.3** In a stable woodland ecosystem, the total biomass of the trees is \(150,000\ \text{kg}\). The total biomass of the herbivorous insects that feed on these trees is \(12,000\ \text{kg}\).
Calculate the percentage efficiency of biomass transfer from the trees to the herbivorous insects. Show your working. [2 marks]
**7.4** Suggest two reasons why some of the biomass from the trees is not transferred to the herbivorous insects. [2 marks]
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解題
**7.1** - Warm temperature (1 mark): increases the rate of enzyme-controlled reactions in decomposers (1 mark) - Moist/wet conditions (1 mark): water is required by decomposers for metabolic reactions and to dissolve nutrients for absorption (1 mark) - (Alternative: Oxygen (1 mark): required for aerobic respiration of decomposers (1 mark))
**7.2** - A = Photosynthesis, B = Plant Respiration, C = Animal Respiration, D = Combustion *(All 4 correct = 2 marks; 2 or 3 correct = 1 mark; 0 or 1 correct = 0 marks)*
**7.4** - Not all parts of the trees are eaten by the insects (e.g., woody bark, roots) (1 mark) - Some biomass is lost as carbon dioxide and water during the plant's own respiration (1 mark) - Some biomass falls off as dead leaves and decays instead of being consumed (1 mark) *(Any 2 reasons for 2 marks)*
評分準則
**7.1** [4 marks] - 1 mark for naming condition 1, 1 mark for explanation - 1 mark for naming condition 2, 1 mark for explanation *(Accept warm temperature, moisture, or oxygen)*
**7.2** [2 marks] - 2 marks for all 4 processes identified correctly - 1 mark if only 2 or 3 processes are identified correctly
**7.3** [2 marks] - 1 mark for correct working shown - 1 mark for correct final answer of 8% (Award 2 marks for correct final answer with no working shown)
**7.4** [2 marks] - 1 mark per valid reason up to 2 marks (e.g., inedible parts, loss to decomposers, plant respiration)
題目 8 · structured
10 分
Organisms adapt to survive in their environment over millions of years.
**8.1** In a population of bacteria, some individuals possess a mutation that makes them resistant to a widely used antibiotic.
Describe how natural selection can lead to the entire population of bacteria becoming resistant to this antibiotic over time. [4 marks]
**8.2** Jean-Baptiste Lamarck proposed a different theory of evolution based on the inheritance of acquired characteristics.
Explain why Lamarck's theory is no longer accepted by modern scientists. [2 marks]
**8.3** In a forest ecosystem, two different species of bird live in the same trees and feed on insects: - Species X has a long, thin beak to reach insects deep within the tree bark. - Species Y has a short, thick beak to crush beetles on the bark surface.
Explain how these different beak shapes reduce competition between the two species. [2 marks]
**8.4** Define the term 'biodiversity' and state one reason why high biodiversity is important for a stable ecosystem. [2 marks]
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解題
**8.1** - Treatment with the antibiotic acts as a selection pressure, killing the non-resistant bacteria (1 mark) - The mutated, resistant bacteria survive the treatment (1 mark) - The surviving resistant bacteria reproduce rapidly (by binary fission) (1 mark) - They pass on the gene/allele for antibiotic resistance to their offspring, increasing the proportion of resistant bacteria in the population (1 mark)
**8.2** - We now know that characteristics acquired during an organism's life do not change their genes/DNA (1 mark) - Therefore, acquired characteristics cannot be passed on to offspring via gametes (1 mark)
**8.3** - The beak shapes adapt each species to feed on different types of insects or in different parts of the tree (different niches) (1 mark) - This reduces direct competition for the same food resource between Species X and Species Y (1 mark)
**8.4** - Biodiversity is the variety of different species of organisms in a particular habitat or ecosystem (1 mark) - High biodiversity increases stability because species are less dependent on a single other species for food or shelter (1 mark)
評分準則
**8.1** [4 marks] - 1 mark for antibiotic acting as a selection pressure / killing non-resistant strain - 1 mark for survival of mutated/resistant bacteria - 1 mark for reproduction of surviving bacteria - 1 mark for passing on the resistance gene/allele to offspring
**8.2** [2 marks] - 1 mark for noting acquired traits do not affect genes/DNA - 1 mark for stating that only genetic information can be inherited
**8.3** [2 marks] - 1 mark for identifying different food niches / different insects eaten - 1 mark for linking this to reduced direct competition
**8.4** [2 marks] - 1 mark for defining biodiversity as the variety of species - 1 mark for explaining stability in terms of reduced interdependence/reliance on a single food source
題目 9 · Short Answer & Structured
10 分
A student investigated the absorption of nitrate ions by the roots of barley plants. They measured the concentration of nitrate ions inside the root hair cells of plants grown in aerated soil (containing plenty of oxygen) and in waterlogged soil (containing very little oxygen). The concentration of nitrate ions in the soil water surrounding both sets of roots was \(0.15\text{ mmol/dm}^3\). Table 1 shows the results.
**01.1** State the process by which nitrate ions are normally absorbed by the root hair cells from the aerated soil. [1 mark]
**01.2** Explain how the data in Table 1 supports your answer to part 01.1. [2 marks]
**01.3** Describe how the structure of a root hair cell is adapted to increase the rate of absorption of water and mineral ions. [2 marks]
**01.4** Calculate the ratio of the concentration of nitrate ions inside the root cells in aerated soil to that in waterlogged soil. Show your working. [2 marks]
**01.5** Explain why the concentration of nitrate ions in the root cells is much lower in waterlogged soil than in aerated soil. [3 marks]
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解題
01.1: Active transport is the process used to absorb ions against a concentration gradient. 01.2: In aerated soil, the concentration inside the cells is \(8.4\text{ mmol/dm}^3\), which is much higher than the soil concentration of \(0.15\text{ mmol/dm}^3\). This indicates the ions are being moved from low to high concentration (against a concentration gradient). 01.3: Root hair cells have long projections (or 'hairs') which significantly increase the surface area of the root in contact with soil water, allowing more absorption. 01.4: Ratio = \(8.4 : 1.2 = 7 : 1\). 01.5: Waterlogged soil lacks oxygen. Root cells require oxygen for aerobic respiration to release energy (ATP). Without sufficient energy, active transport of nitrate ions cannot occur at the normal rate.
評分準則
**01.1** active transport [1 mark] *Do not accept diffusion or osmosis.*
**01.2** concentration of ions is higher inside the cell than in the soil / outside [1 mark] (therefore) ions are moved against a concentration gradient / from a low concentration to a high concentration [1 mark]
**01.3** has a long hair-like extension/projection [1 mark] which increases the surface area (for absorption) [1 mark]
**01.4** \(8.4 \div 1.2\) or \(8.4 : 1.2\) [1 mark] \(7\) or \(7 : 1\) [1 mark] *Allow 2 marks for correct final answer with no working.*
**01.5** waterlogged soil has less oxygen [1 mark] so less (aerobic) respiration occurs in root cells [1 mark] less energy is released/available (for active transport) [1 mark] *Do not accept energy is 'produced' or 'created'.*
卷二 (Genetics, Ecology, Bioenergetics and Enzymes)
Answer all questions in the spaces provided. You must have a ruler and a calculator.
9 題目 · 90 分
題目 1 · Structured
10 分
Figure 1 shows a pedigree chart for a family with a history of a genetic condition called ALX syndrome, which is caused by a recessive allele, a. The dominant allele is A. (1.1) State the difference between genotype and phenotype. [2 marks] (1.2) What is meant by a dominant allele? [1 mark] (1.3) Refer to the pedigree chart: Unaffected parents (Individual 5 and Individual 6) have an affected child (Individual 11). Explain how this proves that the allele for ALX syndrome is recessive. [2 marks] (1.4) Draw a genetic diagram to show how two heterozygous parents can produce an affected offspring. Calculate the probability of this occurring. [4 marks] (1.5) State the expected phenotypic ratio of the offspring from this cross. [1 mark]
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解題
1.1 Genotype is the combination of alleles an organism possesses, whereas phenotype is the physical manifestation of those genes. 1.2 A dominant allele is one that is fully expressed in the organism's phenotype, masking the effect of a recessive allele if present. 1.3 Since both parents (5 and 6) are healthy but produce an affected child (11), they must be carriers (heterozygous, Aa). This means they possess the allele without showing the trait, which is only possible if the trait is recessive. 1.4 Heterozygous cross: Aa x Aa. Gametes: A and a. Offspring genotypes: AA (normal), Aa (carrier), Aa (carrier), aa (affected). Probability of aa = 1/4 = 0.25. 1.5 Phenotypic ratio: 3 normal (unaffected) : 1 affected.
評分準則
1.1 Genotype definition (1 mark) + Phenotype definition (1 mark). 1.2 Correct definition of dominant allele (1 mark). 1.3 Explanation of carrier parents (1 mark) + explanation that a dominant disease would require an affected parent (1 mark). 1.4 Correct parental genotypes/gametes (1 mark) + correct offspring genotypes (1 mark) + identifying affected genotype (1 mark) + correct probability of 0.25 (1 mark). 1.5 Correct ratio of 3:1 (1 mark).
題目 2 · Structured
10 分
A marine ecosystem food chain is represented as: Phytoplankton -> Zooplankton -> Herring -> Gannet. Table 1 shows the estimated biomass at each level: Phytoplankton (12000 g/m"), Zooplankton (1500 g/m"), Herring (120 g/m"), Gannet (6 g/m"). (2.1) Identify the producer and the primary consumer in this food chain. [2 marks] (2.2) Calculate the percentage efficiency of biomass transfer from the Zooplankton to the Herring. Show your working. [2 marks] (2.3) Explain three different ways that biomass is lost between trophic levels. [3 marks] (2.4) Explain why a pyramid of numbers can sometimes be inverted, but a pyramid of biomass is almost always pyramid-shaped. [3 marks]
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解題
2.1 The producer is the phytoplankton (photosynthetic organism) and the primary consumer is the zooplankton (herbivore feeding on phytoplankton). 2.2 Biomass transfer efficiency = (Biomass at higher level / Biomass at lower level) * 100 = (120 / 1500) * 100 = 8%. 2.3 Biomass is lost through: 1. Respiration (carbon dioxide and water vapor lost to the environment), 2. Material not consumed (e.g. bones, hard scales), 3. Excretion (losing urea) and egestion (losing undigested faeces). 2.4 A pyramid of numbers does not take individual size into account; for instance, a single oak tree can feed thousands of caterpillars, creating an inverted base. A pyramid of biomass reflects actual living tissue mass, which always decreases as energy is lost at each progressive trophic step.
評分準則
2.1 Producer identified (1 mark) + Primary consumer identified (1 mark). 2.2 Correct calculation step (1 mark) + final correct percentage of 8% (1 mark). 2.3 Any three valid ways of biomass loss (3 marks). 2.4 Explaining pyramid of numbers variations (1 mark) + defining biomass measurement (1 mark) + explaining why biomass must decrease due to thermodynamic loss (1 mark).
題目 3 · Structured
10 分
A student investigated the effect of light intensity on the rate of photosynthesis in pondweed (Elodea) by placing a lamp at different distances (10 cm, 20 cm, 30 cm, 40 cm) and counting the number of oxygen bubbles released per minute. (3.1) Write the balanced chemical equation for photosynthesis. [2 marks] (3.2) State the independent variable in this investigation. [1 mark] (3.3) Suggest two factors that the student should control to ensure the test is fair. [2 marks] (3.4) Explain how light intensity changes as distance from the lamp increases, referencing the inverse square law. [2 marks] (3.5) The rate of bubble production remains constant when the lamp is moved closer than 10 cm. Explain why. [3 marks]
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解題
3.1 Balanced equation: 6CO2 + 6H2O -> C6H12O6 + 6O2. 3.2 The independent variable is the factor changed by the student, which is the light intensity (or the distance of the lamp). 3.3 Control variables: 1. Temperature of the pondwater (use a heat shield or a water bath), 2. Concentration of carbon dioxide (use a fixed concentration of sodium hydrogen carbonate solution). 3.4 According to the inverse square law, light intensity is inversely proportional to the square of the distance. If distance is doubled, light intensity drops to one-quarter. 3.5 Below 10 cm, the rate plateaus because light is no longer the limiting factor. The process is now restricted by another factor, such as temperature or carbon dioxide concentration.
評分準則
3.1 Correct chemical formulas (1 mark) + correct balancing (1 mark). 3.2 Correctly identifying independent variable (1 mark). 3.3 Two correct control variables identified (2 marks). 3.4 Referencing inverse relationship (1 mark) + explaining the mathematical square relationship (1 mark). 3.5 Explaining that light is no longer limiting (1 mark) + identifying temperature or CO2 as alternative limiting factors (2 marks).
題目 4 · Structured
10 分
A student investigated the rate of lipid digestion by the enzyme lipase at different temperatures. (4.1) Name the substrate, the products, and the organ where lipase is produced in the human digestive system. [3 marks] (4.2) Explain how bile assists in the digestion of lipids. [2 marks] (4.3) Describe and explain the effect of temperature on the rate of lipid digestion between 10 °C and 37 °C. [3 marks] (4.4) At 60 °C, no lipid digestion occurs. Explain why. [2 marks]
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解題
4.1 Substrate: Lipids (fats/oils). Products: Fatty acids and glycerol. Lipase is produced in the pancreas (and small intestine). 4.2 Bile performs two roles: 1. Emulsifies large lipid globules into small droplets to increase the surface area available for lipase, 2. Neutralises hydrochloric acid from the stomach to provide the optimal alkaline environment for pancreatic enzymes. 4.3 Between 10 °C and 37 °C, rising temperature increases the kinetic energy of both the substrate and enzyme molecules. This leads to faster movement, increasing the frequency of successful collisions and rate of active site binding. 4.4 At 60 °C, the enzyme is denatured. The high thermal energy breaks intramolecular bonds, changing the three-dimensional configuration of the active site permanently so the substrate can no longer fit.
評分準則
4.1 Correct substrate (1 mark) + both products (1 mark) + organ of production (1 mark). 4.2 Emulsification role (1 mark) + acid neutralisation / pH optimum role (1 mark). 4.3 Reference to increasing kinetic energy (1 mark) + more frequent collisions (1 mark) + more enzyme-substrate complexes formed (1 mark). 4.4 Term 'denatures' used correctly (1 mark) + explanation of permanent shape change of the active site (1 mark).
題目 5 · Structured
10 分
Decomposition plays a critical role in the carbon cycle. (5.1) Name two main groups of decomposer organisms. [2 marks] (5.2) Explain how decomposers release carbon dioxide back into the atmosphere. [2 marks] (5.3) Suggest and explain two environmental conditions that would increase the rate of decay in a compost bin. [4 marks] (5.4) Explain why compost is added to soil by gardeners. [2 marks]
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解題
5.1 The two primary groups of decomposers are bacteria and fungi. 5.2 Decomposers secrete extracellular digestive enzymes to break down organic matter. They absorb the soluble products and utilize them in aerobic respiration, which generates carbon dioxide as a waste product. 5.3 1. Warmth: Higher temperatures increase the kinetic energy of enzymes, accelerating biochemical reactions of decay. 2. Oxygen availability: Air holes allow oxygen to enter, promoting fast aerobic respiration instead of slow anaerobic decay. 5.4 Compost is rich in recycled plant nutrients and mineral ions (such as nitrates/phosphates) which replenish the soil, supporting healthy plant growth and protein synthesis.
Respiration can occur both aerobically and anaerobically. (6.1) Write the word equation for anaerobic respiration in yeast cells. [1 mark] (6.2) Write the word equation for anaerobic respiration in human muscle cells. [1 mark] (6.3) Compare anaerobic respiration in yeast and human muscle cells by stating two differences and one similarity. [3 marks] (6.4) Explain what is meant by 'oxygen debt' and describe how the body removes lactic acid after intense exercise. [5 marks]
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解題
6.1 Yeast: Glucose -> Ethanol + Carbon dioxide. 6.2 Humans: Glucose -> Lactic acid. 6.3 Differences: 1. Yeast anaerobic respiration produces ethanol, while human muscle cells produce lactic acid. 2. Yeast anaerobic respiration releases carbon dioxide gas, while human anaerobic respiration does not. Similarity: Both pathways break down glucose without utilizing oxygen. 6.4 Oxygen debt is the volume of additional oxygen required after exercise to break down accumulated lactic acid. Lactic acid is carried by blood to the liver, where oxygen is used to convert it back into glucose or glycogen, or completely oxidise it into carbon dioxide and water.
評分準則
6.1 Correct equation for fermentation (1 mark). 6.2 Correct equation for lactic acid pathway (1 mark). 6.3 Two valid differences (2 marks) + one similarity (1 mark). 6.4 Definition of oxygen debt (1 mark) + role of the liver (1 mark) + transport via blood (1 mark) + conversion of lactic acid back to glucose/glycogen (1 mark) + oxidation to CO2 and water (1 mark).
題目 7 · Structured
10 分
All eukaryotic cells divide via the cell cycle. (7.1) State the primary purpose of mitosis in multicellular organisms. [2 marks] (7.2) Describe the three main stages of the cell cycle, including mitosis. [3 marks] (7.3) Explain two key differences between the daughter cells produced by mitosis and those produced by meiosis. [4 marks] (7.4) State the specific location in the human body where meiosis takes place. [1 mark]
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解題
7.1 Mitosis is essential for growth (increasing cell count) and tissue repair (replacing damaged or dead cells). 7.2 Stage 1: Interphase (growth, increase in sub-cellular structures, and duplication of DNA). Stage 2: Mitosis (chromosomes are pulled to opposite poles and nucleus divides). Stage 3: Cytokinesis (cytoplasm and cell membrane divide to form two separate cells). 7.3 Differences: 1. Ploidy: Mitosis yields diploid (2n) daughter cells, while meiosis yields haploid (n) cells. 2. Variation: Mitosis produces genetically identical clones, whereas crossing over/independent assortment in meiosis results in genetically diverse cells. 7.4 Meiosis only occurs within human reproductive organs, i.e., testes in males and ovaries in females.
評分準則
7.1 Growth (1 mark) + repair of tissues / cell replacement (1 mark). 7.2 Stage 1: DNA replication / organelle increase (1 mark) + Stage 2: Mitosis / division of nucleus (1 mark) + Stage 3: Cytokinesis / division of cytoplasm (1 mark). 7.3 Difference 1 (number of chromosomes/ploidy) described (2 marks) + Difference 2 (genetic variation/similarity) described (2 marks). 7.4 Correct location (testes or ovaries) (1 mark).
題目 8 · Structured
10 分
Desert plants, such as cacti, have evolved various adaptations to survive with limited water. (8.1) Name two resources, other than water, that desert plants compete for. [2 marks] (8.2) Explain how a thick waxy cuticle and rolled leaves reduce water loss in plants. [3 marks] (8.3) Compare the adaptive advantages of having deep taproots versus having extensive shallow root systems in desert plants. [3 marks] (8.4) Suggest why opening stomata only at night is highly advantageous for desert plants. [2 marks]
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解題
8.1 Desert plants compete for resources including light, space, and mineral ions (nutrients). 8.2 A thick waxy cuticle is hydrophobic, acting as a physical barrier preventing water from evaporating directly from the leaf surface. Rolled leaves trap a microclimate of humid air around the stomata. This decreases the concentration gradient of water vapour between the inside and outside of the leaf, lowering transpiration. 8.3 Deep taproots can reach deeply stored groundwater reserves, ensuring survival during prolonged droughts. Conversely, widespread shallow roots can capture brief desert rainfall instantly before the water evaporates from the hot soil surface. 8.4 Temperatures at night are significantly cooler and relative humidity is higher. Opening stomata at night allows carbon dioxide uptake while minimizing the rate of water loss through transpiration.
評分準則
8.1 Two valid resources identified (2 marks). 8.2 Role of waxy cuticle (1 mark) + role of rolled leaves in trapping moist air (1 mark) + reduction of transpiration rate (1 mark). 8.3 Advantage of deep roots (1 mark) + advantage of shallow roots (1 mark) + comparative summary linking to water capture (1 mark). 8.4 Cooler/more humid conditions at night explained (1 mark) + linked to reduced rate of water loss (1 mark).
題目 9 · structured
10 分
A student investigated the effect of carbon dioxide concentration on the rate of photosynthesis in *Cabomba* (fanwort) pondweed at two different temperatures.
The student kept the light intensity constant. The rate of photosynthesis was determined by counting the number of oxygen bubbles released by the pondweed per minute.
The results are shown in **Table 1**.
**Table 1**
| Carbon dioxide concentration (%) | Number of bubbles per minute at 15 °C | Number of bubbles per minute at 25 °C | | :---: | :---: | :---: | | 0.00 | 0 | 0 | | 0.02 | 8 | 12 | | 0.04 | 15 | 24 | | 0.06 | 20 | 32 | | 0.08 | 22 | 38 | | 0.10 | 22 | 40 | | 0.12 | 22 | 40 |
**01.1** State two variables that the student should have controlled in this investigation, other than temperature and light intensity. [2 marks]
**01.2** Describe the effect of increasing carbon dioxide concentration on the rate of photosynthesis at 15 °C. Use data from **Table 1** to support your answer. [2 marks]
**01.3** Explain why the rate of photosynthesis was higher at 25 °C than at 15 °C at 0.10% carbon dioxide concentration. [3 marks]
**01.4** Calculate the percentage increase in the rate of photosynthesis at 0.06% carbon dioxide concentration when the temperature was increased from 15 °C to 25 °C. Show your working. [3 marks]
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解題
**01.1** Two control variables could be: 1. The mass, length, or species of the *Cabomba* pondweed piece. 2. The wavelength (or colour) of the light source, or the pH of the water/solution used.
**01.2** - As carbon dioxide concentration increases from 0.00% to 0.08%, the rate of photosynthesis increases (from 0 to 22 bubbles per minute). - Above 0.08% carbon dioxide concentration, the rate remains constant at 22 bubbles per minute.
**01.3** - Photosynthesis is an enzyme-controlled reaction. - Increasing the temperature from 15 °C to 25 °C increases the kinetic energy of the enzymes and substrate molecules. - This results in more frequent successful collisions between enzyme active sites and substrates, increasing the rate of reaction.
**01.4** - At 15 °C, rate = 20 bubbles per minute. - At 25 °C, rate = 32 bubbles per minute. - Increase in rate = \(32 - 20 = 12\) bubbles per minute. - Percentage increase = \(\frac{12}{20} \times 100 = 60\%\).
評分準則
**01.1** - Any two from: - species of pondweed (1) - length / mass / surface area of pondweed (1) - wavelength / colour of light (1) - volume of solution / water (1) - pH of solution / water (1) *(do not accept temperature or light intensity/distance of light)*
**01.2** - rate increases as carbon dioxide concentration increases up to 0.08% (1) - rate remains constant / levels off at 22 bubbles per minute above 0.08% (1)
**01.3** - photosynthesis is controlled by enzymes (1) - molecules / enzymes / substrates have more kinetic energy (1) - leading to more frequent / more successful collisions (1)
**01.4** - correct extraction of data from table: 20 and 32 (1) - correct calculation of change and relative division: \(\frac{32 - 20}{20}\) or \(\frac{12}{20}\) (1) - correct calculation of percentage: 60 (%) (1) *(award 3 marks for the correct final answer of 60% with no working shown)*
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