Cambridge IGCSE · Thinka 原創模擬試題

2025 Cambridge IGCSE Biology (0610) 模擬試題連答案詳解

Thinka Jun 2025 (V3) Cambridge International A Level-Style Mock — Biology (0610)

80 分75 分鐘2025
分析模擬試題
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge International A Level Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

甲部

Answer all questions. Write your answers in the spaces provided on the question paper.
6 題目 · 80 分
題目 1 · structured
15 分
This question is about human nutrition and the processes within the alimentary canal. (a) Complete the following information by identifying the missing names of the enzymes and their substrates: (i) Name the enzyme secreted in the mouth. (ii) Name the substrate of the enzyme secreted in the mouth. (iii) Name the specific protease secreted in the stomach. (iv) Name the substrate that trypsin acts upon in the duodenum. [4 marks] (b) Describe how proteins are digested as they pass through the stomach and the small intestine. In your answer, refer to the enzymes involved, the locations of digestion, and the conditions required for these enzymes to work effectively. [5 marks] (c) Explain the roles of the liver, the gall bladder, and bile in the digestion of fats. [4 marks] (d) Define the terms ingestion and egestion to show the difference between them. [2 marks]
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解題

Part (a): (i) Amylase is secreted by salivary glands in the mouth. (ii) Amylase breaks down starch into maltose. (iii) Pepsin is the specific protease enzyme functioning in the stomach. (iv) Trypsin is a protease secreted by the pancreas into the duodenum to act on proteins or polypeptides. Part (b): Protein digestion starts in the stomach with pepsin under highly acidic conditions created by hydrochloric acid. Once the food (chyme) moves to the duodenum, it is met with trypsin (secreted by the pancreas), which works in neutral to alkaline conditions. Peptidases finish the digestion of peptides into amino acids at the lining of the small intestine. Part (c): The liver synthesises bile, which is stored in the gall bladder. Bile neutralises acidic chyme and emulsifies fats, multiplying their surface area for efficient chemical breakdown by lipase. Part (d): Ingestion is the entry point (mouth), whereas egestion is the exit of indigestible waste (anus) which was never absorbed into the body cells.

評分準則

(a) [Total: 4 marks, 1 mark for each correct answer] (i) Amylase (ii) Starch (iii) Pepsin (accept: gastric protease) (iv) Protein / polypeptide / peptide. (b) [Total: 5 marks, max 5 from the following points] - Protein digestion starts in the stomach [1] - Stomach secretes hydrochloric acid which provides acidic conditions / optimum pH (pH 1.5-2.5) [1] - Pepsin breaks down protein to polypeptides [1] - Pancreas secretes trypsin into the duodenum [1] - Duodenum has neutral/alkaline conditions (due to bile / sodium hydrogencarbonate) [1] - Trypsin breaks down proteins / polypeptides to peptides [1] - Peptidases (on cell membranes of the small intestine) break down peptides to amino acids [1]. (c) [Total: 4 marks, max 4 from the following points] - Liver produces bile [1] - Gall bladder stores bile (and releases it into the duodenum) [1] - Bile emulsifies fats / breaks large fat globules into smaller droplets [1] - This increases surface area for lipase action / speeds up chemical digestion [1] - Bile is alkaline / neutralises hydrochloric acid (from stomach) to provide optimum pH for lipase [1]. (d) [Total: 2 marks] - Ingestion: taking substances / food / drink into the body through the mouth [1] - Egestion: passing out of undigested / unabsorbed food as faeces through the anus [1] (Reject: definition of excretion).
題目 2 · structured
10 分
An investigation was carried out to study the effect of running speed on the breathing rate and tidal volume (the volume of air breathed in or out during a single breath) of a healthy teenager. The results are shown below:

Table 1.1
Running speed / km/h | Breathing rate / breaths per minute | Tidal volume / dm³
0 (at rest) | 12 | 0.5
6 (jogging) | 18 | 1.2
12 (running) | 28 | 2.5

(a) Calculate the volume of air breathed in one minute (minute ventilation) at 12 km/h. Show your working and state the unit. [2]

(b) Explain why both breathing rate and tidal volume increase during running. [4]

(c) Describe three features of the gas exchange surface in the human lungs that allow for rapid diffusion. [3]

(d) State the name of the muscle group located between the ribs that contracts to pull the ribcage upwards and outwards during inhalation. [1]
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解題

(a) To calculate the volume of air breathed in one minute (minute ventilation) at 12 km/h:
Minute ventilation = Breathing rate × Tidal volume
At 12 km/h, breathing rate = 28 breaths/min and tidal volume = 2.5 dm³.
Calculation: \(28 \times 2.5 = 70\).
Unit: dm³/min (or dm³ min⁻¹).

(b) During exercise/running:
1. Muscle cells contract more frequently and vigorously, which requires more energy.
2. This increases the rate of aerobic respiration in muscle cells.
3. Consequently, more oxygen is required and more carbon dioxide is produced as a waste product.
4. The concentration of carbon dioxide in the blood increases, lowering blood pH.
5. The brain (medulla) detects this change and sends faster/stronger nerve impulses to the diaphragm and intercostal muscles, increasing the rate and depth of breathing to supply more oxygen and remove carbon dioxide rapidly.

(c) Adaptations of the gas exchange surface (alveoli) include:
1. Millions of alveoli provide a very large surface area.
2. Alveolar walls are extremely thin (one cell thick) which provides a short diffusion distance.
3. A dense network of capillaries maintains a steep concentration gradient by constantly removing oxygenated blood and bringing carbon-dioxide-rich blood.
4. A moist lining allows gases to dissolve before diffusing across the membrane.

(d) The external intercostal muscles contract to pull the ribcage upwards and outwards during inhalation (inspiration).

評分準則

(a) [Maximum 2 marks]
- Award 1 mark for correct working: 28 x 2.5 = 70.
- Award 1 mark for correct unit: dm³/min or dm³ per minute (accept dm³ min⁻¹).

(b) [Maximum 4 marks]
- muscles contract more / require more energy [1]
- increased rate of aerobic respiration [1]
- increased demand for oxygen / increased production of carbon dioxide [1]
- increased CO2 concentration lowers pH of blood [1]
- detected by the brain [1]
- brain sends impulses to diaphragm / intercostal muscles to contract faster / harder [1]

(c) [Maximum 3 marks]
- Large surface area (due to many alveoli) [1]
- Thin walls / one cell thick (short diffusion distance) [1]
- Rich capillary network / good blood supply (maintains concentration gradient) [1]
- Moist lining (allows gases to dissolve) [1]

(d) [Maximum 1 mark]
- External intercostal muscles [1] (Reject: 'intercostal muscles' alone, Reject: 'internal intercostal muscles')
題目 3 · structured
12 分
Homeostasis is a critical process in mammals. One example of homeostasis is the regulation of blood glucose concentration.

(a) Define the term homeostasis. [2]

(b) Describe how the pancreas and liver work together to lower blood glucose concentration when it rises above normal levels, such as after eating a carbohydrate-rich meal. [6]

(c) Type 1 diabetes is a disease in which the body is unable to regulate blood glucose levels effectively.
(i) State two symptoms of Type 1 diabetes. [2]
(ii) Describe two ways in which Type 1 diabetes is managed or treated. [2]
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解題

(a) Homeostasis is defined as the maintenance of a constant internal environment.

(b) When blood glucose concentration increases above normal levels:
- The change is detected by cells in the pancreas.
- The pancreas responds by secreting the hormone insulin into the blood.
- Insulin is transported in the blood plasma to its target organ, the liver (and muscle cells).
- Insulin stimulates the liver cells to absorb glucose from the blood.
- Inside the liver cells, glucose is converted into the insoluble storage carbohydrate, glycogen.
- This reduces the concentration of glucose in the blood, returning it to the set point (negative feedback).

(c)(i) Symptoms of Type 1 diabetes include:
- Extreme thirst / high water intake
- Frequent urination / glucose present in urine
- Unexplained weight loss
- Severe fatigue / tiredness

(c)(ii) Management and treatment of Type 1 diabetes include:
- Regular insulin injections (the dosage is matched to food intake and activity levels).
- Frequent monitoring of blood glucose concentration using a blood glucose meter.
- A carefully controlled diet, particularly regulating the intake of simple sugars and carbohydrates.

評分準則

(a) [Max 2 marks]
- maintenance of a constant / stable; [1]
- internal environment; [1]

(b) [Max 6 marks]
- increase in blood glucose is detected by the pancreas; [1]
- pancreas secretes / releases insulin; [1]
- insulin travels in the blood / plasma; [1]
- insulin binds to receptors on liver cells / targets the liver; [1]
- liver cells absorb / take up glucose from the blood; [1]
- glucose is converted into glycogen; [1]
- glycogen is stored (in liver/muscles); [1]
- blood glucose concentration decreases back to normal / negative feedback; [1]

(c)(i) [Max 2 marks]
- Any two from: high blood glucose levels / glucose in urine / frequent urination / polyuria / extreme thirst / polydipsia / weight loss / fatigue / tiredness / lethargy; [2]

(c)(ii) [Max 2 marks]
- Any two from: regular insulin injections / insulin therapy; monitoring of blood glucose concentration (using test strips/sensor); controlled diet / restricting simple sugar intake / regulating carbohydrate consumption; [2]
題目 4 · structured
17 分

A student investigated the effect of light intensity on the rate of photosynthesis in an aquatic plant, Cabomba caroliniana.

The student set up a lamp at various distances from the plant and measured the volume of gas collected in a gas syringe over a period of 5 minutes at each distance. The results are shown in Table 4.1.

Table 4.1

Distance of lamp from plant / cmVolume of gas collected in 5 minutes / cm3Rate of gas production / cm3 per minute101.500.30201.200.24300.750.15400.300.06500.150.03600.15X

(a) State the balanced chemical equation for photosynthesis, using chemical symbols. [3]

(b) (i) Calculate the value of X in Table 4.1. Show your working and state the units. [2]

(b) (ii) Describe the relationship between the distance of the lamp and the rate of gas production shown in Table 4.1. [2]

(b) (iii) Explain the trend in the rate of gas production as the distance increases from 10 cm to 50 cm, in terms of limiting factors. [3]

(c) (i) Explain how the structure of the palisade mesophyll layer of a leaf is adapted to maximize photosynthesis. [3]

(c) (ii) Describe how the structure and behavior of stomata control gas exchange and transpiration in a leaf during a hot, dry day. [4]

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解題

(a) Balanced chemical equation:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light / chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) (i)
Working: \(\frac{0.15\text{ cm}^3}{5\text{ min}} = 0.03\)
Value of X: 0.03
Unit: \(\text{cm}^3/\text{min}\) or \(\text{cm}^3\text{ min}^{-1}\)

(b) (ii)
As the distance of the lamp increases, the rate of gas production decreases. The decrease is steep between 10 cm and 50 cm, and then the rate remains constant/levels off at 0.03 \(\text{cm}^3/\text{min}\) between 50 cm and 60 cm.

(b) (iii)
As distance increases, light intensity decreases. Light intensity is the limiting factor in this range. Consequently, less light energy is absorbed by chlorophyll, resulting in a lower rate of photosynthesis and less oxygen being produced.

(c) (i)
Palisade mesophyll cells are column-shaped and closely packed together near the upper surface of the leaf to intercept the maximum amount of sunlight. They also contain a very high density of chloroplasts to absorb light energy efficiently.

(c) (ii)
On a hot, dry day, guard cells lose water by osmosis and become flaccid. This causes the stomatal pore to close, which reduces water loss via transpiration. However, this closure also limits the diffusion of carbon dioxide into the leaf, thereby decreasing the rate of photosynthesis.

評分準則

(a) [Max 3 marks]

  • Correct chemical formulas for reactants (\(\text{CO}_2\) and \(\text{H}_2\text{O}\)) [1]
  • Correct chemical formulas for products (\(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\)) [1]
  • Correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) [1]
Note: Accept light and chlorophyll over the arrow, but do not penalise if absent. Reject word equations.

(b) (i) [Max 2 marks]

  • Correct calculation / numerical value: 0.03 [1]
  • Correct unit: \(\text{cm}^3/\text{min}\) or \(\text{cm}^3\text{ min}^{-1}\) [1]

(b) (ii) [Max 2 marks]

  • As distance increases, rate of gas production decreases [1]
  • Reference to the rate levelling off / becoming constant from 50 to 60 cm [1]

(b) (iii) [Max 3 marks]

  • Increasing distance reduces light intensity [1]
  • Light intensity is the limiting factor (between 10 cm and 50 cm) [1]
  • Less light energy absorbed by chlorophyll, leading to a lower rate of photosynthesis/oxygen production [1]

(c) (i) [Max 3 marks]

  • Cells are elongated/column-shaped and closely packed together [1]
  • Positioned near the upper surface of the leaf to receive maximum sunlight [1]
  • Contain many chloroplasts / high concentration of chlorophyll to absorb light energy [1]

(c) (ii) [Max 4 marks]

  • Guard cells lose water by osmosis [1]
  • Guard cells become flaccid [1]
  • Stomatal pore closes [1]
  • Reduces water loss / transpiration [1]
  • Award max 1 mark for explaining that closing stomata also limits carbon dioxide entry / decreases photosynthesis

題目 5 · structured
13 分
Penicillin is an antibiotic produced on an industrial scale using the fungus Penicillium chrysogenum in a fermenter. (a)(i) State the kingdom to which Penicillium belongs. [1] (a)(ii) State one feature of organisms in this kingdom that distinguishes them from plants. [1] (b) Explain why the fermenter and its growth medium must be sterilized using superheated steam before inoculation. [3] (c) The temperature and pH inside the fermenter are monitored continuously. Explain why it is important to maintain these two conditions at optimum levels during the production of penicillin. [4] (d) Describe the role of the stirrer (paddles) and the air supply in the fermenter. [4]
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解題

(a)(i) Fungi / Fungus. (a)(ii) Chitin cell wall (instead of cellulose) / Do not carry out photosynthesis / Lack chlorophyll / Feed saprophytically (or parasitically). (b) To kill any existing/unwanted microorganisms or pathogens; to prevent contamination of the penicillin product; to prevent competition between the Penicillium fungus and other microbes for glucose/nutrients/oxygen. (c) Temperature and pH directly affect the activity of the enzymes in the fungus. At optimum temperature and pH, enzymes operate at their maximum rate, leading to the highest yield of penicillin. If the temperature is too high or pH is too extreme, the enzymes will denature as their active sites change shape, preventing substrates from binding and stopping the metabolic reactions. (d) The stirrer/paddles mix the contents to ensure a uniform temperature throughout the fermenter, prevent the fungus from settling at the bottom, and evenly distribute nutrients and oxygen. The air supply provides oxygen which is essential for aerobic respiration of the fungus, releasing energy needed for growth and penicillin synthesis.

評分準則

Award up to 13 marks. (a)(i) Fungi [1]. (a)(ii) Chitin cell wall / lack chlorophyll / do not photosynthesise / saprophytic nutrition [1] (Accept any valid structural or nutritional difference; Reject: 'has no cell wall'). (b) Any three from: to kill/destroy unwanted bacteria/microorganisms [1]; to prevent contamination of the final product [1]; to prevent competition for nutrients/oxygen [1]; to prevent toxins from being produced by other microbes [1] (Max [3]). (c) Any four from: temperature and pH affect enzyme activity / enzymes have an optimum temperature and pH [1]; maintaining optimum conditions ensures maximum rate of reaction/product yield [1]; extreme pH or high temperature denatures enzymes [1]; active site changes shape [1]; substrate can no longer fit into the active site / metabolic reactions stop [1] (Max [4]). (d) Any four from: [Stirrer] mixes/distributes nutrients / oxygen / heat evenly [1]; prevents fungus from settling at the bottom [1]; [Air supply] provides oxygen [1]; for aerobic respiration [1]; to release energy for fungal growth/penicillin production [1] (Max [4]).
題目 6 · structured
13 分
Fig. 1.1 describes some of the feeding relationships in a marine ecosystem: Phytoplankton are microscopic organisms that photosynthesise. Krill and zooplankton feed on phytoplankton. Herring feed on both krill and zooplankton. Cod feed on herring and krill. Harbour seals feed on cod and herring. (a) Define the term 'trophic level'. [2 marks] (b) Use the information provided to: (i) Identify one organism from the description that acts as both a secondary consumer and a tertiary consumer. [1 mark] (ii) Construct a complete food chain containing exactly four trophic levels. Use arrows to show the direction of energy transfer. [2 marks] (c) Explain why ecosystems rarely support food chains with more than five trophic levels. [4 marks] (d) Decomposers are vital for the continuous recycling of nutrients in this marine ecosystem. (i) Name one group of organisms that act as decomposers. [1 mark] (ii) Explain the role of decomposers in making nitrogen-containing ions available to producers. [3 marks]
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解題

(a) A trophic level is the position of an organism in a food chain, food web, or pyramid. (b)(i) Cod (since phytoplankton -> krill -> cod makes it a secondary consumer, and phytoplankton -> krill -> herring -> cod makes it a tertiary consumer). (b)(ii) Phytoplankton -> Krill -> Herring -> Cod (or Phytoplankton -> Zooplankton -> Herring -> Cod, or Phytoplankton -> Krill -> Herring -> Harbour seal). (c) Energy is lost at each trophic level through respiration (as heat), excretion, egestion, and because some parts of organisms are uneaten. Only about 10% of the energy is transferred to the next level, meaning there is insufficient energy left at the top to support another trophic level. (d)(i) Bacteria or Fungi. (d)(ii) Decomposers break down dead organic matter and waste products, digesting proteins and urea to release ammonium ions. Nitrifying bacteria then convert these ammonium ions into nitrates, which can be absorbed by producers to synthesize amino acids and proteins.

評分準則

(a) [Max 2] Award 1 mark for: position of an organism in a food chain / food web / pyramid. Award 1 mark for: ref to feeding / energy transfer. (b)(i) [Max 1] Cod (accept Harbour seal). (b)(ii) [Max 2] Award 1 mark for any correct 4-organism chain in the correct sequence (e.g., Phytoplankton -> Krill -> Herring -> Cod). Award 1 mark for all arrows pointing in the correct direction (from food source to consumer). (c) [Max 4] Award 1 mark for each of the following points, up to 4: Energy is lost at each trophic level / energy transfer is inefficient; Only approximately 10% of energy is transferred to the next trophic level (or 90% lost); Energy is lost as heat from respiration; Energy is lost via excretion / egestion / metabolic waste (e.g. feces, urea); Some parts of organisms are not eaten (e.g. bones, teeth, shells) or die without being eaten; Insufficient energy remains at higher trophic levels to support another viable population. (d)(i) [Max 1] Bacteria / Fungi (Reject: earthworms / detritivores). (d)(ii) [Max 3] Award 1 mark for each of the following points, up to 3: Decomposers break down / decay dead organic matter / animal waste; They digest / decompose proteins / amino acids / urea; This process releases ammonium ions / ammonia; Nitrifying bacteria convert ammonium ions to nitrites and then to nitrates; Nitrates are absorbed by producers / phytoplankton to build proteins / chlorophyll.

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