Cambridge IGCSE · thinka 原创模拟试题

2025 Cambridge IGCSE Biology (0610) 模拟试题及答案详解

Thinka Jun 2025 (V3) Cambridge IGCSE-Style Mock — Biology (0610)

80 分75 分钟2025
分析模拟试题
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge IGCSE 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.
7 题目 · 97 分
题目 1 · structured
15 分
1 (a) Some organs of the human digestive system are responsible for the physical and chemical breakdown of food.

Table 1.1 shows information about three of these structures.

Complete Table 1.1 by filling in the missing names of structures, functions, or the representative letters based on your knowledge of the alimentary canal.

**Table 1.1**
| Name of the structure | Function | Representative letter on a diagram |
| :--- | :--- | :---: |
| Stomach | ........................................................................ | **S** |
| ......................................... | produces pancreatic juice containing amylase, trypsin, and lipase | **P** |
| Gallbladder | stores ................................................................ | **G** |
| ......................................... | produces bile | **L** |
[4]

(b) (i) Describe the role of physical digestion in the alimentary canal. [3]

(ii) State the soluble product of complete protein digestion. [1]

(iii) Name the specific region of the alimentary canal where these digested products are absorbed into the blood. [1]

(iv) Explain why pepsin actively breaks down protein molecules in the stomach but cannot break them down in the duodenum. [6]
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解题

**1 (a)**
* **Stomach Function**: Churns food (mechanical digestion) OR secretes gastric juice / hydrochloric acid / pepsin.
* **Structure P**: Pancreas (produces pancreatic amylase, trypsin, and lipase).
* **Gallbladder Function**: Stores bile (which emulsifies fats).
* **Structure L**: Liver (produces bile).

**1 (b) (i)**
Physical digestion is the breakdown of food into smaller pieces without chemical alteration of the food molecules. It:
1. Increases the surface area of food for chemical digestion by enzymes.
2. Mixes food with digestive juices.
3. Makes food easier to swallow and pass through the alimentary canal via peristalsis.

**1 (b) (ii)**
Amino acids.

**1 (b) (iii)**
Ileum (accept small intestine / villi).

**1 (b) (iv)**
1. The stomach contains hydrochloric acid, creating an acidic environment (low pH) which is the optimum pH for pepsin activity.
2. In the duodenum (small intestine), bile and pancreatic juice (containing sodium hydrogencarbonate) are secreted.
3. These secretions are alkaline, neutralising the stomach acid and raising the pH to neutral or slightly alkaline.
4. This change in pH denatures the pepsin enzyme.
5. Denaturation alters the shape of pepsin's active site.
6. The active site is no longer complementary in shape to the protein substrate, meaning enzyme-substrate complexes cannot form.

评分标准

**1 (a)**
Four correct inputs to complete the table (one mark per correct row) [4]:
* **Row 1**: churns food / mixes food / secretes gastric juice / secretes hydrochloric acid / secretes pepsin
* **Row 2**: pancreas
* **Row 3**: bile
* **Row 4**: liver

**1 (b) (i)**
Any three from [3]:
* breaks down food into smaller pieces / breaks down food without chemical change / physical reduction in size;
* increases surface area (for enzyme action / chemical digestion);
* makes food easier to swallow / lubricates food;
* increases ease of passage / movement through the alimentary canal / aids peristalsis;
* mixes food with digestive enzymes / bile;

**1 (b) (ii)**
* amino acids [1]

**1 (b) (iii)**
* ileum / small intestine / villi [1]
*(Reject: large intestine, duodenum, stomach)*

**1 (b) (iv)**
Any six from [6]:
* stomach has gastric juice / hydrochloric acid / low pH / acidic pH;
* (stomach pH) is the optimum pH for pepsin;
* bile / pancreatic juice / pancreatic secretions are released into the duodenum / small intestine;
* these secretions are alkaline / contain sodium hydrogencarbonate;
* neutralise (stomach / gastric) acid / raise pH;
* (higher / alkaline pH in duodenum) denatures pepsin;
* alters the shape of the active site (of pepsin);
* active site is no longer complementary to the substrate / protein molecule;
* enzyme-substrate complexes can no longer form / cannot form / pepsin becomes inactive;
*(Accept: reverse argument (ora) for pancreatic proteases / trypsin where appropriate)*
题目 2 · 結構題
10 分
Table 1.1 shows the percentage of oxygen and carbon dioxide in inspired air and expired air of a healthy human.

$$\begin{array}{|l|c|c|} \hline \textbf{Gas} & \textbf{Inspired air / \%} & \textbf{Expired air / \%} \\ \hline \text{oxygen} & 21.0 & 16.0 \\ \hline \text{carbon dioxide} & 0.04 & 4.00 \\ \hline \end{array}$$

**Table 1.1**

(a) (i) Describe and explain the difference in the percentage of oxygen between inspired and expired air shown in Table 1.1. [2]

(ii) Describe and explain the difference in the percentage of carbon dioxide between inspired and expired air shown in Table 1.1. [2]

(b) The lining of the trachea contains specialized cells that protect the gas exchange system from pathogens. Explain the role of goblet cells and ciliated cells in this protective process. [3]

(c) Explain why a person continues to breathe deeply and rapidly for a period of time after they have finished a session of vigorous exercise. [3]
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解题

(a) (i)
- Description: The percentage of oxygen decreases from 21.0% in inspired air to 16.0% in expired air.
- Explanation: Oxygen is absorbed by diffusion from the alveoli into the blood to be used in aerobic respiration by body cells.

(a) (ii)
- Description: The percentage of carbon dioxide increases from 0.04% in inspired air to 4.00% in expired air.
- Explanation: Carbon dioxide is produced as a waste product of respiration in body cells, transported in the blood, and diffuses into the alveoli to be excreted.

(b)
- Goblet cells produce and secrete mucus, which traps pathogens, bacteria, and dust particles inhaled into the airways.
- Ciliated cells have tiny hair-like structures (cilia) that beat in a coordinated rhythm to sweep the trapped mucus upwards, away from the lungs towards the throat, where it can be swallowed.

(c)
- During vigorous exercise, muscles may respire anaerobically because insufficient oxygen is supplied, leading to a build-up of lactic acid in the muscles and blood.
- This build-up of lactic acid creates an oxygen debt.
- Continued deep and rapid breathing after exercise is required to supply the extra oxygen needed to transport, break down, and oxidize the lactic acid in the liver.

评分标准

(a) (i) Max 2 marks:
- Description: percentage of oxygen is lower in expired air / decreases from 21.0% to 16.0% [1]
- Explanation: oxygen diffuses from alveoli into the blood / is used in respiration [1]

(a) (ii) Max 2 marks:
- Description: percentage of carbon dioxide is higher in expired air / increases from 0.04% to 4.00% [1]
- Explanation: carbon dioxide is produced in respiration / diffuses from blood into alveoli [1]

(b) Max 3 marks:
- goblet cells produce / secrete mucus [1]
- mucus traps pathogens / bacteria / dust / dirt [1]
- ciliated cells / cilia beat / sweep mucus upwards / away from lungs / to throat [1]

(c) Max 3 marks:
- anaerobic respiration occurred (during exercise) [1]
- lactic acid is produced / builds up [1]
- reference to oxygen debt [1]
- extra oxygen is needed to break down / oxidize / remove lactic acid [1]
题目 3 · structured
12 分
3 (a) Define the term *homeostasis*.

...........................................................................................................................................
........................................................................................................................................... [2]

(b) Table 3.1 shows the sweat rate of an adult athlete at different ambient temperatures.

Table 3.1

| Ambient temperature / °C | Sweat rate / g per hour |
| :--- | :--- |
| 15 | 10 |
| 20 | 20 |
| 25 | 50 |
| 30 | 130 |
| 35 | 250 |

(i) Describe the relationship between ambient temperature and sweat rate shown in Table 3.1.

...........................................................................................................................................
........................................................................................................................................... [2]

(ii) Calculate the percentage increase in sweat rate when the ambient temperature rises from 25 °C to 35 °C.

Show your working.

...........................................................................................................................................
...........................................................................................................................................

Percentage increase = ............................................................ % [3]

(c) Explain how sweating helps to lower body temperature.

...........................................................................................................................................
........................................................................................................................................... [2]

(d) Describe how the blood vessels in the skin respond when body temperature increases, and explain how this helps to cool the body.

...........................................................................................................................................
...........................................................................................................................................
........................................................................................................................................... [3]
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解题

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

(b) (i) As the ambient temperature increases, the sweat rate increases. The increase is non-linear, rising much more rapidly at temperatures above 25 °C.
(ii) Change in sweat rate = 250 - 50 = 200 g per hour.
Percentage increase = (200 / 50) * 100 = 400%.

(c) Water in sweat is released onto the skin surface. As this water evaporates, it absorbs and removes heat energy from the skin and underlying blood vessels, which lowers body temperature.

(d) Arterioles in the skin dilate (vasodilation), allowing more blood to flow through the capillaries closer to the skin surface. This increases the loss of heat energy from the blood to the surroundings by radiation, conduction, and convection.

评分标准

3 (a)
maintenance of constant / stable / steady (internal conditions);
internal environment; [2]

(b) (i)
as temperature increases, sweat rate increases;
the increase is non-linear / increases slowly at first then more rapidly / use of paired data points from Table 3.1 with units; [2]

(ii)
correct subtraction: 250 - 50 = 200 (g per hour);
correct division and multiplication: (200 / 50) * 100;
400 (%); [3]

(c)
sweat / water is released onto the skin surface and evaporates / turns into gas;
(evaporation) absorbs / removes heat energy from the skin / body / blood; [2]

(d)
arterioles dilate / widen / undergo vasodilation;
more blood flows through capillaries closer to the skin surface;
more heat is lost to surroundings / air by radiation / conduction / convection; [max 3]
题目 4 · structured
17 分

4 Marram grass, Ammophila arenaria, is a xerophytic plant that grows on coastal sand dunes.

(a) Fig. 4.1 shows a diagram of a cross-section of a rolled leaf of Ammophila arenaria.

(i) Identify two structural adaptations of Ammophila arenaria visible in the leaf cross-section that serve to minimize the loss of water. [2]

(ii) Explain the mechanism by which rolled leaves decrease the rate of transpiration in this species. [3]

(b) An investigation was conducted to measure the speed of transpiration in Ammophila arenaria under varying air movement velocities using a potometer.

Table 4.1 shows the results of the investigation.

Wind speed %% m/s Rate of water loss %% mg per min per cm² 0.51.8 1.03.2 1.54.5 2.05.4 2.55.8 3.05.9

(i) Determine the percentage rise in the rate of moisture release as the velocity of wind accelerates from 0.5 m/s to 2.0 m/s. Show your working. [3]

(ii) Describe the pattern of moisture loss shown in Table 4.1 and explain these observations in terms of physical processes and leaf adaptations. [6]

(c) Water is required for photosynthesis, which takes place in the mesophyll cells of the leaf.

(i) Write the balanced chemical equation representing the process of photosynthesis. [2]

(ii) Identify the specific mineral nutrient needed by plant cells to synthesize chlorophyll. [1]

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解题

(a)(i) Visible xerophytic features of Ammophila arenaria include rolled leaves, a thick waxy cuticle on the outer leaf surface, stomata sunken in pits, and epidermal hairs on the inner curled surface.

(a)(ii) Rolled leaves trap moist, humid air inside the curled leaf structure. This increases the humidity immediately surrounding the stomata, which decreases the water vapour concentration gradient between the inside of the leaf and the external microenvironment. Consequently, diffusion of water vapour out of the stomata is significantly slowed down, reducing overall transpiration.

(b)(i) At 0.5 m/s, the rate of water loss is 1.8 mg per min per cm².
At 2.0 m/s, the rate of water loss is 5.4 mg per min per cm².
Percentage increase = ((5.4 - 1.8) / 1.8) * 100 = (3.6 / 1.8) * 100 = 200%.

(b)(ii) As wind speed increases from 0.5 m/s to 2.0 m/s, the rate of water loss increases rapidly from 1.8 to 5.4 mg per min per cm². At wind speeds above 2.5 m/s, the rate of water loss plateaus, reaching a maximum of 5.9 mg per min per cm².
This occurs because wind carries away water vapour that accumulates outside the stomata, preventing a humid boundary layer from forming. This maintains a steep water vapour concentration gradient, promoting rapid diffusion. At high wind speeds, the boundary layer is completely swept away, meaning further wind increases do not affect the gradient, or stomata close to prevent excessive water loss.

(c)(i) The balanced chemical equation is: 6CO2 + 6H2O → C6H12O6 + 6O2

(c)(ii) Magnesium is the essential mineral ion required by plants to synthesize chlorophyll.

评分标准

(a)(i) [max 2]
- Rolled / curled leaf lamina;
- Presence of epidermal hairs on the inner surface;
- Sunken stomata / stomata located in pits;
- Thick waxy cuticle on the outer epidermis;
- Presence of hinge cells;

(a)(ii) [max 3]
- Leaves roll to trap a layer of moist air / water vapour inside the curled space;
- This reduces the water vapour concentration gradient between the inside and outside of the stomata;
- Decreasing the rate of diffusion / transpiration of water out of the leaf;

(b)(i) [max 3]
- Selection of correct values from table: 1.8 and 5.4 [1];
- Correct calculation process: ((5.4 - 1.8) / 1.8) * 100 [1];
- Correct final value: 200 (%) [1];

(b)(ii) [max 6]
Description (max 3):
- Rate of moisture release increases as wind speed increases;
- Linear / rapid increase at lower wind speeds (0.5 to 2.0 m/s);
- Rate of moisture release levels off / plateaus at higher wind speeds (above 2.5 m/s);
Explanation (max 3):
- Moving air removes the humid boundary layer of water vapour from around the leaf surface;
- This maintains a steep water vapour concentration gradient between the leaf interior and atmosphere;
- Accelerating the diffusion of water vapour through the stomata;
- At very high wind speeds, stomata may close to protect the plant from desiccation / water vapour removal rate reaches its physical limit;

(c)(i) [max 2]
- Correct formulas of reactants and products: 6CO2 + 6H2O → C6H12O6 + 6O2 [1];
- Correct balancing [1];

(c)(ii) [1]
- Magnesium (ion) / Mg2+;

题目 5 · structured
17 分

4 Marram grass, Ammophila arenaria, is a xerophytic plant that grows on coastal sand dunes.

(a) Fig. 4.1 shows a diagram of a cross-section of a rolled leaf of Ammophila arenaria.

(i) Identify two structural adaptations of Ammophila arenaria visible in the leaf cross-section that serve to minimize the loss of water. [2]

(ii) Explain the mechanism by which rolled leaves decrease the rate of transpiration in this species. [3]

(b) An investigation was conducted to measure the speed of transpiration in Ammophila arenaria under varying air movement velocities using a potometer.

Table 4.1 shows the results of the investigation.

Wind speed / m/sRate of water loss / mg per min per cm²0.51.81.03.21.54.52.05.42.55.83.05.9

(i) Determine the percentage rise in the rate of moisture release as the velocity of wind accelerates from 0.5 m/s to 2.0 m/s. Show your working. [3]

(ii) Describe the pattern of moisture loss shown in Table 4.1 and explain these observations in terms of physical processes and leaf adaptations. [6]

(c) Water is required for photosynthesis, which takes place in the mesophyll cells of the leaf.

(i) Write the balanced chemical equation representing the process of photosynthesis. [2]

(ii) Identify the specific mineral nutrient needed by plant cells to synthesize chlorophyll. [1]

查看答案详解

解题

(a)(i) Visible xerophytic features of Ammophila arenaria include rolled leaves, a thick waxy cuticle on the outer leaf surface, stomata sunken in pits, and epidermal hairs on the inner curled surface.

(a)(ii) Rolled leaves trap moist, humid air inside the curled leaf structure. This increases the humidity immediately surrounding the stomata, which decreases the water vapour concentration gradient between the inside of the leaf and the external microenvironment. Consequently, diffusion of water vapour out of the stomata is significantly slowed down, reducing overall transpiration.

(b)(i) At 0.5 m/s, the rate of water loss is 1.8 mg per min per cm².
At 2.0 m/s, the rate of water loss is 5.4 mg per min per cm².
Percentage increase = ((5.4 - 1.8) / 1.8) * 100 = (3.6 / 1.8) * 100 = 200%.

(b)(ii) As wind speed increases from 0.5 m/s to 2.0 m/s, the rate of water loss increases rapidly from 1.8 to 5.4 mg per min per cm². At wind speeds above 2.5 m/s, the rate of water loss plateaus, reaching a maximum of 5.9 mg per min per cm².
This occurs because wind carries away water vapour that accumulates outside the stomata, preventing a humid boundary layer from forming. This maintains a steep water vapour concentration gradient, promoting rapid diffusion. At high wind speeds, the boundary layer is completely swept away, meaning further wind increases do not affect the gradient, or stomata close to prevent excessive water loss.

(c)(i) The balanced chemical equation is: 6CO2 + 6H2O → C6H12O6 + 6O2

(c)(ii) Magnesium is the essential mineral ion required by plants to synthesize chlorophyll.

评分标准

(a)(i) [max 2]
- Rolled / curled leaf lamina;
- Presence of epidermal hairs on the inner surface;
- Sunken stomata / stomata located in pits;
- Thick waxy cuticle on the outer epidermis;
- Presence of hinge cells;

(a)(ii) [max 3]
- Leaves roll to trap a layer of moist air / water vapour inside the curled space;
- This reduces the water vapour concentration gradient between the inside and outside of the stomata;
- Decreasing the rate of diffusion / transpiration of water out of the leaf;

(b)(i) [max 3]
- Selection of correct values from table: 1.8 and 5.4 [1];
- Correct calculation process: ((5.4 - 1.8) / 1.8) * 100 [1];
- Correct final value: 200 (%) [1];

(b)(ii) [max 6]
Description (max 3):
- Rate of moisture release increases as wind speed increases;
- Linear / rapid increase at lower wind speeds (0.5 to 2.0 m/s);
- Rate of moisture release levels off / plateaus at higher wind speeds (above 2.5 m/s);
Explanation (max 3):
- Moving air removes the humid boundary layer of water vapour from around the leaf surface;
- This maintains a steep water vapour concentration gradient between the leaf interior and atmosphere;
- Accelerating the diffusion of water vapour through the stomata;
- At very high wind speeds, stomata may close to protect the plant from desiccation / water vapour removal rate reaches its physical limit;

(c)(i) [max 2]
- Correct formulas of reactants and products: 6CO2 + 6H2O → C6H12O6 + 6O2 [1];
- Correct balancing [1];

(c)(ii) [1]
- Magnesium (ion) / Mg2+;

题目 6 · structured
13 分
Protease enzymes can be produced on an industrial scale using the bacterium *Bacillus subtilis* grown in a fermenter.

**(a)**
(i) State the name of the kingdom to which *Bacillus subtilis* belongs. [1]
(ii) State **two** reasons, other than their rapid reproduction rate, why bacteria are useful organisms to use in biotechnology. [2]

**(b)**
The temperature in the fermenter must be carefully controlled.
(i) Explain why the temperature of the fermenter would increase if it is not cooled, and how the water jacket helps to control this temperature. [3]
(ii) State the name of the gas that must be supplied to the fermenter to allow the bacteria to respire aerobically. [1]
(iii) Explain the roles of the stirrer (paddles) in the fermenter. [2]

**(c)**
The protease enzymes harvested from the fermenter can be added to biological washing powders to help remove stains.
(i) Explain how protease enzymes in washing powder remove blood stains from clothes. [3]
(ii) Some species of *Bacillus* are pathogenic. State the definition of a pathogen. [1]
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解题

**(a)**
(i) Prokaryote / Bacteria
(ii) Any two from: ability to make complex molecules, lack of ethical concerns, genetic code shared with all other organisms, presence of plasmids.

**(b)**
(i) Aerobic respiration of the bacteria releases heat energy. The water jacket allows cold water to flow around the outer wall of the fermenter, absorbing heat to maintain the optimum temperature and prevent enzymes from denaturing.
(ii) Oxygen
(iii) Stirring ensures that nutrients, oxygen, and temperature are distributed evenly throughout the fermenter, and prevents the bacteria from settling at the bottom.

**(c)**
(i) Blood stains contain insoluble proteins (such as haemoglobin). Protease enzymes break down these large, insoluble proteins into small, soluble amino acids, which can then easily dissolve in water and be washed away.
(ii) A disease-causing organism.

评分标准

**(a) (i)**
Prokaryote / Prokaryotidae / Bacteria [1]

**(a) (ii)**
Any two from:
- ability to make complex molecules [1]
- lack of ethical concerns [1]
- genetic code shared with all other organisms [1]
- presence of plasmids [1]
*(Ignore: rapid reproduction rate - given in stem)*

**(b) (i)**
1. (Aerobic) respiration releases heat / thermal energy; [1]
2. Cold water circulates / flows through the water jacket; [1]
3. To absorb heat / maintain optimum temperature / prevent denaturation of enzymes; [1]

**(b) (ii)**
Oxygen (gas); [1]

**(b) (iii)**
Any two from:
- Distributes nutrients / oxygen / heat evenly / maintains uniform temperature; [1]
- Prevents bacteria from settling / keeps bacteria in suspension; [1]
- Improves contact between bacteria and nutrients / oxygen; [1]

**(c) (i)**
1. Protease breaks down / digests / hydrolyses proteins (in the blood stain); [1]
2. Into soluble amino acids / peptides; [1]
3. Which dissolve in water / are washed away easily; [1]

**(c) (ii)**
Disease-causing organism; [1]
题目 7 · structured
13 分
A mangrove swamp is a unique coastal ecosystem. Fig. 6.1 shows a food web for a mangrove swamp.

$$\text{mangrove leaves} \longrightarrow \text{fiddler crabs} \longrightarrow \text{egrets} \longrightarrow \text{crocodiles}$$
$$\text{mangrove leaves} \longrightarrow \text{mosquitoes} \longrightarrow \text{tree frogs} \longrightarrow \text{egrets}$$
$$\text{mangrove leaves} \longrightarrow \text{mosquitoes} \longrightarrow \text{small fish} \longrightarrow \text{large fish} \longrightarrow \text{crocodiles}$$

**Fig. 6.1**

**(a) (i)** Identify from the food web in Fig. 6.1:
- the producer: .......................................................................
- a tertiary consumer: ............................................................. [2]

**(a) (ii)** State the name of the organism in Fig. 6.1 that feeds at more than one trophic level.
............................................................................................................................................. [1]

**(b)** Only a small proportion of the energy in the mangrove leaves is transferred to the crocodiles.
Explain why energy is lost at each trophic level in a food chain.
.............................................................................................................................................
.............................................................................................................................................
............................................................................................................................................. [3]

**(c) (i)** Mangrove forests are often cleared for coastal development and aquaculture.
Suggest two undesirable effects of clearing mangrove forests on the local ecosystem.
1. .........................................................................................................................................
2. ......................................................................................................................................... [2]

**(c) (ii)** Explain what is meant by managing a resource, such as a forest, in a *sustainable* way.
.............................................................................................................................................
............................................................................................................................................. [2]

**(d)** A conservation group plans to protect a rare bird species living in the mangrove swamp.
Describe three methods, other than habitat protection, that can be used to conserve endangered animal species.
1. .........................................................................................................................................
2. .........................................................................................................................................
3. ......................................................................................................................................... [3]
查看答案详解

解题

**(a) (i)**
- The producer in any food web is the photosynthetic organism that makes its own organic nutrients using light energy. In Fig. 6.1, this is **mangrove leaves**.
- A tertiary consumer feeds on secondary consumers (at the fourth trophic level). In the path: leaves (1st) -> mosquitoes (2nd) -> tree frogs (3rd) -> **egrets** (4th), the egrets act as a tertiary consumer. In the path: leaves (1st) -> mosquitoes (2nd) -> small fish (3rd) -> **large fish** (4th) -> crocodiles (5th), the **large fish** is the tertiary consumer and **crocodiles** are quaternary consumers.

**(a) (ii)**
**egrets** feed on fiddler crabs (which eat mangrove leaves, making egrets secondary consumers at the 3rd trophic level) and on tree frogs (making egrets tertiary consumers at the 4th trophic level).

**(b)**
Energy is lost at each transfer along a food chain due to several biological processes:
- Energy is released during respiration and lost as heat to the environment.
- Some energy remains in excretory waste products (urine) or egested materials (faeces) which are not passed to the next consumer.
- Not all parts of an organism are consumed or can be digested by the predator (e.g., bones, shells).
- Energy is used by the organism itself for movement, active transport, and metabolic maintenance.

**(c) (i)**
Clearing mangroves has severe environmental impacts:
- **Habitat loss**: Many marine and terrestrial organisms rely on mangroves as nursery grounds and shelters.
- **Coastal instability**: Mangroves stabilise coastal soils; clearing them leads to rapid coastal erosion and increased flooding from storms.

**(c) (ii)**
Sustainable management means using a natural resource at a rate that allows it to regenerate naturally so that it never runs out (e.g., replanting harvested trees, establishing logging quotas).

**(d)**
Methods to conserve endangered species beyond habitat protection include:
- Captive breeding and release.
- Legislation and international treaties banning hunting and trade.
- Public education to reduce human-wildlife conflict.

评分标准

**(a) (i)** [2 marks total]
- producer: mangrove leaves [1]
- tertiary consumer: egrets / large fish / crocodiles [1]

**(a) (ii)** [1 mark total]
- egrets [1]

**(b)** [3 marks total]
Any three from:
- energy lost as heat (from respiration) ; [1]
- not all parts of the organism are eaten / some parts are indigestible ; [1]
- energy lost in excretory products (urine / urea) / egested material (faeces) ; [1]
- energy used by the organism for movement / growth / active transport ; [1]

**(c) (i)** [2 marks total]
Any two from:
- loss of habitat (for coastal / marine organisms) ; [1]
- disruption to food webs / reduction in biodiversity / extinction of species ; [1]
- coastal erosion / increased risk of flooding ; [1]
- less carbon dioxide removed from atmosphere (less photosynthesis) / global warming ; [1]

**(c) (ii)** [2 marks total]
- resource is produced / replaced as rapidly as it is removed ; [1]
- so that it does not run out / remains available for the future / replanting trees ; [1]

**(d)** [3 marks total]
Any three from:
- captive breeding programmes ; [1]
- ban hunting / poaching / trade of the species (or laws / legislation) ; [1]
- education (of local people / public) ; [1]
- monitoring population sizes / warden patrols ; [1]
- seed banks / gene banks / zoos / wildlife parks ; [1]

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