📚 Organisms 2.1.3 – Mechanism of Breathing Exam Practice | 生物 2.1.3 呼吸机制真题精练
Welcome to your focused revision session on the mechanism of breathing. This article covers typical exam questions and model answers for CIE/IAL Biology Topic 2.1.3. By the end, you will be confident in explaining inspiration, expiration, pressure changes, spirometry, and related calculations using precise biological terminology.
欢迎进行呼吸机制的专项复习。本文针对 CIE / IAL 生物课题 2.1.3 精选了典型考题与标准答案。学完本文，你将能自信地使用准确的生物学术语解释吸气、呼气、压强变化、肺量测定以及相关计算。
1. Key Structures for Ventilation | 通气关键结构
The diaphragm is a dome-shaped sheet of muscle that forms the floor of the thoracic cavity. The external intercostal muscles run between the ribs and are essential for lifting the rib cage. The internal intercostal muscles lie deeper and assist mainly during forceful expiration. Pleural membranes line the lungs and the chest wall, creating a fluid-filled pleural cavity that reduces friction and sticks the lungs to the rib cage.
膈肌是一片穹顶状的肌肉，构成胸腔的底部。外肋间肌在肋骨之间走行，对上提胸廓至关重要。内肋间肌位置更深，主要在用力呼气时提供帮助。胸膜覆盖在肺脏与胸壁表面，形成的胸膜腔充满液体，既能减少摩擦又能使肺紧贴胸廓。
2. The Process of Inspiration | 吸气过程
Quiet inspiration is an active process. The diaphragm contracts and flattens, moving downwards. At the same time, the external intercostal muscles contract, pulling the ribs upwards and outwards. These movements increase the volume of the thoracic cavity. As a result, the pressure inside the lungs drops below atmospheric pressure, and air flows in through the airways to equalise the pressure difference.
平静吸气是一个主动过程。膈肌收缩变平、向下移动。同时外肋间肌收缩，将肋骨向上向外牵拉。这些动作增大了胸腔的容积，使肺内压强降至大气压以下，空气便经气道流入以平衡压力差。
During deep or forced inspiration, accessory muscles such as the sternocleidomastoid and scalenes also contract to raise the sternum and first two ribs even further, maximising thoracic volume.
在深呼吸或用力吸气时，胸锁乳突肌和斜角肌等辅助肌肉也会收缩，进一步上提胸骨与前两根肋骨，将胸腔容积最大化。
3. The Process of Expiration | 呼气过程
Quiet expiration is typically a passive process. The diaphragm and external intercostal muscles relax. The diaphragm returns to its domed shape, and the rib cage moves down and inwards due to gravity and the elastic recoil of lung tissue. This decreases thoracic volume, so the pressure inside the lungs becomes higher than atmospheric pressure, forcing air out.
平静呼气通常是一个被动过程。膈肌和外肋间肌舒张，膈肌恢复穹顶状，胸廓在重力和肺组织弹性回缩作用下向下向内移动。胸腔容积减小，导致肺内压强高于大气压，气体被挤出。
During forced expiration, the internal intercostal muscles contract to pull the ribs down and in, and the abdominal muscles contract to push the diaphragm further upwards. This produces a more rapid and complete expulsion of air.
用力呼气时，内肋间肌收缩将肋骨向下向内牵拉，腹肌收缩进一步向上推挤膈肌，从而更快、更彻底地呼出气体。
4. Pressure–Volume Relationship (Boyle’s Law) | 压力–体积关系（波义耳定律）
Breathing relies on the inverse relationship between pressure and volume at constant temperature, described by Boyle’s Law. When the volume of the thoracic cavity increases, the pressure inside the lungs decreases, and vice versa.
呼吸依赖于恒温下压强与体积的反比关系，即波义耳定律。当胸腔容积增大时，肺内压强降低；反之亦然。
P₁V₁ = P₂V₂ (at constant temperature)
P₁V₁ = P₂V₂（恒温条件下）
A typical exam question asks: ‘If the thoracic volume increases from 2.5 dm³ to 3.2 dm³ and the initial intrapulmonary pressure is 100 kPa, calculate the new pressure, assuming no temperature change.’ Using Boyle’s Law: 100 kPa × 2.5 dm³ = P₂ × 3.2 dm³. Therefore, P₂ = (100 × 2.5) ÷ 3.2 = 78.1 kPa. The pressure drop allows air to enter the lungs.
一道典型考题会问：“如果胸腔容积从 2.5 dm³ 增加到 3.2 dm³，初始肺内压为 100 kPa，假设温度不变，计算新的压强。”利用波义耳定律：100 kPa × 2.5 dm³ = P₂ × 3.2 dm³。因此，P₂ = (100 × 2.5) ÷ 3.2 = 78.1 kPa。压力下降使空气得以进入肺部。
5. Lung Volumes: Definitions | 肺容积定义
Tidal volume (TV) is the volume of air breathed in or out during a normal, quiet breath. In a typical adult, TV is approximately 0.5 dm³ (500 cm³).
潮气量 (TV) 是平静呼吸时每次吸入或呼出的气体量。成人潮气量约为 0.5 dm³ (500 cm³)。
Inspiratory reserve volume (IRV) is the extra volume of air that can be forced in beyond a normal inspiration. Expiratory reserve volume (ERV) is the extra volume of air that can be forced out after a normal expiration.
补吸气量 (IRV) 是在平静吸气末再用力吸气所能额外吸入的气体量。补呼气量 (ERV) 是在平静呼气末再用力呼气所能额外呼出的气体量。
Residual volume (RV) is the air that remains in the lungs even after a maximal expiration; this cannot be measured by a simple spirometer. Vital capacity (VC) is the total volume of air that can be moved in and out of the lungs (VC = TV + IRV + ERV).
残气量 (RV) 是最大呼气后仍残留在肺内的气体量，无法用普通肺量计测出。肺活量 (VC) 为能进出肺部的最大气体总量（VC = TV + IRV + ERV）。
6. Interpreting a Spirometer Trace | 解读肺量计曲线
A spirometer produces a graph of volume changes over time. A typical trace shows regular waves during quiet breathing. The vertical distance between a peak (end of inspiration) and the following trough (end of expiration) represents tidal volume.
肺量计会生成一条容积随时间变化的曲线。典型曲线在平静呼吸期间呈现规则的波形。波峰（吸气末）与后续波谷（呼气末）之间的垂直距离即为潮气量。
To determine inspiratory reserve volume from a trace, identify the highest peak during forced inspiration from a normal breath and measure the extra volume beyond the tidal peak. Similarly, expiratory reserve volume is the extra volume exhaled beyond the normal tidal trough during forced expiration.
要从曲线确定补吸气量，可在一次正常呼吸后尽力吸气的最高峰处，测量超出潮气波峰的额外容积。同理，补呼气量是在尽力呼气时超出正常潮气波谷的额外容积。
Vital capacity is read as the total vertical distance between a forced maximal inspiration peak and a forced maximal expiration trough. The trace also allows you to calculate breathing rate by counting the number of breathing cycles in a known time interval.
肺活量可读取为最大吸气峰与最大呼气谷之间的总垂直距离。曲线还可用于计算呼吸频率：只需数出已知时间间隔内的呼吸周期数即可。
7. Calculating Minute Ventilation | 计算每分通气量
Minute ventilation (pulmonary ventilation) is the total volume of air moved into the lungs per minute. It is calculated using the formula:
每分通气量（肺通气量）是每分钟进入肺部的气体总量，计算公式为：
Minute ventilation = Tidal volume × Breathing rate
每分通气量 = 潮气量 × 呼吸频率
Example: A person has a tidal volume of 0.45 dm³ and a breathing rate of 14 breaths per minute. Calculate their minute ventilation.
例题：某人的潮气量为 0.45 dm³，呼吸频率为每分钟 14 次。计算其每分通气量。
Minute ventilation = 0.45 dm³ × 14 min⁻¹ = 6.3 dm³ min⁻¹.
每分通气量 = 0.45 dm³ × 14 min⁻¹ = 6.3 dm³ min⁻¹。
If the person begins to exercise and both tidal volume and breathing rate increase, minute ventilation rises sharply. For instance, if TV becomes 2.0 dm³ and rate becomes 28 breaths min⁻¹, minute ventilation becomes 56 dm³ min⁻¹, delivering more oxygen to working muscles.
如果此人开始运动，潮气量和呼吸频率都增加，每分通气量会急剧上升。例如，TV 变为 2.0 dm³，频率变为 28 次/分钟，每分通气量则达到 56 dm³ min⁻¹，为工作肌肉输送更多的氧气。
8. Exam-style Data Analysis | 真题风格数据分析
The table below shows data from a student at rest and during two levels of exercise.
下表是某学生在静息和两种运动强度下的数据。
| Activity | Tidal volume / dm³ | Breathing rate / min⁻¹ | Minute ventilation / dm³ min⁻¹ |
|---|---|---|---|
| Rest | 0.50 | 12 | 6.0 |
| Light jog | 1.20 | 22 | 26.4 |
| Sprinting | 2.10 | 35 | 73.5 |
Explain the changes in minute ventilation as exercise intensity increases.
解释随着运动强度增加，每分通气量的变化。
From rest to sprinting, minute ventilation increases about 12‑fold. This happens because the demand for oxygen rises and more carbon dioxide must be removed. Both tidal volume and breathing rate increase to achieve higher ventilation. The data show TV more than quadruples, while breathing rate almost triples. The combination gives a dramatic increase in air flow.
从静息到冲刺跑，每分通气量增大约 12 倍。这是因为需氧量升高、更多的二氧化碳需要被清除。潮气量和呼吸频率双双增加，以实现更高的通气量。数据显示潮气量增到四倍多，呼吸频率接近三倍，共同带来气流的急剧提升。
Under nervous or hormonal control (e.g., adrenaline), the respiratory centre in the medulla oblongata sends more frequent impulses to the diaphragm and intercostal muscles, raising both tidal volume and rate.
在神经或激素（如肾上腺素）的控制下，延髓中的呼吸中枢更频繁地向膈肌和肋间肌发送冲动，从而同时提升潮气量和频率。
9. Common Mistakes to Avoid | 常见错误避坑
Mistake 1: ‘Expiration is an active process.’ In reality, quiet expiration is passive and relies on elastic recoil. Only forced expiration is active.
错误 1：“呼气是主动过程。”实际上，平静呼气是被动的，依赖弹性回缩。只有用力呼气才是主动的。
Mistake 2: ‘The lungs expand on their own.’ The lungs are passive; they follow the movements of the thoracic wall because of the cohesion created by the pleural fluid.
错误 2：“肺会自己扩张。”肺是被动器官，它们因胸膜液产生的附着力而跟随胸壁运动。
Mistake 3: Confusing the roles of internal and external intercostal muscles. External intercostals are for inspiration (lift the ribs up), internal intercostals are mainly for forced expiration (pull the ribs down).
错误 3：混淆内、外肋间肌的作用。外肋间肌用于吸气（上提肋骨），内肋间肌主要用于用力呼气（下拉肋骨）。
Mistake 4: Ignoring units in calculations. Always check whether volumes are given in dm³ or cm³. Consistency is key: 1 dm³ = 1000 cm³. If you mix units, answers will be wrong.
错误 4：计算时不注意单位。务必检查体积单位是 dm³ 还是 cm³。保持单位一致是关键：1 dm³ = 1000 cm³。若混用单位，答案必然错误。
10. Quick Practice Quiz | 快速练习小测
1. Which muscles contract during quiet inspiration? (a) Internal intercostals only, (b) Diaphragm and external intercostals, (c) Diaphragm and internal intercostals, (d) Abdominal muscles only.
1. 平静吸气时哪些肌肉收缩？(a) 仅内肋间肌，(b) 膈肌和外肋间肌，(c) 膈肌和内肋间肌，(d) 仅腹肌。
Answer: (b) Diaphragm and external intercostals contract during quiet inspiration, increasing thoracic volume.
答案：(b) 膈肌和外肋间肌收缩，增大胸腔容积。
2. State whether quiet expiration is an active or passive process and explain why.
2. 判断平静呼气是主动还是被动过程，并解释原因。
It is passive. The diaphragm and external intercostals relax, allowing elastic recoil of the lungs and rib cage to decrease thoracic volume, pushing air out.
是被动过程。膈肌和外肋间肌舒张，肺与胸廓的弹性回缩使胸腔容积减小，将空气排出。
3. During an asthma attack, the airways narrow, making expiration difficult. Explain why forced expiration becomes an active process.
3. 哮喘发作时气道变窄，呼气困难。解释为什么此时用力呼气成为主动过程。
To overcome increased airway resistance, the internal intercostal muscles and abdominal muscles must contract strongly to generate higher pressure in the thoracic cavity and expel air.
为克服增加的气道阻力，内肋间肌和腹肌必须强力收缩，在胸腔内产生更高的压力以挤出空气。
4. A spirometer trace shows 5 complete breaths in 20 seconds. Calculate the breathing rate.
4. 肺量计曲线显示在 20 秒内有 5 次完整呼吸。计算呼吸频率。
20 s = 1/3 minute. Breathing rate = 5 breaths ÷ (1/3 min) = 15 breaths min⁻¹.
20 秒 = 1/3 分钟。呼吸频率 = 5 次 ÷ (1/3 min) = 15 次/分钟。
5. True or False: Pleural fluid allows the lungs to slide past the rib cage with minimal friction and helps keep the lungs inflated.
5. 判断对错：胸膜液能使肺在紧贴胸廓的同时以最小摩擦滑动，并有助于保持肺的扩张状态。
True. The pleural fluid lubricates the surfaces and the negative pressure in the pleural cavity keeps the lungs expanded against the rib cage.
对。胸膜液起润滑作用，胸膜腔内的负压则使肺紧贴胸廓保持扩张。
Published by TutorHao | Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导