Chapter 14: Breathing and Exchange of Gases Case Study Questions | biology Class 11 CBSE

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Chapter 14: Breathing and Exchange of Gases Case Study Questions | biology Class 11 CBSE | ByteNotes.in

Case Study

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Case Set

Case Set 1

Case Analysis

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Rohan, a 16-year-old swimmer, undergoes a pulmonary function test before joining the school swim team. The doctor uses a spirometer and records the following: Tidal Volume = 500 mL, Inspiratory Reserve Volume = 3000 mL, Expiratory Reserve Volume = 1100 mL, Residual Volume = 1200 mL. The doctor notes that Rohan has an excellent lung profile for a swimmer. She explains that the various volumes can be combined to derive capacities that are used in clinical diagnosis. She emphasises that one particular volume can never be measured directly by the spirometer but must be estimated indirectly, and that its presence in the lungs between breaths is physiologically crucial for maintaining continuous gas exchange.

Question 1: From the spirometric data given, calculate Rohan's Vital Capacity (VC).

Vital Capacity (VC) = ERV + TV + IRV
VC = 1100 + 500 + 3000 = 4600 mL
VC represents the maximum volume of air Rohan can breathe in after a forced expiration, indicating excellent lung function suitable for competitive swimming.

Question 2: Which volume did the doctor say cannot be measured directly by the spirometer, and why is its presence physiologically important?

The Residual Volume (RV = 1100–1200 mL) cannot be measured directly by a spirometer because it is the air remaining in the lungs even after a forcible expiration and cannot be voluntarily expelled.
Its presence prevents complete collapse of the alveoli between breaths, ensuring the respiratory surface remains open for continuous gas exchange.
RV also contributes to Functional Residual Capacity (FRC = ERV + RV), which maintains relatively stable partial pressures of O2 and CO2 in the alveoli between breaths.

Question 3: Calculate Rohan's Total Lung Capacity (TLC) and Functional Residual Capacity (FRC). Explain why TLC is always greater than Vital Capacity.

TLC = RV + ERV + TV + IRV = 1200 + 1100 + 500 + 3000 = 5800 mL; FRC = ERV + RV = 1100 + 1200 = 2300 mL.
TLC is always greater than VC because TLC = VC + RV; it includes the Residual Volume that can never be voluntarily expelled.
VC represents only the air exchangeable by voluntary effort, whereas TLC represents the absolute maximum air the lungs can hold — the difference (RV) is the permanent air reservoir that keeps alveoli patent and facilitates uninterrupted diffusion of O2 and CO2.

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Case Set 2

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A first-year medical student is studying a table of partial pressures of respiratory gases at different sites in the body. The table shows: Atmospheric air — pO2: 159 mm Hg, pCO2: 0.3 mm Hg; Alveoli — pO2: 104 mm Hg, pCO2: 40 mm Hg; Deoxygenated blood — pO2: 40 mm Hg, pCO2: 45 mm Hg; Oxygenated blood — pO2: 95 mm Hg, pCO2: 40 mm Hg; Tissues — pO2: 40 mm Hg, pCO2: 45 mm Hg. The student is asked to trace the movement of both gases through the complete respiratory cycle and explain why the diffusion membrane is ideally designed for this exchange. The student also notices that despite CO2 having a smaller partial pressure gradient at the alveoli, it is exchanged efficiently.

Question 1: Using the partial pressure data, state the direction of O2 movement at the alveoli and at the tissues.

At the alveoli: O2 moves from alveolar air (pO2 = 104 mm Hg) into deoxygenated blood (pO2 = 40 mm Hg) — down a gradient of 64 mm Hg.
At the tissues: O2 moves from oxygenated blood (pO2 = 95 mm Hg) into tissues (pO2 = 40 mm Hg) — down a gradient of 55 mm Hg.
In both cases O2 diffuses from high to low partial pressure, establishing a unidirectional flow from atmosphere to metabolising cells.

Question 2: Why does CO2 diffuse efficiently across the alveolar membrane despite having a smaller partial pressure gradient compared to O2?

The partial pressure gradient for CO2 at the alveoli is only 5 mm Hg (blood: 45 mm Hg → alveoli: 40 mm Hg), much smaller than the 64 mm Hg gradient for O2.
However, the solubility of CO2 is 20–25 times higher than that of O2, so the amount of CO2 that can diffuse per unit difference in partial pressure is much higher than for O2.
This high solubility more than compensates for the smaller pressure gradient, ensuring rapid and efficient CO2 clearance from blood at the alveoli.

Question 3: Describe the three layers of the diffusion membrane and explain how each structural feature contributes to efficient gas exchange.

The diffusion membrane consists of: (i) the thin squamous epithelium of the alveoli — being only one cell thick, it offers minimal diffusion distance for gases passing from alveolar air into the tissue layer.
(ii) The basement substance — a thin matrix layer between the epithelium and endothelium that provides structural support while remaining highly permeable to respiratory gases.
(iii) The endothelium of alveolar capillaries — a single endothelial cell layer that is the final barrier before gases enter the blood; its thinness ensures that O2 rapidly enters RBCs and CO2 rapidly exits.
The total thickness of all three layers combined is much less than a millimetre, minimising diffusion distance and maximising the rate of exchange for both O2 and CO2.

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Case Set 3

Case Analysis

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Dr. Meera is treating a 55-year-old retired construction worker who spent 30 years working in a stone-crushing plant without respiratory protection. He now presents with progressive shortness of breath, reduced exercise tolerance, and a chronic cough. A spirometry test reveals markedly elevated Residual Volume and reduced Vital Capacity. A chest X-ray shows dense fibrous tissue throughout both lung fields. Dr. Meera also notes in her patient history that the man smoked cigarettes from age 20 to 50. She suspects two concurrent respiratory disorders. She explains to a medical student accompanying her that dust exposure and smoking affect the lungs through different but equally damaging mechanisms.

Question 1: Name the two respiratory disorders Dr. Meera suspects and identify the primary cause of each.

The first suspected disorder is Occupational Respiratory Disorder (pulmonary fibrosis) caused by long-term exposure to excessive stone dust in the grinding/crushing industry, which overwhelmed the body's defence mechanisms leading to inflammation and fibrosis.
The second suspected disorder is Emphysema, caused by cigarette smoking — one of the major causes identified in the chapter — which damages alveolar walls and reduces the respiratory surface.
Both disorders are chronic and progressive, explaining the patient's long history of worsening respiratory symptoms.

Question 2: Explain the mechanism by which long-term dust exposure leads to fibrosis and lung damage.

In industries involving grinding or stone-breaking, so much dust is produced that the body's defence mechanisms cannot fully cope with continuous particle inhalation.
Long-term exposure causes persistent inflammation of the lung tissue as the immune system repeatedly attempts to clear the particles.
This chronic inflammation triggers fibrosis — the proliferation of fibrous (connective) tissue — which stiffens and scars the lungs, reducing compliance and causing serious, often irreversible lung damage.

Question 3: Analyse how emphysema leads to the elevated Residual Volume and reduced Vital Capacity seen in the spirometry results.

In emphysema, the walls of the alveoli are damaged and destroyed; smaller alveolar units merge into larger, less elastic air spaces, reducing the elastic recoil of the lungs.
With reduced elastic recoil, the lungs cannot fully deflate during expiration — air becomes trapped in the damaged alveolar spaces, causing Residual Volume to increase significantly above normal (1100–1200 mL).
The destruction of alveolar walls also reduces the total Expiratory Reserve Volume (since the damaged lung cannot expel as much air), and the overall Vital Capacity (VC = ERV + TV + IRV) falls because ERV decreases.
Additionally, the reduced respiratory surface from alveolar wall destruction impairs gas exchange, worsening the patient's shortness of breath and exercise intolerance beyond what the spirometry volumes alone indicate.

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Case Set 4

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During a biology practical class, students are shown an animation of the human respiratory system. The animation depicts air entering through external nostrils, travelling through various structures, and finally reaching tiny balloon-like sacs deep in the lungs. The teacher pauses the animation and asks students to note that the trachea and larger airways are supported by rings of a particular tissue, while the final gas-exchange structures are not. She also explains that the lungs are housed inside an air-tight cavity whose walls are made of bone, cartilage, and muscle, and that the lungs themselves cannot be directly controlled — they depend entirely on this cavity for volume changes. She then asks students to explain what would happen to gas exchange if the thoracic cavity developed a leak.

Question 1: What tissue supports the trachea and bronchi, and why is this support necessary?

The trachea, primary, secondary, and tertiary bronchi, and initial bronchioles are supported by incomplete cartilaginous rings.
This cartilaginous support prevents the larger airways from collapsing inward during inspiration when intra-pulmonary pressure is less than atmospheric pressure, ensuring uninterrupted air passage to the alveoli.
Terminal bronchioles and alveoli lack this support because they need to expand and contract with breathing; rigid cartilage would prevent the flexibility required for gas exchange.

Question 2: Describe the anatomical boundaries of the thoracic chamber and explain why it must remain air-tight for breathing to occur.

The thoracic chamber is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs, and on the lower side by the dome-shaped diaphragm.
The lungs are situated within this anatomically air-tight chamber; because pulmonary volume cannot be directly altered, any change in thoracic volume is reflected as an equal change in pulmonary volume.
The air-tight nature is essential: if the thoracic cavity were not sealed, changes in diaphragm and intercostal muscle positions would not create the pressure gradients needed to move air in and out of the lungs.

Question 3: If the thoracic cavity developed a hole (pneumothorax), explain step by step how breathing and gas exchange would be disrupted.

A hole in the thoracic cavity breaks the air-tight seal; outside air enters the thoracic cavity directly, equalising the pressure between the outside and the pleural space around the lungs.
Without the sealed cavity, contraction of the diaphragm and intercostal muscles can no longer reduce intra-pulmonary pressure below atmospheric pressure — no pressure gradient is generated for inspiration.
Air stops flowing into the lungs during inspiration attempts; the lung on the affected side collapses (atelectasis) because the mechanical coupling between thoracic expansion and lung expansion is lost.
With collapsed or non-expanding lungs, the alveolar surface for gas exchange is drastically reduced; O2 cannot diffuse from alveolar air into blood and CO2 cannot be cleared, leading to respiratory failure if not treated.

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Case Set 5

Case Analysis

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Priya is a physiologist studying the oxygen-carrying properties of blood. She draws blood samples from a volunteer at rest, notes the pO2 and pCO2 in pulmonary venous blood and systemic venous blood, and plots an oxygen dissociation curve. She observes that when blood is exposed to conditions mimicking active muscle tissue — elevated CO2, increased H+ concentration, and higher temperature — the curve shifts and haemoglobin releases more O2 than under resting conditions. She also notes from the data that 97% of O2 is carried bound to haemoglobin while 3% is dissolved in plasma, and she records that every 100 mL of oxygenated blood delivers approximately 5 mL of O2 to tissues under normal physiological conditions.

Question 1: State the pO2 values in oxygenated blood (pulmonary venous blood) and in the tissues, and explain what drives O2 from blood to tissues.

pO2 in oxygenated blood (systemic arteries/pulmonary veins) = 95 mm Hg; pO2 in tissues = 40 mm Hg.
The partial pressure gradient of 55 mm Hg (95 − 40) drives O2 to diffuse from the oxygenated blood into the tissues by simple diffusion — from high to low partial pressure.
This diffusion occurs across the capillary endothelium and tissue cell membranes, and is facilitated by the thin membranes involved and the continuous consumption of O2 in cellular respiration, which maintains a low tissue pO2.

Question 2: Explain the shape of the oxygen dissociation curve and why it is described as sigmoid.

The oxygen dissociation curve is obtained by plotting percentage saturation of haemoglobin with O2 against pO2; it has a characteristic sigmoid (S-shaped) pattern.
The sigmoid shape reflects cooperative binding: as the first O2 molecule binds to haemoglobin, the affinity of the remaining subunits for O2 increases, producing the steep rising portion of the curve at intermediate pO2 values.
At very high pO2 (alveoli), haemoglobin becomes nearly fully saturated — the curve plateaus; at low pO2 (tissues), haemoglobin releases O2 steeply — the curve drops sharply — making it highly efficient at both loading and unloading O2.

Question 3: Analyse how elevated CO2, increased H+ concentration, and higher temperature in active muscle tissue shift haemoglobin's behaviour, and explain the physiological advantage of this shift.

In active muscle: cellular catabolism produces large amounts of CO2, raising pCO2 and driving the carbonic anhydrase reaction to produce more H+, lowering pH; metabolic heat also raises local temperature.
Each of these factors — high pCO2, high H+ concentration, and elevated temperature — individually and collectively reduces haemoglobin's affinity for O2, shifting the oxygen dissociation curve to the right (the Bohr effect).
This rightward shift means haemoglobin releases more O2 at the same pO2; in highly active muscle, this ensures a greater O2 supply exactly where metabolic demand is highest — an elegant self-regulating mechanism.
The physiological advantage is that O2 delivery is automatically matched to metabolic rate: the more active a tissue, the more CO2 and H+ it produces, the more O2 haemoglobin releases to it — without requiring neural or hormonal signals.

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Chapter sections

Chapter 14: Breathing and Exchange of Gases | Case Study Questions

Passage

Question 1: From the spirometric data given, calculate Rohan's Vital Capacity (VC).

Question 2: Which volume did the doctor say cannot be measured directly by the spirometer, and why is its presence physiologically important?

Question 3: Calculate Rohan's Total Lung Capacity (TLC) and Functional Residual Capacity (FRC). Explain why TLC is always greater than Vital Capacity.

Passage

Question 1: Using the partial pressure data, state the direction of O2 movement at the alveoli and at the tissues.

Question 2: Why does CO2 diffuse efficiently across the alveolar membrane despite having a smaller partial pressure gradient compared to O2?

Question 3: Describe the three layers of the diffusion membrane and explain how each structural feature contributes to efficient gas exchange.

Passage

Question 1: Name the two respiratory disorders Dr. Meera suspects and identify the primary cause of each.

Question 2: Explain the mechanism by which long-term dust exposure leads to fibrosis and lung damage.

Question 3: Analyse how emphysema leads to the elevated Residual Volume and reduced Vital Capacity seen in the spirometry results.

Passage

Question 1: What tissue supports the trachea and bronchi, and why is this support necessary?

Question 2: Describe the anatomical boundaries of the thoracic chamber and explain why it must remain air-tight for breathing to occur.

Question 3: If the thoracic cavity developed a hole (pneumothorax), explain step by step how breathing and gas exchange would be disrupted.

Passage

Question 1: State the pO2 values in oxygenated blood (pulmonary venous blood) and in the tissues, and explain what drives O2 from blood to tissues.

Question 2: Explain the shape of the oxygen dissociation curve and why it is described as sigmoid.

Question 3: Analyse how elevated CO2, increased H+ concentration, and higher temperature in active muscle tissue shift haemoglobin's behaviour, and explain the physiological advantage of this shift.

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