Respiratory System

Question 1 (10)

(a) (4)

• At sea level, alveolar $P_{O_2}$ ~100 mmHg lies on the flat upper plateau → saturation ~97–98% (1).
• At $P_{O_2}$ ~45 mmHg the value falls onto the steep part of the sigmoid curve → saturation drops sharply to roughly 75–80% (1).
• Less O₂ is loaded onto haemoglobin, so arterial oxygen content and tissue delivery fall (1).
• Low blood/tissue O₂ (hypoxia) causes breathlessness/hyperventilation as the body attempts to raise oxygen uptake (1).

(b) (4)

• Increased 2,3-BPG binds haemoglobin and lowers its affinity for O₂ (1).
• This shifts the dissociation curve to the right (1).
• At the tissues, the rightward shift means O₂ is released more readily (unloaded at a given $P_{O_2}$) (1).
• So even though loading in the lungs is slightly reduced, tissues still receive adequate O₂ because unloading is enhanced — a beneficial compensation (1).

(c) (2)

• Peripheral chemoreceptors in the carotid and aortic bodies detect low arterial $P_{O_2}$ (1).
• Central chemoreceptors respond mainly to CO₂/H⁺ (pH), not to low O₂; at altitude CO₂ is actually washed out by hyperventilation, so only the peripheral (O₂-sensitive) receptors drive the initial response (1).

Question 2 (10)

(a) (4)

• Lung inflation depends on negative (sub-atmospheric) intrapleural pressure which keeps the lung pulled against the chest wall (1).
• The stab lets air into the pleural space, so intrapleural pressure rises to atmospheric (1).
• The pressure gradient holding the lung open is lost, so the elastic recoil of the left lung makes it collapse (1).
• The right pleural cavity is a separate sealed compartment, so its negative pressure is intact and that lung continues to inflate normally (1).

(b) (4)

• Diaphragm contracts (flattens/moves down) and external intercostals contract, raising the ribcage up and out → thoracic volume increases (1).
• Increased volume lowers intrapleural pressure (more negative) (1).
• This lowers intra-alveolar pressure below atmospheric (1).
• Air flows down the pressure gradient into the lungs (1).

(c) (2)

• Air must be removed so intrapleural pressure becomes sub-atmospheric (negative) again (1), restoring the gradient that holds the lung expanded (1).

Question 3 (10)

(a) (6) — award 2 marks each factor (mechanism + effect):

• CO₂: high CO₂ forms carbonic acid → raises H⁺; this lowers haemoglobin's O₂ affinity → curve shifts right → more O₂ released (2).
• Lactic acid / low pH: increased H⁺ directly reduces Hb affinity for O₂ (Bohr effect) → more unloading (2).
• Temperature: higher local temperature lowers Hb O₂ affinity → rightward shift → more O₂ delivered to hard-working muscle (2). (Overall: right shift = more O₂ unloaded exactly where metabolic demand is highest.)

(b) (4)

• In the alveoli CO₂ diffuses out into the air, lowering blood CO₂ and H⁺ → pH rises (1).
• Higher pH / lower CO₂ increases Hb affinity for O₂ (leftward shift / reverse Bohr) (1).
• High alveolar $P_{O_2}$ plus increased affinity → O₂ loads readily onto Hb (1).
• As O₂ binds, it promotes CO₂ release (Haldane effect) allowing CO₂ to be exhaled (1).

Question 4 (10)

(a) (4) — any two, 2 marks each:

• Large surface area: many filaments and lamellae maximise area for the low O₂ content of water (2).
• Counter-current flow: water flows opposite to blood, maintaining a diffusion gradient along the whole length so more O₂ is extracted (2).
• Thin lamellae / short diffusion distance or continuous one-way ventilation to move viscous dense water efficiently (2).

(b) (4)

• Tracheae rely on diffusion of gas to reach cells directly (1); diffusion is only efficient over short distances (1).
• A large insect would have tissues too far from tracheoles for adequate O₂ supply, limiting body size (1).
• The fish has a circulatory system with haemoglobin that actively transports O₂ throughout the body, so distance is not a limit and larger size is possible (1).

(c) (2)

• A very thin, moist exchange surface / short diffusion distance (1) with a large surface area maximising diffusion (1). (Either point stated as the shared feature earns the marks.)

Question 5 (10)

(a) (4)

• Reaction: $CO_2 + H_2O \underset{\text{carbonic anhydrase}}{\rightleftharpoons} H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$ (2).
• HCO₃⁻ diffuses out of the red blood cell into plasma; to maintain electrical balance Cl⁻ moves in — the chloride shift (2).

(b) (4)

• With carbonic anhydrase inhibited, HCO₃⁻ formation slows, so the bicarbonate % falls (1).
• More CO₂ stays dissolved / bound to Hb, so those fractions rise (1).
• Less buffering of H⁺ via bicarbonate route changes handling of acid; CO₂ conversion is impaired so CO₂ removal at lungs is reduced (CO₂ retained) (1).
• Retained CO₂ forms more acid → tendency toward lower blood pH (respiratory acidosis) (1).

(c) (2)

• Bicarbonate = 70% of 52 mL = $0.70 \times 52 = 36.4$ mL per 100 mL blood (2).

[
  {"claim":"Bicarbonate CO2 volume = 70% of 52 mL = 36.4 mL","code":"total=52; frac=0.70; result = (frac*total == 36.4)"},
  {"claim":"CO2 fractions sum to 100%","code":"result = (70+23+7 == 100)"},
  {"claim":"Dissolved plasma CO2 volume = 7% of 52 = 3.64 mL","code":"result = (0.07*52 == 3.64)"}
]