Level 3 — ProductionRespiratory System

Respiratory System

45 minutes60 marksprintable — key stays hidden on paper

Level 3: Production (From-Scratch Derivations & Explain-Out-Loud)

Time Limit: 45 minutes Total Marks: 60

Instructions: Answer all questions. Diagrams should be clearly labelled. Where "explain out loud" is indicated, write your reasoning as a continuous logical chain, not bullet fragments.


Question 1 — Pathway & Mechanics (10 marks)

(a) From memory, trace the complete pathway of a single oxygen molecule from the external nostril to the site of gas exchange, naming every structure in order. (6)

(b) A student claims "we breathe in because our lungs pull air in like a pump." Correct this statement by explaining, from scratch, the pressure-based mechanism of inhalation. Reference the diaphragm, external intercostals, thoracic volume and Boyle's Law. (4)


Question 2 — Pressure Changes Derivation (10 marks)

Using Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2), derive quantitatively why air flows into the lungs during inhalation.

(a) State Boyle's Law and explain what it predicts qualitatively when thoracic volume increases. (3)

(b) Suppose at rest intrapulmonary pressure equals atmospheric pressure of 101.3101.3 kPa at a lung volume of 2.42.4 L. On inhalation the thoracic cavity expands so lung volume becomes 2.52.5 L before any air enters. Calculate the new intrapulmonary pressure and hence the pressure gradient driving airflow. State the direction of airflow. (5)

(c) Explain why this calculated pressure difference is only momentary. (2)


Question 3 — Alveolar Gas Exchange (Explain Out Loud) (10 marks)

Explain, as a continuous reasoning chain, how the structure of an alveolus is adapted for efficient gas exchange. Your answer must connect at least four structural features to Fick's principle of diffusion (ratesurface area×concentration differencediffusion distance\text{rate} \propto \frac{\text{surface area} \times \text{concentration difference}}{\text{diffusion distance}}). (10)


Question 4 — Gas Transport & The Bohr Effect (12 marks)

(a) Describe the three ways carbon dioxide is transported in the blood, giving the approximate proportion carried by each. (6)

(b) Write the chemical equation for the reversible reaction that occurs in red blood cells and name the enzyme that catalyses it. (3)

(c) Explain out loud how this reaction underlies the Bohr effect, and state its physiological benefit at actively respiring tissues. (3)


Question 5 — Oxygen–Haemoglobin Dissociation Curve (10 marks)

(a) Sketch the oxygen–haemoglobin dissociation curve. Label both axes and indicate typical loading (lungs) and unloading (tissues) points. (4)

(b) Explain from scratch why the curve is sigmoidal (S-shaped) rather than a straight line. Reference cooperative binding. (3)

(c) On your sketch, draw and label the curve shift produced by the Bohr effect, and explain what this means for oxygen delivery. (3)


Question 6 — Comparative Respiration & Regulation (8 marks)

(a) Compare the fish gill and the insect tracheal system as respiratory surfaces. Address the transport medium, the role (or absence) of blood, and countercurrent flow. (5)

(b) Explain how an increase in blood CO2_2 during exercise is detected and how the body responds to restore homeostasis. Name the control centre and the type of receptor. (3)

Answer keyMark scheme & solutions

Question 1 (10 marks)

(a) Pathway (6 marks) — 1 mark per correctly ordered group (max 6): Nostril/nasal cavity → pharynx → larynx → trachea → bronchi → bronchioles → alveolar ducts → alveoli. Why: air is filtered/warmed/moistened in the nose, the larynx/trachea are held open by cartilage rings, and branching increases into terminal alveoli where exchange occurs. Award 1 mark each for: nasal cavity, pharynx, larynx, trachea, bronchi/bronchioles, alveoli.

(b) Correcting the misconception (4 marks)

  • Lungs are passive; they do not actively pull (1).
  • Diaphragm contracts and flattens; external intercostals contract raising the ribs up and out → thoracic volume increases (1).
  • By Boyle's Law, increased volume → decreased intrapulmonary pressure below atmospheric (1).
  • Air flows down the pressure gradient into the lungs — inflation is a consequence, not a cause (1).

Question 2 (10 marks)

(a) Boyle's Law: at constant temperature, pressure is inversely proportional to volume (P1/VP \propto 1/V) (1). If thoracic volume increases, pressure of gas in the lungs decreases (2).

(b) Using P1V1=P2V2P_1V_1 = P_2V_2: P2=P1V1V2=101.3×2.42.5=97.25 kPaP_2 = \frac{P_1 V_1}{V_2} = \frac{101.3 \times 2.4}{2.5} = 97.25 \text{ kPa} (3) Pressure gradient = 101.397.25=4.05101.3 - 97.25 = 4.05 kPa (1). Since intrapulmonary pressure < atmospheric, air flows into the lungs (1).

(c) As air enters, lung pressure rises back toward atmospheric, abolishing the gradient — so equilibrium is quickly re-established and flow stops (2).


Question 3 (10 marks)

Marking: up to 2 marks per feature correctly linked to Fick (max 8) + 2 for coherent Fick framing.

Model chain:

  • Fick states diffusion rate rises with surface area and concentration gradient and falls with distance (2).
  • Millions of alveoli provide an enormous total surface area (~70 m²) → increases area term → faster diffusion (2).
  • Wall one cell thick (squamous epithelium) + thin capillary endothelium → very short diffusion distance → increases rate (2).
  • Dense capillary network + continuous blood flow maintains a steep concentration gradient by constantly removing oxygenated blood/bringing deoxygenated blood (2).
  • Moist thin film of surfactant/fluid allows gases to dissolve and diffuse; ventilation refreshes alveolar air keeping the gradient high (2).

Question 4 (12 marks)

(a) CO₂ transport (6 marks) — 2 marks each:

  • As bicarbonate ions (HCO3\text{HCO}_3^-) in plasma — ~70% (2).
  • Bound to haemoglobin as carbaminohaemoglobin — ~23% (2).
  • Dissolved directly in plasma — ~7% (2).

(b) (3 marks) CO2+H2OH2CO3H++HCO3\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- (2) Enzyme: carbonic anhydrase (1).

(c) Bohr effect (3 marks)

  • The reaction releases H⁺, lowering pH in respiring tissue (1).
  • Increased H⁺/CO₂ decreases haemoglobin's affinity for oxygen (shifts curve right), promoting O₂ release (1).
  • Benefit: more O₂ is unloaded exactly where metabolic demand is highest (1).

Question 5 (10 marks)

(a) Axes: x = partial pressure of O₂ (kPa), y = % saturation of Hb (2). S-shaped curve with lungs point at high pO₂/~98% and tissues point at lower pO₂/lower saturation labelled (2).

(b) Binding of the first O₂ to a haem group changes Hb conformation, increasing affinity of remaining sites (cooperative/positive cooperativity) (2). This makes uptake slow at first then rapid, producing the sigmoid shape rather than linear binding (1).

(c) Bohr shift: curve shifts to the right (increased CO₂/H⁺/temperature) (1). At any given pO₂ saturation is lower (1), so more oxygen is released to actively respiring tissues (1).


Question 6 (8 marks)

(a) Gills vs tracheae (5 marks)

  • Fish gills: exchange with water (medium) (1); O₂ diffuses into blood which transports it around the body (1); countercurrent flow of water and blood maintains a gradient along the whole lamella maximising uptake (1).
  • Insect tracheae: air-filled tubes deliver O₂ directly to tissues via tracheoles (1); no blood involved in gas transport — diffusion straight to cells, no countercurrent system (1).

(b) Regulation (3 marks)

  • Rising CO₂ lowers blood pH, detected by chemoreceptors (in medulla/aorta/carotid bodies) (1).
  • The medulla oblongata (respiratory centre) increases the rate and depth of breathing via the phrenic/intercostal nerves (1).
  • More CO₂ is exhaled, pH returns to normal — negative feedback (1).

[
  {"claim":"Q2b intrapulmonary pressure after expansion is 97.25 kPa","code":"P1=101.3; V1=2.4; V2=2.5; P2=P1*V1/V2; result = abs(P2-97.248)<0.01"},
  {"claim":"Q2b pressure gradient is about 4.05 kPa","code":"P1=101.3; V1=2.4; V2=2.5; P2=P1*V1/V2; grad=P1-P2; result = abs(grad-4.052)<0.01"},
  {"claim":"Q4a CO2 transport percentages sum to 100","code":"result = (70+23+7)==100"},
  {"claim":"Boyle inverse relation: volume increase lowers pressure","code":"P1=101.3; V1=2.4; V2=2.5; P2=P1*V1/V2; result = P2<P1"}
]