Level 5 — MasteryRespiratory System

Respiratory System

⏱ 75 minutes60 marksprintable — key stays hidden on paper

Mastery Examination (Level 5)

Time limit: 75 minutes Total marks: 60 Instructions: Answer ALL questions. Show all reasoning, calculations, and derivations. Where computation is required, state the model and assumptions explicitly. Use $...$ for mathematical expressions.

Question 1 — Mechanics, Pressure, and Gas Flow (24 marks)

A physiologist models a single breath. During quiet inhalation the diaphragm and external intercostals contract; treat the thoracic cavity as a sealed cylinder of gas at body temperature (constant $T$ ).

(a) Explain the causal chain linking (i) contraction of the diaphragm and external intercostals, (ii) thoracic volume change, (iii) intrapleural/intra-alveolar pressure change, and (iv) bulk air flow. Reference Boyle's Law explicitly. (5)

(b) At the start of inhalation the alveolar gas volume is $V_1 = 2.40\text{ L}$ at intra-alveolar pressure $P_1 = 101.3\text{ kPa}$ (equal to atmospheric, no flow). The thorax then expands so that, before any air enters, the alveolar volume would momentarily become $V_2 = 2.55\text{ L}$ . Assuming isothermal behaviour, compute the transient intra-alveolar pressure $P_2$ and hence the pressure gradient $\Delta P = P_{atm} - P_2$ driving inflow. (4)

(c) Air flows down this gradient. Model airflow as laminar Poiseuille flow through the trachea (radius $r = 9.0\text{ mm}$ , length $L = 0.12\text{ m}$ , air viscosity $\eta = 1.9\times10^{-5}\text{ Pa·s}$ ). Using the $\Delta P$ from part (b), compute the volumetric flow rate $Q = \frac{\pi r^4\,\Delta P}{8\eta L}$ in L/s, and comment on whether this exceeds a resting tidal requirement of about $0.5\text{ L}$ delivered over a $2\text{ s}$ inhalation. (5)

(d) A patient's airway narrows so that effective radius falls by 20%. Assuming the same driving $\Delta P$ , determine the percentage change in $Q$ , and explain physiologically why bronchoconstriction is so dangerous relative to the small radius change. (4)

(e) Trace the full pathway of an oxygen molecule from external nostril to the site of gas exchange, naming at least seven structures in correct order, and state at which structure gas exchange with blood occurs. (6)

Question 2 — Oxygen Transport and the Bohr Effect (24 marks)

Oxygen binding to haemoglobin (Hb) is cooperative and can be modelled by the Hill equation: $Y = \frac{P^n}{P_{50}^{\,n} + P^n}$ where $Y$ is fractional saturation, $P$ is the partial pressure of $O_2$ (in kPa), $P_{50}$ is the $P$ at half-saturation, and $n$ is the Hill coefficient.

(a) Explain, in molecular terms, why $n > 1$ for haemoglobin but $n = 1$ for myoglobin, and what the sigmoidal shape means physiologically for $O_2$ loading in lungs vs unloading in tissues. (5)

(b) Take $n = 2.8$ and $P_{50} = 3.5\text{ kPa}$ (arterial/normal conditions).

(i) Compute $Y$ at the alveolar $P_{O_2} = 13.3\text{ kPa}$ . (2)
(ii) Compute $Y$ at a resting tissue $P_{O_2} = 5.3\text{ kPa}$ . (2)
(iii) Hence compute the fraction of bound $O_2$ released to resting tissue during one pass. (2)

(c) The Bohr effect: actively respiring muscle produces $CO_2$ and lowers pH. Model this as a rightward shift of $P_{50}$ to $5.0\text{ kPa}$ (same $n$ ). Recompute $Y$ at tissue $P_{O_2} = 5.3\text{ kPa}$ , and hence the new fraction of $O_2$ released (still loading at 13.3 kPa where you may reuse (b)(i) with the shifted curve — recompute it). Quantify the extra oxygen delivered due to the Bohr shift, and explain the biochemical cause of the shift. (7)

(d) Write a short, self-contained Python function hill(P, n, p50) returning $Y$ , then describe (in words or pseudocode) how you would use it to numerically locate $P_{50}$ from a dataset of $(P, Y)$ pairs where $P_{50}$ is unknown. State one validity check you would apply to confirm your fitted $n$ and $P_{50}$ are physiologically sensible. (6)

Question 3 — Comparative Respiratory Surfaces & Regulation (12 marks)

(a) Fish gills use countercurrent exchange while human alveoli use a concurrent/pool arrangement with air. Explain, using the concept of a maintained diffusion gradient (Fick's law $J \propto A\,\Delta C / d$ ), why countercurrent flow can extract up to ~80% of dissolved $O_2$ whereas a concurrent system is limited to ~50%. (5)

(b) Insects use tracheae delivering air directly to tissues. State two consequences this has for (i) body size limits and (ii) the role of the circulatory system in gas transport. (3)

(c) Describe how a rise in blood $CO_2$ during exercise is detected and how it increases ventilation rate, naming the chemoreceptors, the sensed variable, and the effector muscles. (4)

Answer keyMark scheme & solutions

Question 1

(a) (5 marks)

Diaphragm contracts → flattens/moves down; external intercostals contract → ribs move up and out (1).
Thoracic (and hence lung) volume increases (1).
By Boyle's Law ( $PV = \text{const}$ at constant $T$ ), increased volume → decreased pressure; intrapleural pressure becomes more negative and intra-alveolar pressure drops below atmospheric (2).
Air flows down the pressure gradient from high (atmosphere) to low (alveoli) = bulk inflow (1).

(b) (4 marks) Isothermal → $P_1V_1 = P_2V_2$ (1). $P_2 = \frac{P_1 V_1}{V_2} = \frac{101.3 \times 2.40}{2.55} = \frac{243.12}{2.55} = 95.34\text{ kPa}$ (2) $\Delta P = 101.3 - 95.34 = 5.96\text{ kPa} \approx 5.96\times10^{3}\text{ Pa}$ (1)

(c) (5 marks) $r^4 = (9.0\times10^{-3})^4 = 6.561\times10^{-9}\text{ m}^4$ (1). $Q = \frac{\pi (6.561\times10^{-9})(5.96\times10^{3})}{8(1.9\times10^{-5})(0.12)}$ Numerator $= \pi \times 6.561\times10^{-9} \times 5960 = 1.228\times10^{-4}$ (1). Denominator $= 8 \times 1.9\times10^{-5} \times 0.12 = 1.824\times10^{-5}$ (1). $Q = \frac{1.228\times10^{-4}}{1.824\times10^{-5}} = 6.73\text{ m}^3/\text{s}$ — but converting: $Q \approx 6.73\times10^{3}\text{ L/s}$ (1). This vastly exceeds the required $0.5/2 = 0.25\text{ L/s}$ ; the idealised Poiseuille value is enormous because real breathing uses a far smaller $\Delta P$ , turbulent/branched flow, and the airway resistance of the whole tree, not a single wide tube. Conclusion: the gradient is more than adequate; airway resistance and elastic recoil dominate, not gradient scarcity (1).

(d) (4 marks) $Q \propto r^4$ . New radius $= 0.8r$ , so $Q_{new}/Q = 0.8^4 = 0.4096$ (2). Percentage change $= (0.4096 - 1)\times100 = -59.0\%$ (a ~59% drop) (1). A mere 20% narrowing cuts flow by ~59% because of the fourth-power dependence; bronchoconstriction therefore causes disproportionately severe reductions in airflow, explaining acute distress in asthma (1).

(e) (6 marks) Any correct ordered pathway with ≥7 structures (½ each up to 6): Nostril/nasal cavity → pharynx → larynx → trachea → primary bronchus → bronchioles → terminal/respiratory bronchioles → alveoli. Gas exchange occurs at the alveoli (alveolar–capillary membrane) (must be stated for full marks).

Question 2

(a) (5 marks)

Hb has four subunits; binding of $O_2$ to one subunit induces a conformational change (T→R state) raising affinity of the others = positive cooperativity → $n>1$ (2).
Myoglobin has a single binding site, no cooperativity possible → $n=1$ , hyperbolic curve (1).
Sigmoidal shape means Hb is near-saturated at high lung $P_{O_2}$ (efficient loading) yet steeply unloads over the physiological tissue range (efficient delivery where $P_{O_2}$ falls) (2).

(b) (i) $P=13.3$ : $P^n = 13.3^{2.8}$ , $P_{50}^n = 3.5^{2.8}$ . $Y = \dfrac{13.3^{2.8}}{3.5^{2.8}+13.3^{2.8}} \approx 0.982$ (2). (ii) $P=5.3$ : $Y = \dfrac{5.3^{2.8}}{3.5^{2.8}+5.3^{2.8}} \approx 0.740$ (2). (iii) Fraction released $= 0.982 - 0.740 = 0.242 \approx 24\%$ (2).

(c) (7 marks) Shifted $P_{50}=5.0$ . Loading at 13.3: $Y = \dfrac{13.3^{2.8}}{5.0^{2.8}+13.3^{2.8}} \approx 0.955$ (1). Unloading at 5.3: $Y = \dfrac{5.3^{2.8}}{5.0^{2.8}+5.3^{2.8}} \approx 0.552$ (2). Released with Bohr shift $= 0.955 - 0.552 = 0.403 \approx 40\%$ (1). Extra $O_2$ delivered vs part (b): $0.403 - 0.242 = 0.161$ , i.e. ~16 percentage points more oxygen unloaded (1). Cause: metabolic $CO_2$ + $H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$ ; the $H^+$ (and $CO_2$ binding to Hb) stabilises the T (low-affinity) state, raising $P_{50}$ and shifting the curve right — Hb releases more $O_2$ precisely where metabolism is highest (2).

(d) (6 marks)

def hill(P, n, p50):
    return P**n / (p50**n + P**n)

(2 for correct function.) Fitting method (3): given data pairs, define residual $r_i = Y_i^{meas} - \text{hill}(P_i,n,P_{50})$ ; minimise $\sum r_i^2$ over $(n,P_{50})$ using nonlinear least squares (e.g. scipy.optimize.curve_fit). Alternatively linearise via the Hill plot: $\log\!\big(\tfrac{Y}{1-Y}\big) = n\log P - n\log P_{50}$ ; slope $=n$ , and $P_{50}=10^{(\text{intercept}/-n)}$ i.e. where $\log(Y/(1-Y))=0$ . Validity check (1): confirm fitted $n\in[2.5,3.2]$ and $P_{50}\approx 3–4$ kPa for normal Hb (physiologically plausible), and that residuals are small/unbiased.

Question 3

(a) (5 marks)

Fick: flux depends on maintained concentration difference $\Delta C$ across the surface (1).
Countercurrent: water and blood flow in opposite directions, so blood always meets water of higher $O_2$ than itself along the entire length → a gradient is sustained the whole way, allowing uptake toward the incoming-water concentration (~80%) (2).
Concurrent/pool: both approach a common equilibrium; once blood and water $O_2$ equalise (~halfway value), the gradient vanishes and no further uptake occurs, capping extraction near ~50% (2).

(b) (3 marks) (i) Diffusion through tracheae is slow over long distances, so oxygen supply limits maximum body size in insects (1.5). (ii) Gas transport is decoupled from blood; the circulatory (haemolymph) system plays little/no role in $O_2$ transport, carrying nutrients not oxygen (1.5).

(c) (4 marks)

Rising blood $CO_2$ lowers pH; detected by central chemoreceptors (medulla) sensing CSF $H^+$ / $P_{CO_2}$ and peripheral chemoreceptors (carotid & aortic bodies) (2).
Sensed variable is primarily $P_{CO_2}$ / $[H^+]$ (not directly $O_2$ ) (1).
The respiratory centre increases firing to effector muscles — diaphragm and intercostals — raising rate and depth of breathing (1).

[
  {"claim":"P2 from Boyle's law is 95.34 kPa (approx)","code":"P2=101.3*2.40/2.55; result = abs(P2-95.34)<0.05"},
  {"claim":"Flow reduced to 0.8**4 = 0.4096 i.e ~59% drop","code":"ratio=0.8**4; result = abs(ratio-0.4096)<1e-4 and abs((ratio-1)*100+59.04)<0.1"},
  {"claim":"Normal release fraction b(iii) approx 0.242","code":"import sympy as sp\nY1=sp.Rational(133,10)**sp.Float(2.8)/(sp.Float(35,1)**0); \nfa=lambda P,p50:(P**2.8)/(p50**2.8+P**2.8)\nrel=fa(13.3,3.5)-fa(5.3,3.5); result = abs(float(rel)-0.242)<0.01"},
  {"claim":"Bohr-shifted release fraction approx 0.403 and extra ~0.16","code":"fa=lambda P,p50:(P**2.8)/(p50**2.8+P**2.8)\nnormal=fa(13.3,3.5)-fa(5.3,3.5)\nbohr=fa(13.3,5.0)-fa(5.3,5.0)\nresult = abs(bohr-0.403)<0.01 and abs((bohr-normal)-0.161)<0.01"}
]