Level 5 — MasteryDigestive System

Digestive System

⏱ 75 minutes60 marksprintable — key stays hidden on paper

Mastery Examination (Level 5): Cross-Domain Analysis & Modeling

Time limit: 75 minutes Total marks: 60 Instructions: Answer ALL questions. Show all reasoning, calculations, and code logic. Use $...$ notation for mathematical expressions. Diagrams may be described in words.

Question 1 — Absorptive Surface Optimization (Villi & Microvilli) [22 marks]

The efficiency of the small intestine depends on maximizing surface area for absorption (subtopics 4.1.6, 4.1.9).

(a) Describe how each of the three anatomical levels of folding (circular folds, villi, microvilli) contributes to increasing the absorptive surface area, and explain why surface area is the limiting factor for absorption rate. [5]

(b) Model the small intestine as a smooth cylinder of length $L = 6.0\text{ m}$ and internal radius $r = 1.25\text{ cm}$ . (i) Compute the internal surface area $A_0$ of the smooth cylinder (in $\text{m}^2$ ). [3] (ii) Circular folds (plicae circulares) multiply the area by a factor of $3$ , villi by a further factor of $10$ , and microvilli by a further factor of $20$ . Compute the total effective surface area $A_{\text{eff}}$ . [3] (iii) State the ratio $A_{\text{eff}}/A_0$ and comment on its biological significance. [2]

(c) Absorption of glucose across the epithelium follows a diffusion-plus-transport model. For a simplified passive component, Fick's law gives flux $J = \frac{D \cdot A \cdot \Delta C}{\Delta x}$ where $D = 6.0\times10^{-10}\ \text{m}^2\,\text{s}^{-1}$ , $\Delta C = 5.0\ \text{mol}\,\text{m}^{-3}$ , membrane thickness $\Delta x = 1.0\times10^{-6}\ \text{m}$ , and $A = A_{\text{eff}}$ from part (b). (i) Compute the diffusive flux $J$ (in $\text{mol}\,\text{s}^{-1}$ ). [4] (ii) Explain, using this model, why active transport (Na⁺/glucose symport) is also required rather than diffusion alone. [2]

Question 2 — Reaction Kinetics of a Digestive Enzyme [20 marks]

Pepsin (subtopics 4.1.5, 4.1.8) hydrolyses proteins in the stomach.

(a) Explain the roles of HCl and pepsinogen in gastric juice, and describe why the enzyme's activity depends on pH. [4]

(b) Enzyme kinetics follow the Michaelis–Menten equation: $v = \frac{V_{\max}[S]}{K_m + [S]}$ Given $V_{\max} = 12\ \mu\text{mol}\,\text{min}^{-1}$ and $K_m = 2.0\ \text{mmol}\,\text{L}^{-1}$ : (i) Compute the reaction rate $v$ when $[S] = 6.0\ \text{mmol}\,\text{L}^{-1}$ . [3] (ii) Determine the substrate concentration $[S]$ at which $v = \tfrac{1}{2}V_{\max}$ , and confirm its relationship to $K_m$ . [3] (iii) Derive an expression for $[S]$ in terms of $v$ , $V_{\max}$ , $K_m$ , and hence find $[S]$ when $v = 9\ \mu\text{mol}\,\text{min}^{-1}$ . [4]

(c) Write pseudocode (or Python) for a function rate(S, Vmax, Km) that returns $v$ , and describe how you would use it in a loop to generate a saturation curve. State what happens to $v$ as $[S] \to \infty$ and justify biologically. [6]

Question 3 — Peristalsis as a Propagating Wave [18 marks]

Peristalsis (subtopic 4.1.11) moves a food bolus along the alimentary canal.

(a) Describe the mechanism of peristalsis, naming the muscle layers involved and explaining how the coordinated contraction/relaxation produces net forward movement. [5]

(b) A peristaltic contraction wave travels down the oesophagus modeled as a displacement wave $y(x,t) = A\sin(kx - \omega t)$ with wavelength $\lambda = 0.08\ \text{m}$ and wave speed $v_w = 0.04\ \text{m}\,\text{s}^{-1}$ . (i) Compute the wavenumber $k$ and angular frequency $\omega$ . [4] (ii) A bolus must travel the full oesophageal length $L = 0.25\ \text{m}$ . Compute the time taken. [3]

(c) Trace the complete path of food from mouth to anus, naming every major structure in order, and state at which structure the peristaltic wave from part (b) operates. [6]

Answer keyMark scheme & solutions

Question 1

(a) [5]

Circular folds — permanent ridges of mucosa/submucosa that force chyme into spiral flow and add area (1)
Villi — finger-like projections of the mucosa, each containing a capillary network and lacteal, greatly increasing area (1)
Microvilli — brush border on the apical membrane of epithelial cells, the finest level of folding (1)
Surface area is limiting because absorption (diffusion/transport) rate is proportional to area — more area = more transporters/membrane in contact with chyme (1)
Thin epithelium + rich blood supply maintain the concentration gradient, but flux still scales with $A$ (1)

(b)(i) [3] Cylinder lateral surface area $A_0 = 2\pi r L$ . $A_0 = 2\pi (0.0125)(6.0) = 0.15\pi \approx 0.4712\ \text{m}^2$ (2 for setup, 1 for value)

(b)(ii) [3] Total factor $= 3\times10\times20 = 600$ . $A_{\text{eff}} = 600 \times 0.4712 = 282.7\ \text{m}^2$ (≈ 283 m²) (1 factor, 2 value)

(b)(iii) [2] Ratio $= 600$ (1). Significance: a modest tube achieves a huge absorptive area (~size of a room), enabling rapid, complete nutrient uptake within transit time (1).

(c)(i) [4] $J = \frac{D A \Delta C}{\Delta x} = \frac{(6.0\times10^{-10})(282.7)(5.0)}{1.0\times10^{-6}}$ Numerator $= 6.0\times10^{-10}\times282.7\times5.0 = 8.48\times10^{-7}$ ; divide by $10^{-6}$ : $J \approx 0.848\ \text{mol}\,\text{s}^{-1}$ (setup 2, arithmetic 2)

(c)(ii) [2] Passive diffusion only works while lumen concentration exceeds cell concentration; once blood/cell glucose is comparable or higher, $\Delta C \to 0$ and flux stops. Active Na⁺/glucose symport moves glucose against its gradient, allowing near-complete absorption regardless of luminal depletion (2).

Question 2

(a) [4]

HCl denatures dietary proteins and provides low pH (~2) (1)
HCl converts inactive pepsinogen → active pepsin (autocatalytic) (1)
Pepsinogen secreted as inactive zymogen to protect the chief cells from self-digestion (1)
Enzyme activity is pH-dependent because pH affects ionization of the active-site residues / 3D shape; pepsin's optimum is acidic (~2), outside which it denatures (1)

(b)(i) [3] $v = \frac{12\times6.0}{2.0+6.0} = \frac{72}{8.0} = 9.0\ \mu\text{mol}\,\text{min}^{-1}$ (setup 1, sub 1, value 1)

(b)(ii) [3] Set $v=\tfrac12 V_{\max}$ : $\tfrac12 V_{\max} = \frac{V_{\max}[S]}{K_m+[S]} \Rightarrow K_m+[S] = 2[S] \Rightarrow [S] = K_m = 2.0\ \text{mmol L}^{-1}$ This confirms $K_m$ is the substrate concentration giving half-maximal velocity (setup 2, conclusion 1).

(b)(iii) [4] Rearrange: $v(K_m+[S]) = V_{\max}[S] \Rightarrow vK_m = [S](V_{\max}-v) \Rightarrow [S] = \frac{vK_m}{V_{\max}-v}$ (derivation 2). Substitute $v=9$ : $[S] = \frac{9\times2.0}{12-9} = \frac{18}{3} = 6.0\ \text{mmol L}^{-1}$ (consistent with part (i)) (value 2)

(c) [6]

def rate(S, Vmax, Km):
    return Vmax * S / (Km + S)      # (2)

Loop over range of S values, call rate(), store (S, v) pairs, plot to get saturation curve (2)
As $[S] \to \infty$ , $v \to V_{\max}$ (curve plateaus) (1)
Biological justification: all enzyme active sites become saturated; rate limited by enzyme quantity/turnover, not substrate (1)

Question 3

(a) [5]

Two smooth-muscle layers: inner circular, outer longitudinal (1)
Behind the bolus: circular muscle contracts, longitudinal relaxes → constriction pushing bolus forward (1)
Ahead of the bolus: circular relaxes, longitudinal contracts → receiving segment widens (1)
The alternating contraction/relaxation forms a travelling wave (1)
Coordinated by the enteric nervous system, producing net one-way propulsion independent of gravity (1)

(b)(i) [4] $k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{0.08} = 25\pi \approx 78.54\ \text{rad m}^{-1}$ (2) $\omega = v_w k = 0.04\times78.54 = \pi \approx 3.14\ \text{rad s}^{-1}$ (or $\omega = 2\pi v_w/\lambda = 2\pi(0.5) = \pi$ ) (2)

(b)(ii) [3] $t = \dfrac{L}{v_w} = \dfrac{0.25}{0.04} = 6.25\ \text{s}$ (setup 1, value 2)

(c) [6] Path (1 mark per correct block, max 5): mouth → pharynx → oesophagus → stomach → small intestine (duodenum, jejunum, ileum) → large intestine (colon) → rectum → anus. The wave in part (b) operates in the oesophagus (1).

[
  {"claim":"Q1b A0 = 2*pi*r*L for r=0.0125,L=6 ≈0.4712","code":"A0=2*pi*Rational(125,10000)*6; result = abs(float(A0)-0.4712389)<1e-4"},
  {"claim":"Q1b Aeff = 600*A0 ≈282.7","code":"A0=2*pi*Rational(125,10000)*6; Aeff=600*A0; result = abs(float(Aeff)-282.743)<1e-2"},
  {"claim":"Q1c J = D*A*dC/dx ≈0.848","code":"A0=2*pi*Rational(125,10000)*6; Aeff=600*A0; J=Integer(6)*10**-10*Aeff*5/(Integer(1)*10**-6); result = abs(float(J)-0.8482)<1e-3"},
  {"claim":"Q2 v at S=6: 9.0","code":"S,Vmax,Km=6,12,2; v=Vmax*S/(Km+S); result = v==9"},
  {"claim":"Q2biii S at v=9 is 6.0","code":"v,Vmax,Km=9,12,2; S=v*Km/(Vmax-v); result = S==6"},
  {"claim":"Q3b k=25pi, omega=pi, t=6.25","code":"lam=Rational(8,100); vw=Rational(4,100); k=2*pi/lam; om=vw*k; t=Rational(25,100)/vw; result = (k==25*pi) and (om==pi) and (t==Rational(25,4))"}
]