Excretory System & Homeostasis

Question 1

(a) (4)

• Inulin is freely filtered at the glomerulus (1).
• It is neither reabsorbed nor secreted in the tubule (1).
• Therefore all inulin appearing in urine came only from filtration (1).
• So the volume of plasma cleared of inulin per minute = volume filtered per minute = GFR (1).

(b) (3) $C_{inulin} = \dfrac{U \cdot V}{P} = \dfrac{26.0 \times 1.2}{0.25} = \dfrac{31.2}{0.25} = 124.8\ \text{mL/min}$

• Correct substitution (1), arithmetic (1), units mL/min (1). GFR ≈ 125 mL/min (normal).

(c) (5) $C_{urea} = \dfrac{18.0 \times 1.2}{0.30} = \dfrac{21.6}{0.30} = 72\ \text{mL/min}$ (2).

• Filtered urea rate $= P \times GFR = 0.30 \times 124.8 = 37.44$ mg/min (1).
• Excreted urea rate $= U \times V = 18.0 \times 1.2 = 21.6$ mg/min (1).
• Fraction reabsorbed $= \dfrac{37.44 - 21.6}{37.44} = \dfrac{15.84}{37.44} = 0.423 \approx 42\%$ (1).
• (Equivalently $1 - C_{urea}/GFR = 1 - 72/124.8 = 0.423$.)

(d) (4)

• Glucose is freely filtered so filtration ≠ 0 (1).
• In the proximal convoluted tubule it is completely reabsorbed by Na⁺-glucose cotransporters (SGLT) (1); none reaches urine so clearance = 0 (1).
• Above the renal threshold the transporters saturate (transport maximum, $T_m$), so excess glucose is excreted → clearance rises from 0 (glycosuria, e.g. diabetes) (1).

(e) (4) For freely filtered, non-secreted, non-reabsorbed solute: amount excreted rate = amount filtered rate: $U_X V = GFR \cdot P_X$. Then $C_X = \dfrac{U_X V}{P_X} = \dfrac{GFR \cdot P_X}{P_X} = GFR$ (2).

• This is independent of $P_X$ (1).
• Assumption: filtrate concentration equals plasma concentration (free filtration) and GFR is constant/independent of $P_X$ (1).

Question 2

(a) (6)

• Sweating loses water → plasma osmolarity rises / volume falls (1).
• Osmoreceptors in the hypothalamus detect the rise (1).
• ADH is synthesised in hypothalamus (supraoptic/paraventricular nuclei), stored & released from posterior pituitary (1).
• ADH binds V2 receptors on collecting-duct principal cells (1).
• Triggers insertion of aquaporin-2 channels into the apical membrane (1).
• Water reabsorbed → concentrated (low-volume) urine, osmolarity restored (negative feedback) (1).

(b) (7) Steady state: $0 = k_{in} - k_{out}[ADH](\Pi^* - \Pi_0)$. Take $k_{out}=1$ (absorbed into $a$) and $[ADH]=a(\Pi^*-\Pi_{set})$: $k_{in} = a(\Pi^* - \Pi_{set})(\Pi^* - \Pi_0)$ Substitute $k_{in}=2, a=0.01, \Pi_{set}=290, \Pi_0=280$: $2 = 0.01(\Pi^* - 290)(\Pi^* - 280)$ $200 = (\Pi^* - 290)(\Pi^* - 280)$ Let $x=\Pi^*$: $x^2 - 570x + 81200 = 200 \Rightarrow x^2 - 570x + 81000 = 0$ (quadratic, 3 marks). $x = \frac{570 \pm \sqrt{570^2 - 4\cdot81000}}{2} = \frac{570 \pm \sqrt{324900 - 324000}}{2} = \frac{570 \pm \sqrt{900}}{2} = \frac{570 \pm 30}{2}$ $x = 300$ or $x = 270$. Physical root ($\Pi^*>\Pi_{set}=290$): $\Pi^* = 300$ mOsm/L (2). (Reject 270 as it violates the ADH-on branch, 2).

(c) (4) $f(\Pi) = k_{in} - a(\Pi-\Pi_{set})(\Pi-\Pi_0)$. $f'(\Pi) = -a[(\Pi-\Pi_0)+(\Pi-\Pi_{set})]$ (2). At $\Pi^*=300$: $f' = -0.01[(300-280)+(300-290)] = -0.01(30) = -0.30 < 0$ (1). Negative derivative ⇒ perturbations decay ⇒ equilibrium is stable; the feedback is stabilising (1).

(d) (5) (1 each row, max 5)

Question 3

(a) (6)

• Toxicity: ammonia > urea > uric acid (2).
• Water needed: ammonia (most) > urea > uric acid (least) (2).
• ATP cost per N: uric acid > urea > ammonia (least) (1).
• Habitat link: aquatic fish excrete ammonia (abundant water dilutes toxicity); terrestrial mammals make urea (water-economy compromise); birds/reptiles excrete uric acid (paste, conserves water, safe in shelled eggs) (1).

(b) (5)

• Moles urea/day $= 30/60 = 0.5$ mol (1).
• ATP cost $= 0.5 \times 4 = 2.0$ mol ATP-equivalents/day (2).
• N atoms: each urea has 2 N ⇒ $0.5 \times 2 = 1.0$ mol N atoms $= 6.02\times10^{23}$ atoms/day (1).
• Order-of-magnitude comment: ~2 mol ATP/day is small relative to total daily ATP turnover (~tens of mol), so ureagenesis is metabolically affordable (1).

(c) (7) Pseudocode (Euler) (3):

T = T0
dt = 0.1
for step in range(N):
    dTdt = (Hgen - k*(T - Tenv)) / C
    T = T + dTdt*dt
    record(t, T)

Steady state: set $dT/dt=0$: $H_{gen} = k(T_{ss}-T_{env})$ ⇒ $T_{ss} = T_{env} + \frac{H_{gen}}{k}$ (2). Physiology (2):

• Change $k$ (heat loss): vasodilation/vasoconstriction of skin vessels; sweating/panting (evaporative loss). (also piloerection, behavioural).
• Change $H_{gen}$ (heat production): shivering thermogenesis and non-shivering (brown fat/metabolic) thermogenesis.

[
  {"claim":"Inulin clearance (GFR) = 124.8 mL/min","code":"U,V,P=26.0,1.2,0.25; result = abs(U*V/P - 124.8) < 1e-9"},
  {"claim":"Urea fraction reabsorbed approx 0.4231","code":"GFR=124.8; Curea=18.0*1.2/0.30; filt=0.30*GFR; exc=18.0*1.2; frac=(filt-exc)/filt; result = abs(frac-0.42307692307692313) < 1e-6"},
  {"claim":"Osmolarity equilibrium Pi*=300 (physical root)","code":"x=symbols('x'); sols=solve(Eq((x-290)*(x-280),200),x); result = 300 in sols and 270 in sols"},
  {"claim":"Stability derivative f'(300) = -0.30 < 0","code":"a=0.01; Pi=300; fp=-a*((Pi-280)+(Pi-290)); result = abs(fp+0.30)<1e-9 and fp<0"},
  {"claim":"Daily urea ATP cost = 2.0 mol and 0.5 mol urea","code":"mol=30/60; atp=mol*4; result = abs(mol-0.5)<1e-9 and abs(atp-2.0)<1e-9"}
]