Endocrine System

Question 1 (24 marks)

(a) [6 marks]

• $+p_3 G$: rising glucose stimulates pancreatic β-cells to secrete insulin — insulin secretion is driven by glucose concentration. (2)
• $-p_2 I$: insulin promotes glucose uptake (GLUT4 in muscle/fat) and suppresses hepatic output, so higher insulin lowers glucose. (2)
• Together these two coupled terms form a negative feedback loop: glucose ↑ → insulin ↑ → glucose ↓. Constants: $p_1$ = insulin-independent glucose clearance, $p_2$ = insulin sensitivity/effectiveness, $p_3$ = β-cell responsiveness to glucose, $p_4$ = insulin degradation/clearance rate. (2)

(b) [5 marks] $A = \begin{pmatrix} -p_1 & -p_2 \\ p_3 & -p_4 \end{pmatrix}$ (2)

• Trace $= -p_1 - p_4 < 0$ since both positive. (1.5)
• Determinant $= (-p_1)(-p_4) - (-p_2)(p_3) = p_1 p_4 + p_2 p_3 > 0$. (1.5)

(c) [7 marks] Characteristic equation: $\lambda^2 - (\text{tr})\lambda + \det = 0$, i.e. $\lambda^2 + (p_1+p_4)\lambda + (p_1p_4 + p_2p_3)=0$.

• $\text{tr} = -(0.03+0.25) = -0.28$; $\det = (0.03)(0.25)+(0.8)(0.02) = 0.0075 + 0.016 = 0.0235$. (2)
• Discriminant $D = (p_1+p_4)^2 - 4\det = 0.28^2 - 4(0.0235) = 0.0784 - 0.094 = -0.0156 < 0$. (2)
• Eigenvalues: $\lambda = \dfrac{-0.28 \pm \sqrt{-0.0156}}{2} = -0.14 \pm 0.0625\,i$ (approx). (2)
• Real part negative + non-zero imaginary part ⇒ stable spiral (damped oscillation). (1)

(d) [6 marks]

• Complex eigenvalues with negative real part ⇒ glucose returns to baseline via damped oscillations (overshoot/undershoot before settling). (2)
• Healthy response: fast, well-damped return; diabetic (Type 2): reduced $p_2$ (insulin resistance) or $p_3$ (β-cell failure) ⇒ slower/sustained high glucose, weaker feedback. (2)
• Glucagon (antagonist to insulin) raises glucose when it falls too low, preventing hypoglycaemic undershoot — completing the bidirectional homeostatic control. (2)

Question 2 (22 marks)

(a) [6 marks] (1.5 per row, table):

(b) [6 marks] Mass balance: $R_T = [R] + [HR]$ ⇒ $[R] = R_T - [HR]$. (2) From $K_d = \dfrac{[H][R]}{[HR]}$: $[HR]K_d = [H](R_T - [HR])$. (2) $[HR]K_d + [H][HR] = [H]R_T \Rightarrow [HR] = \dfrac{[H]R_T}{K_d + [H]}$. $\boxed{\theta = \frac{[HR]}{R_T} = \frac{[H]}{K_d + [H]}}$ (2)

(c) [4 marks] At $[H]=K_d$: $\theta = \dfrac{K_d}{K_d+K_d} = \dfrac{1}{2} = 0.5$. ✓ (2) At $[H]=3K_d$: $\theta = \dfrac{3K_d}{K_d+3K_d} = \dfrac{3}{4} = 0.75$. (2)

(d) [6 marks]

def occupancy(H, Kd):
    return H / (Kd + H)

(2) — correct formula. Explanation (4): Even at low $\theta$ (few receptors bound), each activated receptor triggers a G-protein → many cAMP molecules → each activates many kinases → cascade. This enzymatic amplification means a low hormone concentration (small $[H]$, low occupancy) produces a large intracellular response — hence tiny circulating hormone amounts suffice (subtopic 4.5.4).

Question 3 (14 marks)

(a) [6 marks]

Hypothalamus ──TRH──▶ Anterior Pituitary ──TSH──▶ Thyroid
                                                     │
                                              T3 & T4 (thyroxine)
                                                     │
      ◀─────── (−) negative feedback ─────────◀──────┘
      (T3/T4 inhibit both hypothalamus TRH and pituitary TSH)

Marks: TRH (1), TSH (1), T3/T4 (1), correct sequence hyp→pit→thyroid (1), feedback arrow to pituitary (1), feedback arrow to hypothalamus (1).

(b) [8 marks]

• Model: $\text{TSH} = c/[T_4]$ and $[T_4] = k\cdot\text{TSH}$. (1)
• Substitute: $[T_4] = k\cdot(c/[T_4]) \Rightarrow [T_4]^2 = kc \Rightarrow [T_4] = \sqrt{kc}$. (3)
• Single positive root ⇒ unique stable set-point; if $T_4$ rises above it, TSH falls (less stimulation) driving $T_4$ back down, and vice versa — self-correcting. (2)
• Primary hypothyroidism: thyroid gain $k$ falls (gland fails), so for any TSH, $[T_4]$ is low. Low $T_4$ removes negative feedback ⇒ TSH rises markedly (high TSH, low T4) — the diagnostic signature. (2)

[
  {"claim": "Eigenvalues of A have negative real part and nonzero imaginary part (stable spiral)",
   "code": "p1,p2,p3,p4=0.03,0.8,0.02,0.25\nA=Matrix([[-p1,-p2],[p3,-p4]])\nevs=list(A.eigenvals().keys())\nre_neg=all(re(e)<0 for e in evs)\nim_nz=any(im(e)!=0 for e in evs)\nresult=bool(re_neg and im_nz)"},
  {"claim": "Discriminant of characteristic polynomial is negative (-0.0156)",
   "code": "p1,p2,p3,p4=0.03,0.8,0.02,0.25\ntr=p1+p4\ndet=p1*p4+p2*p3\nD=tr**2-4*det\nresult=bool(abs(D-(-0.0156))<1e-9)"},
  {"claim": "Receptor occupancy: theta=0.5 at H=Kd and theta=0.75 at H=3Kd",
   "code": "H,Kd=symbols('H Kd',positive=True)\ntheta=H/(Kd+H)\nt1=theta.subs(H,Kd)\nt2=theta.subs(H,3*Kd)\nresult=bool(simplify(t1-Rational(1,2))==0 and simplify(t2-Rational(3,4))==0)"},
  {"claim": "HPT set-point T4 = sqrt(k*c) satisfies both loop equations",
   "code": "k,c,T4=symbols('k c T4',positive=True)\nsetpt=sqrt(k*c)\nTSH=c/setpt\ncheck=simplify(k*TSH-setpt)\nresult=bool(check==0)"}
]