Immune System

Question 1

(a) Epidemic cannot grow when each infection produces on average <1 new infection: $R_e < 1$ (1 mark, meaning: infection chain dies out). Set $R_0(1-p) < 1 \Rightarrow 1-p < 1/R_0 \Rightarrow p > 1 - \dfrac{1}{R_0}$ (2 marks). Therefore the critical threshold: $\boxed{p_c = 1 - \frac{1}{R_0}}$ (1 mark).

(b) $p_c = 1 - 1/15 = 14/15 = 0.9\overline{3}$ (2 marks). Immune fraction $p = \varepsilon v$, require $\varepsilon v \ge p_c$: $v \ge \frac{p_c}{\varepsilon} = \frac{0.9333}{0.95} = 0.9825 \approx 98.2\%$ (2 marks). Comment: ~98% coverage is very high and hard to sustain; because $R_0$ is large and efficacy <100%, even universal vaccination is barely sufficient — this is why measles resurges when coverage dips (1 mark).

(c) (5 marks)

• Active immunity: the individual's own immune system responds to antigen, produces antibodies AND memory cells; slow onset but long-lasting (1).
• Passive immunity: pre-formed antibodies transferred (e.g. mother→fetus, antiserum); immediate but temporary, no memory (1).
• A vaccine provides active (artificial) immunity — it presents antigen (attenuated/inactivated pathogen/subunit) triggering the person's own response (1).
• Enables durable herd immunity because it generates immunological memory (memory B and T cells) (1) → immunity persists for years, keeping the immune fraction $p$ high and stable over time, unlike passive immunity which would wane in weeks (1).

(d) (6 marks)

def herd_threshold(R0, eff):
    pc = 1 - 1/R0            # critical immune fraction
    if pc <= 0:              # R0 <= 1, no vaccination needed
        return 0.0
    v = pc / eff
    if v > 1:               # required coverage exceeds 100%
        return "impossible"
    return v

(function logic 3 marks) Impossible condition: need $v = p_c/\varepsilon \le 1 \Rightarrow \varepsilon \ge p_c = 1 - 1/R_0$. So herd immunity is unachievable by vaccination when $\varepsilon < 1 - \frac{1}{R_0} \iff R_0 > \frac{1}{1-\varepsilon}$ (2 marks). Justification: even vaccinating everyone ($v=1$) only makes $p=\varepsilon$; if $\varepsilon < p_c$ then $R_e = R_0(1-\varepsilon) > 1$ and the epidemic still grows (1 mark).

Question 2

(a) (6 marks) For $t \ge \tau_{\text{lag}}$: $\dfrac{dA}{dt} + \delta A = k$ (linear first-order). Integrating factor $e^{\delta t}$ (1). $\dfrac{d}{dt}(A e^{\delta t}) = k e^{\delta t} \Rightarrow A e^{\delta t} = \frac{k}{\delta}e^{\delta t} + C$ (2). General: $A = \frac{k}{\delta} + Ce^{-\delta t}$ (1). Apply $A(\tau_{\text{lag}})=0$: $C e^{-\delta\tau_{\text{lag}}} = -\frac{k}{\delta}$ (1). Substituting and letting $u = t-\tau_{\text{lag}}$: $A(t) = \frac{k}{\delta}\left(1 - e^{-\delta(t-\tau_{\text{lag}})}\right)\ \checkmark$ Plateau ($t\to\infty$): $A_{ss} = \dfrac{k}{\delta}$ (1).

(b) (5 marks) Plateau $= k/\delta = 8/0.2 = 40$ AU (2). 90% of plateau: $1 - e^{-\delta u} = 0.9 \Rightarrow e^{-\delta u} = 0.1$ (1). $u = \dfrac{\ln 10}{\delta} = \dfrac{2.3026}{0.2} = 11.51$ days (2).

(c) (6 marks) Ratio of plateaus $= \dfrac{k'/\delta}{k/\delta} = \dfrac{5k}{k} = 5$ (2 marks). Cellular/molecular basis (4 marks, 1 each + 1):

• Shorter lag: memory B cells already exist as an expanded clone specific to the antigen — no need to start clonal selection from a rare naïve cell, so plasma-cell differentiation is rapid.
• Higher magnitude ($k'=5k$): larger memory clone → more plasma cells → greater secretion rate.
• Higher affinity: affinity maturation (somatic hypermutation + selection in germinal centres during primary response) yields higher-affinity B-cell receptors.
• Class switching: secondary response is dominated by IgG (vs IgM in primary), giving better tissue penetration and longer half-life.

(d) (5 marks) IgG monomer = 2 heavy + 2 light chains; the two Fab arms each carry one antigen-binding site formed by $V_H+V_L$ → valency 2 (2). IgM is a pentamer (5 monomers joined by a J chain), $5\times2 = 10$ potential binding sites (1). High valency increases avidity and cross-links antigens on many cells/particles → efficient agglutination, so IgM is a powerful early agglutinin despite lower individual-site affinity (2).

Question 3

(a) (6 marks — 1 each)

(b) (7 marks)

• No co-stimulation ($x_1=1,x_2=0$): exponent $= 4(1)+4(0)-6 = -2$; $P = \dfrac{1}{1+e^{2}} = \dfrac{1}{1+7.389} = 0.119$ (2).
• Full ($x_1=1,x_2=1$): exponent $= 4+4-6 = 2$; $P = \dfrac{1}{1+e^{-2}} = \dfrac{1}{1.1353} = 0.881$ (2).
• Interpretation: signal 1 alone (MHC-peptide) gives low activation (~12%) → the T cell is effectively not activated / rendered anergic; this two-signal requirement enforces self-tolerance because self-antigen-presenting cells usually lack co-stimulation (2).
• Autoimmunity: if co-stimulation is inappropriately provided against self (e.g. infection/tissue damage supplying signal 2, or defective regulatory checkpoints), self-reactive T cells cross the threshold and attack self tissue → autoimmune disease (1).

(c) (5 marks) Probability both alleles distinct = 1 − P(both same). Drawing two of $m$ equally frequent alleles (with replacement / large pool): P(same) $= 1/m$, so $H = 1 - \frac{1}{m}$ (2). For $m=20$: $H = 1 - 0.05 = 0.95$ (1). Advantage: a heterozygote presents peptides on two different MHC variants, covering a broader range of pathogen peptides → detects more pathogens; heterozygote advantage (overdominant selection) maintains high MHC polymorphism in populations (2).

[
  {"claim": "Measles pc = 1 - 1/15 = 14/15", "code": "R0=15; pc=1-Rational(1,R0); result = (pc==Rational(14,15))"},
  {"claim": "Required vaccination coverage v = pc/eff approx 0.9825", "code": "pc=1-Rational(1,15); v=pc/Rational(95,100); result = (abs(float(v)-0.98246)<1e-3)"},
  {"claim": "Antibody plateau k/delta = 40 AU", "code": "k=8; d=Rational(2,10); result = (Rational(k,1)/d==40)"},
  {"claim": "Time to 90% plateau = ln(10)/0.2 approx 11.513 days", "code": "d=Rational(2,10); t=ln(10)/d; result = (abs(float(t)-11.5129)<1e-3)"},
  {"claim": "Secondary plateau ratio = 5", "code": "k=symbols('k'); d=symbols('d'); ratio=(5*k/d)/(k/d); result = (simplify(ratio)==5)"},
  {"claim": "P_activate no co-stim = 1/(1+e^2) approx 0.1192", "code": "P=1/(1+exp(2)); result = (abs(float(P)-0.11920)<1e-4)"},
  {"claim": "P_activate full signal = 1/(1+e^-2) approx 0.8808", "code": "P=1/(1+exp(-2)); result = (abs(float(P)-0.88080)<1e-4)"},
  {"claim": "MHC heterozygosity H = 1-1/20 = 0.95", "code": "m=20; H=1-Rational(1,m); result = (H==Rational(19,20)) and (float(H)==0.95)"}
]