Population & Community Ecology

Question 1 (12 marks)

(a) Derivation of logistic form (4)

• Start: $\frac{dN}{dt}=rN$; per-capita rate $\frac{1}{N}\frac{dN}{dt}=r$ (constant). (1)
• Biological reasoning: per-capita rate should decline as $N\to K$. Impose a linear decline: $r_{\text{eff}}=r\left(1-\frac{N}{K}\right)$, which equals $r$ at $N=0$ and $0$ at $N=K$. (2)
• Substitute: $\boxed{\dfrac{dN}{dt}=rN\left(1-\dfrac{N}{K}\right)}$. (1)

(b) Solve the ODE (6)

• Separate: $\dfrac{dN}{N\left(1-\frac{N}{K}\right)}=r\,dt$. (1)
• Partial fractions: $\dfrac{1}{N(1-N/K)}=\dfrac{1}{N}+\dfrac{1/K}{1-N/K}$. (2)
• Integrate: $\ln N-\ln\left(1-\frac{N}{K}\right)=rt+C \Rightarrow \ln\frac{N}{1-N/K}=rt+C$. (1)
• Apply $N(0)=N_0$: $C=\ln\frac{N_0}{1-N_0/K}$. Exponentiate and solve for $N$. (1)
• Result: $\boxed{N(t)=\dfrac{K}{1+\left(\frac{K-N_0}{N_0}\right)e^{-rt}}}$. (1)

(c) Maximal growth rate (2)

• Let $g(N)=rN(1-N/K)$. $g'(N)=r(1-2N/K)=0\Rightarrow N=K/2$. (1)
• $g''=-2r/K<0$ ⇒ maximum. So growth rate peaks at $N=K/2$. (1)

Question 2 (8 marks)

(a) Explain out loud (4)

• Early: $N$ small, so absolute rate $rN(1-N/K)\approx rN$ ≈ exponential; as $N$ rises the number of reproducing individuals grows, so absolute rate climbs. (2)
• Later: the $(1-N/K)$ brake grows stronger as $N\to K$; even though per-individual $r$ is fixed, the fraction of realised growth shrinks, so absolute rate falls back toward 0. The peak is where these opposing trends balance. (2)

(b) Computation (4)

• Max absolute rate at $N=K/2=5000$. (2)
• $\frac{dN}{dt}=rN(1-N/K)=0.8\times5000\times(1-0.5)=0.8\times5000\times0.5=2000$ cells/hr. (2)

Question 3 (10 marks)

(a) Distinguish (6)

• Density-dependent: effect intensity rises with population density. Examples: competition for food (per-capita resource falls as $N$ rises); disease/parasite transmission (spreads faster at high density); predation (predators focus on dense prey). Mechanism: feedback tied to crowding. (3)
• Density-independent: effect is unrelated to density. Examples: floods, drought, cold snaps, fire. Mechanism: abiotic events kill a fraction regardless of how many are present. (3)

(b) Link to $(1-N/K)$ (4)

• As $N$ rises, density-dependent factors (resource depletion, waste, disease) reduce per-capita birth and raise per-capita death. (2)
• When $N=K$, these feedbacks bring births = deaths (net zero growth), exactly the biological meaning of the $(1-N/K)$ factor going to 0. Thus $K$ is the density at which density-dependent regulation balances the population. (2)

Question 4 (10 marks)

(a) $l_x$ and classification (5)

• $l_x=n_x/n_0$: $1.0,\ 0.5,\ 0.25,\ 0.125,\ 0.062,\ 0.031$. (2)
• Constant ratio (~0.5 survive each interval) ⇒ constant per-capita mortality per age class. On a log scale $\ln l_x$ is linear. (2)
• ⇒ Type II survivorship curve. (1)

(b) Link to selection (5)

• Type II ⇒ mortality risk roughly equal at all ages. (1)
• Type III (heavy early death, few survive) associates with r-selected species: many offspring, little parental care. Type I (low early death) associates with K-selected. Type II is intermediate. (2)
• Reasoning: species with constant mortality invest moderately per offspring; the pattern reflects a strategy where survival is not concentrated at a life stage, unlike the "many cheap offspring" r-strategy that gives Type III. (2)

Question 5 (10 marks)

(a) Lotka–Volterra (6)

• Prey: $\frac{dV}{dt}=aV-bVW$; Predator: $\frac{dW}{dt}=cVW-dW$. (2)
• $a$=prey growth, $b$=predation rate, $c$=conversion efficiency, $d$=predator death. (2)
• Out-of-phase reasoning: abundant prey → predators eat/reproduce → predators rise → prey crash → predators starve/decline → prey recover → cycle repeats. Predator peak lags prey peak by ~quarter cycle. (2)

(b) Keystone species (4)

• A keystone species has an ecological impact disproportionately large relative to its abundance. (2)
• Removing a keystone predator (e.g., Pisaster) releases a dominant competitor (e.g., mussels) from predation; that competitor monopolises space, out-competing/excluding other species ⇒ competitive exclusion ⇒ biodiversity falls. (2)

Question 6 (10 marks)

(a) Symbiosis notation (3)

• Mutualism $(+/+)$ — e.g., pollinator & flower. (1)
• Commensalism $(+/0)$ — e.g., barnacle on whale. (1)
• Parasitism $(+/-)$ — e.g., tapeworm in host. (1)

(b) Biodiversity & stability (4)

• Functional redundancy: multiple species perform similar roles, so loss of one is buffered. (2)
• Response diversity/insurance: diverse species respond differently to disturbance, keeping some functional under stress; more stable productivity. (2)

(c) Resource partitioning (3)

• Competing species specialise on different niches (e.g., feeding times, prey sizes, microhabitats), lowering niche overlap and competition intensity. (2)
• Reduced competition allows coexistence rather than exclusion ⇒ more species persist ⇒ higher biodiversity. (1)

[
  {"claim": "Logistic growth rate is maximal at N = K/2", "code": "N,r,K = symbols('N r K', positive=True); g = r*N*(1-N/K); crit = solve(diff(g,N),N); result = (crit==[K/2])"},
  {"claim": "Max absolute growth rate = 2000 cells/hr for r=0.8,K=10000,N=5000", "code": "r=Rational(8,10); K=10000; N=K/2; rate = r*N*(1-N/K); result = (rate==2000)"},
  {"claim": "Survivorship l_x values give constant ratio 0.5 (Type II)", "code": "n=[1000,500,250,125]; lx=[Rational(v,1000) for v in n]; ratios=[lx[i+1]/lx[i] for i in range(len(lx)-1)]; result = all(rr==Rational(1,2) for rr in ratios)"},
  {"claim": "Logistic solution satisfies the ODE dN/dt = rN(1-N/K)", "code": "t=symbols('t'); r,K,N0=symbols('r K N0',positive=True); N=K/(1+((K-N0)/N0)*exp(-r*t)); lhs=diff(N,t); rhs=r*N*(1-N/K); result = simplify(lhs-rhs)==0"}
]