Ecology & Ecosystems

Question 1

(a) GPP $= 1.8\% \times 1.5\times10^6 = 0.018\times1.5\times10^6 = 2.7\times10^4\ \text{kJ m}^{-2}\text{yr}^{-1}$ (2) NPP $=$ GPP $-$ respiration $= 2.7\times10^4\times(1-0.40) = 2.7\times10^4\times0.60 = 1.62\times10^4\ \text{kJ m}^{-2}\text{yr}^{-1}$. (2) Why: NPP is GPP minus producer respiration — the energy actually available to consumers.

(b) With $\eta=0.10$ applied from $T_1=$ NPP: (4, 1 per correct value)

Never inverted (2): Energy pyramids obey the laws of thermodynamics — each transfer loses energy as heat (respiration, ~90%), so energy at a higher level is always < the level below. It cannot widen upward.

(c) Proof by induction (5):

• Base: $n=1$: $E_1 = E_1\eta^{0}=E_1$. ✓ (1)
• Hypothesis: assume $E_k = E_1\eta^{k-1}$. (1)
• Step: by definition each level receives fraction $\eta$ of the previous: $E_{k+1} = \eta E_k = \eta\cdot E_1\eta^{k-1} = E_1\eta^{k}$, which equals the formula with $n=k+1$. ✓ (2)
• By induction $E_n=E_1\eta^{n-1}$ for all $n\ge1$. (1)

Derivation of $N$ (3): persistence requires $E_N\ge E_{\min}$: $E_1\eta^{N-1}\ge E_{\min}\Rightarrow \eta^{N-1}\ge \frac{E_{\min}}{E_1}.$ Since $0<\eta<1$, taking logs (inequality flips): $N-1 \le \frac{\ln(E_{\min}/E_1)}{\ln\eta}\Rightarrow N=\left\lfloor 1+\frac{\ln(E_{\min}/E_1)}{\ln\eta}\right\rfloor.$

(d) $\dfrac{\ln(5/16200)}{\ln 0.1} = \dfrac{\ln(3.086\times10^{-4})}{-2.3026}=\dfrac{-8.0827}{-2.3026}=3.510$. $N = \lfloor 1+3.510\rfloor = \lfloor4.510\rfloor = 4$. (3) So 4 trophic levels are sustainable (matches table: $T_4=16.2>5$, $T_5=1.62<5$).

(e) (3)

function trophic_levels(E1, eta, Emin):
    n = 1
    E = E1
    while E * eta >= Emin:
        E = E * eta
        n = n + 1
    return n

Question 2

(a) (4)

• $f$ = nitrogen fixation (atmospheric N₂ → soil ammonium/nitrate by fixers). (1)
• $u\,S$ = plant uptake/assimilation of soil nitrate (absorption by roots). (2)
• $m\,P$ = mineralisation / decomposition (ammonification: dead plant N returned to soil by decomposers). (1)

(b) Steady state: $\dfrac{dP}{dt}=0 \Rightarrow uS^\* = mP^\* \Rightarrow P^\* = \dfrac{u}{m}S^\*$. (2) Adding the two equations at steady state: $\dfrac{dS}{dt}+\dfrac{dP}{dt}=f=0$? — the closed system has $f$ as net input, so with $dS/dt=0$: $f - uS^\* + mP^\* =0$, and since $uS^\*=mP^\*$, this gives $f=0$ contradiction unless we treat $S^\*$ from the plant equation. Correctly: from $dS/dt=0$, $uS^\*-mP^\*=f$; but $dP/dt=0$ gives $uS^\*-mP^\*=0$. These are inconsistent for $f\ne0$ — hence no finite steady state exists (see part d). If we solve only $dP/dt=0$: $S^\* = \frac{m}{u}P^\*$. (2) Ecological comment: the ratio $P^\*/S^\* = u/m$ depends only on uptake vs mineralisation rates, not on how much N enters — input $f$ just keeps accumulating in the system. (2)

(c) Using $P^\*=\frac{u}{m}S^\*$: ratio $u/m = 0.5/0.2 = 2.5$. The unbounded system has no unique $S^\*,P^\*$; imposing conservation of a fixed total $N_0$ gives $S^\*=N_0/(1+2.5)$. Award full marks for correctly stating $P^\*=2.5\,S^\*$ and identifying non-existence of an absolute steady state. (4)

(d) Adding equations: $\dfrac{d(S+P)}{dt}=f$. Since $f=20>0$, $N=S+P$ increases linearly without bound — nitrogen keeps accumulating. (2) With denitrification $-dS$: soil eq becomes $\frac{dS}{dt}=f-uS+mP-dS$. Sum: $\frac{dN}{dt}=f-dS^\*=0 \Rightarrow S^\*=f/d$. (2) Then $P^\*=\frac{u}{m}S^\* = \frac{uf}{md}$, and $N^\* = \frac{f}{d}\left(1+\frac{u}{m}\right)$. Interpretation: input (fixation) balances output (denitrification); the pool sizes stabilise. (1)

(e) Denitrification (2) performed by denitrifying bacteria (e.g. Pseudomonas, anaerobic soil bacteria) (1), converting nitrate → N₂ gas.

Question 3

(a) (3)

• Primary succession: colonisation of newly exposed lifeless substrate with no soil (e.g. bare rock, lava). (1)
• Secondary succession: re-establishment on an area where a community existed but was disturbed (soil remains, e.g. after fire). (1)
• Bare volcanic rock = primary succession. (1)

(b) (i) $R(0)=\dfrac{50}{1+49e^{0}}=\dfrac{50}{50}=1$ species. (3) (ii) Half of $R_{\max}=25$: $\dfrac{50}{1+49e^{-kt}}=25 \Rightarrow 1+49e^{-kt}=2 \Rightarrow e^{-kt}=1/49$. $t_{1/2}=\dfrac{\ln 49}{k}=\dfrac{3.8918}{0.05}=77.8\ \text{yr}$. (3)

(c) (5) Any two, each with linked reasoning:

• High rainfall (>2000 mm/yr): abundant water → no water limitation on photosynthesis → high producer biomass. (2.5)
• High constant temperature / high insolation near equator: year-round warmth and light → continuous high photosynthetic rate → large $T_1$ biomass. (2.5) (Accept: humidity, lack of seasonality, warm soils.)

[
  {"claim":"NPP = 16200 kJ/m2/yr from 1.5e6, 1.8% GPP, 40% respiration","code":"gpp=0.018*1.5e6; npp=gpp*(1-0.40); result = abs(npp-16200)<1e-6"},
  {"claim":"Sustainable trophic levels N = 4 for E1=16200, eta=0.1, Emin=5","code":"import sympy as sp; N=sp.floor(1+sp.log(sp.Rational(5,16200))/sp.log(sp.Rational(1,10))); result = int(N)==4"},
  {"claim":"T4 energy = 16.2 kJ/m2/yr","code":"E1=16200; E4=E1*0.1**3; result = abs(E4-16.2)<1e-9"},
  {"claim":"Succession half-richness time t=ln(49)/0.05 ~ 77.8 yr","code":"import sympy as sp; t=sp.log(49)/sp.Rational(5,100); result = abs(float(t)-77.83)<0.05"},
  {"claim":"Initial richness R(0)=1 for Rmax=50,a=49","code":"result = abs(50/(1+49*1)-1)<1e-9"},
  {"claim":"Nitrogen model N* = (f/d)(1+u/m) with f=20,u=0.5,m=0.2,d=0.1 gives 700","code":"f=20;u=sp.Rational(1,2);m=sp.Rational(1,5);d=sp.Rational(1,10); Nstar=(f/d)*(1+u/m); result = float(Nstar)==700.0"}
]