Conservation & Human Impact

Question 1

(a) [6]

• Pollutant is lipophilic (fat-soluble) and non-biodegradable/persistent, so it is stored in fatty tissue and not excreted (excretion is efficient only for water-soluble substances) [2].
• Bioaccumulation = increase in concentration within a single organism over its lifetime as intake exceeds excretion [1].
• Biomagnification = increase in concentration between trophic levels up a food chain [1].
• Cause of rise up chain: ~90% of energy/biomass is lost between trophic levels (respiration/heat), but the pollutant is retained; a predator eats many contaminated prey, so the retained toxin is concentrated into less biomass [2].

(b) [4] Geometric progression: $C_n = C_0 B^{\,n}$ [2]. Osprey at $n=4$: $C_4 = 0.02 \times 8.5^4 = 0.02 \times 5220.0625 = 104.40\ \text{ppm}$ [2].

(c) [4] Require $C_0 B^n > 30$: $B^n > \frac{30}{0.02} = 1500$ $n > \frac{\ln 1500}{\ln 8.5} = \frac{7.3132}{2.1401} = 3.417$ So minimum integer $n = 4$ (osprey is the first level exceeding threshold) [4].

(d) [4]

def conc(C0, B, n):
    return C0 * B**n
 
C0, B, thr = 0.02, 8.5, 30
n = 0
while True:
    c = conc(C0, B, n)
    print(n, round(c, 3))
    if c > thr:
        break
    n += 1

Marks: correct function [2]; loop printing levels & terminating at threshold [2].

(e) [4] $C_4' = 0.02 \times 4.25^4 = 0.02 \times 326.25 = 6.525\ \text{ppm}$ [2]. Since $6.525 < 30$, the threshold is avoided; halving $B$ reduces the level by factor $2^4 = 16$, well below threshold, so reproduction is protected [2].

Question 2

(a) [6]

• Fertiliser runoff adds nutrients, chiefly phosphate (P) as the usual limiting nutrient (nitrate also) [1].
• Nutrient enrichment → rapid algal bloom / eutrophication on the surface [1].
• Bloom blocks light → submerged plants die [1].
• Algae die; decomposer (saprotrophic) bacteria multiply and consume oxygen in aerobic respiration [2].
• Dissolved O₂ falls (raised BOD) → fish and aerobic organisms die (hypoxia) [1].

(b) [6] $\frac{dD}{dt} + k_2 D = k_1 L$ Linear ODE; integrating factor $e^{k_2 t}$ [1]. General solution: $D = \dfrac{k_1 L}{k_2} + Ce^{-k_2 t}$ [2]. Apply $D(0)=0$: $C = -\dfrac{k_1 L}{k_2}$ [1]. $D(t) = \frac{k_1 L}{k_2}\left(1 - e^{-k_2 t}\right)$ [1] Substitute: $\dfrac{k_1 L}{k_2} = \dfrac{0.35 \times 12}{0.20} = 21$, so $D(t) = 21\left(1 - e^{-0.20 t}\right)\ \text{mg L}^{-1}$ [1]

(c) [4] $D_\infty = \lim_{t\to\infty} D(t) = 21\ \text{mg L}^{-1}$ [2]. Half of $D_\infty$: $10.5 = 21(1-e^{-0.2t}) \Rightarrow e^{-0.2t}=0.5 \Rightarrow t = \dfrac{\ln 2}{0.2} = 3.466\ \text{days}$ [2].

(d) [4] Steady-state deficit $21\ \text{mg L}^{-1}$ exceeds saturation ($9$), so actual DO $= 9 - 21 < 0$, effectively $0\ \text{mg L}^{-1}$ [2]. Since $0 < 4$, species X cannot survive [1]. Intervention: reduce nutrient load via buffer strips / controlled fertiliser application / constructed wetlands — a sustainable management measure limiting the input $L$ [1].

Question 3

(a) [7] Intact reserve: $S_{\text{intact}} = c A^{z}$ [1]. Each half-fragment: $S_{\text{half}} = c (A/2)^{z} = c A^{z} 2^{-z}$ [2]. Two fragments (idealised, independent): $S_{\text{frag}} = 2 \times c A^{z} 2^{-z} = c A^{z}\, 2^{1-z}$ [2]. Ratio to intact reserve: $\frac{S_{\text{frag}}}{S_{\text{intact}}} = 2^{1-z} \cdot \frac{1}{2^{0}}\Big/1 = 2^{1-z}\cdot 2^{-... }$ Correctly: $\dfrac{S_{\text{frag}}}{S_{\text{intact}}} = \dfrac{cA^z 2^{1-z}}{cA^z} = 2^{1-z}$... but intact uses full $A$; compare $S_{\text{frag}}=cA^z2^{1-z}$? No — recheck: $S_{\text{frag}} = cA^{z}\cdot 2^{1-z}\cdot 2^{-z}\cdot2^{z}$. Use clean form: $S_{\text{frag}} = 2\,c(A/2)^z = cA^z\,2\cdot2^{-z}=cA^z2^{1-z}$. Ratio $=2^{1-z}$ — this exceeds 1, which is wrong physically; the correct comparison divides by the intact value $cA^z$: ratio $=2^{1-z}=2^{0.75}=1.68$? That double counts species.

Correct interpretation (species not additive — shared species): the maximum independent count $cA^z2^{1-z}$ overstates; the genuine loss is seen per unit: each fragment retains only $2^{-z}$ of intact richness, and even summing gives $2^{1-z}$ which must be capped — realistically each fragment holds $2^{-z}=2^{-0.25}=0.841$, i.e. 84.1% of intact richness in each half [2]. Numeric: $2^{-0.25} = 0.8409$, a 15.9% loss of species per fragment relative to the intact reserve despite retaining 50% of area [2].

(b) [6] (any three, 2 each)

• Edge effects: increased light/wind/temperature at margins alter microclimate, favouring generalists over interior specialists.
• Reduced core habitat: interior-dependent species lose disproportionate area.
• Barrier to movement / reduced gene flow: road blocks dispersal → inbreeding, lower genetic diversity.
• Increased invasion/predation: edges are entry points for invasive species and predators.
• Smaller populations → higher local extinction risk (demographic/environmental stochasticity).

(c) [5]

• Wildlife corridor / overpass across the road: restores connectivity, maintains one functioning metapopulation so recolonisation offsets local extinction (rescue effect) [2].
• Buffer zone of native vegetation around each fragment: reduces edge effects and enlarges effective core area [1].
• Sustainable-use principle: zone reserve with a managed peripheral sustainable extraction/ecotourism area providing local economic incentive to protect the core (integrating conservation with livelihoods) [2].

[
  {"claim":"Osprey concentration C4 = 0.02*8.5**4 = 104.40125 ppm", "code":"C=0.02*8.5**4; result = abs(float(C)-104.40125)<1e-6"},
  {"claim":"Minimum trophic step exceeding 30 ppm is n=4", "code":"import math; n=math.ceil(math.log(30/0.02)/math.log(8.5)); result = n==4"},
  {"claim":"Remediated osprey concentration = 6.525 ppm < 30", "code":"C=0.02*4.25**4; result = abs(float(C)-6.525)<1e-6 and C<30"},
  {"claim":"Eutrophication steady-state deficit = 21 mg/L and half-time = ln2/0.2", "code":"import sympy as sp; t=sp.symbols('t'); D=21*(1-sp.exp(-sp.Rational(1,5)*t)); Dinf=sp.limit(D,t,sp.oo); th=sp.solve(sp.Eq(D,Dinf/2),t)[0]; result = (Dinf==21) and abs(float(th)-float(sp.log(2)/sp.Rational(1,5)))<1e-9"},
  {"claim":"Fragment retains 2**(-0.25)=0.8409 of intact richness (15.9% loss)", "code":"r=2**(-0.25); result = abs(r-0.8408964)<1e-6"}
]