201. A rare mammal is difficult to observe directly. Researchers record its faecal pellets along standardised transects each year and find values of \(6\), \(9\), \(14\) and \(18\) pellet groups per kilometre over four successive surveys. Which interpretation is most justified?
ⓐ. The standardised indirect index suggests an increasing population trend
ⓑ. The exact population size in the fourth year is \(18\)
ⓒ. Faecal evidence cannot provide any information about population density
ⓓ. The mammal population has increased exactly threefold
Correct Answer: The standardised indirect index suggests an increasing population trend
Explanation: Indirect signs can function as relative indices when survey method and effort are kept comparable. The rising number of pellet groups per kilometre supports an inference of increasing relative abundance. Sampling effort and detection probability must be controlled before treating differences in counts as differences in abundance. The biological signal in a relative index can be separated from methodological noise only by keeping the sampling basis comparable. Relative indices are interpretable only when sampling effort, gear, area and detectability are sufficiently comparable. It does not provide an exact census or justify converting the index directly into a precise number of animals.
202. A population starts with \(N_t=750\). During the interval, \(B=90\), \(D=55\) and \(E=25\). At the end of the interval, \(N_{t+1}=800\). What was immigration \(I\)?
ⓐ. \(20\)
ⓑ. \(40\)
ⓒ. \(30\)
ⓓ. \(50\)
Correct Answer: \(40\)
Explanation: The population increased by \(800-750=50\). Using \[ N_{t+1}=N_t+[(B+I)-(D+E)] \] We obtain \[ 50=(90+I)-(55+25) \] \[ 50=10+I \] So \(I=40\). In the demographic balance, \(B+I\) contributes individuals while \(D+E\) removes them. Net population change is the difference between total additions and total losses. The immediate change in population density is found by comparing total additions \(B+I\) with total losses \(D+E\). Whether density rises or falls depends on the sign of \((B+I)-(D+E)\). Births and immigration increase the population; deaths and emigration reduce it over the same interval. Comparing \(B+I\) with \(D+E\) determines both the direction and magnitude of the immediate population change. This illustrates how an unmeasured demographic process can be inferred when the remaining components and the net change are known.
203. A population has a carrying capacity of \(K=1000\). At one stage its density is \(N=250\), and later its density rises to \(N=750\). According to the logistic factor \(\frac{K-N}{K}\), which statement is correct?
ⓐ. The factor rises from \(0.25\) to \(0.75\), so limitation becomes weaker
ⓑ. The factor remains \(0.50\), so limitation remains unchanged
ⓒ. The factor falls from \(0.75\) to \(0.25\), so limitation becomes stronger
ⓓ. The factor becomes greater than \(1\), so growth becomes unrestricted
Correct Answer: The factor falls from \(0.75\) to \(0.25\), so limitation becomes stronger
Explanation: In logistic growth, density-dependent limitation is represented by
\[
\frac{K-N}{K}.
\]
When \(N=250\),
\[
\frac{K-N}{K}
=
\frac{1000-250}{1000}
=
\frac{750}{1000}
=
0.75.
\]
When \(N=750\),
\[
\frac{K-N}{K}
=
\frac{1000-750}{1000}
=
\frac{250}{1000}
=
0.25.
\]
Thus, as population density rises toward carrying capacity, the logistic factor decreases from \(0.75\) to \(0.25\). A smaller value of this factor means that finite resources impose a stronger limitation on further population growth. Therefore, option C is correct.
204. Population P has \(N=200\) and \(r=0.30\), while Population Q has \(N=600\) and \(r=0.08\). Under exponential-growth conditions, which population has the greater instantaneous absolute rate of increase?
ⓐ. Both populations, because \(rN=60\) for each
ⓑ. Population Q, because \(rN=48\)
ⓒ. Population P, because \(rN=60\)
ⓓ. Neither population, because \(rN=0\) without \(K\)
Correct Answer: Population P, because \(rN=60\)
Explanation: For P, \[ \frac{dN}{dt}=0.30\times200=60 \] For Q, \[ \frac{dN}{dt}=0.08\times600=48 \] P has the larger instantaneous absolute increase despite having the smaller current population. Resource limitation must be weak for the exponential relation to remain an appropriate description of population increase. The relation \(rN\) separates per-capita growth potential from the absolute number of individuals added per unit time. Exponential increase is an idealised growth pattern that omits the slowing effect of finite carrying capacity. A positive \(r\) is a per-capita property, while the realised absolute change at any instant is \(rN\). The model is most applicable during intervals when competition for essential resources has not yet imposed strong density dependence. Exponential growth is multiplicative: more current individuals produce a larger absolute increment at the same intrinsic rate. Both \(r\) and \(N\) must be considered.
205. Two populations have the same \(r\) and \(K=1000\).
| Population | \(N\) |
|---|
| P | \(100\) |
| Q | \(950\) |
Which statement best compares their logistic limitation?
ⓐ. P experiences stronger limitation because smaller populations always grow more slowly
ⓑ. Q has stronger density limitation because \(\frac{K-N}{K}\) is smaller
ⓒ. Both experience identical limitation because \(K\) is identical
ⓓ. Neither experiences limitation until \(N\) becomes exactly equal to \(K\)
Correct Answer: Q has stronger density limitation because \(\frac{K-N}{K}\) is smaller
Explanation: For P, the density-dependent factor is \(\frac{1000-100}{1000}=0.90\). For Q, it is \(\frac{1000-950}{1000}=0.05\). Q is much more strongly constrained by carrying capacity. When \(N\) is close to \(K\), the logistic factor approaches zero and the net growth rate becomes small. Resource limitation is represented mathematically by multiplying \(rN\) by \(\frac{K-N}{K}\). Environmental improvement or deterioration can alter \(K\) by changing the number of individuals the habitat can sustain. The density-dependent multiplier \(\frac{K-N}{K}\) converts the exponential term into a logistic growth rate. As \(N\) approaches \(K\), the factor \(\frac{K-N}{K}\) becomes smaller and realised growth slows. The supportable population level represented by \(K\) is species- and condition-dependent rather than permanently fixed. Logistic growth scales the exponential term \(rN\) by the density-dependent factor \(\frac{K-N}{K}\). Logistic limitation increases progressively as population density approaches \(K\).
206. A population follows logistic growth. At time P, \(N\) is far below \(K\); at time Q, \(N\) is close to \(K\). Which change between P and Q is expected?
ⓐ. The factor \(\frac{K-N}{K}\) increases, producing progressively faster growth
ⓑ. The factor \(\frac{K-N}{K}\) decreases, reducing net population increase
ⓒ. \(r\) necessarily becomes zero because population density has increased
ⓓ. Carrying capacity becomes identical to current population density at every stage
Correct Answer: The factor \(\frac{K-N}{K}\) decreases, reducing net population increase
Explanation: As \(N\) increases toward \(K\), the difference \(K-N\) becomes smaller. \(\frac{K-N}{K}\) declines and increasingly suppresses the realised growth rate. The density-dependent multiplier declines toward zero as population density nears carrying capacity. In the logistic model, \(rN\) is reduced by \(\frac{K-N}{K}\), which represents the effect of finite carrying capacity. Carrying capacity summarises the sustainable limit imposed by the habitat, and that limit can shift when habitat quality changes. The factor \(\frac{K-N}{K}\) modifies the exponential increase according to how close the population is to \(K\). This produces the decelerating portion of the sigmoid logistic curve.
207. An exponentially growing population doubles every \(5\ \text{years}\). Assuming the conditions remain unchanged, approximately how long will it take to increase from \(N_0\) to \(8N_0\)?
ⓐ. \(5\ \text{years}\)
ⓑ. \(10\ \text{years}\)
ⓒ. \(40\ \text{years}\)
ⓓ. \(15\ \text{years}\)
Correct Answer: \(15\ \text{years}\)
Explanation: Increasing from \(N_0\) to \(8N_0\) requires three doublings: \[ N_0\rightarrow2N_0\rightarrow4N_0\rightarrow8N_0 \] At \(5\ \text{years}\) per doubling, the required time is \(3\times5=15\ \text{years}\). Under exponential conditions, the instantaneous absolute change is proportional to current population size through \(\frac{dN}{dt}=rN\). Unlimited-resource conditions allow the intrinsic rate of natural increase to operate without the logistic reduction factor. The quantity \(e^{rt}\) is the growth factor applied to the starting population \(N_0\). A persistent J-shaped curve requires that limiting resources do not become important over the interval considered. Because \(\frac{dN}{dt}=rN\), the same \(r\) produces a larger absolute increase in a larger population. Multiplicative exponential growth is captured by \(N_t=N_0e^{rt}\), not by adding a fixed number each interval. This follows directly from exponential doubling behaviour under constant \(r\).
208. Two habitats support the same species. Habitat P has abundant food and space, whereas Habitat Q has severe resource limitation. If both populations begin at the same small \(N\), which prediction is most appropriate?
ⓐ. P may show near-exponential growth; Q should show stronger density limitation
ⓑ. Both must follow identical trajectories because species identity alone determines growth
ⓒ. Q must have a larger carrying capacity because its resources are scarcer
ⓓ. P must show a sigmoid curve immediately even if resources remain effectively unlimited
Correct Answer: P may show near-exponential growth; Q should show stronger density limitation
Explanation: Population growth depends on environmental conditions as well as the species' intrinsic potential. Abundant resources allow a population to express more nearly exponential growth, while limiting resources strengthen density-dependent constraints characteristic of logistic growth. Exponential growth links absolute increase directly to the product of intrinsic rate and current population size. Under exponential conditions, growth is not reduced merely because population density has become larger. Population membership is defined by individuals of the same species within the geographical area relevant to the investigation. Finite resources eventually make pure exponential growth unrealistic for most natural populations over long periods. Carrying capacity reflects resource support and would generally not be expected to increase when resources become more restricted.
209. A population produces many very small offspring under one habitat condition. After environmental change, survival becomes strongly associated with greater parental investment per offspring. Which evolutionary shift is most consistent with natural selection?
ⓐ. Fewer, well-provisioned offspring when fitness rises
ⓑ. Toward even more offspring regardless of their survival
ⓒ. Toward permanent cessation of reproduction
ⓓ. Toward identical life-history traits in all species
Correct Answer: Fewer, well-provisioned offspring when fitness rises
Explanation: Life-history traits are shaped by selection acting on reproductive fitness. If greater parental investment substantially improves offspring survival, a strategy producing fewer but better-provisioned offspring can become favoured. Life-history traits are favoured according to their effect on lifetime reproductive fitness, not by maximising any single trait in isolation. Natural selection acts on the fitness consequences of reproductive trade-offs, so no one breeding strategy is optimal in every habitat. Offspring number, offspring size and breeding frequency represent trade-offs whose value depends on survival and environmental conditions. No one reproductive strategy is universally optimal.
210. A predator population becomes so efficient that its prey density declines to extremely low levels. Which sequence is most plausible if the predator depends almost exclusively on that prey?
ⓐ. Prey decline → predator dependence on that prey decreases without any change in diet
ⓑ. Prey decline → more food → predator growth
ⓒ. Prey decline → density-dependent limitation on prey weakens → predator food supply increases
ⓓ. Prey decline → food limitation for predator → predator decline
Correct Answer: Prey decline → food limitation for predator → predator decline
Explanation: Predators depend on prey as a food resource. Excessive exploitation may initially benefit the predator but can ultimately deplete its food base. The strength of predation can shape prey abundance, and changes in prey abundance may then propagate through the community. Predation supplies energy to the consumer while imposing mortality that may keep prey populations below otherwise attainable densities. A reduction in predator abundance can increase prey survival and alter subsequent population growth. Predator-mediated mortality is one ecological process capable of changing the density of a prey population. Consumption by predators is a direct route through which interspecific interaction becomes a change in prey demography. Predator pressure influences the number of prey surviving to reproduce and can thereby affect future population size. Once prey become very scarce, predator survival and reproduction can decline, illustrating why unrestricted overexploitation is not sustainable.
211. A prey species has two defensive traits: individuals are difficult to detect against their background, and those that are detected taste unpleasant to predators. These traits operate respectively by
ⓐ. reducing detection and discouraging predators from feeding
ⓑ. reducing digestion and increasing prey visibility
ⓒ. increasing predator attraction and reducing camouflage
ⓓ. increasing prey birth rate and predator mortality directly
Correct Answer: reducing detection and discouraging predators from feeding
Explanation: Camouflage acts before or during detection by making prey difficult to distinguish from the background. Distastefulness acts after detection by discouraging predators from consuming or repeatedly attacking the prey. Trophic transfer and population regulation occur together when predator feeding removes prey from the population. Where predators are effective, prey abundance reflects not only births and resources but also predator-imposed mortality. Predation can influence population density directly through mortality and indirectly through changes in competition among prey species. Predation can reduce prey abundance while transferring energy from prey to higher trophic levels. The two mechanisms reduce predation at different stages of the predator-prey interaction.
212. A plant chemical causes no immediate mortality in a herbivore but reduces its fertility after repeated feeding. Why can this still function as an effective anti-herbivore defence?
ⓐ. Plant defence is restricted to compounds that kill instantly
ⓑ. Any chemical affecting reproduction must be a plant hormone rather than a defence compound
ⓒ. Fertility effects cannot influence population interactions
ⓓ. Reducing herbivore reproduction can lower herbivore fitness and future feeding pressure
Correct Answer: Reducing herbivore reproduction can lower herbivore fitness and future feeding pressure
Explanation: Chemical defence can operate through several mechanisms, including illness, feeding deterrence, digestive interference, reproductive disruption and mortality. By consuming prey, predators both obtain trophic energy and exert mortality that can regulate prey populations. Plants and animal prey use multiple defence routes, including structural barriers, behavioural avoidance and toxic or distasteful compounds. Concealment acts before capture by reducing detection; chemical unpalatability acts after encounter by reducing the value of consumption. A compound that reduces herbivore fertility decreases herbivore fitness and can reduce the future abundance or impact of the herbivore population even without causing immediate death.
213. Species P and Q consume the same resource. The amount of resource remains high, yet P's feeding rate drops whenever Q is present. When Q is physically separated while the resource remains shared, P's feeding rate recovers. Which conclusion is strongest?
ⓐ. Q is a predator of P
ⓑ. P and Q cannot be competitors because food is abundant
ⓒ. Q lowers P's feeding efficiency through interference
ⓓ. The interaction must be commensal because the resource remains available
Correct Answer: Q lowers P's feeding efficiency through interference
Explanation: P's feeding efficiency recovers when physical separation prevents direct interference by Q, even though both species still have access to the same abundant resource. This identifies the presence or behaviour of Q as the cause of the reduced feeding performance. Such a pattern is characteristic of interference competition. Competition does not always require complete depletion of food or space; one species can reduce another species' fitness by directly interfering with its ability to use an otherwise abundant resource.
214. Species P has \(r=0.16\) when grown alone but \(r=0.05\) when grown with species Q. Species Q has \(r=0.12\) alone and \(r=0.04\) with P. Which conclusion is best supported?
ⓐ. P benefits while Q is harmed
ⓑ. Both species experience reduced fitness in each other's presence
ⓒ. The two species are in commensalism because both retain positive \(r\)
ⓓ. Competition is absent because neither \(r\) becomes negative
Correct Answer: Both species experience reduced fitness in each other's presence
Explanation: Competition is demonstrated by a reduction in fitness, not necessarily by negative population growth. Both species show lower \(r\) values when the competitor is present. Competition should be inferred from depressed fitness, not from mere proximity, similarity or resource overlap. Experimental exclusion tests the competitive mechanism directly; natural-history associations may remain consistent with several alternative causes. Competition is demonstrated by a reduction in fitness caused by another species, not merely by co-occurrence or taxonomic similarity. Evidence for competition is strongest when competitor presence is changed experimentally while other major conditions remain comparable. The interaction is consistent with competition even though both populations may still have positive net growth.
215. A laboratory study shows that species P excludes species Q when both depend on one limiting resource. In a complex natural habitat, however, both species coexist. Which explanation is most consistent with population-competition principles?
ⓐ. Laboratory evidence becomes invalid whenever field coexistence occurs
ⓑ. Field coexistence proves that the two species never compete
ⓒ. Gause's principle requires that every pair of competitors coexist in nature
ⓓ. Resource partitioning can promote coexistence in a complex habitat
Correct Answer: Resource partitioning can promote coexistence in a complex habitat
Explanation: Competitive exclusion is clearest when species depend strongly on the same limiting resource under controlled conditions. Natural environments may provide multiple resources, spatial differences or behavioural opportunities that reduce direct overlap. Controlled exclusion or removal experiments provide stronger causal evidence than historical associations in which several ecological factors may have changed together. Shared habitat or shared food does not by itself prove competition; the presence of one species must reduce the fitness of the other. Removing a superior competitor can produce competitive release, whereas differences in feeding place, time or method can reduce overlap through resource partitioning. Resource partitioning can permit coexistence without implying that competitive effects are absent.
216. A superior competitor is removed from part of a habitat. The inferior competitor initially expands into the vacated region, but after several generations it begins using a different microhabitat even when the superior competitor is experimentally reintroduced. Which two processes are represented respectively?
ⓐ. Amensalism followed by competitive exclusion
ⓑ. Parasitism followed by mutualism
ⓒ. Predation followed by brood parasitism
ⓓ. Competitive release followed by partitioning
Correct Answer: Competitive release followed by partitioning
Explanation: Expansion after removal of a competitor is competitive release. Later use of a different microhabitat that reduces overlap with the returning competitor represents resource partitioning. The decisive evidence for competition is a negative effect on survival, growth or reproduction attributable to another species. Competitive release is revealed when a restricted species expands after a competitor is removed; resource partitioning instead promotes coexistence by separating resource use. Manipulating a competitor and observing the response of the other species can isolate causation more clearly than a historical correlation. Competition is a fitness interaction, so simple overlap in habitat or resource use is insufficient unless performance is reduced. Experimental removal provides direct evidence of competitive effects, whereas historical co-occurrence or decline usually supports a more cautious inference. These are related but distinct ecological processes: one reveals the effect of removing competition, while the other reduces competition during coexistence.
217. A parasite living within a host loses elaborate sensory structures but develops powerful attachment structures and produces very large numbers of reproductive stages. Which interpretation best integrates these features?
ⓐ. High reproductive output makes attachment mechanisms unnecessary
ⓑ. Parasitism requires increasing complexity of every organ system
ⓒ. Host life reduces unneeded structures but favours attachment and transmission
ⓓ. Structural simplification shows that the parasite no longer depends on the host
Correct Answer: Host life reduces unneeded structures but favours attachment and transmission
Explanation: Parasitic adaptation is selective rather than uniformly simplifying or complicating the organism. Features that are unnecessary in the host environment can be reduced, while specialised attachment structures and high reproductive capacity improve persistence and transmission. The host can remain alive for a long period and still be harmed if parasite exploitation lowers its fitness. Successful parasites may be highly host-specific and transmission-efficient even though some independent sensory or digestive structures become simplified. Parasite success is achieved at the host's expense, distinguishing parasitism from commensalism or mutualism. These traits reflect functional adaptation to parasitic life.
218. A parasite normally follows the sequence: snail → fish → definitive host. A control programme removes most fish but leaves snail populations unchanged. Which effect is most direct?
ⓐ. The parasite no longer requires transmission between hosts
ⓑ. Development in snails is sufficient for normal completion even when the fish stage is absent
ⓒ. Snail development may occur, but loss of the fish stage blocks normal life-cycle completion
ⓓ. Loss of the fish stage increases transmission directly from snails to the definitive host
Correct Answer: Snail development may occur, but loss of the fish stage blocks normal life-cycle completion
Explanation: Removing a required later intermediate host does not necessarily prevent earlier stages from occurring, but it interrupts progression through the normal life cycle. When a parasite requires successive hosts or a vector, failure at one necessary transmission step prevents normal completion even if earlier development occurs. Reduced host condition, fecundity or survival can all express the negative side of a parasitic interaction. The parasite-host relationship is defined by unequal fitness consequences rather than by a requirement that the host be killed. Complex life cycles are chain-dependent: interruption of one obligatory host or vector stage can prevent the parasite from reaching its final stage. Host exploitation provides resources to the parasite while diminishing the host's ability to survive or reproduce. The fish stage is required for successful onward development, so reducing fish availability creates a transmission bottleneck.
219. A brood-parasitic bird population initially lays eggs closely matching those of its host. Over generations, host eggs become much more speckled because hosts increasingly reject poorly matching eggs. Which change in the brood parasite would most strongly restore its success?
ⓐ. Evolution toward larger body size without changing eggs
ⓑ. Evolution toward more similar egg speckling
ⓒ. Elimination of egg mimicry
ⓓ. Reduction in host-nest searching
Correct Answer: Evolution toward more similar egg speckling
Explanation: Host rejection places selection on the brood parasite to improve egg resemblance. If host egg appearance changes, parasite eggs that more closely match the new pattern are less likely to be detected and ejected. A living host may continue to support a parasite even while suffering cumulative losses in growth and reproductive output. Parasitism is identified from the combination of parasite benefit and host harm, not from the duration of the association alone. Fitness loss to the host is the essential negative consequence that accompanies the parasite's gain. The host may show subtle costs such as slower growth or fewer offspring even when mortality is unchanged. This illustrates reciprocal evolutionary pressure between host defence and parasite counter-adaptation.
220. An epiphytic plant gains support from a tree without taking nutrients from it. Later, evidence shows that the epiphyte's dense growth consistently reduces light reaching the tree's leaves and lowers the tree's growth. How should the interaction now be interpreted?
ⓐ. The tree is harmed, so the interaction is no longer commensalism
ⓑ. It remains commensalism because the epiphyte still gains support
ⓒ. It becomes mutualism because both species are physically associated
ⓓ. It becomes competition only if the epiphyte dies
Correct Answer: The tree is harmed, so the interaction is no longer commensalism
Explanation: Commensalism requires benefit to one species with no detectable effect on the other. Once reliable evidence shows a cost to the supporting tree, the relationship no longer fits \((+,0)\). One-sided benefit with no measurable response in the partner separates commensalism from mutualism and parasitism. Commensalism requires a measurable benefit to one species with no detectable benefit or harm to the other. The \((+,0)\) pattern describes one-sided benefit while the second species shows no measurable fitness effect. Ecological classification depends on measured consequences rather than on the names traditionally assigned to superficially similar associations.
