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Questions on populations and their limits
Population, community and habitat as separate terms; the sigmoid growth curve and its lag, log, slowing and stationary phases; carrying capacity; density-dependent against density-independent limiting factors; intraspecific and interspecific competition; the competitive exclusion principle and the ecological niche; predator-prey cycles and the lag between the two peaks; and estimating the size of a mobile population by mark, release and recapture with the Lincoln index.
15 original questions · 54 marks · the populations and their limits notes · Ecology, populations and environmental change
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Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the points are much easier to see when you have something of your own to compare against.
A land manager claims that the carrying capacity of a grouse moor is a single fixed number, that can be measured once and then used to set the shooting quota every year afterwards. Evaluate this claim.
Mark scheme
- B1 for: at any given time the moor genuinely does support only a limited population, set by the food, nesting sites and other resources available, so there is a real ceiling that can be measured
- B1 for: a measured value gives managers a starting reference for setting a quota, which is better than setting one with no measurement to refer to at all
- B1 against: carrying capacity is not fixed, because the resources available change with the weather, so a good breeding season or a drought raises or lowers the ceiling from one year to the next
- B1 against: a population can temporarily rise above the capacity present at the time before falling back after a delay, so a single count taken during an overshoot could overstate what the habitat can sustain in the long term
- B1 judgement: a single fixed number measured once is not a sound basis for every future year's quota; the quota should instead be reviewed each year against a current measure of resources or population trend
Describe the four phases of a sigmoid population growth curve in terms of birth rate and death rate.
Mark scheme
- B1 lag phase: numbers barely move because there are few individuals to reproduce, so even a birth rate above the death rate produces very few offspring, and the organisms are still acclimatising
- B1 log phase: resources are plentiful and little is in the way, so birth rate greatly exceeds death rate and the population multiplies by a constant factor in each equal interval
- B1 slowing phase: food per individual falls, waste accumulates and disease spreads more easily between crowded hosts, so birth rate falls, death rate rises and the gap between them narrows
- B1 stationary phase: birth rate and death rate match, and the population fluctuates about a level rather than sitting on it, because the factors setting that level vary with the weather and the season
Explain why the peak in a predator population arrives after the peak in the population of its prey, and explain what that lag shows about the claim that predators control prey numbers.
Mark scheme
- B1 when prey are abundant the predators are well fed and rear more young that survive
- B1 predator abundance responds through reproduction and survival, or through immigration, and each of those takes time, so the rise in predator numbers appears months or years after the rise in prey
- B1 by the time predator numbers have risen the prey are being eaten faster than they breed and fall, so predators then starve and fall in turn, and with predation relieved the prey recover and the cycle repeats
- B1 the lag shows that predator numbers respond to prey abundance; it does not rule out predators controlling prey, since predators that do reduce prey still peak after them, so that claim needs other evidence: the Yukon experiments on snowshoe hares raised density about threefold by adding food, about twofold by excluding predators and about elevenfold by doing both, so the two act together
A researcher catches 150 field voles in a wood of area 3.0 hectares, marks them and releases them. A week later she catches 120 voles, of which 18 carry a mark. The estimated population size = (number marked in the first sample × number caught in the second sample) ÷ number of marked animals in the second sample. Calculate the estimated size of the population and calculate its density.
Mark scheme
- M1 (150 × 120) ÷ 18 = 18 000 ÷ 18
- A1 an estimated population of 1000 voles
- M1 density = estimated population ÷ area = 1000 ÷ 3.0
- A1 about 330 voles per hectare, with the unit
Larvae of two barnacle species settle across the whole of a rocky shore, but adults of one species are found only high up. When the other species is removed, the first survives much lower down. Explain these observations in terms of niche and competition.
Mark scheme
- B1 a niche is the role a species plays in its community, meaning all the conditions it can tolerate and all the resources it uses, and it is not the same thing as a habitat
- B1 the range of conditions the first species can tolerate is wider than the range in which it is found, because its larvae settle and survive low on the shore when the other species is absent
- B1 lower on the shore the second species grows faster and undercuts or smothers the first, so interspecific competition removes it from that part of the shore without any direct attack
- B1 competition therefore trims the conditions a species could occupy down to the conditions it does occupy, and where two similar species coexist they usually differ in some dimension of the niche instead
Give the carrying capacity of the culture from the graph, and calculate the mean rate of increase in the number of cells between day 8 and day 16, giving the unit.
The size of a yeast population over 24 days, constructed for this question rather than measured. The dashed line is drawn where the curve levels off. Mark scheme
- B1 the carrying capacity is 5000 cells per cm³, read from where the curve levels off
- M1 reads about 3000 cells per cm³ at day 8 and about 5000 at day 16, accepting values to the nearest 500
- M1 mean rate = change in number ÷ change in time = (5000 − 3000) ÷ 8
- A1 about 250 cells per cm³ per day, with the unit, accepting 220 to 280 from readings inside the range above
Gause grew two species of Paramecium, single-celled organisms that feed on bacteria, in tubes of the same bacterial culture. Grown separately, each species increased and then levelled off at a steady population. Grown together in one tube, P. aurelia again reached a steady population, while P. caudatum declined until none were left. Neither species preys on or attacks the other. Explain these results.
Mark scheme
- B1 grown separately, each population grows until it is limited by its food supply, the bacteria, and then births balance deaths at the carrying capacity
- B1 grown together, the two species compete for the same food, so there is interspecific competition between species with the same niche
- B1 P. aurelia must obtain the food more effectively or reproduce faster, so it takes the larger share and leaves too little for P. caudatum, whose death rate then exceeds its birth rate
- B1 two species with the same niche cannot coexist indefinitely in one place, so the better competitor excludes the other
Compare intraspecific competition with interspecific competition, referring to which organisms are competing, why one is usually the more intense, and the effect each has on a population.
Mark scheme
- B1 intraspecific competition is between members of the same species, whereas interspecific competition is between members of different species that need some of the same resource
- B1 intraspecific competition is usually the more intense of the two, because individuals of one species need very nearly the same resources in very nearly the same way, whereas different species usually overlap less completely
- B1 intraspecific competition regulates the size of a single population, holding it around a carrying capacity, whereas interspecific competition reduces the population sizes of both species involved
- B1 intraspecific competition does not restrict the range of conditions a species can occupy, whereas severe interspecific competition can exclude a species from part of the range of conditions it could otherwise tolerate
Explain how resource partitioning allows two similar species to coexist in the same habitat, rather than one excluding the other by competitive exclusion.
Mark scheme
- B1 species that appear to share a niche usually differ in some dimension of it, such as feeding height, activity time or the size of prey or seed taken, rather than overlapping completely
- B1 that difference reduces how much the two species actually compete for the same resource, even though their niches overlap in other respects
- B1 each species can then obtain resources the other does not fully exploit, so neither drives the other to local extinction
- B1 this resource partitioning, rather than one species occupying the whole of its fundamental niche unchallenged, is the pattern expected where two similar species are found living together
Outline the assumptions that the mark, release and recapture method relies on for its population estimate to be reliable.
Mark scheme
- B1 the marked individuals released back into the population mix back in fully at random before the second sample is taken
- B1 the mark stays attached and stays visible throughout the study, without fading, being rubbed off or lost at a moult
- B1 the mark does not affect the survival of the marked individual, for example by attracting predators or being toxic
- B1 the population is closed between the two samples, with no births, deaths, immigration or emigration changing its size in the meantime
A student estimates the number of woodlice in a patch of leaf litter, but between her two samples several marked animals moult and lose their marks. Suggest how this affects her estimate, and suggest two changes to her method that would reduce the problem.
Mark scheme
- B1 fewer marked animals are found in the second sample, and that number is the denominator of the index, so a smaller denominator makes the estimated population too high
- B1 she should leave a shorter interval between the samples: long enough for the marked animals to mix back in, but not long enough for many of them to moult
- B1 she should mark only adults, which moult less often than growing juveniles, and give the estimate as one of adults; or sample in cooler weather, when moulting is less frequent
A density-dependent factor is one whose effect on a population, such as the proportion of individuals it kills, increases as the population becomes more crowded; a density-independent factor affects the same proportion whatever the density. A gardener lists six factors affecting a population of aphids on his roses: a hard frost in April, an increase in ladybirds that eat the aphids, an insecticide spray, competition for young shoots, a virus that spreads faster between crowded aphids, and a hailstorm. Identify the three factors that are density-dependent.
Mark scheme
- B1 predation by ladybirds is density-dependent, because predators find and eat prey more easily as the density of aphids rises
- B1 competition for young shoots is density-dependent, because the resource available to each aphid falls as more aphids compete for the same limited shoots
- B1 the virus is density-dependent, because a disease that spreads between crowded hosts transmits faster and further as the population becomes denser
A biologist estimates a population of crabs in a rock pool by mark and recapture. Some marked crabs learn to enter the baited trap again because it offers food, so they turn up in the second sample more often than an average crab. Suggest what this does to the population estimate, and suggest one change to the method that would reduce the problem.
Mark scheme
- B1 trap-happy crabs are over-represented among the marked crabs recaught, so the proportion of the second sample found to carry a mark is inflated
- B1 that proportion is the denominator of the Lincoln index, so an inflated denominator produces a population estimate that is too low
- B1 using unbaited traps, or a different capture method for the second sample, would stop learned behaviour making a marked crab more likely than an unmarked one to be caught
State what is meant by the carrying capacity of a habitat, and state why a population can rise above it for a time.
Mark scheme
- B1 the population size that a particular environment can support over a sustained period, given the resources available in it
- B1 the damage done by too many individuals takes time to feed back as deaths and failed breeding, so the population overshoots and is usually followed by a crash to below the carrying capacity
State what distinguishes a population from a community, and state which of the two terms includes every species present in a wood.
Mark scheme
- B1 a population is all the organisms of one species living in the same place at the same time, able to interbreed, whereas a community is every population of every species living there together
- B1 the community is the term that includes every species present, since a population by definition refers to only one of them
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