Questions on succession and environmental change

Question 1

A moorland has been monitored at the same twelve sites since 1975. Over that period the nitrogen deposited from the air has risen and the number of plant species per square metre has fallen. A report concludes that nitrogen deposition caused the loss of species. Evaluate that conclusion.

Accepted Answer

B1 for: the series is long, so a genuine trend can be separated from a run of unusual years, and the fall is measured against a baseline recorded before it began rather than against an impression B1 for: a mechanism can be stated. Added nitrogen favours a few vigorous grasses which grow tall and shade the smaller species out, and that is the same competition for light that drives succession, so the claim is more than bare correlation B1 against: the two series were measured over the same period, and almost everything else changed over that period as well, so temperature, rainfall, grazing intensity and burning are alternative explanations that have not been excluded B1 against: the question does not say how the twelve sites were chosen, so they may not be a random sample of the moor, and there is no untreated plot to compare them against B1 judgement: the monitoring cannot establish the cause on its own; what would settle it is a manipulation, with plots assigned in advance to receive added nitrogen and control plots that do not, recorded by the same method. The honest claim from the monitoring alone is that richness fell at this rate over this period, with nitrogen deposition as one candidate cause

Question 2

Describe how you would investigate the effect of light intensity on the distribution of bluebells from inside a wood out into the neighbouring field.

Accepted Answer

B1 lay a belt transect from inside the wood out into the field, across the change in light, with a quadrat at regular intervals along it B1 use quadrats of the same size throughout and record the percentage cover, or the number, of bluebells in each B1 measure light intensity at each quadrat with a light meter held at the same height, taking all readings over a short period at the same time of day and using a mean of repeat readings B1 repeat with further transects started from randomly chosen points along the edge of the wood, and calculate the mean abundance at each distance B1 state a null hypothesis and use Spearman's rank correlation between light intensity and abundance, comparing the value with the critical value at p = 0.05

Question 3

Explain why the community occupying a bare rock surface changes in a predictable direction rather than at random.

Accepted Answer

B1 each community alters the abiotic conditions of the place it occupies, and alters them in ways that suit some other species better than itself B1 lichens trap dust, hold water and secrete acids that break the rock down, and their remains add organic matter, so a thin soil forms, which is what a moss needs and a lichen does not B1 as the soil deepens it holds more water and more mineral ions, so larger plants with deeper roots can establish and add far more organic matter when they die B1 the mechanism at each handover is competition, usually for light: a taller species shades a shorter one, which then photosynthesises below its compensation point, grows less and is lost from the community, which is why the sequence runs small to large

Question 4

Explain what a plagioclimax is, and explain why a nature reserve on chalk downland is grazed rather than fenced and left alone.

Accepted Answer

B1 a plagioclimax is a community held at a stage short of its climax by a repeated interference, usually grazing, mowing, burning or ploughing B1 grazing animals eat the seedlings of hawthorn and other woody plants, so the sere is held at a grassland stage instead of running on to scrub and then woodland B1 the many small light-demanding herbs of chalk grassland, and the orchids and butterflies that depend on them, persist only while nothing is shading them B1 if the grazing stops, woody seedlings survive and grow and shade the herbs, which then photosynthesise below their compensation point and are lost, so richness falls

Question 5

Compare a pioneer community with a climax community, referring to species richness, net primary productivity and stability.

Accepted Answer

B1 a pioneer community holds very few species, tolerant of extreme abiotic conditions, whereas a climax community holds far more species in a complex, structured community B1 net primary productivity is low in a pioneer community, whereas it is higher in a climax community, though not as high as in the fastest-growing middle stages, because a mature climax community carries a large mass of non-photosynthetic tissue to support B1 biomass and soil organic content are very low in a pioneer community and high in a climax community, since each successive generation has added litter that decomposers have converted into humus B1 a climax community is more stable than a pioneer community, because its more complex food web, with many links between species, can absorb the loss of one species where a simple pioneer community cannot

Question 6

Describe how ecologists use a chronosequence to study succession, and describe the assumption on which the method depends.

Accepted Answer

B1 a set of sites that differ in age but are otherwise similar is identified, such as stations at increasing distance inland along a transect from a dune's strandline B1 each site is treated as representing a different stage of the same sere, so the sequence of stages is read from the differences between the sites rather than watched over time at any one of them B1 abiotic and biotic measurements, such as soil organic content and the number of plant species present, are recorded at each site and compared along the sequence B1 the method assumes that the sites differ only in age and not in some other factor, such as distance inland also changing exposure to salt spray, which is an assumption rather than an observation

Question 7

A landowner argues that the best thing conservation can do for a species-rich chalk grassland reserve is to fence it off and leave it alone, since deciduous woodland is the natural climax vegetation of lowland Britain. Discuss this argument.

Accepted Answer

B1 for: leaving land unmanaged requires no continued labour or cost, and allows succession to proceed towards the community that the local climate and soil would naturally support B1 for: it is true that deciduous woodland is the usual climax vegetation of lowland Britain, so leaving the site alone would eventually produce a stable, self-sustaining community B1 against: the grassland's high species richness, and the orchids and butterflies associated with it, persist only because grazing has held the site at a plagioclimax; without continued management, woody seedlings survive and shade out the light-demanding herbs, and most of those species are lost as scrub and then woodland develop B1 judgement: leaving the reserve alone is a management decision with a foreseeable consequence rather than the absence of one, so if the reserve was designated to conserve the grassland species, active management such as grazing is needed to hold the community at that stage

Question 8

The table shows the age and the percentage organic content of the soil at four sites along a chronosequence on ground exposed by a retreating glacier, constructed for this question rather than measured. Calculate the mean rate of increase in organic content between the 40-year site and the 120-year site, giving the unit.

Accepted Answer

M1 mean rate of increase = change in organic content ÷ change in age, using the 40-year value of 1.8% and the 120-year value of 6.6% M1 change in organic content = 6.6 − 1.8 = 4.8%, over a change in age of 120 − 40 = 80 years A1 4.8 ÷ 80 = 0.060 A1 about 0.060% of organic content per year, with the unit given

Question 9

Explain why the muscles of a mammal become rigid a few hours after death.

Accepted Answer

B1 after death respiration stops being aerobic because the cells receive no oxygen, and ATP runs out once anaerobic respiration ends B1 without ATP, calcium ions are not pumped back into the sarcoplasmic reticulum, so they stay in the sarcoplasm and the binding sites on actin remain exposed B1 myosin heads bind to actin, forming cross-bridges B1 ATP is needed to detach the myosin heads from actin, so the cross-bridges cannot break and the muscle is locked rigid

Question 10

A body is found in a room at 10 °C with a core temperature of 25 °C. In a model of cooling, the core stays at 37 °C for a plateau of 2 hours and then follows T = T_room + (37 − T_room)e^−kt, where t is the time in hours after the plateau and k = 0.05 per hour. Calculate the estimated time since death.

Accepted Answer

M1 e^−kt = (25 − 10) ÷ (37 − 10) = 15 ÷ 27 = 0.556 M1 −0.05t = ln 0.556 = −0.588 A1 t = 0.588 ÷ 0.05 = 11.8 hours after the plateau A1 time since death = 11.8 + 2 = 13.8 hours, about 14 hours

Question 11

In an exercise with invented values, a blowfly species needs 1800 degree hours above a base temperature of 10 °C to develop from egg to third-instar larva. Third-instar larvae are the oldest stage on a body. Calculate the minimum time since death if the mean temperature at the scene was 19 °C, and again if it was 14 °C.

Accepted Answer

M1 at 19 °C, degree hours gained each hour = 19 − 10 = 9 A1 1800 ÷ 9 = 200 hours, about 8.3 days M1 at 14 °C, degree hours gained each hour = 14 − 10 = 4 A1 1800 ÷ 4 = 450 hours, about 18.8 days

Question 12

Describe the succession of insects on a body exposed on land, and describe how it is used to estimate the time since death.

Accepted Answer

B1 blowflies arrive first, within minutes to hours, and lay eggs in the moist openings and wounds; flesh flies follow B1 predatory beetles and parasitoid wasps arrive to feed on the fly larvae B1 as the remains dry, beetles that eat dried skin and ligament, such as hide beetles, replace the flies, followed by mites and moths that eat hair B1 the groups present show which stage the body has reached, compared with reference data for the same region, season and setting, giving an estimate over weeks to months

Question 13

A body is found outdoors in summer. Its core temperature is the same as the air temperature, the limbs are no longer rigid, the abdomen is swollen with gas, and blowfly larvae in their second and third instars are present but there are no pupae. Suggest an approximate time since death, using each piece of evidence.

Accepted Answer

B1 core temperature equal to the surroundings means the body has finished cooling, so death was more than about a day ago B1 rigor has developed and passed off, which takes a day or two or more, especially in warm weather B1 bloating shows bacterial decomposition is under way but the soft tissues have not collapsed, which fits days rather than weeks B1 third-instar larvae with no pupae show the oldest insects have not finished feeding, giving a minimum of several days at summer temperatures; together the evidence suggests roughly three days to a week

Question 14

A student ran a belt transect from a dry bank down into a wet hollow. In each of eight quadrats she measured the water content of the soil and counted the soft rush plants. The data are idealised. Water content (%): 12, 18, 21, 25, 30, 34, 41, 47. Number of soft rush plants: 1, 3, 2, 6, 5, 9, 8, 12. Use r_s = 1 − [6Σd² ÷ n(n² − 1)] to calculate Spearman's rank correlation coefficient, and use the critical value of 0.738 for n = 8 at p = 0.05 to write a conclusion.

Accepted Answer

M1 both variables ranked, giving squared rank differences of 0, 1, 1, 1, 1, 1, 1 and 0, so Σd² = 6 M1 r_s = 1 − (6 × 6) ÷ (8 × 63) = 1 − 36 ÷ 504 A1 r_s = 0.93; accept 0.929 B1 0.93 is greater than 0.738, so the null hypothesis is rejected: there is a significant positive correlation between soil water content and the number of soft rush plants

Question 15

Two students estimated the number of daisies on a playing field 60 m by 40 m, each using twenty quadrats of 0.25 m². One placed the quadrats at random coordinates and counted 46 daisies in total. The other placed each quadrat where he could see daisies and counted 130 in total. The data are idealised. Calculate the estimate of the population from the random quadrats, and explain which estimate is more valid.

Accepted Answer

M1 mean of 46 ÷ 20 = 2.3 daisies per quadrat, so a density of 2.3 ÷ 0.25 = 9.2 daisies per m² A1 9.2 × (60 × 40) = 9.2 × 2400 = 22 080, about 22 000 daisies B1 the quadrats placed by eye fell where daisies were easy to see, so they over-represent dense patches and overestimate the population B1 the random estimate is more valid, because every part of the field had an equal chance of being sampled, so the sample is not biased by the observer's choice

Question 16

Explain why secondary succession on an abandoned field reaches woodland far sooner than primary succession on bare rock.

Accepted Answer

B1 the soil already exists, so no part of the sequence has to be spent building it from a bare surface, which is the slowest step of all B1 that soil already holds nutrients, water and decomposers, so conditions are tolerable to larger plants from the start B1 it also holds a seed bank of seeds, roots and rhizomes that survived the clearance, so the plants of middle seral stages appear within a year or two rather than waiting to arrive as windblown seeds or spores

Question 17

A chalk grassland held 38 plant species per square metre while it was grazed. Twenty-six years after grazing stopped it held 12 species per square metre. Calculate the percentage decrease in the number of species per square metre.

Accepted Answer

M1 percentage change = (change ÷ original value) × 100, and the original value is the grazed figure of 38 M1 the decrease is 38 − 12 = 26 species per square metre, so 26 ÷ 38 = 0.684 A1 a decrease of 68.4 per cent, or 68 per cent to two significant figures

Question 18

Outline the sequence of plant communities in primary succession on a sand dune, from bare sand near the strandline to a stage with shrubs.

Accepted Answer

B1 sea rocket colonises the bare, mobile sand nearest the strandline, tolerating salt spray and occasional burial by blown sand B1 marram grass follows, growing up through accumulating sand and binding it, with nitrogen-fixing plants such as legumes also establishing early because nitrogen is scarce on new, undeveloped ground B1 as organic matter and nutrients accumulate in the developing soil, grasses and herbs are followed by shrubs, in a sequence of increasing height and decreasing exposure to salt and burial

Question 19

Explain the difference between the facilitation model of succession and the inhibition model, and state which of the two describes the classic sere from bare rock to woodland.

Accepted Answer

B1 in the facilitation model, each community alters the abiotic conditions of the place it occupies in ways that make those conditions more suitable for the species that replace it B1 in the inhibition model, early-arriving species occupy the ground and resist replacement, so a later species establishes only once disturbance removes the earlier occupants, rather than because conditions have been made more suitable for it B1 the facilitation model describes the classic sere from bare rock to woodland, in which pioneers such as lichens begin the soil that later species need; it is one model among several, since not every replacement in nature is one species preparing the ground for the next

Question 20

Alders and legumes, both of which carry nitrogen-fixing bacteria in root nodules, are often among the first shrubs to establish on bare ground left by a retreating glacier. Suggest why.

Accepted Answer

B1 nitrogen is the scarce element on newly exposed ground, since none has yet been added by the ammonification of dead organic matter B1 a plant carrying nitrogen-fixing bacteria in root nodules can obtain nitrogen from the atmosphere rather than depending on nitrate already present in the thin, undeveloped soil B1 this gives it access to a resource other early colonists lack, so it can establish and grow where a plant without that association would be limited by the shortage of nitrogen

Question 21

Explain why the net primary productivity of a climax woodland is somewhat lower than that of the middle stages of succession that preceded it, even though the climax holds far more biomass.

Accepted Answer

B1 net primary productivity is gross photosynthesis minus the respiration of the producers themselves, not simply the rate of photosynthesis B1 a mature woodland's producers include an enormous mass of non-photosynthetic tissue, mostly the wood of trunks, branches and roots, which is living and must be kept alive by respiration even though it captures no light B1 that respiration subtracts a large amount from gross photosynthesis, so despite having the greatest biomass of any stage, the climax does not have the greatest net primary productivity, which instead comes at a stage where growth is fast and there is less non-photosynthetic tissue to support

Question 22

A nature reserve has resurveyed the same twenty permanent quadrats every five years since 1960, to track succession directly. Suggest one advantage this method has over a chronosequence, and suggest one limitation of the method itself.

Accepted Answer

B1 because the same quadrats are followed through time, the sites are guaranteed to differ only in time and not in some other factor, unlike a chronosequence, which assumes without observing it that its sites differ only in age B1 the record only continues for as long as somebody keeps returning to survey the same quadrats, so a change of staff, funding or ownership can end a series decades before its full value is realised B1 repeated visits to mark and record the same quadrats can themselves disturb the ground being studied, for example by trampling, which is a source of change the method introduces rather than observes

Question 23

A student lowers the pins of a point quadrat 50 times in a grassland. Cocksfoot grass is hit 35 times, red fescue 18 times and white clover 12 times. Calculate the percentage cover of each species, and explain why the three values add up to more than 100 per cent.

Accepted Answer

M1 percentage cover = number of hits ÷ number of pins lowered × 100 A1 cocksfoot 35 ÷ 50 × 100 = 70 per cent; red fescue 36 per cent; white clover 24 per cent B1 a pin can touch more than one species where leaves overlap in layers, so one pin can give several hits and the total, 130 per cent, exceeds 100

Question 24

Explain one advantage and two disadvantages of using the ACFOR scale to compare the abundance of seaweeds on two rocky shores.

Accepted Answer

B1 advantage: it is quick and needs no counting or cover estimate, so a large area or many quadrats can be surveyed, which suits seaweeds that are hard to count B1 disadvantage: it is subjective, so two recorders may give the same quadrat different letters and results from the two shores may not be comparable B1 disadvantage: it gives categories rather than measured numbers, so the difference between the shores cannot be analysed with a statistical test that needs measured data

Question 25

Suggest the most suitable sampling technique for each of the following, giving a reason each time: the cover of grasses in a dense, layered meadow; the number of limpets on a rock face; the abundance of lichens on the trunks of trees across a large woodland.

Accepted Answer

B1 point quadrat, because grasses grow as a mat with overlapping layers and pins give an objective measure of cover B1 individual counts in a frame or gridded quadrat, because limpets are distinct individuals that can be counted B1 the ACFOR scale, because lichens are hard to count and the area is large, so a quick subjective estimate allows many trees to be surveyed

Question 26

Two bodies with the same core temperature are found at the same time. One is a large adult found clothed indoors; the other is a small adult found lightly dressed in a cold stream. Explain why they are unlikely to have died at the same time.

Accepted Answer

B1 a body cools towards the temperature of its surroundings, and cools faster the colder they are and in water, which conducts heat away much faster than air B1 the small body has a larger surface area to volume ratio and less insulating fat, and light clothing insulates less, so it loses heat faster B1 so the body in the stream reached the same temperature much sooner, meaning it died more recently than the body indoors

Question 27

State what is meant by a pioneer species, and state what pioneers do to a bare surface.

Accepted Answer

B1 the first species to colonise a bare surface, adapted to survive there when almost nothing else can B1 they stabilise the surface and hold water, and when they die their remains add the first organic matter, so a thin soil begins to form

Question 28

Explain why forensic entomology gives a minimum time since death rather than the time of death itself.

Accepted Answer

B1 it measures the age of the oldest insects, which dates when the eggs were laid rather than when the person died B1 flies may not lay eggs for some time after death, for example at night, in heavy rain, or if the body was indoors, wrapped or buried

Age of site / years	Organic content of soil / %
5	0.3
40	1.8
120	6.6
300	9.4