Questions on limiting factors and primary productivity

Question 1

Two groups of pigs are given the same food. Per pig per day, each ingests 30 000 kJ and loses 9 000 kJ in faeces and urine. Respiration uses 18 600 kJ in pigs kept outdoors and 16 800 kJ in pigs kept in a heated shed with restricted movement. Calculate the net production of each group as a percentage of the energy ingested, and explain the difference. The figures are constructed for this question.

Accepted Answer

M1 outdoors: N = I − (F + R) = 30 000 − (9 000 + 18 600) A1 2 400 kJ, which is 8.0 per cent of the energy ingested M1 indoors: N = 30 000 − (9 000 + 16 800) A1 4 200 kJ, which is 14 per cent of the energy ingested B1 in the heated shed the pigs lose less heat to their surroundings, so less respiration is needed to keep their body temperature up B1 with movement restricted, less energy is used in muscle contraction, so with F unchanged the energy saved from respiration goes into growth

Question 2

Describe how you would investigate the effect of carbon dioxide concentration on the rate of photosynthesis of pondweed.

Accepted Answer

B1 pondweed such as Elodea, the same length or mass each time, placed cut end uppermost in a boiling tube of sodium hydrogencarbonate solution B1 a range of at least five concentrations of sodium hydrogencarbonate as the source of carbon dioxide B1 the lamp kept at a fixed distance in a darkened room, with an LED or a heat shield so it does not warm the water B1 the tube kept in a water bath at a fixed temperature, checked with a thermometer B1 the plant left for a few minutes at each new concentration before measuring, so the rate settles B1 gas collected for a fixed time in a photosynthometer and its volume found from the bubble length, repeated at each concentration and a mean rate calculated

Question 3

A population of caterpillars in a field ingests 8 400 kJ m⁻² year⁻¹ of leaf material. Of this, 5 200 kJ m⁻² year⁻¹ is lost in faeces, and a further 1 900 kJ m⁻² year⁻¹ is lost in respiration. Calculate the net production of the caterpillars, and calculate what percentage of the energy they ingested it represents.

Accepted Answer

M1 N = I − (F + R), so add the two losses together first: 5 200 + 1 900 = 7 100 kJ m⁻² year⁻¹ M1 N = 8 400 − 7 100 A1 N = 1 300 kJ m⁻² year⁻¹ M1 percentage = net production ÷ energy ingested × 100 = 1 300 ÷ 8 400 × 100 A1 15 per cent, to two significant figures

Question 4

The table gives constructed values for the plankton of an area of open water. Calculate the percentage of energy transferred from the phytoplankton to the zooplankton, and explain why a pyramid of biomass for this water is inverted while its pyramid of energy is not.

Accepted Answer

M1 960 ÷ 8 000 × 100 A1 12 per cent of the energy is transferred B1 the pyramid of biomass is measured at one moment, so it shows only the standing biomass at that time B1 phytoplankton reproduce very quickly and are eaten about as fast as they grow, so a small standing biomass produces many times its own mass over a year B1 the pyramid of energy measures energy flow over a year; each level receives its energy only from the level below and loses most of it, so it can never be inverted

Question 5

A student used a photosynthometer with a capillary tube of internal diameter 0.80 mm to collect the gas given off by pondweed for 5 minutes at each lamp distance. The table gives constructed results. Calculate the rate of gas production at 20 cm in mm³ min⁻¹, and use the data to evaluate the student's claim that the rate is proportional to light intensity.

Accepted Answer

M1 radius 0.40 mm and volume = πr²l = π × 0.40² × 15.4 A1 volume of the bubble = 7.7 mm³ A1 7.7 ÷ 5 = 1.5 mm³ min⁻¹ B1 moving the lamp from 20 cm to 10 cm makes the light four times as intense, by the inverse square law, but the bubble length rises only from 15.4 mm to 28.6 mm, less than double, so the rate is not proportional to light intensity across the range B1 at the higher intensities another factor, such as carbon dioxide concentration or temperature, is becoming limiting; accept that the gas collected is not pure oxygen, or that there are no repeats

Question 6

A graph shows the rate of photosynthesis against light intensity, with three curves obtained at three different carbon dioxide concentrations. The curves lie on top of one another at low light intensity and level off at different heights. Explain how to identify the limiting factor in each region of this graph.

Accepted Answer

B1 on the steep part at low light intensity the rate rises as light intensity rises, so light is the limiting factor there B1 the three curves coincide in that region, which shows the extra carbon dioxide is doing nothing, because there is not enough light energy to use what is already present B1 where a curve levels off, more light no longer raises the rate, so light has stopped being limiting and something else has taken over B1 the curves plateau at different heights, and since carbon dioxide concentration is what differs between them, carbon dioxide is the factor limiting the rate on the plateaux of the two lower curves, while the graph cannot show what limits the highest one (accept another factor such as temperature); a plateau marks the point where the factor on the horizontal axis stopped mattering, not an absolute maximum

Question 7

In a grassland the gross primary productivity is 18 000 kJ m⁻² year⁻¹ and the producers lose 6 000 kJ m⁻² year⁻¹ in respiration. The primary consumers accumulate 1 080 kJ m⁻² year⁻¹. Calculate the net primary productivity and the percentage of it transferred to the primary consumers.

Accepted Answer

M1 net primary productivity = gross primary productivity minus the energy lost in respiration by the producers A1 18 000 − 6 000 = 12 000 kJ m⁻² year⁻¹, with both parts of the unit kept M1 percentage transfer = energy in the consumers ÷ net primary productivity × 100 = 1 080 ÷ 12 000 × 100 A1 9.0 per cent; dividing by the gross figure would be wrong, because the energy the producers had already respired away was never available to a herbivore

Question 8

Explain why only about ten per cent of the energy in one trophic level reaches the level above it.

Accepted Answer

B1 not all of the material in the level below is eaten: roots, wood and dead litter are left behind B1 not all of what is eaten is digested, and the indigestible part leaves the body as faeces B1 some of what is absorbed is lost in excretion, as urea in mammals B1 a large share of the rest is respired to power movement and, in birds and mammals, to maintain a constant body temperature, so only what is left builds new tissue; transfer to a bird or mammal is often only 5 to 10 per cent, while transfer to an insect or a fish can be higher

Question 9

A grower runs carbon dioxide enrichment continuously in a glasshouse through a dull week in December and finds no increase in yield at all. Suggest why the enrichment failed to work, and suggest how the system should be run instead.

Accepted Answer

B1 enrichment raises the rate only while carbon dioxide is genuinely the limiting factor B1 through a dull December week the light intensity is low, so light is limiting and the extra carbon dioxide cannot be used: there is not enough energy to drive the light-dependent stage and supply the Calvin cycle B1 the rate is set by whichever factor is in shortest supply, so raising any other factor changes nothing and the cost of the gas is spent for no return B1 the enrichment should be dosed on a light sensor, so gas is released only when the light is bright enough for carbon dioxide to become limiting, and supplementary lamps are the change that would help on the dull days

Question 10

Explain why a graph of the rate of photosynthesis against temperature rises to a peak and then falls, whereas a graph against light intensity or carbon dioxide concentration only rises and then levels off.

Accepted Answer

B1 light intensity and carbon dioxide concentration are the energy and the raw material the reaction itself needs, so supplying more of either lets the existing enzymes work faster up to the point where something else becomes limiting, at which point the rate levels off rather than falling B1 temperature instead acts on the enzymes of the Calvin cycle, rubisco among them, rather than being a raw material B1 below the optimum, raising temperature increases the rate as usual, since molecules have more kinetic energy and collide more often and more successfully A1 above the optimum the enzymes, rubisco among them, begin to denature, so the rate falls even though light and carbon dioxide may both still be plentiful; accept also that respiration, or the share of rubisco's reactions taking oxygen, rises with temperature, or that stomata close

Question 11

Describe how the energy content of a sample of plant biomass is measured, from collecting the sample to obtaining a value in kilojoules.

Accepted Answer

B1 the fresh sample is dried in an oven at around 80 °C and reweighed repeatedly until its mass stops falling, because fresh mass is mostly water and would otherwise vary with recent weather B1 that constant mass is recorded as the dry mass of the sample B1 a known dry mass is then burned in oxygen inside a calorimeter, a sealed vessel surrounded by a known volume of water A1 the temperature rise of that water is used to calculate the energy released, and dividing by the mass burned gives the energy per gram, which is then scaled up to the kJ m⁻² year⁻¹ productivity figures quote

Question 12

On a still, overcast day a plant's net carbon dioxide uptake becomes measurably positive only two hours after sunrise, whereas on a bright, clear day it becomes positive within twenty minutes of sunrise. Suggest why, using the concept of the compensation point.

Accepted Answer

B1 respiration continues at a similar rate throughout, day or night, using up organic matter and releasing carbon dioxide regardless of how bright it is outside B1 the plant only shows a net uptake of carbon dioxide once its rate of photosynthesis exceeds its rate of respiration, which is the compensation point B1 on the overcast day light intensity rises far more slowly after sunrise than on the clear day, so it takes longer for photosynthesis to catch up with and then overtake the constant rate of respiration A1 on the clear day light intensity rises quickly, so the compensation point is reached, and passed, much sooner after sunrise

Question 13

Compare gross primary productivity with net primary productivity, and compare the situation of a fast-growing crop plant with that of a large forest tree in terms of the proportion of GPP that is available as NPP.

Accepted Answer

B1 gross primary productivity is the total energy fixed by photosynthesis, whereas net primary productivity is what remains once the plants' own respiration has been subtracted, so NPP is always less than GPP B1 NPP, not GPP, is what is available to be eaten by consumers or to accumulate as new plant material B1 a large forest tree has a substantial mass of non-photosynthetic tissue, wood, bark and roots, to respire and maintain, so a large share of its GPP, commonly nearer 50 per cent, is lost in its own respiration A1 a fast-growing crop plant carries much less non-photosynthetic tissue relative to its leaf area, so a smaller share of its GPP is spent on respiration and a correspondingly larger share is left as NPP

Question 14

Compare a pyramid of biomass with a pyramid of energy as ways of representing the same ecosystem.

Accepted Answer

B1 a pyramid of biomass shows dry mass per unit area at one time, in g m⁻², whereas a pyramid of energy shows energy per unit area per year, in kJ m⁻² year⁻¹ B1 both allow for the size of the organisms, and both need organisms to be killed, by drying, and for energy also by burning in a calorimeter B1 a pyramid of biomass can be inverted because it ignores how quickly each level is replaced, whereas a pyramid of energy is never inverted B1 a pyramid of energy is harder to obtain, needing measurements through a whole year, whereas biomass needs one sample; but equal masses can hold different amounts of energy, so biomass is a less reliable guide to energy transfer

Question 15

Explain how spraying a wheat crop with a herbicide and an insecticide increases the efficiency of energy transfer from sunlight to humans.

Accepted Answer

B1 the herbicide kills weeds, which compete with the wheat for light, water and mineral ions B1 so more of the light falling on the field is absorbed by the wheat, and more of the field's primary productivity is in the crop B1 the insecticide kills insects that eat the crop, so less of the crop's energy passes into food chains that do not end in humans B1 the food web is simplified, so a larger share of the net primary productivity ends up in the harvested grain that humans eat

Question 16

Equal numbers of algal beads were put into two tubes of hydrogencarbonate indicator, which was orange-red at the start. Tube A was lit and tube B was wrapped in foil. After an hour tube A was purple and tube B was yellow. Explain these results and the purpose of tube B.

Accepted Answer

B1 in tube A the rate of photosynthesis was greater than the rate of respiration, so there was a net uptake of carbon dioxide from the solution B1 less carbon dioxide raises the pH, which turns the indicator purple B1 in tube B there was no light, so the algae only respired; the carbon dioxide released lowered the pH and turned the indicator yellow B1 tube B is a control showing that the change in tube A was caused by light, through photosynthesis, rather than by the beads or the indicator alone

Question 17

Explain what is happening to a plant at its compensation point, and explain why a shade plant on a woodland floor can survive there while a sun plant of open ground would not.

Accepted Answer

B1 every living plant cell respires continuously, day and night, while photosynthesis happens only in the light, so what is measured from outside is the difference between the two rates B1 at the compensation point the rate of photosynthesis exactly equals the rate of respiration, so there is no net exchange of carbon dioxide or of oxygen and no organic matter accumulates B1 a shade plant reaches its compensation point at a low light intensity, so under a closed canopy it still spends much of the day above it and gains, whereas a sun plant needs far more light to break even and would spend the day running at a loss on its stores

Question 18

Explain why a single group of plants can only be used once to obtain a productivity figure by the biomass and calorimetry method, rather than being measured repeatedly over a season.

Accepted Answer

B1 obtaining a dry mass requires drying the sample to constant mass, which kills and dehydrates the plant material completely B1 obtaining its energy content then requires burning that dry mass in a calorimeter, which destroys the sample entirely A1 neither step leaves anything of the original plants to measure again, so a productivity figure is measured once from a given batch of organisms and a fresh batch must be collected for every further measurement

Question 19

A student calculates a consumer's net production as N = I − F + R and obtains a value larger than the energy the consumer ingested. Explain what has gone wrong, and explain what the correct form of the equation should give instead.

Accepted Answer

B1 faeces and respiration are both energy the consumer has lost from what it ingested, so both must be subtracted from the intake, not just one of them B1 writing N = I − F + R instead adds the respiratory loss back on rather than subtracting it, which is why the student's answer came out larger than the energy ingested in the first place A1 the correct form keeps both losses inside a bracket subtracted together, N = I − (F + R), so that net production can never exceed the energy ingested

Question 20

In a lake, the primary consumers have a net productivity of 640 kJ m⁻² year⁻¹, and the secondary consumers that eat them accumulate 76.8 kJ m⁻² year⁻¹. Calculate the percentage of energy transferred from the primary consumers to the secondary consumers.

Accepted Answer

M1 percentage transfer = energy in the level above ÷ energy in the level below × 100 M1 76.8 ÷ 640 × 100 A1 12 per cent

Question 21

Explain why a pyramid of numbers for a food chain based on a single oak tree can be inverted, and give one advantage of a pyramid of numbers.

Accepted Answer

B1 a pyramid of numbers counts individuals and takes no account of their size or mass B1 one oak tree is very large and supports thousands of small herbivores such as caterpillars, so the producer bar is narrower than the primary consumer bar above it B1 advantage: it is quick and easy to obtain, needing no special equipment, and the organisms are not killed

Question 22

A lamp is moved from 10 cm to 25 cm away from a piece of pondweed. Calculate the light intensity reaching the pondweed at 25 cm as a percentage of the intensity at 10 cm.

Accepted Answer

M1 light intensity is proportional to 1/d², the inverse square law M1 (1/25²) ÷ (1/10²) = 100 ÷ 625 A1 0.16, so 16 per cent

Question 23

A student investigates the effect of the colour of light on the rate of photosynthesis of pondweed using coloured filters. Explain why the student must measure the light intensity reaching the plant through each filter, and predict which colours give the highest and lowest rates.

Accepted Answer

B1 filters of different colours transmit different amounts of light, so intensity would change with colour and the experiment would compare brightness; intensity is measured with a light meter and matched, for example by moving the lamp B1 blue and red light give the highest rates, because chlorophyll and the accessory pigments absorb them strongly B1 green light gives the lowest rate, because little of it is absorbed; most is reflected or transmitted

Question 24

State what is meant by a limiting factor, and state the units in which primary productivity is measured.

Accepted Answer

B1 a limiting factor is the factor, of all those a process requires, whose shortage is holding the rate back at that moment, and the only one whose increase would speed the process up B1 productivity is measured as energy per unit area per unit time, such as kJ m⁻² year⁻¹ or kJ ha⁻¹ year⁻¹

Question 25

State what is meant by gross primary productivity, and state the equation linking it to net primary productivity.

Accepted Answer

B1 gross primary productivity is the total chemical energy fixed by photosynthesis in a given area in a given time, before any of it is used B1 net primary productivity = gross primary productivity − respiration by the producers, that is NPP = GPP − R

Trophic level	Dry biomass at one time / g m⁻²	Energy / kJ m⁻² year⁻¹
Phytoplankton	5	8 000
Zooplankton	15	960
Small fish	3	110