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Limiting factors, glasshouses and the energy that gets through

One factor holds the rate back at a time, and the shape of a graph tells you which. That idea earns a grower money and it explains why a food chain runs out of energy after four or five links. It also explains the compensation point, which is the moment in the morning when a plant stops running at a loss.

Before this The light-dependent stage · The Calvin cycle · Respiration and its rate

Before you start

Plants photosynthesise during the day and respire at night. It is a tidy picture and it is false. Every living plant cell respires continuously, day and night, because it needs ATP continuously — root cells never see light at all and respire around the clock. What changes with the light is photosynthesis, and in daylight it usually runs faster than respiration, so the plant takes in more carbon dioxide than it gives out. The apparent switch is a net figure, not a switch.

What you should be able to do

One factor at a time

A process made of many steps runs at the pace its slowest step allows. Photosynthesis needs light energy, carbon dioxide as raw material and a workable temperature for its enzymes, and at any moment one of the three is holding the rate back. That one is the limiting factor: the factor in shortest supply, and the only one whose increase would speed things up.

Two different things happen to these curves. Raising carbon dioxide or temperature moves the height at which each levels off; it does nothing at all to the steep part on the left, where all three curves lie on top of one another.

That overlap on the left is the whole argument. Where the curves coincide the rate depends only on how much light there is, so light is limiting and the extra carbon dioxide is doing nothing — there is not enough energy to use what is already there. Where the curves flatten and separate, more light no longer helps, and whatever is different between the curves is what is now limiting.

Read a plateau carefully. A plateau does not mean the plant has hit some absolute maximum; it means the factor on the x-axis has stopped being the one that matters. Raise something else and the plateau moves up. If raising carbon dioxide lifts the plateau, carbon dioxide was limiting there; if raising the temperature lifts it further, temperature was limiting after that.

Temperature behaves differently from the other two, and it is worth saying why. Light intensity and carbon dioxide concentration feed the reaction; temperature affects the enzymes catalysing it. So a temperature graph rises to an optimum and then falls, because beyond roughly 40 °C the enzymes of the Calvin cycle — rubisco among them — begin to denature. Light and carbon dioxide curves flatten out; they do not come back down.

Limiting factor
The factor, of all those a process requires, whose shortage is holding the rate back at that moment.
Plateau
The region of a rate graph where increasing the factor on the horizontal axis no longer increases the rate, because something else has become limiting.

The compensation point

A plant is doing two opposite things at once. Photosynthesis takes carbon dioxide in and gives oxygen out; respiration does the reverse, and never stops. What you measure from outside is the difference between them.

The gap between the dashed line and the solid one is the respiration rate, and it is the same gap at every light intensity. The plant only gains once the solid line is above zero.

At dawn, light intensity is low and photosynthesis is slower than respiration, so the plant releases carbon dioxide overall: it is running at a loss, living off stores. As the light strengthens, photosynthesis speeds up until it exactly matches respiration. At that intensity there is no net gas exchange at all — that is the compensation point — and above it the plant begins to accumulate organic matter.

Compensation point
The light intensity at which the rate of photosynthesis exactly equals the rate of respiration, so there is no net exchange of carbon dioxide or oxygen.
Compensation period
The time taken from first light for a plant to reach its compensation point.

Species differ in where their compensation point sits, and it explains a good deal about where they grow. A shade plant on a woodland floor reaches compensation at a low intensity, so it can survive under a closed canopy; a sun plant of open ground needs far more light before it breaks even and would starve in the same spot. It also explains why seedlings under dense weeds die: they never spend enough of the day above their compensation point to build anything.

Glasshouses, with actual numbers

Growers apply all of this commercially, and the arithmetic decides whether it is worth doing. Three things get adjusted.

Carbon dioxide. Outdoor air is about 0.04 per cent carbon dioxide, which is well below what rubisco could use. Enriching a glasshouse to around 0.10 per cent typically lifts yield by a fifth to a third in tomatoes and cucumbers. The gas often comes from burning natural gas or propane, which conveniently supplies heat at the same time. Above roughly 0.15 per cent the benefit tails off and the plants may be damaged, so more is not better indefinitely.

Temperature. Most glasshouse crops are run somewhere between 20 and 25 °C. Warmer speeds the enzymes of the Calvin cycle, but it also speeds respiration, which spends the sugars the crop is supposed to accumulate, and past about 30 °C the stomata begin to close and rubisco takes up more oxygen. There is an optimum and it is not simply 'as hot as possible'.

Light. Supplementary lamps extend the effective day through a northern winter, which is why glasshouse tomatoes are available in February. Lamps are the expensive option, and a grower will only run them while the extra crop is worth more than the electricity.

Is the enrichment worth it?

A grower's 400 m² glasshouse yields 24 kg of tomatoes per square metre per year. Enriching the air to 0.10 per cent carbon dioxide raises the yield by 25 per cent, and the enrichment costs £3200 a year to run. Tomatoes sell for £1.60 per kilogram. Should the grower do it?

Work out the extra crop first. The increase is 25 per cent of 24, so 6 kg per square metre per year, and over 400 m² that is 6 × 400 = 2400 kg of extra tomatoes.

Now the money: 2400 × £1.60 = £3840 of extra income against £3200 of cost, leaving £640. On these figures the grower should do it, but the margin is 20 per cent of the cost, which is thin.

The biology behind the answer is worth stating in an exam. Enrichment only pays while carbon dioxide is genuinely the limiting factor. Run it on a dull winter afternoon when light is limiting instead and the yield does not rise at all, and the £3200 is spent for nothing — which is exactly why commercial systems dose carbon dioxide on a light sensor rather than continuously.

Productivity, and what reaches the next mouth

Step back from the single leaf to a whole field, and the question becomes how much chemical energy the plants there capture in a year. Gross primary productivity is the total: everything fixed by photosynthesis. But the plants respire a substantial share of it themselves — commonly 20 to 50 per cent — and what is left is what actually accumulates as new plant material.

NPP = GPP − RNet primary productivity is gross primary productivity minus the energy lost in respiration by the producers. NPP is what is available to the organisms that eat them.

Two separate inefficiencies are shown here. Less than three per cent of the arriving sunlight is captured at all, and only about a tenth of what does get captured survives each step up the chain.

Units matter, and they are marked. Productivity is a rate per unit area, so it is quoted as energy per square metre per year: kJ m⁻² year⁻¹. Leave off the area or the time and the figure means nothing — a number of kilojoules alone cannot be compared between a field and a forest.

Most of the sunlight falling on a field is never captured. Some misses the leaves entirely and lands on soil; some is reflected off the waxy cuticle; some is transmitted straight through; much of it is at wavelengths the pigments do not absorb, the green band among them. What is absorbed usefully typically fixes one to three per cent of the incoming energy.

percentage transfer = (energy in the level ÷ energy in the level below) × 100Use the same units on top and bottom, and be clear which two levels you are comparing before you divide.

Between trophic levels the losses have four familiar causes. Not all of the plant is eaten — roots, wood and litter are left behind. Not all of what is eaten is digested; the indigestible part leaves as faeces. Some of what is absorbed is excreted, as urea in mammals. And a large share is respired to power movement, and in birds and mammals to maintain body temperature. Only what is left builds new tissue and is available to the next level.

That last cause explains a pattern worth quoting. Transfer to a mammal or a bird is often only 5 to 10 per cent, because keeping a constant body temperature is expensive; transfer to an insect or a fish can be considerably higher. It is also the argument behind eating lower down the chain: feeding grain to people rather than to cattle avoids one ninety-per-cent loss.

TRY IT — Working with productivity figures

In a grassland, gross primary productivity is 24 000 kJ m⁻² year⁻¹ and the plants lose 8 400 kJ m⁻² year⁻¹ in respiration. The primary consumers accumulate 1 248 kJ m⁻² year⁻¹. Calculate the net primary productivity and the percentage of it transferred to the primary consumers, and suggest two reasons why the percentage is not higher.

Check your answer

Net primary productivity is gross minus respiration: 24 000 − 8 400 = 15 600 kJ m⁻² year⁻¹. Keep the units on the answer; they are part of it.

The transfer is 1 248 ÷ 15 600 × 100 = 8.0 per cent. Divide by the NPP, not by the GPP: the energy the plants have already respired away was never available to a herbivore.

Two reasons from the standard list, each phrased as a loss: much of the plant material is not eaten at all, including roots and dead litter, and much of what is eaten is not digested and passes out as faeces. Respiration by the consumers themselves is a third, and if they are mammals, maintaining body temperature accounts for a large part of it.

The commonest error here is dividing by 24 000, which gives 5.2 per cent and loses the marks. Decide which productivity the question means before you press the button.

In the exam

Check yourself

A grower measures the rate of photosynthesis of a tomato crop through a clear spring day. The rate rises through the morning, levels off from about 11 am to 3 pm despite the light continuing to brighten until 1 pm, and then falls through the afternoon. Explain the shape of this curve, and suggest one change the grower could make to raise the midday part of it.

Answer

Through the morning the rate rises because light intensity is increasing and light is the limiting factor: more light means more photoionisation, more ATP and reduced NADP, and a faster Calvin cycle.

From about 11 am the rate levels off even though the light is still brightening, so light has stopped being limiting. Something else has taken over — most likely the carbon dioxide concentration around the leaves, which the crop itself has been depleting all morning, or the temperature if it is high enough to be closing stomata.

The afternoon fall follows the light back down: once the intensity drops far enough, light becomes limiting again and the rate falls with it.

To raise the middle of the curve the grower should increase the factor that is limiting there, which means carbon dioxide — enriching the air to about 0.10 per cent. Adding more light during the plateau would change nothing at all, which is the whole point of identifying the limiting factor before spending money on it.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

Question 14 marks

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.

Mark scheme
  1. B1 on the steep part at low light intensity the rate rises as light intensity rises, so light is the limiting factor there
  2. 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
  3. 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
  4. 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 at the plateau; a plateau marks the point where the factor on the horizontal axis stopped mattering, not an absolute maximum

Question 24 marks

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.

Mark scheme
  1. M1 net primary productivity = gross primary productivity minus the energy lost in respiration by the producers
  2. A1 18 000 − 6 000 = 12 000 kJ m⁻² year⁻¹, with both parts of the unit kept
  3. M1 percentage transfer = energy in the consumers ÷ net primary productivity × 100 = 1 080 ÷ 12 000 × 100
  4. 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 34 marks

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

Mark scheme
  1. B1 not all of the material in the level below is eaten: roots, wood and dead litter are left behind
  2. B1 not all of what is eaten is digested, and the indigestible part leaves the body as faeces
  3. B1 some of what is absorbed is lost in excretion, as urea in mammals
  4. 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 44 marks

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.

Mark scheme
  1. B1 enrichment raises the rate only while carbon dioxide is genuinely the limiting factor
  2. 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
  3. 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
  4. 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 53 marks

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.

Mark scheme
  1. 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
  2. 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
  3. 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 62 marks

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

Mark scheme
  1. 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
  2. B1 productivity is measured in kJ m⁻² year⁻¹, that is energy per unit area per unit time

Worth remembering

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