A2 12 Energy and respiration | A Level Biology 9700

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🎯What you need to be able to do

Outline the need for energy in living organisms and describe the features that make ATP the universal energy currency.
State that ATP is made by substrate-linked phosphorylation and by chemiosmosis in mitochondria and chloroplasts.
Explain the relative energy values of carbohydrates, lipids and proteins as respiratory substrates.
Define the respiratory quotient and calculate RQ values from equations for respiration.
Describe and carry out respirometer investigations to determine RQ and the effect of temperature on rate.
State where each stage of aerobic respiration occurs and outline glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation.
Relate mitochondrial structure to function, outline anaerobic respiration in mammals and yeast, explain the difference in energy yield, and explain how rice is adapted to submerged roots.
Describe investigations using DCPIP and methylene blue to follow the rate of respiration in yeast.

📚The biology

Why ATP

Energy is needed for active transport, movement (muscle contraction, cilia, cytoplasmic streaming) and anabolic reactions such as DNA replication and protein synthesis. ATP is the universal currency because:

it releases a small, usable quantity of energy in one step, so little is wasted — unlike glucose, which would release far too much at once;
the reaction is a single-step hydrolysis of the terminal phosphate, so it is fast;
it is soluble and moves easily within the cell;
it is readily regenerated from ADP and phosphate;
it cannot leave the cell, so it is not lost.

ATP is not a long-term store — a cell holds only seconds' worth and recycles it constantly. It is made in two ways: by transfer of phosphate in substrate-linked reactions, and by chemiosmosis in the membranes of mitochondria and chloroplasts.

Respiratory substrates and RQ

Lipids yield the most energy per gram, because they contain the most hydrogen per unit mass — and it is hydrogen, carried by reduced NAD and FAD to the electron transport chain, that ultimately generates ATP. Carbohydrates yield less, proteins least of the three in practice (and are used only when carbohydrate and lipid are exhausted).

The respiratory quotient is

\[ \mathrm{RQ} = \frac{\text{molecules of CO}_2\ \text{produced}}{\text{molecules of O}_2\ \text{used}} \]

For glucose, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, so RQ = 6/6 = 1.0. Lipids are less oxidised to begin with, so they need more oxygen per carbon: RQ is about 0.7. Proteins give about 0.9. An RQ above 1.0 means carbon dioxide is being produced without oxygen being used — anaerobic respiration is occurring alongside aerobic.

The four stages

Glycolysis — in the cytoplasm.
Link reaction — in the mitochondrial matrix.
Krebs cycle — in the mitochondrial matrix.
Oxidative phosphorylation — on the inner membrane of the mitochondrion.

Glycolysis. Glucose is phosphorylated using 2 ATP — an investment — to form fructose 1,6-bisphosphate (6C). This splits into two triose phosphate molecules (3C), which are oxidised to pyruvate (3C), producing 4 ATP (so a net gain of 2) and 2 reduced NAD. No oxygen is used at this stage.

Link reaction. When oxygen is available, pyruvate enters the mitochondrion. It is decarboxylated (losing CO₂) and dehydrogenated (reducing NAD), leaving a 2C acetyl group which coenzyme A carries into the Krebs cycle as acetyl coenzyme A.

Krebs cycle. Oxaloacetate (4C) accepts the 2C fragment from acetyl coenzyme A to form citrate (6C). Through a series of small steps citrate is converted back to oxaloacetate, and in doing so undergoes decarboxylations (releasing CO₂) and dehydrogenations that reduce NAD and FAD. One ATP is made per turn by substrate-linked phosphorylation. The cycle turns twice per glucose, because glycolysis produced two pyruvate.

Oxidative phosphorylation. This is where almost all the ATP is made:

Reduced NAD and FAD deliver hydrogen atoms to carriers in the inner membrane; the hydrogen atoms split into protons and energetic electrons.
Electrons pass along the electron transport chain, releasing energy at each step.
That energy is used to transfer protons across the inner membrane, from matrix to intermembrane space, building a proton gradient.
Protons return to the matrix by facilitated diffusion through ATP synthase, and the energy released drives ATP synthesis. This is chemiosmosis.
Oxygen is the final electron acceptor, combining with electrons and protons to form water.

You are not required to know the individual carriers or the structural detail of ATP synthase.

Why removing oxygen stops everything, not just the last step. Without oxygen there is nothing to accept electrons at the end of the chain, so electrons back up, the carriers stay reduced, and reduced NAD cannot be reoxidised. With no supply of oxidised NAD, the link reaction and Krebs cycle halt too — even though neither uses oxygen directly. That chain of reasoning is worth three or four marks and is the most commonly asked application in this topic.

Mitochondrial structure

A double membrane. The inner membrane is folded into cristae, giving a large surface area for the electron transport chain and ATP synthase. The intermembrane space is narrow, so a proton gradient is established quickly with relatively few protons. The matrix contains the enzymes of the link reaction and Krebs cycle, plus circular DNA and 70S ribosomes. Cells with a high energy demand — muscle, the proximal convoluted tubule — have many mitochondria with densely packed cristae, and that is the observation exam questions ask you to explain.

Anaerobic respiration

Without oxygen, only glycolysis continues, giving a net 2 ATP per glucose. The problem is regenerating oxidised NAD so glycolysis can keep going, and the two solutions differ:

Mammals — lactate fermentation. Pyruvate accepts hydrogen from reduced NAD, forming lactate. Reversible: lactate is later oxidised back to pyruvate when oxygen is available.
Yeast — ethanol fermentation. Pyruvate is decarboxylated to ethanal, which accepts hydrogen from reduced NAD to form ethanol. Carbon dioxide is released. Irreversible.

Aerobic respiration yields far more ATP because the hydrogen carried by reduced NAD and FAD from the link reaction and Krebs cycle can be passed to the electron transport chain, and the glucose is completely oxidised to carbon dioxide and water. In anaerobic conditions the substrate is only partly broken down and much of the energy remains locked in lactate or ethanol. A detailed ATP tally is not required.

Rice

Rice grows with its roots submerged, where oxygen is scarce. Three adaptations are named:

Aerenchyma — air spaces in the roots and stems through which oxygen diffuses down from the shoot to the submerged tissue.
Ethanol fermentation in the roots, with a tolerance of the ethanol produced, allowing some ATP generation when oxygen runs out.
Faster growth of stems in response to rising water, so leaves stay above the surface and gas exchange with the air continues.

Measuring respiration

A respirometer holds living material in a sealed tube with soda lime or potassium hydroxide to absorb carbon dioxide, connected to a capillary tube containing a manometer fluid. Oxygen consumed reduces the volume of gas, so the fluid moves; the rate of movement gives oxygen uptake. A control tube containing glass beads of the same volume, and a water bath to hold temperature constant, are both essential.

To find RQ you take two readings on the same material: one with the carbon dioxide absorbent, which gives oxygen used, and one without, which gives the net change — oxygen used minus carbon dioxide produced.

Redox indicators offer a different route. DCPIP and methylene blue are blue when oxidised and colourless when reduced. Added to a yeast suspension, they accept hydrogen from the dehydrogenation reactions of respiration, so the time taken to decolourise is a measure of the rate of respiration — the faster the decolourisation, the faster the rate. Use this to compare temperatures or substrate concentrations, with a colorimeter if a numerical result is wanted.

✏️Worked example

Germinating seeds were placed in a respirometer at 20 °C. With soda lime present, the manometer fluid moved 48 mm towards the respirometer in 10 minutes. With the soda lime removed and the experiment repeated on fresh seeds of the same mass, the fluid moved 14 mm towards the respirometer in 10 minutes. The capillary tube has a cross-sectional area of 1.5 mm². (a) Calculate the volume of oxygen used and the volume of carbon dioxide produced, per minute. (b) Calculate the RQ and identify the substrate. (c) Explain the purpose of the control tube and why the water bath matters.

(a) With soda lime: all carbon dioxide produced is absorbed, so the volume change is due to oxygen used alone.

\[ V = 48 \times 1.5 = 72\ \text{mm}^{3}\ \text{in 10 min} = 7.2\ \text{mm}^{3}\ \text{min}^{-1} \]

Without soda lime: carbon dioxide produced partly replaces the oxygen used, so the movement measures the difference, oxygen used minus carbon dioxide produced:

\[ 14 \times 1.5 = 21\ \text{mm}^{3}\ \text{in 10 min} = 2.1\ \text{mm}^{3}\ \text{min}^{-1} \]

So carbon dioxide produced = 7.2 − 2.1 = 5.1 mm³ min⁻¹.

(b)

\[ \mathrm{RQ} = \frac{\mathrm{CO_2}\ \text{produced}}{\mathrm{O_2}\ \text{used}} = \frac{5.1}{7.2} = 0.71 \]

An RQ of about 0.7 indicates that lipid is the main respiratory substrate. That is entirely expected in germinating seeds, many of which store food as lipid because it yields more energy per gram and is stored without water.

(c) The control tube contains glass beads of the same volume as the seeds, with the same soda lime, and is connected to the other arm of the manometer. Gases expand and contract with changes in atmospheric pressure and temperature, and such a change would move the fluid whether or not the seeds were respiring. The control experiences the same physical changes but no respiration, so the manometer records only the difference — the volume change caused by the seeds.

The water bath keeps temperature constant, for two reasons: the gas in the tube would expand if the apparatus warmed, giving a false reading, and respiration is enzyme-controlled, so its rate depends strongly on temperature. Without control, temperature would be an uncontrolled variable affecting the very thing being measured.

Check it. RQ must fall in a sensible range. For an organism respiring aerobically it lies between about 0.7 (lipid) and 1.0 (carbohydrate); 0.71 sits at the lipid end and is plausible. An RQ above 1.0 would mean carbon dioxide produced without oxygen used, indicating some anaerobic respiration alongside aerobic — possible, but it should make you re-read the data. An RQ above about 1.5, or a negative value, means an arithmetic slip: most likely you subtracted the wrong way round in part (a).

Treating the second reading as the carbon dioxide produced. Without soda lime the manometer shows the net volume change, not the carbon dioxide output. Using 2.1 directly gives RQ = 2.1/7.2 = 0.29, which is not a possible respiratory quotient for any substrate and should be caught by the sanity check. Note also that the fluid moving towards the respirometer means volume is decreasing, so oxygen use exceeds carbon dioxide production — consistent with an RQ below 1.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain why the complete absence of oxygen stops the Krebs cycle, even though the Krebs cycle does not use oxygen.

The Krebs cycle depends on a supply of oxidised NAD and FAD to accept hydrogen from its dehydrogenation reactions. These coenzymes are reoxidised by passing their hydrogen to the electron transport chain on the inner mitochondrial membrane. Oxygen is the final electron acceptor of that chain, combining with electrons and protons to form water. Without oxygen, electrons cannot leave the chain, so the carriers remain reduced and cannot accept more; reduced NAD and FAD therefore cannot be reoxidised. With no oxidised coenzyme available, the dehydrogenation steps of the Krebs cycle and the link reaction cannot proceed, so both stop. Only glycolysis continues, and only because fermentation provides an alternative way of reoxidising NAD.

2. Calculate the RQ for the respiration of a fatty acid, given the equation C₁₈H₃₆O₂ + 26O₂ → 18CO₂ + 18H₂O, and explain what the value tells you.

RQ = molecules of CO₂ produced ÷ molecules of O₂ used = 18 ÷ 26 = 0.69, usually quoted as about 0.7. The value is well below 1.0 because a fatty acid is much less oxidised than a carbohydrate: it contains proportionally far more hydrogen and far less oxygen, so more oxygen must be taken in from outside to oxidise it fully. That extra hydrogen is exactly why lipids yield more energy per gram — more hydrogen delivered to the electron transport chain by reduced NAD and FAD means more protons pumped and more ATP made. So a low RQ and a high energy value are two consequences of the same chemical fact.

3. Describe how a proton gradient is used to make ATP in a mitochondrion.

Reduced NAD and FAD deliver hydrogen atoms to carriers in the inner mitochondrial membrane. The hydrogen atoms split into protons and energetic electrons. The electrons pass along the electron transport chain, releasing energy as they move from carrier to carrier. This energy is used to actively transfer protons from the matrix across the inner membrane into the intermembrane space, so a proton concentration gradient (and an electrical gradient) builds up across the membrane. Protons then return to the matrix by facilitated diffusion through the enzyme ATP synthase, and the energy released as they move down the gradient drives the phosphorylation of ADP to ATP. This process is chemiosmosis. Finally, oxygen acts as the final electron acceptor, combining with the electrons and protons to form water and so keeping the chain clear.

4. Compare lactate fermentation in mammals with ethanol fermentation in yeast.

Similarities: both occur in the cytoplasm when oxygen is absent; both exist to reoxidise reduced NAD so that glycolysis can continue; both yield only the net 2 ATP from glycolysis; both begin from pyruvate. Differences: in mammals, pyruvate directly accepts hydrogen from reduced NAD to form lactate, with no carbon dioxide released, and the process is reversible — lactate is later transported to the liver and oxidised back to pyruvate when oxygen is available. In yeast, pyruvate is first decarboxylated to ethanal, releasing carbon dioxide, and ethanal then accepts the hydrogen to form ethanol; this is irreversible, and ethanol accumulates until it becomes toxic to the yeast.

5. Explain how rice is adapted to grow with its roots submerged in water.

Submerged soil is waterlogged and low in oxygen, since oxygen diffuses far more slowly through water than air. Rice has three adaptations. (i) Aerenchyma — large air spaces running through the stem and roots, forming a continuous pathway down which oxygen diffuses from the shoot above water to the submerged root tissue, allowing aerobic respiration to continue there. (ii) Root cells carry out ethanol fermentation when oxygen does run short, and are unusually tolerant of the ethanol produced, which would poison most plants; this yields a small amount of ATP anaerobically. (iii) Rapid stem elongation in response to rising water levels, so that leaves and the tops of the stems remain above the water surface and gas exchange with the atmosphere is maintained.

6. A student uses methylene blue with a yeast suspension to compare respiration at 20 °C and 35 °C. Describe the expected result and explain the principle, then state two variables to control.

Methylene blue is blue when oxidised and colourless when reduced. Added to a yeast suspension, it acts as an artificial hydrogen acceptor, taking hydrogen from the dehydrogenation reactions of respiration in place of NAD. The faster the rate of respiration, the faster it is reduced and the shorter the time to decolourise. At 35 °C respiration is faster than at 20 °C, since the enzymes are nearer their optimum and molecules have more kinetic energy, so the suspension should decolourise in a shorter time. Controlled variables: the concentration and volume of yeast suspension; the concentration and volume of methylene blue; the concentration and volume of respiratory substrate (glucose); the time allowed for the yeast to equilibrate at each temperature; and the same person judging the end point — better still, use a colorimeter to fix the end point objectively. The tubes should also be sealed or covered with a layer of oil to exclude oxygen, which would otherwise reoxidise the indicator.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Any well-made animation of the electron transport chain and chemiosmosis — the proton gradient is far easier to watch than to read
Nuffield Foundation practical biology — respirometer protocols, including the RQ determination and the control tube
Khan Academy — cellular respiration, if you want the stages at a slower pace than the syllabus summary allows