Pearson Edexcel International A Level Biology: Aerobic and Anaerobic Respiration (YBI11)

This guide teaches aerobic and anaerobic respiration from scratch for the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. It covers statements 7.1 to 7.8 of Topic 7, including Core Practicals 15 and 16. Topics 1 to 4 are International AS (IAS) and Topics 5 to 8 are International A2 (IA2), so this material is A2. There are no tiers. Use it with the revision notes and the practice questions. The course hub and printable checklist show the rest of the course, and the free diagnostics show where to start. Mitochondrial structure is revised in cell ultrastructure and microscopy, and the matching energy-capture topic is photosynthesis.

What this guide covers

Statement	What you must be able to do
7.1 (i)	Describe respiration as splitting the substrate to release carbon dioxide and reuniting hydrogen with oxygen
7.1 (ii)	Explain that respiration is stepped, each step catalysed by a specific intracellular enzyme
7.2	Describe glycolysis: phosphorylation, ATP by substrate level phosphorylation, reduced coenzyme, pyruvate, lactate
7.3	Describe the link reaction and Krebs cycle, and where they happen
7.4	Explain oxidative phosphorylation, chemiosmosis and ATP synthase
7.5	Describe what happens to lactate after anaerobic respiration in animals
7.6	Define and calculate respiratory quotient (RQ)
7.7	Core Practical 15: a redox indicator to investigate respiration in yeast
7.8	Core Practical 16: a simple respirometer to find rate of respiration and RQ

7.1 The overall reaction, and why it is stepped

(i) Glucose is the usual respiratory substrate. The carbon skeleton is split up and released as carbon dioxide, a waste product. The hydrogen atoms are removed from the substrate and, in the end, joined to oxygen from the air to make water. The union of hydrogen with oxygen releases a large amount of energy, and some of it is used to make ATP.

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Do not say the oxygen you breathe in ends up in carbon dioxide. It ends up in water.

(ii) The reaction happens as a series of small steps. Each step is controlled and catalysed by a specific enzyme inside the cell. You do not need the names of the enzymes. The advantage is that energy is released in small amounts, so more of it can be captured in ATP and less is lost as heat. Burning glucose in one step would release the energy as heat.

7.2 Glycolysis

Glycolysis happens in the cytoplasm and does not need oxygen. It is the first stage of both aerobic and anaerobic respiration. You do not need intermediate compounds.

1. Phosphorylation. Glucose is phosphorylated using two molecules of ATP. This gives a hexose bisphosphate, which is more reactive and cannot leave the cell.
2. Splitting. The hexose bisphosphate splits into two three-carbon molecules.
3. Oxidation. Each three-carbon molecule loses hydrogen (dehydrogenation). The hydrogen is accepted by the coenzyme NAD, giving reduced NAD.
4. ATP production. Four molecules of ATP are made by substrate level phosphorylation: a phosphate group is transferred from a substrate molecule directly to ADP.
5. Product. Two molecules of pyruvate.

Net yield per glucose: 2 ATP (4 made, 2 used), 2 reduced NAD and 2 pyruvate.

When oxygen is absent

Reduced NAD must be turned back into NAD, or glycolysis would stop for lack of NAD. In animal cells, pyruvate accepts the hydrogen from reduced NAD and becomes lactate. This regenerates NAD so that glycolysis can go on making a small amount of ATP. No further ATP is made by this step, and the lactate is a waste product that builds up in the cell and the blood. Yeast does the same job by forming ethanol and carbon dioxide.

With oxygen present, pyruvate enters the mitochondria instead.

7.3 The link reaction and the Krebs cycle

These take place in the matrix of the mitochondrion. Glycolysis, by contrast, is in the cytoplasm.

Link reaction. Pyruvate enters the mitochondrion. It is decarboxylated (carbon dioxide removed) and dehydrogenated (hydrogen removed and accepted by NAD). The two-carbon acetyl group is carried by coenzyme A into the Krebs cycle. Per glucose, the link reaction happens twice.

Krebs cycle. The acetyl group joins a four-carbon compound to make a six-carbon compound. Then, in a cycle of enzyme-controlled steps:

• carbon dioxide is released by decarboxylation;
• hydrogen is removed by dehydrogenation and passed to NAD and FAD, giving reduced NAD and reduced FAD;
• one ATP per turn is made by substrate level phosphorylation;
• the four-carbon compound is regenerated.

You do not need the names of the other compounds.

Worked example: totals for one glucose

Stage	CO₂	Reduced NAD	Reduced FAD	ATP (substrate level)
Glycolysis	0	2	0	2 net
Link reaction (×2)	2	2	0	0
Krebs cycle (×2 turns)	4	6	2	2
Total	6	10	2	4

All six carbon atoms of glucose have become carbon dioxide. The ten reduced NAD and two reduced FAD carry the hydrogen to the next stage.

7.4 Oxidative phosphorylation

This takes place on the inner mitochondrial membrane, which is folded into cristae. The folds give a large surface area that holds many electron carriers and ATP synthase molecules. The intermembrane space is small, so the H⁺ concentration changes quickly when protons are pumped in.

1. Reduced NAD and reduced FAD deliver hydrogen. They split into protons and electrons and are reoxidised to NAD and FAD, which return to the matrix cycles.
2. Electrons pass along a chain of carriers. At each transfer the electrons lose energy.
3. The energy released drives pumps that move H⁺ out of the matrix, through the inner membrane and into the gap between the two membranes. An electrochemical gradient builds up.
4. H⁺ flows back into the matrix down the gradient through the channel in ATP synthase. The enzyme uses the flow to join ADP and Pi, making ATP. The use of a proton gradient to make ATP is chemiosmosis.
5. At the end of the chain, oxygen is the final electron acceptor: it takes up the electrons together with H⁺, and water is the product.

Without oxygen, electrons cannot leave the last carrier. The chain backs up, reduced NAD and reduced FAD are not reoxidised, and the Krebs cycle and link reaction stop too. That is why the Krebs cycle needs oxygen indirectly even though no oxygen is used in it.

Worked example: ATP yield

Suppose the data given are 2.5 ATP per reduced NAD and 1.5 ATP per reduced FAD passing through the chain. The totals above give:

• from reduced NAD: 10 × 2.5 = 25 ATP
• from reduced FAD: 2 × 1.5 = 3 ATP
• by substrate level phosphorylation: 4 ATP
• total: 25 + 3 + 4 = 32 ATP per glucose

Textbooks quote different ratios, so use the values in the question. Real yields are lower because protons leak across the membrane.

7.5 What happens to lactate

After a period of vigorous exercise, muscle contains lactate. It diffuses into the blood and is carried to the liver. There, with oxygen, it is converted back to pyruvate. Pyruvate can then be oxidised in the link reaction and Krebs cycle, or converted to glucose and glycogen. Oxygen is needed for this, so breathing rate stays high after exercise. The extra oxygen used is sometimes called the oxygen debt.

7.6 Respiratory quotient

RQ = volume (or amount) of carbon dioxide produced ÷ volume (or amount) of oxygen consumed, over the same time.

RQ depends on the substrate because substrates differ in how much oxygen they need per carbon dioxide made.

Substrate	Approximate RQ
Lipid	about 0.7
Protein	about 0.9
Carbohydrate	1.0
Mixed aerobic and anaerobic respiration (yeast)	more than 1

Worked example: calculating RQ from an equation

Palmitic acid: C₁₆H₃₂O₂ + 23O₂ → 16CO₂ + 16H₂O.

RQ = 16 ÷ 23 = 0.70 (2 s.f.). For glucose, 6 ÷ 6 = 1.0.

Worked example: interpreting measurements

Germinating oil-rich seeds absorb 36 mm³ of oxygen and release 25 mm³ of carbon dioxide in 30 minutes.

RQ = 25 ÷ 36 = 0.69, so lipid is the main respiratory substrate.

A yeast culture producing 38 mm³ of carbon dioxide while taking up 30 mm³ of oxygen gives 38 ÷ 30 = 1.27. A value above 1 means some carbon dioxide was released without oxygen being used, from anaerobic respiration.

7.7 Core Practical 15: redox indicator with yeast

The specification asks you to use an artificial hydrogen carrier (a redox indicator) to investigate respiration in yeast. Methylene blue or DCPIP are common choices. Both are blue when oxidised and colourless when reduced. They accept hydrogen from the dehydrogenation reactions in place of NAD, so the time taken to lose the colour shows how fast hydrogen is being released.

Method.

1. Prepare a yeast suspension in a glucose solution.
2. Warm tubes of suspension in water baths set to the temperatures you are testing.
3. Add the same volume of indicator to each tube, start a stopwatch and mix.
4. Record the time for the blue colour to disappear.
5. Calculate rate as 1000 ÷ time, which gives a value per 1000 seconds.

Controls. Keep the volume and concentration of yeast, glucose and indicator the same. Include a tube of boiled yeast to show that the colour change needs living cells. Repeat each temperature and calculate a mean.

Worked example: effect of temperature

Temperature (°C)	Time to decolourise (s)	Rate (1000 ÷ time)
20	420	2.38
30	210	4.76
40	120	8.33
50	300	3.33
60	no change in 900	0

Between 20 and 40 °C the rate rises because enzyme and substrate molecules have more kinetic energy and collide more often. At 50 °C the rate falls as enzymes begin to denature. At 60 °C the yeast is killed and the indicator stays blue.

7.8 Core Practical 16: simple respirometer and RQ

A simple respirometer is a sealed tube holding the organisms, connected to a capillary tube with a drop of coloured fluid. Germinating seeds or small invertebrates such as woodlice are suitable.

Oxygen uptake. Put a carbon dioxide absorber (soda lime or potassium hydroxide solution) in the tube, with the organisms held above it on gauze. The carbon dioxide produced is removed, so the fluid moves towards the tube by the volume of oxygen used. Volume = πr² × distance moved.

Controls. Place the apparatus in a water bath to keep temperature constant, because gas volume changes with temperature. Allow time for equilibration. Use a control tube with an equal volume of inert material, such as glass beads, to show any change in pressure not caused by respiration. Wear eye protection with potassium hydroxide.

RQ. Repeat without the absorber. Now the movement is the difference between oxygen used and carbon dioxide produced.

Worked example

The capillary has a radius of 0.5 mm. In 10 minutes, with absorber, the fluid moves 24 mm. Without absorber it moves 6 mm. The seeds weigh 1.5 g.

• Oxygen used = π × 0.5² × 24 = 18.8 mm³
• Rate = 18.8 ÷ 10 ÷ 1.5 = 1.26 mm³ g⁻¹ min⁻¹
• Net movement without absorber = π × 0.5² × 6 = 4.71 mm³
• Carbon dioxide produced = 18.8 − 4.71 = 14.1 mm³
• RQ = 14.1 ÷ 18.8 = 0.75

Common errors

• Writing that the Krebs cycle “makes lots of ATP”. It makes only one ATP per turn by substrate level phosphorylation. Most ATP comes from oxidative phosphorylation.
• Saying lactate is made from glucose directly. It is made from pyruvate.
• Saying oxygen is used in the Krebs cycle.
• Forgetting that the link reaction occurs twice per glucose.
• Dividing oxygen by carbon dioxide when working out RQ.
• Describing the respirometer as measuring “gas produced” when the absorber means it measures oxygen used.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology (YBI11) specification, Issue 2, published by Pearson Education Limited.