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

These notes condense statements 7.1 to 7.8 of Topic 7 (aerobic and anaerobic respiration, including Core Practicals 15 and 16) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. Topics 1 to 4 are International AS (IAS) and Topics 5 to 8 are International A2 (IA2); this is A2 content and there are no tiers. Read the study guide for full explanations and test yourself with the practice questions. The course hub, checklist and free diagnostics cover the rest. For the structure of the mitochondrion, see cell ultrastructure and microscopy.

Key definitions

Term	Meaning
Respiration (7.1)	Stepped, enzyme-controlled splitting of a substrate; carbon is lost as CO₂ and hydrogen is joined to oxygen, releasing energy used to make ATP
Substrate level phosphorylation	ATP made by transferring a phosphate group from a substrate directly to ADP
Oxidative phosphorylation	ATP made using energy from electrons passing along the electron transport chain, with oxygen as final acceptor
Decarboxylation	Removal of CO₂ from a substrate
Dehydrogenation	Removal of hydrogen from a substrate, usually passed to NAD or FAD
Chemiosmosis	Flow of H⁺ down an electrochemical gradient through ATP synthase, driving ATP synthesis
Respiratory quotient	CO₂ produced ÷ O₂ consumed

Location and products at a glance

Stage (statement)	Where	In	Out
Glycolysis (7.2)	Cytoplasm	Glucose, 2 ATP, NAD	2 pyruvate, 4 ATP (net 2), 2 reduced NAD
Link reaction (7.3)	Matrix	Pyruvate, NAD, coenzyme A	Acetyl group, CO₂, reduced NAD
Krebs cycle (7.3)	Matrix	Acetyl group	2 CO₂, 3 reduced NAD, 1 reduced FAD, 1 ATP per turn
Oxidative phosphorylation (7.4)	Inner membrane	Reduced NAD, reduced FAD, O₂, ADP, Pi	NAD, FAD, H₂O, ATP

Per glucose, the link reaction and Krebs cycle each run twice. Per glucose the whole pathway gives 6 CO₂, 10 reduced NAD, 2 reduced FAD and 4 ATP from substrate level phosphorylation.

7.1 Stepped process

• Substrate loses carbon as CO₂ (waste) and hydrogen is reunited with atmospheric oxygen to make water.
• Many small steps, each catalysed by a specific intracellular enzyme (names not needed).
• Small steps mean energy is released gradually and more is captured in ATP, less lost as heat.

7.2 Glycolysis in steps

1. Phosphorylate glucose using 2 ATP.
2. Split the hexose bisphosphate into two three-carbon molecules.
3. Dehydrogenate: hydrogen goes to NAD, giving reduced NAD.
4. Make 4 ATP by substrate level phosphorylation.
5. Two pyruvate formed.

Without oxygen in animals: pyruvate + reduced NAD → lactate + NAD. The NAD allows glycolysis to continue. In yeast, ethanol and CO₂ are formed instead.

7.3 Link reaction and Krebs cycle

• Link: pyruvate is decarboxylated and dehydrogenated, and the acetyl group is carried by coenzyme A.
• Krebs: acetyl group joins a four-carbon compound; decarboxylation and dehydrogenation steps follow; one ATP by substrate level phosphorylation; four-carbon compound regenerated.
• Both occur in the mitochondrial matrix. Glycolysis does not.
• Neither uses oxygen directly. Both depend on oxygen because the electron transport chain must reoxidise NAD and FAD.

7.4 Oxidative phosphorylation in steps

1. Reduced NAD and reduced FAD give up hydrogen (protons and electrons).
2. Electrons pass along carriers in the inner membrane, losing energy.
3. That energy drives H⁺ out of the matrix and into the gap between the membranes.
4. H⁺ gradient forms.
5. H⁺ flows back through ATP synthase: ADP + Pi → ATP.
6. Oxygen accepts electrons and H⁺ to make water.

Cristae give surface area for carriers and ATP synthase. The small intermembrane space lets the gradient build quickly.

7.5 Lactate after anaerobic respiration

Muscle → blood → liver. In the liver lactate is converted to pyruvate. Pyruvate is then oxidised through the Krebs cycle or converted to glucose or glycogen. This needs oxygen, so oxygen uptake stays above resting level after exercise.

7.6 RQ

Substrate	RQ
Lipid	about 0.7
Protein	about 0.9
Carbohydrate	1.0
Yeast, aerobic plus anaerobic	above 1

Method for RQ from an equation: balance it, take the CO₂ coefficient and divide by the O₂ coefficient.

Core Practical 15: redox indicator and yeast (7.7)

• Redox indicator (for example methylene blue) accepts hydrogen and loses its blue colour.
• Time to colourless shows how fast hydrogen is released; rate = 1 ÷ time.
• Vary one factor (such as temperature or glucose concentration); control the others.
• Boiled yeast is the negative control.
• Reduced rate at high temperature: enzymes denature.

Core Practical 16: respirometer (7.8)

• Sealed tube with organisms on gauze, CO₂ absorber below, capillary with coloured drop.
• With absorber: movement = O₂ used. Volume = πr² × distance.
• Without absorber: movement = O₂ used minus CO₂ produced.
• CO₂ produced = O₂ used minus net movement. RQ = CO₂ ÷ O₂.
• Control tube with inert material, water bath, equilibration time, same mass of organism in each run.

Worked reminder

Radius 0.5 mm, drop moves 16 mm in 8 minutes with absorber. Volume = π × 0.25 × 16 = 12.6 mm³, so 1.57 mm³ per minute.

Worked reminder: ATP from a single stage

Suppose the data given are 2.5 ATP per reduced NAD and 1.5 ATP per reduced FAD. The Krebs cycle for one glucose gives 6 reduced NAD, 2 reduced FAD and 2 ATP directly.

• 6 × 2.5 = 15
• 2 × 1.5 = 3
• 2 by substrate level phosphorylation
• total from the cycle and its coenzymes: 15 + 3 + 2 = 20 ATP

Always use the ratios supplied in the question and show each product on a separate line.

Worked reminder: rate from a redox indicator

A yeast tube decolourises in 250 s. Rate = 1000 ÷ 250 = 4.0 per 1000 s. A tube that takes 125 s has twice the rate, 8.0.

Aerobic and anaerobic compared

Feature	Aerobic	Anaerobic (animal)
Oxygen needed	Yes, as final electron acceptor	No
Stages	Glycolysis, link, Krebs, oxidative phosphorylation	Glycolysis, then pyruvate to lactate
Final product	CO₂ and water	Lactate
ATP per glucose	Many	Net 2
Reduced NAD reoxidised by	Electron transport chain	Pyruvate

Why the mitochondrion is suited to its job

Feature	Function
Inner membrane folded into cristae	Large area for carriers and ATP synthase
Small intermembrane space	H⁺ gradient builds quickly
Matrix	Contains enzymes for the link reaction and Krebs cycle
Double membrane	Separates the matrix from the cytoplasm and allows a proton gradient

Distinctions to keep clear

• Substrate level vs oxidative phosphorylation. The first transfers a phosphate from a substrate. The second uses a proton gradient.
• Decarboxylation vs dehydrogenation. Carbon dioxide is lost in one, hydrogen in the other.
• NAD vs FAD. Both are coenzymes. Reduced FAD enters the chain after reduced NAD and drives less ATP synthesis.
• Lactate vs pyruvate. Pyruvate is the end of glycolysis. Lactate is formed from pyruvate when oxygen is lacking.

Quick self-test

1. Where in the cell does glycolysis take place?
2. For the glycolysis of 7 glucose molecules, state the ATP made by substrate level phosphorylation, the net ATP and the reduced NAD.
3. How many molecules of carbon dioxide are released in the link reaction and Krebs cycle for each glucose?
4. How many reduced NAD come from the Krebs cycle alone for 5 glucose molecules?
5. Which enzyme in the inner membrane makes ATP?
6. What is the final electron acceptor?
7. Why does the Krebs cycle stop in the absence of oxygen?
8. Where does lactate go after exercise?
9. Calculate the RQ of stearic acid, C₁₈H₃₆O₂ + 26O₂ → 18CO₂ + 18H₂O.
10. A tissue releases 54 mm³ of carbon dioxide and takes up 60 mm³ of oxygen. Calculate the RQ and suggest the substrate.
11. In a respirometer with a capillary of radius 1 mm, the drop moves 12 mm. What volume of oxygen was used?

Answers

1. The cytoplasm.
2. ATP made is 7 × 4 = 28, net ATP is 28 − 14 = 14, and reduced NAD is 7 × 2 = 14.
3. 6: 2 from the link reaction and 4 from the Krebs cycle.
4. 5 × 6 = 30.
5. ATP synthase.
6. Oxygen.
7. Reduced NAD and reduced FAD are not reoxidised, so there is no NAD or FAD to accept hydrogen.
8. It is carried by the blood to the liver and converted to pyruvate.
9. 18 ÷ 26 = 0.69.
10. 54 ÷ 60 = 0.90, which suggests protein or a mixture of carbohydrate and lipid.
11. π × 1² × 12 = 37.7 mm³.

Where marks are usually lost

• Saying glycolysis makes “2 ATP” without showing 4 made and 2 used.
• Placing glycolysis in the mitochondrion.
• Stating that oxygen is “used in” the Krebs cycle or link reaction.
• Describing chemiosmosis without the proton gradient across the inner membrane.
• Omitting that oxygen accepts electrons and H⁺ to form water.
• Writing that lactate is “removed” with no destination or conversion.
• Reversing the RQ ratio.
• Writing “time” as the rate in a redox indicator experiment instead of 1 ÷ time.
• Giving the respirometer reading as oxygen used when no absorber was present.

Official syllabus

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