IB DP Biology: Cell respiration: ATP, glycolysis, Krebs cycle and the electron transport chain (DP Biology) – Revision Notes

These revision notes condense IB Diploma Programme Biology topic C1.2 (Cell respiration), as set out in the IB Biology guide, first assessment 2025. C1.2.1 to C1.2.6 are SL and HL. C1.2.7 to C1.2.17 are additional higher level and are marked “HL only”. For full explanations and worked examples, read the study guide.

Then test yourself with the practice set. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics.

Key definitions

Term	Meaning
ATP	Adenosine triphosphate, a nucleotide; the energy currency of the cell
Cell respiration	Release of energy from carbon compounds in cells to make ATP
Gas exchange	Physical movement of oxygen in and carbon dioxide out across a surface
Oxidation	Loss of electrons; removal of hydrogen from a substrate is dehydrogenation (HL only)
Reduction	Gain of electrons (HL only)
NAD	Coenzyme that carries hydrogen; becomes reduced NAD when it accepts it (HL only)
Decarboxylation	Removal of carbon dioxide from a compound (HL only)
Chemiosmosis	ATP synthesis driven by protons flowing down a gradient through ATP synthase (HL only)

ATP in one page (C1.2.1 to C1.2.3)

• ATP is soluble, small, mobile, releases a usable amount of energy on hydrolysis, and is rapidly regenerated.
• Hydrolysis: ATP + water → ADP + phosphate, releasing energy.
• Synthesis: ADP + phosphate → ATP, needing energy.
• You do not need the energy value in kilojoules.
• Three uses to quote: active transport, anabolism (for example protein synthesis), movement of the cell or its parts (for example chromosomes in mitosis).

Respiration versus gas exchange (C1.2.4)

• Respiration: chemical, inside cells, makes ATP.
• Gas exchange: physical, across a surface, supplies oxygen and removes carbon dioxide.
• Main substrates are glucose and fatty acids; many other organic compounds can be used.

Aerobic and anaerobic in humans (C1.2.5)

	Aerobic	Anaerobic
Substrate	Carbohydrate, lipid, others	Carbohydrate only
Oxygen	Required	Not required
ATP yield	Large	Small
Waste	Carbon dioxide, water	Lactate
Site	Cytoplasm and mitochondria	Cytoplasm

Word equations: glucose + oxygen → carbon dioxide + water (+ ATP); glucose → lactate (+ ATP).

Measuring rate (C1.2.6)

Method in steps for a respirometer:

1. Place the organisms in a sealed tube with a carbon dioxide absorbent.
2. Set up a control tube with inert beads of the same volume.
3. Record the movement of the bead or coloured drop over a set time.
4. Calculate volume = πr² × distance moved.
5. Divide by time, then by mass, to give a rate per gram per minute.
6. Repeat and find a mean.

Other rate indicators: carbon dioxide released (gas syringe, pH change, data logger), heat released. Variables that change the rate: temperature, substrate concentration, oxygen availability, pH, inhibitors, tissue type and mass.

Rate formula: rate = amount of change / time taken. Percentage change = (new − old) / old × 100.

Redox and NAD (C1.2.7, HL only)

• Oxidation and reduction occur together.
• Removing H (a proton plus an electron) from a substrate oxidises it.
• NAD accepts the hydrogen and is reduced.
• Reduced NAD carries the hydrogen to the electron transport chain.

The pathway at a glance (HL only)

Stage	Where	In	Out
Glycolysis	Cytoplasm	Glucose, 2 ATP, NAD	2 pyruvate, 4 ATP (net 2), 2 reduced NAD
Link reaction	Matrix	Pyruvate, NAD, coenzyme A	Acetyl-CoA, carbon dioxide, reduced NAD
Krebs cycle	Matrix	Acetyl group, oxaloacetate	Carbon dioxide, ATP, reduced NAD; oxaloacetate regenerated
Electron transport chain	Inner membrane	Reduced NAD, oxygen, ADP, phosphate	NAD, water, ATP

Glycolysis ideas (C1.2.8): phosphorylation, lysis, oxidation, ATP formation. Every step needs a different enzyme. No intermediates required.

Krebs ideas (C1.2.12): only name citrate (6C) and oxaloacetate (4C). Citrate forms when an acetyl group joins oxaloacetate. Each turn has four oxidations (dehydrogenations) and two decarboxylations.

Per glucose: carbon dioxide released in the link reaction plus Krebs = 6. Reduced NAD from glycolysis, link and Krebs = 2 + 2 + 6 = 10, using the usual three per turn.

Anaerobic pathways (C1.2.9 and C1.2.10, HL only)

	Humans	Yeast
Hydrogen acceptor	Pyruvate	Ethanal
Products	Lactate	Ethanol and carbon dioxide
NAD regenerated	Yes	Yes
Net ATP per glucose	2	2

The pathway is identical up to pyruvate. Regenerating NAD lets glycolysis continue. Baking: carbon dioxide raises the dough and ethanol evaporates. Brewing: ethanol is retained in a sealed vessel.

Electron transport chain (C1.2.13 to C1.2.16, HL only)

1. Reduced NAD passes a pair of electrons to the first carrier and becomes NAD.
2. Electrons flow along carriers on the inner mitochondrial membrane.
3. The energy released moves protons out of the matrix, into the intermembrane space.
4. Protons flow back through ATP synthase. This couples the energy release to phosphorylation of ADP.
5. Oxygen takes electrons and protons from the matrix, forming water. This lets the chain keep flowing.

If oxygen is absent, the chain backs up, reduced NAD builds up, and the link reaction and Krebs cycle slow.

Lipids versus carbohydrates (C1.2.17, HL only)

• Lipids give more energy per gram: less oxygen, more oxidisable hydrogen and carbon.
• Fatty acids are broken into 2C acetyl groups and enter as acetyl-CoA, bypassing glycolysis.
• Glycolysis and anaerobic respiration need carbohydrate.

Small worked reminders

Energy per gram. Using supplied values of 17 kJ g⁻¹ for carbohydrate and 37 kJ g⁻¹ for lipid, the mass needed for 1000 kJ is 1000 / 17 = 58.8 g of carbohydrate and 1000 / 37 = 27.0 g of lipid. Lipid needs less than half the mass.

Describing a rate graph. Quote the starting and finishing values with units, say where the rate peaks, then give a biological reason for each change. For respiration versus temperature, the rise reflects more frequent enzyme–substrate collisions, and the fall reflects denaturation.

Writing a pathway answer. Name the stage, its location, what is oxidised, what is reduced, and what is released. For example: in the link reaction, in the matrix, pyruvate is oxidised and decarboxylated, NAD is reduced, and an acetyl group joins coenzyme A.

Must-know distinctions

• Respiration versus gas exchange.
• Oxidation (electron or hydrogen loss) versus reduction (gain).
• Decarboxylation (carbon dioxide lost) versus dehydrogenation (hydrogen lost).
• Substrate-level ATP in glycolysis and Krebs versus ATP from chemiosmosis.
• Humans: pyruvate to lactate. Yeast: pyruvate to ethanal to ethanol.
• Where each stage occurs: glycolysis in the cytoplasm; link and Krebs in the matrix; chain on the inner membrane.

Quick self-test

1. State the full name of ATP and the type of molecule it is.
2. Name the two organic substrates that are the main respiratory fuels.
3. A drop moves 30 mm in 12 minutes in a capillary of radius 0.45 mm, with 2.0 g of maggots. Calculate the oxygen uptake in mm³ g⁻¹ min⁻¹.
4. The rate rises from 0.95 to 1.80 cm³ min⁻¹. Calculate the percentage increase.
5. Why does anaerobic respiration in humans produce only a small amount of ATP? (HL only)
6. State the net yield of glycolysis per glucose. (HL only)
7. Which molecule is regenerated when pyruvate becomes lactate, and why does that matter? (HL only)
8. How many carbon dioxide molecules are released from one glucose in the link reaction plus Krebs cycle? (HL only)
9. Name the 6C compound that forms when an acetyl group joins oxaloacetate. (HL only)
10. What is the role of oxygen in the chain? (HL only)
11. Explain why a gram of lipid releases more energy than a gram of carbohydrate. (HL only)

Answers

1. Adenosine triphosphate; a nucleotide.
2. Glucose (carbohydrate) and fatty acids (lipid).
3. Volume = π × 0.45² × 30 = 19.1 mm³. Per minute = 1.59. Per gram = 0.795 mm³ g⁻¹ min⁻¹ (3 s.f.).
4. (1.80 − 0.95) / 0.95 × 100 = 89.5%.
5. Only glycolysis operates, so there is no Krebs cycle or chain; the net gain is 2 ATP per glucose.
6. 2 ATP (net), 2 reduced NAD and 2 pyruvate.
7. NAD. It lets glycolysis carry on, so ATP production continues.
8. 6 (2 from the link reaction, 4 from the Krebs cycle).
9. Citrate.
10. It is the terminal electron acceptor: it takes electrons and protons from the matrix to form water, so electrons keep flowing.
11. Lipids contain less oxygen and more oxidisable hydrogen and carbon per gram, so more can be oxidised.

Where marks are usually lost

• Using “breathing” or “gas exchange” where the answer needs respiration.
• Saying ATP is “made” from energy without mentioning ADP and phosphate.
• Giving lactate as a product in yeast, or ethanol in humans.
• Leaving out the units or the “per gram” in a rate calculation.
• Giving the gross glycolysis yield of 4 ATP when the question asks for net.
• Placing oxygen use in the Krebs cycle itself.
• Writing “electrons make ATP” when the energy is used to pump protons first.
• Saying oxygen “accepts hydrogen” without electrons and protons.
• Describing a trend in a graph without quoting figures with units.
• Stating that fat can be used in anaerobic respiration.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Topic C1.2 (Cell respiration), statements C1.2.1 to C1.2.17; C1.2.7 to C1.2.17 are additional higher level.