OCR A Level Biology A: Chemiosmosis, anaerobic respiration and respiratory quotient (H420) – Revision Notes

These revision notes condense OCR Level 3 Advanced GCE in Biology A (H420), specification Version 4.1 (April 2026), section 5.2.2 Respiration, learning outcomes (h) to (l). For full explanations, start with the study guide. Then test yourself with the practice questions, the free 10-minute diagnostics, the course hub and the printable checklist.

Key terms

Term	Meaning
Chemiosmosis	Synthesis of ATP driven by the flow of protons down an electrochemical gradient through ATP synthase
Electron transport chain	Carriers in a membrane that pass electrons down energy levels; energy released pumps H⁺
ATP synthase	Enzyme and channel in the membrane; H⁺ flow through it joins ADP and inorganic phosphate
Oxidative phosphorylation	ATP made using electrons from reduced NAD and FAD, with oxygen as final acceptor
Photophosphorylation	ATP made using the electron transport chain driven by light energy
Fermentation step	Reaction that turns reduced NAD back into NAD so glycolysis can continue
Respiratory quotient (RQ)	CO₂ produced ÷ O₂ consumed

5.2.2(h) Chemiosmosis in steps

Oxidative phosphorylation (inner mitochondrial membrane, cristae):

1. Reduced NAD and reduced FAD donate electrons to the electron transport chain.
2. Electrons move down the carriers. Released energy pumps H⁺ from the matrix to the intermembrane space.
3. An electrochemical gradient forms: more H⁺ in the intermembrane space.
4. H⁺ flows back to the matrix through ATP synthase. ADP + Pi → ATP.
5. Oxygen accepts electrons and H⁺ at the end of the chain, forming water.

Photophosphorylation (thylakoid membrane): light excites electrons in chlorophyll, the chain pumps H⁺ from the stroma into the thylakoid space, and H⁺ flows back to the stroma through ATP synthase.

Must-know distinctions

	Mitochondrion	Chloroplast
H⁺ pumped into	Intermembrane space	Thylakoid space
Electrons come from	Reduced NAD and FAD	Chlorophyll excited by light
Final acceptor	Oxygen	NADP (non-cyclic)
ATP made in	Matrix	Stroma

If oxygen is absent: electrons cannot leave the chain, reduced NAD is not reoxidised, the gradient is not built, and ATP synthesis by chemiosmosis stops.

Worked reminder. A pH difference of 1.0 across a membrane is a ten-fold difference in H⁺ concentration (10¹ = 10), and 0.5 is about 3.2-fold (10^0.5).

5.2.2(i) Anaerobic respiration

	Mammals	Yeast
Starting point	Pyruvate from glycolysis	Pyruvate from glycolysis
Steps	Pyruvate + reduced NAD → lactate + NAD	Pyruvate → ethanal + CO₂, then ethanal + reduced NAD → ethanol + NAD
Enzyme (first step)	Lactate dehydrogenase	Pyruvate decarboxylase
Gas released	None	CO₂
Reversible?	Lactate can be converted back to pyruvate (in the liver)	Ethanol formation is not reversible
Net ATP	2 (glycolysis only)	2 (glycolysis only)

Why the NAD step matters: glycolysis uses NAD. Without a way to regenerate it, glycolysis stops and ATP production stops.

Benefits: ATP can be produced quickly with no oxygen, such as in muscle during sprinting; yeast and some other organisms can live in low oxygen conditions.

Why the yield is lower:

• Pyruvate is not oxidised any further, so the link reaction and Krebs cycle do not happen.
• Reduced NAD is used to make lactate or ethanol, so it does not supply the electron transport chain.
• Energy remains in lactate or ethanol.
• Only substrate-level phosphorylation in glycolysis makes ATP: 2 per glucose, compared with about 30 aerobically.

Yeast practical. Oil on the surface excludes oxygen. Rate can be measured from gas volume, bubble count or colour change of a redox indicator. Control: same yeast mass, temperature and glucose concentration, with boiled yeast as a negative control.

5.2.2(j) Energy values

Substrate	Approx. energy / kJ g⁻¹	Route in
Carbohydrate	17	Glycolysis
Lipid	37	Glycerol to triose phosphate; fatty acids to acetyl CoA
Protein	17	Deamination, then pyruvate, acetyl CoA or Krebs cycle intermediate

Lipid gives more energy per gram because it has more hydrogen per molecule, so more reduced NAD and FAD are made and more H⁺ is pumped.

5.2.2(k) Respiratory quotient

RQ = CO₂ produced ÷ O₂ consumed. No units.

RQ	Suggests
About 1.0	Carbohydrate
About 0.9	Protein
About 0.7	Lipid
Above 1.0	Some anaerobic respiration, or an organic acid substrate

Method in steps for an equation

1. Write the balanced equation.
2. Take moles of CO₂ made and moles of O₂ used.
3. Divide CO₂ by O₂.

Worked reminder. Glucose: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. RQ = 6 ÷ 6 = 1.0.

5.2.2(l) Respirometer practicals

Method in steps

1. Put the organisms on gauze above soda lime or potassium hydroxide solution, which absorbs CO₂.
2. Connect a capillary tube with a drop of coloured fluid. Use a control tube with beads or dead organisms.
3. Equilibrate in a water bath, seal, and record the drop’s position at set times.
4. Repeat and calculate a mean.

Calculations

• Volume of O₂ used = π r² × distance moved by the drop (r is the capillary radius, not the diameter).
• Rate = volume ÷ time. Divide by mass for rate per gram.
• RQ = (movement with absorber − movement without absorber) ÷ movement with absorber.

Worked reminder. The drop moves 25 mm in 5 min in a capillary of radius 0.4 mm. Volume = π × 0.4² × 25 = 12.6 mm³. Rate = 12.6 ÷ 5 = 2.5 mm³ min⁻¹.

Factors: temperature (rate rises then falls as enzymes denature), substrate concentration (rate rises then levels off), type of substrate (compare using RQ).

Explaining inhibitors and changes

Questions often change one thing and ask for the effect. Work through the chain in order.

Change	Effect on proton gradient	Effect on ATP and oxygen use
Electron transport chain blocked	Gradient falls as no more H⁺ is pumped	Less ATP; less oxygen used; reduced NAD builds up
ATP synthase blocked	H⁺ cannot flow back, so the gradient rises and slows pumping	Less ATP; less oxygen used
Membrane made leaky to H⁺	Gradient cannot be kept	Oxygen use continues but little ATP is made; more energy is lost as heat
No oxygen	Chain backs up, so no pumping	Aerobic ATP stops; fermentation takes over

Reading a rate from a graph. Rate is the gradient. Take two points far apart on a straight section, and divide the change in volume by the change in time. For a curve, draw a tangent at the time asked. Data supplied: if gas collected rises from 4.2 cm³ to 10.2 cm³ over 6 min on a straight section, rate = 6.0 ÷ 6 = 1.0 cm³ min⁻¹. Give units every time.

Explain-question frame. Name the structure or stage, say what happens there, and then say what the result is for ATP. For example: “Oxygen is the final acceptor, so electrons keep flowing, so H⁺ keeps being pumped, so ATP synthase keeps making ATP.”

Quick self-test

1. In which part of a mitochondrion is ATP synthase found?
2. What is the final electron acceptor in oxidative phosphorylation, and what is formed?
3. Which side of the inner mitochondrial membrane has the higher H⁺ concentration?
4. Why does glycolysis stop if lactate is not formed in a working muscle?
5. Name the two products of anaerobic respiration in yeast.
6. Why does anaerobic respiration give a much lower ATP yield than aerobic respiration?
7. A person needs 555 kJ from a lipid. Taking lipid as 37 kJ g⁻¹, what mass is needed?
8. Calculate the RQ of seedlings that take in 1.2 cm³ of oxygen and release 0.9 cm³ of carbon dioxide in the same time.
9. What does an RQ of 0.7 suggest about the substrate?
10. In a respirometer, why is potassium hydroxide solution or soda lime included?
11. A drop moves 18 mm in 6 min in a capillary of radius 0.5 mm. State the volume of oxygen used per minute, to 2 significant figures.
12. Why can a yeast culture growing only by fermentation not be given an RQ?

Answers

1. The inner mitochondrial membrane (it is embedded in the cristae).
2. Oxygen; it combines with electrons and H⁺ to form water.
3. The intermembrane space side.
4. NAD is not regenerated, so glycolysis cannot continue.
5. Ethanol and carbon dioxide.
6. Only glycolysis makes ATP (2 per glucose); pyruvate is not oxidised further, and reduced NAD is used to make lactate or ethanol rather than to drive the electron transport chain.
7. 555 ÷ 37 = 15 g.
8. 0.9 ÷ 1.2 = 0.75.
9. Mainly lipid.
10. It absorbs the carbon dioxide produced, so the movement of the drop shows oxygen used only.
11. π × 0.5² × 18 = 14.1 mm³; 14.1 ÷ 6 = 2.4 mm³ min⁻¹.
12. It takes in no oxygen, so dividing carbon dioxide by oxygen would mean dividing by zero.

Where marks are usually lost

• Saying “protons” where the question wants the idea of a proton gradient across a membrane, or writing “electrons flow through ATP synthase”.
• Not saying which side of the membrane the H⁺ is pumped to.
• Writing that oxygen is “used up” in the Krebs cycle.
• Saying anaerobic respiration “produces ATP from lactate” instead of explaining NAD regeneration.
• Leaving out the units of the rate, or not dividing by the mass or the time.
• Using the capillary diameter in π r².
• Writing the RQ ratio upside down.
• Giving a “mixture of substrates” without saying why the RQ lies between 0.7 and 1.0.

Official syllabus

OCR Level 3 Advanced GCE in Biology A (H420), specification Version 4.1 (April 2026), first assessment 2025, published by OCR. These notes follow section 5.2.2 Respiration, learning outcomes (h) to (l).