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

This study guide is aligned to the OCR Level 3 Advanced GCE in Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. It covers section 5.2.2 Respiration, learning outcomes (h) to (l): the chemiosmotic theory, anaerobic respiration in eukaryotes, the energy values of respiratory substrates, the respiratory quotient and respirometer practicals.

Links: course hub, printable checklist. Next, use the revision notes, the practice questions and the free 10-minute diagnostics.

What this unit covers

Outcome	What you must be able to do
5.2.2(h)	Explain the chemiosmotic theory: the electron transport chain, proton gradients and ATP synthase, in oxidative phosphorylation and photophosphorylation
5.2.2(i)(i)	Describe anaerobic respiration in mammals and yeast, the benefits of respiring anaerobically, and why the ATP yield is much lower than aerobic
5.2.2(i)(ii)	Describe practical investigations into respiration rates in yeast under aerobic and anaerobic conditions
5.2.2(j)	Compare the relative energy values of carbohydrates, lipids and proteins as respiratory substrates
5.2.2(k)	Use and interpret the respiratory quotient (RQ), including calculating it
5.2.2(l)	Describe practical investigations into the effect of factors such as temperature, substrate concentration and different respiratory substrates on the rate of respiration, including respirometers

5.2.2(h) The chemiosmotic theory

The theory says that the energy released as electrons pass along a chain of carriers is used to build a proton (H⁺) gradient across a membrane, and that the flow of protons back down that gradient drives the synthesis of ATP.

In oxidative phosphorylation

Oxidative phosphorylation happens on the inner mitochondrial membrane, folded into cristae for a large surface area.

1. Reduced NAD and reduced FAD, made in glycolysis, the link reaction and the Krebs cycle, give up their electrons and hydrogen to the first carriers of the electron transport chain. NAD and FAD are regenerated.
2. The electrons pass from carrier to carrier, each step to a lower energy level. Energy released at some steps is used by proton pumps to move H⁺ from the matrix into the space between the two membranes (the intermembrane space).
3. This builds an electrochemical gradient: a higher concentration of H⁺ outside the matrix, and a charge difference across the inner membrane. The inner membrane is impermeable to H⁺.
4. H⁺ flows back into the matrix down its gradient, but only through channels in ATP synthase. The flow turns part of the enzyme and makes ADP and inorganic phosphate join to form ATP.
5. At the end of the chain, oxygen is the final electron acceptor. In the matrix, it combines with electrons and H⁺ to form water.

Without oxygen, the last carrier cannot pass its electrons on, the chain backs up, reduced NAD and FAD are not reoxidised, and chemiosmosis stops.

In photophosphorylation

The same principle works in thylakoid membranes. Light excites electrons in chlorophyll; they pass along an electron transport chain and the energy released pumps H⁺ from the stroma into the thylakoid space. H⁺ then flows back to the stroma through ATP synthase, and ATP is made from ADP and inorganic phosphate.

Feature	Oxidative phosphorylation	Photophosphorylation
Membrane	Inner mitochondrial membrane	Thylakoid membrane
Source of energy for electrons	Reduced NAD and reduced FAD	Light, absorbed by chlorophyll
H⁺ pumped from	Matrix	Stroma
H⁺ accumulates in	Intermembrane space	Thylakoid space
H⁺ flows back through ATP synthase into	Matrix	Stroma
Final electron acceptor	Oxygen	NADP in non-cyclic; none in cyclic

Worked example: size of a proton gradient

Data supplied: the pH of the intermembrane space is 7.0 and the pH of the matrix is 7.8. Calculate how many times greater the H⁺ concentration is outside the matrix.

pH is a log scale, so a difference of 0.8 pH units is a ratio of 10 to the power 0.8. 10^0.8 = 6.3. The H⁺ concentration in the intermembrane space is about 6.3 times that in the matrix.

5.2.2(i) Anaerobic respiration in eukaryotes

Anaerobic respiration is glycolysis followed by a step that regenerates NAD, with no link reaction, Krebs cycle or oxidative phosphorylation. Glycolysis gives 2 pyruvate, a net 2 ATP and 2 reduced NAD. Glycolysis needs NAD, and the cell has a small supply. The fermentation step turns reduced NAD back into NAD.

In mammals: lactate

Pyruvate accepts hydrogen from reduced NAD and becomes lactate, catalysed by lactate dehydrogenase. NAD is regenerated, so glycolysis goes on making ATP. Lactate is carried in the blood to the liver, where it can be converted back to pyruvate and then glucose, or oxidised when oxygen is available. Lactate build-up lowers pH and contributes to muscle fatigue.

In yeast: ethanol and carbon dioxide

Pyruvate is decarboxylated to ethanal, releasing carbon dioxide. Ethanal accepts hydrogen from reduced NAD and becomes ethanol, regenerating NAD. Ethanol formation is not reversible.

Benefits of respiring anaerobically

• Muscle can keep producing ATP at a high rate during short bursts of intense activity, when oxygen supply cannot keep up.
• Yeast and other microorganisms can grow where oxygen is scarce.

Why the ATP yield is much lower

The products, lactate or ethanol, still contain much of the energy of glucose. Reduced NAD is used to make those products, so it is not passed to the electron transport chain, and no proton gradient is built from it. Only the 2 ATP from glycolysis are made, compared with about 30 per glucose in aerobic respiration (textbooks give figures between about 30 and 32).

Worked example. Using 2 and 30 ATP per glucose, anaerobic respiration gives 2 ÷ 30 × 100 = 6.7% of the aerobic yield. The same ATP needs about 15 times as much glucose (30 ÷ 2).

Yeast practical investigations (5.2.2(i)(ii))

Yeast is grown in a glucose solution, and the rate of respiration is found from the carbon dioxide made.

• Aerobic conditions: the suspension is left open to air, or air is bubbled through it.
• Anaerobic conditions: a layer of oil or liquid paraffin on the surface keeps oxygen out. Boiled and cooled solution has less dissolved oxygen.
• Measuring the rate: collect the gas in a gas syringe or over water, count bubbles in a set time, or use a redox indicator such as methylene blue, which loses its colour faster when the yeast respires faster.
• Controls: same mass of yeast, temperature (water bath) and glucose concentration, plus a tube of boiled yeast.

Worked example: rate from data. Data supplied for anaerobic yeast: carbon dioxide collected is 0 cm³ at 0 min, 1.8 at 2, 3.4 at 4, 4.6 at 6 and 5.2 at 8. Mean rate from 0 to 4 min = 3.4 ÷ 4 = 0.85 cm³ min⁻¹. Mean rate from 6 to 8 min = (5.2 − 4.6) ÷ 2 = 0.30 cm³ min⁻¹. The rate falls because glucose is used up and ethanol accumulates.

5.2.2(j) Relative energy values of substrates

Respiration can use carbohydrate, lipid and protein. The energy released per gram differs.

Substrate	Approximate energy per gram	Entry to respiration
Carbohydrate	about 17 kJ g⁻¹	Glucose goes into glycolysis
Lipid	about 37 kJ g⁻¹	Glycerol is converted to triose phosphate and enters glycolysis. Fatty acids are broken into two-carbon units that form acetyl CoA
Protein	about 17 kJ g⁻¹	Amino acids are deaminated; the carbon skeleton enters as pyruvate, acetyl CoA or a Krebs cycle intermediate

These are approximate; use values given in a question. Lipids give more than twice the energy per gram because their molecules have many more hydrogen atoms for each oxygen atom. More hydrogen means more reduced NAD and reduced FAD per gram, more H⁺ pumped across the inner membrane and more ATP made. Protein is used mainly when carbohydrate and lipid are short, as in starvation; deamination produces ammonia, converted to urea in the liver.

Worked example. Data supplied: carbohydrate 17 kJ g⁻¹, lipid 37 kJ g⁻¹. What mass of lipid gives the same energy as 100 g of carbohydrate? Energy = 100 × 17 = 1700 kJ. Mass of lipid = 1700 ÷ 37 = 46 g (2 s.f.).

5.2.2(k) Respiratory quotient

The respiratory quotient is the ratio of carbon dioxide produced to oxygen consumed, over the same period.

RQ = CO₂ produced ÷ O₂ consumed

It has no units. Calculate it from an equation (moles, or gas volumes at the same temperature and pressure) or from measurements.

Substrate	Typical RQ
Carbohydrate	about 1.0
Protein	about 0.9
Lipid	about 0.7

Lipids contain little oxygen, so more oxygen is needed to oxidise them, which lowers the RQ.

Worked example 1: from an equation. Oleic acid is C₁₈H₃₄O₂. Two molecules react with oxygen: 2C₁₈H₃₄O₂ + 51O₂ → 36CO₂ + 34H₂O. RQ = 36 ÷ 51 = 0.71, close to the typical lipid value.

Worked example 2: malic acid. C₄H₆O₅ + 3O₂ → 4CO₂ + 3H₂O. RQ = 4 ÷ 3 = 1.3. A respiratory substrate that is an organic acid can give an RQ above 1.

Interpreting RQ

• About 1.0: mainly carbohydrate.
• About 0.7: mainly lipid. About 0.9: mainly protein, or a mixture.
• Between 0.7 and 1.0: a mixture of substrates.
• Above 1.0: some anaerobic respiration, or an organic acid as substrate.
• Below 0.7: can happen when lipid is being converted to carbohydrate.
• Yeast respiring only anaerobically takes in no oxygen, so RQ cannot be calculated.

5.2.2(l) Respirometers and rate investigations

A respirometer measures the oxygen taken in by living material (germinating seeds, small invertebrates, yeast) in a sealed tube.

Set-up. The organisms sit on gauze above a carbon dioxide absorber, such as soda lime or potassium hydroxide solution. The tube is joined to a capillary tube containing a drop of coloured liquid, and the apparatus is in a water bath. As oxygen is used, carbon dioxide is absorbed, so gas volume falls and the drop moves toward the organisms.

Method.

1. Set up a control tube of the same volume with glass beads or dead organisms.
2. Equilibrate the apparatus in the water bath before sealing, and check the joints are airtight.
3. Record the drop’s position at the start and at regular intervals.
4. Repeat to calculate a mean.
5. Handle potassium hydroxide with care: it is corrosive, so wear eye protection, and keep it from touching the organisms.

Calculating the rate. The volume of oxygen used = π r² × distance moved by the drop, where r is the radius of the capillary tube.

Worked example. Capillary diameter = 1.0 mm, so r = 0.5 mm. The drop moves 34 mm in 10 minutes for 2.5 g of germinating seeds. Volume = π × 0.5² × 34 = 26.7 mm³. Rate = 26.7 ÷ 10 = 2.67 mm³ min⁻¹. Rate per gram = 2.67 ÷ 2.5 = 1.1 mm³ g⁻¹ min⁻¹.

Measuring RQ. Run one tube with the absorber and one without.

• With the absorber, the drop moves because of oxygen use only.
• Without it, carbon dioxide made replaces some of the oxygen used, so the drop moves by (oxygen used − carbon dioxide produced).
• Carbon dioxide produced = movement with absorber − movement without.

Worked example. Same capillary, same organisms. With the absorber the drop moves 46 mm. Without it, 11 mm. Carbon dioxide produced is proportional to 46 − 11 = 35 mm. RQ = 35 ÷ 46 = 0.76. The capillary area cancels, because both volumes use the same area.

Factors you can investigate:

• Temperature: water baths at different temperatures. Rate rises with enzyme activity, then falls as enzymes denature.
• Substrate concentration: vary glucose concentration for yeast. Rate rises, then levels off when enzymes or transport are saturated.
• Different substrates: give yeast different sugars, or compare starch-rich and oil-rich seeds using RQ.

Common errors

• Saying oxygen is used in the Krebs cycle. It is used at the end of the electron transport chain.
• Saying electrons, not protons, flow through ATP synthase.
• Writing RQ upside down.
• Using diameter instead of radius in π r².

Official syllabus

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