Anaerobic Respiration, Substrates & RQ
When oxygen is absent, the four-stage aerobic pathway shuts down — but cells can still extract small amounts of ATP through fermentation, which regenerates NAD to keep glycolysis turning. Lactate (mammals), ethanol (yeast and plant roots), and the rice plant's specific adaptations to flooded soils. Beyond glucose: lipids and proteins as alternative substrates — differing in their hydrogen content, their energy yield per gram, and the ratio of CO₂ produced to O₂ consumed (the respiratory quotient). Respirometers and redox indicators give the experimental tools to measure all of it.
Anaerobic respiration
When oxygen is unavailable — vigorously contracting muscle, yeast in a sealed brewing vessel, plant roots in flooded soil — the electron transport chain stops, the link reaction and Krebs cycle halt, and aerobic respiration cannot proceed. But cells must still produce some ATP. Anaerobic respiration uses only the first stage of respiration — glycolysis — and converts pyruvate to a different end-product, simply to regenerate NAD so glycolysis can continue.
Why glycolysis must continue — the NAD problem
Glycolysis (Topic 12A) requires NAD to accept hydrogens during the oxidation of triose phosphate. Each glucose produces 2 reduced NAD by the time pyruvate is formed. In aerobic conditions, these reduced NAD molecules are re-oxidised on the inner mitochondrial membrane by donating electrons to the ETC. In anaerobic conditions, the ETC is shut down — reduced NAD piles up, NAD becomes scarce, and glycolysis stops.
Fermentation is NOT primarily a way to produce ATP — it is a way to regenerate NAD. The "purpose" of converting pyruvate to lactate or ethanol is to dispose of the hydrogens carried by reduced NAD, so that NAD becomes available for the next round of glycolysis. The small ATP yield (just the 2 net ATP from glycolysis itself) is a side benefit.
Lactate fermentation — mammalian muscle
In mammalian muscle cells under intense activity (when O₂ supply cannot meet demand), pyruvate is converted directly to lactate:
- Glycolysis proceeds normally in the cytoplasm: glucose → 2 pyruvate, with net 2 ATP and 2 reduced NAD per glucose
- Pyruvate stays in the cytoplasm (does NOT enter mitochondria)
- Reduced NAD transfers its hydrogens to pyruvate, catalysed by lactate dehydrogenase
- Pyruvate (3C) + reduced NAD → lactate (3C) + NAD
- The regenerated NAD returns to glycolysis to accept hydrogens from the next round — allowing continued ATP production
Net yield per glucose: 2 ATP, 0 reduced NAD remaining (all are recycled), 2 lactate
Lactate accumulating in muscle is reversibly toxic and contributes to the burning sensation of intense exercise. After exercise, blood transports lactate to the liver, where it is converted back to pyruvate and either oxidised aerobically or used to re-synthesise glucose (gluconeogenesis). The oxygen consumed during this recovery is the oxygen debt — explaining elevated breathing and heart rate after vigorous activity. Lactate fermentation is therefore reversible in mammals.
Ethanol fermentation — yeast and plant cells
In yeast and in some plant tissues (e.g. roots in waterlogged soil), pyruvate is converted to ethanol in two enzyme-catalysed steps:
- Glycolysis proceeds normally: glucose → 2 pyruvate, with net 2 ATP and 2 reduced NAD per glucose
- Step 1 — decarboxylation: pyruvate (3C) loses CO₂ (catalysed by pyruvate decarboxylase), forming ethanal (2C) — sometimes called acetaldehyde
- Step 2 — reduction: ethanal accepts hydrogens from reduced NAD (catalysed by ethanol dehydrogenase), forming ethanol (2C); NAD is regenerated
- Regenerated NAD returns to glycolysis
Per glucose: 2 ATP, 2 ethanol (2C each), 2 CO₂ released, 0 reduced NAD remaining
- Brewing and winemaking: yeast ferments sugars from grain or grapes; ethanol is the product, CO₂ is released
- Bread baking: yeast in dough produces CO₂ that creates air pockets (bread rises); ethanol evaporates during baking
- Ethanol is toxic above ~14% concentration — this limits the alcohol content of naturally fermented beverages; yeast cells die when ethanol accumulates beyond their tolerance
- Ethanol fermentation is irreversible in plants — ethanol cannot be converted back to pyruvate, so it must be excreted or tolerated
Aerobic vs anaerobic — why aerobic is so much more productive
| Feature | Aerobic respiration | Anaerobic respiration |
|---|---|---|
| Oxygen needed? | Yes (final electron acceptor) | No |
| Stages used | All four (glycolysis + link + Krebs + oxidative phosphorylation) | Glycolysis only |
| Mechanism of ATP synthesis | Substrate-linked + chemiosmotic (the latter dominates) | Substrate-linked only |
| End product of glucose | CO₂ and H₂O (complete oxidation) | Lactate (mammals) OR ethanol + CO₂ (yeast/plants) — energy still trapped |
| ATP yield per glucose | Much greater — the majority comes from chemiosmotic mechanism on inner mitochondrial membrane | 2 ATP (glycolysis only) |
| Site | Cytoplasm + mitochondria | Cytoplasm only |
The 9700 syllabus asks you to explain the difference, not simply state numbers. The reasoning:
- Anaerobic respiration uses only glycolysis — just 2 ATP per glucose by substrate-linked phosphorylation
- The link reaction and Krebs cycle do not run, so additional reduced NAD and FAD are not produced — the chemiosmotic mechanism cannot operate
- Glucose is not fully oxidised — lactate, ethanol, and CO₂ still contain considerable chemical energy that has not been captured as ATP
- Without O₂ as the final electron acceptor, even the existing reduced NAD (from glycolysis) cannot be used to drive proton pumping
Note on language: the syllabus explicitly states "a detailed account of the total yield of ATP from the aerobic respiration of glucose is not expected" — so don't quote 30 or 32 ATP. Comparison should be qualitative and mechanistic.
Rice plants — adapted for life in flooded soil
Rice (Oryza sativa) is grown in flooded paddies. The roots are submerged in waterlogged, oxygen-poor soil. The 9700 syllabus specifies three adaptations that allow rice to survive these anaerobic conditions:
Aerenchyma is a tissue containing large, interconnected air-filled spaces running the length of stems and roots. Oxygen diffuses from the leaves and stem (above water) down through the aerenchyma to the submerged root cells, providing some O₂ for aerobic respiration despite the waterlogged environment.
When O₂ supply via aerenchyma is insufficient, root cells switch to ethanol fermentation. Rice tissues tolerate ethanol better than most plant species, allowing the roots to continue producing ATP from glycolysis while submerged. Ethanol is excreted into the surrounding water.
When water levels rise, rice plants respond with rapid elongation of the stem and leaves — this lifts the photosynthetic and gas-exchanging tissues above the water surface, restoring access to atmospheric O₂ and avoiding prolonged anoxic conditions. This is achieved through hormone-driven cell elongation in the internodes.
Rice is the only specific named example in the 9700 syllabus for plant adaptations to anaerobic conditions. Memorise all three adaptations — aerenchyma + ethanol fermentation + faster stem growth. A "discuss adaptations" question on rice is straightforward marks if you have these three points. Conversely, mentioning only "ethanol fermentation" without aerenchyma and stem growth costs two-thirds of the available marks.
Why must pyruvate be converted to lactate in vigorously contracting muscle?
- A. Lactate is a more efficient energy store than pyruvate
- B. Pyruvate cannot enter the mitochondria when O₂ is low
- C. The conversion regenerates NAD, allowing glycolysis to continue producing ATP
- D. Lactate is needed by the Krebs cycle as a starting substrate
Rice plants are grown in flooded paddies where the roots are submerged in waterlogged soil.
(a) Explain why submerged rice roots cannot rely on normal aerobic respiration. [2]
(b) Describe THREE adaptations that allow rice plants to survive these conditions. [3]
(c) Explain why the energy yield from anaerobic respiration in rice roots is much less than the energy yield from aerobic respiration in rice leaves. [4]
(a) Why aerobic respiration is limited in waterlogged roots [2 marks]
- Waterlogged soil contains very little dissolved O₂ available to root cells [1]
- Without O₂ as the final electron acceptor, the electron transport chain stops, so the link reaction, Krebs cycle, and oxidative phosphorylation cannot proceed [1]
(b) Three rice adaptations [3 marks]
- Aerenchyma in roots (and stems) — air-filled spaces allow O₂ to diffuse down from leaves/stems above water to submerged roots [1]
- Ethanol fermentation in root cells when O₂ supply is still insufficient — allows ATP production via glycolysis; rice tolerates ethanol relatively well [1]
- Faster growth (elongation) of stems and leaves when water levels rise — lifts photosynthetic tissue above water, restoring atmospheric O₂ access [1]
(c) Why anaerobic yield is much less [4 marks]
- Anaerobic respiration uses only glycolysis — producing only 2 ATP per glucose by substrate-linked phosphorylation [1]
- The link reaction and Krebs cycle cannot operate without O₂ — so additional reduced NAD and FAD are not produced [1]
- Without reduced NAD/FAD donating electrons to the ETC, no proton gradient is built across the inner mitochondrial membrane — chemiosmotic ATP synthesis cannot occur [1]
- Aerobic respiration captures additional ATP from chemiosmotic phosphorylation, which produces the majority of cellular ATP; ethanol and CO₂ produced by fermentation still contain considerable chemical energy that is not captured [1]
Substrates, RQ & respirometer practicals
Glucose is the textbook respiratory substrate — but cells can also respire lipids and (when other supplies are exhausted) proteins. The amount of energy released per gram and the ratio of CO₂ produced to O₂ consumed differ between substrates — differences that can be measured experimentally with a respirometer to identify which substrate an organism is using.
Three respiratory substrates
All three substrate classes ultimately enter the respiratory pathway, but at different points and with different energy yields:
Glucose enters glycolysis directly. Glycogen (animal liver/muscle) and starch (plants) are storage carbohydrates that are mobilised first when energy is needed. Carbohydrates are fastest to mobilise — less metabolic processing required — so they are the body's default rapid-energy source.
Lipids (triglycerides) are hydrolysed to fatty acids and glycerol. Glycerol enters glycolysis; fatty acids are broken down to 2C acetyl groups (a process called β-oxidation, not assessable in detail) which feed directly into the Krebs cycle as acetyl-CoA. Used as a long-term energy store.
Used only when carbohydrate and lipid reserves are inadequate (e.g. starvation, prolonged exercise). Amino acids are deaminated (NH₂ group removed and excreted as urea, Topic 14) and the remaining carbon skeleton enters the pathway at various points (often as pyruvate or as a Krebs intermediate). Catabolising body protein progressively damages muscle and organ tissue.
Relative energy values — lipids highest per gram
The 9700 syllabus requires you to explain why lipids release more energy per gram than carbohydrates or proteins. The reasoning depends on the hydrogen content of the substrate:
- The energy harvested in respiration ultimately comes from hydrogen atoms — their electrons are passed to NAD/FAD, then to the ETC, where they drive chemiosmotic ATP synthesis
- The more H atoms a molecule contains per gram, the more reduced NAD and reduced FAD are produced — and the more ATP can be made by oxidative phosphorylation
- Lipids are highly reduced (lots of C–H bonds, very few C–O bonds) — high H content per gram
- Carbohydrates are partially oxidised already (every other C is bonded to O, as in –CH–OH) — medium H content per gram
- Proteins are intermediate — H content varies with the amino acid composition
Approximate energy values (kJ per gram):
- Lipids: ~37 kJ g⁻¹ — the highest, ~2× carbohydrate
- Proteins: ~17 kJ g⁻¹ — intermediate
- Carbohydrates: ~16 kJ g⁻¹ — lowest
Lipids store roughly twice as much energy per gram as glycogen or starch. They also store without water — carbohydrate stores are hydrated (glycogen attracts ~3× its mass in water in living cells), so the practical energy density of lipid storage is much higher. A small mammal with body fat reserves can survive weeks without food; equivalent mass of glycogen would last days.
Trade-off: lipids are slower to mobilise (lipase digestion + β-oxidation + transport), so they are not suitable as the rapid-energy substrate for short bursts of activity. Hence the metabolic strategy: glycogen first, lipids second, protein last.
The respiratory quotient (RQ)
The respiratory quotient (RQ) is the ratio:
RQ = molecules of CO₂ produced ÷ molecules of O₂ consumed
(Equivalent to volume of CO₂ produced / volume of O₂ consumed, since at constant temperature and pressure equal volumes of gas contain equal numbers of molecules.)
RQ varies depending on the respiratory substrate, because different substrates require different amounts of O₂ for complete oxidation and produce different amounts of CO₂ per molecule respired.
RQ calculation from respiration equations
The 9700 syllabus requires you to calculate RQ values from respiration equations. Three classic worked examples:
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
RQ = 6 CO₂ ÷ 6 O₂ = 1.00
All carbohydrate respiration gives RQ = 1.0 because the molecule already contains enough internal O to balance C and H exactly.
C₅₇H₁₀₄O₆ + 80 O₂ → 57 CO₂ + 52 H₂O
RQ = 57 CO₂ ÷ 80 O₂ = 0.71
Lipid respiration gives RQ ≈ 0.7 because lipids are very reduced (lots of H atoms, very little internal O); a lot of external O₂ is needed to oxidise all those C and H atoms, but the amount of CO₂ released is limited by the carbon count alone.
The exact equation depends on the amino acid composition, but a typical protein respiration equation looks like:
2 C₃H₇NO₂ + 6 O₂ → (NH₂)₂CO + 5 CO₂ + 5 H₂O
RQ values for proteins typically range from 0.8 to 0.9 depending on amino acid composition.
RQ — substrate identification table
| Respiratory state | Typical RQ | Interpretation |
|---|---|---|
| Pure carbohydrate respiration | 1.0 | Complete oxidation of glucose; equal molecules CO₂ and O₂ |
| Pure lipid respiration | ~0.7 | Lipids highly reduced; need much O₂ to oxidise all H |
| Pure protein respiration | 0.8–0.9 | Intermediate, varies with amino acid composition |
| Mixed substrate (typical mammal at rest) | ~0.85 | Suggests roughly equal use of carbohydrate and lipid |
| Anaerobic respiration in yeast | Not normally expressed | CO₂ produced (from ethanol fermentation) but O₂ consumed = 0; RQ is not normally used in this context. If treated mathematically, CO₂/O₂ would be extremely high / undefined. |
| Anaerobic respiration in mammals | Cannot be calculated | No O₂ consumed AND no CO₂ produced (lactate fermentation); both numerator and denominator are zero |
| RQ > 1 in aerobic conditions | >1.0 | Indicates anaerobic respiration is occurring alongside aerobic; CO₂ from fermentation adds to aerobic CO₂ without extra O₂ consumption |
Respirometer practical — measuring O₂ consumption and RQ
A respirometer measures the rate at which an organism (germinating seeds, small invertebrates such as woodlice, blowfly larvae, etc.) consumes O₂ and/or produces CO₂. Combined with an alkali to absorb CO₂, it can also be used to determine RQ.
- A sealed chamber containing the test organism
- A capillary tube connected to the chamber, containing a small drop of coloured fluid (the manometer fluid) — movement of this drop indicates volume change in the apparatus
- A tap or three-way valve allowing the system to be reset between readings
- A water bath at constant temperature — gas volume changes with temperature would otherwise dominate any biological signal
- A control tube (no organism, otherwise identical conditions) to detect any non-biological volume changes
- Place an alkali — KOH (or sodium hydroxide, or soda lime) — in the chamber alongside the organism, separated from the organism by a wire mesh
- The alkali absorbs all CO₂ produced by the organism: 2 KOH + CO₂ → K₂CO₃ + H₂O
- Therefore, any decrease in gas volume in the chamber is due to O₂ consumption alone (CO₂ produced is removed; only O₂ reduction is detected)
- Track the movement of the manometer drop over a fixed time interval; calculate rate of O₂ consumption = volume change / time
To find RQ, run the same organism in two respirometers under identical conditions, differing only in the use of alkali:
- Tube 1 (with KOH): gas volume change = O₂ consumed (CO₂ absorbed by KOH)
- Tube 2 (no KOH; water instead): gas volume change = O₂ consumed − CO₂ produced (the net change — CO₂ produced partially offsets O₂ consumed)
- Calculate CO₂ produced = (volume change in Tube 1) − (volume change in Tube 2)
- Calculate RQ = CO₂ produced / O₂ consumed
This two-tube comparison turns the respirometer into a substrate-identification tool.
A respirometer with germinating seeds gives the following readings over 10 minutes:
- Tube 1 (with KOH): manometer drop moves a distance corresponding to 5.0 cm³ reduction in volume
- Tube 2 (no KOH): manometer drop moves a distance corresponding to 1.5 cm³ reduction in volume
O₂ consumed = 5.0 cm³ (from Tube 1)
CO₂ produced = 5.0 − 1.5 = 3.5 cm³
RQ = 3.5 / 5.0 = 0.70 — suggesting predominant lipid respiration (consistent with germinating oily seeds e.g. sunflower or maize germinating with stored lipid reserves)
The same apparatus is used to investigate how temperature affects respiration. Place the respirometer in water baths at different temperatures (e.g. 10, 20, 30, 40 °C); allow equilibration; record the rate of manometer-drop movement. Plot rate vs temperature.
Expected pattern: rate increases with temperature up to ~35–40 °C (Q₁₀ effect on enzyme activity); then drops sharply as enzymes denature. The shape of the curve depends on the species — ectotherms typically show a steeper response than endotherms whose body temperature is regulated.
Redox indicators — DCPIP and methylene blue for yeast respiration
Direct measurement of O₂ or CO₂ in yeast suspensions is awkward at small scale. The 9700 syllabus offers an alternative: redox indicators — coloured dyes that change colour when reduced.
- The yeast cell respires substrate; reduced NAD and reduced FAD accumulate
- A redox indicator (an artificial alternative electron acceptor) added to the mixture intercepts electrons that would otherwise go to O₂ or to fermentation
- The indicator becomes reduced — its colour changes (typically from coloured to colourless, or vice versa)
- The time taken for the colour to change is inversely proportional to the rate of respiration: faster respiration → more reduced coenzymes → faster colour change
Blue when oxidised, colourless when reduced. Add to a yeast suspension with a respiratory substrate (e.g. glucose). Monitor the time for blue colour to disappear — faster decolourisation indicates faster respiration.
Blue when oxidised, colourless when reduced. Used the same way as DCPIP but with slightly different reduction potential. Reversibly oxidised by air at the surface of the suspension — sometimes a layer of oil is added to exclude O₂ and prevent re-oxidation that would mask the result.
- Prepare yeast suspensions at standard concentration with a glucose substrate
- Add a fixed volume of redox indicator (DCPIP or methylene blue) to several test tubes
- Place tubes in water baths at different temperatures (e.g. 20, 30, 40, 50 °C); allow equilibration
- Add yeast to each tube simultaneously; start a timer
- Record the time for colour change (e.g. blue to colourless) at each temperature
- Calculate rate of respiration ≈ 1/time
- Plot rate vs temperature; expected: increasing rate up to optimum, then sharp decline as enzymes denature
- Use yeast suspensions at constant concentration and constant temperature
- Vary the glucose concentration (e.g. 0.05, 0.1, 0.2, 0.5, 1.0 mol dm⁻³)
- Add a fixed volume of redox indicator to each tube
- Record the time for colour change; calculate rate as 1/time
- Plot rate vs substrate concentration: rate rises with concentration up to a saturation level (similar to Topic 3 enzyme kinetics — consistent with rate-limiting enzymes in glycolysis)
An organism's respiratory quotient is measured at 0.7. What is the most likely substrate?
- A. Pure carbohydrate
- B. Pure lipid
- C. Pure protein
- D. Lactate from anaerobic respiration
A student investigates the respiration of germinating peas Both contain identical numbers of peas at 25 °C, but only Tube 1 contains potassium hydroxide solution. After 5 minutes, the manometer fluid moves the following distances:
• Tube 1 (with KOH): drop moves 8.0 cm corresponding to 4.0 cm³ volume reduction
• Tube 2 (no KOH): drop moves 6.0 cm corresponding to 3.0 cm³ volume reduction
(a) Explain why a control tube containing no organisms is essential. [2]
(b) Calculate the rate of O₂ consumption in cm³ min⁻¹. [1]
(c) Calculate the rate of CO₂ production in cm³ min⁻¹. [2]
(d) Calculate the RQ and identify the most likely substrate. [3]
(a) Why a control tube [2 marks]
- The control tube allows correction for non-biological volume changes — e.g. small leaks, ambient temperature fluctuations, atmospheric pressure changes [1]
- Any volume change observed in the control is subtracted from each experimental tube to isolate the change due to the organism alone, increasing accuracy [1]
(b) O₂ consumption [1 mark]
- Tube 1 with KOH absorbs all CO₂ produced — volume reduction = O₂ consumed only
Rate = 4.0 cm³ / 5 min = 0.80 cm³ min⁻¹ [1]
(c) CO₂ production [2 marks]
- Tube 2 (no KOH) volume reduction = O₂ consumed − CO₂ produced
So CO₂ produced = (Tube 1 reduction) − (Tube 2 reduction) = 4.0 − 3.0 = 1.0 cm³ in 5 min [1] - Rate of CO₂ production = 1.0 / 5 = 0.20 cm³ min⁻¹ [1]
(d) RQ and substrate identification [3 marks]
- RQ = CO₂ produced / O₂ consumed = 1.0 / 4.0 = 0.25 [1]
- However, the calculated value is much lower than the typical lipid RQ of 0.7 — this is below biological norms; either the data contain anomaly, the peas are absorbing CO₂ for biosynthesis (e.g. into amino acids during germination), or some CO₂ is being absorbed by the surrounding solution [1]
- Realistic interpretation: a normal value would be RQ ~0.7 (lipid), as germinating oily seeds (e.g. sunflower) respire stored lipids; for typical pea seeds with stored carbohydrate, RQ ≈ 1.0 would be expected. The data either reveal a methodological problem or an unusual respiration pattern; this evaluation point itself can score [1]
Note: Realistic Paper 4 calculation questions usually give cleaner numbers yielding RQ near 0.7, 0.85, or 1.0. This question is constructed to also test ACE-style evaluation of unexpected results.
Topic 12B Practice — Comprehensive
Mixed practice covering anaerobic respiration, substrates, RQ, and respirometer practical work.
Which statement correctly describes ethanol fermentation in yeast?
- A. Pyruvate is reduced directly to ethanol in a single step, producing 2 ATP
- B. Pyruvate is decarboxylated to ethanal, then reduced to ethanol; reduced NAD is regenerated to NAD
- C. Pyruvate enters the mitochondrial matrix where it is converted to ethanol
- D. Glucose is converted directly to ethanol without forming pyruvate
Lipids release approximately twice as much energy per gram as carbohydrates. Which biochemical reason best explains this?
- A. Lipids enter the Krebs cycle directly without going through glycolysis
- B. Lipids contain more carbon atoms per molecule
- C. Lipids contain more hydrogen atoms per gram, producing more reduced NAD/FAD and so more ATP via oxidative phosphorylation
- D. Lipids respire faster than carbohydrates due to higher enzyme affinity
Stearic acid is a fatty acid with the molecular formula C₁₈H₃₆O₂. Its complete oxidation in respiration is described by:
C₁₈H₃₆O₂ + 26 O₂ → 18 CO₂ + 18 H₂O
(a) Calculate the RQ for stearic acid. Show your working. [2]
(b) Compare your calculated value with the typical RQ for carbohydrate respiration (1.0). What does this tell you about the H content of stearic acid relative to glucose? [2]
(a) RQ calculation [2 marks]
- RQ = CO₂ produced / O₂ consumed = 18 / 26 [1]
- = 0.69 (or 0.7 to 1 sf) [1]
(b) Comparison and interpretation [2 marks]
- The RQ for stearic acid (0.69) is much lower than for glucose (1.0) — meaning more O₂ is consumed per CO₂ produced when respiring stearic acid [1]
- This is because stearic acid is more reduced than glucose: it contains many C–H bonds and very little internal oxygen. Carbohydrates already contain a high proportion of oxygen within their structure, so less external O₂ is needed per CO₂ produced. Lipids, with minimal internal oxygen, require far more external O₂ to fully oxidise each carbon, raising O₂ consumption relative to CO₂ output and giving a lower RQ [1]
Key principle: stearic acid has very little oxygen relative to its carbon and hydrogen atoms, so much more external O₂ is needed to oxidise all its hydrogen to water. Glucose already contains substantial oxygen within the molecule (C₅H₁₂O₅), so less external O₂ is needed per CO₂ produced. This difference in internal oxygen — not H:C ratio — is why lipid RQ is lower than carbohydrate RQ.
Rice (Oryza sativa) is grown commercially in flooded paddies where the roots are submerged in water for much of the growing season.
(a) Describe the role of aerenchyma in rice plants. [2]
(b) Explain how ethanol fermentation in rice root cells differs from aerobic respiration in terms of (i) ATP yield and (ii) end products. [4]
(c) Explain why rice plants exhibit faster stem growth when water levels rise. [2]
(a) Aerenchyma role [2 marks]
- Aerenchyma is a tissue containing large, interconnected air-filled spaces running through stems and roots [1]
- Allows oxygen to diffuse from photosynthetic tissues (above water) down through the stem to the submerged root cells, supplying some O₂ for aerobic respiration despite the waterlogged soil [1]
(b) Ethanol fermentation vs aerobic [4 marks]
- (i) ATP yield: ethanol fermentation produces only 2 ATP per glucose (from glycolysis only); aerobic respiration produces much more ATP — the majority via chemiosmotic phosphorylation in the mitochondrion [2]
- (ii) End products: ethanol fermentation produces ethanol (2C) and CO₂ per glucose, with no water; aerobic respiration produces only CO₂ and water (with no organic end-product retaining energy) [2]
(c) Faster stem growth [2 marks]
- Rapid stem elongation (driven by hormone-stimulated cell elongation in internodes) lifts the photosynthetic and gas-exchanging leaves above the water surface [1]
- This restores access to atmospheric O₂ for the leaves and provides O₂ via aerenchyma to the roots; reduces dependence on inefficient ethanol fermentation [1]
A student plans to investigate the effect of glucose concentration on the rate of yeast respiration using methylene blue as a redox indicator.
(a) State the role of methylene blue in this experiment. [2]
(b) State THREE controlled variables and how each should be controlled. [3]
(c) Suggest TWO improvements that would increase the reliability of the data. [2]
(a) Methylene blue's role [2 marks]
- Methylene blue is a redox indicator — an artificial electron acceptor that intercepts electrons from reduced NAD/FAD produced during respiration [1]
- Methylene blue is blue when oxidised; it becomes colourless when reduced; the time taken for the colour to disappear is inversely proportional to the rate of respiration [1]
(b) Controlled variables [3 marks]
- Temperature — held constant in a thermostatically controlled water bath at a defined value (e.g. 30 °C) [1]
- Yeast concentration — same volume of yeast suspension at the same starting concentration in each tube [1]
- Volume of methylene blue solution — same volume and same starting concentration in each tube; same amount of mixing [1]
- (Other accepted answers: pH of the suspension, time/duration of equilibration before adding yeast, volume of glucose solution to maintain total volume across concentrations) [1]
(c) Improvements [2 marks; any two]
- Use a layer of oil on top of the yeast suspension — prevents O₂ from re-oxidising methylene blue from atmosphere, ensuring colour change reflects only yeast respiration [1]
- Use a colorimeter rather than judging colour change by eye — gives quantitative measurement of decolourisation, eliminating subjective judgement of end-point [1]
- Use multiple replicates at each glucose concentration to identify anomalies and calculate means [1]
- Use a wider range or finer intervals of glucose concentrations to map the rate-vs-concentration curve more precisely [1]
Topic 12B — Common Mistakes
- 🥵Saying "lactate fermentation produces ATP from pyruvate"Wrong. The conversion of pyruvate to lactate does NOT produce ATP — it just regenerates NAD. The 2 ATP per glucose come entirely from glycolysis (which precedes fermentation). The function of lactate fermentation is purely to recycle NAD, NOT to make ATP itself.
- 🍤Saying ethanol fermentation produces "ethanol and water"Ethanol fermentation produces ethanol AND CO₂, not water. The CO₂ comes from the decarboxylation step (pyruvate → ethanal). The CO₂ is what makes bread rise and gives sparkling drinks their bubbles. Forgetting CO₂ loses marks.
- 🌉Confusing reversibility: lactate fermentation IS reversible; ethanol fermentation is NOTMammals can convert lactate back to pyruvate in the liver (gluconeogenesis or oxidation), repaying the oxygen debt. Plants and yeast cannot reverse ethanol fermentation — ethanol must be excreted. Don't claim the reverse.
- 🍺Saying "yeast respires anaerobically and aerobically"Imprecise. Yeast can do both, but the choice depends on conditions: with O₂ available, yeast respires aerobically; without O₂, yeast switches to ethanol fermentation. The same cell can do both at different times, but not simultaneously to a meaningful degree.
- 🥯Mentioning only ethanol fermentation when discussing rice adaptationThe 9700 syllabus specifies THREE adaptations: aerenchyma, ethanol fermentation, faster stem growth. Mentioning only ethanol fermentation typically loses two-thirds of the marks. Memorise all three; recognise that the syllabus wording is "limited to" these three — nothing else is required.
- 🍔Saying "lipids respire faster than carbohydrates"Reversed. Carbohydrates are FASTER to mobilise (glycolysis can use glucose immediately); lipids are slower (need lipase digestion + β-oxidation transport). What lipids offer is energy density per gram, not speed. Hence the metabolic strategy: carbohydrates for short bursts, lipids for endurance.
- 🧐Defining RQ as "O₂ consumed / CO₂ produced"The definition is the OTHER way around: RQ = CO₂ produced / O₂ consumed. Putting it backwards gives reciprocal values: lipid would appear as 1.4 instead of 0.7. Memorise the order: CO₂ on top.
- 🧬Saying "RQ > 1 means anaerobic respiration"Imprecise. RQ > 1.0 means CO₂ is being produced WITHOUT additional O₂ consumption — consistent with anaerobic respiration occurring alongside aerobic. It does NOT mean fully anaerobic. Pure anaerobic in mammals (lactate) gives no measurable RQ at all (no O₂ or CO₂ from fermentation). Pure anaerobic in yeast (ethanol) produces CO₂ but no O₂ is consumed — RQ is not normally expressed in this context; mathematically it would be undefined/extremely high.
- 🧸Forgetting the KOH role in respirometerKOH (or any strong alkali) absorbs CO₂ produced by the organism. Therefore the volume reduction in a respirometer with KOH = O₂ consumed only (CO₂ is removed before it can offset the volume change). WITHOUT KOH, the volume change reflects O₂ consumed minus CO₂ produced (net change). The two-tube method gives both values.
- ⚙Forgetting the control tube in respirometer experimentsA control tube (no organism, otherwise identical) is essential to correct for non-biological volume changes — ambient temperature drift, atmospheric pressure changes, small leaks. Without it, you cannot tell if observed volume changes are due to the organism or to environmental factors. Examiner-flagged frequent loss.
- 🎱Treating DCPIP and methylene blue as "stains"They are redox indicators, not stains. They change colour when reduced (gain electrons), not because they bind to anything. A stain (e.g. iodine for starch) sticks to the molecule it identifies. A redox indicator participates in the chemistry — intercepting electrons that would otherwise go to NAD or O₂.
- ❗Quoting "32 ATP" or "30 ATP" as the aerobic totalRepeating from Topic 12A: the 9700 (2025-2027) syllabus does NOT require a specific aerobic ATP total. Use "the majority of cellular ATP from oxidative phosphorylation" instead. This applies to anaerobic comparison answers too — describe aerobic as producing "much more" ATP without quoting numbers.
Topic 12B completes Topic 12 by adding the anaerobic pathway, alternative substrates, RQ, and the experimental tools to measure them. Highest-yield items: both fermentation pathways with regenerated NAD as the central purpose, the three rice adaptations (aerenchyma + ethanol fermentation + faster stem growth), the relative-energy explanation via H content, RQ definition CO₂/O₂, RQ values for the three substrates (1.0 / ~0.7 / 0.8–0.9), respirometer with KOH for O₂-only and the two-tube method for RQ, and DCPIP/methylene blue as redox indicators (not stains) for yeast respiration rate. Calculation questions on RQ from respiration equations are very common — practise reading off coefficients of CO₂ and O₂ in given equations. Synoptic links to Topic 12A (aerobic), Topic 14 (deamination of amino acids), Topic 7 (transport of glucose).