Respirometer with germinating seeds: malonate, RQ and a dead-seed control
Respirometry with germinating seeds. A student uses a closed manometric respirometer at 20 °C. Each vessel contains 5.0 g of germinating pea seeds (starchy) and, for the O₂ readings, a CO₂ absorber in the side arm; a parallel vessel without absorber allows CO₂ to be measured. Readings are taken over 10 min.
| Condition | O₂ used (mL) | CO₂ released (mL) |
|---|---|---|
| no inhibitor | 1.50 | 1.47 |
| 10 mmol/L malonate | 0.45 | 0.44 |
| 10 mmol/L malonate + 20 mmol/L succinate | 1.42 | 1.39 |
| seeds boiled 10 min | 0.03 | 0.02 |
A second run with 5.0 g of germinating sunflower seeds (lipid-rich) gives 1.60 mL O₂ used and 1.15 mL CO₂ released in 10 min. Malonate is a structural analogue of succinate. The student writes in her report: “Malonate destroys the enzyme, which is why respiration stops.”
1. Calculate the O₂ consumption rate in µL min⁻¹ g⁻¹ fresh mass for the control and for the malonate condition, and state the percentage inhibition. Show your working.
Read the worked answer
Working step by step avoids the usual unit trap:
- Control: 1.50 mL per 10 min = 0.150 mL min⁻¹ for the whole 5.0 g sample.
- Per gram: 0.150 ÷ 5.0 = 0.030 mL min⁻¹ g⁻¹ = 30 µL min⁻¹ g⁻¹.
- Malonate: 0.45 mL per 10 min = 0.045 mL min⁻¹; ÷ 5.0 = 0.009 mL min⁻¹ g⁻¹ = 9 µL min⁻¹ g⁻¹.
- Inhibition: (30 − 9)/30 = 0.70, i.e. 70 %.
For orientation: the boiled control gives 0.03 mL per 10 min = 0.003 mL min⁻¹ → 0.6 µL min⁻¹ g⁻¹, about 2 % of the control. This is not respiration but leakage or thermal drift, and it defines the detection limit against which the 9 µL min⁻¹ g⁻¹ must be judged - malonate still leaves 15 times the blank, so respiration is slowed, not stopped.
A strong answer also notes that the rate is an average over 10 min; if the inhibitor acts gradually the initial rate would be higher and the inhibition at a given moment is underestimated. Repeating with readings every 2 min would resolve this.
2. The student's conclusion is that malonate “destroys the enzyme”. Use the succinate-rescue data to refute this and explain the inhibition at the molecular level. Predict the effect of doubling the succinate concentration at unchanged malonate.
Read the worked answer
If malonate destroyed succinate dehydrogenase (complex II), then adding succinate could not restore the rate - the enzyme would be absent. The rate returns to 1.42 mL O₂ per 10 min (95 % of the uninhibited value), so the enzyme is present and active. Malonate is a reversible competitive inhibitor: it binds the succinate site but cannot be dehydrogenated, so it occupies the site without turnover.
Kinetic reasoning:
- Competitive inhibition raises the apparent Kₘ (more substrate is needed for half-maximal rate) but leaves Vmax unchanged, because sufficiently high substrate displaces the inhibitor.
- With 10 mmol/L malonate, 20 mmol/L succinate already restores 95 % of the control rate, which tells you the inhibitor's affinity for the site is comparable to, or lower than, that of succinate at these concentrations.
- The inhibition also occurs at a single point in the cycle, so the residual 9 µL min⁻¹ g⁻¹ in the malonate-only vessel reflects flux through the rest of the chain fed by other substrates (e.g. 2-oxoglutarate, isocitrate) and by any succinate produced endogenously.
Prediction for doubling succinate at the same malonate: the rate rises further, toward the control value; because inhibition is competitive, no fixed concentration of malonate can suppress the rate to zero once succinate is saturating. If the student instead doubled malonate at fixed succinate, the rate would fall. Non-competitive inhibitors would behave differently (Vmax falls, and extra substrate cannot rescue), so the rescue experiment is precisely the diagnostic test.
3. Calculate the respiratory quotient for both seed types and interpret each value. What would an RQ above 1.0 indicate, and why is the boiled-seed value not exactly zero?
Read the worked answer
RQ = CO₂ released ÷ O₂ consumed.
- Peas: 1.47 ÷ 1.50 = 0.98 - close to 1.0, as expected when hexose is fully oxidised (C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O). Germinating starch-rich seeds respire mainly carbohydrate.
- Sunflower: 1.15 ÷ 1.60 = 0.72 - typical of lipid oxidation, where the substrate is more reduced and more O₂ is needed per CO₂ released. Lipid RQ values lie around 0.7.
Interpretive caution: an RQ between the two values does not prove a mixture of substrates. Conversion of fat to carbohydrate consumes O₂ without releasing CO₂ and drives the measured RQ below 0.7, while partially anaerobic metabolism of carbohydrate (lactate or ethanol production, no O₂ used for that fraction) drives the whole-seed RQ above 1.0. An RQ above 1.0 therefore indicates that some CO₂ is being produced without corresponding O₂ uptake - anaerobic glycolysis or organic-acid decarboxylation.
The boiled-seed control reads 0.03 mL O₂ and 0.02 mL CO₂, not zero. These volumes are about 2 % of the control and arise from thermal equilibration, small pressure changes and manometer reading error; the RQ calculated from such values (0.67) is meaningless because both numbers are within the noise. Reporting the blank as a percentage of the experimental value is the correct way to present it.