Respiration Rates in Germinating Seeds
A student places 10 g of germinating pea seeds in a sealed respirometer at 25 °C and measures O2 consumption and CO2 production over 20 minutes. Results: O2 consumed = 42 µmol; CO2 produced = 50 µmol. The seeds are then placed under nitrogen gas; CO2 production continues at 18 µmol/20 min while O2 consumption falls to zero. A separate control with dead (boiled) seeds shows no gas exchange. The student also notes that the respiratory quotient (CO2/O2) for pure carbohydrate oxidation is about 1.0 and for fat oxidation is about 0.7.
1. Calculate the respiratory quotient for the aerobic trial and interpret it in terms of the substrates being oxidized.
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RQ = CO2 produced / O2 consumed = 50/42 ≈ 1.19. A value above 1.0 cannot be explained by carbohydrate alone (≈1.0) or fat (≈0.7). An RQ > 1 typically indicates that some CO2 is being produced without corresponding O2 consumption - e.g., partial anaerobic fermentation occurring alongside aerobic respiration, or oxidation of more reduced organic acids. So the seeds are likely respiring a mix of substrates with a contribution from fermentation, not pure carbohydrate.
2. Explain the nitrogen-gas result and why CO2 production persists when O2 is absent.
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Under N2, oxidative phosphorylation cannot run because there is no terminal electron acceptor, so O2 consumption is zero. However, glycolysis continues, and pyruvate is converted to ethanol (or lactate) via fermentation, regenerating NAD+ so glycolysis can keep producing ATP. That fermentation releases CO2 (in ethanolic fermentation, acetaldehyde → ethanol + CO2), explaining the 18 µmol/20 min. The boiled-seed control confirms the gas exchange is biological, not chemical.
3. Predict how the ATP yield per glucose changes between the two conditions and justify quantitatively.
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Aerobically, complete oxidation of one glucose yields roughly 30–32 ATP (via glycolysis, pyruvate oxidation, Krebs cycle, and oxidative phosphorylation). Anaerobically, only the net 2 ATP from glycolysis are gained, since fermentation does not generate ATP beyond regenerating NAD+. So ATP per glucose drops by roughly an order of magnitude (from ~30 to ~2). This means the seeds must consume far more glucose anaerobically to sustain the same ATP output, which is why fermentation rates can be high despite low efficiency.