Elodea and the Lamp: When the Inverse-Square Test Fails at One Point
A student investigates photosynthesis in Elodea using a lamp placed at four distances from a tube of pondweed in hydrogencarbonate solution. At each distance she allows five minutes for equilibration, then collects the oxygen evolved in a capillary for five minutes and records the volume.
Her results: at 10 cm, 41 mm³ min⁻¹, water at 34 °C; at 20 cm, 12.0 mm³ min⁻¹, water at 26 °C; at 30 cm, 5.5 mm³ min⁻¹, water at 23 °C; at 40 cm, 3.0 mm³ min⁻¹, water at 22 °C. The capillary is graduated in 1 mm³ divisions.
She writes in her report: 'The rate of photosynthesis is not proportional to 1/d², because the result at 10 cm does not fit the pattern, so the plant must have been damaged.' The teacher replies that the plant was healthy and asks her to check the whole experimental arrangement before blaming the organism.
1. Use the 20 cm result as the reference and test the inverse-square prediction at each other distance. What exactly is anomalous?
Read the worked answer
Predicted rate = 12.0 × (20/d)² mm³ min⁻¹.
At 10 cm: 12.0 × 4 = 48.0 predicted; 41 observed, i.e. 85% of prediction, a shortfall of 7 mm³ min⁻¹. At 30 cm: 12.0 × (20/30)² = 5.33 predicted; 5.5 observed, 3% above prediction. At 40 cm: 12.0 × (20/40)² = 3.00 predicted; 3.0 observed, exactly on prediction.
So three points follow the inverse-square law almost perfectly, and the anomaly is confined to 10 cm, where the light is strongest. The student's phrase 'does not fit the pattern' is too broad: it is one point, not the relationship, that departs. Interpretation must also respect resolution: at 40 cm, 3 mm³ min⁻¹ over five minutes is 15 divisions, so counting error is a few percent, whereas at 10 cm the deficit is far larger than that.
2. Give two competing explanations for the 10 cm point that do not require a damaged plant, and say what distinguishes them.
Read the worked answer
Explanation one: the lamp heats the water (34 °C at 10 cm against 22 °C at 40 cm), and the collected gas volume understates oxygen production because oxygen is less soluble in warm water. Solubility falls by roughly 20% between 22 °C and 34 °C, which is the right order of magnitude for a 15% shortfall.
Explanation two: at 34 °C the light-harvesting apparatus is under stress, so the true rate of oxygen production falls, whether through photoinhibition at high intensity or through thermal effects on the thylakoid membrane.
These make opposite predictions about dissolved oxygen. If the plant is producing at the predicted rate and only the collection is affected, a dissolved-oxygen probe will read close to 48 units' worth. If production has genuinely fallen, the probe will also read low. Holding the tube in a water bath at 22 °C and using a heat filter between lamp and tube removes the confounding variable altogether.
3. Redesign the single control that would most improve the experiment, and state the result that would support the inverse-square conclusion.
Read the worked answer
Keep the pondweed tube in a thermostatted water bath at 22 °C and place a transparent heat filter (a water tank) between the lamp and the tube, so that distance changes light intensity but not temperature. Measure the same variable the same way at every distance, and use longer collection times at 30 and 40 cm to keep the counted volume well above the 1 mm³ resolution.
If the inverse-square law then holds at 10 cm as well, yielding about 48 mm³ min⁻¹, the earlier anomaly was thermal. If the 10 cm rate stays around 41 despite constant temperature, then intensity itself is limiting above some threshold, for example through photoinhibition or because another factor, such as CO₂ supply, has become rate-limiting. Either outcome is informative, which is the mark of a well-designed control.