Question 1 Report
Fig. 4.1 shows the results of a study of water pollution in a river. A factory releases untreated waste containing sugars and other nutrients into the river at one point. Scientists measured the concentration of dissolved oxygen in the water and the number of a pollution-sensitive animal, the stonefly larva, at four sampling points. Point 1 is upstream of the factory and points 2, 3 and 4 are increasing distances downstream of the factory. The scientists used the data in the table to explain the effect of the waste on living organisms in the river.
| Sampling point | Dissolved oxygen / arbitrary units | Number of stonefly larvae |
|---|---|---|
| 1 (upstream) | 95 | 26 |
| 2 | 38 | 4 |
| 3 | 52 | 9 |
| 4 | 84 | 21 |
(a) State how the concentration of dissolved oxygen changes between point 1 and point 2. [1]
(b) Explain why the dissolved oxygen concentration falls just downstream of the factory. [3]
(c) Using the data, explain the link between the dissolved oxygen and the number of stonefly larvae. [2]
This question tests interpretation of water-pollution data and the process of eutrophication / oxygen depletion caused by organic waste.
(a) Between point 1 and point 2 the dissolved oxygen decreases / falls (from 95 to 38 arbitrary units) [1].
(b) The oxygen falls just downstream because the waste contains sugars and nutrients that bacteria feed on [1]; the bacteria multiply rapidly and respire aerobically [1], and this respiration uses up the dissolved oxygen in the water [1]. This oxygen demand created by decomposing organic matter is why untreated sugary waste is so damaging [3].
(c) The data show that where the oxygen is low the number of stonefly larvae is low too, so the two follow the same pattern (oxygen 38 with 4 larvae at point 2, rising back to 84 with 21 larvae at point 4) [1]; this is because the larvae need oxygen for respiration and cannot survive where oxygen is low, which is why they are used as pollution indicators [1]. [2]
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