Upstream and downstream of a sewage outflow: reading the macroinvertebrates
Students survey macroinvertebrates at two sites on a river. Site A is 200 m upstream of a sewage treatment outflow; site B is 500 m downstream. At each site they take five kick-samples and identify all animals. Site A contains 12 taxa: mayfly nymphs 84, stonefly nymphs 31, caddis larvae 40, midge larvae 12, and other taxa 24. Site B contains 6 taxa: midge larvae 210, tubificid worms 95, leeches 18, mayfly nymphs 4, caddis larvae 3, and one other taxon with 6 individuals. The students use a simple score in which mayfly = 3, stonefly = 3, caddis = 2, midge = 1, worm = 1 and leech = 1, and they multiply each score by the number of individuals and add the totals. Site A scores much higher than site B. They conclude that site A has better water quality. A local angler says the sewage works is irrelevant because the river is self-cleaning and the difference is just natural.
1. Evaluate the students' method and conclusion.
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Method
The score multiplies a tolerance score by the number of individuals. This makes the total depend mainly on abundance, not on diversity: a site with many midges can score highly even if it has few taxa. The two sites may also differ in sampling effort, substrate and flow, so the comparison is not controlled.
What the data actually show
Site A has 12 taxa, including mayfly and stonefly nymphs, which are sensitive to organic pollution. Site B has only 6 taxa, dominated by midge larvae (210) and tubificid worms (95), which tolerate low oxygen and organic enrichment. That pattern is consistent with organic pollution downstream of the sewage outflow.
Conclusion
The direction of the students' conclusion is plausible, but their method does not prove it. They should use equal sampling effort, identify to the same taxonomic level, and calculate an average score per taxon (total score divided by number of taxa) or a diversity index. They also need replicate samples, several seasons and physicochemical data such as dissolved oxygen, BOD and ammonia.
2. Explain how you would test whether the sewage outflow causes the difference, given that you cannot manipulate the outflow.
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Design
The outflow cannot be manipulated, so use a comparative design that controls for natural variation:
- Sample several reference sites upstream and several sites downstream at increasing distance, for example 50 m, 200 m, 500 m and 1 km.
- Sample repeatedly in different seasons and before and after any change at the works, using a before-after control-impact design if possible.
- Measure dissolved oxygen, BOD5, ammonia, phosphate, flow, temperature and substrate at each site.
- Keep sampling effort identical and analyse with site as a random factor.
A single upstream site and a single downstream site cannot separate the effect of the outflow from natural downstream change. A consistent drop in sensitive taxa and a rise in tolerant taxa along the distance gradient, matched by oxygen and ammonia data, would be strong evidence of a causal effect.
3. Explain what 'self-cleaning' means and why it does not mean that the discharge has no impact.
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'Self-cleaning' means that a river can break down organic matter and reoxygenate downstream through turbulence and photosynthesis, so pollution effects diminish with distance. It describes the river's assimilative capacity; it does not mean that the discharge has no impact.
Near the outfall, microbial decomposition of organic matter consumes oxygen and can create an oxygen sag. Ammonia can be toxic, and nutrients can cause eutrophication. The community at 500 m is already dominated by low-oxygen-tolerant midges and tubificid worms, which indicates that recovery is incomplete at that point. The angler is right that the river may recover further downstream, but wrong to conclude that the sewage works is irrelevant to the measured difference.