QUESTION 119 (6 marks)
Equal-sized invertebrate samples are taken from two ponds using the same technique. Use . The species counts are shown.
| Species | P | Q |
|---|---|---|
| A | 12 | 32 |
| B | 12 | 2 |
| C | 12 | 1 |
| D | 0 | 1 |
a) Calculate SDI for each sample to three decimal places. Show your working. [3 marks]
b) Compare species richness and evenness in the two samples. [2 marks]
c) Explain why the larger species count in Q does not establish that Q has the higher SDI. [1 mark]
Practice marking scheme
Answer
P: richness 3, SDI 0.686. Q: richness 4, SDI 0.211. P is more even and has the higher SDI; Q has greater richness.
Working
Each sample has N = 36. P: 1 − 3(12)(11)/(36×35) = 0.685714. Q: 1 − [32(31)+2(1)]/1260 = 0.211111. Q contains more species but is strongly dominated by A, so richness and the diversity index rank the ponds differently.
Marking criteria
- Uses N = 36 and denominator 1260. [1 mark]
- Calculates P SDI 0.686. [1 mark]
- Calculates Q SDI 0.211. [1 mark]
- States richness 3 in P and 4 in Q. [1 mark]
- States that P is more even. [1 mark]
- Explains that dominance in Q lowers its index despite greater richness. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.