QUESTION 153 (5 marks)
Two lineages differ at six sites in a conserved sequence. Use the supplied model , with independent changes in both lineages and no repeated substitutions. Calibration constrains each lineage’s rate to differences per million years. Treat this as a model range, not a statistical confidence interval.
a) Calculate the minimum and maximum divergence times under the supplied rate bounds. [2 marks]
b) Calculate the estimate using r = 0.25. [1 mark]
c) Explain why the slower rate produces the older estimate. [1 mark]
d) State one assumption that limits application of this model to real sequences. [1 mark]
Practice marking scheme
Answer
The model range is 10–15 million years. A rate of 0.25 gives 12 million years. A slower rate implies a longer time for the same number of differences.
Working
At r = 0.30, t = 6/(2×0.30) = 10. At r = 0.20, t = 6/(2×0.20) = 15. The central rate gives 6/0.50 = 12. These bounds depend on the supplied clock assumptions; variation in rate through time or repeated substitutions would require a more detailed model.
Marking criteria
- Calculates minimum 10 million years at r = 0.30. [1 mark]
- Calculates maximum 15 million years at r = 0.20. [1 mark]
- Calculates 12 million years. [1 mark]
- Explains that slower accumulation takes longer to generate six differences. [1 mark]
- Identifies constant/calibrated rate, independence, no repeated substitutions, or a relevant sequence/population timing limitation. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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