Given the following binary character matrix for five plant species, use the principle of maximum parsimony to infer the most parsimonious cladogram. Characters: C₁ (flower colour), C₂ (leaf arrangement), C₃ (seed type). Data: Species A: 0 0 1, Species B: 0 1 1, Species C: 1 0 0, Species D: 1 1 0, Species E: 0 0 0. Explain each step and give the final tree.
Answer
Cladogram: ((A,B),E,(C,D)) – A and B form a sister group, C and D form another sister group, and E branches off next to the (A,B) clade.
Step-by-step solution
- List the character states for each species.
- Identify taxa that share the most characters: A and B share C₁ = 0 and C₃ = 1, differing only in C₂.
- C and D share C₁ = 1 and C₃ = 0, differing only in C₂.
- Species E shares C₁ = 0 and C₂ = 0 with A, but differs in C₃ (0 vs 1).
- Construct a provisional tree grouping (A,B) and (C,D) as sister pairs.
- Place E on the branch that minimizes additional changes. Attaching E to the node joining A and B requires only one change (C₃: 1 → 0).
- Count total character changes for the tree: C₁ changes – 1 (between the two major clades); C₂ changes – 2 (within each sister pair); C₃ changes – 2 (A–B vs E and C–D vs E). Total = 5 changes.
- Check alternative arrangements; any other placement of E or different pairings increase the total number of changes beyond 5.
- Therefore the tree ((A,B),E,(C,D)) is the most parsimonious.
Common mistake: Counting the same character change multiple times or ignoring the need to minimize total changes.
Concept tested: Maximum parsimony phylogeny
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