Given the following binary character matrix for five insect species, use the principle of maximum parsimony to infer the most parsimonious cladogram. Characters: C₁ (wing presence), C₂ (metamorphosis type), C₃ (mouthpart type). Data: Species A: 1 0 1, Species B: 1 1 1, Species C: 0 0 0, Species D: 0 1 0, Species E: 1 0 0. Explain each step and give the final tree.
Answer
The most parsimonious tree groups (A,B) together, (C,D) together, with E branching off the (A,B) clade; total tree length = 3 changes.
Step-by-step solution
- List all possible unrooted trees for five taxa (there are 15).
- For each tree, calculate the minimum number of character changes required for each character using Fitch’s algorithm.
- Summarize the total changes for each tree; the tree with the smallest sum is the most parsimonious.
- The tree ((A,B),(C,D),E) requires: C₁ changes – wing loss once (from 1 to 0) and gain once; C₂ changes – metamorphosis type changes twice; C₃ changes – mouthpart type changes once. Total = 3 changes.
- All other trees need ≥4 changes, so the tree ((A,B),(C,D),E) is the most parsimonious.
Common mistake: Counting the same character change on multiple branches instead of the minimum required
Concept tested: Maximum parsimony phylogeny
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