Using the principle of maximum parsimony, construct the most parsimonious phylogenetic tree for the following species based on binary characters: 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.

Step-by-step solution

1. Step 1: List character states for each species. C1: A=0,B=0,C=1,D=1,E=0. C2: A=0,B=1,C=0,D=1,E=0. C3: A=1,B=1,C=0,D=0,E=0.
2. Step 2: Calculate the number of changes required for each possible grouping. Grouping (AB)(CD) with E as outgroup requires 2 changes in C1, 2 in C2, and 2 in C3 – total 6 changes.
3. Step 3: Test alternative groupings. Grouping (A,B,E) vs (C,D) requires 3+4=7 changes. Grouping (C,D,E) vs (A,B) also requires 7 changes. Grouping all together in a star requires 8 changes.
4. Step 4: The grouping with the least changes (6) is the most parsimonious. Thus the tree is: ((A,B),(C,D)),E.
5. step 5: Conclude that the final tree places A and B as sister taxa, C and D as sister taxa, and E as the sister to the (AB)(CD) clade.

Concept tested: Phylogenetic inference via parsimony

Published 7 October 2026 · How this solution was prepared and checked