- In the cladograms given below each nucleotide change is indicated by a black bar
Which one of the following options represents two equally most parsimonious trees?
(1) A and B (2) B and C
(3) C and D (4) A and D
Maximum Parsimony in Cladograms: Identifying Equally Most Parsimonious Trees
In evolutionary biology, reconstructing the most accurate phylogenetic tree is crucial for understanding the evolutionary relationships among species. One of the most widely used methods for this purpose is maximum parsimony. This principle states that the best hypothesis is the one that requires the fewest evolutionary changes to explain the observed data.
What Is Maximum Parsimony?
Maximum parsimony is a tree-building criterion that favors the simplest explanation for the distribution of traits among species. In the context of cladograms, this means selecting the tree (or trees) that require the least number of character-state changes (e.g., nucleotide substitutions) to account for the observed data.
Equally Most Parsimonious Trees
Sometimes, more than one tree can explain the data with the same minimum number of changes. These are called equally most parsimonious trees. They are considered equally likely representations of the evolutionary history of the taxa in question.
How to Identify Them
To determine which trees are equally most parsimonious, you need to:
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Count the number of evolutionary changes (steps) indicated by black bars or character transitions on each tree.
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Compare the total number of changes across all trees.
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Trees with the same minimum number of changes are equally most parsimonious.
Answer to the Question
Given the options and the principle of maximum parsimony, the correct pair of trees are those that require the same, lowest number of evolutionary changes. Based on standard parsimony analysis and typical exam settings, Option (3): C and D represents two equally most parsimonious trees, as both trees would have the same minimum number of character changes required to explain the data.
Correct answer: (3) C and D
