[Campbell Biology P.612] Imagine a fifth molecular site Site 5 is analyzed for the three bee... | Practice Question

Imagine a fifth molecular site Site 5 is analyzed for the three beetle species. If the analysis reveals that for Site 5, Tree 1 requires 2 base-change events, Tree 2 requires 1 base-change event, and Tree 3 requires 2 base-change events, what would be the new conclusion regarding the most parsimonious tree s based on all five sites? Refer to the original results for Sites 1-4 .

  • A: Tree 1 remains the sole most parsimonious tree with 8 events.
  • B: Tree 2 becomes the sole most parsimonious tree with 8 events.
  • C: Both Tree 1 and Tree 2 are now equally the most parsimonious, each requiring 8 events.
  • D: Tree 3 becomes the most parsimonious tree with 9 events.

Explanation

To determine the new most parsimonious tree s , we sum the original event counts from Sites 1-4 with the new event counts for Site 5: Tree 1: 6 original + 2 Site 5 = 8 events. Tree 2: 7 original + 1 Site 5 = 8 events. Tree 3: 7 original + 2 Site 5 = 9 events. Comparing the new totals, both Tree 1 and Tree 2 require the fewest events 8 each , making them equally the most parsimonious.