Building a Phylogeny: Homology, Shared Derived Characters and Parsimony · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Comparing phylogenetic trees

Science · Genetics & Evolution · ages 19-21
Name ______________________   Date ____________
  1. A tree places A and B as sisters, with C outside their pair. The most recent common ancestor of all three taxa has state 2. A and B have state 3; C has state 2. What is the minimum number of changes for this character?

    Answer: ______________

  2. Whale flipper bones and human arm bones were inherited from a common ancestor. Bird and insect wings evolved separately as flight structures. Which conclusion matches this evidence?

    • Both comparisons show homology because the features have similar uses.
    • The bones show homology; the wings show separate evolutionary origins.
    • Both comparisons show convergence because the features have different structures.
  3. Under equal-weight parsimony, gains count as changes but losses do not.

    Circle one:   True   False

  4. An ancestor has exactly three descendant lineages. Which proposed group is a clade containing that ancestor?

    • The ancestor and two descendants, leaving out the third lineage.
    • The ancestor and three similar but unrelated lineages.
    • The ancestor and every one of its three descendant lineages.
  5. A state was present in the ancestor of a studied group and remains in every sampled member. What does this state tell you about branching within that group?

    • It cannot distinguish the branching order within this sampled group.
    • It identifies the earliest split within the group.
    • It identifies the closest pair because every member retains the ancestral state.
  6. Backbones occur in mammals and other vertebrates. Can backbone presence alone identify mammals as a distinct subgroup?

    • Yes; inheritance makes the trait unique to mammals.
    • No; backbone presence also occurs in other vertebrate groups.
    • Yes; a trait shared throughout mammals must distinguish them from every other vertebrate.
  7. The ancestor of A, B and C had a tail. A retains it; B and C lack it. Which statement about tail absence is supported?

    • It is ancestral because the common ancestor had a tail.
    • It proves B and C are sisters, even if their tails were lost independently.
    • It is a derived state that can support B and C as a group.
  8. For two equally weighted characters, Tree P needs minimum counts of 2 and 3 changes. Tree Q needs 4 and 2. Which tree has the lower total, and what is that total?

    • Tree P: a total of 5 changes.
    • Tree Q: a total of 6 changes.
    • A tie: a total of 6 changes.
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Comparing phylogenetic trees W1-mt_GFmiVZ8g-F-s1

  1. 1 · One change from state 2 to state 3 on the shared ancestral branch of A and B explains both tips. C retains state 2.
  2. The bones show homology; the wings show separate evolutionary origins. · The corresponding limb bones are homologous because of common ancestry. The stated separate origins of the flight structures indicate convergence.
  3. False · Equal-weight parsimony counts both gains and losses. Either changes a character state along a branch.
  4. The ancestor and every one of its three descendant lineages. · A clade includes the chosen ancestor and all its descendants. It cannot omit a descendant or add an unrelated lineage.
  5. It cannot distinguish the branching order within this sampled group. · Every member retains the same ancestral state. It therefore does not distinguish the group's internal branching arrangements.
  6. No; backbone presence also occurs in other vertebrate groups. · Backbones are ancestral within mammals and occur beyond mammals. Their presence does not distinguish mammals from other vertebrates.
  7. It is a derived state that can support B and C as a group. · Absence differs from the stated ancestral presence, so it is derived. A shared loss may support B and C as a group, although independent losses could mislead.
  8. Tree P: a total of 5 changes. · Tree P totals 2 + 3 = 5 changes. Tree Q totals 4 + 2 = 6. Equal-weight parsimony prefers P for these data.
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