User talk:Peter coxhead/Work/Phyletic terminology

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Ancestor to descendant or vice versa[edit]

"It's clearly not correct to omit "nearest common" and just use "ancestor" since ancestors older than the nearest common ancestor may have descendants outside the group."

The word "nearest" is redundant here: "common" will do. We can't (by the definition) include any earlier common ancestor of the group without also including all its descendents, so the new common ancestor then becomes the nearest common ancestor. Dendropithecus (talk) 09:46, 4 October 2011 (UTC)[reply]

I think this isn't correct as a matter of strict logic. Consider the tree:
X

A

B

C

D

E

Consider this statement: "the group C+D+E is a clade (=monophyletic group) because its members are the only descendants of their common ancestor". This is not correct; X is a common ancestor, and its descendants A and B are omitted, so according to this statement the group C+D+E isn't a clade.
Whereas if we say "the group C+D+E is a clade because its members are the only descendants of their nearest common ancestor", this is clearly correct.
I think this makes my point about "descendant --> ancestor" definitions.
Here's what I think is the decision process for deciding the type of a group of species. Given a group of species:
  1. Find the nearest common ancestor of all the members of the group
  2. Find all the descendants of that nearest common ancestor
  3. Does the group have exactly the same members as the set of all these descendants?
    "Yes" – the group is monophyletic
    "No" – some of the descendants are omitted from the group; go to 4
  4. Are all the sister groups of the omitted descendants included in the group?
    "Yes" – the group is paraphyletic
    "No" – the group is polyphyletic
"Nearest" is essential if you start from a given group and work back to the ancestor and then forwards again. Peter coxhead (talk) 10:42, 4 October 2011 (UTC)[reply]
I agree that, if you want a formula for finding a clade, then your prescription works. Deciding whether or not any given set is a clade is a different matter. There seems to be some confusion over whether or not the common ancestor must by part of the group. According to your definition, and mine (below), it must be.
Define a group of species as a clade if it contains one common ancestor and all the descendants of this common ancestor (and no other species).
If we look at the set C+D+E+X, this does contain one common ancestor, X, but it does not contain all the descendants of X, so it is not a clade according to this definition, nor is it according to yours. To make it into a clade, by either definition, it would have to include A+B.
C+D+E is not a clade, by either definition, because it does not contain any common ancestor. To make it into a clade, we could include a species Y, ancestral to C+D+E but not ancestral to A+B. If we insert this hypothetical Y either on the node below, which isn't named (nodes never are, I notice), or anywhere below this but above the node where A+B branch off, then Y+C+D+E is a clade, by either definition, so my definition does pass as equivalent to yours by this test.
(I tried to redraw your tree, with Y in place, but I've run out of time) Dendropithecus (talk) 12:31, 4 October 2011 (UTC)[reply]
Ok, you're absolutely right that I've slipped between what Podani calls "monoclady" and "monophyly" (excluding or including the ancestor). This was careless of me, but shows the difficulty of getting this exactly right!
If the ancestors must be included then it's quite correct to say, as you did, that "a group of species [is] a clade if it contains one common ancestor and all the descendants of this common ancestor (and no other species)."
The problem with this as the only definition is that it does not start from the leaves of the tree, but these are very often the only things we know about. For example, I've just been editing the article on Roscoea. To decide whether this genus was monophyletic, researchers constructed a phylogenetic tree for 15 extant species. They then concluded "The results suggest that Roscoea is monophyletic ... Roscoea itself is divided into two sister clades which correlate with geography." So when they use the terms "monophyletic" and "clade", what do they mean?
Their phylogenetic tree (reproduced in the article) does not attempt to show ancestors: it only shows the likely branching pattern of the extant species. (By the way this answers your point above ["nodes never are [named], I notice"]: most cladograms in the literature do not represent actual entities at the nodes; the nodes only show the hypothesized branching pattern.) So we need a way of defining a clade starting from the leaves. Roscoea is monophyletic because the branching pattern of the cladogram shows that all of its species are descended from their nearest common ancestor and there are no other known descendants from that hypothetical common ancestor. If you want you can say that all the hypothetical ancestors of the extant species of Roscoea belong to the genus as well, but this is completely irrelevant to the way that the terms "monophyly" and "clade" are used in real articles like this one: they refer to the entities actually used in the cladistic analysis (whether these entities are extant or extinct), which then form the leaves of the tree. Now Podani would say that they should have used the term "monoclady", but they didn't. "Monophyly" is regularly used to cover groups which either include or exclude the hypothetical ancestors.
Anyway, thanks very much for your comments, because you've clarified for me that in sorting this out in my own mind, I need to get the "-phyly" / "-clady" issue right first. Peter coxhead (talk) 13:17, 4 October 2011 (UTC)[reply]
On rewriting the page, I've decided that you are absolutely right about not needing "nearest" in the definition of "monophyly"; the definitions can be written without needing to use this word (although implicitly the common ancestor will be the nearest). This word or an equivalent restriction is needed, though, in defining "paraphyly". Peter coxhead (talk) 15:20, 4 October 2011 (UTC)[reply]