3TA (correction)

[Richard Jensen]

Interestingly, an analysis using Jaccard's coefficient and ignoring the outgroup
(which is redundant; if two taxa have identical characters, they are the same in
the context of the analysis!) gives the "intuitive" solution: (A),(B,C,D).  Using
the simple matching coefficient or Manhattan distance does not yield the
untuitive grouping.  If the outgroup is included (which would not be the case for
a phenetic analysis), Jaccard's coefficient is undefined for O and A and neither
the simple matching coefficient nor Manhattan distance recover the intuitive
group.

Cheers,

Dick


>
> When analyzed properly with an outgroup, Davids example indeed holds:
> parsimony does not reconstruct a group (BCD). Using parsimony, there is no
> character support for this group. That raises an interesting point: what
> makes the presence of this group so glaringly obvious to most people
> looking at this matrix:
>
> Data:
> O 00
> A 00
> B 01
> C 10
> D 11
>
> Intuitively, we appear to argue that B and C are both closer to D than A is
> to D, so they must be in the same cluster. Is our intuition wrong, or is
> cladistic parsimony all wrong?
>

--
Richard J. Jensen              | tel: 574-284-4674
Department of Biology      | fax: 574-284-4716
Saint Mary's College         | e-mail: rjensen at saintmarys.edu
Notre Dame, IN 46556    | http://www.saintmarys.edu/~rjensen