3TA (correction)

[Richard Zander]

The chance of two synapomorphies occurring for a particular two terminal
taxa is 88%. This assumes the character states represent independent
evolutionry events. Thus, none of the solutions possible through parsimony
analysis of this data set are reliable at a 95% confidence level.

The problem is nonsense. The cladogram should be collapsed into a bush, if
reliability were taken into account.

R.Z.

-----Original Message-----
From: P.Hovenkamp
To: TAXACOM at USOBI.ORG
Sent: 4/11/2003 4:09 AM
Subject: Re: [TAXACOM] 3TA (correction)

At 10:48 AM 4/10/03 +0200, I wrote:
>At 08:55 AM 4/10/03 +0100, David Williams wrote:
>
>(...)
>>My interest in three-item data has nothing to do with any of the above
and
>>is almost painfully simple. What might be the answer to the simple two
>>character problem AB(CD) + AC(BD)? As far as I am aware most (all?)
methods
>>yield no result, or return the same two trees, as if the solution to 2
+ 2
>>= 2 + 2. I see the answer as A(BCD) [2 + 2 = 4]. I need no method, no
>>matrix, no program (all my MSc students suggest the same answer as
well,
>>before they learn that computer programs know better). Now, to me,
>>something is wrong.
>
>Yes - but where?
>When I run this data through Nona using all the usual trappings
(heuristic
>search settings etc.) I get the following results: two most
parsimonious
>trees (corresponding with the two trees alluded to by David), consensus
of
>which is A(BCD). What most people would take to be the result of the
>analysis is exactly what David would accept as the result of the
analysis.
>So why is he not satisfied with the result being produced by a strict
>consensus?

Off-list David informed me that I had, indeed, made a beginner's
mistake.
Oops. Sorry.
I had mistaken the resolution of the basal node for a real supported
resolution. In fact, it is only supported by the decision to use an
outgroup - which defines the in-group.

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?

David (and the other 3TA-proponents - this matrix is a derivation of an
example given by Nelson in his "Nullius in Verba" paper - sorry, I don't
have the reference) seem to prefer intuition. Others prefer parsimony,
on
grounds that have been debated extensively.

Peter Hovenkamp