Cladistics looks for the relationships between taxa in
terms of an evolutionary cost. By minimizing this cost, it
is expected to find an evolutionary scenario that closely
matches the hierarchical diversification process through
transmission with modification. In other words, inheritance
of innovations from common ancestors is the simplest way to
explain diversity.
To understand why this can only be achieved by using
character (or parameter) matrix instead of distances that
measures global similarities, consider the case of a journey
between two cities.
Looking at a map, you can easily measure the distance
between the two cities with a rule. This might be ok if you
fly, but this is not very useful if you travel by car
because then you have to take into account the landscape and
the existing roads among other things.
First you have to look at a precise roadmap, and compute
for each road and considering all possible bifurcations, the
true number of kilometers you will have to travel. Note that
there is no trick to avoid looking at all possible paths.
Then you can decide to choose the shortest way according to
the parsimony criterion.
But cost might not be measured by kilometers only. You may
consider the time it takes. Highways are certainly faster,
but you should consider the probability of traffic jams or
that of slow trucks or animals on smaller roads. It is
common wisdom that the quickest ways are not necessarily the
shortest or the most direct ones.
You can also think about money with the fuel you will burn.
Depending on your car, depending on the slopes for the
different roads, the cost can be quite different in every
case.
Lastly, you can consider the pleasure or the comfort of the
journey. This is certainly less quantitative and objective,
but still important.
We have here a typical multivariate problem, and defining
an evolutionary cost is not always straightforward. As the
above should illustrate, character-based methods like
cladistics explore an unknown landscape with a metrics which
is defined by the choice of the multivariate cost. Indeed,
for living organisms or for galaxies, there is no roadmap…
Distance-based approaches assume a metrics and do not care
very much on the cost (and even on the landscape). To
understand further the difference, let us be more precise
and consider the following parameter or character matrix:
| p1 | p2 | p3 | p4 | |
| O | 0 | 0 | 0 | 0 |
| A | 1 | 0 | 0 | 0 |
| B | 0 | 1 | 1 | 0 |
| C | 0 | 1 | 1 | 1 |
If you have href="https://astrocladistics.org/cladistics/constructing-a-tree/">learned
to build a tree, you are able to find that the most
parsimonious tree rooted with O is:
data-permalink="https://astrocladistics.org/2012/01/31/evolutionary-cost/treeclad1rooted/"
data-orig-file="https://astrocladistics.files.wordpress.com/2012/01/treeclad1rooted.png?w=620"
data-orig-size="144,135" data-comments-opened="1"
data-image-meta="{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":""}"
data-image-title="treeclad1rooted"
data-image-description=""
data-medium-file="https://astrocladistics.files.wordpress.com/2012/01/treeclad1rooted.png?w=620?w=144"
data-large-file="https://astrocladistics.files.wordpress.com/2012/01/treeclad1rooted.png?w=620?w=144"
class="aligncenter size-full wp-image-228"
title="treeclad1rooted"
src="cid:part2.03080401.01040408-at-panix.com" alt="">Now,
from the parameters, you can compute a distance, the most
common being the euclidian distance. The corresponding
distance matrix (showing here the square of the euclidian
distance) is:
| O | A | B | C | |
| O | 0 | 1 | 2 | 3 |
| A | 1 | 0 | 3 | 4 |
| B | 2 | 3 | 0 | 1 |
| C | 3 | 4 | 1 | 0 |
One could also compute the “edit” or Levenshtein distance,
which measures the number of substitution (here 0-1)
occuring in the full set of parameters between two objects.
The matrix distance in the present case is identical to the
one above. Note that even though it might look like
cladistics because it compares the changes in parameter
values, it is a distance and thus measures these changes
globally.
From any character matrix you can compute a distance
matrix, but the reverse is most generally untrue. Hence,
somehow, when we use distances, we loose some information.
From a distance matrix, we can build a hierarchical tree
representing the relative distances between the objects.
Using hclust in R, this gives:
data-permalink="https://astrocladistics.org/2012/01/31/evolutionary-cost/hclust_tree_1-3/"
data-orig-file="https://astrocladistics.files.wordpress.com/2012/01/hclust_tree_12.png?w=620"
data-orig-size="288,288" data-comments-opened="1"
data-image-meta="{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":""}"
data-image-title="hclust_tree_1"
data-image-description=""
data-medium-file="https://astrocladistics.files.wordpress.com/2012/01/hclust_tree_12.png?w=620?w=288"
data-large-file="https://astrocladistics.files.wordpress.com/2012/01/hclust_tree_12.png?w=620?w=288"
class="aligncenter size-full wp-image-234"
title="hclust_tree_1"
src="cid:part4.02090701.00040708-at-panix.com" alt=""
srcset="https://astrocladistics.files.wordpress.com/2012/01/hclust_tree_12.png
288w,
https://astrocladistics.files.wordpress.com/2012/01/hclust_tree_12.png?w=150
150w" sizes="(max-width: 288px) 100vw, 288px">From
this tree, one concludes that there are two groups: (O,A)
and (B,C). The distance within the two members of each group
is 1 while the minimum distance between the groups is 2. So
the two methods agree that B and C are very close to each
other and could form a group, but cladistics do not see any
reason to put O and A in a same group. Indeed the cladogram
is easier to interpret because it must be read in terms of
the evolutionary cost.
In general, it appears that character-based and
distance-based analysis in phylogeny give very close
results. I think this is because if only synapomorphies are
used, which ideally should be the case for cladistics, then
the landscape is not too much tortuous so that the metrics
assumed by distance-based approaches is more or less
adequate.
by dranorter on July
3, 2013 - 07:10
You say there is no trick to avoid looking at all
possible paths. However, I think you are thinking
of the Traveling Salesman Problem, where we must
visit every node in an efficient manner. When
traveling from point A to point B there certainly
are useful tricks and shortcuts. Dijkstra’s
Algorithm is a common approach, and is much faster
than looking at all possibilities.
Of course, the ‘multivariate’ case gets more
complicated. I’m only disagreeing with that one
sentence.