Phenetics


Consider these taxa, characters and states:

internalatrial two temporalpedicillate
Taxonamnionlegsscalesbloodnostrilsseptumfenestrationshemipenesgizzardteethfeatherswingsvertebrae
perchnonoyescoldnonononononononoyes
coelocanthnonoyescoldyesyesnonononononoyes
salamandernoyesnocoldyesyesnononoyesnonoyes
frognoyesnocoldyesyesnononoyesnonoyes
turtleyesyesyescoldyesyesnonononononoyes
humanyesyesnowarmyesyesnonononononoyes
geckoyesyesyescoldyesyesyesyesnonononoyes
snakeyesnoyescoldyesyesyesyesnonononoyes
alligatoryesyesyescoldyesyesyesnoyesnononoyes
budgyyesyesnowarmyesyesyesnoyesnoyesyesyes

The preceding matrix can be represented numerically (for convenience) as:
12345678910111213
perch0000000000000
coelocanth0000110000000
salamander0110110001000
frog0110110001000
turtle1100110000000
human1111110000000
gecko1100111100000
snake1000111100000
alligator1100111010000
budgy1111111010110

And, then we can construct a similarity matrix as follows...

The percent similarity for any two taxa is simply the number of characters alike for the two taxa divided by the number of characters:
perchcoelocanthsalamanderfrogturtlehumangeckosnakealligatorbudgy
perch856262695454625423
coelocanth7777856969776938
salamander100777754466246
frog777754466246
turtle8585768554
human69626969
gecko928554
snake7646
alligator69
budgy
All of the phenetic methods we will review begin with this matrix and begin by connecting the two most similar taxa. In this case, there is one pair that is identical (salamander, frog) so we can draw:
          100
           :
           ,==== salamander
           |
           `==== frog

Now, how do we connect the next taxon? One must caculate the similarity of each as-yet un-grouped taxon to the two groups already formed above, and also the similarity of the two groups to each other.

In the Weighted Pair Group Method of Arithmetic averages (WPGMA), the similarity between any single taxon and some group is simply the average of the all of the involved pairwise similarities. That is, for example, the similarity between perch and (salamander, frog) is

(perch:salamander+perch:frog+salamander:frog)/3, or
= (62+62+100)/3
= 75
and then, too, the similarity between (salamander, frog) and gecko is then
(salamander:gecko+frog:gecko+ salamander:frog)/3
= (54+54+100)/3
= 69
Thus we can convert the matrix above to the following with a new column and row for the Salamander+Frog group group we created above:
perchcoelocanthturtlehumangeckosnakealligatorbudgysalamander+frog
perch85695454625423 75
coelocanth856969776938 75
turtle8585768554 85
human69626969 85
gecko928554 69
snake7646 64
alligator69 75
budgy 64

Now, the most similar set of taxa is (snake, gecko) and we can draw:

      92 100
       :  :
       :  ,==== salamander
       :  |
       :  `==== frog
       :
       ,======= snake 
       |
       `======= gecko

perchcoelocanthturtlehumanalligatorbudgySaFrSnGe
perch85695454237569
coelocanth856969387569
turtle8585548584
human69698574
alligator697584
budgy6464
SaFr65
SnGe
For the last entry in this table we needed to calculate the average similarity across two pre-exisitng groups (salamander frog) and (snake gecko). This is then the average of the 6 pairwise similarity values:

(100 + 92 + 54 + 54 + 46 + 46) / 6 = 65

This presents us with a bit of a problem because we have two possible things we could do with the resulting similarities: either connect turtle to Salamander + Frog or connect Human to Salamander + Frog, or create two separate equally optimal solutions and continue for each:

85  92 100
 :   :  :
 :   :  ,==== salamander
 ,======|
 |   :  `==== frog
 |   :
 `=========== turtle
     :
     ,======= snake 
     |
     `======= gecko
OR
85  92 100
 :   :  :
 :   :  ,==== salamander
 ,======|
 |   :  `==== frog
 |   :
 `=========== human
     :
     ,======= snake 
     |
     `======= gecko

If you break ties 'systematically' (sensu PAUP*), that is according to the order of appearance in the matrix, you'd get the UPGMA tree on the left if you completed this procedure. If you broke ties randomly, you might get the tree on the right here.