Page 220 - Applied Probability

10. Molecular Phylogeny
206
G

G G

1 G A3 2 G A4 1G A 3 2 G A4
(a) (b)
FIGURE 10.4. Another Maximum Parsimony Assignment

result of sequencing four contemporary taxa 1, 2, 3, and 4 at a particular
DNA site; Figure 10.3 (b) represents an assignment of bases to the inter-
nal nodes that leads to only one disagreement between neighboring nodes
in the given evolutionary tree. This assignment, which is not unique, is a
maximum parsimony assignment. Figures 10.4 (a) and 10.4 (b) depict a
maximum parsimony assignment to a different evolutionary tree. In this
case, the minimum number of base changes is two. Hence, these two evo-
lutionary trees can be distinguished given the bases observed on the four
contemporary taxa. When many sites are considered, each rooted tree is
assigned a maximum parsimony score at each site. These scores are then
added over all sites to give a maximum parsimony criterion for a rooted
tree.
One flaw in this scheme is that several rooted trees will possess the same
maximum parsimony score. This fact can be appreciated by considering the
role of the root. The root is unique in having exactly two neighbors. All
other internal nodes have three neighbors, and the tips have one neighbor.
If, on one hand, the bases at the two neighbors of the root agree, then the
root will be assigned their shared base. If, on the other hand, the bases
at the two neighbors of the root disagree, then the root will be assigned a
base agreeing with one neighbor and disagreeing with the other neighbor. In
either case, omitting the root leaves the maximum parsimony score assigned
to the rooted tree unchanged. Thus, rooted trees that lead to the same
unrooted tree are indistinguishable under maximum parsimony. Figure
10.5 illustrates how two different rooted trees can collapse to the same
unrooted tree. The unrooted tree with the minimum maximum parsimony
sum is declared the best unrooted tree.
Let us now demonstrate in detail how maximum parsimony operates
[4, 8, 10]. Assignment of bases to nodes is done inductively starting with
the tips, at which the bases are naturally fixed. Now suppose that we have