# Which nucleotide as start point for maximum parsimony?

Background
Maximum parsimony is used to build an evolutionary tree by choosing the "simplest" one. I see a lot of examples about how this should be done. For example: Question
However it's not clear for me WHICH nucleotide should be used as starting point? Let's consider the first tree(from the figure above): Nucleotide G was chosen as starting point resulting in one mutation, however if we choose T we would have two mutations and if we choose A we would have three. So what determines the nucleotide which should be used to start with?

• Both the text on your diagram and the WP link explain this don't they? The chosen ancestral residue is the one which, in order to fit the tree, needs the fewest total changes. If this doesn't answer your question then maybe you need to rephrase it. – Alan Boyd May 28 '17 at 8:59
• For every tree the number of mutation should be counted to determine the simplest tree rigth? So let's say we have 4 sequences --> 3 unrooted trees --> write every tree down --> for every (informative) position fill in the nucleotides and count the number of mutation needed. But to count them you need to know the "ancesteral nucleotide". So as you suggest you should make the trees as simple as possible (by selecting the best fitting nucleotide as ancestor) and than from these trees againg select the simplest one (the one with the fewest mutations)?@AlanBoyd – KingBoomie May 28 '17 at 10:02
• Yes, so you need to try the tree with each possible ancestral nucleotide in order to find which gives maximum parsimony (lowest number of substitutions). – Alan Boyd May 28 '17 at 10:48
• Hmm so actually you can make 3x4 trees in the case of 4 sequences? and than select the simplest? @AlanBoyd – KingBoomie May 28 '17 at 13:20