Question

Given the distance matrix in the table below, construct a parsimonious tree. Species 1 Species 2 Species 3 Species 4 Species 5 Species 6 Species 7 19 Species 1 18 9 17 7 8 Species 2 19 18 1 17 16 Species 3 18 4 -- 20 5 19 17 Species 4 9 18 20 16 5 Species 5 17 1 5 16 19 20 Species 6 7 17 19 5 19 2 16 Species 7 17 4 20 -- Upload Choose a File

Answer 1

A distance matrix comes from a character matrix, this means that the highest values represent the larger differences. We have to chose the lowest value and propose the involved species as sister species, we are then going to make another table but now with these 2 species together and using mean values when required. We will repeat this process until we have related all the species, let us beggin:

The lowest value here is 1 between 2 and 5, we have (2-5). The next table is:

	1	2-5	3	4	6	7
1	0
2-5	18	0
3	18	4.5	0
4	9	17	20	0
6	7	18	19	5	0
7	8	18	17	4	2	0

Now the lowest value is 2 between 6 and 7, now we have (2-5) and (6-7), the next table is:

	1	2-5	3	4	6-7
1	0
2-5	18	0
3	18	4.5	0
4	9	17	20	0
6-7	7.5	18	18	4.5	0

Now the lowest value is 4.5 and occurs two times, between 2-5 and 3 forming ((2-5)3) and between 6-7 and 4 forming ((6-7)4). The next table is:

	1	((2-5)3)	((6-7)4)
1	0
((2-5)3)	18	0
((6-7)4)	8.25	18.25	0

Now the lowest value is 8.25 between 1 and ((6-7)4), that forms (((6-7)4)1) and we already had ((2-5)3).

We could make another table but that would only link our two final groups which we already know have to be linked. The final tree then is:

2536 6 7 4 1

Remember that nodes can rotate in a tree and the relationships do not get affect by that.