Question

Given the morphological characteristics in the table below, construct a parsimonious tree.

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 Species 1 11 18 2 19 17 3 Species 2 11 17 9 18 19 10 Species 3 18 17 18 2 4 17 Species 4 2 9 18 20 5 4 Species 5 19 18 2 20 7 17 Species 6 17 19 4 5 7 21 Species 7 3 10 17 4 17 21

Answer 1

Here, the parsimonious tree is constructed on the basis of distance mapping techniques called as UPGMA- Unweighted pair group method with arthiemetic mean. The whole process with step by step diagrammatic illustration are given below.

as we know that parsimonious hypothesis means the progeny with less dissimilarity and evolution and having more similar characteristics.

Here, you should focus one thing that where th distance was not same we are supposed to write the average of two.

Geinen distance Matrin : fpecies 1 2 3 18 17 2. 3. 4 5 6 4 5 16 17 (2 1917 3 9 18 g 10 1824 17. 5 4. 7 17 120 4 5 6.17 10 17 5 -- 5 - Here, we can identify the species with list difference. ice 2- differences & make group. 0 ♡ Reconstruction of distance matrin 1/4 2 14 10 18 113-5323.5 9 17 118 9 3 124 7 17 NOW, 6 hew matrix 121 again with the grouping 385 Speclis 2/4 3/5 2 315 26 1/4 18.75 10 11 (35) ► 3/5 117.55.5 7 2 9. to 6 121 3 7 4t with least difference of 3.5 c species of

3 n Next matrix now with grouping of /r/z Species es 7 6. 11-146.797.75 3/5 17-555 2 g 6. Now, we have the phylogentic tore (4) as Again by combining 2/4/7/2 & making distance matrix, we have + 15 6 24/7/2 3/5 14328038 5.5. 6. * Now, 1/4/7/2 have less distance between Species ( rather than 3/s, do final Answees 0 > 4 4 final, phylogenetic th distance between toel showing 1/4/7/2/6 group s 2 6 with 3/5 group >3 of speced