Download presentation
Presentation is loading. Please wait.
1
WPGMA Input: Distance matrix Dij; Initially each element is a cluster. nr- size of cluster r Find min element Drs in D; merge clusters r,s Delete elts. r,s, add new elt. t with Dit=Dti=nr/(nr+ns)•Dir+ ns/(nr+ns) • Dis Repeat
2
The distance table dog bear raccon weasel seal sea lion cat chimp 32 48 51 50 98 148 26 34 29 33 84 136 42 44 92 152 38 86 142 24 89 90
3
The distance table dog bear raccon weasel seal sea lion cat chimp 32 48 51 50 98 148 26 34 29 33 84 136 42 44 92 152 38 86 142 24 89 90
4
Starting tree Distance between these two taxa was 24, so each branch has a length of 12. ss 12 12 seal sea lion We call the father node of seal and sea lion “ss”.
5
Removing the seal and sea-lion rows and columns,
and adding the ss row and columns dog bear raccon weasel ss cat chimp 32 48 51 ? 98 148 26 34 84 136 42 92 152 86 142 89
6
Computing dog-ss distance
bear raccon weasel seal sea lion cat chimp 32 48 51 50 98 148 Here, i=seal, j=sea lion, k = dog. n(i)=n(j)=1. D(ss,dog) = 0.5D(sea lion,dog) + 0.5D(seal,dog) = 49.
7
The new table. Starting second iteration…
dog bear raccon weasel ss cat chimp 32 48 51 49 98 148 26 34 31 84 136 42 44 92 152 41 86 142 89
8
Starting tree Distance between bear and raccoon was 26, so each branch has a length of 13. seal sea lion 12 ss br 13 13 bear raccoon We call the father node of seal and sea lion “ss”.
9
Computing br-ss distance
dog bear raccon weasel ss cat chimp 49 31 44 41 89.5 142 Here, i=raccoon, j=bear, k = ss. n(i)=n(j)=1. D(br,ss) = 0.5D(bear,ss)+0.5D(raccoon,ss)=37.5.
10
The new table. Starting second iteration…
dog br weasel ss cat chimp 40 51 49 98 148 38 37.5 88 144 41 86 142 89
11
Starting tree Distance between br and ss was 37.5, so each branch has a length of But this is the distance from brss to the leaves. The distance brss to ss is =6.75. The distance between brss to br is =5.75 brss 6.75 5.75 seal sea lion 12 ss br 13 13 bear raccoon
12
Computing dog-brss distance
weasel ss cat chimp 40 51 49 98 148 Here, i = br, j = ss, k = dog. n(i)=n(j)=2. D( brss , dog ) = 0.5D( br , dog ) + 0.5D( ss , dog )=44.5.
13
The new table. Starting second iteration…
dog brss weasel cat chimp 44.5 51 98 148 39.5 88.75 143 86 142
14
Starting tree Distance between brss and w was 39.5, so wbrss is mapped to the line The distance to brss, is thus, 1 wbrss brss 19.75 18.75 br 13 ss 12 seal sea lion bear raccoon weasel
15
Computing dog-wbrss distance
weasel cat chimp 44.5 51 98 148 Here, i = brss, j = weasel, k = dog. n(i)=4, n(j)=1. D( wbrss , dog ) = 0.8D( brss , dog ) + 0.2D( weasel , dog )= 44.5*8/10+51*2/10 = ( )/10=45.8
16
45.8 98 148 88.2 142.8 The new table. Starting second iteration… dog
wbrss cat chimp 45.8 98 148 88.2 142.8
17
Starting tree Distance between wbrss and dog was 45.8, so dwbrss is mapped to the line 22.9 The distance to wbrss, is thus, 3.15 dwbrss 22.9 wbrss brss 19.75 18.75 br 13 ss 12 seal sea lion bear raccoon weasel dogl
18
89.833 143.66 148 The new table. Starting second iteration… dwrbss cat
chimp 89.833 143.66 148
19
Starting tree Distance between dwbrss and cat was , so cdwbrss is mapped to the line The distance to dwbrss, is thus, cdwbrss dwbrss 22.9 wbrss brss 19.75 18.75 br 13 ss 12 cat seal sea lion bear raccoon weasel dog
20
The new table. Starting second iteration…
cdwrbss chimp
21
Starting tree 72.14 Distance between cdwbrss and chimp was , so THE ROOT is mapped to the line The distance to dwbrss, is thus, cdwbrss dwbrss 22.9 wbrss brss 19.75 18.75 br 13 ss 12 cat seal sea lion bear raccoon weasel dog chimp
22
Neighbor Joining Algorithm Saitou & Nei, 87
Input: Distance matrix Dij; Initially each element is a cluster. Find min element Drs in D; merge clusters r,s Delete elts. r,s, add new elt. t with Dit=Dti=(Dir+ Dis – Drs)/2 Repeat Present the hierarchy as a tree with similar elements near each other
Similar presentations
