wpgma
DESCRIPTION
WPGMA. Input: Distance matrix D ij; Initially each element is a cluster. n r - size of cluster r Find min element D rs in D; merge clusters r,s Delete elts. r,s, add new elt. t with D it =D ti = n r /(n r +n s )• D ir + n s /(n r +n s ) • D is Repeat. The distance table. - PowerPoint PPT PresentationTRANSCRIPT
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
The distance table
dog bear raccon weasel seal sea lion
cat chimp
dog 0 32 48 51 50 48 98 148bear 0 26 34 29 33 84 136raccon 0 42 44 44 92 152weasel 0 44 38 86 142seal 0 24 89 142sea lion 0 90 142cat 0 148chimp 0
The distance table
dog bear raccon weasel seal sea lion
cat chimp
dog 0 32 48 51 50 48 98 148bear 0 26 34 29 33 84 136raccon 0 42 44 44 92 152weasel 0 44 38 86 142seal 0 24 89 142sea lion 0 90 142cat 0 148chimp 0
Starting tree
seal sea lion
We call the father node of seal and sea lion “ss”.
12 12
Distance between these two taxa was 24, so each branch has a length of 12.
ss
Removing the seal and sea-lion rows and columns,and adding the ss row and columns
dog bear raccon weasel ss cat chimp
dog 0 32 48 51 ? 98 148bear 0 26 34 ? 84 136raccon 0 42 ? 92 152weasel 0 ? 86 142ss 0 89 142cat 0 148chimp 0
Computing dog-ss distance
dog bear raccon weasel seal sea lion
cat chimp
dog 0 32 48 51 50 48 98 148
),())()(
)((),()
)()(
)(()),(( kjD
jnin
jnkiD
jnin
inkijD
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.
The new table. Starting second iteration…
dog bear raccon weasel ss cat chimp
dog 0 32 48 51 49 98 148bear 0 26 34 31 84 136raccon 0 42 44 92 152weasel 0 41 86 142ss 0 89 142cat 0 148chimp 0
Starting tree
We call the father node of seal and sea lion “ss”.
Distance between bear and raccoon was 26, so each branch has a length of 13.
seal sea lion
12 12
ss
bear raccoon
13 13
br
Computing br-ss distance
dog bear raccon weasel ss cat chimp
ss 49 31 44 41 0 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.
),())()(
)((),()
)()(
)(()),(( kjD
jnin
jnkiD
jnin
inkijD
The new table. Starting second iteration…
dog br weasel ss cat chimp
dog 0 40 51 49 98 148br 0 38 37.5 88 144weasel 0 41 86 142ss 0 89 142cat 0 148chimp 0
Starting tree
Distance between br and ss was 37.5, so each branch has a length of 18.75. But this is the distance from brss to the leaves. The distance brss to ss is 18.75-12=6.75. The distance between brss to br is 18.75-13=5.75
seal sea lion
12 12
ss
bear raccoon
6.75
13
brss
br
5.75
13
Computing dog-brss distance
dog br weasel ss cat chimp
dog 0 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.
),())()(
)((),()
)()(
)(()),(( kjD
jnin
jnkiD
jnin
inkijD
The new table. Starting second iteration…
dog brss weasel cat chimp
dog 0 44.5 51 98 148brss 0 39.5 88.75 143weasel 0 86 142cat 0 148chimp 0
Starting tree
Distance between brss and w was 39.5, so wbrss is mapped to the line 19.75. The distance to brss, is thus, 1
seal sea lion
0
ss
bear raccoon
brss
br 1312
19.7518.75
weasel
wbrss
Computing dog-wbrss distance
dog brss weasel cat chimp
dog 0 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 = (356+102)/10=45.8
),())()(
)((),()
)()(
)(()),(( kjD
jnin
jnkiD
jnin
inkijD
The new table. Starting second iteration…
dog wbrss cat chimp
dog 0 45.8 98 148wbrss 0 88.2 142.8cat 0 148chimp 0
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
seal sea lion
0
ss
bear raccoon
brss
br 1312
22.9
18.75
weasel
dwbrss
19.75
dogl
wbrss
The new table. Starting second iteration…
dwrbss cat chimp
dwrbss 0 89.833 143.66cat 0 148chimp 0
Starting tree
Distance between dwbrss and cat was 89.833, so cdwbrss is mapped to the line 44.9165The distance to dwbrss, is thus, 22.0165
seal sea lion
0
ss
bear raccoon
brss
br 1312
44.9165
18.75
weasel
cdwbrss
19.75
dog
wbrss22.9
cat
dwbrss
The new table. Starting second iteration…
cdwrbss chimp
cdwrbss 0 144.2857chimp 0
Starting tree
Distance between cdwbrss and chimp was 144.2857, so THE ROOT is mapped to the line 72.14285The distance to dwbrss, is thus, 27.22635
seal sea lion
0
ss
bear raccoon
brss
br 1312
72.14
18.75
weasel
dwbrss
19.75
dog
wbrss22.9
cat
cdwbrss44.9165
chimp
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