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WPGMA

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.

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WPGMA

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  1. WPGMA • Input: Distance matrix Dij; Initially eachelement 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

  3. The distance table

  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

  6. Computing dog-ss distance 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…

  8. ss 12 12 seal sea lion Starting tree Distance between bear and raccoon was 26, so each branch has a length of 13. br 13 13 bear raccoon We call the father node of seal and sea lion “ss”.

  9. Computing br-ss distance 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…

  11. ss 12 12 seal sea lion 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 brss 6.75 5.75 br 13 13 bear raccoon

  12. Computing dog-brss distance 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…

  14. 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 wbrss brss 19.75 18.75 br 13 ss 12 0 weasel seal sea lion bear raccoon

  15. Computing dog-wbrss distance 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

  16. The new table. Starting second iteration…

  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 0 weasel seal sea lion bear raccoon dogl

  18. The new table. Starting second iteration…

  19. 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 cdwbrss 44.9165 dwbrss 22.9 wbrss brss 19.75 18.75 br 13 ss 12 0 cat weasel seal sea lion bear raccoon dog

  20. The new table. Starting second iteration…

  21. Starting tree 72.14 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 cdwbrss 44.9165 dwbrss 22.9 wbrss brss 19.75 18.75 br 13 ss 12 0 cat weasel seal sea lion bear raccoon dog chimp

  22. Neighbor Joining Algorithm Saitou & Nei, 87 • Input: Distance matrix Dij; Initially eachelement 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

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