1. Self-organizing map numeric vectors and sequence motifs
Xuhua Xia

2. Co-expressed genes
Fig A subset of six co-expressed genes from yeast gene expression data (Cho et al., 1998). These genes have similar expression profile and form a tight cluster in a gene expression tree built from distances that measure differences in expression profiles.
Xia Bioinfomatics and the cell. Springer
Slide 2

3. Distances and scale effect
Euclidian distance:
Mahalanobis distance = Euclidian distance after data standardization:
Slide 3

4. Clustering approaches
Genes whose expressions changes synchronously have short distances
We need to identify these genes by clustering them together
Two approaches
Conventional approach (single-linkage, complete-linkage, average linkage). UPGMA is an averega linkage linkage algorithm)
Artificial neural network approach (SOM)
Slide 4

5. UPGMA
Gene1
Gene1 Gene2 Gene3 Gene4 Gene5 Gene Gene Gene Gene Gene5
D12,3 = (D1,3 + D2,3)/2 = D12,4 = (D1,4 + D2,4)/2 = D12,5 = (D1,5 + D2,5)/2 = 0.189
Gene12 Gene3 Gene4 Gene5 Gene Gene Gene Gene5
Gene2
Gene3
Gene4
Gene5
Gene3
Gene4
Gene5
Gene1
Gene2
(1,2),(3,4,5)
Gene4
Gene5
Gene3
Gene1
Gene2
((1,2),3),(4,5)
Xuhua Xia

6. UPGMA
Gene1 Gene2 Gene3 Gene4 Gene5 Gene Gene Gene Gene Gene5
D123,4 = (D1,4 + D2,4 + D3,4)/3 = D123,5 = (D1,5 + D2,5 +D3,5)/3 = 0.185
Gene123 Gene4 Gene5 Gene Gene Gene5
D1234,5 = (D1,5 + D2,5 +D3,5 + D4,5)/4 = 0.184
Gene4
Gene5
Gene3
Gene1
Gene2
Gene5
Gene4
Gene3
Gene1
Gene2
(((1,2),3),4),5)
Slide 6
Xuhua Xia

7. Phylogenetic Relationship from UPGMA
Gene1 Gene2 Gene3 Gene4 Gene5 Gene Gene Gene Gene Gene5
Gene12 Gene3 Gene4 Gene5 Gene Gene Gene Gene5
Gene123 Gene4 Gene5 Gene Gene Gene5
Slide 7
Xuhua Xia

8. Branch Lengths D12 = 0.015 ((1,2),(3,4,5))
D12,3 = (D1,3 + D2,3)/2 = (previous slide) D12,4 = (D1,4 + D2,4)/2 = D12,5 = (D1,5 + D2,5)/2 = 0.189
D123,4 = (D1,4 + D2,4 + D3,4)/3 = (previous slide) D123,5 = (D1,5 + D2,5 +D3,5)/3 = 0.185
D1234),5 = (D1,5 + D2,5 +D3,5 + D4,5)/4 = 0.184
((1,2),(3,4,5))
(((1,2),3),(4,5))
((((1,2),3),4),5)
0.0075
Gene1
Gene2
Gene3
Gene4
Gene5
0.019
0.06
((1:0.0075,2:0.0075),(3,4,5))
(((1:0.0075,2:0.0075):0.019,3:0.019),(4,5))
((((1:0.0075,2:0.0075):0.0115,3:0.019):0.041,4:0.06):0.032,5:0.092)
0.092
The reference book (Xia 2007) gave a different example. Make sure to go through it to gain a better understanding
Slide 8
Xuhua Xia

9. UPGMA Result
Slide 9

10. SOM A grid of "artificial neurons" Training data
numeric vectors
sequence motifs
A distance or similarity index
between numeric vectors
between sequence motifs
An algorithm to update the neurons in response to input (the learning process)
Slide 10

11. Data
Gene T0
T10
T20
Sum 1 93 76 87
256 2 80 81 85
246 3 89 88
262 4 69 74 96
239 5 95
277 6 65
237 7
268 8 78
255 9 97
264 10 67 55
218 11 91 90
276 12 72
215 13 79 94
251 14
250 15 66 64 63
193
Slide 11

12. Data and SOM grid Gene T0 T10 T20 Sum 1 93 76 87 256 2 80 81 85 246 389 88
262 4 69 74 96
239 5 95
277 6 65
237 7
268 8 78
255 9 97
264 10 67 55
218 11 91 90
276 12 72
215 13 79 94
251 14
250 15 66 64 63
193
3 by 3 SOM grid with random initial vectors 1 2 3 T0
T10
T20
2.2
11.5
33.4
41.9
27.6
0.8 6
63.8
51.2
28.5
30.2 47
28.7
51.3 9
61.6
38.2
17.9
40.5
76.1
71.2
79.8
94.6
23.2
76.2
40.9
23.9
Slide 12

13. Training
We now randomly choose one gene, and suppose we happen to have chosen Gene 4 with T0, T10 and T20 equal to 69, 74 and 96, respectively (Table 6-1). The Euclidean distances (designated hereafter as d) between this gene and each of the 9 nodes (Table 6-3) show that Gene 4 is closest to node (3,1), with
Gene T0
T10
T20
Sum 1 93 76 87
256 2 80 81 85
246 3 89 88
262 4 69 74 96
239 5 95
277 6 65
237 7
268 8 78
255 9 97
264 10 67 55
218 11 91 90
276 12 72
215 13 79 94
251 14
250 15 66 64 63
193 1 2 3 T0
T10
T20
2.2
11.5
33.4
41.9
27.6
0.8 6
63.8
51.2
28.5
30.2 47
28.7
51.3 9
61.6
38.2
17.9
40.5
76.1
71.2
79.8
94.6
23.2
76.2
40.9
23.9
Winning node
All 9 distances: 1 2 3
111.0
109.0
78.0
77.2
98.5
86.2
37.8
76.4
79.6
Slide 13

14. Updating 1 2 3 T0
T10
T20
2.2
11.5
33.4
41.9
27.6
0.8 6
63.8
51.2
28.5
30.2 47
28.7
51.3 9
61.6
38.2
17.9
40.5
76.1
71.2
79.8
94.6
23.2
76.2
40.9
23.9
Learning rate: use 0.5
Gene 4: 69, 74 and 96 1 2 3 T0
T10
T20
2.2
11.5
33.4
41.9
27.6
0.8
6.0
63.8
51.2
38.7
41.1
59.3
28.7
51.3
9.0
61.6
38.2
17.9
54.7
75.0
83.6
77.1
89.5
41.4
76.2
40.9
23.9
Contrinue with other vectors and repeat until no more updating is possible
Slide 14

15. Trained SOM 1 2 3 T0
T10
T20
80.7
77.6
73.3 74
77.9
67.3
69.2
79.5
72.2
84.4
81.9
82.1
82.5
81.2
81.4
78.5
78.2
89.6
85.4 91
88.1
82.9
91.7
79.9
79.1
89.4
Gene T0
T10
T20
Row
Col d 1 93 76 87 3 2
9.69 80 81 85
4.41 89 88
6.54 4 69 74 96
13.77 5 95
6.80 6 65
17.43 7
4.90 8 78
10.17 9 97
6.12 10 67 55
23.00 11 91 90
6.30 12 72
6.19 13 79 94
4.84 14
13.63 15 66 64 63
16.54
Slide 15

16. Sequence as a matrix
Table 2. A matrix representation of sequence “ACCGTTA” (a). The resulting position weight matrix (b) is obtained after adding a pseudocount of 0.01 to each cell in (a), with background frequencies being 0.3, 0.2, 0.2, and 0.3 for A, C, G, and T, respectively.
(a) 1 2 3 4 5 6 7 A C G T
(b)
1.695
−4.963
−4.379
2.280
Slide 16

17. Learning
Table 3. Updating the node in Table 2a by the new input sequence “GCCATTA” and the resulting position weight matrix (PWM) obtained in (b) after adding a pseudocount of 0.01 to each cell in (a).
(a) 1 2 3 4 5 6 7 A C G T
(b)
0.723
−5.935
1.716
−5.350
2.301
1.308
Slide 17