the cell (2, 1), a search was carried out to the next cell with the

ignment score. The search was in the shaded area shown in Table

cell (3, 2) had the largest score, being three. Therefore the third

of sequence x was aligned with the second residue G of sequence

to the following alignment:

Table 7.10. Backward propagation search from the cell (2, 1).

1

2

3

4

5

A

G

T

C

A

1

C

3

2

2

2

0

2

A

4

2

2

1

1

3

G

3

3

2

1

0

4

G

2

3

2

1

0

5

C

1

1

1

2

0

6

A

1

0

0

0

1

the cell (3, 2), the next cell with the largest alignment score was

nd. The search area was the shaded area shown in Table 7.11.

e, the cell (4, 3) had the largest score, being two. Therefore the

t was shown below:

Table 7.11. Backward propagation search from the cell (3, 2).

1

2

3

4

5

A

G

T

C

A

1

C

3

2

2

2

0

2

A

4

2

2

1

1

3

G

3

3

2

1

0

4

G

2

3

2

1

0

5

C

1

1

1

2

0

6

A

1

0

0

0

1