t. Remember this alignment was moved from the previous gap to

ment, i.e., a NULL alignment to the current alignment between

ue T of sequence x and the residue T of sequence y. Therefore,

onal move is shown below,

൜^ݔ:

െ | T

#GGCT

ݕ:

െ | T

#GCT ^{ൠ⟹cost ൌ0 ൅0 ൌ0}

ng these three moves, the diagonal move was the best one because

least distance, i.e., being zero. Therefore this move was selected

imal alignment for this step for this cell and a zero was filled into

At the same time a note of the best move (D for the diagonal

as denoted for the cell (1, 1) on the right panel of Table 7.3.

7.3. Alignment result for the cell (1, 1). D stands for the diagonal move.

0

1

2

3

4

1

2

3

4

-

T

G

C

T

T

G

C

T

-

0

1

2

3

4

T

1

0

D

G

2

G

3

C

4

T

5

moves as well as their associated distances are shown below for

1, 3). The optimisation resulted in a partial alignment cost as one

the horizontal move for this cell. Table 7.4 shows the optimal

t for the cell (1, 3).

move

൜^ݔ:

െT െ| െ

#GGCT

ݕ:

െT G | C

#T

ൠ⟹cost ൌ1 ൅1 ൌ2

move

൜^ݔ:

െെെ|T

#GGCT

ݕ:

T G C| െ

#T ^{ൠ⟹cost ൌ 3 ൅1 ൌ4}

move

൜^ݔ:

െെ |T

#GGCT

ݕ:

T G | C

#T

ൠ⟹cost ൌ 2 ൅1 ൌ3