Összerendezés előtt:
1. szekvencia: AGGVLIIQVG |||||| 2. szekvencia: AGGVLIQVG
Összerendezés után:
1. szekvencia: AGGVLIIQVG ||||| |||| 2. szekvencia: AGGVL-IQVG
Q---ACFWE---MKVRT---ACFYI---MACP QMKV---WEACF---RTMKV---YIKVF---P
A | C | G | T | |
---|---|---|---|---|
A | 1 | 0 | 0 | 0 |
C | 0 | 1 | 0 | 0 |
G | 0 | 0 | 1 | 0 |
T | 0 | 0 | 0 | 1 |
Egységmátrix nukleotidokra
C S T P A G N D E Q H R K M I L V F Y W C 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 S 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 T 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 P 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 N 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 D 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 E 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 Q 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 H 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 R 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 K 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 M 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 I 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 L 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 V 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 F 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 Y 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 W 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 |
Egységmátrix aminosavakra
Cisztein | C | 12 | |||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Speciális | S | 0 | 2 | ||||||||||||||||||
T | -2 | 1 | 3 | ||||||||||||||||||
P | -3 | 1 | 0 | 6 | |||||||||||||||||
A | -2 | 1 | 1 | 1 | 2 | ||||||||||||||||
G | -3 | 1 | 0 | -1 | 1 | 5 | |||||||||||||||
Asx, Glx | N | -4 | 1 | 0 | -1 | 0 | 0 | 2 | |||||||||||||
D | -5 | 0 | 0 | -1 | 0 | 1 | 2 | 4 | |||||||||||||
E | -5 | 0 | 0 | -1 | 0 | 0 | 1 | 3 | 4 | ||||||||||||
Q | -5 | -1 | -1 | 0 | 0 | -1 | 1 | 2 | 2 | 4 | |||||||||||
Bázikus | H | -3 | -1 | -1 | 0 | -1 | -2 | 2 | 1 | 1 | 3 | 6 | |||||||||
R | -4 | 0 | 1 | 0 | -2 | -3 | 0 | -1 | -1 | 1 | 2 | 6 | |||||||||
K | -5 | 0 | 0 | -1 | -1 | 2 | 1 | 0 | 0 | 1 | 0 | 3 | 5 | ||||||||
Alifás | M | -5 | -2 | -1 | -2 | -1 | -3 | -2 | -3 | -2 | -1 | -2 | 0 | 0 | 6 | ||||||
I | -2 | -1 | 0 | -2 | -1 | -3 | -2 | -2 | -2 | -2 | -2 | -2 | -2 | 2 | 5 | ||||||
L | -6 | -3 | -2 | -3 | -2 | -4 | -3 | -4 | -3 | -2 | -2 | -3 | -3 | 4 | 2 | 6 | |||||
V | -2 | -1 | 0 | -1 | 0 | -1 | -2 | -2 | -2 | -2 | -2 | -2 | -2 | 2 | 4 | 2 | 4 | ||||
Aromás | F | -4 | -3 | -3 | -5 | -4 | -5 | -4 | -6 | -5 | -5 | -2 | -4 | -5 | 0 | 1 | 2 | -1 | 9 | ||
Y | 0 | -3 | -3 | -5 | -3 | -5 | -2 | -4 | -4 | -4 | 0 | -4 | -4 | -2 | -1 | -1 | -2 | 7 | 10 | ||
W | -8 | -2 | -5 | -6 | -6 | -7 | -4 | -7 | -7 | -5 | -3 | 2 | -3 | -4 | -5 | -2 | -6 | 0 | 0 | 17 | |
C | S | T | P | A | G | N | D | E | Q | H | R | K | M | I | L | V | F | Y | W |
A log odds mátrix 250 PAM-ra. Pozitív értékek: konzervatív cserék, negatívak: valószínűtlen cserék. Az aminosavak tulajdonságaik szerint csoportosítva vannak felsorolva, ezért az átló közelében lévő pontszámok nagyobbak.
A két szekvencia eltérése százalékban | Evolúciós távolság PAM-ban |
---|---|
1 | 1 |
10 | 11 |
20 | 23 |
30 | 38 |
40 | 56 |
50 | 80 |
60 | 112 |
70 | 159 |
80 | 246 |
(a) Identities = 36/52 (69%), Positives = 47/52 (90%) Query: 214 KMGPGFTKALGHGVDLGHIYGDNLERQYQLRLFKDGKLKYQVLDGEMYPPSV 265 GP+FTK+ HGVDL+HIYG++LERQ +LRLFKDGK+KYQ+++GEMYPP+V Sbjct: 97 ERGPAFTKGKNHGVDLSHIYGESLERQHKLRLFKDGKMKYQMINGEMYPPTV 148 (b) Identities = 36/53 (68%), Positives = 47/53 (89%) Query: 214 KMGPGFTKALGHGVDLGHIYGDNLERQYQLRLFKDGKLKYQVLDGEMYPPSVE 266 + GP FTK HGVDL HIYG++LERQ++LRLFKDGK+KYQ+++GEMYPP+V+ Sbjct: 97 ERGPAFTKGKNHGVDLSHIYGESLERQHKLRLFKDGKMKYQMINGEMYPPTVK 149 |
(a) PAM250 összerendezés, (b) BLOSUM 62 összerendezés. A preferált cseréket jelző + jelek nem ugyanott vannak, ezért a (b) összerendezés eggyel hosszabb is.
Példa:
>bbs|69040 70 kda cyclooxygenase-related protein [mice, Peptide Partial, 80aa] Score = 145 bits (362), Expect = 7e-34 Identities = 66/80 (82%), Positives = 73/80 (90%) Query: 294 LPGLMLYATLWLREHNRVCDLLKAEHPTWGDEQLFQTTRLILIGETIKIVIEEYVQQLSG 353 +PLGM+YAT+WLREHNRVCDLLK EHP WGDEQLFQT+RLILIGETIKIVIE+YVQ LSG Sbjct: 1 VPGLMMYATIWLREHNRVCDLLKQEHPEWGDEQLFQTSRLILIGETIKIVIEDYVQHLSG 60 Query: 354 YFLQLKFDPELLFGVQFQYR 373 Y +LKFDPELLF QFQY+ Sbjct: 61 YHFKLKFDPELLFNQQFQYQ 80 |
M T F R D L L S V S F E G P R P D S S A G G M X T X F X X R X X D X X L X X L X X S X X X X V X S X X X X F X X E X G X X X P X X R X X P X X O S X X X X S X X X X A X G X X X G X X X |
Két azonos szekvencia (sertés lizozim): | |
Nagyon hasonló szekvenciák (rokon fajok lizozimjei): | |
Távoli, de rokon szekvenciák (lizozim és alfa-laktalbumin): |
A | D | L | G | A | V | F | A | L | C | D | R | Y | F | Q | |
A | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
D | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
L | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
G | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
R | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
T | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Q | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
N | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
C | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
D | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
R | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
Y | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
Y | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
Q | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
(Itt Wk a k hosszúságú hézaghoz tartozó gap penalty.)
Tehát mindegyik cella tartalmához hozzáadjuk három érték közül a legnagyobbat. A három érték:
Wk=0 gap penaltyt használva a mátrix félig kitöltve így fest:
A | D | L | G | A | V | F | A | L | C | D | R | Y | F | Q | |
A | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 4 | 3 | 2 | 1 | 1 | 0 |
D | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 4 | 4 | 2 | 1 | 1 | 0 |
L | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 4 | 3 | 2 | 1 | 1 | 0 |
G | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
R | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 5 | 4 | 3 | 3 | 1 | 1 | 0 |
T | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
Q | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 5 | 4 | 3 | 2 | 1 | 1 | 1 |
N | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
C | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 4 | 5 | 3 | 2 | 1 | 1 | 0 |
D | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 3 | 3 | 4 | 2 | 1 | 1 | 0 |
R | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 2 | 2 | 2 | 3 | 1 | 1 | 0 |
Y | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 2 | 2 | 2 | 2 | 2 | 1 | 0 |
Y | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 2 | 1 | 0 |
Q | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
A megjelölt 1-eshez hozzáadjuk az almátrix maximumát (5), így a cellába 6-os kerül.
A | D | L | G | A | V | F | A | L | C | D | R | Y | F | Q | |
A | 9 | 7 | 6 | 6 | 7 | 6 | 6 | 7 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
D | 7 | 8 | 6 | 6 | 6 | 6 | 6 | 6 | 5 | 4 | 4 | 2 | 1 | 1 | 0 |
L | 6 | 6 | 7 | 5 | 5 | 5 | 5 | 5 | 6 | 4 | 3 | 2 | 1 | 1 | 0 |
G | 5 | 5 | 5 | 6 | 5 | 5 | 5 | 5 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
R | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 4 | 3 | 3 | 1 | 1 | 0 |
T | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
Q | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 4 | 3 | 2 | 1 | 1 | 1 |
N | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 4 | 3 | 2 | 1 | 1 | 0 |
C | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 | 1 | 0 |
D | 3 | 4 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 4 | 2 | 1 | 1 | 0 |
R | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 3 | 1 | 1 | 0 |
Y | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 1 | 0 |
Y | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 2 | 1 | 0 |
Q | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
Az útvonalat "leolvasva" adódik az összerendezés:
ADLGAVFALCDRYFQ |||| |||| | ADLGRTQN-CDRYYQ |
x | A | D | L | G | A | V | F | A | L | C | D | R | Y | F | Q | |
x | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
A | 0.0 | |||||||||||||||
D | 0.0 | |||||||||||||||
L | 0.0 | |||||||||||||||
G | 0.0 | |||||||||||||||
R | 0.0 | |||||||||||||||
T | 0.0 | |||||||||||||||
Q | 0.0 | |||||||||||||||
N | 0.0 | |||||||||||||||
C | 0.0 | |||||||||||||||
D | 0.0 | |||||||||||||||
R | 0.0 | |||||||||||||||
Y | 0.0 | |||||||||||||||
Y | 0.0 | |||||||||||||||
Q | 0.0 |
Hi,j = max { Hi-1,j-1 + s( ai, bj); maxk>=1 ( Hi-k,j - Wk ); maxk>=1 ( Hi,j-k - Wk ); 0 }
Tehát kiszámítunk 3 értéket, jelentésük: az (i,j) cellánál végződő lokális összerendezés pontszáma 3 lehetséges esetben:
x | A | D | L | G | A | V | F | A | L | C | D | R | Y | F | Q | |
x | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
A | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 1.0 | 0.0 | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
D | 0.0 | 0.0 | 2.0 | 0.0 | 0.0 | 0.0 | 0.7 | 0.0 | 0.0 | 0.7 | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 0.0 |
L | 0.0 | 0.0 | 0.0 | 3.0 | 0.0 | 0.0 | 0.0 | 0.3 | 0.0 | 1.0 | 0.3 | 0.0 | 0.7 | 0.0 | 0.0 | 0.0 |
G | 0.0 | 0.0 | 0.0 | 0.0 | 4.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.7 | 0.0 | 0.0 | 0.3 | 0.0 | 0.0 |
R | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 3.7 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.3 | 1.0 | 0.0 | 0.0 | 0.0 |
T | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 3.3 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.7 | 0.0 | 0.0 |
Q | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 3.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.3 | 1.0 |
N | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 2.7 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
C | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 2.3 | 1.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
D | 0.0 | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 2.0 | 2.0 | 0.0 | 0.0 | 0.0 | 0.0 |
R | 0.0 | 0.0 | 0.0 | 0.7 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 1.7 | 3.0 | 0.0 | 0.0 | 0.0 |
Y | 0.0 | 0.0 | 0.0 | 0.0 | 0.3 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 1.3 | 4.0 | 0.0 | 0.0 |
Y | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 2.3 | 3.7 | 0.0 |
Q | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 2.0 | 4.7 |
A mátrix a Hi,j = max { Hi-1,j-1 + s( ai, bj); 0} képlet szerint kitöltve. Egyezés pontszáma 1, eltérésé -1/3. Egy tizedesre kerekített értékeket adtunk meg.
Végeredmény:
x | A | D | L | G | A | V | F | A | L | C | D | R | Y | F | Q | |
x | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
A | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 1.0 | 0.0 | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
D | 0.0 | 0.0 | 2.0 | 0.7 | 0.3 | 0.0 | 0.7 | 0.0 | 0.0 | 0.7 | 0.0 | 1.0 | 0.0 | 0.0 | 0.0 | 0.0 |
L | 0.0 | 0.0 | 0.7 | 3.0 | 1.7 | 1.3 | 1.0 | 0.7 | 0.3 | 1.0 | 0.3 | 0.0 | 0.7 | 0.0 | 0.0 | 0.0 |
G | 0.0 | 0.0 | 0.3 | 1.7 | 4.0 | 2.7 | 2.3 | 2.0 | 1.7 | 1.3 | 1.0 | 0.7 | 0.3 | 0.3 | 0.0 | 0.0 |
R | 0.0 | 0.0 | 0.0 | 1.3 | 2.7 | 3.7 | 2.3 | 2.0 | 1.7 | 1.3 | 1.0 | 0.7 | 1.0 | 0.0 | 0.0 | 0.0 |
T | 0.0 | 0.0 | 0.0 | 1.0 | 2.3 | 2.3 | 3.3 | 2.0 | 1.7 | 1.3 | 1.0 | 0.7 | 0.3 | 0.7 | 0.0 | 0.0 |
Q | 0.0 | 0.0 | 0.0 | 0.7 | 2.0 | 2.0 | 2.0 | 3.0 | 1.7 | 1.3 | 1.0 | 0.7 | 0.3 | 0.0 | 0.3 | 1.0 |
N | 0.0 | 0.0 | 0.0 | 0.3 | 1.7 | 1.7 | 1.7 | 1.7 | 2.7 | 1.3 | 1.0 | 0.7 | 0.3 | 0.0 | 0.0 | 0.0 |
C | 0.0 | 0.0 | 0.0 | 0.0 | 1.3 | 1.3 | 1.3 | 1.3 | 1.3 | 2.3 | 1.0 | 0.7 | 0.3 | 0.0 | 0.0 | 0.0 |
D | 0.0 | 0.0 | 1.0 | 0.0 | 1.0 | 1.0 | 1.0 | 1.0 | 1.0 | 1.0 | 2.0 | 2.0 | 0.7 | 0.3 | 0.0 | 0.0 |
R | 0.0 | 0.0 | 0.0 | 0.7 | 0.7 | 0.7 | 0.7 | 0.7 | 0.7 | 0.7 | 0.7 | 1.7 | 3.0 | 1.7 | 1.3 | 1.0 |
Y | 0.0 | 0.0 | 0.0 | 0.0 | 0.3 | 0.3 | 0.3 | 0.3 | 0.3 | 0.3 | 0.3 | 0.3 | 1.7 | 4.0 | 2.7 | 2.3 |
Y | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 1.3 | 2.7 | 3.7 | 2.3 |
Q | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 1.0 | 2.3 | 2.3 | 4.7 |
A teljesen kitöltött mátrix. A k darab delécióhoz tartozó gap penalty értéke: Wk=1+(1/3)k volt. A megjelölt 1.7-es érték származtatása: az átlósan fölötte lévő érték 2.0, ehhez hozzáadva az eltérés pontszámát (-1/3) kapjuk az 1.7-et. Az elemmel azonos sorban és oszlopban lévő elemek közül a megjelölt 3.7 adja a maximumot, ebből a 3 delécióhoz tartozó gap penaltyt (1+3*(1/3)=2) levonva szintén 1.7-et kapunk, így a cellába 1.7 kerül.
ADLGAVFALCDRYFQ ADLGRTQNC-DRYYQ |
Összerendezendő szekvenciák: TTGACACCCTCCCAATTGTA és ACCCCAGGCTTTACACAT Globális összerendezés: TTGACACCCTCC-CAATTGTA :: :: :: : ACCCCAGGCTTTACACAT--- Lokális összerendezés: ---------TTGACACCCTCCCAATTGTA TTGACAC :: :::: vagyis :: :::: ACCCCAGGCTTTACACAT----------- TTTACAC |