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
|