The weights attached to an edge should be the 'sum of pairs' weights. Specifically, each edge is labelled by 3 characters, say x_1, x_2, x_3, which may be residues or gap characters. There are 3 different unordered pairs of these characters, namely (x_1, x_2), (x_1, x_3) and (x_2, x_3). The weight attached to the edge is then the sum of the 3 pair weights, where the weight of (x_i, x_j) is
>sp|P01308|INS_HUMAN Insulin OS=Homo sapiens GN=INS PE=1 SV=1 MALWMRLLPLLALLALWGPDPAAAFVNQHLCGSHLVEALYLVCGERGFFYTPKTRREAEDLQVGQVELGGGPGAGSLQPLALEGSLQKRGIVEQCCTSICSLYQLENYCN >tr|Q25C78|Q25C78_BUBBU Insulin (Fragment) OS=Bubalus bubalis OX=89462 GN=bpi PE=2 SV=1 FVNQHLCGSHLVEALYLVCGERGFFYTPKARREVEGPQVGALELAGGPGAGGLEGPPQKRGIVEQCCASVCSLYQLENYCN >tr|F6QRS1|F6QRS1_XENTR Insulin OS=Xenopus tropicalis OX=8364 GN=fbxw8 PE=3 SV=3 MLPLSSIMALWMQCLPLVLVLLFSTPNTEALANQHLCGSHLVEALYLVCGDRGFFYYPKIKRDIEQAMVNGPQDNELDGMQLQPQEYQKMKRGIVEQCCHSTCSLFQLESYCNusing the BLOSUM62 score matrix for the pairwise scores:
A R N D C Q E G H I L K M F P S T W Y V B Z X * A 4 -1 -2 -2 0 -1 -1 0 -2 -1 -1 -1 -1 -2 -1 1 0 -3 -2 0 -2 -1 0 -4 R -1 5 0 -2 -3 1 0 -2 0 -3 -2 2 -1 -3 -2 -1 -1 -3 -2 -3 -1 0 -1 -4 N -2 0 6 1 -3 0 0 0 1 -3 -3 0 -2 -3 -2 1 0 -4 -2 -3 3 0 -1 -4 D -2 -2 1 6 -3 0 2 -1 -1 -3 -4 -1 -3 -3 -1 0 -1 -4 -3 -3 4 1 -1 -4 C 0 -3 -3 -3 9 -3 -4 -3 -3 -1 -1 -3 -1 -2 -3 -1 -1 -2 -2 -1 -3 -3 -2 -4 Q -1 1 0 0 -3 5 2 -2 0 -3 -2 1 0 -3 -1 0 -1 -2 -1 -2 0 3 -1 -4 E -1 0 0 2 -4 2 5 -2 0 -3 -3 1 -2 -3 -1 0 -1 -3 -2 -2 1 4 -1 -4 G 0 -2 0 -1 -3 -2 -2 6 -2 -4 -4 -2 -3 -3 -2 0 -2 -2 -3 -3 -1 -2 -1 -4 H -2 0 1 -1 -3 0 0 -2 8 -3 -3 -1 -2 -1 -2 -1 -2 -2 2 -3 0 0 -1 -4 I -1 -3 -3 -3 -1 -3 -3 -4 -3 4 2 -3 1 0 -3 -2 -1 -3 -1 3 -3 -3 -1 -4 L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4 -2 2 0 -3 -2 -1 -2 -1 1 -4 -3 -1 -4 K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5 -1 -3 -1 0 -1 -3 -2 -2 0 1 -1 -4 M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5 0 -2 -1 -1 -1 -1 1 -3 -1 -1 -4 F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6 -4 -2 -2 1 3 -1 -3 -3 -1 -4 P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7 -1 -1 -4 -3 -2 -2 -1 -2 -4 S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4 1 -3 -2 -2 0 0 0 -4 T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5 -2 -2 0 -1 -1 0 -4 W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11 2 -3 -4 -3 -2 -4 Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7 -1 -3 -2 -1 -4 V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4 -3 -2 -1 -4 B -2 -1 3 4 -3 0 1 -1 0 -3 -4 0 -3 -3 -2 0 -1 -4 -3 -3 4 1 -1 -4 Z -1 0 0 1 -3 3 4 -2 0 -3 -3 1 -1 -3 -1 0 -1 -3 -2 -2 1 4 -1 -4 X 0 -1 -1 -1 -2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -2 0 0 -2 -1 -1 -1 -1 -1 -4 * -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 1 Gap penalty: -6
Then run your WDAG program from homework #4 [you may have to modify it] on this file to produce a highest scoring path (and its score) giving a highest-scoring alignment of the three sequences. You should adjust the program output such that each of the edge labels (corresponding to an aligned column of residues) appears on a separate line.
Your program should output the following:
Answer to 2013 HW5 [in the current format]
Answer to 2012 HW5 [in the current format]