[CHANDA7]

If it is about number of different ways those 2 bonds can form between the 4 residues, then it should be 3 ways (i think so...not sure though...) 6 is number of combinations of residue pairs possible, while there is one way of bonding per 2 pairs of residues....so it should be 3 ways of bonding...

a-------a---------a---------a

way1: a1-a2 a3-a4

way2: a1-a3 a2-a4

way 3: a2-a3 a1-a4

This exhausts all possible combinations...

3 ways it should be...

Okkay.... Thanks....