Generally cysteine residues form disulfide linkages - so how many combinations are possible out of (say) six residues. Also can cysteine form bonds with all the residues?
This is a mathematical problem rather than biological.
the refined question would be this : we have 6 items and we want to pack them in 3 groups of 2.(We assume that all six cysteines do form a disulfide bond with each other and none of them are ungrouped or left out.Let me know if you have the question about the latter situation.) How many different sorting are possible?
Disclaimer: I am terrible at these types of mathematics . but I believe I have found the right answer.
Method A :
Write all of them!
six items are called A,B,C,D,E,F respectively
AB/CD/EF AB/CE/DF AB/CF/DE
AC/BD/EF AC/BE/DF AC/BF/DE
AD/BC/EF AD/BE/CF AD/BF/CE
AE/BC/DF AE/BD/CF AE/DC/BF
AF/BC/DE AF/BE/DC AF/BD/CE
THE ANSWER IS 15
you could also use C(n,r) function for better understanding but I'm going with this:
we have 6 items. Split them into 3 groups of 2. Group 1 ,group 2 , group 3.
calculate number of different combinations.
group 1 : 6 * 5 /2 = 15 (you can reverse the order of the first 2 objects and the group would still remain the same)
group 2 : 4 * 3 /2 =6 (you can reverse the order of the first 2 objects and the group would still remain the same) (Two of the items are used in the previous group)
group 3 : 2 * 1 /2 =1
(you can reverse the order of the first 2 objects and the group would still remain the same) (Four of the items are used in the previous groups)
You have to multiply these numbers . But do not forget that combination between group 1,2,3 does not matter(you can call any of them 1 , any of them 2 ,etc and the result would NOT change. It is just naming.)
we have 6 repetitious/same groups counted in each unique combination . Therefore devide the whole amount by 6
15 * 6 * 1 / 6 = 15
About the second question the answer is no. Cysteine form disulfide bonds only with itself as far as I know