Supplementary material Case 1, calculation of the number of possible different states of the interactome for n different proteins The calculation is simplified in many ways but provides a realistic estimate of that the number of possible different states of the interactome. Basically, this is a simple combinatorial task. Our basic simplification of the interactome is that it is composed of pair-­‐wise complexes of all the proteins in the proteome, with one possible complex between two proteins. The combinatorial question is: how many combinations of pairs can be selected from n different elements? The first pair can be selected in n × (n − 1)
2
different ways. The second pair can be selected (n − 2) × (n − 3)
2
different ways, and so on to the last one (which is either a solitary protein or the complex of the last two proteins), which can be selected 2 ×1
2
different ways. Because the interactome is the combination of all possible different pairs, we need to multiply these different ways of selecting pairs, which is: n!
2
n
2
However, this way we distinguish all interactomes that correspond to the same pattern of pairs but have them in different orders (e.g., in the case of four proteins A through D, we consider the interactome A-­‐B + C-­‐D different from C-­‐D + A-­‐B). Thus, we have to divide the number obtained by the number of the possible order of n/2 pairs, i.e., n
! 2
Thus, the total number of the possible realizations of the interactome is: n!
2
n
2
n
× !
2
If n = 4500, n! = 2 x 1014487 and the number of possible different interactomes is about 107240. (1) Tompa & Rose Supplementary material, Page 2
Case 2, calculation of the number of possible different states of the interactome for n different proteins with m different interfaces each At the first approximation, this problem is very similar to the previous one, but the number of entities that can interact which each other is n x m, because each interface can interact with each other (e.g. between two proteins having 10 interfaces each, there are as many as 100 possible complexes that are different in a molecular sense). We may then simply take the formula (1) derived under Case 1, and replace n with n x m: (n × m)!
n ×m
n×m
2
2
×
!
2
(2) If n = 4500, and m = 3540, we may use the Stirling approximation (ln n! = n ln n -­‐ n) for very large 8
factorials, which yields (n x m)! = 101.1 x 10 , with the number of possible different interactomes about 7
105.4 x 10 . Case 3, calculation of the number of possible different states of the interactome for n different proteins of k copies, with m different interfaces each Because each protein is present in k (typically more than one) copies, the number of entities in real is even higher than in the previous case, n x k x m. Because k of these are the same in a molecular sense, we cannot simply substitute n with n x k x m in formula (1), but have to proceed as follows. We may start selecting all patterns of pairs (n × k × m)!
n ×k ×m
n×k×m
2
2
×
!
2
times. Many of these, however, are the same due to counting a pair formed by different copies of the same two proteins different (e.g., in the case of four proteins A, B, C, D, each of two copies A1, A2 through D1, D2, we consider all the interactomes including the pair A1-­‐B1, A1-­‐B2, A2-­‐B1, A2-­‐B2, different, whereas in a molecular sense they are the same). Because in this model we have each protein in k copies, each pair is present k2 times in the interactome. Because the interactome has (n x k x m)/2 protein pairs, each interactome constituted of complexes of pairs of different copies of the same two proteins should count the same. With the above numbers, this represents 2
(k )
n×k×m
2
= k n×k×m interactomes that are actually the same. Dividing the above formula with these, the actual number of distinct interactomes in this model is: (n × k × m)!
n ×k×m
n
×
k
×
m
n ×k×m
2
k
×2
×
!
2
Tompa & Rose Supplementary material, Page 3
11
If n = 4500, m = 3540, and k = 3000, (n x k x m)! = 104.9 x 10 , and the number of possible interactomes 10
is about 107.9 x 10 .