If a protein contains four cysteine residues, the number of different ways they can simultaneously form two intra-molecular disulphide bonds is ______.
The question asks for the number of distinct ways to form two simultaneous intra-molecular disulphide bonds from four cysteine residues. A disulphide bond forms between two cysteine residues.
Let the four cysteine residues be denoted as C1, C2, C3, and C4. We need to pair them up to form two disulphide bonds. This is equivalent to finding the number of ways to partition a set of 4 elements into 2 subsets of 2 elements each.
We can use combinations to solve this:
The calculation is as follows: Number of ways = $ \frac{\binom{4}{2} \times \binom{2}{2}}{2!} $
So, the total number of ways is $ \frac{6 \times 1}{2} = 3 $.
Let's list the possible pairings explicitly to confirm:
There are exactly 3 distinct ways to form two simultaneous intra-molecular disulphide bonds from four cysteine residues.
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