Question

The number of different possible ways of forming five intramolecular disulfide bonds with ten cysteine residues of a protein is ________________.

Hint:

Pairing \(2n\) distinct items into \(n\) unordered pairs uses \((2n-1)!!=\dfrac{(2n)!}{2^n n!}\).

Answer 1

Correct Answer: 945

Step 1: Forming five disulfide bonds from ten cysteines is equivalent to pairing the ten distinct residues into five unordered pairs.
Step 2: The number of pairings is \((10-1)!!=9!!=9\times7\times5\times3\times1\). Alternatively, \[ \frac{10!}{2^5\,5!}=\frac{3628800}{32\times120}=945. \] Hence, the number of possible ways is \(\boxed{945}\).