GATE 2023 BT – Question 52
The number of different possible ways of forming five intramolecular disulfide bonds with ten cysteine residues of a protein is ___________.
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Correct answer: 945
Explanation
Ten cysteines form five disulfide bonds, which means splitting the 10 residues into 5 unordered pairs.
**Method.** Pick a partner for the first cysteine (9 ways), then pair up the remaining 8 (pick a partner for the next free one in 7 ways), and so on:
$$9\times7\times5\times3\times1=945 .$$
**Check with the formula** $\dfrac{10!}{2^5\,5!}=\dfrac{3\,628\,800}{32\times120}=945$.
The number of ways is **945**.