Q.52 The number of different possible ways of forming five intramolecular disulfide bonds with ten cysteine residues of a protein is ___________.

Q.52 The number of different possible ways of forming five intramolecular disulfide
bonds with ten cysteine residues of a protein is ___________.

Number of Ways to Form Five Intramolecular Disulfide Bonds with Ten Cysteine Residues

The number of different possible ways to form five intramolecular disulfide bonds from ten distinct cysteine residues is 945.

Problem Explanation

Ten labeled cysteine residues must pair into five disulfide bonds.
Each bond connects exactly two cysteines, and bond order does not matter.
This is equivalent to counting perfect matchings in 10 items.

The total number of pairings for 2n distinct elements forming n disulfide bonds is:

(2n − 1)!! = (2n)! / (2ⁿ n!)

For n = 5:

(10 − 1)!! = 9!! = 945

Step-by-Step Pairing Calculation

Select pairs sequentially:

(10 choose 2) = 45
(8 choose 2) = 28
(6 choose 2) = 15
(4 choose 2) = 6
(2 choose 2) = 1
Product = 45 × 28 × 15 × 6 × 1 = 113400

Divide by 5! because bond order doesn’t matter:

113400 / 120 = 945

Final Answer

945 possible disulfide bonding patterns

Biological Context

Disulfide bonds help stabilize tertiary protein structure.
With ten cysteines available, many incorrect pairings can form, so cellular disulfide isomerases rearrange mispaired cysteines to reach the correct fold.

Introduction

Proteins containing ten cysteine residues may form five intramolecular disulfide bonds.
Counting the number of distinct ways these bonds can pair is a classic GATE BT and IIT JAM exam question.
The answer, 945, highlights the combinatorial complexity of folding and the role of disulfide isomerases in achieving correct structural alignment.

Disulfide Bond Formation in Proteins

Two cysteines oxidize to form a disulfide:

2 R-SH → R-S-S-R + 2H⁺ + 2e⁻

Functions of disulfides:

  • Increase protein stability
  • Stabilize extracellular and secreted proteins
  • Protect against proteolysis

Detailed Combinatorics Proof

General pairing formula:

(2n)! / (2ⁿ n!)

For n = 5:

(10)! / (2⁵ × 5!) = 3628800 / (32 × 120) = 945

Matches the sequential pairing product and confirms the result.

Exam Relevance

Appears often in:

  • GATE Biotechnology
  • IIT JAM Biotechnology
  • Biochemistry and folding thermodynamics

Common mistakes:

  • Not dividing by 5!
  • Treating cysteines as indistinguishable

Final Key Takeaway: The number of ways = 945

 

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