Q.33 A protein has seven cysteine residues. The maximum number of disulfide bonds of
different combinations that can possibly be formed by these seven cysteine residues
is _______ (in integer).
The maximum number of different disulfide bond combinations possible from seven cysteine residues in a protein is 105.
Problem Analysis
Seven cysteine residues can form a maximum of three disulfide bonds, leaving one cysteine unpaired since each bond requires two cysteines. The question asks for the total number of distinct pairing combinations achievable with this maximum.
Calculation Method
The number of ways to form k disulfide bonds from 2k cysteines equals the double factorial (2k-1)!!, representing perfect matchings: 1 way for 2 cysteines, 3 ways for 4, and 15 ways for 6. For seven cysteines, select any one of the seven to remain free (7 choices), then pair the remaining six into three bonds in 15 ways: 7 × 15 = 105.
Step-by-Step Derivation
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Choose the unpaired cysteine: 7 options.
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Pair remaining 6 cysteines:
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First cysteine pairs with any of 5 others: 5 ways.
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Next unpaired pairs with any of 3: 3 ways.
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Last two automatically pair: 1 way.
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Total for 6: 5 × 3 × 1 = 15 (or (5!!)).
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Final: 7 × 15 = 105 combinations.
Verification Table
| Cysteines | Max Bonds | Combinations |
|---|---|---|
| 2 | 1 | 1 |
| 4 | 2 | 3 |
| 6 | 3 | 15 |
| 7 | 3 | 105 |
| 8 | 4 | 105 |
Introduction: Disulfide Bonds and Cysteine Pairing in Proteins
Maximum disulfide bonds from seven cysteine residues is a key combinatorics problem in protein biochemistry, frequently appearing in CSIR NET Life Sciences and GATE Biotechnology exams. Disulfide bonds stabilize protein tertiary structure by covalently linking cysteine thiol groups, but with an odd number like seven cysteines, one remains unpaired. This article breaks down the exact calculation yielding 105 different combinations, essential for competitive exam success.
Why 105? The Mathematical Foundation
For an even number 2n cysteines forming n bonds, combinations follow the double factorial formula: (2n-1)!! = (2n-1) × (2n-3) × … × 1. With seven cysteines (2×3 + 1), maximum bonds = 3:
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Ways to pair 6 cysteines: (5!!) = 5 × 3 × 1 = 15.
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Multiply by 7 choices for the free cysteine: 105.
Practical Examples for Exam Prep
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2 cysteines: 1 bond (1 way).
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4 cysteines: 2 bonds (3 ways: 1-2&3-4, 1-3&2-4, 1-4&2-3).
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6 cysteines: 3 bonds (15 ways).
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7 cysteines: 105 ways (as calculated).
This matches protein folding studies where cysteine pairing complexity explodes with count.
CSIR NET/GATE Application
In real proteins, not all combinations occur due to spatial constraints, but exams test theoretical maximums. Remember: odd cysteines always leave one free; use n × (2n-3)!! for 2n-1 residues.
