Question

A protein has eight cysteine residues. How many possible ways are there to make four disulfide bridges in its structure?

• A. 105
• B. 4096
• C. 24
• D. 1680

Hint:

Number of ways for n disulfide bonds: \(\frac{(2n)!}{2^n \times n!}\).
For n = 4, it's 105.
Disulfide bonds are important for protein structure and stability.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Disulfide bridges are covalent bonds formed between two cysteine residues in a protein.
They play a crucial role in stabilizing the tertiary structure of proteins.
The number of ways to form disulfide bonds among cysteine residues is a combinatorial problem.

Step 2: Key Formula or Approach:
For a protein with 2n cysteine residues forming n disulfide bonds, the number of possible combinations is:
\[ \text{Number of ways} = \frac{(2n)!}{2^n \times n!} \] Here, 2n = 8 (so n = 4).

Step 3: Detailed Explanation:
Substitute n = 4 into the formula:
\[ \frac{(8)!}{2^4 \times 4!} = \frac{40320}{16 \times 24} = \frac{40320}{384} = 105 \] Thus, there are 105 possible ways to form four disulfide bridges.
Option B (4096) is \(2^8\) (incorrect).
Option C (24) is 4! (incorrect).
Option D (1680) is incorrect.

Step 4: Final Answer:
Thus, the number of possible ways is 105, which corresponds to option (A).
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