If a protein contains four cysteine residues, the number of different ways they can simultaneously form two intra-molecular disulphide bonds is ______.
Calculating Disulphide Bond Combinations
The question asks for the number of distinct ways to form two simultaneous intra-molecular disulphide bonds from four cysteine residues. A disulphide bond forms between two cysteine residues.
Combinatorial Approach
Let the four cysteine residues be denoted as C1, C2, C3, and C4. We need to pair them up to form two disulphide bonds. This is equivalent to finding the number of ways to partition a set of 4 elements into 2 subsets of 2 elements each.
We can use combinations to solve this:
- Choose 2 residues out of 4 for the first bond: This can be done in $\binom{4}{2}$ ways.
- Choose 2 residues out of the remaining 2 for the second bond: This can be done in $\binom{2}{2}$ ways.
- Since the order in which we form the two bonds does not matter (e.g., forming a bond between C1-C2 and C3-C4 is the same as forming C3-C4 and C1-C2), we divide by the number of ways to order the bonds, which is $2!$.
The calculation is as follows: Number of ways = $ \frac{\binom{4}{2} \times \binom{2}{2}}{2!} $
- $ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 $
- $ \binom{2}{2} = \frac{2!}{2!(2-2)!} = 1 $
- $ 2! = 2 $
So, the total number of ways is $ \frac{6 \times 1}{2} = 3 $.
Identifying Possible Pairings
Let's list the possible pairings explicitly to confirm:
- Bond 1: C1-C2, Bond 2: C3-C4
- Bond 1: C1-C3, Bond 2: C2-C4
- Bond 1: C1-C4, Bond 2: C2-C3
There are exactly 3 distinct ways to form two simultaneous intra-molecular disulphide bonds from four cysteine residues.
Important Questions from Combinations
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(Answer in integer)
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- The number of ways in which a supervisor can choose four workers out of 10 equally competent workers is ________.
