All Exams Test series for 1 year @ ₹349 only
Question

The number of different possible ways of forming five intramolecular disulfide bonds with ten cysteine residues of a protein is ________

Understanding Disulfide Bond Formation

The question asks for the number of ways to form 5 intramolecular disulfide bonds using 10 cysteine residues. A disulfide bond forms between two cysteine residues. Since we have 10 cysteines and need 5 bonds, all cysteines are involved in forming pairs.

Calculating Possible Pairings

This is a combinatorial problem of pairing up 10 distinct items (cysteine residues). We can calculate this using the formula for partitioning $2n$ items into $n$ pairs, which is given by the double factorial $(2n-1)!!$.

In this case, $N = 10$ cysteine residues, so $n = N/2 = 5$ pairs (disulfide bonds).

The number of ways is calculated as $(2 \times 5 - 1)!! = 9!!$.

Double Factorial Calculation

The double factorial $9!!$ is calculated as:

$ 9!! = 9 \times 7 \times 5 \times 3 \times 1 $

Performing the multiplication:

$ 9 \times 7 = 63 $ $ 63 \times 5 = 315 $ $ 315 \times 3 = 945 $ $ 945 \times 1 = 945 $

Therefore, there are 945 different possible ways to form five intramolecular disulfide bonds with ten cysteine residues.

Alternative Combinatorial Calculation

Alternatively, we can choose pairs sequentially and account for overcounting:

  • Choose the first pair from 10 cysteines: $\binom{10}{2}$ ways.
  • Choose the second pair from the remaining 8: $\binom{8}{2}$ ways.
  • Choose the third pair from the remaining 6: $\binom{6}{2}$ ways.
  • Choose the fourth pair from the remaining 4: $\binom{4}{2}$ ways.
  • Choose the fifth pair from the remaining 2: $\binom{2}{2}$ ways.

The initial calculation is $\binom{10}{2} \times \binom{8}{2} \times \binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}$.

Calculating the values:

$ \frac{10 \times 9}{2} \times \frac{8 \times 7}{2} \times \frac{6 \times 5}{2} \times \frac{4 \times 3}{2} \times \frac{2 \times 1}{2} = 45 \times 28 \times 15 \times 6 \times 1 = 113,400 $

Since the order of forming the 5 pairs does not matter, we divide by $5!$ (the number of ways to order the pairs):

$ \frac{113,400}{5!} = \frac{113,400}{120} = 945 $

Both methods confirm that there are 945 possible ways.

Was this answer helpful?

Important Questions from Combinations

  1. How many ways are there to pack six copies of the same book into four identical boxes, where a box can contain as many as six books ?

  2. Bob is studying the effect of coral and sponge species on reef ecosystems using experiments in artificial square tanks. In each tank, he places 3 species of corals and 2 species of sponges. If there are 6 species of corals and 5 species of sponges to choose from, the minimum number of tanks required to test all combinations of 3 coral and 2 sponge species is _______ 

    (Answer in integer)

  3. There are nine species of Impatiens (balsams) found in laterite plateaus of the northern Western Ghats, each with a distinct colour. If a plateau has exactly 6 species, then the number of possible colour combinations in the plateau is __________. (Answer in integer)
  4. Ten teams participate in a tournament. Every team plays each of the other teams twice. The total number of matches to be played is
  5. The number of ways in which a supervisor can choose four workers out of 10 equally competent workers is ________.
Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App