Disulfide bond formation A protein, shown schematically in...

Disulfide bond formation A protein, shown schematically in Figure $6.30,$ may contain several cysteines (represented as light colored balis), which may pair together to form disulfide bonds. For six cysteines, three disulfide bonds can form; the figure shows the pairing arrangement $1-6,2-5,3-4 .$ How many different disulfide pairing arrangements are possible? Derive the general formula for the number of different pairing arrangements when there are $n$ cysteines (and $n$ is even). (Adapted from problem 1.11 of $\mathrm{K}$. Dill and S. Bromberg, Molecular Driving Forces, 2 nd ed. Garland Science, 2011 .)

Disulfide bond formation
A protein, shown schematically in Figure $6.30,$ may contain several cysteines (represented as light colored balis), which may pair together to form disulfide bonds. For six cysteines, three disulfide bonds can form; the figure shows the pairing arrangement $1-6,2-5,3-4 .$ How many different disulfide pairing arrangements are possible? Derive the general formula for the number of different pairing arrangements when there are $n$ cysteines (and $n$ is even). (Adapted from problem 1.11 of $\mathrm{K}$. Dill and S. Bromberg, Molecular Driving Forces, 2 nd ed. Garland Science, 2011 .)

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Perfect Matching

A perfect matching is a way of partitioning a set with an even number of elements into pairs such that each element is used exactly once. This concept is crucial in combinatorial problems where one must count the number of ways to form distinct pairs from a collection of items.

Double Factorial

The double factorial, denoted as (n - 1)!! for an odd n, is the product of all odd integers from 1 up to n - 1. It naturally emerges in counting problems involving pairings as it reflects the sequential reduction in the number of available choices when forming pairs.

Overcount Correction in Combinatorial Counting

When calculating the number of pairings, it is common to initially count arrangements where the order of pairs and the order within each pair is considered. Correcting for this overcounting by dividing by the appropriate factors, such as 2^(n/2) for pair order and (n/2)! for the order of pairs, yields the correct formula for the number of disjoint pairings.

Transcript

Please give Ace some feedback