Exploring the Math of Glucose Polymers: Understanding 6^n

[bo reddude]

Homework Statement

not homework, just curious

Relevant Equations

(CH2O)_n

Let's say you have n glucose monomers. (C6H12O6) n

You want to find out how many possible polymers exist in combining those n number of glucose molecules randomly.

So glucose_1 has 6 OHs that can combine with glucose_2 which also has 6 OHs. Starting with glucose_1's first carbon C1, at that position, you can have 6 different OHs from glucose_2 attaching to it.

Since glucose_1 has 5 other OHs, and each of those OHs can have 6 different OH from glucose_2 attaching to it, you have total of 36 different configuration of glucose dimers.
what happens if you were to think about all possible combination of glucose polymer of arbitrary length n?

It seems like 6^ n, but I can't work out the details in trying to explain it. What's the mathematical formalism involved in this?

Thanks for any help.

 

[Borek]

This is more math and combinatorics than chemistry.

In chemistry glucoses don't react at random -OH, in the dominating hemiacetal structure some positions are much more prone to react during polymerization.

 

[bo reddude]

Is there a way to cross post to the math forum on here?

 

[jim mcnamara]

Thread moved

 

[bo reddude]

thank you

 

[Borek]

To all math people wondering what the question is:

This star (let's call it a "monomer") can be connected to identical stars by linking any of the ends with any other end of any other star (this would be roughly what chemists call "condensation", and the product is "polymer" - don't treat these terms too seriously as chemical terms here, I am using them for brevity and ignoring details). Assume each star can connect only to two others. Question is, how many different "polymers" can be produced from n "monomers".

Supposedly OP can add some details, the question was originally posted in chemistry - but that's not how the glucose really react, so could be there are some additional limits on how these abstract representation can behave.

 

[bo reddude]

no one wants to try and answer this question?