Joined: Feb. 2008
|Quote (raguel @ Nov. 29 2012,17:20)|
|Quote (Jerry Don Bauer @ Nov. 29 2012,16:25)|
|They aren't all going to be enantiomers. They may be isomers, and possibly even stereoisomers. Some of those stereoisomers will be enantiomers of each other (which may or not be itself), but the rest will be diasteriomers (that may also have a mirror image).|
I can only imagine you still don't get it at this point. Let's try an example and see what happens. Suppose there's a chiral compound A(s), and it's mirror image A® and they can form a dimer. A(s)-A® and A(s)-A(s) are not enantiomers. They are diasteriomers. Do you agree or disagree?
And if they aren't all going to be enantiomers or the same product then you don't know ahead of time the odds of any one of them forming.
Why would I agree or disagree, it's all irrelevant to anything I've stated.
But, I'm curious...... all of that means.....exactly what to you? How is it relevant to you? I'll reserve comment until you expand on your point....
LOL. Well, let's start with something simple then. You originally wrote this:
|The odds against assembling a protein chain consisting of only left-handed amino acids by chance is 2 to the “n” th power. And “n” is the number of attached amino acids in the protein.|
But you didn't show how you came up with that equation. I assumed you came to the above conclusion because you think, for reasons you alone can verify, that each possible reaction at each step has the same chance. (For example, say that there's already a polypeptide P and one is reacting it with a racemic mixture of arginine. You are claiming that there's an equal chance of P-A(l) and P-A(d) forming.) If that's not what you meant,show us how you determined that probability. If you did mean that, please explain why you are assuming that each possibility has an equal chance.
Any thoughts on this yet?Remember to show your work