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Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: madscientist on March 03, 2007, 12:31:00 AM

Title: formation of phosphodiester bond
Post by: madscientist on March 03, 2007, 12:31:00 AM

Hi all,

I am having trouble comprehending the bonding when the phosphodiester
linkage forms. i understand that the energy released by the 'coupled'
formation rxn of a pyrophosphate molecule 'supplies' the energy for the
endergonic rxn? is the pyrophosphate formation a hydrolysis rxn? is the
bonding between the 5' alpha phospate group to the 3' hydroxyl group a
dehydration rxn?

any advice appreciated

Title: Re: formation of phosphodiester bond
Post by: Yggdrasil on March 04, 2007, 05:53:17 PM

The reaction is a substitution reaction.  Basically, the electrophile is the 5' alpha phosphate, the nucleophile is the 3' hydroxyl, and the leaving group is the pyrophosphate (the gamma and beta phosphates).  As in most other substitution reactions, the nucleophile displaces the leaving group on the electrophile.

Why is this reaction favorable?  Well, breaking the bond between the alpha and beta phosphates does not release energy.  Rather, this bond is very weak so the energy released by forming a bond between the 3' OH and the 5' alpha phosphate is greater than the energy needed to break the alpha-beta phosphate bond.  Furthermore, the pyrophosphate is very favorably hydrolyzed into two phosphates.  This leaves a low steady state concentration of pyrophosphate which helps make the formation of phosphodiester bonds more favorable.

Title: Re: formation of phosphodiester bond
Post by: madscientist on March 04, 2007, 06:45:17 PM

That is a top notch explaination Yggdrasil! Thankyou very much for that, you should take pleasure in the fact that you have made one other person in this world a lil less stupid :P

cheers,

madscientist