- November 23, 2010 at 4:30 pm #14173
I have to predict the positions of the glucose molecule that would become labeled from the 1-14C pyruvate (i.e, the label in the carboxyl group of pyruvate).
I first thought that carbons 1 and 6 were the correct answers, but they are not. Then I thought that 1,3,and 6 were correct but they are not. I know that pyruvate has 3 carbons and glucose has 6, but I know two how to approach this problem correctly. Any help will be appreciate it.
Thank you and happy holidays!
- November 23, 2010 at 6:05 pm #102412JackBeanParticipant
look on the reactions, which take place in gluconeogenesis and their mechanism. It’s quite simple to trace the labeled atom…
- November 23, 2010 at 7:10 pm #102413
So for this kind of problem you look at where the new carbon is being attached to, correct?
- November 24, 2010 at 8:20 am #102429
- November 24, 2010 at 7:34 pm #102448USFBiomedParticipant
3,4… its a head to head bonding, not tail to tail, so 1,6 would be the reverse…
- November 24, 2010 at 7:55 pm #102451
Thank you USFBiomed. So since pyruvate is 3 carbons and glucose is 6 I connect the last of pyruvate with the first incoming carbon atom so 3 and 4 like you said. Is this explanation correct? Thanks again for your help and happy holidays!
- April 24, 2015 at 11:46 pm #115707MoleculeusParticipant
When the two triosephosphates condense head-to-head to form a hexose, carbon 1 of the triose (originating from the carboxyl of pyruvate) forms carbons 3 and 4 of the hexose, carbon 2 of the triose forms carbons 2 and 5 of the hexose, and carbon 3 of the triose forms carbons 1 and 6 of the hexose.
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