Friday, October 05, 2007
Are You as Smart as a Third Year University Student? Q4
Question 1
Question 2
Question 3The open-chain form of fructose 1,6-bisphosphate is shown as the substrate for the aldolase reaction. Why?
a) the open-chain form is more abundant inside the cell
b) cyclic molecules destabilize the transition state
c) the product of the previous reaction in glycolysis
is the open-chain form
d) the open-chain form is thermodynamically more stable
and this contributes to the positive standard Gibbs free
change for the reaction
e) the active site of the enzyme can’t accommodate the
furanose or pyranose forms
I'm going to say c) the addition of the phosphate groups increases the thermodynamic barrier for cyclisation , effectively trapping the fructose molecule in the open state, while also sequestering inside the cell due to the large negative charges on the phosphates.
ReplyDeleteNick, "c" can't be right because students will have already seen the previous reaction in glycolysis—the one catalyzed by phosphofructokinase 1—and they will know that the substrates and products are cyclized furanose forms.
ReplyDeleteFurthermore, students should realize that aldolase evolved as an enzyme that catalyzes the reverse reaction so there was unlikely to be selection for using the open-chain form as a substrate.
(e)
ReplyDeleteIn the cell, there is an equilibrium between cyclic and open-chain forms. Although cyclic is highly favoured, open-chain is still present, allowing it to be used by the enzyme (which can only accomadate the open-chain form)
Whoops, forgot some basic organic chemistry there. The addition of the extra phosphate doesn't effect the re-cyclisation to the furanose form as there is an enol group still in the open molecule...
ReplyDeleteMy problem here though was ignorance over the evolutionary history of aldolase, I should have remembered and realised that if the citric acid cycle ran in the opposite direction, a good deal of the gycolysis pathway would have also run that way. Thus producing C3 to C6 units for use elsewhere, but primarily for neoglucogenesis.
But back to the question, It appears from a quick look at the PyMol render of the human aldolase B enzyme, that the active site can accommodate the cyclic form. However it looks like it wouldn't bind particularly as strongly as the open form. Thus it could be that the open form is more thermodynamically stable in the active site than the closed form. So d) is the probable answer going on the above lines of logic...
Nick,
ReplyDelete"d" can't be right because we know that at equilibrium the furanose form predominates in vivo.
Mathew got it right. Remember that the aldolase reaction catalyzes fusion of two 3-carbon molecules to create fructose. It stands to reason that the primary product of the reaction is the open-chain form and that form must be accommodated in the active site of the enzyme.
The question only becomes difficult to answer if the students are fixated on glycolysis and forget about gluconeogenesis. Since I spend a large part of the course teaching them that biosynthesis pathways are always more important than degradation pathways—in the big picture of biochemsitry—I expect them to get this right.
In case you're interested, about 70% of my second year students choose the correct answer.