Something more to ask.
mmuurrttuuzzaa at hotmail.com
Sun Feb 3 12:09:27 EST 2002
Sorry if I am using your board to ask so many questions. I am
reading the article on allosteric regulation of catalysis for Fructose
The question I have is that the article mentions that the enzyme has
two states the R and T. The R-state has an engaged/disordered
conformation. The free energy between these two states is very small,
which enables the enzyme to catalyze the hydrolysis of
Fructose-1,6-bisphosphate to Fructose-6-phosphate.
1) In the article it suggests that if the active site loop is
always engaged than (meaning not in engaged/disordered conformation of
R-state) it is a "dead-end complex".
Please explain what "dead end complex" means, what I can
extrapolate is that the enzyme will not catalyze the substrate because
even if the substrate(key) fits in the "lock"(enzyme), the
enzyme(lock) has no way to turn the "key" to catalyze it. Is my
understanding valid or not. Please note that if you think the article
is wrong in its assumption of engaged/disordered conformation please
Now when AMP binds to the enzyme it causes the conformation to
switch to the disengaged T-state. Than the article continues to say
that AMP inhibits catalysis because in the T-state there isn't an
"engaged/disordered" like conformation analogus to the R-state. And
that the free energy between the R and T state is very great to enable
the AMP, to cause catalysis.
2) So Article suggest that by having mutation that destabilize the
disengaged conformation the free energy difference between the R-T
state is less allowing larger catalysis to occur because the
population of R-state is greater even after AMP inhibition.
"Biphasic AMP inhibition should appear whenever a mutation selectively
destabilizes the disengaged loop conformation of the T-state"
Please explain what "Biphasic" means?
My understanding is simply that the presence of both R and T
state even after AMP inhibition. Is this correct?
3) When the article talks about free energy does it mean
[R-state Enzyme] + AMP -----------> [T-state Enzyme]
For wild type the equilibrium lies further to the left.
[R-state Enzyme] + AMP ------------> [T-state Enzyme]
For mutation the Concentration of T-state has decreased.
so the equlibrium is not as far to the left as wild type.
Thank you for your help.
For those of you interested in the article.
Nelson, S.W., Kurbanov, F.T., Honzatko, R.B., and Fromm, H.J. (2001).
The N-terminal segment of recombinant porcine
fructose-1,6-bisphosphatase participates in the allosteric regulation
of catalysis. J. Biol. Chem. 276, 6119-6124. [PubMed] [JBC]
P.S. One last question what does Porcine mean?
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