Something more to ask.

Phil Harrison arsphys at cc.usu.edu
Mon Feb 4 11:50:17 EST 2002

Frank is right.  The purpose of this list is not to do your schoolwork for 
you.  These questions are the sort you should take to your teacher.  That 
is what he/she is paid for.

Note: any common dictionary can tell  you the meaning of "porcine."  Don't 
waste people's time asking them to do for you things you could easily do 
for yourself.


At 09:09 AM 02/03/2002 -0800, you wrote:

>Hello again,
>   Sorry if I am using your board to ask so many questions. I am
>reading the article on allosteric regulation of catalysis for Fructose
>First Part
>   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
>say so.
>Second Part
>   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
>Wild type:
>   [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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