isomerization of G6P to F6P by PGI

Question 1

PGI

Answer 1

• antiparallel (so that youre picking up active site amino acids from two separate subunits)
• homodimer

Question 2

PGI catalyzes rxn with…

Answer 2

really small negative delta G, which tells us its at equilibrium and is controlled by concentration gradient.
(A lot of S will form P, if a lot of P will form S)

Question 3

What do you have to do before isomerization? What happens during and after isomerization?

Answer 3

• before isomerization, must open 5C + O ring
• once the ring is open, isomerization of the RCHO to a ketone
• finally, the ring is closed to a smaller size which is a 4C + O ring

Question 4

PGI has near absolute stereospecificity

Answer 4

this means its not 100% stereospecific… means you are only forming one product

Question 5

At equilibrium

Answer 5

• can go through a common intermediate, which is the cis Enediolate intermediate.
• common intermediate can sen dyou to 2 separate products due to epimerization (flip orientation of first chiral OH to mannose) or isomerize to fructose. WE DO NOT WANT THE MANNOSE, NO EPIMERIZATION
  ( cann start as any form [glucose, mannose, fructose] and end up as any form, but enzyme (PGI) only selects glucose to fructose

Question 6

PGI selects which pathway

Answer 6

glucose and fructose, does not allow mannose to be used or formed

Question 7

microscopic reversibility

Answer 7

to go backwards, you must follow the same steps as the forward direction.

Question 8

What organic chem reactions are present in the mech of isomerization of G6P to F6P

Answer 8

• step 1: opening a hemiacetal (RCHO)
• step 2: ketoenol taut
• step 3: keto enol taut
• last step: hemiacetal formation (with ketone)

Question 9

Histidine

Answer 9

• in other dimer subunit so active site has AA from both subunits

Question 10

When ring opens to linear form, what happens?

Answer 10

there are key bond rotations:
- C3 and C4 bond rotate to bring C1(Where the carbonyl is already) and C2 (where you want to put the carbonyl) close to Glu for isomerization of RCHO to ketone
- then, there is a re-rotation of the bond to close the ring

Question 11

Why is that bond rotation important?

Answer 11

• Mannose 6 P, if present, it could require a C2 C3 bond rotation instead of C3 C4 bond roation
• this M6P bond rotation is blocked by a glutamine
• blocking M6P from the common intermediate
• how is this connected to microscopic reversibility?
• If the enzyme was fed M6P, unable to form F6P. This means that, due to the rule of microscopic reversibility, Glu is blocking Glucose and Fruc from forming mannose. THis explains near absolute stereospeicifity of G6P, if mannose cannot become intermediate, then the intermediate cannot be mannose. Therefore only G6P and F6P can exist in equilibrium

Question 12

which 4 amino acids

Answer 12

his, glu, lys, gln