more about HK-IV (GK) Flashcards

1
Q

HK-IV must do what

A

must buffer blood glucose levels, so it must have a lower glucose glucose affinity. Why? Activity must increase as blood glucose increases. This is good because you dont want your liver to compete with other places such as the brain for blood glucose
So, NO HK-IV does NOT obey michaelis menten… is it sigmoidal

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2
Q

How is liver glucose usually stored/used and why

A

Why: liver usually uses fats for energy
Stores it as glycogen or uses it in the pentose phosphate pathway

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3
Q

How is HK-IV regulated

A

NOT regulated by G6P, it is instead regulated by protein-protein interactions (GKRP glucokinase regulatory protein)

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4
Q

Low glucose in liver cell..

A

HK-IV is sequestered in the nucleus bound to GKRP (Not the cytosol) means liver glycolysis is going to be operating SLOWLY, because keeping the enzyme that owuld be bringing glucose into the cell is not in the cytosol…. being held in nucleus instead

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5
Q

High glucose in liver cell

A

HK-IV is not sequestered or inhibited

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6
Q

What must happen for GKRP to bind to HK-IV?

A

We need to have F6P also bound to GKRP.

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7
Q

Two ways to activate HK-IV?

A
  1. glucose can change the HKIV conformation
  2. Conformation change of HK-IV due to the molecule F1P
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8
Q

How is F1P different from F6P?

A

F1P forms a pyranose ring while f6P is a furanose ring… ring size is different with F1P (as well as phosphate placement)

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9
Q

F1P and F6P competitiion

A

compete for a common binding site of GKRP.
If F1P present, allow GKRP and HK-IV to break apart (GKRP and HV-IV bound hydrophobically), when this happens HK-IV will leave.

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10
Q

Cytosolic HK-IV and flux controller

A

Complexes with dephospho forms of PFK-2/FBPase-2
Flux controller has two forms… Legs together hands apart and hands together legs apart.

These have two different functions. Legs apart = can bind HK-IV, form multi protein complex. This is the dephospho form, get additional activation of both proteins (flux and HK-IV) by protein-protein interactions which are going to involve things like hydrogen bonding or hydrophobic interactions etc

Legs together = phospho

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11
Q

HK-IV binding increases PFK-2 activity, so….

A

(increase in the complex activity) there is an increase in F26P… the more F26P present, PFK-1 activity may increase!
So if PFK1 more active, glycolysis will be greater.

This is another protein involved in making F26P

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11
Q

Why shoudl HK-IV stick to PFK/FBPase-2?

A

lots of extra glucose = liver pull it out to use it by increasing glycolysis.

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12
Q

Where is F1P coming from?

A

fructose from your diet

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13
Q

HK-IV structure characteristics

A

Monomeric, but exhibits cooperativity. (if i do fast thing before slow thing, i get sigmoidal kinetics which gives you cooperativity)

When glucose is phosphorylated to G6P, it is much quicker to bind the next glucose than it is to convert to inactive conf.
So When GKRP binds, it binds to the super open conformation (Down in the slow box, which makes you stuck in the super open conformation )

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14
Q

What is the more stable conformation of HK-IV?

A

More stable conformation is inactive with low affinity and a slow transition to and from the more active (high affinity) and less stable conformation.
Similar to PGK, can do a conf selection and pop shift

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15
Q

F1P details

A

F1P as pyranose form, will bind to GKRP.
(F6P will bind as a furanose ring! Important difference!)
Upon F1P binding, GKRP’s helix (in SIS-1) does a power helix interaction with F1P’s Phosphate. (All amides aligned that created partial positive on one end and partial negative on the other end)

16
Q

GKRP’s lid

A

Lid interacts with F1P: Positively charge Lysine in GKRP’s lid offsets the negative charge of F1P’s phosphate.
Power helix not full positive charge, need more coming from somewhere else.

17
Q

F6P details

A

WHen F6P is bound to GKRP, the H bonds between THE LID’S argenine, and a histidine in GKRP break, which causes an change in conformation and a more favorable HK-IV interaction with GKRP

18
Q

Is the changing complex between F1P and F6P stable?

A

Not really, there is only 13 Kj/mol of energy released during this process, which is just enough to allow for this regulatory process to occur