October 31, 2024, 09:32:57 PM
Forum Rules: Read This Before Posting


Topic: Glucose activation as first step of glycogenesis  (Read 7641 times)

0 Members and 1 Guest are viewing this topic.

Offline opel65

  • Regular Member
  • ***
  • Posts: 14
  • Mole Snacks: +0/-0
Glucose activation as first step of glycogenesis
« on: September 01, 2013, 11:12:43 AM »
Hi, I've biochem retake tomorrow.

I've stumbled upon this question which I don't understand.
What is meant by "glucos activation?"
I know that the alpha-D-Glucose is the source of  glucose in glycogenesis.
I know also that this glucose attached to UDP is the UDP-glucose, and it is made from glucose 1-P.

I still don't know what activation means. Every student in our group got wrong on this question.

Please help

EDIT:
I've found this.

Glycogenin can serve as an acceptor of glucose from UDP-glucose. This glucose serves as a site at which the initial glucosyl groups are attached.

Is this correct?
« Last Edit: September 01, 2013, 11:26:26 AM by opel65 »

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Glucose activation as first step of glycogenesis
« Reply #1 on: September 01, 2013, 11:57:14 AM »
In principle, adding a glucose molecule to the end of a glycogen chain should be easy.  The reaction essentially looks like a standard condensation reaction with the -OH group at the end of glycogen as the nucleophile and the -OH group on carbon 1 of the glucose molecule as the leaving group.  So why does the cell use a much more complicated mechanism for making glycogen?

One major problem with the above reaction is that it is not thermodynamically favorable.  Specifically, an -OH group is not a very favorable leaving group.  So, in order to get the reaction to proceed, the cell must activate the glucose molecule by attaching a good leaving group to it.  Here, UDP plays that role.

Similar processes occur throughout biology, where monomers of a particular polymer must be activated by attaching them to a good leaving group.  The nucleotides are activated by phosphorylation prior to incorporation into DNA or RNA (which attaches a pyrophosphate leaving group).  Similarly, during protein synthesis, attachment of the amino acid to the tRNA acts as the activation step.

Essentially, all of these reactions create an unstable intermediate with a bond that can be easily broken.  This unstable bond makes it very easy for enzymes inside the cell to break that bond and replace it with a bond to another molecule.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5688
  • Mole Snacks: +329/-24
Re: Glucose activation as first step of glycogenesis
« Reply #2 on: September 02, 2013, 10:06:59 AM »
I agree with Yggdrasil, but I would stress that there is a kinetic component to the meaning of activation, as well as a thermodynamic one.  If you write out the reaction, to add glucose to a growing strand of glycogen, you will see that there is a poor leaving group (which suggests that the reaction will be slow).  On the other hand, if you write out the reaction to add a glucosyl group from UDP-glucose, there is now a good leaving group.  The best way to see this is to look at the conjugate acids of the two leaving groups in question.  The stronger conjugate acid is associated with the weaker conjugate base, and weaker bases make better leaving groups than strong bases do, all else held equal. 

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Glucose activation as first step of glycogenesis
« Reply #3 on: September 02, 2013, 03:54:09 PM »
While in typical organic chemistry reactions, activation may act primarily by altering the kinetics of a particular reaction, in biochemistry, activation steps act primarily to alter the thermodynamics of a reaction.

For example, in the case of glycogen synthesis, the enzyme glycogen phosphorylase catalyzes the joining of an unactivated glucose-1-phosphate to glycogen, using inorganic phosphate as the leaving group:

glucose-1-phosphate + glycogen  :lequil: glycogen+1 + Pi

Of course, because the concentration of inorganic phosphate in the cell is so high, what actually occurs when glycogen phosphorylase gets activated is the breakdown of glycogen (i.e. the reverse of the reaction written above).  However, if one were to put put glycogen with the enzyme in the presence of high concentrations of glucose-1-phosphate and low concentrations of inorganic phosphate, the enzyme would likely have no trouble catalyzing the synthesis of glycogen at a reasonable rate. 

Thus in this case. the kinetics of the reaction is not the problem, and activation of glucose-1-phosphate with UDP is required in order for glycogen synthesis to be thermodynamically favored.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5688
  • Mole Snacks: +329/-24
Re: Glucose activation as first step of glycogenesis
« Reply #4 on: September 02, 2013, 05:57:07 PM »
How good a leaving group phosphate is depends on its protonation state; in spite of this ambiguity, it should still be a better leaving group than hydroxide in aqueous solution.  Also, we seem to be looking at two slightly different polymerizations.  I was considering the idea of polymerizing glucose into glycogen, and you are examining polymerizing glucose 1-phosphate into glycogen.

More generally the trouble with the question that we are discussing (is activation a kinetic issue or a thermodynamic one?) is that I cannot think of examples of adding a group to a molecule that change one property but not the other.

Offline opel65

  • Regular Member
  • ***
  • Posts: 14
  • Mole Snacks: +0/-0
Re: Glucose activation as first step of glycogenesis
« Reply #5 on: September 04, 2013, 06:28:48 PM »
Thank you both for these elaborate answers.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5688
  • Mole Snacks: +329/-24
Re: Glucose activation as first step of glycogenesis
« Reply #6 on: September 05, 2013, 09:03:01 AM »
What answer was your instructor looking for?

Sponsored Links