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In glycolysis, glucose is converted to glucose 6-phosphate so it can not diffuse out of the membrane. Then it is converted to fructose 6-phosphate.

Why is this? Perhaps it makes it less stable so it is easier to break down into pyruvate?

That is just a guess, is anyone able to provide more information about this?

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    $\begingroup$ Don't guess. Please do some research before posting basic questions that can be answered by reading a text book of biochemistry. For example Chapter 16 of Berg et al. online. $\endgroup$ – David Aug 30 '17 at 21:20
  • $\begingroup$ However as I do not think either of the answers (including the one you accepted) are adequate, the point is less obvious than I imagined. I have therefore provided my own answer. $\endgroup$ – David Sep 6 '17 at 13:45
  • $\begingroup$ In glycolysis, free energy (sequestered in the form of ATP) is derived from the splitting of glucose. One mechanistic explanation for the conversion of glucose to fructose is that it facilitates splitting of glucose via (reverse) aldol condensation (in the aldolase reaction) as aldol condensations are are 'facilitated' by having a carbonyl group next to the site of cleavage.. See this great article for a much better description than what I have given: weizmann.ac.il/plants/Milo/sites/plants.Milo/files/publications/… $\endgroup$ – user1136 Sep 6 '17 at 19:51
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Avoiding diffusion is one reason to phosphorylate glucose, the other is that it is removed from the osmotic balance between inside and outside of the membrane, so it can be transported at a high rate.

The Glucose-6-phosphate can then be used as a substrate for different pathways, namely glycolysis and the pentose phosphate way, and (depending on the organism) also be converted into glycogen and starch for further storage.

The reason for the phosphorylation lies further downstream in glycolysis: The isomerization by the glucose phosphate isomerase and the subsequent second phosphorylation into Fructose-1,6-biphosphate make the conversion and dedication of the molecule into the glycolysis irreversible.

Fructose-1,6-biphosphate is then cleaved by the aldolase into two C3-units: Dihydroxyacetone phosphate (DHAP) and Glyceraldehyde 3-phosphate (GA-3-P). GA-3-P is converted into DHAP, so that downstream only one metabolite needs to be processed.

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  • $\begingroup$ Very good response thank you, in addition I finded out that fructose 1,6 biphosphate is somewhat more "simetrical" than glucose-6-phosphate, if it wasn't isomerized it would only form 1 usable substrate, the other would be slightly off shape and wouldn't fit in enzymes further in the chain, I not sure if this is accurate, but it sounds logical. $\endgroup$ – Rafael Franco Aug 30 '17 at 15:02
  • $\begingroup$ @RafaelFranco - please don't thank people - upvote the answer if it's useful and accept it when it answers your question. That way other users can immediately assess the impact of the answer. $\endgroup$ – AliceD Aug 30 '17 at 22:35
  • $\begingroup$ So what part of your answer is to the question of “why fructose?”. I quote: “Then it [glucose 6-phosphate] is converted into fructose-6-phosphate, why is this?” $\endgroup$ – David Sep 6 '17 at 13:48
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If there is no glucose there is no need for glycolysis: I deduce from this truism that – at some early stage in the evolution of metabolism – a pathway resembling gluconeogenesis must have arisen before glycolysis. This is just another way of stating the obvious fact that autotrophy must have preceded heterotrophy.

If the aldolase reaction for triose → hexose evolved around glyceraldehyde 3-phosphate/dihydroxyacetone phosphate then perhaps the involvement of fructose 1,6-bisphosphate was dictated by the underlying chemistry of the aldolase condensation, and the steps to glucose followed from there. Note that the main carbon fixation pathway in photoautotrophs does indeed use this process: carbon fixation by RuBisCO generates 3-phosphoglycerate and the subsequent action of triosephosphate isomerase and aldolase generates fructose 1,6-bisphosphate.

So, my main point is that seeking the underlying design principles in the glycolytic pathway may be futile: glycolysis represents the catabolic use of a sequence of reactions that originally evolved for the purpose of anabolism.

Disclaimer: I have probably read this idea somewhere but I have no recollection of where. If anyone can point to an authoritative version of this please do so.

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    $\begingroup$ The anonymous downvoter strikes again. $\endgroup$ – Alan Boyd Aug 31 '17 at 6:39
  • $\begingroup$ Alan, regardless of whether gluconeogenesis or glycolysis came first (interesting — you should post this as a separate question) the question is remains — “why is there a fructose hexose stage between the triose and the glucose hexose?” That is the essence of what the poster is asking, even if he is coming at it from glycolysis. $\endgroup$ – David Sep 6 '17 at 13:51
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Clarifying the question

The pathway of glycolysis starts a hexose (glucose), but at a certain point — the aldolase reaction — two molecules of a triose are generated, then interconverted to the same triose, after which the pathway continues to pyruvate, generating net ATP. These triose molecules each carry a phosphate, which, as the poster mentions, prevents them diffusing through the cell membrane, so the hexose precursor needs to carry two phosphates. Thus the question “why is fructose formed?” can be restated as:

“Why is the hexose precursor of the two triose phosphates fructose 1,6-bis phosphate, and not simply glucose 1,6-bis phosphate?”

Short Answer

The chemical structure of glucose is not suitable to form a bis-phosphate compound that can yield interconvertable triose phosphates in the biochemical reactions the cell employs.

Fuller Answer with Chemical Details

First consider the linear and cyclized structures of relevant phosphate derivatives of glucose and fructose:

Hexoses and their phosphates

Glucose 1,6-bis phosphate may be unfamiliar: it is found in cells in low amounts and may have some regulatory function. Inspection shows that, because the aldehydic group is at the 1-position of glucose, a phosphate can only be added at that position after cyclization generates an alcohol there. Once glucose 1,6-bis phosphate has been formed it cannot convert to a straight-chain structure, unlike fructose 1,6-bis phosphate where the ketonic group is in the 2-position. (Hence none is shown.) This is the key point.

Now let us turn to the triose–hexose interconversion, which is necessary for both glycolysis and gluconeogenesis, which ever originated first. This has two requirements:

  1. That the triose phosphate molecules can react with one another to form the hexose carbon backbone (easier to consider the reaction in this direction).
  2. That if different triose molecules are required they can easily interconvert from (or to) a single triose.

Requirement 1 is satisfied most easily in biological systems by an aldol condensation such as the aldolase reaction (also found in the Calvin cycle) in which the nucleophilic carbon of an enolate attacks the carbon of an aldehyde.

Requirement 2 can be satisfied by the isomerization of an aldehyde (like glyceraldehyde 3-phosphate) to a ketone (like dihydroxyacetone phosphate).

Triose phosphate interconversions

The fact that a ketone like dihydroxyacetone phosphate has the potential to form an enolate (shown above) explains the particular suitability of these trioses in relation to the aldose reaction:

Aldolase reaction

[Taken from Chemistry Libre Texts.]

The initial product of this reaction is, of necessity, a linear hexose 1,6 bis-phosphate — with the fructose as the hexose as its carbonyl is at C2. After the fructose 1,6-bis phosphate has lost the phosphate at the C1 position it can isomerize to glucose 6-phosphate (with the C1 position free to form an aldehyde) in a similar manner to the triose phosphate isomerization.

So it is not just the symmetry of fructose (noted by the poster in a comment) that is important — but the structure of glucose itself. Glucose 1,6-bis phosphate is not, of course, the only sugar that can become ‘locked’ in its cyclic form. For example, the deoxy-ribose rings in the backbone of DNA cannot linearize because the bases are attached to the C1ʹ where the aldehydic group was originally.

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  • $\begingroup$ I would be amazed if there are no typos or chemical errors in this and would glad to be notified of any that are found. $\endgroup$ – David Sep 6 '17 at 14:01

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