Like @Karsten, I am very wary of “why?” questions in biochemistry, which often come from individuals who see themselves as some sort of evolutionary God. And their answers tend to reflect their own scientific mindset: “to the man with a hammer, everything looks like a nail”.
Nevertheless, it is not unreasonable for a chemist to look at the reactions in metabolic pathways from a chemical point of view and try to identify an overall ‘strategy’, as long as one resists the temptation to conclude that this is the only, or even the best, strategy. It may be the first one that worked and became frozen as metabolism elaborated around it, or it may just take advantage of other pathways that had already developed when it evolved. Glyceraldehyde 3-phosphate is the product of the light reactions of photosynthesis and of gluconeogenesis. Dihydroxyacetone phosphate is the immediate precursor of glycerol 3-phosphate, needed for triglyceride synthesis. Fructose 6-phosphate is a product of the pentose phosphate shunt.
In addition, one may be able to rationalize the rejection of an alternative pathway on several grounds, so who is to say that one is the ‘reason’ for rejection?
What is the question?
The problem here is that the poster’s instructor (I assume the context is student–instructor) meant one thing by “why” and the poster assumed, quite reasonably, that a mechanistic/molecular answer was required and gave a correct one. The substrate specificity of the glycolytic enzyme aldolase (strictly fructose 1,6-bisphosphate aldolase) is such that it does not ‘act’ on glucose 6-phosphate in glycolysis.
What did the instructor actually mean? The “waits until the formation of fructose 1,6-bisphosphate” suggests to me “What strategic advantage can you see in an aldolase acting on the bisphosphate form of the hexose rather than the 6-phosphate form?”, for which I provide a rationale below. But it would seem from his answer in terms of (methyl?) glyoxal formation that his concern was the reactivity of the aldehyde group in one of the triose products that would be produced instead of dihydroxyacetone phosphate.
The obvious objection to this and answer to my interpretation of the question is that the strategy of the first part of glycolysis is to convert a hexose to two molecules of a triose phosphate and to do this a hexose bis-phosphate is formed (see diagram). Not the only way, but one that works and, in the case of fructose 1,6-bisphosphate, allows easy and rapid isomerization between the two trioses. (Glyceraldehyde 3-phosphate is the form that is further metabolized to pyruvate.)
Unfair to the Instructor?
Bringing in the monophosphate, glucose 6-phosphate seems rather a pretext to introduce a story about glyoxal. But perhaps I am being unfair. Let us presume he is well aware of the need for two molecules of triose phosphate and is taking it for granted that the triose resulting from the “top” half of the hexose (C1 to C3) will be phosphorylated after an appropriate aldolase reaction. I am not going to say this would be unviable or inferior (although that is possible) but if such a strategy had been developed, surely the triose phosphate formed would no longer be suitable for conversion to glyoxal.
Don’t ask questions in answers
But just this once, the focus on glucose 6-phosphate raises another question, why have an extra step to produce fructose 1,6-bisphosphate from glucose 1,6-bisphosphate? I presume that the aldolase reaction on the latter would result in triose molecules that were chemically ‘unsuitable’ for the subsequent steps, but my chemistry is too rusty to be sure. Is this so? And would the aldolase products of fructose 6-phosphate convert to glyoxal? (Answers provided in the comments will be used to improve this answer.)
Observations on the toxicity of Glyoxal
Although the toxicity of glyoxal is not in question, the ability of the cell to deal with this needs to be considered.
First I would like to ask whether the supposition is actually correct: that delaying the aldolase reaction until the fructose 1,6-bisphosphate stage per se prevents the formation of toxic glyoxal?
In Berg et al. Biochemistry I read the following:
“…second TIM (triose phosphate isomerase) suppresses an undesirable side reaction, the decomposition of the enediol intermediate into methyl glyoxal and orthophosphate…This labile intermediate is trapped in the active site by the movement of a loop of ten residues…”
Second I would point out that certain bacteria and even human cells synthesize glyoxal, for reasons that in some cases are not entirely clear. (As far as the chemistry goes, they use hydroxyacetone phosphate as the precursor.) However, at least in bacteria, detoxification enzymes have developed (Arch Microbiol (1998) 170 : 209–219).