Friday, October 05, 2007
Are You as Smart as a Third Year University Student? Q5
Question 1
Question 2
Question 3
Question 4
The standard Gibbs free energy change for the aldolase reaction in the direction of cleavage is +28 kJ mol-1. What does this tell you about the properties of this reaction in yeast cells that are actively producing ATP via glycolysis?
a) flux through this particular reaction will be
in the direction of gluconeogenesis
b) the activity of this enzyme must be regulated
c) there must be another enzyme in yeast that bypasses this reaction
d) this is the rate limiting reaction in glycolysis
e) the concentration of FBP will be very much higher than
the concentration of G3P
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9 comments :
a) I would think, it depends ultimately on the fates of G3P and DAP, which if they're being used it will "drive" the equilibrium of the adolase reaction to the production of them and vis versa if I remember my biochemistry right... But at +28 kJ mol-1 to towards cleavage the thermodynamic bias, idealised, is towards the formation of fructose.
Nick, it can't be "a" in cells that are actively converting glucose to pyruvate, can it?
Did you read, The Aldolase Reaction and the Steady State" That's where the answer is. This question is designed to test whether students really understood the meaning of standard Gibbs free energy and near-equilibrium reactions.
I strongly suspect it's a question that most biochemistry students around the world would be unable to answer because most of metabolism these days is being taught by Professors who don't understand it. (Perhaps because their research interests don't require them to.)
Ah, in that case it will be e), as mentioned in the adolase reaction link, the equilibrium position of the reaction will lie strongly on the FBP side, resulting in much higher concentrations of FBP compared to DAP and G3P.
Which I should have actually known given I covered enzyme kinetics last semester.
One question though, can the reaction be driven by removal of DAP and G3P, or is does it run close to Vmax when at equilibrium already?
Nick asks,
One question though, can the reaction be driven by removal of DAP and G3P, ...
If the pathway is running in the direction of glycolysis then the removal of pyruvate will have the effect of (eventually) reducing the concentrations of DAP and G3P. This will result in more F1,6P being converted to DAP and G3P to restore the equilibrium values.
The flux in the pathway will be determined by the rate at which pyruvate is drawn off and by the amount of glucose present in the cell. It's better to talk about flux in a pathway that to refer to an individual reaction being "driven" in one direction.
... or is does it run close to Vmax when at equilibrium already?
The maximum velocity of an enzyme reaction will only approach Vmax at high substrate concentrations. No enzymes operate at maximum velocity inside the cell.
In order to appreciate this you need to think about what the equilibrium constant (Keq) means. It's the ratio of the forward and reverse reactions. In the case of the aldolase reaction if you start out with equal concentrations of substrates and products then the rate of conversion of DAP & G3P to F1,6P will be very high initially and the reverse reaction rate will be lower. Over time the rate of formation of F1,6P drops as the concentrations of DAP and G3P fall. At equilibrium this rate is very much lower than the maximum possible rate under high concentrations of DAP & G3P.
If all enzymes were operating at maximum velocity then the flux in a pathway would only be limited by the concentrations of enzyme and the cell could not respond quickly to changes. This would mean, for example, that the amount of ATP being formed in your muscle cells is as high as it gets even when you are fasting. In that case, pumping more glucose into the blood stream would have no effect.
The correct answer is "None of the above", but that isn't given as an option.
(a) is obviously wrong: the cells are "actively producing ATP via glycolysis".
(b) is irrelevant. The conditions stated say nothing about regulation, and thermodynamics isn't about regulation.
(c) might conceivably apply to a more complex example, but not to this one.
(d) is again irrelevant: thermodynamics isn't about rates.
(e) is also wrong, but for more subtle reasons. The standard Gibbs energy of the reaction refers to standard conditions, i.e. to 1 mol/L concentrations of all reactants (not counting water or protons), but the concentrations in the cell are nowhere near 1 mol/L. This doesn't matter much for most enzyme-catalysed reactionsv with the same numbers of molecules on both sides of the chemical equation, but it matters a lot for aldolase which has one molecule of fructose 1,6-bisphosphate on one side and two molecules of triose phosphates on the other. In the standard state the equilibrium constant strongly favours formation of fructose 1,6-bisphosphate, so the reaction could only proceed towards triose phosphates with efficient removal of the products. However, cellular concentrations are very much lower, and at these concentrations the equilibrium constant favours the formation of triose phosphates (albeit not very strongly, and the reaction is quite capable of going the opposite way in gluconeogenesis).
athel,
Your reasoning is flawed, "e" is the correct answer.
Most cellular reactions are near-equilibrium reactions and that includes the aldolase reaction. Thus, under normal steady-state conditions inside the cell the concentration of fructose 1,6-biphosphate will be much higher than the concentrations of DAP or G3P. This will be true if flux is in the direction of glycolysis or if it's in the direction of gluconeogenesis.
The relative concentrations of substrates and products can be calculated directly from the standard Gibbs free energy change and all biochemistry students should know how to do this.
The one little wrinkle is that the steady-state concentrations of DAP and G3P will not be equal because of the equilibrium of the triose phosphate isomerase reaction. It doesn't change the answer to this question.
I thought you might say that, and I realized that (e) was supposed to be the right answer, but it isn't.
The equilibrium constant we can calculate from the standard Gibbs energy is about 6.7 x 10^-5 mol/L. So the concentrations of triose phosphates in equilibrium with 1 mol/L fructose 1,6-bisphosphate are indeed very small. (Like you, I ignore the triose phosphate isomerase reaction, which complicates the issue without adding anything important to the principle at issue.)
But who cares what happens to 1 mol/L fructose 1,6-bisphosphate, which is far higher than anything the cell will see? If we put both triose phosphate concentrations to 50 micromol/L, for example, we find the equilibrium concentration of fructose 1,6-bisphosphate to be 37 micromol/L, i.e. the equilibrium slightly favours the glycolytic direction (see Biochem. Educ. 9, 133-137 (1981)). Depending on what you mean by “directly” it is just not true that “the relative concentrations of substrates and products can be calculated directly from the standard Gibbs free energy change”; you need to specify what the total concentration is. The sort of straightforward calculation you have in mind works fine for reactions with the same numbers of molecules on both sides of the equation, but it doesn’t work for reactions like aldolase.
Unless by “normal steady-state conditions” you mean concentrations of the order of 1 mol/L it is also not true that "under normal steady-state conditions inside the cell the concentration of fructose 1,6-biphosphate will be much higher than the concentrations of DAP or G3P”.
athel,
I understand your concern. You are correct to say that I was wrong when I claimed that you can "directly" calculate the relative concentrations of reactants and products from the equilibrium constant. It's true that this "direct" calculation does not apply to reactions where the number of reactants and products are unequal.
I should have been more careful to generalize the statement, especially when dealing with an exception like aldolase. The correct statement is that if you know the concentrations of either reactant(s) or product(s) then you can calculate the other from Keq.
The problem with multiplying small numbers comes up time and time again in biochemistry. The concentration calculations are very sensitive to the standard Gibbs free energy change and Keq.
This is a good example. You used a standard Gibbs free energy change of 18 kJ mol-1 to get your Keq value of 6.7 x 10-5. Assuming that the concentrations of DAP and G3P are 50 μM, when you plug your Keq value into the equation it's true that the concentration of F1,6P is lower than 50 μM.
However, the actual standard Gibbs free energy change is 28 kJ mol-1 and Keq is close to 10-5. Using those values, when the 3-carbon compounds are at 50 μM the concentration of F1,6P is 250 μM—or five times higher.
Since 50 μM is probably close to the low end of in vivo concentrations, it's still correct to say that there's more F1,6P than either G3P or DAP. However, I should be specifying a cellular concentration in order to be more clear. I will do so from now on.
I didn't realize how close we were to the range of concentrations that would make my statement (answer "e") incorrect.
Thanks.
I'm glad we now agree about the main points.
You are probably right that 50μM is on the low side. I don't remember why I chose that value (more than a quarter of a century ago!). Probably I asked a physiologically-minded colleague what he thought the concentration of fructose 1,6-bisphosphate in the cell was likely to be, and he almost certainly said "it's not known" (because that was the answer I always got to that sort of question) and then probably suggested 50μM as a reasonable guess.
So far as the discrepancy of values for the ΔG° are concerned: you are right, I used the numbers I looked up in 1980, and was too lazy to check whether they agree with the values one can find today.
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