Question 2PDB's (Protein Data Bank) molecule of the month for September is citrate synthase, one of the enzymes of the citric acid cycle. Read the PDB website to find out more about this interesting enzyme. [Hat Tip: Philip J at Biocurious]
Citrate synthase catalyzes the following reaction,
One of the most interesting things about this reaction is that the standard Gibbs free energy change of the reaction (ΔG°′) is −31.5 kJ mol-1. Here's a question that I ask of my second year students.
Like most reactions in vivo, the actual Gibbs free energy change for this reaction is zero. Normally you might expect that such a large negative standard Gibbs free energy change would indicate that the forward reaction is coupled to the synthesis of ATP. Indeed, the hydrolysis of the similar thioester bond in succinyl CoA (Step 5 of the citric acid cycle) is coupled to the synthesis of GTP (or ATP). However, in the case of the citrate synthase reaction, the available energy is used for a different purpose. What is this purpose?
10 comments :
The available energy is used to produce FADH and NADH, that are then part of the Electron Transport Chain that pumps H+ ions (actually Hydronium, H3O+, I think) across the membrane of the mitochondrion, storing potential energy in the form of a chemical gradient.
Wait, most reactions in vivo have a delta G of zero? Doesn't that imply equilibrium? Cells are certainly not at equilibrium even though some reactions can be usefully thought of as at equilibrium.
Ouch! Disclosure: I had to peek on the web (well, Wikipedia), so I don't win anything except knowledge.
Citrate synthase is the first pace-maker enzyme in this cycle, so presumably some energy goes into release of a competitive inhibitor such as too much product or produced NADH.
(And I can't help but think that this steady state regulation somehow connects to bob's question on delta G = 0.)
A related question: citrate synthase uses conformational change to drive the reaction in the desired direction. Assuming the first change is made by the first molecule attached, do we get all the energy back when releasing the product or is there somehow a heat loss to the environment? I'm asking, because I have seen energy potential diagram for reactions, but I haven't seen one for conformational changing molecules. (IANAC. :-P)
There's a difference between a dynamic and static equilibrium.
Also, the available energy is used to drive the reaction towards citrate.
This is bringing out the worst of my gambling instincts (it's in the family); that along with Torbjorn being tempted to cheat by looking on the web, it's clear that the "IDiots" have a point if they claim Sandwalk is a corrupting influence.
Dunbar, assuming your answer was directed at me, it's not a difference between dynamic and static equilibrium. I'm talking about thermodynamic equilibrium and most processes in cells are not in thermodynamic equilibrium, they are more like non-equilibrium steady states (of course even that is not quite true).
Most metabolic reactions inside the cell take place under conditions where the concentrations of substrates and products are at near-equilibrium concentrations. The ones that don't are catalyzed by regulated enzymes whose activity is controlled so their reactions don't reach equilibrium.
Regulated enzymes are only a small subset of all enzymes. As a general rule, a major pathway will only have one regulated enzyme.
The reaction catalyzed by citrate synthase is a near-equilibrium reaction in in vivo. The steady-state equilibrium concentrations of citrate are very much greater than the concentration of oxaloacetate. The equilibrium ratio of substrate and products (Keq) is related to, and can be calculated from, the standard Gibbs free energy change.
The large negative Gibbs free energy change for the citrate synthase reaction ensures that it will proceed when the concentrations of oxaloacetate are very low. Inside the cell, the concentation of oxaloacetate is, indeed, very low.
If the reaction were coupled to synthesis of ATP, the over all standard Gibbs free energy change would not be sufficient to drive the reaction in the direction of citrate in the presence of a low oxaloacetate concentration.
Ok ok, let's see. The amount of OAA in tends to be a very low in the body, and thus the very large gibbs free energy is used essentially to drive the cycle forward. How'd I do?
Ah, thanks! I can't think as a chemist. (Probably because I learned TD afterwards. I would need to get to a lab and learn how it works that way.)
So the problem is usually not to get too much product (and invoke pace-maker feedback), but to get enough product against a concentration difference. Neat!
[Btw, Larry, if you don't read my late "Paradox" comment: Thanks for the book reference!]
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