ATP contains energy in its high energy bond

[biophysics]

I understand that ATP contains energy in its high energy bond. However, there are two high energy bonds in ATP. Thus ADP still has 1 high energy bond.

Is this bond used for energy? Can you give an example where the energy of ADP is used?

 

[Moonbear]

Yes, that bond is used for energy too, with the conversion from ADP to AMP. As for an example, I can't think of one off the top of my head, but I know I've encountered some along the way.

 

[Andy Resnick]

It's not correct to think that ATP stores energy in chemical bonds. In actuality, the energy capability ATP derives from the nonequilibrium concentration imbalance between ATP and ADP- the energy is the Gibbs free energy, not electronic bond energy. This is the essence of the chemiosmotic theory.

And there's nothing special about adenosine, either- some molecules work by converting GTP and GDP.

Nucleotide synthesis involves conversion of ATP to AMP.

 

[biophysics]

Hi,

thanks for the reply

Can you give me an example where the energy of a molecule IS due to bond energy and NOT due to an equilibrium imbalance?

I believe that a high bond energy is manifested in an equilibrium where ADP is favored. Thus it is the high bond energy which gives ATP/ADP a skewed equilibrium. The two (high bond energy + equilibrium) are related

 

[Ygggdrasil]

A system at equilibrium can never do work (ΔG for any process at equilibrium is zero), so all chemical transformations from which energy can be harnessed must necessarily be out of equilibrium. For any chemical process, the maximum amount of work that can be produced by that process (ΔG) is given by:

ΔG = ΔG^o + RTlnQ

Where Q is the reaction quotient. Thus, two factors play into the fact that ATP + H₂O --> ADP + PO₄^3- produces a lot of energy. First, the free energy of the products is much lower than the free energy of the reactants, leading to a very negative ΔG^o (remember, a negative ΔG means the process releases free energy). Second, the cell maintains a low [ADP]/[ATP] ratio which makes Q small (Q < 1), so RTlnQ is also very negative. Thus, both the bond energy of ATP and the [ADP]/[ATP] ratio contribute to the energy obtained from ATP hydrolysis.

 

[biophysics]

thanks for the reply,

Lets assume that we have a reaction which releases 0 energy. If we put this reaction in disequilibrium, we will not see any release of energy even as reactions occur to bring the system into equilibrium. Thus:

It is only the bond energy that contributes to the energy released and the disequilibrium simply allows harvesting of this energy.

 

[Renge Ishyo]

I have read that there is no difference between the energy in a phosphate bond in ATP and a phosphate bond in any other compound. The term "high energy bond" represents a sort of confusion that Biochemists have when trying to explain the hydrolysis of ATP. For example, in R. Chang's "Physical Chemistry for the Chemical and Biological Sciences" the author writes on the topic:

The large decrease in the Gibbs energy that results from the hydrolysis of ATP and ADP prompted some biochemists to use the term "high energy phosphate bond" in describing these compounds. Unfortunately, this term wrongly implies that the P-O bond in these molecules is somehow different from the normal covalent bond, which is untrue.

The author goes on the explain that the large decrease in the standard Gibbs is due both to an easing of the electrostatic repulsion of neighboring phosphate groups and an increase in entropy from the reactants due to the generation of two resonance stabilized products as opposed to one.

Some others above have said that ATP useage in the cell is concentration dependent which is true (although this doesn't account for the large negative standard Gibbs Free energy change since the concentration of all reactants and products for the standard Gibbs are the same: 1M).

Also, another thing to keep in mind is that the rate of hydrolysis is greatly assisted by enzyme architecture. For example, many enzymes have magnesium bind to two out of the three phosphates, easing the removal of the terminal phosphate and making this reaction proceed much faster than the reactions removing the other phosphates. There are also enzymes that also push forward the hydrolysis of both phosphates, such as in the formation of the second messenger cyclic AMP (which only has one phosphate left). In this way the enzyme makes it so that hydrolysis of two phosphates can be "selected" for as opposed to hydrolysis of the single phosphate (the speed of the formation of cyclic AMP in this case would be far faster than removal of a single phosphate and formation of ADP...).

Hope this helped.

 

[biophysics]

(although this doesn't account for the large negative standard Gibbs Free energy change since the concentration of all reactants and products for the standard Gibbs are the same: 1M

Thanks for replying.

You say in the above that the reason that concentration disequilibrium does not affect the amount of energy released is because standard gibbs concentrations are the same.

I thought: The concentration disequilibrium isn't what releases energy in ANY case. It is the entropy and alleviation of electrostatic repulsion SOLELY that contribute to the energy release.

Looking forward to your comment on this

 

[Renge Ishyo]

The concentration disequilibrium isn't what releases energy in ANY case. It is the entropy and alleviation of electrostatic repulsion SOLELY that contribute to the energy release.

No, it IS the concentration disequilibrium that releases the energy as some have said in more detail above (it is rather the nature of the phosphate bond that in and of itself is nothing special).

The statement I made about the standard Gibbs might be a bit confusing so I will explain it better (or try to anyway). What the negative value for the standard Gibbs tells you is that the equilibrium condition for ATP in water will have more ADP relative to ATP (not equal amounts of each like what you started with since ADP formation is favored over ATP for the entropy and steric reasons described). After the start of the measurement, energy will be released as ATP is converted into ADP until the equilibrium amounts of each are reached. Once equilibrium is reached that is it, the Gibbs for the overall reaction will be zero and you will get no net energy/work out of the system. One way to think of the standard Gibbs is as the amount of energy that will be released as equal concentrations of each reactant adjust to their equilibrium values at standard conditions of equal concentrations and standard temperature/pressure (using the physical chem definition of the Gibbs here, the Biochem standard Gibbs has another tweak applied to it). Useful work can only be done as the concentrations adjust far from equilibrium (this is why the body is trying desperately to stay away from equilibrium at all times...it needs to be away from equilibrium so that the chemical conversions can do useful work).

This is one of the reasons why "high energy phosphate bond" doesn't make sense. Thinking of it that way might confuse you into thinking that the reaction of ATP to ADP is unidirectional (or downhill) at all times. It is not; in fact if you start with a high enough concentration of ADP relative to ATP you can actually get the spontaneous forward reaction to be in favor of ATP formation (you would get an entropy payoff in such a situation by increasing the amount of ATP in the mixture...the body does do this at times too; remember it has to form ATP as well as break it down so don't be scared by the idea that these bonds can be broken and reformed reversibly).

 

[Ygggdrasil]

thanks for the reply,

Lets assume that we have a reaction which releases 0 energy. If we put this reaction in disequilibrium, we will not see any release of energy even as reactions occur to bring the system into equilibrium. Thus:

It is only the bond energy that contributes to the energy released and the disequilibrium simply allows harvesting of this energy.

You can harvest energy from a process for which ΔG^o is zero. A prime example drives the production of ATP:

H⁺_(outside) --> H⁺_(inside)

Obviously, there is nothing different about the hydrogen ion whether its inside or outside the mitochondrial matrix. When there are equal concentrations of hydrogen ions on either side of the mitochondrial inner membrane, ΔG from the transport is zero and no energy can be harvested. However, once you have a higher concentration of H⁺ in the inter-membrane space than in the mitochondrial matrix (i.e. your hydrogen concentrations are out of equilibrium), you can harvest energy from the transport of protons across the mitochondrial inner membrane.

The basic principle at play: in order to take a system away from equilibrium, you must perform work. Therefore, you can harvest energy from a system returning to equilibrium.


Re: Renge Ishyo, I do agree that people often misinterpret the statement that ATP contains a high energy bond, however, I do not agree that the term is fundamentally flawed. Most introductory biochemistry textbooks make it seem as if releasing the gamma phosphate of ATP powers biological processes. This is untrue; it is the transfer of the gamma phosphate to another molecule that is important. What makes the phosphodiester bond in ATP special is that it is an especially weak bond that is easily transferred to another molecule. In the language of organic chemistry, the gamma phosphate of ATP is a good leaving group that can be easily transferred to another molecule to activate that molecule (i.e. convert the molecule from an unreactive species into a reactive species).

Because the terminal phosphate is easily transferred, we can think of it as an unstable (i.e. high energy) bond. Therefore, I still think it is fair to consider the gamma phosphodiester bond of ATP as a high energy bond.

 

[biophysics]

thanks for the above responses-

In the hydrogen ion example- I understand that a system moves toward equalizing concentrations to maximize entropy, and hence minimize the energy of the sytem.

If we apply this principle to ATP/ADP, then should the system not move towards 100% ADP
to minimize the combined energy of all the molecules. Yet the equilibrium rests at a point where some small amount ATP is present. Why can this be true?

It is mentioned above that if the concentration of ATP is lower than its equilibrium concentration, ATP synthesis is favored. How does this increase entropy?

thanks

 

[Renge Ishyo]

If we apply this principle to ATP/ADP, then should the system not move towards 100% ADP
to minimize the combined energy of all the molecules. Yet the equilibrium rests at a point where some small amount ATP is present. Why can this be true?

It is mentioned above that if the concentration of ATP is lower than its equilibrium concentration, ATP synthesis is favored. How does this increase entropy?

Would a system with 100% ADP be more ordered or disordered than a system with some ATP mixed in with ADP? Entropy favors a mixture which answers both of the above questions. More ATP is present in the equilibrium mixture than you might originally guess...think about it, if it was so hard to react ADP and P to reform ATP how would your body ever be able to make any new ATP once the molecule was used up? The equilibrium is skewed a bit towards the formation of more ADP relative to ATP at a given moment, but it is not so far off the mark that with a modest input of energy more ATP molecules cannot readily be generated...it is chosen for this balance. If the goal was simply to drive a reaction forward in only one direction with no thought given to reusing the molecule once you were finished with it, then there are many more molecules with far more negative standard Gibbs energies that the body could use...

Ygggdrasil, I am clearly outvoted when it comes to the use of "high energy phosphate bond" and I know it will continue to be used. I just don't like the confusion it seems to cause (although I guess the problem here is that a more accurate explanation can be even more confusing...).