Question about ATP/ADP Watch

DoubleDoors
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Hi all,

I'm currently preparing notes for the nucleic acids section of the OCR AS course, module 2. I am looking at some information on phosphorylation of nucleotides, and I am a bit confused with some of the theory behind it.

The theory is that breaking ATP down into ADP will release energy, but I understand that a way of representing this direction is as follows:

ATP -> ADP + P

So, my question is, why is energy released when the phosphate bond is broken? I had always thought that bond breaking was endothermic.
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spiritless98
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I've seen a similar question posted on yahoo- I'll copy and paste that here because it was a great explanation.


Great great question, and I wish biologists would learn how to balance chemical reaction equations so that people don't have to ask it anymore.

Because you haven't balanced your reaction, and it actually involves a lot more than just breaking one bond.

Breaking chemical bonds is always ENDOthermic, requiring energy going INTO the system. Formation of chemical bonds is always EXOthermic, releasing energy FROM the system into the surroundings. Just like you've said.

Although hardly any chemical reactions actually follow such a pathway, you can imagine any chemical reaction as occurring by first breaking all the bonds in the reactants (uphill, energy in) and then forming all the bonds in the products (downhill, energy out). If the bonds in the reactants are weaker than those in the products, it takes less energy in to break (a little bit uphill) than comes back out to make (a lot more downhill), so that reaction releases a net output of energy, and the reaction exothermic (downhill) overall.

Reactions are exothermic if the products have stronger bonds than the reactants did. Which is the case for ATP hydrolysis.

ATP hydrolysis is often written as ATP => ADP + Pi, which gives the impression that all you've done is break one bond. But the reaction isn't balanced. It's really ATP + HOH ==> ADP + Pi. You break a P-O (in ATP) and an O-H (in water), and you also form a P-O (in Pi) and an O-H (in ADP). However, the bonds you form are very slightly stronger, so the reaction is slightly exothermic. (And keep in mind that this is NOT a lot of energy we're talking about. 30 kJ/mol is a small amount. Why they keep getting called "high energy" bonds is beyond me. As well, this is a lousy example to illustrate the principle, because most of the driving force has more to do with entropy and solvation of the free phosphate than the bonds you make and break, not unexpected given that you make and break the same kinds of bonds. It's not really all that exoTHERMIC, with a negative delH, it's exERGONIC, with a negative delG, which includes the entropy.)

Breaking the P-O bond in ATP is exothermic because you make new bonds in ADP and Pi that are stronger, as becomes blear if you balance the reaction. The same apparent trouble shows up in catabolism: it's described as "bond breaking", and C-C bonds are broken, but catabolic reactions are exothermic because of the products, which have stronger bonds than the ones you break.


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DoubleDoors
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(Original post by spiritless98)
I've seen a similar question posted on yahoo- I'll copy and paste that here because it was a great explanation.


Great great question, and I wish biologists would learn how to balance chemical reaction equations so that people don't have to ask it anymore.

Because you haven't balanced your reaction, and it actually involves a lot more than just breaking one bond.

Breaking chemical bonds is always ENDOthermic, requiring energy going INTO the system. Formation of chemical bonds is always EXOthermic, releasing energy FROM the system into the surroundings. Just like you've said.

Although hardly any chemical reactions actually follow such a pathway, you can imagine any chemical reaction as occurring by first breaking all the bonds in the reactants (uphill, energy in) and then forming all the bonds in the products (downhill, energy out). If the bonds in the reactants are weaker than those in the products, it takes less energy in to break (a little bit uphill) than comes back out to make (a lot more downhill), so that reaction releases a net output of energy, and the reaction exothermic (downhill) overall.

Reactions are exothermic if the products have stronger bonds than the reactants did. Which is the case for ATP hydrolysis.

ATP hydrolysis is often written as ATP => ADP + Pi, which gives the impression that all you've done is break one bond. But the reaction isn't balanced. It's really ATP + HOH ==> ADP + Pi. You break a P-O (in ATP) and an O-H (in water), and you also form a P-O (in Pi) and an O-H (in ADP). However, the bonds you form are very slightly stronger, so the reaction is slightly exothermic. (And keep in mind that this is NOT a lot of energy we're talking about. 30 kJ/mol is a small amount. Why they keep getting called "high energy" bonds is beyond me. As well, this is a lousy example to illustrate the principle, because most of the driving force has more to do with entropy and solvation of the free phosphate than the bonds you make and break, not unexpected given that you make and break the same kinds of bonds. It's not really all that exoTHERMIC, with a negative delH, it's exERGONIC, with a negative delG, which includes the entropy.)

Breaking the P-O bond in ATP is exothermic because you make new bonds in ADP and Pi that are stronger, as becomes blear if you balance the reaction. The same apparent trouble shows up in catabolism: it's described as "bond breaking", and C-C bonds are broken, but catabolic reactions are exothermic because of the products, which have stronger bonds than the ones you break.


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Thanks a lot! I suspected something had been missing and as I do chemistry as well I was considering that there was a negative enpalthy change, but none of the sources specified that it was a hydrolysis reaction so I had never picked up on that. Thanks again.
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