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Posted

okay so i loosely understand this is the energy currency of the cells/body whatever but my friend said its the "high energy phosphate bonds" where the energy comes from, so its actually not the molecule itself but the energy in the bonds between the phosphate groups? arnt these just single covalent? so where does all this "energy" come from? thanks

Posted

In a few words, coupling phosphate bond to substrate molecules makes the substrate more reactive, because as your friend said, phophate bond are high energy bonds (their breakage is exergonic), also the phosphate group is a good leaving group.

Posted

ok cool thanks dude interesting...

 

also if i could just ask another question. - enzymes=proteins right? in the stomach we have hcl and pepsin. if one were to take pancreatic enzymes hcl+pepsin would destroy the enzymes/proteins is this correct? thanks

Posted

ok cool thanks dude interesting...

 

also if i could just ask another question. - enzymes=proteins right? in the stomach we have hcl and pepsin. if one were to take pancreatic enzymes hcl+pepsin would destroy the enzymes/proteins is this correct? thanks

 

Theoretically, yes. Pepsin is a protease that would break down pancreatic enzymes in an acidic environment.

  • 3 months later...
Posted

The energy in the high energy bonds is measured by the energy (actually it is delta G) you obtain by splitting the molecule via the reaction with water precisely at the bond in question (This is called hydrolysis). In the hydrolysis reaction the high energy bond is not split into two (producing radicals), but it is moving from one atom of the original molecule to an atom of the water molecule.

 

Both sides of the bond are relevant for the energy. So if both sides are acids (e.g. Phosho-group on both sides like in ATP), the energy is higher than if one side is an acid (e.g. phospho group) and the other an alcohol group (-OH). There is a clear correlation between the energy of a bond and the acid strength (pKa) of the two molecules you obtain after hydrolysis.

Posted

While it is commonly understood that ATP contains a "high-energy" phosphate bond that allows it to do useful work in the cell, in reality this is only part of the story. The spontaneity of the reaction ATP --> ADP + Pi is very much dependent upon the ratio of ATP/ADP that exists within the cell. A high ratio means that ATP hydrolysis is very favorable, and the delta-G is quite negative, and work can be done by coupling ATP hydrolysis to various reactions that require energy input to drive them forward. So, essentially, it is not a property of the phosphate bond in the ATP that allows it to be a useful energy currency, but rather its relative abundance compared to the hydrolyzed form ADP. The cell has to be sure to maintain the proper ratio of ATP/ADP in order for the ATP to be useful at all.

Posted (edited)

That is right.

A factor of 10 (e.g. between the concentration of ATP to the concentration of ADP) changes deltaG by -5.7 kJ/mol.

[ATP] / [ADP] is typically 10

However, since the standard deltaG°' assumes 1 M and the hydrolysis of ATP has more products than educts that actual deltaG value is much bigger than the deltaG°' value:

 

deltaG = deltaG°' + RT ln ( ( [ADP] [Pi] ) / [ATP] )

 

so [ADP] / [ATP] beeing 1/10 changes deltaG by -5.7 kJ/mol

and [Pi] beeing more around 1 mM instead of 1 M changes deltaG by 3 * -5.7 kJ/mol = -17.1 kJ/mol

This means the effect of the poor chosen standard (there are no chemicals at 1 M in cells) is bigger than the ATP/ADP ratio.

 

So if we assume Pi to be about 4 mM you can say the actual deltaG (and hence the energy value) of ATP is -20 kJ/mol more than the standard deltaG.

 

-----------------------------------------

For those who want to have all the details:

Actually there are the same amount of educts (ATP and H2O) as products (ADP and phosphate), but all the standard deltaG values are calculated under the assumption of a fix H2O concentration (which is a good assumption). Actually you always have to divide by the standard concentration. So the formula is precisely:

 

deltaG = deltaG°' + RT ln ( ( ( [ADP]/1M ) ( [Pi]/1M ) ) / ( [ATP]/1M ) ( [H2O]/55M ) )

Edited by Jens

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