I can purchase ATP powder from a supplier and resuspend the ATP in water at mM c

Question

I can purchase ATP powder from a supplier and resuspend the ATP in water at mM concentrations without safety concerns. This might be surprising given that ATP hydrolysis is highly exergonic under these conditions and releases considerable energy (AGo -7.3 kcal/mol). How is this possible? Why doesn't the tube immediately combust upon addition of water? ?N ORDERPORIT TO COMEUST IT WOULD NEED TO BE CHTf LY2E EVENT THE ReTz ar UNDERboES (3pts) If ADP and Pi were held constant (see table above), at what ATP concentration - ADP phosphorylation would become exergonic? ?NOPORA Tb 1NOUCE nnetas.NE REACT?ON ron hop pHospmoetyemey The RER cau 2pts) The SAALLER TO nteryou calculated above should be very small why do youtinkitse Lag bee THE CELL NT O HEEP TH5 PONTANEITY TO BE BETTER ENEBbY PRODUCTION ONA CONTIU OR

Explanation / Answer

Page 1:

Question 1 and 2-Your calculations of Keq and delta G are right.

Question 3- ATP hydrolysis is Exergonic and spontaneous because the calculated delta G for the given concentrations of ATP, ADP, and Pi is less than zero.

Page 2:

Question 1- Your justification is right, hydrolysis of ATP is highly exergonic meaning that energy states of reactants are higher than the energy states of products, and hence the reaction is favorable and spontaneous. But for something to combust, it needs to cross an energy barrier known as Activation Energy, which can be overcome by the introduction of a catalyst that lowers the Activation Energy of combustion of ATP.

Question 2- Please not, Equilibrium means that the reaction is not favoring any particular direction (neither forward nor reverse), and in such a scenario deltaG should be equal to zero.

Also note, if delta G > 0, the reaction favors the forward direction i.e. reactants will tend to produce products.

           And if delta G < 0, the reaction favors the reverse direction i.e. products will tend to produce reactants.

      And if delta G= 0, the reaction will be at equilibrium i.e. it will not tend to go in either direction.

Question 3:

You can take the help of above discussion to answer this question. You reasoning of reaction becoming Endogonic (i.e. delta G > 0) is right but I think your calculations are wrong. (Its not clear which table the question refers to ) If you take the concentrations you used in calculating Keq and deltaG i.e. [ATP] = 3.4 mM, [ADP] = 1.3 mM, and [Pi] = 4.8 mM. You can proceed in the following way

We know for the reaction to favor reverse direction delta G should be greater than zero.

We know, deltaG = -7300 cal/mol + RT *ln Keq.

So for deltaG to be greater than zero, you must have the term RT*ln Keq to greater than 7300cal/mol.

Now, you should plug in the values and calculate for the [ATP] from the equation

ln{ [ADP]*[Pi]/[ATP]} = 7300/RT. This will be your concentration of ATP at which the reaction will start to switch direction.

Question 4: Your reasoning is right here.

Thanks!

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