The hydrolysis of acetyl phosphate to give acetate and hydrogen phosphate ion has ∆G = -10.3 kcal/mol (-43.1 kJ/mol). Combine the equations and ∆G values to determine whether coupling of this reaction with phosphorylation of ADP to produce ATP is favorable. (You need give only compound names or abbreviations in the equations.)

Question

Accepted Answer

Welcome back everyone. The hydrolysis of phosphoenolpyruvate to pyro and phosphate has delta G of negative 14.8 kg calories per mole or negative 61.9 kilojoules per mole. While the hydrolysis of A P to AD P has delta G of negative 7.3 kg calories per mole or negative 30.5 kilojoules per mole determine whether coupling this reaction with the phosphorylation of AD P to produce A TP is energetically favorable. So we essentially want to identify the two reactions. The first one is the hydrolysis of phosphoenolpyruvate. Essentially, we can use an abbreviation pe P. And according to the problem, we are performing hydrolysis. So we're adding water and we are producing pyro and phosphate. So first of all, let's introduce pyro, then we are going to have phosphate. And specifically, in this case, we have to recall that it's actually hydro phosphate. So we would have HPO or to negative as one of our products. And then essentially what we want to do is introduce H plus this would be the required equation. And now for the second one, well, we are essentially given the hydrolysis of at TP to AD P. So a TP on the reactant side, essentially, since we have another hydrolysis, we're introducing water. Now, we are producing AD P. And for this reaction, we have to recall that we're also producing hydro phosphate. He 42 negative as well as H plus hydro num and were given their delta G values. So for the first one, well, essentially, we have a negative 14.8 kilo calories per more or the second one, the delta G value given to us is negative 7.3 kg calories from all. However, the problem states that we are coupling this reaction with the phosphorylation of AD P to produce a TP, right. So we essentially have to reverse the second equation to couple it with the first one. Let's see what we are getting. So first of all, we are just rewriting the first equation. Nothing changes. Delta G remains the same negative 14.8 kilo calories per more, but we are reversing the second one. So the products become the reactants ad P reacts with HPO 42 negative. And we have hydro and now on the product side, A TP and water according to Jesus law, whenever we are reversing an equation, the sign of delta G flips. So now we have 7.3 kilo calories per mole. He paused the valley if we are coupling them. Well, we simply want to add them, right? What we notice is that water is a common species on opposite sides. So we can cancel it out, we can cancel out hydro num and we can cancel out HBO 42 negative. And this gives us the final equation pe P plus ad P produces fire of eight plus A TP. Since we added those equations, we also want to add the delta G values negative 14.8 plus 7.3. And this gives us negative 7.5 kilo calories per more. Based on this value, we can tell that it's negative. And we want to recall that whenever delta G is negative, this is a spontaneous reaction. So yes, this coupled reaction is favorable because it has a negative delta G value. Thank you for watching.

Verified video answer for a similar problem:

Key Concepts

Gibbs Free Energy (∆G)

Coupled Reactions

Phosphorylation of ADP