AAMC The Official MCAT Practice Question phy/chem section Q #4

[m25]

---

Members don't see this ad.

Why is the answer B? I thought the Keq=1 given in the passage only applies to the reaction shown in Figure 2, not Figure 1. I interpret Figure 2 as only showing part of the intermediate steps in the overall aspartate transaminase reaction. Just because part of the reaction is Kep=1, it doesn't mean the overall reaction also have Kep=1. Thus I thought the answer was A because the passage mention that aspartate transamination is a catabolic reaction, and from my understanding, catabolic reaction is usually a spontaneous reaction that produces ATP. Am I missing some kind of basic concept here? Attached below is the relevant passage and answer.

 

[azor ahai]

It states that AA catabolism in many cases proceeds through a transamination step. It doesn't state that the transamination step itself is catabolic. I agree that they could have specified that Keq ~ 1 for figure 1 instead of figure 2, or both. The fundamental thing to remember here is the importance of the relative concentrations of reactants and products which is represented by Keq. The formula given expresses that as well, but think intuitively: if the reaction is at equilibrium, the reverse and forward rates are equal so there is no change in Gibb's Free Energy.

Also note that no ATP is required or produced - something which is necessary for either catabolism or anabolism. This should give you a hint that while the passage is talking about AA catabolism, this particular step does not have to be catabolic.

 

[m25]

Thanks for answering!
When you mentioned:

The fundamental thing to remember here is the importance of the relative concentrations of reactants and products which is represented by Keq. The formula given expresses that as well, but think intuitively: if the reaction is at equilibrium, the reverse and forward rates are equal so there is no change in Gibb's Free Energy.

were you referring to a specific part in the passage besides the Keq=1 part?
Actually, do the arrow in the diagram

say anything about the spontaneity of the reaction, or is it a completely different matter that only tells us that it's a reversible reaction?

 

[azor ahai]

It only tells you that it is reversible. In fact, that could have helped you determine that Fig 1. has Keq ~ 1 also.

That part you quoted, I mean in general you should be using the dG = -RT(lnKeq) formula. This shows you how important Keq is to spontaneity. As well as think intuitively, if a reaction is in equilibrium between the products and reactants then there is no particular direction it will favor. The forward reaction will not result in a lower free energy state just as the reverse wouldn't.

 

[m25]

It only tells you that it is reversible. In fact, that could have helped you determine that Fig 1. has Keq ~ 1 also.

That part you quoted, I mean in general you should be using the dG = -RT(lnKeq) formula. This shows you how important Keq is to spontaneity. As well as think intuitively, if a reaction is in equilibrium between the products and reactants then there is no particular direction it will favor. The forward reaction will not result in a lower free energy state just as the reverse wouldn't.

Oh okay, thanks!

So regarding the arrow in the diagram, are you saying that reversible reaction tells us that it has Keq around 1? But I thought reversibility and Kep were separate concept...?

 

[azor ahai]

Oh okay, thanks!

So regarding the arrow in the diagram, are you saying that reversible reaction tells us that it has Keq around 1? But I thought reversibility and Kep were separate concept...?

Wait, what exactly do you mean by Kep? All this time I thought you just meant Keq.

Yeah the reversible arrows could indicate that. But keep in mind, a lot of times you will see the reversible arrows but with another thing happening, like the oxidation of NADH or something. That's different.

 

[m25]

Wait, what exactly do you mean by Kep? All this time I thought you just meant Keq.

Opps, sorry, it's a typo. I meant Keq.

Yeah the reversible arrows could indicate that. But keep in mind, a lot of times you will see the reversible arrows but with another thing happening, like the oxidation of NADH or something. That's different.

So are you saying that having the reversible arrow could be an indicator that the reaction is Keq=1 unless it's coupled with other reactions such as NADH? Sorry that got me really confused...

Also, I just remembered how the equation dG = -RT(lnKeq) only applies when the reaction is happening under standard condition according to: http://forums.studentdoctor.net/thr...if-g-is-a-standard-gibbs-free-energy.1122141/
Since the catabolism of amino acids are likely occurring in human body which is definitely NOT under standard, condition, isn't using the dG = -RT(lnKeq) equation invalid to begin with?

 

[azor ahai]

Opps, sorry, it's a typo. I meant Keq.


So are you saying that having the reversible arrow could be an indicator that the reaction is Keq=1 unless it's coupled with other reactions such as NADH? Sorry that got me really confused...

Also, I just remembered how the equation dG = -RT(lnKeq) only applies when the reaction is happening under standard condition according to: http://forums.studentdoctor.net/thr...if-g-is-a-standard-gibbs-free-energy.1122141/
Since the catabolism of amino acids are likely occurring in human body which is definitely NOT under standard, condition, isn't using the dG = -RT(lnKeq) equation invalid to begin with?

yeah, for anything not in the standard state use dG = standard dG + RTlnKeq

 

[m25]

yeah, for anything not in the standard state use dG = standard dG + RTlnKeq

So why is the answer choice explanation not using the dG = standard dG + RTlnKeq equation?