Official AAMC C/P Section Bank Thread

[andelJ94]

I figured it would be better to have all the Section Bank Qs easy to find since I have seen the same Q's come up multiple times and this will make it easier to find them all.

I am having trouble figuring out the reasoning behind this Q, #46 in the AAMC C/P Section Bank

Q 46

They say the answer is D, and I went with B. I am trying to figure out the proper reasoning, since the AAMC explanations suck so much as to be useless.

The only information given in the passage is this equation which relates the Kapp of a substrate for an enzyme to its K value via comparison with the K value of another substrate for that same enzyme.

In this Equation Kapp refers to the apparent (from the experiment) association constant of NAG for the enzyme, KNAG3 represents the affinity between the same enzyme and NAG3, and KNAG represents the actual (if i got that right) affinity of NAG for the enzyme.

Rearranging the equation I can get Kapp = (KNAG3) / (1 + KNAG[NAG])


The Q wants us to determine what would lower Kapp. Looking at equation, 1 I can lower the Kapp by several ways:

Lower the affinity of the enzyme for the substrate, NAG3 (i.e. KNAG3)
Raise the affinity of the enzyme for substrate NAG (i.e. KNAG)
Raise the concentration of [NAG]

Thus, by adding more of the 2nd substrate, NAG, I can lower Kapp. But I am nto sure if this is the reasoning I am supposed to use. The explanation does not mention the equation at all!

So, I am looking for alternative reasons to explain why D is correct given the question information.

The way I read this is:

Experiment 1 = Substrate 1 (aka compound 1) + Enzyme (Protein A)
K1 is resulting affinity value

Experiment 2 = Substrate 2 (aka compound 2) + Enzyme (protein A)
Kapp is the resulting affinity value, and Kapp = K1


We need a way to lower Kapp, so aside from the equation, which says if I add more of the 2nd substrate (NAG in the original equation) I will lower the Kapp value for the substrate, how else does adding more of Substrate 2 to the 2nd titration lower its Kapp?


Am I wrong, or did the AAMC just chose to ignore the equation in its explanation for some reason? Does Kapp in the Q stem NOT refer to the apparent association constant between compound 2 and the protein? Does the 2nd titration include protein A, compound 2 AND compound 1? The text says "under identical conditions but in the...." so maybe they mean protein A and compound 1 again with compound 2 added?

That I could also see why adding more of compound 2 could interfere with the Kapp for compound 1, but I thought Kapp refers to the apparent affinity of compound 2 for the protein.

This Q has me baffled if the equation is NOT the way to the right answer. Seems the AAMC would try to at least explain the correct answer, lord knows they cannot be bothered to describe the wrong answers.

I thank the MCAT gods for non-AAMC materials.


Rant over, thank you all!

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[5words]

OMG, this question took me a whole day, in the passage CITC was defined in the passage as a Competitve ITC. So both cmpd1 and 2 are present and Compound two is acting as the inhibitor... In competitive inhibition, Km increase as the concentration of the inhibitor increases.. So because K1 is the enzyme affinity for Compound 1 , and that you only get K apparent (learned this today) when two substances competing for the same active site. So, they are three ways to solve this:

-The first is to know that Km differ from Ka, Km is the substrate concentration at 1/2 Vmax hence isnt the same as Ka. Once you understand this, it becomes easy, since High Km = low affinity whereas Low Km = High affinity. In the K stem, it states that upon adding 10mM Compd2 (minuscule amount) which is competing with compound 1 for the E active site, kapp = k1, Hence to decrease the Kapp which is the affinity that compound1 has for Protein A in presence of compd2 (the inhibitor), is to increase the apparent Km. For competitive inhibitor, Km increases as the increases , in this case [compound2]

-The other way was to look at figure 2 in the passage, and undertsand that the basics of CITC described in the second paragraph. So, NAG the second compound has a low affinity (ka not Km) for HEW enzyme, hence it's being added to compete with NAG3. So, NAG is the inhibitor same as compd 2. The graph show that KNAG3/Kapp increase as [NAG] increases... Note that KNAG3 here is the K1 (not km) that is the affinity of the Compound 1 for Protein A. Thus, in our case, Kapp and [compound2] are inversely proportional... Thus by increasing one, you;re decreasing the other.

- The third was to use the equation, knowing that [NAG] or the compound with the weakest affinity was our compound two because, the first sentence of Paragraph two states that "NAG monomer interacts weakly with HEW lysosomes, and it is not possible to measure KNAG or deltaH of its interaction directly ".. In the question stem, coumpond2 , unlike 1, wasnt directly mixed/ or didnt directly interact with Protein A.


I guess the take home point is that Km is not a direct measure of an enzyme affinity, but its reciprocal. Hence KM =/= Ka.

 

[andelJ94]

OMG, this question took me a whole day, in the passage CITC was defined in the passage as a Competitve ITC. So both cmpd1 and 2 are present and Compound two is acting as the inhibitor... In competitive inhibition, Km increase as the concentration of the inhibitor increases.. So because K1 is the enzyme affinity for Compound 1 , and that you only get K apparent (learned this today) when two substances competing for the same active site. So, they are three ways to solve this:

-The first is to know that Km differ from Ka, Km is the substrate concentration at 1/2 Vmax hence isnt the same as Ka. Once you understand this, it becomes easy, since High Km = low affinity whereas Low Km = High affinity. In the K stem, it states that upon adding 10mM Compd2 (minuscule amount) which is competing with compound 1 for the E active site, kapp = k1, Hence to decrease the Kapp which is the affinity that compound1 has for Protein A in presence of compd2 (the inhibitor), is to increase the apparent Km. For competitive inhibitor, Km increases as the increases , in this case [compound2]

-The other way was to look at figure 2 in the passage, and undertsand that the basics of CITC described in the second paragraph. So, NAG the second compound has a low affinity (ka not Km) for HEW enzyme, hence it's being added to compete with NAG3. So, NAG is the inhibitor same as compd 2. The graph show that KNAG3/Kapp increase as [NAG] increases... Note that KNAG3 here is the K1 (not km) that is the affinity of the Compound 1 for Protein A. Thus, in our case, Kapp and [compound2] are inversely proportional... Thus by increasing one, you;re decreasing the other.

- The third was to use the equation, knowing that [NAG] or the compound with the weakest affinity was our compound two because, the first sentence of Paragraph two states that "NAG monomer interacts weakly with HEW lysosomes, and it is not possible to measure KNAG or deltaH of its interaction directly ".. In the question stem, coumpond2 , unlike 1, wasnt directly mixed/ or didnt directly interact with Protein A.


I guess the take home point is that Km is not a direct measure of an enzyme affinity, but its reciprocal. Hence KM =/= Ka.

@5words Thanks! This is what I figured, but I had no idea how I was supposed to conclude that BOTH molecules were used in the 2nd experiment of the passage or the 2nd titration of the Q. The passage to me seemed to be reworded to be deliberately vague, and the actual paper the passage is based on is behind a pay wall.

 

[DaAlienist]

OMG, this question took me a whole day, in the passage CITC was defined in the passage as a Competitve ITC. So both cmpd1 and 2 are present and Compound two is acting as the inhibitor... In competitive inhibition, Km increase as the concentration of the inhibitor increases.. So because K1 is the enzyme affinity for Compound 1 , and that you only get K apparent (learned this today) when two substances competing for the same active site. So, they are three ways to solve this:

-The first is to know that Km differ from Ka, Km is the substrate concentration at 1/2 Vmax hence isnt the same as Ka. Once you understand this, it becomes easy, since High Km = low affinity whereas Low Km = High affinity. In the K stem, it states that upon adding 10mM Compd2 (minuscule amount) which is competing with compound 1 for the E active site, kapp = k1, Hence to decrease the Kapp which is the affinity that compound1 has for Protein A in presence of compd2 (the inhibitor), is to increase the apparent Km. For competitive inhibitor, Km increases as the increases , in this case [compound2]

-The other way was to look at figure 2 in the passage, and undertsand that the basics of CITC described in the second paragraph. So, NAG the second compound has a low affinity (ka not Km) for HEW enzyme, hence it's being added to compete with NAG3. So, NAG is the inhibitor same as compd 2. The graph show that KNAG3/Kapp increase as [NAG] increases... Note that KNAG3 here is the K1 (not km) that is the affinity of the Compound 1 for Protein A. Thus, in our case, Kapp and [compound2] are inversely proportional... Thus by increasing one, you;re decreasing the other.

- The third was to use the equation, knowing that [NAG] or the compound with the weakest affinity was our compound two because, the first sentence of Paragraph two states that "NAG monomer interacts weakly with HEW lysosomes, and it is not possible to measure KNAG or deltaH of its interaction directly ".. In the question stem, coumpond2 , unlike 1, wasnt directly mixed/ or didnt directly interact with Protein A.


I guess the take home point is that Km is not a direct measure of an enzyme affinity, but its reciprocal. Hence KM =/= Ka.

@5words Thanks! This is what I figured, but I had no idea how I was supposed to conclude that BOTH molecules were used in the 2nd experiment of the passage or the 2nd titration of the Q. The passage to me seemed to be reworded to be deliberately vague, and the actual paper the passage is based on is behind a pay wall.

So are the AAMC explanations really that bad? Taking so much time to understand a single questions does not instill much confidence in their ability to explain their own answers and wrong answers. Can someone show a screenshot of how they tried to explain this? Does the AAMC bother to explain their passages?

 

[5words]

So are the AAMC explanations really that bad? Taking so much time to understand a single questions does not instill much confidence in their ability to explain their own answers and wrong answers. Can someone show a screenshot of how they tried to explain this? Does the AAMC bother to explain their passages?

Yeah, their explanations are def bad!

 

[DoctorCrush]

I started the B/B section a while back and certain explanation is complete trash -____- I literally had to reddit so many passages cause I did not understand how the heck certain things came about. Especially weird experiments I've never seen.

 

[DaAlienist]

I started the B/B section a while back and certain explanation is complete trash -____- I literally had to reddit so many passages cause I did not understand how the heck certain things came about. Especially weird experiments I've never seen.

Yeah, the AAMC explanations suck. I am using the NS materials and they have videos that take you through all the AAMC practice MCATs, in detail. This has made my review so much easier and allowed me to learn a whole lot more than I could from the AAMC "explanations." I would certainly check them out if you are frustrated by the AAMC.