C/P Section Bank #46 (Experimental setup)

[theonlytycrane]

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Hoping to talk this one out 🙂 In the question stem, compound 2 had inhibition on Protein A. Do we assume that adding more may lower the affinity constant?

Also, is the titration curve supposed to be read like a saturation plot? Enzyme is converting substrate, but I'm not fully understanding the heat interactions.

 

[aldol16]

I'm not familiar with the specific question, but if you know that Compound 2 is a competitive inhibitor of the protein, then adding more of it would decrease the Kapp (affinity for compound 1) because compound 2 would be competing for binding sites, thereby resulting in fewer binding sites available for the actual substrate, compound 1.

 

[theonlytycrane]

@aldol16 Thanks, this helps. I was confused by the question stem because it says that 10mM of the inhibitor resulted in Kapp = Km. Maybe there just wasn't enough inhibitor added yet to decrease the apparent affinity constant?

 

[aldol16]

@aldol16 Thanks, this helps. I was confused by the question stem because it says that 10mM of the inhibitor resulted in Kapp = Km. Maybe there just wasn't enough inhibitor added yet to decrease the apparent affinity constant?

Yeah, it's definitely possible. If you have too little inhibitor, your Kapp is not going to shift much from your K1 and so you would observe no change. I'm not sure if the question stem states explicitly it's a competitive inhibitor though.

 

[drobgyn32]

I'm not familiar with the specific question, but if you know that Compound 2 is a competitive inhibitor of the protein, then adding more of it would decrease the Kapp (affinity for compound 1) because compound 2 would be competing for binding sites, thereby resulting in fewer binding sites available for the actual substrate, compound 1.

But doesn't a competitive inhibitor INCREASE Km?

 

[laczlacylaci]

But doesn't a competitive inhibitor INCREASE Km?

Here is my thought process:

Kapp=apparent association constant=(comp1)^2/(comp2) If you increase Kapp, this means increased affinity to comp1.
so as comp2 increases, the Kapp should decrease. (because its an inhibitor) I think you have Kapp and Km mixed up.

ITC exp: Comp1+ProtA results in large K1
CITC exp: Comp1+ProA+10mMComp2(competitive inhibitor) results in Kapp=K1 (the reason why we observe this is, I think like @theonlytycrane said is that: we may not have a very concentrated comp2 (inhibitor)) If we did have concentrated comp2 (enough to compete over comp1, then we would expect Kapp<K1)

The Q-stem asks you to increase K1 to become greater than Kapp, or in other words, decrease Kapp, so we can choose D=increase comp2 concentration.

Let me know if that helps @drobgyn32

 

[laczlacylaci]

@theonlytycrane

What is a good way to understand the difference between K1 and Kapp?

I understand that a decrease in Kapp results in decreased affinity of com1 and increase affinity of comp2 on protA.
It seems that K1 is the affinity constant between comp1 & protA.

Is Kapp basically an integration of comp2 in the picture, correct?

 

[theonlytycrane]

@drobgyn32 A competitive inhibitor increases Km, but Km is a dissociation constant in the classic M.M. model.

In this case the "K"s are affinity constants. That's the most important thing to take note of in any enzyme kinetics passage- if we're talking about a Kd or Ka (affinity).

@laczlacylaci I think you're onto it. Compound 1 and protein A have some affinity for one another = K1. We add 10 mM of Compound 2 (sounds like a typical inhibitor type of setup), and we measure again but the new K (Kapp) still is the same as K1. So to change the affinity between Compound 1 and protein A, we just add more inhibitor.

It's basically just a complicated way to just ask about the typical enzyme kinetics model.

 

[drobgyn32]

Sorry, I'm still confused.
Expt 1: Protein A and Cpd1; K1
Expt 2: Protein A and Cpd1 and Cpd2; Kapp = K1

So here, the Kapp or K1 - are they the same as Kd and/or Km? The MCAT regards Kd and Km as the same.

The experiment done is CITC which makes Cpd2 a competitive inhibitor of Cpd1. Correct? So, to decrease Kapp, why would I add more Cpd2? Competitive inhibitors INCREASE Km or Kapp.

Could you please re-explain 🙁 I'd really appreciate it

 

[drobgyn32]

Here is my thought process:

Kapp=apparent association constant=(comp1)^2/(comp2) If you increase Kapp, this means increased affinity to comp1.
so as comp2 increases, the Kapp should decrease. (because its an inhibitor) I think you have Kapp and Km mixed up.

ITC exp: Comp1+ProtA results in large K1
CITC exp: Comp1+ProA+10mMComp2(competitive inhibitor) results in Kapp=K1 (the reason why we observe this is, I think like @theonlytycrane said is that: we may not have a very concentrated comp2 (inhibitor)) If we did have concentrated comp2 (enough to compete over comp1, then we would expect Kapp<K1)

The Q-stem asks you to increase K1 to become greater than Kapp, or in other words, decrease Kapp, so we can choose D=increase comp2 concentration.

Let me know if that helps @drobgyn32

Kapp=apparent association constant=(comp1)^2/(comp2)

Where is this equation coming from?

 

[theonlytycrane]

Sorry, I'm still confused.
Expt 1: Protein A and Cpd1; K1
Expt 2: Protein A and Cpd1 and Cpd2; Kapp = K1

So here, the Kapp or K1 - are they the same as Kd and/or Km? The MCAT regards Kd and Km as the same.

The experiment done is CITC which makes Cpd2 a competitive inhibitor of Cpd1. Correct? So, to decrease Kapp, why would I add more Cpd2? Competitive inhibitors INCREASE Km or Kapp.

Could you please re-explain 🙁 I'd really appreciate it

Km = Kd. These are dissociation constants and this is separate from the question stem.

The question stem is talking about an affinity constant, not a dissociation constant.

 

[laczlacylaci]

Sorry, I'm still confused.
Expt 1: Protein A and Cpd1; K1
Expt 2: Protein A and Cpd1 and Cpd2; Kapp = K1

So here, the Kapp or K1 - are they the same as Kd and/or Km? The MCAT regards Kd and Km as the same.

The experiment done is CITC which makes Cpd2 a competitive inhibitor of Cpd1. Correct? So, to decrease Kapp, why would I add more Cpd2? Competitive inhibitors INCREASE Km or Kapp.

Could you please re-explain 🙁 I'd really appreciate it

Maybe it'll help to understand the difference between the different constants.

K1=affinity constant

higher K1 or Ka means higher affinity, hence the name.
Kapp=affinity constant=relating to inhibition or competition for the enzyme.
Kd=Km=dissociation rate constant

(notice how this is the inverse) lower Kd or Km means higher affinity

In the Q-stem, it first introduces comp1 & protA, their affinity is K1. They stated that its pretty high.
Then they introduce comp2, and from the experimental name, we know that it's a competitive inhibitor, this is where Kapp comes into play. (or you can think of this as K2)

comp1-protA complex measures K1
comp1-protA+comp2 measures K2 (we assumed that addition of an inhibitor, should lower K2/app, because this K is still the measurement of comp1-protA)

Comp2 is an inhibitor for comp1, meaning it binds at the active site of protA to inhibit binding of comp1 to protA.
The q-stem then says that upon addition of 10mM of comp2, the Kapp=K1. This means the addition of the comp2 didn't really do anything.


The last sentence then asks what can we adjust to make Kapp<K1. Meaning how can we make Kapp smaller. Just think that you can't change K1 anymore. To decrease Kapp=to decrease the affinity between comp1-protA w/ inhibitor, we increase comp2.

 

[kinzane]

But doesn't a competitive inhibitor INCREASE Km?

A higher Km means lower affinity for its substrate!