Aamc 9 #131

[LaurenMarie]

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This question is making me so mad! (Kaplan says the answer is decreases) At first I thought decreases, but then I changed it to remains the same. The question asks for the change in concentration of osteoblastin. I get that the competitive inhibitor would decrease the rate of production but why would the concentration decrease?

If anything, the concentration would increase because peptide A would still produce some osteoblastin because peptide B is only a competitive inhibitor.



Any thoughts??

 

[piggle]

The last paragraph of the passage states that "Peptide B has no effect on OB, does not stimulate osteoblatin production, and competitively inhibits Peptide A"

So think about it this way: you've got a set amount of Peptide A and you keep adding Pep B. We know that Pep A + OB = osteoblastin. But since we're adding more and more and more of Pep B, eventually there will be more Pep B binding to OB than Pep A binding to OB. And since Pep B does NOT stimulate osteoblastin production, the overall amount of osteoblastin decreases. Hope that helps!

 

[AdrianVeidt]

The last paragraph of the passage states that "Peptide B has no effect on OB, does not stimulate osteoblatin production, and competitively inhibits Peptide A"

So think about it this way: you've got a set amount of Peptide A and you keep adding Pep B. We know that Pep A + OB = osteoblastin. But since we're adding more and more and more of Pep B, eventually there will be more Pep B binding to OB than Pep A binding to OB. And since Pep B does NOT stimulate osteoblastin production, the overall amount of osteoblastin decreases. Hope that helps!

I understand your rationale, but if peptide B does not stimulate production, wouldn't that mean that the concentration would stay constant? Lack of stimulation =/= inhibition.

Remains constant looks like the right answer to me, but I'm clearly wrong.

 

[piggle]

Well how I thought of it (and how I think the testmakers want us to think of it) is because you're adding increasing amounts of Peptide B to the mix, there will be a point where Peptide A won't even get a chance to bind. That means as time goes on, we get decreasing amounts of osteoblastin.

 

[AdrianVeidt]

Well how I thought of it (and how I think the testmakers want us to think of it) is because you're adding increasing amounts of Peptide B to the mix, there will be a point where Peptide A won't even get a chance to bind. That means as time goes on, we get decreasing amounts of osteoblastin.

OOOH I get what your saying now. The constant amount of peptide A will eventually be unable to combine with the OB because of the increasing amount of peptide B, so the lack of production over time will decrease the concentration of osteoblastin.

I feel kind of silly now.

 

[piggle]

👍

yep you got it! haha, I should have made that clearer the first time 😛

 

[curt656]

One has to assume that once inhibition reaches the point where production of osteoblatin is less than osteoblatin consumption, the concentration will decrease. The key is that the osteoblatin is being used (or degraded) somewhere, not just floating aimlessly in the solution.

 

[LaurenMarie]

Thanks everyone! That makes sense I guess I was just over thinking it 🙄

 

[sciencebooks]

One has to assume that once inhibition reaches the point where production of osteoblatin is less than osteoblatin consumption, the concentration will decrease. The key is that the osteoblatin is being used (or degraded) somewhere, not just floating aimlessly in the solution.

I was waiting for somebody to say this. We have to work with the fact that our body does generally use the things it produces.