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NBME 20 Answers

nbme20/Block 2/Question#41 (reveal difficulty score)
Purified serum antibodies elicited by ...
Proteins X and Y express the same epitopes ๐Ÿ” / ๐Ÿ“บ / ๐ŸŒณ / ๐Ÿ“–
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 +11  upvote downvote
submitted by โˆ—usmile1(154)
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If you look at Uworld question ID 12299 it has a wonderful explanation for this. If they share the same epitopes, it will have a downward slope. If they share none of the same epitopes, the line will be horizontal across the graph (indicating no change as the amount of Y added increases)

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eacv  omg YES!! thanks Uworld I got it correct! exactly this qx asked the exact opposite thing! Hahaha I loved it !! +9
pg32  Even after reading the UWorld explanation, I am still not sure how the answer that reads, "Protein Y expresses all of the epitopes expressed by protein X, but protein X does not..." is incorrect. Based on the graph, I don't see a way we can rule out that answer choice and it sounds more likely than both X and Y having the EXACT SAME epitopes. Can anyone explain? What would the graph look like if the quoted answer choice was correct? +3
69_nbme_420  If you make up an example with numbers, it really helps! โ€œProtein Y expresses all of the epitopes expressed by X, but protein X does not express all of epitopes expressed by Protein Y.โ€ If we say protein Y has epitopes 1, 2, and 3. Then Protein X has epitopes 1 and 3. Then we can clearly see the relationship the AMOUNT of Y added relative to X bound would NOT be linear. Stated another way โ€“ we need an exponentially more amount of Y to COMPLETELY unbind X and therefore there would not be a one to one depiction in the graph Similar logic applies for the answer choice that states "protein X expresses all of the epitopes expressed by protein Y, but protein Y does not express all of the epitopes expressed by protein X. E.g. If protein Y has epitopes 1 and 2. And protein X has epitopes 1, 2, and 3. Here again, we have satisfied the answer choices condition, and no matter how much we increase protein Y, protein X will still have epitope 3 bound in this case. +5
69_nbme_420  Just to clarify for the first scenario: We have 3 epitopes on Y, and 2 epitopes on X. That means, assuming the epitopes are all present in equal amounts, if I add 300 grams of protein Y to the solution - only 200 grams will bind protein X. AND ONLY 200 grams of protein X can be unbound. Hope the numbers help! +1
fruitkebabs  For anybody still stuck on "Protein Y expresses all of the epitopes expressed by protein X, but protein X does not," although this statement may be true, there is not enough information in the question to prove this. We know for fact that because the Amount of labeled X bound reaches 0, at the very least, protein X and Y express the same epitopes since at a certain concentration, Y is able to completely displace all X from the system. This doesn't exclude the possibility that there may be extra epitopes on Y, but it doesn't prove it either. +4



 +2  upvote downvote
submitted by โˆ—amarousis(27)
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so how would the graph look for protein x expresses all of the epitopes expressed by protein y, but protein y does not express all of the epitopes expressed by protein x?

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drw  then some anti-X cannot relocate to Y even Y is added at whatever high dose. at this condition, the line can never touch the axis-Y. on the contrary, if Y express all epitopes on the X, but X does not express all epitopes on the Y, that means some Y epitopes are not seen on X. at this condition, I don't know what will be the line looked like. +1
medbound57  I think that the line's downward slope would be steeper, since Y has more sites that the antibody would bind to. +

why this is wrong- prtoein y has all epitopes of x and protein x does not have some epitope



 +1  upvote downvote
submitted by โˆ—usmleboy(19)
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An antibody can only recognize a single epitope. Since we see the more Y added leads to less X bound, then you can reason they share the same binding sites, and Y is overpowering X.

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srmtn  OMG this is the reason why! thank you!!! +
srmtn  but there are antibodies that can interact with 10 epitopes...example IgM... sorry got lost again :( +
chj7  For anyone still dwelling on this, the process they talk about in the question stem is possibly polyclonal (thus the issue of multiple epitopes). +



 +1  upvote downvote
submitted by rhizopusmucor(1)
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I found this, I think it gives kinda of an explanation: If the regression of log ((A'/A) -1) on -log B is linear with a slope of -1, then this indicates that the antagonism is competitive and by definition the agonist and antagonist act at the SAME recognition sites. So basically the more you add Y, the less X is bound, which means they have a same structural component (epitope) and must act on the same site? Don't know if this makes sense...

http://facpub.stjohns.edu/~yoburnb/pages/dictimages/schild1.html

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 +0  upvote downvote
submitted by โˆ—stepwarrior(29)
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this question was utter BS, but the way I justified "same epitopes" was that it said "in the context of anti-serum X," i.e., which epitopes would have been developed in the context of the antibodies present. It's unlikely that anti-X antibodies would bind to epitopes on Y that aren't on X. But you could argue there are theoretical epitopes on Y that could be bound by another antibody that doesn't exist in this scenario.

Very drawn out and took me 5 minutes to actually figure out on the test, honestly a waste of time in my opinion, but there you go.

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 +0  upvote downvote
submitted by โˆ—peachespeaches(2)
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Because it is a straight downward slope, you can also tell that Y and X are bound in similar ways by antibodies. What differentiates the two, then, isn't epitope binding capability, but the concentration. All else equal with binding sites, more Y with an unchanging amount of X will lead to less X bound, in a 1:1 manner.

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 +0  upvote downvote
submitted by โˆ—ilikecheese(46)
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High X is bound when low Y is added, and low X is bound when Y is added. So MAYBE they are competing for the same binding spot/epitope due to this relationship (epitope= antibody binding site) ??????????

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 +0  upvote downvote
submitted by โˆ—breis(56)
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your guess is as good as mine.....................................................................

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nala_ula  I spent so long on this question and same... hahaha +



 +0  upvote downvote
submitted by am.cassandre(0)
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Ran across a question similar to this in another question bank. In that question, there was no change as the concentration of the new protein was added, which meant that the two proteins did not share similar epitopes. In this question, the antibody is just trying to bind its epitope, so adding more or y means less binding of protein x because the antibody favors the proteins the same.

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This question tests the basis of competitive inhibition. if they express the same epitope, they will bind the same receptors. Adding more X will mean less binding of Y, hence the downward slope.



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