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NBME 20 Answers

nbme20/Block 2/Question#41 (58.7 difficulty score)
Purified serum antibodies elicited by ...
Proteins X and Y express the same epitopes🔍,📺
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 +9 
submitted by usmile1(119),
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If you kloo ta dwoUrl tqusonie ID 12992 ti sah a neldoufrw oeannptilxa fro .tihs I f yeht arhes het amse osppiee,t it will aevh a rdwdanwo opfleI.s yhte rshea nneo fo hte easm topseeip, het lien lliw eb tlraizhoon acrsso the pghar gciniadni(t on hnegac as teh mtnauo of Y dedad csresean)i

eacv  omg YES!! thanks Uworld I got it correct! exactly this qx asked the exact opposite thing! Hahaha I loved it !! +8  
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! “Protein Y expresses all of the epitopes expressed by X, but protein X does not express all of epitopes expressed by Protein Y.” 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 – 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. +2  



 +1 
submitted by amarousis(24),
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so woh uodwl hte rphag kloo fro piernot x ssxsreeep lal of the ippeoest derespxes by ionterp y, but npeotir y sedo nto spresxe all fo teh iosepetp desspeerx yb ireonpt ?x

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. +  



 +1 
submitted by usmleboy(13),

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.

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 :( +  



 +1 
submitted by rhizopusmucor(1),
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I ufdno h,sit I tnkhi ti iegsv indak of na aenoinxpl:ta fI teh isogrneres of log A'()A/( -1) on lg-o B is nilaer wtih a elpos fo -,1 tnhe thsi inceatdsi ttah the inaogstman is toepiivmetc adn yb dientnfoii eth gnstaio and oannsttgia tca at het EMAS itogncineor siest. oS blsaiclya hte omre yuo dad Y, the lses X si obun,d cwihh neams htye aveh a aems uasurclttr oennpmoct pet(pe)oi nda tmus cat no hte sema siet? on'tD nkwo fi isht kasem .seens..

.cifuuma/hshgpays/bnt/cla/setbbsitd~d1tisde:oppo/ng.cuemt/hj.rhl




 +0 
submitted by stepwarrior(23),

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.




 +0 
submitted by ilikecheese(47),
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Hhig X is nobdu nhwe wol Y si ,ddeda dna olw X is bnduo when Y si de.dSda o YMBEA htye aer ipgcemont rof eht smea binnigd pstoitp/eepo eud to stih ihroilntpsae pe(tepio= aiynbodt dinigbn e??)????s ?i?t??




 +0 
submitted by breis(53),
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yuro esugs si as dgoo sa n.................................i........e.................m...........

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



 +0 
submitted by am.cassandre(0),
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aRn ocarss a inetosqu imlrisa ot shti ni enahtro oteqsuni kanb. nI ttha outneq,si rteeh wsa on engahc sa het nttanrienocco of teh nwe ieprtno saw d,aded hcwih etanm thta hte tow tpineros ddi tno ehsra iraslmi eetppi.so In siht ueq,ontsi het nbtdyiao is tjus yngtri to bidn sit ptpe,oie os ndgiad mero ro y aesmn ssle gnidnib of irotnep x acubees eht nytaodib ovfsra the soirnept het m.ase