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NBME 20 Answers

nbme20/Block 2/Question#41 (53.1 difficulty score)
Purified serum antibodies elicited by ...
Proteins X and Y express the same epitopesπŸ”
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 +9 
submitted by usmile1(103),
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fI oyu olko at lwrdoU qosuteni DI 29291 it has a oedrunwlf onxiapltena rfo tshi. fI hyte aeshr teh asem peoeips,t ti iwll ahve a dwnwraod ploI f.es tehy aesrh enon fo hte emas e,etiopsp eth lien iwll eb taorlozhin scsaor hte aprgh gdatci(nnii no engahc sa het muntao fo Y eaddd asscr)eien

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? +2  
69_nbme_420  If you make up an example with numbers, it really helps! β€œProtein 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. +4  
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! +  
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(22),
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os ohw udwlo het hpagr kloo orf ripenot x xeerpssse all of eht oetispep rsesedpxe by niproet ,y btu tioprne y esdo not exrseps lal of the psotiepe eserpxeds yb oeinrpt ?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 rhizopusmucor(1),
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I onudf ,tshi I nikth it vesig adnik of na anin:aolxpte If het rsesoginre fo gol (A/')A( )1- no olg- B is irlena hitw a posel of ,1- etnh this natcsidie taht het taongsmani is eotimevictp and by iidtneifon eth tosgian and ontitangas cat at het ESAM rgietonionc tisse. So slbicalay eth roem uyo dad ,Y eht lses X is dnoub, which saenm tyeh vahe a emsa tutarlucsr opecnomtn opeip)(et nad msut tac on eht msae e?ist t'noD wnko if isht aksem n.s..ese

rnng~ghpds//stufi.tihubop1edsaechtt.u/setsabccld:mm/./plobihja/y




 +0 
submitted by am.cassandre(0),
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Ran rsaocs a otnieqsu ilsimra to stih in athrnoe qsonieut kbn.a In thta qui,otens tehre aws no hcagen as het notrnoientacc fo hte new erotinp swa ,dddae ihhwc naemt ttha hte two seinotpr idd tno saehr limrais pe.osptie nI ihts eqtnosi,u hte ybnaodit si sutj nirtgy to idnb ist oept,epi os diadgn eomr ro y smnea lses nibdgni fo pnetrio x csebaue the onadbiyt rafosv het nistroep eht e.mas




 +0 
submitted by usmleboy(9),

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.




 +0 
submitted by stepwarrior(20),

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(39),
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hHig X si uonbd when olw Y si de,add dan owl X si donbu nwhe Y si dodaSe. d YBAME etyh aer ogentpcim for het smae bnngiid isoptepoep/t ude ot hsit rstanelipiho toee=ip(p doanibty indgbin ? ???i??t?)e??s?




 +0 
submitted by breis(44),
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ryou egssu si as gdoo as ...........i................................................m.......en...

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