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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
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 +20 
submitted by โˆ—thomasalterman(175),
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orgcnidcA ot FA, het nrlepansole nltmgiea sonctnai eht aitl of hte senarpac dan eht elinspc rrtyea ;m&ap vine


 +11 
submitted by โˆ—usmile1(147),
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fI uoy loko ta rUwdol suqioent ID 92912 it ash a frwnoledu aenpxaltoin orf t.sih If yhet aserh the same eipp,steo ti iwll eahv a wronwdad ospIl f.e etyh serah neon of teh msae soeeptp,i teh elni iwll eb zlinoortha sscaor het hpagr actigdnni(i on cnaheg sa the mtunao fo Y ddeda ece)ssniar

eacv  gom YS!!E thnska wlorUd I otg ti !ortecrc laeytxc sthi qx dskea eht teaxc oseotppi ht!ign Hahaah I vedol it !! +8  
pg32  vneE rafte eagnird eth dUolrW nipxotaen,al I am slitl nto ures owh het rsanew tath daes,r eP"trion Y ssrpeeexs lla of het osepiept edepsxres yb tenropi ,X btu ptrnioe X osde ".t.n.o si icrot.ncre esadB on eht rgpa,h I tond' ese a wya we acn erlu uot hatt snware ocheic dna ti nuossd rmoe lkylie atnh obht X dna Y vniahg het CTEXA SEAM pie.eopts Can naneoy alip?enx thaW duwlo het rhpga oolk ilek fi eth teoqud senwra ecihoc asw c?terocr +3  
69_nbme_420  If uyo meak up an meplxae htiw eun,srmb ti rlaeyl hpesl! enitPoโ€œr Y xseesrpse all fo the sipeotep rsxsdpeee yb X, tub terinop X sedo not srexesp lla fo ptspeieo espexsder yb treiPon .IYfโ€ we ays troinpe Y ash tiepoeps 1, ,2 dan .3 hTen reioPtn X sha oipesept 1 dan 3. heTn ew acn yclaler ese hte oatpseihnlri the MUTAON fo Y deadd viealter to X budno dlouw TNO be .nlreia edttaS oanhert yaw โ€“ ew ened na etoiyllaxenpn orme unoatm fo Y ot YMEPECLLTO dinbnu X and thoerrefe eerht udlwo nto eb a eno ot eno ipnoidcte ni hte i Sapilarmrgh gloci sapelpi ofr eht ewasnr ohecci htta ttessa "prioetn X xsersepse lla fo teh iopspeet dxsreespe by ipeonrt ,Y tbu ionrpet Y esdo otn rsexspe lla fo teh pisepteo rdeexspes by pitoern Eg. X.. fI nirteop Y sah eitospep 1 dna .2 Adn pirnoet X has poitseep ,1 2, dan .3 Hree n,agai we hvae dtsafisei eht ranswe cchoies dn,ciiootn dna no trteam who ucmh we eniresac ptenoir ,Y etprino X illw listl eahv pieopte 3 nubod ni hsit acs.e +5  
69_nbme_420  Jtus to clifray rof hte strif cien:sroa We aveh 3 pietepos no ,Y nad 2 tsppioee no X. hTat e,nsam nsmiasgu eht teosppie era all etsnper ni equla ,amsuont if I dad 003 rmasg of epintro Y to het oiountsl - ylno 020 amgrs lilw dnbi rpotnie .X AND NYLO 020 rgmas fo ontprie X cna eb uu.nobdn poHe eht rsmbune ehl!p +1  
fruitkebabs  roF odbyayn tisll ustck on oie"Pnrt Y epeexsrss lal fo het oipsepte deexresps yb eitropn ,X utb pnioetr X deos ",ton uoaghthl tsih metenstta aym be ure,t eetrh si tno hgenou ioamifonrtn ni hte nqueoist ot evpor hsti. We wokn fro tcfa thta ucaesbe hte uomntA fo belaeld X nbodu rcaeshe ,0 at the eyvr l,sate eitpnor X dna Y eesrxsp eht amse potpseei nscie ta a enacrti tec,ntncoairon Y is albe ot ecymlptoel sdailcep lla X rofm hte smtyes. isTh od'snet ulxeedc the lsitsyiiobp hatt ehter aym be xarte ptpoeesi no ,Y tbu ti neosdt' orvep ti eiteh.r +3  



 +2 
submitted by โˆ—amarousis(26),
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os who wodul hte aprhg kool ofr intorep x rpsesseex lal fo the iepopest edrpeesxs yb ptinero ,y but renptio y dose otn xrepsse lla fo teh tpsiepoe xesdseerp by ipnreto ?x

drw  hnet soem itn-Xa nncato acoeerlt ot Y vene Y is added at vwerhtea hhig o.des ta tshi tncdonoi,i het ilne nac vrnee otuhc eth -ixaY.s n o the ,otyracrn fi Y sesxper lla peepitos no hte X, tub X odes ton esersxp lla potiepes on the ,Y htat nsaem mseo Y epopsiet are not eesn on X. at ihts iocodnint, I tn'do kwon thaw will eb eht nlie deloko .ikle +1  
medbound57  I tnhik ttha eht lsien' daowrnwd polse uldow be er,estpe scien Y ash oemr teiss ttah hte aibdotyn ldouw dnib .to +  

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

+1/- prudvi96(0),


 +1 
submitted by โˆ—usmleboy(18),
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An bnaoytid acn ylno cieroenzg a eglnsi eppto.ei nScie we ees eth rmeo Y addde ldesa to esls X u,obdn ethn yuo can snreao yhte earsh het emas bgdnnii s,tise dna Y si eenigrovworp .X

srmtn  MOG itsh si the asnore w!yh anhkt y!uo!! +  
srmtn  utb rthee aer idtniaobes ttah can ettarcin htiw 01 ote.eli.m.pxesppea .I.gM. yrosr tog tosl aigna (: +  
chj7  roF eonyan itsll niegldwl no i,hts teh rscsope teyh tkla uabot in the nusetqio mste si sypiolsb yollnlaopc hut(s teh ssieu of elmpiltu .spiopet)e +  



 +1 
submitted by rhizopusmucor(1),
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I nufod tish, I tnikh ti egvis ikadn of na onnapait:exl fI eht gsrnreesoi fo gol )A((/'A 1-) on -ogl B si arelin wthi a elosp of ,-1 hnte hits cneasdtii atth teh sngmaoitna is tveictoimpe adn yb inifonietd eth naoistg adn totansnaig act at eht AMSE tecrgiioonn eitss. So lyclbaasi teh meor yuo add Y, the sels X is uob,nd wichh samne hyte avhe a esma raulsrctut tnnmpeoco eteo)ppi( and smtu cta on hte mase its?e otDn' know fi shit mksea .s.s.nee

lncpgnetg.stmapi.fhj1boaiuu:pscibsh~e///edrlmsht.h//ydt/sodtacub




 +0 
submitted by โˆ—stepwarrior(28),
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hsti esotqiun aws urtet BS, tbu teh way I jdeiuistf e"sam ppioste"e aws tath ti idas "in het cxnetto fo semn-rtuia X," ,i.e. hwhic pieseopt oulwd have enbe vdpoedlee in eht tcoxnet of eht iitoasdenb rept.sen 'stI lenkyliu hatt -iXatn adoensibit olwdu ndbi ot ioppetse no Y tath re'tna on .X Btu oyu udloc auerg herte rea licotatrehe ppsiteoe on Y ahtt ludco eb nudob yb hronaet itobaynd htat os'nedt tesxi in stih icnoe.rsa

ryVe rdnaw out and otok me 5 mnuites to latyclua egfuir tuo no het te,st toenlhsy a tsewa fo iemt ni ym ipno,ino but ehetr yuo .og




 +0 
submitted by โˆ—peachespeaches(1),

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.




 +0 
submitted by โˆ—ilikecheese(46),
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ihgH X si bodun hwen owl Y si e,addd nda wlo X is noudb wnhe Y is oedda.dS MYBEA tyhe era enpcgotmi for eht smae bnidnig eptio/pseotp ude ot sith eilanotrisph e(oepp=it antoydbi gbinndi ?se ????i??t???)




 +0 
submitted by โˆ—breis(56),
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oyur uegss is as ogod sa .m...........................i...e...............................n.......

nala_ula  I tsepn os lnog no isht nutieqso dan ...amse aahahh +  



 +0 
submitted by am.cassandre(0),
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nRa soascr a qoeitsun mriails to tshi ni onrthea oqnisetu knab. In that e,nisuqot htree swa on geachn sa eth nrocttcinnoea of hte new eronpti asw ,dddae whcih amnte htta eht wto itpoensr ddi ont ahesr siarlim .pieepsot nI ihts teniuoqs, eht iytoanbd is tujs nrgtiy to inbd tis ,ioteepp os gniadd meor ro y asnme elss ndnbiig of ntoepir x baesecu the aobyndti orafvs hte nrpoitse the aesm.


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.

+5/- latdorsi(4),


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