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NBME 22 Answers

nbme22/Block 3/Question#23 (reveal difficulty score)
A 45-year-old woman comes to the physician ...
25% ๐Ÿ” / ๐Ÿ“บ / ๐ŸŒณ
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 +16 
submitted by โˆ—lamhtu(135),
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camhetrioHsooms aacoedtiss iwth LAH,-3A tbu teh aippeartrop inesnegrc ttse is umers rnenfrrasit iatnaourst nad ntieirfr seelvl


 +7 
submitted by โˆ—b1ackcoffee(89),
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I hgthtou endgragri ihts uesitoqn as HLA eegn tinurisbodti (ne kb.ocl)

nvEe if we adricds iyhegtervn adn tsuj itnhk ubtao eht tlas seennetc tui(q)os,en rfo yna ge,en cachen fo igsinlb iahngv axcte saem llleae si %52

seusopp ereth aer 4 lllaese fro a gnee ni nasepr,t ,a,A ,B dan b tahw( I enma ot asy is all 4 rae reetinfdf in btoh s)ran.pte

oNw uppeoss eno hgdtarue sha aB oeTt ypoge.n aevh taexc maes lsleae,l teorh ghutdrae tusm ehav

0%5 cahnce fo a omang( a nda A rmof tmdphuoeiilmrte )l y%5b 0 enchca of B na(gmo b dan B rfmo ftr)hae

eoimbncd iblpayotirb for yna gnee lodwu be 5%.2

e,saleP notip tuo nya lfuat ni hte no.gnerisa

cjdinurdreamz  i used the same exact reasoning +  
skonys  Suffering from success over here because I just read it as there one locus "6*" and shes homozygous aa. Thus her sisters has a 25% chance of being aa too because punnet square. Most probably the wrong logic but hey, we out here. +  



 +4 
submitted by โˆ—bubbles(78),
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naC soeneom pliexan peylropr ohw we wokn atht htsi aittr sllofow deanMniel eistcnge nda si asuolatom sviesecer dna rreurofthem woh hte setnrpa were gtehouoyz?rse

I segsdeu a olt no tihs iunotseq dna ogt ulkcy (:

niboonsh  Autosomal Dominant disorders usually present as defects in structural genes, where as Autosomal Recessive disorders usually present as enzyme deficiencies. P450 is an enzyme, so we are probably dealing with an autosomal recessive disorder. furthermore, the question states there was a "homozygous presence of p450.....". In autosomal recessive problemos, parents are usually heterozygous, meaning that 1/4 of their kiddos will be affected (aka homozygous), 1/2 of the kids will be carriers, and 1/4 of their kids will be unaffected. +36  
nwinkelmann  Is this how we should attack this probelm?: First clue stating endoxifen is active metabolite of Tamoxifen should make us recognize this undering first pass hepatic CYP450 metabolism? Once we know that, the fact that the metabolite is decrease suggests an enzyme defect, which is supported by patient's homozygous enzyme alleles. Then use the general rule that enzyme defects are AR whereas structural protein defects are AD inheritance patters. Once we know the pattern, think that most common transmission of AR comes from two carrier parents. So offspring alleles = 25% homozygous normal, 50% heterozygous carrier, and 25% homozygous affected, thus sister has a 25% of having the same alleles as patient (i.e. homozygous CYP450 2D6*4)? +6  
impostersyndromel1000  we had the exact same thought process, so i too am hoping this is the correct way to approach it get reasoning friend +  
ajss  thanks for this explanation, I totally forgot about AR patterns are most likely enzymes deficiencies, this kind of make the question easier if you approach it that way, thanks +  



 +3 
submitted by โˆ—dentist(83),
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  • zemeyn ndseeycfcii = AR

  • mozuooyghs ecpeenrs fo .PCY"..

:EQD uyzghsooom + AR = 2%5




 +1 
submitted by โˆ—its_raining_jimbos(27),
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oxinTefma hsa ot eb mdbtiozelea via itfsr asps ebtamisoml to na tvacie ateblmoite iefnnx).doe( hTe eipantt sah cdesderae ctntisrcanoeno of het doaleiebmzt ocdtupr idncnitiga atht het aโ€™inetstp airp fo otcheycmor 540P lalelse erโ€™nat leiiotgnzabm emxftioan roeyl.tccr Teh qnesouit is niagks awth het ccesnha rea eth isesrt sah the saem yept,geno ichwh ldowu be 5%2 g-&;-t /21 * 12/ = 4/1

medschul  How do we know the parents are not homozygous +8  
yotsubato  Chances are they are not unless they had or are incestuous +  



 +0 
submitted by 07chess(2),
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ikgehvnnOitr srcdeew m.e The ioqtuens emts kass fro a accehn rof a ritsse ot teg eth saem alsele,l nto taht ehs si shzgmoyuoo ihtw mteh. oS I deikcp 7%5 eitnsda of . %A25t teh pot fo ti, eyht upt D(*I642 ugthhto htta msean tath pt. ash 4 pcsioe of (wlseleaolt morf ahce atrepn ikel L TAe.)yHh aveh ot amek noeiustsq lisiobmpse to nrn.uddetsa




 +0 
submitted by g8427(0),
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If oems oen can hple me dsdeaunnrt cb im a tib .dnucesof I utnedrsdna the uthhtog cospres nad I ezildrae thta sthi aws na AR adsisee dna I asol got eth /41 ec,aeftfd 12/ rcriaer dna 1/4 ef.datnfceu uBt I cohse %0 cb I ugedfir fi it asw na RA ssdieea eht 1 ldcih deryaal diseeasd asw zyooghmuso fadetfce 4/(1 ceef)tf.ad hchiW aeld em to tkihn that the othre isrset saw theire a crirrea or otn acfdtfee ta a.ll mA I jtus orev ktiinhgn ihts ro am I tno yllfu tnrgsdinaendu ahtsw gigno on?

rush  you have to think about each child individually, doesn't matter what the siblings have. The question states what are the odds of the child getting the disease. So regardless of the other siblings it still is Mom (1/2) dad (1/2) which makes it 1/4 AR +  
titanesxvi  But how do we know that the parents are heterozygous for the mutation +  
need_answers  we know that the parents have to be heterozygous Aa X Aa because on a 2x2 table, the only way the daughter could be homozygous for an AR is by having both parents be carriers (Aa) so the question was asking what are the chances the sister has the same alleles (aa) and there is only a 25% of having the same alleles. +  



 +0 
submitted by timekeeper36(0),
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So I haet sith qsnutioe useabce oyu ct'na ellt waht het 'anprset eptsoeygn .are sThi mezyen detefc sit'n tnsgihome ahtt yuo anc ellt yoceitllhopagn os ew cn'ta sutj msseua eth nrteaps rea esohre.enegotu

tI hyltsoen setd'o enve tartem if it si ouolstama ercessevi ro ton isnec we nloy eahv to owyrr buaot zgmsyoouho off isngprs. Here si the ahtm.

saePnrt sbsoeilp tpneogyse:

A - loNmr;a a - lrman.oba

M:mo Aa D:ad Aa what( eth notueisq is siugnas)m

m:Mo aA :daD aa

:moM aa :daD Aa

M:om aa D:da aa oh(w is to ysa sith nit's pbiols)e?s

(omM AA ro adD AA si psliiesomb ceins het sogfirpnf ac'nt eb )aa

rFom tohse pgneotye tsopsiiliesib ew anc uaactcell trebet bto.aiisperilb

nceiS si't bsoeplis for esrnpta ot aevh ayn of ethes 4 tegeyosnp we tjsu esumsa ahtt caeh spboyitilis si 4.1/

oMm: aA + Da:d Aa gt;-& iognpfrsf aa = 41/ iplmyltu( hits yb /41 nicse wre'e agmuissn ulqae spiisybotil ofr ertho opneteys)g &;g-t 16/1

:Mmo aa + D:da aA g-&t; norfsgpif aa = 1/2 g&-;t */1/214 = 18/

:omM aA + a:dD aa &-g;t soipfngrf aa = /12 &;tg- *412/1/ = 18/

:oMm aa + da:D aa t-&;g fspgiornf aa = 1 ;-gt& 1/1*4 = /14

nehT ew dad lla het aipbitoreslib gtote /e1hr61 + 18/ + 8/1 + /14 = 69/1

I nwok the sno'ueqtis wranes is ..5%2. os I ugses tpu 5%2 on teh st.et UBT fi uoy nwta to acltaluy taeusdrn,dn sthi is a bterte appoi.mxniorta heT lyon lmprboe is we tndo' konw teh geen bodtistuniir fo teh eaeslll or I cna vgei tterbe htma P:

rrSoy is't nogl dan ugsnifcno




 +0 
submitted by fukprometric(3),
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Teehr saw a uolrdw intuoqes no siht. HAL are pdsesa dnwo ot fpfgrsoin in tlspohya.ep trenaP X has A nad B ,eotpphayl eaprtn Y sha C adn D lh.oaeppyt sTeeh avhe eednMlani e,cineihatrn ie.. AR. If eht nptetia si ozmoousyhg rof C,yD2p6 htta eamsn ttah arPetn X and Pteanr Y othb ahev a paheltoyp htiw .cY2PD6 .ei: A oyptpalhe ash c,yD2p6 C hlyappote sah c26Dyp

rFo erh itrs,es eethr is a 0%5 cecnah mom sspsae odwn erh A ehpaol,tpy nad a 05% ncecha eth add epssas dwon sih C ephltp.oya /12 x 2/1 = 4/1




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