NBME 20 1/25 (NBME Answers)

Retired NBME 20 Answers

nbme20/Block 3/Question#1 (reveal difficulty score)
The frequency of an autosomal recessive ...
1/25 🔍 / 📺 / 🌳 / 📖
tags: biostats

+16 upvote downvote
submitted by jotajota94(14)

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Ues het gbreHdyWa-inre iauenoqt

aeTk teh ueqsra oort of 0,1601/ nad htta liwl vgie uoy eth yferqcenu of eth cvsreesei llelae = .4/10
tulaCcale eht ynqeuecfr of hte niandotm leleal hwit q=p,1+ ciwhh si =p .0597.
yThe era tlnilge uoy ot ualelactc eht furnyeqce fo hte seaidse rrsrei,ca hihwc si tiwh eth atunoiqe .q2p
Tehy nawt onyl eht sdaeise rsricare in wihch etinolde si .eensrtp To telaluacc ti,sh use eht q veaul )401(/ adn utipyllm by 0%8 ni hits odslhu giev uyo .200.
l,yFnali tuealcacl rof 2q 2P .)09=)70(20(5. .040 = 15.2/

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yex i!cNe ..adn. we are sesouppd ot read teh tmes nad od all thsi ni a ntumei or o?s -/: +28

charcot_bouchard leAell erneyqufc /410. os cairrre rqfe 12/.0 0%8 fo 21/0 si 5/12 0/8010( x /21)0 +15

dickass hA kf,ce 2pq otg em +1

hello_planet A adnhy tthorcsu orf rrW-nbedeiHagy si htta yuo ernllygae nac asuesm p =~ 1 if q if ayirlf ol.w It laos dstne ot eb reasei ot rowk in nractfois fi the aneswr cehisco ear in cftanrosi so ouy todn' ahev ot enbouc ckba and rthfo etbwnee nfirsacot nda .idclemas So ihwt taht, yuo dens pu whit pq2 = 2 * 1 * 1/05 = /205 = 25/1. +9

topgunber teah ihst welho ameclrsb :thign nI one line: 2 * q * Ti h08s.% is for edsedisa dlisdianuvi t(ow q ll.lasee)I = 0116/0 = ^qTh e2 ynrecuqef fo q = o41 0/wN eacrisrr si 2 p .q P si scole to noe numsisag WH mqe q1)=.+p( sTeerh' an dndailatoi etps ni hsti ueiqtson due to eth otw tdeefnrfi sno.umosiatt 2(q) = /12.0 80% fo hsete ciarerrs aer dolisente os ltilmupy /120 * .08 = /251 +

@usmlecrasher

1/40 = .025 = q P + q = 1 -> P + .025 = 1 -> P= 0.975

+3/- alexp1101(11)

+2 upvote downvote
submitted by ∗topgunber(68)

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ahet sith ewlho slramcbe hg:nit nI oen nei:l 2 * q * 08%.

ihsT is rof dsedseai davidsnuiil (wot q e.Islal)el = /10601 = q2 hTe^ fcuqnerye fo q = 0/14

Now recirras si 2 p .q P si loecs ot eno sagiusmn WH eqm p(1=)+q. hTrese' na adniodalit tpes ni hits nsetuqio due ot teh otw tdienerff oisntmua.t

so (q2) = .0/12 08% fo tseeh crreisar ear lnoiseedt so llmtuyip 01/2 * 8.0 = 5/12

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+0 upvote downvote
submitted by thechillhill(1)

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p + q = 1 2^p + p2q + 2^q = 1 if q2^ = 61100/ = 003.006 ehnt q = t)qr2(s^q = 05.02 osle v for p to egt p = 1 - r = 1 - .0205 = 09.75 teh hrozoestygue serriacr = pq2 = 1 - p2^ = 1 - .059 = .50 ^2 q acn eb dodpepr cb/ it's mhuc almesrl thna p.2^ Teh enodeilt si brnosspieel for %08 of the tani.stoum .8 0 x .50 = 004. = 10/40 = 152/

herTe hmtgi eb an eaiesr awy ot od sit,h tub it rdokwe rfo e.m

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thechillhill oS ppaltryena I no'dt wkno who to rmtofa evyr .well rSy!ro +1

pakimd oS ueebcsa i unodlct dnpse rmeo hatn a tmnieu no itsh siouneqt nad nytolhse ddtni acller eht WanHr-rydbeegi ionuqtea itsh is owh i oesdlv ti uednr a eotm unis: uyo ownk in a gevni naiuoptolp hfal fo etmh lwil eb ircerras escni ist an taaulmoos evicresse esedsia Aa A=a AA aa Aa os aA fo ttha half %80 aer edu to dnoletei moasnuitt dna 2%0 rea due noitp symbttuno ai atth cgoil 8%0 fo alfh iont 0%2 fo flha lwli gevi yuo 2/15 +1

draykid .80 x 50. is 4.0 +

topgunber taeh htsi lhoew embaclsr igh:tn nI eno n:lei 2 * q * 0.%8iThs si for iedesdsa saduvdiinil two( q lIa.lees)l = 01160/ = 2 Th^qe enyucfrqe of q = 4/o wN01 srarerci si 2 p .q P is elosc to one magissnu WH qem p)=1+.q( 'Tehsre an odiidlnaat pste ni tshi eosutqin ued to eth otw fefetrdin uit ssootnm.a (2)q = 201./ 08% of ethse raercrsi are eltediosn so iltumpyl /120 * 0.8 = 12/5 +

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