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Retired NBME 20 Answers

nbme20/Block 3/Question#1 (reveal difficulty score)
The frequency of an autosomal recessive ...
1/25 ๐Ÿ” / ๐Ÿ“บ / ๐ŸŒณ / ๐Ÿ“–
tags: biostats

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 +16  upvote downvote
submitted by jotajota94(14)
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esU the Weye-Hirbdrgna qieoutan

  1. Teak the rauesq otor fo 001/6,1 dna htta lwil veig ouy eth qcnryeefu fo eht sersicvee laeell = 4/1.0
  2. tuaCcella eht qnerfeycu fo the mindotna lellae ihwt 1p,=+q chhwi si p= 59.0.7
  3. eThy are neliglt uoy ot ctlcaleua eht yneuqecrf fo eht idesase r,acrirse chihw si htwi hte eaqituno p2q.
  4. Teyh watn ylon teh siaedse arrricse in hchwi olinetde si .nsptree To ltuaaccel is,th ues eht q ealvu /10)4( nad uilyptlm by %80 in hits dulhos gvie uyo .0.20
  5. F,ilanly tellaccua fro 22Pq 059020)=(.).(7 .004 = 15/.2
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yex  Nc!ie .nda.. we ear upsspeod to eard the tesm nda do lla thsi ni a ienutm ro os? :-/ +28
charcot_bouchard  Aelell qnfyucere /10.4 so icrerra rfqe 120./ %08 fo /120 is 25/1 000/18( x /2)10 +15
dickass  Ah ke,cf qp2 tgo me +1
hello_planet  A ahnyd coushrtt rof HWrngbieer-day is ttha oyu alnyeergl anc meuass p =~ 1 fi q fi ryalif .lwo tI oals etsdn ot be esaier ot krwo in tonfcsiar fi teh rwenas ihoscce era ni sntcaiofr so ouy t'odn evah ot ebunco bkca dna rthof teewebn incsfraot and easdm.cli S o with ta,th you sden up hiwt q2p = 2 * 1 * /150 = 502/ = 1/25. +9
topgunber  etha hsit owhel rlcmseab ntg:hi In eon enil: 2 * q * Ti.h08%s is for dieesads iiauldidnvs two( q aelsl.eI)l = 11/006 = 2h^qTe ecneyuqrf of q = 401oN w/ seriracr is 2 p q. P si csoel to eon miusgasn WH meq .p1()q+= heersT' an aontldiiad setp ni shit qnetoius deu ot hte wto ifenrfetd .ismotaon tus 2()q = 0.1/2 80% of hetse rrriasec ear intseoled so itulmply /201 * .08 = /152 +

@usmlecrasher

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

+3/- alexp1101(11)


 +2  upvote downvote
submitted by โˆ—topgunber(68)
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tahe iths ehowl cblmrsae it:hgn nI oen n:eil 2 * q * 08%.

hTsi is ofr adieseds nsdliiuaivd (wot q lselIla)e. = 1/0160 = 2 ^qehT nuceryeqf fo q = 0/14

Now erirrcas is 2 p .q P si ocsel to eon ganmiuss WH emq p=q+1.() 'sreThe an atlnidodai tpse ni hsit unsoiqet deu to hte owt neefitrfd .nimttsoua

so 2)q( = 21/.0 %08 of sehte rceiarrs are dnteoisle so plmulyti /210 * 0.8 = 1/25

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 +0  upvote downvote
submitted by thechillhill(1)
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p + q = 1 2p ^ + 2qp + ^2q = 1 if ^2q = /00161 = 60.3000 h ten q = t2(rq)^sq = 00.52 sel vo fro p to teg p = 1 - r = 1 - 5200. = 709.5 eth ezsoeorhutgy isracrer = 2qp = 1 - ^p2 = 1 - 5.09 = 0.5 2^q acn eb rdodepp /cb sti' chum arselml hnta p.2^ T eh tndleieo is eepsloinsrb rfo %08 fo het ui.atmntos 08. x 05. = .040 = /0401 = 51/2

erheT mhgit eb an ieesar yaw ot od ,sith btu ti redwok orf .em

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thechillhill  So nlaaepyptr I 'ntod wokn ohw to atmfro vyer .wlel rySr!o +1
pakimd  So uebcsae i udoctnl psedn ermo naht a umeitn no tish iosnteuq dna tylsnoeh tindd alcrel eth Wrbyieg-nrdeaH anieouqt isth is owh i voldes ti dunre a uimonts: e oyu nowk ni a vgien ptupaoonil lhaf fo mhte lilw be scrrriae iecns sti na oalsmutoa icersseve ssadiee Aa =Aa AA aa aA Asao fo taht lafh 08% ear edu to dlnieeot stnimuota dan 2%0 aer eud otpin b montyauist taht icolg 08% fo flah tnio %20 fo ahfl liwl vieg ouy /152 +1
draykid  80. x .05 is 4.0 +
topgunber  htea hsit weohl blamersc ighnt: nI eon eni:l 2 * q * s8% Th.i0 is fro iedsdaes anldiviisdu o(tw q eesa.Ill)l = 011/06 = Tq he2^ yfunqreec of q = Nw4/ 0o1 icrrasre si 2 p .q P si scloe to eno nausmgsi WH qme p1(=q).+ sh'reeT na didltaanoi pest ni isth oneuisqt ude ot het wto fidntrefe oi uo.nmtatss )(2q = 01/2. 0%8 fo sehte irercras rae edsteloni os tmiplluy 0/12 * .80 = /251 +



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