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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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eUs eht -egHrabyierWnd teanuoiq

  1. eTka eht sreuqa toro fo 0/,1016 nad thta lwli give ouy the erqcneyuf fo het reescsvie alllee = .0/41
  2. acClltuea hte ueyncqefr of eth oanimdtn laleel hiwt +,p=q1 hhcwi si =p 7.095.
  3. Thye rea lelgnti oyu ot tlacuelac the cnqeufeyr of teh edassie rcrr,eais cwihh is twih teh qionteua .qp2
  4. eyTh ntaw loyn eth sedsaie isrearrc in chiwh leneotdi si .tnrspee To cluecaatl s,iht eus eht q elauv 41(/0) and mltlpyui by %80 ni shit ushdol vige you .002.
  5. anliylF, ccalatuel rfo 22Pq 9..2=)00(70()5 004. = 2.1/5
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yex  N!eic ...nda ew era opsspued ot aedr het tsme nad do lal shit ni a numtie or so? /:- +28
charcot_bouchard  Aeelll rcnfqyeue 1.4/0 so rcirear qfer 201/. 08% fo 21/0 is 1/25 8/0010( x 02)/1 +15
dickass  Ah k,cfe qp2 tgo em +1
hello_planet  A ndayh rtstohuc orf erbagieWH-rdny is tath you elgyaerln anc mesuas p ~= 1 fi q fi yaflir wlo. It sloa ndets to eb aeisre to kowr ni cntrafiso if the rsanwe secicoh aer in orncaisft os oyu on'td eavh to nbuoec ackb dna frtho teewnbe stcaronfi dna ieadcsml. oS whti ,that oyu nsed up ihwt q2p = 2 * 1 * 105/ = 502/ = /.512 +9
topgunber  hate tshi ewlho amlrsebc nit:hg I n eno el:in 2 * q * i8%hT 0.s si orf asiedsde ndiliiudsav wot( q easll.e)Il = 0/0611 = 2^qehT fyruenqce fo q = w04o/1N rrieacsr is 2 p .q P si ecsol ot noe assngmui HW mqe p1(+q=.) Tesr'eh an ilaadntdoi espt ni itsh uesnioqt edu to the wot eenfrftid a.stoonsim ut ()q2 = .1/02 %08 fo tseeh irsrrace era oeedstinl os tmlulipy 012/ * .80 = 125/ +

@usmlecrasher

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

+3/- alexp1101(11)

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 +2  upvote downvote
submitted by โˆ—topgunber(67)
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ehat tish loewh mlcbsare nithg: nI eno inel: 2 * q * 80%.

hsTi is fro aeddsies ddniaiislvu ot(w q eaI.lesll) = 6110/0 = h2q^eT ucnreqefy of q = /014

wNo ierrcars is 2 p q. P si eclso ot neo msnusiga WH qme +(=.p)1q 'rhesTe an oddtiailan pets ni itsh qtoiunes ude to het wto fdtnrefie us.tintamo

os )2(q = 0/21. 0%8 of heset serrrcai rea doietelsn so lpytilum 02/1 * 8.0 = 52/1

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 +0  upvote downvote
submitted by thechillhill(1)
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p + q = 1 p2 ^ + 2qp + ^2q = 1 if q^2 = /10106 = .360000 ht ne q = t2^s(qq)r = 250.0 veosl ofr p to etg p = 1 - r = 1 - .0025 = 957.0 teh hseetugooyzr rrecisra = 2qp = 1 - ^p2 = 1 - 50.9 = 50. q2 ^ anc be rdpdoep /bc sit' umhc llsrema tnha p^2. heT oleitend si eebslsiporn rof %08 fo hte is.utmaton 08. x .50 = 0.40 = 410/0 = 251/

hTree tmgih eb na eraesi ayw ot od ih,ts tbu ti woekdr orf em.

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thechillhill  oS plaprteyan I d'nto kwno hwo to otafmr very .wlel Sy!orr +1
pakimd  oS ebucase i cludtno sdpne mero anth a tmuien no hits nieqsout dna tosenyhl ndtdi llcera teh iyarg-nbWeHdre oeiuqtna hits si how i svodel ti uredn a m s:enuoit uyo knwo in a egvni npaiplootu lhaf of mthe illw be cearrirs sncei sit na somualato eesrvscie esaseid aA a=A AA aa Aa aAso fo that fahl 8%0 ear edu ot dnoeitel otmiunats nda %20 aer due ontpi notyatumbsi taht licgo 0%8 of hfla ntio %20 of lafh lwil iveg ouy 2/51 +1
draykid  80. x .05 is 4.0 +
topgunber  athe sith wlohe abremlcs h:gtni In eno :leni 2 * q * 0%Ti8sh. si orf eedssida uddaliniivs t(wo q Ialles.el) = 06110/ = ^hTq e2 cfneruqey of q = 0/1Nwo 4 rrcsiera si 2 p .q P is cosle ot neo igusamsn WH mqe +q(1)=.p T'erhse na alnoiitdad epst in ihst oqieutns deu ot hte owt itedenrff atn.isto sumo )(q2 = .1/02 0%8 fo eseth ecrarsri ear ieslnotde so miyullpt 210/ * 0.8 = /251 +


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