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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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Use teh neaeiyrgd-rbWH tinaquoe

  1. Take eth squaer toor fo 0,110/6 and ahtt lwil igve uoy het eefcnruqy fo hte esesvicre elella = 40/.1
  2. lacauCelt het necyfurqe fo het namdonti lelale wtih 1q+,p= ihhwc si p= 579.0.
  3. heyT rea ietnlgl you to llectcaau teh fnceeyqru fo eht sdeaise rearsc,ir cwihh is htiw het uaeotniq p.q2
  4. eThy anwt nylo the esdiesa recasrri ni ihhcw eldotine is setern.p To tlacauelc i,hts sue hte q auelv /41)0( nda yltmiupl by %80 ni htsi dhsuol vgei ouy ..200
  5. n,llayFi ctclaalue for 2qP 2 20(5..0)()97=0 .400 = 5/.12
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yex  Neic! ad.n.. we aer poduepss to dera the mste dna od lal iths in a mtneui ro ?os -:/ +28
charcot_bouchard  ellAle eyerfcqun 1./40 so crriaer rqef /1.20 8%0 fo 1/02 si 152/ 180(0/0 x 0/1)2 +15
dickass  hA fc,ke qp2 gto me +1
hello_planet  A dyhan hrsttuoc rfo nybieagdrWeHr- is htat oyu aereylngl can eusasm p ~= 1 fi q fi fariyl w.lo It osal desnt ot eb reiesa ot okrw in ntifocras if hte wernas cohseic ear in isfnrctoa so uyo no'dt evha ot euconb bakc and torhf neebewt froctanis adn el.sdmica So ihtw tt,ha uoy esnd pu wtih 2qp = 2 * 1 * 1/05 = 520/ = 2.5/1 +9
topgunber  aeht htsi owlhe ersclabm htgi:n In neo i:len 2 * q * hT.si 8%0 is rfo eedasids dainiudlivs wt(o q lIa.eesl)l = 11600/ = e2qTh^ qeuerycfn of q = 0 Now14/ rsacrier si 2 p q. P is eolcs to noe igsnsmau HW qem )qp(=+.1 r'hsTee an iainodadtl epts in htsi otquseni due to eht otw nfdiertfe oasnitut.oms )(q2 = 210/. 0%8 of seteh irasrcer are oeitdlesn os pmytllui /201 * 08. = 2/15 +

@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(68)
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eaht shit eohwl mearsclb nt:hgi In noe ni:el 2 * q * 80%.

ihsT si orf dieessda vludaiindis (wot q l.eesIa)ll = 61010/ = h^q2 Te rncefqeyu of q = 410/

oNw rrrisace is 2 p .q P is coels to oen isuamsng HW qme p1)+=(.q 'Tsrhee an aiodlitdan tpes in shit tnqesiuo deu to the wot dfirnteef no.itmutsa

so 2(q) = 0.1/2 %80 fo tseeh isarcerr ear nieetdlos so tpymulil 02/1 * 80. = /215

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 +0  upvote downvote
submitted by thechillhill(1)
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p + q = 1 ^p2 + p2q + q2^ = 1 f i q2^ = /01601 = 30.0006 nhte q = sqqt2r()^ = 002.5 eolvs for p to egt p = 1 - r = 1 - 5002. = .5709 hte oeusygztreoh ricrraes = p2q = 1 - ^p2 = 1 - 50.9 = 0.5 ^q 2 cna be pdoerpd c/b 'sti hucm amllrse than p.2^ T he idenleto si benlispsoer rfo 80% fo the tunasm.toi 80. x 5.0 = 004. = 0/401 = 1/52

ereTh gthim be an iserea wya ot do itsh, ubt ti wrkoed fro m.e

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thechillhill  oS ppnatreyal I o'tdn kwon hwo to famtor very wl.le !Soryr +1
pakimd  oS euaecsb i uoltndc npdse rmoe athn a enitum no thsi qinuoset nad sheynolt nidtd aelrcl hte e-dgynrriabWeH aqintuoe tish si how i dselov ti runde a ouietmn: s you okwn in a negiv unoolpaipt flah of mteh lwil eb aciersrr senic its an mouasaotl sisecveer isedsae aA =aA AA aa Aa sa Ao of that lfha %80 rea due ot itdolene aoitusmtn dna %02 are due pinot nmatyb tsoui atht loigc %80 fo alfh iont %20 of ahlf wlli gvei yuo 512/ +1
draykid  .80 x 50. si 04. +
topgunber  etha isht lohew blmcreas ighnt: In neo nei:l 2 * q * 8s.i %0hT is rof aeiddsse disuiavilnd otw( q a)Ilsell.e = /10061 = e^ q2hT cqreyenuf fo q = Nw14 /0o crsriaer si 2 p q. P is esolc ot neo mauisnsg HW mqe p+q=.)1( eserT'h an tilddaaion tsep ni ihts istqneou ued to hte tow dtiernfef osaits mtuno. (q)2 = .210/ 0%8 fo eetsh sricrrea rae ndotisele os lyipultm 201/ * 8.0 = /251 +


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