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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 eth eWnHrr-ediabyg teaoqnui

  1. kTea teh ruaseq ootr of ,0116/0 nad tath llwi igve oyu eht erueqcfyn of eht sreeveisc lleael = 40/1.
  2. aClucltea hte cnfqryeeu of hte aitdnmon leeall iwht =1,+pq hhicw is =p .079.5
  3. Tehy are eilglnt yuo ot tecaaucll teh uneyrceqf of teh aesdsie ea,crsrir ciwhh si tiwh hte uoeaitnq .p2q
  4. hyeT twan ylon eth desasei raicrrse in wihhc eetnloid si rp.esnet To alctuelca shit, ues hte q vleua 10)4(/ nda ytpulmli yb 80% in iths doshul iveg ouy .00.2
  5. yilFanl, tcalluaec fro 2q2 P 0=(5.))720(90. 00.4 = 51/2.
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yex  cieN! ..nad. ew are dspoeups to read eht tmse dan od lal tshi in a inmuet ro so? /:- +28
charcot_bouchard  lllAee cuynreeqf .41/0 os arrrcei rqfe .201/ %08 of 1/20 si /215 /801(00 x /)021 +15
dickass  Ah cfe,k p2q tgo em +1
hello_planet  A ndhay strucoht orf beyWrgn-adHrei is atth uyo yleaegrnl acn smusea p ~= 1 if q if iyralf wl.o tI saol tndes to eb sariee ot okwr ni frsaction if eth sanwer eisocch are ni sfitonacr so ouy 'ndot veha ot eobcnu cbka dna fhotr wnbetee snocfrtai nad m.esilcad S o twih att,h oyu sden pu hiwt pq2 = 2 * 1 * 10/5 = 250/ = ./251 +9
topgunber  tahe sith lohwe raecslbm t:gihn nI eno neil: 2 * q * s8i%.h0 T si for ddasiees laidnsdivui (otw q lee.Ialsl) = 0011/6 = ^e2h Tq rcfeeuyqn of q = 1N4/0wo sriecrar si 2 p .q P si loces ot eon asgsnumi HW meq p(1.+)q= hee'rTs na aaliitndod spet in iths inoteusq ude to eht wto ifdtreefn sn tms.oitoua 2)q( = 10./2 80% fo eehts reasrric rea tlseeidon so liuptyml 0/21 * 8.0 = 1/25 +

@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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athe htis leohw cemblsar gnith: nI eno ilen: 2 * q * %08.

hsiT is ofr saidsdee usiadiidlvn ow(t q leIe.l)sla = /01106 = qeT^ 2h ufyenqcre of q = 04/1

oNw serracir is 2 p .q P si ecslo ot eno nmsugais WH eqm 1p)(.q=+ 'reehsT na dtlnoaiadi pset in siht teqiunos ude ot het two eiftfernd a.umitnost

os 2)(q = /102. %80 fo seeht acrreirs ear neldeiost so pltyulmi 02/1 * .08 = 215/

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 +0  upvote downvote
submitted by thechillhill(1)
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p + q = 1 p2 ^ + 2pq + ^2q = 1 fi q2^ = 60110/ = .300600 hnet q = r2tq)qs^( = 002.5 ev sol ofr p ot egt p = 1 - r = 1 - 20.05 = 7950. te h utohroygzsee rcerrisa = 2pq = 1 - p2^ = 1 - 90.5 = .50 q2^ can be oeppddr bc/ sti' much lemrsla tnha .p^2 Teh nteeildo is brnplsieose rfo 08% fo teh ttauinms.o .80 x 0.5 = 4.00 = /4001 = 215/

eTreh ithgm be an eserai ywa ot od hi,ts tub it kowrde orf em.

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thechillhill  oS aypatrpeln I 'tdon kwon ohw to ftramo rvey l.elw S!oyrr +1
pakimd  oS ucebesa i dunlcot pdnse mreo htan a uiemtn no ihts uqnoteis adn tohnysel dtdin calerl teh Wdgbr-anreeyHi iqenotua tsih is how i lsvoed it edrnu a msi: teuon you kwno in a nvegi pntauoloip lahf fo emht will eb saicerrr nesci sti an saluomato eiecrssve ssdiaee aA Aa= AA aa Aa Aoas of atht lhaf 08% are ued ot iolednet asnomiutt adn 0%2 rea due otnpi mbitunay tos htta oiglc 80% of half toin 0%2 fo alfh lilw vige oyu /125 +1
draykid  08. x 5.0 si .04 +
topgunber  htea shti ohelw lsrmaebc i:hgnt I n noe einl: 2 * q * i%0.Th8s si ofr sedidsea lividusnadi wt(o q )elI.alesl = 1001/6 = eq2T ^h ueyrecnfq of q = 41N0 ow/ crreaisr is 2 p .q P si celos to eon mgsaiusn WH qem .q)+=(p1 esrThe' na aiddonliat step in siht eioqntsu ued to eth two erfdfetin iutsamo snt.o q()2 = 1.0/2 8%0 fo sehte riesrrca rea eintedols os limtlpuy 12/0 * 08. = 25/1 +


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