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NBME 22 Answers

nbme22/Block 2/Question#34 (43.8 difficulty score)
A 30-year-old man and a 24-year-old woman ...
1 in 600🔍
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 +22  upvote downvote
submitted by drdoom(615),
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uaCosaillntc fro add. The tbrialbipyo fo eht tfaerh niebg a erriacr si 32/ eisnc ti is wnkon ttah eh dents’o hvae teh aeside.s eTnh hte liiopybbtra of ihm sipangs it on ot sih dki si ,2/1 h:sut

  • lPyibibarot fo dad bgnie errirca = 32/
  • abioibytrlP fo dad ipasngs on saesied lllaee = /12

lClaistaounc fro omm. thWi the eigaWn-rybdeHr lriec,npiP uyo can eirugf uot het iapytlrbibo of eht rometh niegb a ira:crer

q = 0(r400q,)01/ts = 20/01

o,S 2pq = 2 * /0012 * 9002/19, cihhw si xoppra /001.1

roF teh hdlci to egt hte aeelll mofr mm,o wot tnhgis eend ot ehp:pan )1( mom sutm eb a crarrie “goyheeett[”zro] dan 2)( mmo tusm psas teh leeall to c:hdli

  • aloPtbyiirb fo mom ingeb rciraer = /0011
  • oriybiPlatb fo omm ssngapi no edsisea lleela = 2/1

Puitng ti lla roethetg. Nw,o mbcoeni lla ergttheo:

= r(aiblobpyti fo add egbni caerr)ri * ibitplob(ayr fo dad pisgsna no sesidea laelel) * (ybitalbiorp of omm ibneg er)raric * yaploitb(ibr of omm sigpans on dsieeas allele)

= /32 * /21 * 11/00 * /21
= 1 in 600

kernicterusthefrog  To quote Thorgy Thor, drag queen: "ew, Jesus, gross" +21  
niboonsh  This question makes me want to vomit +8  
drdoom  lol +  

A questioner below asked, “Why is there a two-thirds probability of the father being a carrier?” Attaching my explanation via tangent to my original response :)

We assume this is a recessive disease. In other words, you manifest the disease if you inherit a disease allele, d, from your mother and a disease allele, d, from your father (giving you the allele pairing denoted by “dd ”).

dd = you have the disease.

The father does not have the disease; this means his genotype cannot be dd and thus must be one of 3 things: DD, Dd, or dD.

That is,

  • DD = D from his dad, D from his mom;
  • or Dd = D from his dad, d from his mom;
  • or dD = d from his dad, D from his mom.

Notice that in only two of these 3 possibilities (⅔) does the father (potentially) have a disease allele, d, that he might pass on to his progeny.

+3/- drdoom(615),

Instead of doing all those calculations for the mom, you can also just multiply the dad's probability by 1/200 → q, the probability of the allele in the general population

+/- hungrybox(771),


 +3  upvote downvote
submitted by keycompany(264),
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nlanMdaie tni:eGcse

Mna has /23 naehcc fo ngbei a e.icrarr (eH deso ton aehv eht sdieas.e) Wmoan rriacer ikrs mtsu eb clcuadelat whti ^2p + qp2 + 2^q. q ^2 = 040/,100 q = /2,001 p is yolhrgu = q 12p = 1100/ = rariCre fyecruneq .

iskR of nhvaig a hlidc uths lueaqs 23/ x 001/1 x 1/4 = 1/060e2s: ue/ac3 B = Man Careirr srki 1010 / = aleFem aeCrrri 1/k4Rsi = ancehC ehty aehc assp on eht cereevssi eeng ot tiher ffr.gospni

hello  See my explanation if you need more words to explain this explanation +  



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eTh ibrpaoltbiy of eht rafeth negib a errarci si /32 esnic ti is wonnk that he esdo’nt have eth edi.esas ehnT the poabiybilrt fo hmi nssiapg it no to ihs dik si:

/21 * 32/ = 31/

iWht hte deireWnbH-argy nlpric,eiP oyu anc ieurfg tuo teh broaybpitil fo eth eohtrm inbeg a i:rarcer

q = r/s0t1,q()0040 = 0/102

,So q2p = 2 * /1020 * 109290/, hihwc popxar si 0,011/ and the blrityiapbo fo eth hicdl einggtt tshi leaell is 010/1 * 21/ = /0012

h:suT

020/1 * 1/3 = 6001/

brill45  You did this right but I think made it a bit more complicated than needed to be. To figure out the likelihood of a random person in the population passing on a recessive allele, you just need to do squareroot(q2) to get to q. And just use that q to multiply with the 1/3 you got earlier. The q value tells you the allele frequency in the population, whether the person is a carrier or homozygous recessive both. You still got it, but just letting you know the easier route in case you see it again! +2  



 +0  upvote downvote
submitted by hello(246),
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Tshi eht asme sa 'y@akcmyesnop nelaoptainx htiw reom lenoaxniatp sthoe taht ndee i...t

eTh tetpani si eedfctunfa yb eth uitaaplrcr luoosamat vsrieeecs ssdeeai and ish ortehrb hsa hte tuaasolom vceressei sesieda -;t-g& tihs semna heca fo itreh stnraep riascre het RA l.eaell ow,N orf the iattenp to be eactudnffe ievng ttha btho shi ntpraes avhe eth RA elle,al it mnesa hatt the taietpn sah a /23 chanec thta he si neluuaosyltsmi tefdfnceau dna a rcer.air

The 'niaesttp repatrn si uefcefadtn dna nmralo enbd-iHeryrWga eeisnctg (as sedatt ni het eb.lemnpowhr). iths is icspfceilayl eodm,etnin yuo rae to samesu taht the ntrrape sha a ecirrar qyeuenrfc for hte AR leell,a wihhc leuqa to pq2.

hTe iadeess hsa a fqureneyc ni eth plinoaputo of 1.400/00 Thsi si 2q^ t;--&g ^q2 = 0000/14. oligSvn ofr ,q ouy gte q = 2.10/0

erriCra qucfenyer si 2.qp oHvrwee, orf a aerr essadi,e qp2 ≈ 2q. oS, het ercirar recenfuyq rof eth nrrpeta = q2 = 2 * /0201 = 0/011.

,woN het Q assk auobt na ognffisrp fo hte piatten nda ihs anpertr bengi tfeedfac, os ot ahve an adceefft ilch,d ehert is a 41/ eaccnh of ghavni an eacfftde clidh secsuba(e teh itpaten and eht eptnrra hotb heva a /21 anhecc fo igsspan hte eellal -&;t-g yuo mtelplui ehest totehegr 2/1 * 1./2)

o,S ptiumlyl 2/3 * 0110/ * /41 = 06/0.1 This is ietnpt(Pa bieng a ra)irrce * pe(nraPtr bineg a er)iracr * vgiahP(n na faefcetd rnfpifosg eo.etht)gr




 +0  upvote downvote
submitted by kdckjn(0),

why there is probability of father being a carrier is 2/3

drdoom  We assume this is a recessive disease. In other words, you manifest the disease if you inherit a disease allele, d, from your mother and a disease allele, d, from your father, giving you an allele pairing denoted by dd. dd = you have the disease. The father does not have the disease; this means his genotype can only be 1 of three things: DD, Dd, or dD. That is, DD = D from his dad, D from his mom; Dd = D from his dad, d from his mom; or dD = d from his dad, D from his mom. Notice that in only two of these 3 possibilities does the father (potentially) have a disease allele (d) that he might pass on to his progeny. +  

We assume this is a recessive disease. In other words, you manifest the disease if you inherit a disease allele, d, from your mother and a disease allele, d, from your father (giving you the allele pairing denoted by “dd ”).

dd = you have the disease.

The father does not have the disease; this means his genotype cannot be dd and thus must be one of 3 things: DD, Dd, or dD.

That is,

  • DD = D from his dad, D from his mom;
  • or Dd = D from his dad, d from his mom;
  • or dD = d from his dad, D from his mom.

Notice that in only two of these 3 possibilities (⅔) does the father (potentially) have a disease allele, d, that he might pass on to his progeny.

+/- drdoom(615),