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nbme21/Block 1/Question#47 (60.1 difficulty score)
A 26-year-old woman (III-2) comes to the ...
50% in females but near 0 in malesπŸ”
tags: genetics pedigree 

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submitted by drmantistoboggan4(17),
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tI dais ti asw ftala to esmal in t,reuo and eht onueqits dakes abotu live bonr g.niprsfof ienSc the maels tare’n einbg brno in eth tifrs ,cpeal I sida 0%5 sefamel nda 0% ealms.

hungrybox  fuck i got baited +31  
jcrll  "live-born offspring" ← baited +23  
sympathetikey  Same :/ +  
arkmoses  smh +  
niboonsh  why is it 50% females tho? +2  
imgdoc  felt like an idiot after i figured out why i got this wrong. +2  
temmy  oh shit! +  
suckitnbme  This isn't exactly right as males can still be born as evidenced by individuals III 6,9,11. This basically an x-linked recessive disease. A carrier mother can still pass her normal X chromosome to a son (50% chance). It's just that the other 50% chance of passing an affected X chromosome results in death of the fetus in utero. Thus all males actually born will not be affected. +2  
makinallkindzofgainz  @suckitnbme, Correct, but if you're a live-born male, you 100% for sure do NOT have the disease, so the chance of a live-born male "being affected" is 0. +3  
spow  @suckitnbme it's not X-linked recessive, otherwise every single son would be affected and therefore have died in utero. It's X-linked dominant +3  
qball  Jail-baited +  
srmtn  correct @spow affected females= X linked Dominant +  

submitted by nwinkelmann(293),
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edigPere = XDL (not lla sogeeninrta tfedaefc = kXdl,i-ne ftasfec eslma dan eamslfe sliarimly = omn.dn)ati

ctfedfAe rheasft = %001 iisnsontsram to sa,tehgurd 0% tmsosainsirn ot
dteceA ff mrhsoet = 5%0 to ut,dhasreg %05 rotmissainsn to oshsB. not rteapns dfaftece = %100 ssrotimniasn ot tsuerdhga e(ud to rtfs'ahe X ro,omose)cmh %05 nmtsoisisnar to osns e(du to h'mtreos X eo) chhmBtsoomro. ptseran ftafdcee hcae tmnnirtiatsg obht to hrduegat hougymzso(o aue)hdrtg = 0%5 nad oemr e.resve

If oniitcdno is iflrmoynu ltaaf ot lasme ni ueo,rt etnh the 50% feedcaft besda no iimntrasosn (as boev)a lwil die in t,oure dan teh %05 ton teeaffdc lliw vhea eliv t.irhsb shTi me,san skri fo lefeam neigb aecdffet = %50 dan kirs fo ieovn-lrb smlea gbnie ecfefdta = 0.%

divakhan  I believe its not XLD, had it been there would still be 25% chance of males to be normal according to the Punnett square. (FA 2020 Page 59. +  
divakhan  ^ 50% chance of being normal.. which is NOT in the answer, it said 0% chance in males! +  
divakhan  Plus you'd see disease in EVERY generation if it was Dominant. For it to be XLD, the father I,1 would have transmitted it to his daughter. Instead the parents are carrier in Gen I. so its XLR disease (II, 3 being a homozygous XLR female) +  
plzhelp123  This is X-linked dominant (think Rett syndrome), if it were x-linked recessive, a female would need to have 2 affected x chromosomes to be symptomatic, which would mean her father would HAVE to be symptomatic as he has only one X chromosome, which is impossible as he would have perished in-utero. The reason the answer is 0 in males is because the question asks specifically about live-born males. If it had asked what is the risk of the fetus being affected then it would be 50% in males (but that 50% would not survive to birth) +2  

submitted by ragacha(14),
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submitted by tsp(0),
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Is the mwano in eht itfsr oanrtginee )(21 ton facfteed bacusee of ecneiomplt a)(epte?nnc12er ahs ot evha the ttair rfo teh sdeiase to olevdpe in eht rutfue e.retianogn

b1ackcoffee  disease developed due to germline mosaicism (mutation in of the oocyte) +  

submitted by an_improved_me(13),

Honest question: Why does it matter what pattern of inheritance it has? For all we know, its multi-factorial/can't be determined.

I feel like the easiest way to answer it is: acknowledging it is "uniformly fatal to males in-utero" = 0% chance for males; and legit just counting the fraction of females in generation III that have the disease = 4/6 ~50%.

submitted by divakhan(14),
This is not XLD, its XLR disease (FA 2020 page 59)

Here, the mother is homozygous for the mutation therefore she displays the disease.

We see that,

  1. It skips generations
  2. Males are more severely affected (die in utero)
  3. Females are affected (only when homozygous)

Here we see, that mother can pass defective X chromosome to 50% females & other 50% will be carriers (healthy X chromosome from father, defective from mother) For males, Mother passes defective X chromosome so they die (only 1 X chromosome)

For the males who are living (III 6,11 shown in pedigree), they have been lucky to survive due to mosaicism! ;)

b1ackcoffee  don't you think you are going through too many hoops (assumptions)? male survive due to mosaicism? Better chance is disease started due to germline mosaicism and it IS XLD. btw the disease is incontinentia pigmenti (not necessary to know). your second last para doesn't make sense, read again with fresh mind and perspective. +1