Genetics Question

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BMW M3

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My friend passed this question on to me. Can anyone see this as a valid question for step 1?

A 34 year old woman (P:1, G:1) has a 46XY down syndrome child. Genetic analysis revealed she has a t(14q21q) translocation. Assuming her partner is genetically normal, what are the odds that her next child will be phenotypically normal?

A) 1/6
B) 1/4
C) 1/3
D) 1/2
E) 2/3















I'm thinking (D) 2/3 with the assumption that monosomy 21 and monosomy 14 will terminate as abortions.

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Dunno how great this website is, but they say the recurrence risk for DS with a balanced Robertsonian translocation is ~12% if the mom's the carrier. 88% isn't even close to any of the possible answer choices though...:confused:

A carrier parent can have a chromosomally normal child, or a child who is a balanced carrier like the parent, or a DS child. Curiously, the chance that such a couple will have another DS child depends on which parent is the carrier. When the mother is a balanced carrier of a t(14;21), there is about a 12% risk for another DS child to be born in each subsequent pregnancy. When the father is the carrier, the observed risk drops to about 3% for DS.
 
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Ah... i came across this website:

it is 1/2 chance for a normal child with a t(13q21q)

7trobEng.GIF

source: http://atlasgeneticsoncology.org/Educ/PolyMecaEng.html




In terms of the t(14q21q), from the question above, the nih website gives this explanation:
ch2f23.gif

source:
http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=hmg.figgrp.219

I guess it all comes down to what the viability of the trisomy/monosomy is. Does anyone know if monosomy 14, monosomy 21 or trisomy 14 are viable?
 
Correct me if I am wrong, but I don't know if the 2x2 table works for this because of independant assortment of chromosomes. For example, if you get the robertsonian 14, you only have 50% chance of getting the maternal 21.

I also don't see how the answer can be 1/2 if you have 3 ways of getting a viable child:
1) Normal
2) Robertsonian 14 (balanced carrier)
3) Robertsonian Downs

therefore, with 3 viable options and 3 non viable options (as per the NIH picture above) I think the answer is: there will be a 2/3 chance that the future child will be phenotypically normal.
 
My friend passed this question on to me. Can anyone see this as a valid question for step 1?

A 34 year old woman (P:1, G:1) has a 46XY down syndrome child. Genetic analysis revealed she has a t(14q21q) translocation. Assuming her partner is genetically normal, what are the odds that her next child will be phenotypically normal?

A) 1/6
B) 1/4
C) 1/3
D) 1/2
E) 2/3

I'm thinking (D) 2/3 with the assumption that monosomy 21 and monosomy 14 will terminate as abortions.

If the parent has a balanced robertsonian translocation -> the theoretical recurrence risk is 1 in 3. Odds her child would be normal is 2/3.

and yes, I could see this as a valid question for step 1. why not?
 
Correct me if I am wrong, but I don't know if the 2x2 table works for this because of independant assortment of chromosomes. For example, if you get the robertsonian 14, you only have 50% chance of getting the maternal 21.

I also don't see how the answer can be 1/2 if you have 3 ways of getting a viable child:
1) Normal
2) Robertsonian 14 (balanced carrier)
3) Robertsonian Downs

therefore, with 3 viable options and 3 non viable options (as per the NIH picture above) I think the answer is: there will be a 2/3 chance that the future child will be phenotypically normal.

you are exactly correct
 
Same Q in BRS path (genetics chapter, p 62, number 2). The answer is 1/3 that child will have disease (answer from BRS: theoretically a person who carriers a robertsonian translocation with chromosome 21 and a second acrocentric chromosome has a 1/3 chance of having a child with trisomy 21...)
 
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