Genetics Question

Discussion in 'Step I' started by BMW M3, May 14, 2008.

  1. BMW M3

    BMW M3 Senior Member
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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.
     
  2. DragonWell

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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:

     
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  3. alpha06

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    I believe the answer is 1/2.
     
  4. DwyaneWade

    DwyaneWade Reiging *** Cynic
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    EDIT: I am wrong.
     
  5. OP
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    BMW M3

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  6. alpha06

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    No monosomies are viable.

    And the only viable trisomies are 21, 18, 13, and i think 8...i saw it in a qbank.
     
  7. InternationlDoc

    InternationlDoc Imported like a Ferrari
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    i dont think anything is viable other than trisomy 21, 13,18
     
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  8. Tiger26

    Tiger26 Senior Member
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    After a rapid review consult, I think there's 1/2 a chance that the child will have it.
     
  9. Nicholaus

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    I did the 2x2 table, and my answer is 1/2.
     
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  10. alpha06

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    The mosaic form of trisomy 8 and 16 are viable.

    Apparently, there is also a viable trisomy 22 that I wasn't aware of.


    Source: wikipedia.
     
  11. OP
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    BMW M3

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    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.
     
  12. MedLover25

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    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?
     
  13. MedLover25

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    you are exactly correct
     
  14. Noroozi

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    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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