USMLE Forum Archives - USMLE Step 1 - mi4
mi4
babbu5508 - 05-31-11 23:51
Which of the following genotypes would be most likely present in this patient?
/ A. 45,XO
/ B. 47,XXX
/ C. 47,XXY
/ D. Trisomy 18
/ E. Trisomy 21
babbu5508 - 05-31-11 23:51
Which of the following genotypes would be most likely present in this patient?
/ A. 45,XO
/ B. 47,XXX
/ C. 47,XXY
/ D. Trisomy 18
/ E. Trisomy 21
The correct answer and explanation will be available after you answer.
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#1
Re: mi4 dinmenace - 06-01-11 13:17 My answer is: a.
It is a chromosomal abnormality in which all or part of one of the sex chromosomes is absent (unaffected humans have 46 chromosomes, of which two are sex chromosomes). Typical females have two X chromosomes, but in Turner syndrome, one of those sex chromosomes is missing or has other abnormalities.
#2
Re: mi4
babbu5508 - 06-16-11 01:42 The correct answer is A. Turner's syndrome is a sex chromosome abnormality, which in about half of live birth cases, is due to 45,XO karyotype. The remainder are usually mosaics, with 45,XO/46,XX or 45,XO/47,XXX karyotypes. It is thought that approximately 98% of Turner's conceptions die in utero early in pregnancy.
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