Cytogenetics question 2 - waliid

[ArchivalUser]

A baby is born with the clinical characteristics of trisomy 21. C-banding is used to examine the chromosomes of the baby. The father has a C-banding polymorphism that distinguishes both his chromosome 21's from each other, and the mother's chromosome 21s as well. The mother's chromosome 21s cannot be distinguished by this banding technique. When the baby's chromosome 21s are examined in metaphase spreads three uniquely banded chromosomes are identified. This data indicates that the non-disjunction event occurred most likely in which ONE of the following?

A. Paternal meiosis I
B. Maternal meiosis I
C. Meiosis II of either parent
D. During mitosis during early embryogenesis
E. Cannot be determined from the data given

[ArchivalUser]

i think e?

[ArchivalUser]

paternal meiosis 2

[ArchivalUser]

amarah, there is no such option in the answer

A. Paternal meiosis I

[ArchivalUser]

A. is the answer

[ArchivalUser]

wallid/saonew can u pl explain??

[ArchivalUser]

The only way that the child could have three chromosomes 21s which are all distinguishable by C-banding is to have inherited two from the father, and one from the mother (remember, mother's chromosome 21s cannot be distinguished by this banding technique). Thus non-disjunction must have occured paternal line. Inorder for the father
to contain two different chromosomes 21s the non-disjunction event would have to occure meiosis I. If it had occured in meiosis II, then themost likelly the father would have donated two identical chromosomes 21s which would not distigushable by the C-banding.