Here is an interesting but fun question. A female has a Robertsonian translocation between one of her X chromosomes and the 8th chromosome. The translocation wrecks several haplosufficient but essential genes on both chromosomes. What would be the out come of mating with a normal male? (Hints: the translocation X can not be viable by itself and I haf to use a Punnett square!

a. Females: 1 Turner, 1 Triple X, 1 carrier, 1 normal, 2 inviable Males: 1 Klinefelter, 1 normal, 4 inviable.
b. Females: 1 Klinefelter, 1 normal, 4 inviable. Males: 1 Turner, 1 Triple X, 1carrier, 1 normal, 2 inviable.
c. Females: 2 Turner, 1carrier, 2 normal, 1 inviable Males: 1 Klinefelter, 2 normal, 3 inviable.
d. Females: 1 Turner, 1 carrier, 3 normal, 1 inviable Males: 2 Klinefelter, 2 normal, 2 inviable.
e. Females: 4 normal, 2 inviable Males: 4 normal, 2 inviable.
f. Females: 1 Turner, 1 Triple X, 2 carrier, 1 normal, 1 inviable Males: 2 Klinefelter, 3 normal, 1 inviable.
g. Females: 1 Turner, 1 Triple X, 2 carrier, 2 normal, O inviable Males: 1 Klinefelter, 4 normal, 1 inviable.
h. Females: 1 Turner, 1 Triple X, 1 carrier, 2 normal, 1 inviable Males: 1 Klinefelter, 2 normal, 3 inviable.

In Robertsonian translocation we would deduce that , the long arms of two acrocentric chromosomes will be joined to a common centromere via the translocation, generating a metacentric chromosome with two arms that are very short. In order to form the viable gametes , gametes should have all the genes i.e. without duplication and deletion. the correct option is c. Females: 2 Turner, 1carrier, 2 normal, 1 inviable Males: 1 Klinefelter, 2 normal, 3 inviable.