Suppose that nondisjunction resulted in the production of newindividuals with the following chromosomal abnormalities: XO,XXX, XYY, XXXX, XXXY, XXXXY.Indicate the expected phenotypic sex corresponding to each of these chromosomalcombinations if it occurred (a) in a human (b) in a Drosophila. How many Barrbodies would there be in human cells showing each of these combinations?

(a) In humans:

• XO: This is known as Turner syndrome. The individual would be phenotypically female.
• XXX: This is known as Triple X syndrome. The individual would be phenotypically female.
• XYY: This is known as Jacob's syndrome. The individual would be phenotypically male.
• XXXX: This is a rare chromosomal disorder. The individual would be phenotypically female.
• XXXY: This is a form of Klinefelter syndrome. The individual would be phenotypically male.
• XXXXY: This is a rare form of Klinefelter syndrome. The individual would be phenotypically male.

(b) In Drosophila (fruit flies), sex is determined by the ratio of X chromosomes to sets of autosomes, not the presence of a Y chromosome. So:

• XO: The individual would be male.
• XXX: The individual would be female.
• XYY: The individual would be male.
• XXXX: The individual would be female.
• XXXY: The individual would be female.
• XXXXY: The individual would be female.

Barr bodies are inactive X chromosomes in a cell. They are named after discoverer Murray Barr. Humans typically have one less Barr body than the number of X chromosomes. So:

• XO: 0 Barr bodies
• XXX: 2 Barr bodies
• XYY: 0 Barr bodies
• XXXX: 3 Barr bodies
• XXXY: 2 Barr bodies
• XXXXY: 3 Barr bodies