Question

Indicate the expected number of Barr bodies in interphase cells of individuals with Klinefelter syndrome; Turner syndrome; and karyotypes \(47, \mathrm{XYY}, 47, \mathrm{XXX},\) and \(48, \mathrm{XXXX}\)

Short answer

Answer: The expected number of Barr bodies for each condition are: 1. Klinefelter syndrome (XXY): 1 Barr body 2. Turner syndrome (XO): 0 Barr bodies 3. 47, XYY: 0 Barr bodies 4. 47, XXX: 2 Barr bodies 5. 48, XXXX: 3 Barr bodies

Step-by-step solution

Klinefelter syndrome

Klinefelter syndrome (XXY) has two X chromosomes and one Y chromosome. Using our formula, Barr bodies = 2 (number of X chromosomes) - 1 = 1 Barr body.

Turner syndrome

Turner syndrome (XO) has one X chromosome and no Y chromosome. Using our formula, Barr bodies = 1 (number of X chromosomes) - 1 = 0 Barr bodies.

47, XYY

47, XYY has one X chromosome and two Y chromosomes. Using our formula, Barr bodies = 1 (number of X chromosomes) - 1 = 0 Barr bodies.

47, XXX

47, XXX has three X chromosomes and no Y chromosome. Using our formula, Barr bodies = 3 (number of X chromosomes) - 1 = 2 Barr bodies.

48, XXXX

48, XXXX has four X chromosomes and no Y chromosome. Using our formula, Barr bodies = 4 (number of X chromosomes) - 1 = 3 Barr bodies. In conclusion, the expected number of Barr bodies for each condition is: 1. Klinefelter syndrome (XXY): 1 Barr body 2. Turner syndrome (XO): 0 Barr bodies 3. 47, XYY: 0 Barr bodies 4. 47, XXX: 2 Barr bodies 5. 48, XXXX: 3 Barr bodies