A man has a condition where all of his gametes undergo nondisjunction of the sex chromosomes in meiosis I, but meiosis II proceeds normally. He mates with a woman who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Turner syndrome (XO)? Assume all gametes and zygotes are viable. Answer is 1/2. Can you explain why?

Turner syndrome is a monosomy condition - a condition in which the victim lacks a chromosome. The condition is caused by a normal gamete fusing with a gamete that lacks a chromosome as a result of nondisjunction during meiosis.

When the gametes of a man undergo nondisjunction, two types of gametes are produced - OY gametes and XXY gametes. In order to produce a child with Turner syndrome (XO), the OY gamete of the man must have participated with a normal XX female gamete.

OY x XX

XO XO XY XY

Hence, the probability of producing a XO child = 2/4 of 1/2

If the XXY gamete participate in fertilization,

XXY x XX

XXX XXX XY XY

XXX females would be produced instead of XO children.