Meiosis (multiple choice) and probability x-linked questions.

1
In this picture:
http://www.google.gr/imgres?imgurl=http://img.tfd.com/dorland/thumbs/nondisjunction.jpg&imgrefurl=http://medical-dictionary.thefreedictionary.com/Meiotic%2Bnondisjunctions&h=387&w=250&sz=17&tbnid=QAhfw9QCTpOEWM:&tbnh=90&tbnw=58&prev=/search%3Fq%3Dmeiosis%2Bnondisjunction%26tbm%3Disch%26tbo%3Du&zoom=1&q=meiosis+nondisjunction&usg=__DXTbh4ibkW7TDKkUIj3xmW8kYkc=&docid=k12PVVYHv9v01M&hl=en&sa=X&ei=GzvGUPnsN8WztAao8oCwAw&sqi=2&ved=0CCkQ9QEwAQ&dur=1433

A is the normal meiosis
B is a meiosis where a nondisjunction occurred in Anaphase I
C is a meiosis where a nondisjunction occurred in Anaphase II

As you see, if a nondisjunction had occurred in Anaphase I, then you would have two cells with one (or more) extra chromosome(s), and two cells with one (or more) less chromosome(s). So b) is wrong.
If a nondisjunction had occurred in Anaphase II, then you would have two cells with the correct number of chromosomes, a cell with one (or more) extra chromosome(s), and a cell with one (or more) less chromosome(s).

So the correct answer is c.

(here is a picture where all the steps of meiosis are shown: http://en.wikipedia.org/wiki/File:Meiosis_diagram.jpg)

2
A man with hemophilia has a genotype: XaY   (Xa is the pathological allele, and X is the normal allele)

His daughter will receive the Xa allele from her father, and a normal X from her mother.

She marries a normal man, so we have the crossing:
XXa  x  XY

children: 25%XX, 25%XY, 25%XXa, 25%XaY.

The Xa allele is recessive, so only the male child will have hemophilia. (0% of the daughters, 50% of the sons)

The probability of the two sons to have hemophilia is (1/2)*(1/2)=1/4=25%