Further information from another source:
Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A.
Answer:
Maternal meiosis II
Explanation:
Jill has the genotype Aa, and Jack has the genotype aa. Jack can only contribute the a, whereas Jill can contribute A or a. For the child to have 2 copies of the A allele and two copies of the a allele, that means the nondisjunction must have happened in the mother.
As for the stage of meiosis, non-disjunction in meiosis I means that homologous chromosomes fail to separate properly. This would mean that the child would inherit Aa from its mother and a from its father. This is not the case.
Non-disjunction in meiosis II means identical sister chromatids fail to separate properly, which means the child would inherit either aa from its mother, or AA from its mother, and a from its father. This could give the genotype AAa. Therefore, nondisjunction must have occurred in maternal meiosis II
