Answer:
The correct answer is option d.
Explanation:
Based on the given question, the genotype of the male parent is aa, the genotype of the female parent is Aa, and the genotype of the offspring is AAa, showing trisomy characteristics. In the given case, the existence of two A chromosomes shows that the chromosomes non-disjunction at the time of gametogenesis have taken place in the female as she is the only one possessing A allele. Hence, it can be concluded that the disjunction is maternal in characteristic.
The duplication of chromosomes takes place at the time of the S-phase of meiosis. Thus, after going through S-phase, the karyotype of the cell will be AAaa (maternal gametogenesis), this set of the chromosome is termed as homologous pair, and the pair of AA or aa is termed as sister chromatids.
The meiotic disjunction takes place in two phases. First, in meiosis I disjunction, in this separation of homologous chromosomes takes place, that is, the maternal cell AAaa get differentiated into aa and AA daughter cells. Secondly, in meiosis II disjunction, the separation of sister chromatids takes place into two independent chromosomes, that is, the formation of two cells each comprising chromosome A from the AA cell and the formation of two cells each with chromosome a from the aa cell.
Thus, the phenomenon of non-disjunction would have taken place at the time of meiosis II in the case of a female gamete to exhibit two A chromosomes. Hence, the correct answer is option d.