Final answer:
In the case of a female Myrmecia pilosula, a Metaphase I cell, which is after DNA replication, would contain 32 picograms of DNA since each chromosome has replicated into two sister chromatids. The correct answer to the question is 2) 32 picograms.
Step-by-step explanation:
The subject of this question is Biology, specifically cell biology and genetics. When considering the amount of DNA in a Metaphase I cell of a female Myrmecia pilosula, it is important to understand the concepts of diploid and haploid cells.
A G1 somatic cell is in the first phase of the interphase portion of the cell cycle, where it has a diploid number of chromosomes, each with one chromatid. As the cell enters the metaphase of meiosis I, each chromosome has been replicated, so the cell still has the diploid number of chromosomes but now each chromosome consists of two chromatids.
In Myrmecia pilosula, a G1 somatic cell nucleus contains 16 picograms of DNA. By the time the cell has progressed to Metaphase I, the chromosomes have undergone replication during the S phase of the cell cycle. Therefore, there would be twice the amount of DNA because each chromosome is replicated into two sister chromatids still joined at the centromere. Hence, a Metaphase I cell will contain 32 picograms of DNA. Therefore, the correct option is 2) 32 picograms.