The house fly, Musca domestica, has a haploid chromosome number of 6. How many chromatids should be present in a diploid, somatic, metaphase cell?

12

12

24

Explanation:

Musca domestica is the scientific name of an animal popularly known as Housefly. From the question, it is known that this organism has a haploid chromosome number of 6.

In a diploid chromosomes, the number of chromatids that will be present will be twice of that in haploid.

Since haploid (n) = 6, diploid (2n) will be; 2 * 6 = 12

Somatic cells on the other hand also exhibit diploid number of chromosomes, again it means we will have (2n) = 2 * 6 = 12

In metaphase cell, cells do make sure they complete the S'phase of the cell cycle before cellular division. Therefore, the DNA present in the chromatids in the S'phase actively engage in chromatids doubling, as a result, 24 chromatids (i. e 12 * 2 = 24) exists in the metaphase cell.