Final answer:
Epithelial tissue from the small intestine of an animal was analyzed, and cells with 8.4 units of DNA were undergoing mitosis, while cells with 4.2 units were in interphase. The cells with 8.4 units were haploid, while the cells with 4.2 units were diploid.
Step-by-step explanation:
Epithelial tissue from the small intestine of an animal was analyzed, and some cells had 8.4 units of DNA while others had 4.2 units. The cells with 8.4 units were undergoing mitosis, which is a phase of the cell cycle where the cell divides into two identical daughter cells. On the other hand, cells with 4.2 units were in interphase, which is a period of cell growth and DNA replication. The cells with 8.4 units of DNA were haploid, meaning they had a single set of chromosomes, while the cells with 4.2 units were diploid, meaning they had two sets of chromosomes.
When analyzing a sample of epithelial tissue from the small intestine, the quantity of DNA found in the cells helps determine the cell cycle stage. Cells with 8.4 units of DNA are likely in the process of mitosis, specifically in the synthesis (S) phase or potentially in G2 phase just before mitosis, where the DNA content is doubled compared to that of a cell in interphase, which might have 4.2 units of DNA. Since the cells with 4.2 units of DNA are in interphase, this amount represents the normal diploid (2n) content for that organism, meaning the cells with 8.4 units cannot be haploid but instead are diploid cells preparing for division.