Question

Does anyone have the answers to this lab, PLEASE HELP

Answer 1

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Answer 2

The image contains two data tables related to an AP Biology assignment about mitotic division in normal and cancerous chicken stomach cells. The task is to calculate the time spent in each phase of mitosis as a percentage of the total cell cycle.

Let's perform the calculations for both Data Table 2 (Mitosis in Normal Chicken Stomach Cells) and Data Table 3 (Mitosis in Cancerous Chicken Stomach Cells).

The percentage of the cell cycle spent in each phase is calculated by dividing the time spent in each phase by the total time for the cell cycle and then multiplying by 100.

For Data Table 2, the total time for one cell cycle is given as 625 minutes.

For Data Table 3, the total time for one cell cycle is 448 minutes.

I will calculate the percentages for both tables. Let's start with Data Table 2:

1. Interphase:
`\( (440)/(625) * 100 \)`

2. Prophase:
`\( (40)/(625) * 100 \)`

3. Metaphase:
`\( (8)/(625) * 100 \)`

4. Anaphase:
`\( (2)/(625) * 100 \)`

5. Telophase:
`\( (10)/(625) * 100 \)`

And then proceed with Data Table 3:

1. Interphase:
`\( (424)/(448) * 100 \)`

2. Prophase:
`\( (50)/(448) * 100 \)`

3. Metaphase:
`\( (12)/(448) * 100 \)`

4. Anaphase:
`\( (3)/(448) * 100 \)`

5. Telophase:
`\( (11)/(448) * 100 \)`

Let's perform these calculations now.

Here are the detailed calculations for each phase of mitosis as a percentage of the total cell cycle:

For Normal Chicken Stomach Cells (Data Table 2):

- Interphase: 70.4%

- Prophase: 6.4%

- Metaphase: 1.28%

- Anaphase: 0.32%

- Telophase: 1.6%

For Cancerous Chicken Stomach Cells (Data Table 3):

- Interphase: 94.64%

- Prophase: 11.16%

- Metaphase: 2.68%

- Anaphase: 0.67%

- Telophase: 2.46%