Question

In a population of 1000 cells, 200 are in various phase of the M stage. For those in interphase, 400 have 1X amount of DNA, 200 have 2X amount of DNA, and the rest have somewhere in between 1X and 2X amount of DNA. What is the total length of the cell cycle if the S phase has a duration of 5 hours?

Answer 1

The answer is 25 hours

Step-by-step explanation:

Solution

Given that:

The total number of cells = 1000 cells

The M -phase = 200 cells

In interphase it becomes:

1 X DNA = 400 cells

2X DNA = 200 cells

In between 1X and 2X = 200 cells

The duration of S phase = 5 hours

Now

The first step to take is to calculate the mitotic index of cells which is given as:

Mitotic Index = The number of cells in mitosis/Total number of cells * 100

Thus

Mitotic Index =Number of cells in prophase + metaphase +anaphase + telophase (P+M+A+T)/1000 * 100

=200/1000 * 100

=20%

Now from the cell cycle 20% is 5 hours, then 100% which is the length of total cell cycle. is about 25 hours.