Question

3. You are working with wild-type cells that have a 16h doubling time. BrdU is a molecule that can be used to measure the amount of DNA inside cells. When added to cell culture, cells will incorporate BrdU into DNA strands during replication as if it were the DNA base thymidine (T).

Your general experiment for using BrdU is:
1. Add BrdU to the media and let the cells continue to grow for 16h
2. Harvest cells and label them with a fluorescent antibody against BrdU
3. Measure fluorescence intensity in cells which is proportional to total DNA content
4. Plot a histogram which shows the number of cells with a certain fluorescence intensity

A. (5pts) As a control, you perform the above experiment on normally growing, wild-type cells, and find the histogram to the right. Please explain why one peak shows roughly twice the fluorescence intensity as the other, what sets cells are represented by each peak, AND why the high-intensity peak is smaller.

PART B:

B. (5pts) You perform the general experiment with the following addition: when you add BrdU, you also add the drug taxol, which stabilizes microtubules. You find the histogram to the right. Please describe how the features of this histogram have changed from Part A, and explain why adding taxol could cause these changes referring to mechanisms we have discussed in class.

PART C:

C. (4pts) You construct a cell line with a temperature sensitive mutant of the G1-cyclin. You perform the general experiment with the following additions: when you add BrdU, you shift to higher temperatures, which deactivates the G1-cyclin. Please draw what you predict will be the resulting histogram after you measure BrdU on the figure to the right.

(Note: The result from the control is printed as a reference, you should draw your response on the same axes.)

Answer 1

Q3

PART A:

One peak showed roughly twice the fluorescence intensity as the one on the left because the cells were in exponential phase. Each replication cycles doubled the number of bacteria and thus the peak on right was twice than that of the preceding one.

The high-intensity peak was smaller because the cells entered the lag phase. During this phase, the rate of the formation of the cells was nearly equal to the death of the cells and eventually, the death rate surpasses the rate of formation of new cells. The availability of the fresh media becomes the limiting factor at this stage.

PART B.

The drug, Taxol would arrest the mitosis of the cells and thus, either the rate of cell division would slow down or would be arrested altogether. As a result, the peak on the right would be more or less similar to the peak on the left.

PART C.

The temperature-sensitive mutant of the G1-cyclin, when subjected to the higher temperatures, loses the activity of the G1-cyclin. The G1-cyclin levels determine the progression of mitosis. For a cell to commit to undergo mitosis G1/S cyclin CDK complex must be formed. In the absence of the G1 cyclin, the cells will not undergo mitosis. As a result, the fluorescence intensity peak on the right would not change.