The approximate cell cycle duration of fast-dividing PC3 cells (mammalian cells
ID: 206343 • Letter: T
Question
The approximate cell cycle duration of fast-dividing PC3 cells (mammalian cells derived from bone metastasis of prostate cancer cells) is 24 hours. PC3 cells are widely used to study prostate cancer. These cells are routinely grown in research labs in 10 cm (diameter) circular dishes as a monolayer (single layer) of synchronously dividing cells to study the various aspects of prostate cancer. When healthy, these cells appear spherical under a microscope with an average diameter of 10um 5. If 30% of the surface area of the dish is covered by the cells at t-0 hours, approximately how many cells are there on the plate? 6. What percent of the surface area of the plate would be covered by PC3 cells at t-24 hours? Ignore cell death to solve this problem. 7. How long would it take starting from time t-0 hours, for the cells to cover 100 percent area of the 10cm circular dish? Ignore cell death to solve this problem. At time t-0 hours, these cells are treated with a drug that is being tested for prostate cancer treatment. This drug acts at the G2/M checkpoint in the cell cycle and slows down mitotic entry in these cells and thereby increasing the cell cycle duration from 24 hours to 32 hours. 8. If these cells are counted 40 hours post drug treatment how many cells would you estimate to be present in the 10cm dish? These mammalian cells can be detached and removed from the dish using the enzyme trypsin. The detached cells can be then re-plated (added) to a fresh 10cm dish at desired cell density. This fact is routinely employed in research labs to expand the number of cells and to make them available for different experiments 9. If it is desired that 90% of the dish be covered by PC3 cells 24hours post re-plating, how many PC3 cells should be plated in the 10cm dish at time t-0 hours?Explanation / Answer
Q5.
Surface area of the dish= pi r2
Surface area = 3.14 x 5 x 5 = 78.571 cm2
30% of the surface area= 23.571 cm2
Surface area of each cell= 3.14 x 5 x 5 = 78.571 micro m2
OR 78.571x 10-8 cm2
Therefre, the cells at '0' hours in the 30% Area = 23.571x 108/78.571
= 2.99 x107 cells
Q6. The cell cycle of the cells completes in approximately 24 hours. Therefore, the number of the cells would be doubled. Thus, the total area covered would be 60%.
Q7.
It will take 108 cells to cover the whole surface area of the dish.
The generation it take to form 108 cells = [Log (final count)- log (initial count)] / 0.301
Generation time =[ log (108) – log (2.99 x107)] / 0.301
= (8 – 7.477)/0.301
=0.523/0.301 = 1.737 generations
or 24 x 1.737 hours = 41.69 hours
Q8.
After 40 hours, the total cells would be
Initial number x 2generations
In 40 hours, generations would be 40/24 =1.666
Therefore total cells at t[40]= 2.99 x107 x 21.666
= 9.52 x107 cells
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