Cell Cycle Problem

March 18, 2007 at 7:59 pm #7206

Participant

This is a problem that I have been given for a cellular biology class. I think I can figure it out, if I knew at what stage of the cell cycle the BrdU will first show up. I realize it’s a DNA precursor so I’m thinking it will be visible once the Chromosomes condense, or Prophase. Any help at this point would be awesome (I have a microbiology midterm to study for tonight as well). Hopefully someone can point me in the right direction. Thanks

2.A student attempted to determine the length of each phase of the cell cycle in a civet cell line. The student counted the number of cells in a culture and found that the number doubled every 28 hr. BrdU, a DNA precursor that can be detected by immunofluorescence microscopy, was added to the culture medium for 15 minutes and, then, the cells were transferred to BrdU-free culture medium. Cells were fixed immediately after removing the BrdU and at 0.5 hour intervals. The samples were processed for immunofluorescence using anti-BrdU antibodies. The results were:

i) 43 out of 150 cells fixed immediately after removal BrdU were stained by anti-BrdU.
ii) 8.0 hr after labeling, 0 of 12 mitotic* cells stained with anti-BrdU; at 8.5 hr, 0 of 16 mitotic cells were stained; at 9.0 hr 2 of 16 mitotic cells were stained; at 9.5 hr 16 of 16 mitotic cells were stained.
(mitotic cells are all cells in prophase, prometaphase, metaphase and anaphase)
iii) The first BrdU positive cells in late telophase were seen at 10.5 hr after labeling.

2A. Determine the length of G1, S, G2 and mitosis. How these were determined should be clear from your answer. (2 points, show your calculations)

2B. How would you determine the percentage of cells that were not dividing? (2 points; Max. 2 sentences)

March 18, 2007 at 8:51 pm #70171

LilKim

Participant

here’s a HINT:

brdu is thymidine analog and can be incorporated into the newly synthesized strand of DNA during replication.

another hint:
http://en.wikipedia.org/wiki/Cell_cycle (check out the cell cycle stages on that page!)

Hope this is a good start for you…

– kim

March 18, 2007 at 10:45 pm #70172

RastaTings

Participant

Thanks for replying, I think that helps me get started. What I have so far is because the brdU is going to be used during DNA synthesis it will be S-phase cells that should uptake the brdU (because that’s when DNA replication occurs). Based on that here’s what I’m thinking;

1. 1.5 hours elapses from the time the first labeled mitotic cells are seen to the time the first labelled telophase cells are seen (telophase not considered mitotic in this question) which suggests the M-phase is 1.5 hours long.

2. After 8.5 hrs, 0 of 16 mitotic cells are stained, implying these cells must have been in G2 when labelling occured. Since it took them 8.5 hrs to enter mitosis, G2 must be 8.5 hours long.

3. at 9.5 hrs, 16 of 16 mitotic cells were stained, implying the S-M transition takes 9.5 hours. Since 8.5 were spent in G2, the S-phase must be 1 hour long.

4. 17 hours of the 28 for the cell cycle remains, suggesting G1 is 17 hours.

That’s what I get, but this doesn’t seem to make sense to me. I was under the impression that S-phase and G2 were much closer in length then these results are suggesting. I assume I’ve made a mistake or I am totally misinterpreting the results?

March 18, 2007 at 11:02 pm #70173

RastaTings

Participant

Ok I think my above idea was wrong. So I have this second one.

1. It’s an asynchronous culture, so the proportion of cells initially labelled should correspond to the approximate length of S-phase, so in this case 43/150 x 28 = 8 hours for S-phase

2. The time it takes for the first mitotic cells to show up should represent G2, so in this case 9.5 hours

3. Mitosis, becuase of the time it takes for cells in telophase to show up, is 1.5 hours.

4. Remaining time corresponds to the length of G1, so 9 hours.

I like this answer a lot better, does it make more sense?

March 19, 2007 at 4:45 am #70181

LilKim

Participant

The only thing that I disagree with you on is #2.

based on the question it says that 2 of 16 cells Brdu positive cells appear at 9 hrs (these are the first cells).. whereas all of them are Brdu positive by 9.5 hours. Based on that I would think that:

g1= 9.5
s=8
g2=9
M=1.5

Nevertheless, I could be completely wrong… but that’s just my two cents!

by the way … good luck with everything!
– KIM

March 19, 2007 at 4:57 am #70183

RastaTings

Participant

no you’re right I see my mistake. Thank you so much for all your help, you don’t know how much I appreciate it

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