Archive for the ‘developmental biology’ Category

Determining the length of the S-phase from double-labeled cell counts

Posted 18 Jun 2010 — by maxhaussler
Category cell biology, cell cycle, developmental biology

A colleague of mine was recently grappling with the question of how to determine the length of the S-phase of a number of cells, which attracted my attention since the problem is quite similar to those we have in bioinformatics.

You might know that dyes exist that label cells that are currently in S-phase (Brdu, Idu, radioactive, etc) , but since cells divide all the time and the cells in a dish are usually not synchronized, one can not simply add the label and then wait for the label to go on and off, as the cells have to be killed to make the label visible. One could synchronize all cells to start their cell cycle at the same time and kill some every second to find out when the label comes on, but this is a huge amount of work.

And unnecessary: As it turns out, only two basic observations are required to infer the length of S phase. As an analogy, imagine all traffic lights of a town going red and green at different times. There is a technique that allows you to obtain the duration of the green phase from just two aerial images of the whole town, taken at a fixed interval. It took us some time to understand how you can deduce the duration of an event based on just two observations of the whole population.

We came across this in a paper by Martynoga et al and by following the literature, found out that the idea goes back to at least Wimber and Quastler, from the early 60s. It requires that two different labels are available, let’s call them green and red, as they actually are in Martynoga et al (and yes, the analogy with the traffic lights is false here, as there are no dyes for traffic lights that stick). The first injection of the dye into the cell population will mark all cells that are currently in S-phase green, the second injection will mark all cells that are in S-phase at the end of the experiment red. Then the cells are killed and the staining is performed.

The first injection is marking cells “green only”, the second one red, but the second ones will actually be “red and green”: There should be no cells “red only”, as the time between the two injections is set short enough that no cell can go into a new cell cycle between the two injections. So we only have “green only” and “red and green” cells and can count them.

What is marked “red/green” is the number of cells that are currently, at the timepoint of the 2nd injection, in the S-phase. (Although, in Martynoga et al, the labelling agent is present 30 min, they also explain that the dye needs 30 minutes to get into the cells. So the effect of killing the cells after 30 min means that we are pulse-labelling at one single timepoint and only the cells that are in S-phase exactly 30min before killing them will be marked.)

What is marked “green only” is the number of cells that have left the S-phase between the two injection. (This is marking all cells that exit their S phase during this time, as all cells that exit the cell cycle before it will not be marked at all and all cells that exit the cycle afterwards will be red/green.)

There two basic ideas that just have to be combined to determine the length of S-phase now:

a) At any time, the same number of cells is in S phase. If I have 100 cells, an S-phase of 4 hours and a cell cycle time of 10 hours, there will be always 40 cells in S-phase or more generally, 40% of all cells. Therefore, the Proportion of cells in S-phase = Time of S Phase / Time of cell cycle, or PS = Ts / Tc

b) During any duration of time, cells will constantly leave S-phase, the same every hour. If I have 100 cells and a total cell cycle time of 10 hours, 10 cells will leave the S-phase each hour. After four hours, 40 cells will have left S-phase. Therefore, the Proportion of cells leaving S-phase = Time between injections / Time of cell cycle, or PL = Ti / Tc

Dividing one formula by the other gives PS / PL = Ts / Ti. You can rearrange this formula to solve for the Time of S Phase Ts = PS / PL * Ti

Perhaps this will be helpful for someone one day who stumbles over this with Google.