And unnecesary: As it turns out, only two observations are required. 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 is requires that two different labels are available, let's call them green and red, as they actually are in Martynoga et al (and no, you cannot really do this with traffic lights, as their colors don't 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: Proportion of cells in S-phase = Time of S Phase / Time of cell cycle
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.
Proportion of cells leaving S-phase = Time between injections / Time of cell cycle
PL = Ti / Tc
Dividing one formula by the other gives PS / PL = Ts / Ti. You can resolve this to Ts = PS / PL * Ti
Perhaps this will be helpful for someone one day who stumbles over this with Google.
