Stem cell Math/modelling question. Please help!
Henrik Seidel
seidel at osiris.RZ-Berlin.MPG.DE
Thu Jul 16 22:52:34 EST 1998
In article <01bd9805$ba7db1a0$6e29f78c at default.fas.harvard.edu>,
"Sean" <megason at fas.harvard.edu> writes:
> Hi all,
> I'm actually a developmental biologist but I thought people that follow
> this newsgroup might be more helpful in answering my question. I'm trying
> to come up with some equations that describe stem cell proliferation.
> Imagine a stem cell can divide in two ways: it can divide to give rise to
> two new stem cells, or it can give rise to one stem cell and one
> differentiated cell (which can never again divide). Lets call the first
> type of division proliferative and the second type differentiative. Let:
> S=number of stem cells
> Si=the initial number of stem cells
> D=number of differentiated cells
> t=time
> l=cell cycle length
> p=probability of proliferative type of division
>
> Also assume the cell cycle of the population of stem cells is asynchronous
> (random phase but equal length) and the initial number of differentiated
> cells is zero.
>
> What I'm trying to figure out are equations giving the number of stem
> cells, S(t), and differentiated cells, D(t), at a given time.
>
> I have S(t) = Si * 2^(tp/l) but I'm not sure this is correct. I don't
> have anything for D(t).
>
> Any help would be greatly appreciated. Thanks.
>
> Sean Megason
> megason at fas.harvard.edu
Hi Sean,
let us assume for a moment that at time zero all cells are in the same state
immediately after cell division (we will work out how to improve that later).
Then we have
/--- those cells divide into two new prolif. cells
VVVVVVVVVVVVVV
/ --- divide into one prol. and one degen. cell
VVVVVVVVVVVVVV
S(1*l) = S(0*l) * p * 2 + S(0*l) * (1-p) = S(0) * (p+1)
S(2*l) = S(1) * (p+1) = S(0) * (p+1)^2
...
S(n*l) = S(0) * (p+1)^n (1)
and
D(0*l) = 0
D(1*l) = S(0*l) * (1-p) = (1-p) * S(0) * (p+1)^0
D(2*l) = D(1*l) + S(1*l) * (1-p) = D(1*l) + (1-p) * S(0) * (p+1)^1
= S(0) * (1-p) * ((1+p)^0 + (1+p)^1)
...
D(n*l) = S(0) * (1-p) * ((1+p)^0 + (1+p)^1 + ... + (1+p)^(n-1))
This is a geometric series -->
(1+p)^n - 1
D(n*l) = S(0) * (1-p) * -----------
(1+p) - 1
D(n*l) = S(0) * (1/p - 1) * ((1+p)^n - 1) (2)
So far so good. Now we have to deal with the case that the cells have
random phases at the beginning. We assign each proliferative cell a
phase that is 0 immediately after its birth by division and gets 1
immediately before the new cell division. It takes the time l to get
from phase 0 to phase 1. Let as assume that at time 0 the phases are
equally distributed, i.e. the number of cells having some particular
phase phi is the same for all phases (or, in more mathematical terms,
the probability density for phase phi is identical for all phases).
Cells having the phase 0 for time t=0 will divide at time t=l. Before
t=l thoses cells are in their zero-st living cycle, and at t=l they
switch to their first living cycle (we start counting with zero). At
that moment their number will increase by a factor of 2*p + (1-p) = p +
1.
On the other hand, cells having, for example, the phase 0.25 at time t=0
will divide at time 0.75*l. That is, those cells are in their zero-st
living cycle for t<0.75*l and switch to their first living cycle at time
t=0.75*l.
I hope you got the idea: at each time t we have to determine how many
cells are in each living cycle, calculate the corresponding S(t) for
each living cycle and sum them up.
Let's be more specific. Say we are interested in some particular time t.
Let further be n=int(t/l) the integer part of t/l and f=frac(t/l) the
fractional part of t/l (i.e. 0 <= f < 1). Then at time t all cells are
either in living cycle n or in living cycle n+1. To illustrate this, let
us investigate some t with 0<t<l. For those times we have n=0, i.e. part
of the cells is in living cycle 0 and part of the cells is in living
cycle 1. So what part? All cells having a starting phase (at time t=0)
that is (1-t/l) <= phi <= 1 will already have switched to living cycle
1. Since the initial phases are equally distributed, the number of those
cells is (1 - (1-t/l))*S(0) = t/l*S(0) where S(0) is the total number
of cells at t=0. The number of cells still being in its zero-st living
cycle is, of course, (1-t/l)*S(0).
In the example above we restricted ourselves to the case 0<t<l. The
common case is:
There are f * S(0) cells in the (n+1)-st living cycle.
There are (1 - f) * S(0) cells in the n-th living cycle.
(remember that n=int(t/l) and f=frac(t/l)).
Now we are already almost done. The number of proliferative cells at time t
is
S(t) = f * S(0) * (1+p)^(n+1) + (1 - f) * S(0) * (1+p)^n
= S(0) * (1+p)^n * (f * (1+p) + 1 - f)
**********************************************************************
* S(t) = S(0) * (1+p)^n * (pf + 1) *
**********************************************************************
This is, by the way, quite well approximated by the function
**********************************************************************
* S(t) = S(0) * (1+p)^(t/l) *
**********************************************************************
so you might prefer the latter equation.
The number of degenerated cells is
D(t) = f * S(0) * (1/p - 1) * ((1+p)^(n+1) - 1) +
(1-f) * S(0) * (1/p - 1) * ((1+p)^n - 1)
D(t) = S(0) * (1/p - 1) * (f*((1+p)^(n+1) - 1) + (1-f) * ((1+p)^n - 1))
D(t) = S(0) * (1/p - 1) * ((1+p)^n * (f*(1+p) + (1-f)) - f - (1-f))
**********************************************************************
* D(t) = S(0) * (1/p - 1) * ((1+p)^n * (pf + 1) - 1) *
**********************************************************************
or, in other words,
**********************************************************************
* D(t) = (1/p - 1) * (S(t) - S(0)) *
**********************************************************************
I wrote this down quite quickly, so you should check if I didn't make a
mistake. Hope this helps.
Regards
--- Henrik
--
Dr. Henrik Seidel, http://www.mpimg-berlin-dahlem.mpg.de/~seidel/
Max Planck Institute for Molecular Genetics
14195 Berlin, GERMANY, Ihnestrasse 73
tel: ++49-30-8413-1613 fax: ++49-30-8413-1384
see my home page for my public PGP key
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