Stem cell Math/modelling question. Please help!

Larry Hunter hunter at nlm.nih.gov
Thu Jul 16 22:52:26 EST 1998


Magason at fas.harvard.edu asks:

  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).  ... 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

Just a little algebra, really.  

I think it is conceptually easier if you do it in terms of generations, so
let's define g = t/l.  Now, write down your definitions the population sizes
of stem cells at time t in terms of the previous generation:

 S(g) = p*2*S(g-1) + (1-p)*S(g-1)

That is, with probability p we get twice as many cells and with probability
(1-p) we get the same number of cells.  Add these two quantities to get the
expected (average) number of cells.

A little algebra tells us that 

 S(g) = (1+p)*S(g-1)

Repeated multiplication is exponentiation, so

 S(g) = [(1+p)^g]*Si

If you want it in terms of time, then substitute back for g

 S(t) = [(1+p)^(t/l)]*Si*l

D(t) is a bit harder, but not much.  At each generation, D(g) increases by
(1-p)*S(g-1), so add up those terms to get the total:

            g 
 D(g) = Sum   (1-p)*S(j-1) 
          j=1 

The (1-p) factor is constant, so move it out of the sum
                   g 
 D(g) = (1-p) * Sum  S(j-1)
                 j=1

Substituting the result we obtained above for S(j), we have
                  g
 D(g) = (1-p)* Sum [(1+p)^(g-1)]*Si
                j=1

Move the constant outside the sum
                    g
 D(g) = (1-p)*Si*Sum (1+p)^(g-1)
                  j=1

And solve for the exponential series:

 D(g) = (1-p)*Si*([(1+p)^g]-1/p)


We're done.  A more legible version of the above is

                        g
                   (1+p)  -1
 D(g) = (1-p)*Si* ------------
                        p


Again, if you prefer it in terms of t, substitute for g

                          (t/l)
                     (1+p)      -1
 D(t) = l*(1-p)*Si* ----------------
                            p


So, for example, let p= 0.5, Si = 1, l=1 and t=1

 S(t) = [(1 + 0.5)^(1/1)]*1*1 = 1.5^1 = 1.5
 D(t) = 1*(1 - 0.5)*1*[(1+0.5)^(1/1)-1/0.5] = 0.5*[1.5-1/0.5] = 0.5

That is, on average, if there is a 50-50 chance that a stem cell will divide
proliferatively, after one division, the average number of stem cells in the
population is 1.5 and the average number of differentiated cells is 0.5

After 10 generations (p=0.5, Si=1, l=1, t=10), the numbers are:

 S(t) = 1.5^10 = ~57.7
 D(t) = 0.5*[1.5^10-1/.5] = ~56.7

This makes sense, in that if there is a 50-50 chance of either path at each
generation after 0, the average population growth will be equal, and the
only difference would be in the initial conditions.

Now, suppose that the stem cell proliferation probability is 0.9 (Si=1,l=1)

 S(1) = 1.9
 D(1) = 0.1
 S(10) = ~613.1
 D(10) = ~68.0

Or, with p=0.1, Si=1, l=1

 S(1) = 1.1
 D(1) = 0.9
 S(10) = ~2.6
 D(10) = ~14.3

Remember that we are modeling a random process, and are therefore estimating
an average number of cells in the populations.  It is a fairly simple
(and worthwhile) extension to estimate the variances as well.

I hope this wasn't homework....

Larry

-- 
Lawrence Hunter, PhD.
National Library of Medicine               phone: +1 (301) 496-9303
Bldg. 38A, 9th fl, MS-54                   fax:   +1 (301) 496-0673
Bethesda. MD 20894 USA                     email: hunter at nlm.nih.gov




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