1
Supplementary Information
Cancer stem cells as a proportion of the total tumour mass
The mathematical model described in Johnston et al.23,24, as based on Fig. 1, assumes
that the stem (respectively transit) cells either die, differentiate or renew at rates  1 ,  2
and  3 (respectively  1 ,  2 and  3 ), and that the fully-differentiated cells die and are
shed into the lumen at a rate  . The model equations for stem cells ( N 0 ), transit cells
( N 1 ) and fully-differentiated cells ( N 2 ) are given by
dN 0
k N2
 N 0  0 0 ,
dt
1  m0 N 0

k0 N 0
dN 1
k1 N 12
 N1 
 N 0   2 
dt
1  m1 N1
1  m0 N 0

(1)

,

(2)

dN 2
kN 
 N 2  N 1   2  1 1 ,
dt
1  m1 N 1 

(3)
where   3  1   2 and   3  1  2 .  and  represent the net per-capita
growth rates of stem and transit cells,  2 and  2 represent the linear parts of the
k
differentiation rates of stem and transit cells, and the terms involving 0 and k1
represent, respectively, the nonlinear feedback associated with stem and transit cells.
m0
and m1 represent feedback saturation parameters, and  is the per capita rate of
fully-differentiated cell population removal.
In the exponential growth phase of the tumour, when the cell populations are very
large, N 0 , N1 , N 2  1, so that we can approximate (1)–(3) by
dN 0
   k 0 m0 N 0 ,
dt
dN 1
   k1 m1 N 1   2  k 0 m0 N 0 ,
dt
dN 2
 N 2   2  k1 m1 N 1 .
dt
(4)
(5)
(6)
2
These are linear ordinary differential equations that can be solved exactly to produce
N 0  n 0 e ( - k 0
m0 )t
(7)
N 1  Ae ( -k 0
m0 )t
 (n1  A)e (  -k1
N 2  Be ( -k 0
m0 )t
 Ce (  -k1
m1 ) t
m1 ) t
(8)
 (n 2 - B - C)e -t ,
(9)
where n 0 , n1 and n 2 are the initial cell populations of stem, semi-differentiated and
fully-differentiated cells, respectively, and constants A, B, C are given by
A
( 2  k 0 m 0 )n0
(   k1 m1 ) A
(  k1 m1 )( n1  A)
,B  2
and C  2
.
  k 0 m0    k1 m1
  k 0 m0  
  k1 m1  
(10)
We note that numerical solution of the full system of equations (1)-(3) gives good
agreement with the solutions (7)-(9) of the approximated equations (4)-(6). We now
consider the two different cases of whether the cancer stem cell (CSC) originates from
tissue stem or transit cells.
(  k 0
Firstly, when the CSC originates from the tissue stem cells, e
(   k1
greater than e
m1 ) t
m0 ) t
will be much
and so the limiting ratio as t   is given by
N 0 : N1 : N 2  n0 : A : B , which can be expressed as
N0 : N1 : N2  1 :
 2  k0 m0  2  k1 m1 
 2  k0 m0
.(11)
:
  k0 m0    k1 m1   k0 m0    k1 m1   k0 m0   
In particular, this yields the key ratio

N0
 2  k 0 m0
 2  k1 m1
1
1 

, where P 
N 0  N1  N 2 1  P
  k 0 m0    k1 m1    k 0 m0  

. (12)

Secondly, when a transit cell is the CSC (or driving cell of the tumour), the stem
cell population is in a steady state given by N 0*   k 0  m0 . In this case the transit
and fully-differentiated cell populations grow without bound while the last term in (5),
representing stem cell differentiation, remains constant. The ratio of the populations in
the limit as t   is given by
N 0 : N1 : N 2  0 : 1 :
 2  k1 m1
,
  k1 m1  
(13)
3
and, in particular, the key ratio is given by
N1
1

,
N1  N2 1  P
whereP 
2  k1 m1
.
  k1 m1  
(14)
We note that qualitatively these results are unchanged if linear feedback is used instead
of saturating feedback. For the stem cells, linear feedback corresponds to the case
where m0  0 , and saturating feedback occurs when m0  0 .
CSCs from transit cells compared to fully-differentiated cells
Alternatively, we could consider the ratio of the transit cell population to the fullydifferentiated cells during unbounded growth. Since we are only interested in the ratio
N1 / N 2 with the N 1 population driving the cancer, we assume that the stem cell
population is constant to ease computation. We choose the constant
D   2 N 0*  k 0 ( N 0* ) 2 (1  m0 N 0* ) to represent the stem cell differentiation rate. Using
equations (2) and (3), we get
d  N1

dt  N 2

N
N
D
  (    ) 1 
  2  1
N2 N2

 N2
2
2


k1 N 1
N 
 
1  1 .
N 2 (1  m1 N 1 ) 
N2 

(15)
When N 1 and N 2 are large, corresponding to the exponential growth phase, this
equation can be written approximately as
d  N1

dt  N 2
 
N 
k
k  N
     1    1    2  1  1
m1
m1  N 2
 
 N2 
2

 .

(16)
This can be solved directly to give
N1
  k1 m1  

,
N 2  2  k1 m1  Ce (  k1 m1  )t
(17)
4
where C is a constant. Therefore, as long as     k1 m1 (which is true in the
exponential growth phase in the saturating feedback model), as t   the limiting ratio
is
N1   k1 m1  

.
N2
 2  k1 m1
(18)
Hence, the general ratio can be expressed as
N1 Net transi t cell growth rate  Differenti ated cell removal rate

.
N2
Maximum transit cell differenti ation rate
(19)