IV Synthetic genetic switches
Reference:
1. T. S. Gardner, C. R. Cantor, and J. J. Collins. Construction of a genetic toggle switch
in Escherichia coli. Nature 403, 339-342 (2000).
In this paper Gardner et al. engineered a plasmid containing two repressor genes mutually
controlling each others expression. In Box 1 postulate the two equations describing the
toggle switch:
du
α1
=
−u
dt 1+ v β
dv
α2
=
−v
dt 1+ uγ
[IV.1]
Let’s derive this equation and explore which assumptions were made during the
derivation. The fast equilibium reactions for this problem are:
K1
P1 + R2β ←
→ P1R2β
K2
P2 + R1γ ←
→ P2R1γ
[IV.2]
K3
γR1 ←
→ R1γ
K4
βR 2 ←
→ R2β
The first two reactions model the binding of the repressors to the two promoters
inhibiting transcription. We will assume that there is only one binding site for the
repressor (in contrast to the lambda phage promoters). However the repressor monomers
can form multimers. Repressor 1 multimerizes with γ subunits and repressor 2 with β
subunits. Note that K3 and K4 are the effective association constants for multimerization.
The units of K3 and K4 are (M)1-γ and (M)1-β respectively. The last two equations assume
that intermediate states are not allowed. This is the Hill approximation as we discussed
before (see for example [II.22]). Remember that the total number of promoter site is
conserved, and the total concentration of both promoters is identical since they are on the
same plasmid:
[PT ] = [P1T ] = [P1] + [P1R2β ] = [P2T ] = [P2 ] + [P2R1γ ]
21
[IV.3]
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004
Now the rate of synthesis of repressor 1 and 2 can be written as:
Rgen1 = k1[PT ]
Rgen2
1
k1[PT ]
[P1]
T
=
=
k
[P
]
1
1 + K1[R 2β ] 1 + K1K 4 [R2 ]β
[P1] + [P1R2β ]
1
k 2 [PT ]
[P2 ]
T
= k 2 [P T ]
=
=
k
[P
]
2
1 + K 2 [R1γ ] 1 + K 2K 3 [R1]γ
[P2 ] + [P2R1γ ]
[IV.4]
The rates k1 and k2 are the effective synthesis rates (including RNA polymerase binding,
transcription, translation, and folding) of repressor protein 1 and 2, respectively).
Assuming a first order decay process with rate δ, the kinetic equations are:
d[R1]
k1
[PT ]
=
− δ[R1
]
dt
1 + K1K 4 [R2 ]β
d[R 2 ]
k 2
[PT ]
=
− δ[R 2
]
dt
1 + K 2K 3 [R1]γ
[IV.5]
~
When we introduce a dimensionless time t = tδ , equation (50) becomes:
d[R1] 1
k1
[PT ]
=
− [R1
]
~
δ 1 + K1K 4 [R 2 ]β
dt
d[R 2 ] 1
k 2
[PT ]
=
− [R2
]
~
δ 1 + K 2K 3 [R1]γ
dt
[IV.6]
if we use the following dimensionless concentrations:
u = [R1](K 2K 3 )1/γ
[IV.7]
v = [R 2 ](K1K 4 )1/β
Equation [IV.6] becomes:
du 1 k1[PT ](K 2K 3 )1/γ
−u
~=
1+ v β
dt δ
dv 1 k 2
[P T ](K1K 4 )1/β
−v
~=
1+ uγ
dt δ
[IV.8]
Finally by defining α1 and α2 as:
k1[PT ](K 2K 3 )1/γ
α1 ≡
δ
T
k [P ](K1K 4 )1/β
α2 ≡ 2
δ
[IV.9]
we recover the ‘Box’ equation:
22
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004
du
α1
−u
~=
d t 1+ v β
dv
α2
−v
~=
d t 1 + uγ
[IV.10]
In steady state both these equations equal zero:
α1
1+ v β
α2
v=
1+ uγ
u=
23
[IV.11]
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004
V Stability analysis
Consider the following two coupled differential equations:
& = f (x, y )
x
& = g ( x, y )
y
[V.1]
The nullclines are defined as:
& = 0 → f (xo , y o ) = 0
x
& = 0 →
g ( xo , y o ) =
0
y
[V.2]
in order to solve [V.2] we linearize around the fixed point (xo,yo):
~
x ≡ x−x
o
~
y ≡ y − yo
[V.3]
If f(x,y) and g(x,y) are approximated by a first order Taylor expansion, [V.2] can be
written as:
∂f
∂f
≡ a~
x + b~
y
+~
y
x& ≈ ~
x
∂y ( xo , y o )
∂x ( xo , y o )
∂g
y& ≈ ~
x
∂x
∂g
+~
y
≡ c~
x + d~
y
∂
y
( xo , yo )
( xo , yo )
[V.4]
or in matrix notation:
r
r
X& = AX
a b 
A =


c d 
r  x& 
X& =  
 y& 
r
 ~
x
X =  ~
y
[V.5]
The matrix A is characterized by its trace and the determinant:
τ = trace( A) = a + d
[V.6]
∆ = det( A) = ad − bc
Let’s try to find a solution of the convenient form:
v
v
v
v& = λv = Av
24
[V.7]
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004
This vector is called the eigenvector, λ is the corresponding eigenvalue. [V.7] can be
solved by:
a − λ
det 
 c
b 
=0
d − λ 
[V.8]
leading to:
λ1 =
λ2 =
τ + τ 2 − 4∆
2
τ − τ 2 − 4∆
[V.9]
2
or
∆ = λ1λ2
τ = λ1 + λ2
[V.10]
For a stable fixed point both λ1 and λ2 should be negative. Therefore a stable fixed point
is characterized by:
∆>0
τ <0
25
[V.11]
7.81/8.591/9.531 Systems Biology – A. van Oudenaarden – MIT– September 2004