V 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 other’s expression. In Box 1 the authors postulate the two equations describing the toggle switch: du
α1
=
−u
dt 1 + v β
dv
α2
=
−v
dt 1 + uγ
[V.1] Let’s derive this equation and explore which assumptions were made during the derivation. The fast equilibrium reactions for this problem are: K1
⎯→ P1R2β
P1 + R2β ←⎯
K2
⎯→ P2R1γ
P2 + R1γ ←⎯
K3
⎯→ R1γ
γR1 ←⎯
[V.2] K4
⎯→ R2β
βR 2 ←⎯
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: 32
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[PT ] = [P1T ] = [P1] + [P1R2β ] = [P2T ] = [P2 ] + [P2R1γ ] [V.3] [V.4] Now the rate of synthesis of repressor 1 and 2 can be written as: Rgen1 = k1[PT ]
Rgen2
[P1]
1
k1[PT ]
T
=
=
k
[P
]
1
[P1] + [P1R2β ]
1 + K1[R 2β ] 1 + K1K 4 [R2 ]β
[P2 ]
1
k 2 [PT ]
T
= k 2 [P ]
= k 2 [P ]
=
[P2 ] + [P2R1γ ]
1 + K 2 [R1γ ] 1 + K 2K 3 [R1]γ
T
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
[V.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
[V.6] if we use the following dimensionless concentrations: u = [R1](K 2K 3 )1/γ
[V.7] 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
[V.8] [V.9] v = [R 2 ](K1K 4 )1/β
Equation [V.6] becomes: Finally by defining α1 and α2 as: k1[PT ](K 2K 3 )1/γ
α1 ≡
δ
T
k 2 [P ](K1K 4 )1/β
α2 ≡
δ
33
7.32/7.81J/8.591J Systems Biology – A. van Oudenaarden – MIT– October 2009
we recover the ‘Box’ equation: du
α1
−u
~=
d t 1+ v β
dv
α2
−v
~=
d t 1 + uγ
[V.10] [V.11] In steady state both these equations equal zero: α1
1+ v β
α2
v=
1 + uγ
u=
34
7.32/7.81J/8.591J Systems Biology – A. van Oudenaarden – MIT– October 2009
VI Stability analysis Consider the following two coupled differential equations: x& = f ( x, y )
y& = g ( x, y )
[VI.1] [VI.2] [VI.3] The nullclines are defined as: x& = 0 → f ( xo , y o ) = 0
y& = 0 → g ( xo , y o ) = 0
in order to solve [VI.2] we linearize around the fixed point (xo,yo): ~
x ≡ x − xo
~
y ≡ y − yo
If f(x,y) and g(x,y) are approximated by a first order Taylor expansion, [VI.2] can be written as: ∂f
∂f
x& ≈ ~
x
+~
y
≡ a~
x + b~
y
∂x ( xo , y o )
∂y ( xo , y o )
∂g
y& ≈ ~
x
∂x
∂g
+~
y
≡ c~
x + d~
y
∂y ( xo , y o )
( xo , yo )
[VI.4] [VI.5] [VI.6] or in matrix notation: r
r
X& = AX
⎡a b ⎤
A=⎢
⎥
⎣c d ⎦
r ⎡ x& ⎤
X& = ⎢ ⎥ ⎣ y& ⎦
r ⎡~
x⎤
X = ⎢ ~⎥
⎣y⎦
The matrix A is characterized by its trace and the determinant: τ = trace( A) = a + d
Δ = det( A) = ad − bc
Let’s try to find a solution of the convenient form: 35
7.32/7.81J/8.591J Systems Biology – A. van Oudenaarden – MIT– October 2009
v
v
v
v& = λv = Av [VI.7] This vector is called the eigenvector, λ is the corresponding eigenvalue. [VI.7] can be solved by: ⎡a − λ
det ⎢
⎣ c
b ⎤
= 0 d − λ ⎥⎦
[VI.8] [VI.9] [VI.10] leading to: λ1 =
λ2 =
τ + τ 2 − 4Δ
2
τ − τ − 4Δ
2
2
or Δ = λ1λ2
τ = λ1 + λ2
For a stable fixed point both λ1 and λ2 should be negative. Therefore a stable fixed point is characterized by: Δ>0
τ <0
[VI.11] Now let us use the evaluate the stability of the toggle switch (Chapter V) as an example: α1
−u
1+ vβ
α2
v& = g(u,v) =
−v
1+ uγ
u& = f(u,v) =
[VI.12] [VI.13] The fixed points are: α1
1+ vβ
α2
v=
1 + uγ
u=
The matrix A is given by: 36
7.32/7.81J/8.591J Systems Biology – A. van Oudenaarden – MIT– October 2009
⎡
⎢ −1
A=⎢
γ −1
⎢ − α 2 γu
⎢⎣ (1 + u γ ) 2
− α1βvβ−1 ⎤
(1 + vβ ) 2 ⎥
⎥ −1 ⎥
⎥⎦
[VI.14] The trace of this matrix is always negative so the only requirement for stability is that the determinant of this matrix is larger than zero. Let us focus on the conditions for which the determinant equals zero. This would define the boundary in parameter space that separates the bistable from monostable region. Setting the determinant to zero gives: α1βvβ−1 α 2 γu γ −1
= 1 (1 + vβ ) 2 (1 + u γ ) 2
[VI.15] [VI.16] Using the conditions for the fixed points [VI.13] gives: βγvβ+1u γ +1 = α1α 2 To be able to solve [VI.16] we have to make some assumptions. We assume that we are working with strong promoters and therefore α1 and α2 are large and that expression ratio between the ON and OFF state is large. In the case where u >> v , the fixed points are approximated by: u ≈ α1
α2 α1γ
[VI.17] 1
βγα β2α1−βγ = 1 ⇔ log α1 ≈ log α 2 γ
[VI.18] v≈
Condition [VI.16] now becomes: Consistent with Fig. 2c‐d in the toggle switch paper. The other boundary is obtained by realizing that the system is symmetry. By replacing α1 by α2 and γ by β, the other boundary is found: log α1 ≈ β log α 2 37
[VI.19] 7.32/7.81J/8.591J Systems Biology – A. van Oudenaarden – MIT– October 2009