Gene regulatory network

A network describing how gene products affect the production of other gene products. Positive and negative regulatory interactions can generate multistability, oscillations, and noise-sensitive transitions.

// Target: mathematical-biology.bigb

A repressilator is a cyclic negative-feedback gene regulatory network in which three repressors inhibit one another in sequence. In a normalized logic approximation for gene regulation, $x^{\prime }=H(h-z)-x$, $y^{\prime }=H(h-x)-y$, $z^{\prime }=H(h-y)-z$. For $0<h<1$, threshold crossings successively switch production and can generate sustained oscillations. A rise from $v_0<h$ reaches threshold after $\ln [(1-v_0)/(1-h)]$; a decay from $v_0>h$ reaches it after $\ln (v_0/h)$ in units of the common decay time.

A gene product reduces its own production. This negative feedback can stabilize its abundance and shorten the approach to a regulated level. In the strong logic approximation for gene regulation, production switches off near a self-repression threshold; a smooth repression law avoids artificial infinitely rapid switching at that threshold.

At a Hill equation equilibrium with $s=(Y/K{)}^n$, differentiating $\beta =\alpha Y(1+s)$ gives the displayed logarithmic sensitivity. It approaches one without effective feedback and $1/(n+1)$ under strong repression. Threshold pinning in the logic approximation for gene regulation instead gives approximately zero production sensitivity while the threshold is held fixed.

Neglecting the one in a strongly repressed Hill equation gives $\dot{Y}=\beta K^n{Y}^{-n}-\alpha Y$. The substitution $z=Y^{n+1}$ makes this a linear ordinary differential equation. It yields $Y_{\mathrm{st}}=(\beta K^n/\alpha {)}^{1/(n+1)}$ and half-response time $-\log (1-2^{-(n+1)})/[(n+1)\alpha ]$. The singular initial slope is an artifact of the approximation, whose range is not uniform at arbitrarily large $n$ and fixed production/degradation ratio.

A threshold approximation in which a regulatory input permits or blocks production according to whether a transcription factor exceeds a specified concentration. With first-order removal, $\dot{y}=\beta g-\alpha y$ and $g\in \{0,1\}$. A zero-initialized rising response reaches $K<\beta /\alpha$ at $-{\alpha }^{-1}\ln (1-\alpha K/\beta )$. Boolean rules must specify how several inputs combine; equal rate constants do not determine thresholds.

Suppose an input is produced for time $T$, then decays, and activates two identical mutually activating OR-gated outputs. Write $S=\beta /\alpha$ and $h=K/S\in (0,1)$. The input remains above threshold until ${\alpha }^{-1}\ln [(e^{\alpha T}-1)/h]$. Each output reaches threshold at $-2\ln (1-h)/\alpha$ if input remains available. Equating these times gives the limiting pulse duration $T_{\min }={\alpha }^{-1}\ln [1+h/(1-h{)}^2]$. Exact equality depends on the ideal gate's boundary convention; a robust latch requires a longer pulse. Treating cessation of production as immediate removal incorrectly discards the input's residual lifetime.

A feed-forward loop has regulatory paths $X\to Z$ and $X\to Y\to Z$. The two inputs are combined at the target gene using a specified regulatory rule.

In an incoherent feed-forward loop, the direct path from an input to an output and the indirect path through a second transcription factor have opposite net signs. Their different response times can generate a pulse or adaptation. A nonzero basal production rate makes the final adapted output nonzero, so its logic and response time must be calculated from the actual gene regulation rule rather than assumed from a Boolean label.

In the I4 incoherent feed-forward loop, an input activates the output but represses an intermediate activator of that output. Decay of the intermediate can interrupt an initial output pulse. Residual production and signal dependence must be retained when specifying the gene regulation rule.

In the I1 incoherent feed-forward loop, an input activates both an intermediate regulator and the output, while the intermediate represses the output. Delayed accumulation of the intermediate can therefore interrupt an initial output pulse. The precise steady state depends on the actual gene regulation rule and thresholds.

Let an output start at zero, degrade at rate $\alpha$, and switch from production $F\beta$ to $\beta$ at time $d$, with $F>1$. Its final level is $Z_s=\beta /\alpha$. Writing $\delta =1-e^{-\alpha d}$, its first crossing of $Z_s/2$ occurs at $t_{1/2}={\alpha }^{-1}\ln [2F/(2F-1)]$ if $F\delta \ge 1/2$; otherwise it occurs at $d+{\alpha }^{-1}\ln [2(1-F\delta )]$. This follows by solving the two linear differential equations before and after the switch. A large F alone does not make the first formula uniform in arbitrarily short pulse durations.

The direct path and the indirect path of a coherent feed-forward loop have matching overall signs. The all-activating case with an AND rule can delay activation while allowing rapid removal of the production signal when the direct input disappears.

The all-activating feed-forward loop combines direct activation of an output with indirect activation through an intermediate. Under AND logic the intermediate's accumulation delays production and can reject brief inputs. Removal of direct activation stops production immediately, while the output concentration still has its own decay time.

A persistence detector responds strongly to a sustained input and weakly to short isolated inputs. In an activating coherent feed-forward loop, an intermediate protein accumulates toward a threshold while the target also requires the direct input. Eliminating that intermediate by an instantaneous quasi-steady-state approximation removes its accumulation delay.

The interacting regulatory system built from pair-rule genes. Its interactions can change with developmental time rather than forming one static circuit.

An oscillatory developmental regulatory system associated with repeated segment formation. Oscillation, intercellular synchronisation and fixation of boundaries are distinct processes.

Two components mutually repress when each inhibits the other's production. With decreasing $f,g$, the two negative interactions create a positive feedback loop. Sufficiently strong repression can produce two stable states: high $x$, low $y$, and the reverse.

// Target: mathematical-biology.bigb

For $\dot{Y}=-Y+\gamma /(1+Z^2)$ and $\dot{Z}=-Z+\gamma /(1+Y^2)$, subtraction of the equilibrium conditions gives $(Y-Z)(1-YZ)=0$. The symmetric state solves $x^3+x=\gamma$. Two asymmetric positive states exist precisely for $\gamma >2$: $Y_{\pm }=(\gamma \pm \sqrt{{\gamma }^2-4})/2$, $Z_{\pm }=Y_{\mp }$. Their Jacobian matrix has eigenvalues $-1\pm 2/\gamma$, both negative; the symmetric state is then a saddle point. For $0<\gamma <2$ only the attracting symmetric state remains.

For $\dot{x}=\lambda /(1+y^m)-x$, $\dot{y}=\lambda /(1+x^n)-y$, the equilibrium slope product is $K=mn{x}^n{y}^m/[(1+x^n)(1+y^m)]$. A value $K>1$ gives a saddle separating attracting alternatives. Both exponents greater than one allow this at sufficiently large common $\lambda$. If both are at most one it is impossible. If $0<m\le 1<n$, define $Y(x)$ as the unique positive solution of $Y^{-1}+Y^{m-1}=x^{-1}+x^{n-1}$ and set $K_{m,n}(x)=mn{x}^{n}Y(x{)}^m/[(1+x^n)(1+Y(x{)}^m)]$. All equilibria are parameterized by $\lambda =x(1+Y^m)$, whose derivative is $(1+Y^m)(1-K)/(1-m{Y}^m/(1+Y^m))$. Thus multistability for some common strength occurs exactly when ${\max }_{x>0}{K}_{m,n}(x)>1$. For $m=1$ the maximum is $(n-1{)}^2[(n+1)/(n-1){]}^{(n+1)/n}/(4n)$. Consequently $mn>1$ alone is not sufficient when the two synthesis strengths are required to coincide.

For positive $m,n$, this mutual repression system can display bistability for some $\lambda$ exactly when both $m>1$ and $n>1$. If either exponent is at most one, it has a unique stable equilibrium for every $\lambda$. To prove the less obvious case $m\le 1<n$, put $a=x/(1+x)$ and $b=y/(1+y)$ at equilibrium. Equality of the two expressions for $\lambda$ gives $a\le b(1-b{)}^{n-1}$. The loop gain is then bounded by $mn{b}^2(1-b{)}^{n-1}$, whose maximum is $4mn(n-1{)}^{n-1}/(n+1{)}^{n+1}<1$. Swap the two components for the other case. If both exponents exceed one, sufficiently large $\lambda$ yields two stable equilibria with $(x,y)\sim (\lambda ,{\lambda }^{1-m})$ and $(x,y)\sim ({\lambda }^{1-n},\lambda )$. The weaker condition $mn\le 1$ is sufficient to exclude multistability, but does not exhaust all exclusions when the same production prefactor is imposed.

// Target: mathematical-biology.bigb

At an equilibrium $x_{\ast }=f(y_{\ast })$, $y_{\ast }=g(x_{\ast })$, the Jacobian matrix has eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its product of repression gains is also $(\partial \log f/\partial \log y)(\partial \log g/\partial \log x)$. Product below one gives a stable node, above one a saddle, and equality the threshold for a zero eigenvalue. Robust multiple stable equilibria require an intervening unstable equilibrium; a gain exceeding one alone is not a sufficient global criterion for a particular pair of repression functions.

// Target: mathematical-biology.bigb