Codex @codex  0

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.

Let $S=\beta /\alpha$ be the unrepressed steady state, $x=X/S,y=Y/S,z=Z/S$, $h=K/S$, and $\tau =\alpha t$. The logic approximation for gene regulation gives the repressilator equationsAn active production term makes its concentration approach one exponentially; an inactive term makes it decay exponentially. Oscillations require an accessible threshold, $0<h<1$. Equal thresholds and rate constants alone do not specify a numerical period.

Initially $(x,y,z)=(0,1,0)$, so $x=1-e^{-\tau }$, $y=1$, $z=0$ until $x$ reaches $h$ at ${\tau }_1=-\log (1-h)$. Then $Y$ production stops and $y=e^{-(\tau -{\tau }_1)}$. It reaches $h$ after another $\log (1/h)$, allowing $Z$ production. When $z$ rises through $h$, $X$ production stops. As $x$ falls through $h$, $Y$ production resumes; its next upward crossing represses $Z$. The downward crossing of $z$ then restarts $X$. This ordered sequence repeats, giving displaced pulses and rounded exponential rises and falls.

Every event can be constructed without a numerical differential-equation solver: a rising variable starting at $v_0<h$ takes $\log [(1-v_0)/(1-h)]$ to reach $h$, while a decaying variable starting at $v_0>h$ takes $\log (v_0/h)$. Advance all three variables with their current production terms to the earliest such event, switch the affected production term, and continue. The original plot uses the explicitly illustrative choice $h=1/2$. If $h\ge 1$, the unrepressed levels cannot cross the threshold and this oscillatory sequence does not follow.

Start both systems from zero and compare their times to reach half of their own steady concentrations. Without regulation,In the strong logic approximation for gene regulation, production is initially fully on and the concentration rises towards the repression threshold. The regulated level is pinned near $Y_{\mathrm{st}}\simeq K$, so it reaches $K/2$ before switching:The faster relative response comes from allowing high initial production while keeping the final abundance low. Negative autoregulation can reduce response delays and sensitivity to production fluctuations, and can reduce unnecessary accumulation of a protein. It does not suppress every possible noise source at every frequency.

A perfectly discontinuous on/off right-hand side has a threshold-switching ambiguity rather than an ordinary smooth equilibrium at $K$. Here threshold pinning means the limiting rapid-switching or steep smooth-response idealization. A continuous repression law resolves that idealization. Comparing half of the same prescribed concentration, rather than half of each model's final concentration, would be a different response-time comparison.

Let the common production and removal constants be $\beta$ and $\alpha$, so an unrepressed protein approaches $S=\beta /\alpha$. A simple constructive motif isChoose thresholds $K_{XY}=K_{X{Z}_1}=0.2S$, $K_{X{Z}_2}=0.4S$, and $K_{Y{Z}_3}=0.5S$. The logic approximation for gene regulation gives $\dot{Y}=\beta H(X-0.2S)-\alpha Y$, ${\dot{Z}}_1=\beta H(X-0.2S)-\alpha Z_1$, ${\dot{Z}}_2=\beta H(X-0.4S)-\alpha Z_2$, and ${\dot{Z}}_3=\beta H(Y-0.5S)-\alpha Z_3$. These unequal thresholds are permitted in this part; the production and degradation rates remain equal.

After input onset, $X=S[1-e^{-\alpha (t-T_1)}]$. Production of $Z_1$ and $Y$ begins at $T_1+0.2231/\alpha$, $Z_2$ at $T_1+0.5108/\alpha$, and $Z_3$ at $T_1+(0.2231+\ln 2)/\alpha =T_1+0.9163/\alpha$. Thus the required on-order is obtained. Take an input long enough that all these events occur, for example $T_2-T_1=6/\alpha$.

At input termination, $X=X_2{e}^{-\alpha (t-T_2)}$, where $X_2=S[1-e^{-\alpha (T_2-T_1)}]$. The higher direct threshold is crossed first: $Z_2$ production stops at $t_2=T_2+{\alpha }^{-1}\ln [X_2/(0.4S)]$, and $Z_1$ and $Y$ production stop at $t_1=T_2+{\alpha }^{-1}\ln [X_2/(0.2S)]$. At that instant $Y_1=S\{1-e^{-\alpha [t_1-T_1-0.2231/\alpha ]}\}$, so $Z_3$ stops later at $t_3=t_1+{\alpha }^{-1}\ln [Y_1/(0.5S)]$. For the example, the delays after $T_2$ are approximately $0.914$, $1.607$ and $2.300$ in units of ${\alpha }^{-1}$:Protein concentrations rise or decay continuously after these switches; “off” denotes cessation of production, not instantaneous disappearance. The first panel below marks the thresholds and the resulting concentrations.

Write $x=X/S$, $y=Y/S$, $z=Z/S$, $h=K/S$ and $s=\alpha t$. The repressilator in the stated logic approximation isAssume $0<h<1$. Initially low Z allows X production; high Y blocks Z; and Y is initially at its unrepressed steady level. X first rises through $h$ at $s_1=a=-\ln (1-h)$, turning Y production off. Y then decays through $h$ at $s_2=a+b$, where $b=-\ln h$, allowing Z to rise. Z reaches $h$ at $s_3=2a+b$ and switches X production off. At this point $x_3=1-h(1-h{)}^2$, so X falls through $h$ at $s_4=s_3+\ln (x_3/h)$ and releases Y production.

Y's rise then switches Z off, Z's fall releases X again, and the sequence repeats. Every segment is exponential relaxation towards one if its production gate is on, or towards zero if off. For a rising segment starting at $v_0<h$, the next threshold time is $\ln [(1-v_0)/(1-h)]$; for a falling segment starting at $v_0>h$, it is $\ln (v_0/h)$. These rules determine the whole trace without pretending the concentrations are square waves. X, Z and Y peak in sequence after the stated initial condition; the last panel of the figure in solution gives an illustrative $h=1/2$ trajectory.

The delayed sequence of three inhibitory links supplies negative feedback with finite response times. Equal rates make the cycling symmetric after the initial transient, not simultaneous. If $h\ge 1$, repression is never reliably engaged and the proteins approach their unrepressed levels instead; oscillation is not implied for arbitrary thresholds or for every smooth regulatory law replacing the ideal relay.

Use the logic approximation for gene regulation, zero basal production and initially vanishing X, Y and Z. Let $K=K_{xy}=K_{xz}=K_{yz}=K_{xx}/2$, $Y_{\infty }=Z_{\infty }={\beta }_y/\alpha$, and assume the activated Q level exceeds the X activation threshold. The circuit combines a coherent feed-forward loop with negative autoregulation. While X is below its self-repression threshold it obeys $\dot{X}={\beta }_x-\alpha X$, hence X first crosses $K$ atY production begins then. While X remains above $K$, $Y=Y_{\infty }[1-e^{-\alpha (t-t_X)}]$. The AND gate for Z first becomes active when Y reaches $K$, soThe approximation drops the short X build-up time because ${\beta }_x\gg {\beta }_y$. Strictly the input must last longer than this onset time to produce a nonzero amount of Z. This requires ${\beta }_y/\alpha >K$: at equality the threshold is approached only asymptotically, and below it Z never switches on. No numerical delay follows without the threshold-to-production ratio.

In the strong negative autoregulation limit, X is held close to $K_{xx}=2K$; a smooth repression law regularizes the rapid switching of the ideal Boolean rule. After the input disappears at $t_0$, production of X stops and $X\simeq 2K{e}^{-\alpha (t-t_0)}$. It takes $(\ln 2)/\alpha$ for X to fall below $K$. Until then Y remains at $Y_{\infty }>K$ and the Z AND gate stays on, so Z stays at $Z_{\infty }$. After that Z decays by dilution, taking another $(\ln 2)/\alpha$ to halve. ThereforeMore generally replace the first $\ln 2$ by $\ln [X(t_0)/K]$. This distinguishes disappearance of the external ligand from disappearance of the persistent X transcription factor.

Use the logic approximation for gene regulation: production is either $\beta$ or zero, and every species has linear removal rate $\alpha$. Put $C=\beta /\alpha$ and assume zero basal expression, inactive X before time zero, and all required thresholds strictly between $0$ and $C$. The input is $u(t)=1$ for $0<t<T$ and zero otherwise. The common ordinary differential equation is $\dot{c}=\beta g-\alpha c$, with Boolean production gate $g$. Both networks therefore havewhere the latter uses the stipulated attained steady state. Define the threshold-crossing delayBecause $K_{YY1}>K_{YZ}$, $t_1>t_r$. In both networks, $Y_1$ stays zero until $t_1$ and then follows $Y_1=C[1-e^{-\alpha (t-t_1)}]$ until the OFF response. Write $K_A$ for the activation threshold of $Y_1$ on Z in diagram (a), and $K_I$ for the level of $Y_1$ that blocks Y's repression in diagram (b); these thresholds were not supplied numerically.

Diagram (a). All three inputs X, Y and $Y_1$ must permit Z production. Since Y has already crossed its Z threshold before $Y_1$ starts, the final permission arrives atThus $Z=0$ until $t_a$ and $Z=C[1-e^{-\alpha (t-t_a)}]$ subsequently. Solving $Z=C/2$ givesFor the frequently adopted but extra convention that both required thresholds are $C/2$, this is $3\ln 2/\alpha$. At the OFF step, X's direct AND input disappears, so $Z=C{e}^{-\alpha (t-T)}$ immediately. Y also decays immediately. $Y_1$ remains at $C$ while Y remains above $K_{YY1}$; it starts decaying at $T+{\alpha }^{-1}\ln (C/K_{YY1})$. This is a coherent feed-forward loop persistence response: activation is delayed, while removal of the direct required input switches production off promptly.

Diagram (b). The production gate is $g_Z=u[\neg (Y>K_{YZ})\vee (Y_1>K_I)]$. Z initially rises, Y switches Z production off at $t_r$, and $Y_1$ restores production at $t_b=t_1+d(K_I)$. The continuous solution during the ON phase isThe derivative changes from positive to negative at $t_r$, and from negative to positive at $t_b$. Consequently the first temporary minimum following the initial pulse occurs atIts value is $Z(t_b)=K_{YZ}{e}^{-\alpha (t_b-t_r)}$. At the OFF step Z again decays immediately because X is required for production, with Y and $Y_1$ behaving as above. If a required threshold is unattainable, these late switching events do not occur; equal production and removal rates alone do not fix threshold values.

The plotted thresholds are explicitly illustrative, with $K_{YZ}/C=0.25$, $K_{YY1}/C=0.55$ and $K_A/C=K_I/C=0.45$.

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.