Past exam of the biological physics course of the University of Oxford / 2019 / 3

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.

Put $h=K/S$, require $0<h<1$, and use the dimensionless time $s=\alpha t$. While X is produced, $X/S=1-e^{-s}$; afterwards $X/S=(1-e^{-\alpha T})e^{-\alpha (t-T)}$. X crosses the activation threshold on its rising branch atSymmetry keeps $Y=Z$. They initially receive production through X alone and need another interval $t_a$ of active input to reach $K$. If this happens while X is still above threshold, their mutual OR activation sustains production permanently. X remains above $K$ untilThe persistence correction to a gene-circuit pulse threshold uses the limiting latch condition is $t_b=2t_a$, yieldingFor $T>T_{\min }$, $Y$ and $Z$ cross their common threshold and approach $S$ even after X decays away. For $T<T_{\min }$ they either remain zero or make a subthreshold pulse and return to zero. Exact equality is sensitive to the convention for the ideal threshold, so robust switching requires a strict excess. At $h=1/2$, $T_{\min }=\ln 3/\alpha$, whereas $Y$ and $Z$ cross at $2\ln 2/\alpha$ in absolute time.

The shorter required production duration reflects X's persistence after production stops. Requiring $T\ge 2t_a$ would incorrectly discard that tail when all activating ligands remain present. If $K\ge S$, no finite pulse reaches the required feedback-active state. The second panel of the figure in solution contrasts pulses shorter and longer than $\ln 3/\alpha$.

For the dynamically regulated interpretation, take $X(0)=S$, $Y(0)=Z(0)=0$, and let X production respond to Y repression after the ligand appears. Define $h=K/S\in (0,1)$. Initially active X drives both Y and Z, giving $Y=Z=S(1-e^{-\alpha t})$. Y reaches $K$ at $t_1=-\ln (1-h)/\alpha$ and switches X production off. X then decays from $S$ to $K$, reaching it atUntil $t_2$, X remains active and Y keeps being produced. Thus $Y_2=S(1-h+h^2)$. At $t_2$, Y production stops, but Y still activates Z. Its subsequent decay crosses $K$ atImmediately afterwards both inputs to the Z OR gate are below threshold, so Z production stops and its derivative changes from positive to negative. Since Z has been continuously produced from zero until $t_3$,For $h=1/2$, $t_{\max 1}=\ln 6/\alpha$ and $Z_{\max 1}=5S/6$. Y repression is lifted at $t_3$, but X must first rebuild from a subthreshold value, so Z has a genuine first local maximum before any later response.

Wording qualification: if “X is always present at its steady state concentration” is read as an external clamp maintaining $X=S$ after ligand arrival, it overrides Y's repression and changes the problem. Then $Z=S(1-e^{-\alpha t})$ rises monotonically, with no finite first maximum. The finite-peak result uses the consistent initial-steady-state interpretation in which the stated negative feedback can act.

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.

A fluorescent reporter associated with one phase of the repressilator should show repeated rises and falls in a tracked cell, with the same underlying oscillatory period but generally delayed and smoothed relative to instantaneous promoter activity. If mature GFP is fused to a repressor and is sufficiently fast to mature, fluorescence approximately tracks that protein's abundance. If GFP is a separately expressed reporter, its own production, maturation, degradation and dilution introduce an additional filter.

For example, $\dot{G}={\beta }_{G}g(t)-{\delta }_{G}G$ converts a periodic gate $g(t)$ into a periodic, phase-lagged concentration. Very stable GFP can accumulate a large baseline and obscure oscillations rather than reproducing the regulator trace exactly; maturation delay and bleaching also matter. Biological noise and cell-to-cell phase differences can make individual cycles irregular and wash out oscillations in a population average. The appropriate sketch is therefore a delayed, rounded oscillatory fluorescence trace, with amplitude and baseline depending on reporter turnover, rather than an unqualified copy of a Boolean gate.