Past exam of the biological physics course of the University of Oxford / 2015 / 1

The messenger RNA has steady state $X_{R,\ast }={\beta }_R/{\alpha }_R$. Since biological translation produces proteins at rate $r{X}_R$, setting the protein degradation rate equal to this production rate givesThe translational burst size is the expected number of proteins made during one messenger RNA lifetime, whose expected value is $1/{\alpha }_R$. It is dimensionless, rather than a production rate. With independent Poisson process translation and an exponential distribution for the lifetime, the resulting count has a geometric distribution on nonnegative integers with expected value $b$ and variance $b(1+b)$.

With zero initial messenger RNA and protein concentration, solving the two linear ordinary differential equations in sequence givesThe last expression applies after the short messenger RNA transient when ${\alpha }_R\gg {\alpha }_P$. All three mean curves have this shape. At a fixed steady state, a larger translational burst size requires fewer transcription events: ${\beta }_R={\alpha }_P{X}_{P,\ast }/b$. Thus $b=400$ gives infrequent large upward jumps followed by protein degradation, whereas $b=3$ gives many smaller jumps and a smoother trace.

Noise depends on molecule counts. If the fixed cell volume is $V$, write $\bar{N}=V{X}_{P,\ast }$. In the instantaneous-burst limit, independent Poisson process arrivals of geometrically distributed bursts, with count arrival rate $\lambda =V{\beta }_R$, and unit protein degradation giveIndeed, the moment equations are $d\mathbb{E}N/dt=\lambda b-{\alpha }_P\mathbb{E}N$ and $d\mathrm{var}N/dt=\lambda b(1+2b)+{\alpha }_P\mathbb{E}N-2{\alpha }_P\mathrm{var}N$. With zero initial counts the variance becomes $\bar{N}[(1+b)-e^{-{\alpha }_{P}t}-b{e}^{-2{\alpha }_{P}t}]$. These equations explain both the approach to the steady state and the scatter about it.

For illustration only, $V=1\mu {\mathrm{m}}^3$ gives $\bar{N}=10^4$ at the higher concentration: the two values of the coefficient of variation are $0.020$ and $0.200$. At the lower concentration, $\bar{N}=100$ and $b=3$ gives $0.200$. Decreasing the concentration by a factor of 100 therefore increases relative fluctuations tenfold, despite keeping small bursts. Absolute fluctuation sizes cannot be inferred without a cell volume. The original sketches below stipulate that volume; they are model realizations, not source data.^[ref]

Normalize both fluorescent reporter intensities by their respective means, correcting unequal brightness and background, and denote them $c,y$. Extrinsic gene-expression noise changes shared cellular conditions and moves both signals together: the scatter is a long narrow cloud around $y=c$. Intrinsic gene-expression noise comes from distinct random expression events in the two genes and broadens the cloud across that diagonal. Comparable contributions therefore give a noticeably wider, still positively correlated cloud. The two right-hand panels in the preceding original illustration show these two cases.

For matched fluorescent reporters with conditionally independent intrinsic fluctuations and corrected measurement noise,The first quantity measures differential variation; the second measures common variation. A clone's successive observations can remain correlated in time, so they need not be independent samples. Shared activation transients should also be separated from the fluctuations of interest.^[ref]

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.

A green fluorescent protein readout follows the corresponding periodically regulated expression, so an individual cell has repeated peaks in fluorescence. Biological translation, fluorophore maturation, and protein degradation introduce a delay and smooth the peaks. For a simple mature fluorescent reporter with periodic production $s(t)$, $\dot{G}=s(t)-\gamma G$ gives $G(t)=G(0)e^{-\gamma t}+{\int }_0^t{e}^{-\gamma (t-u)}s(u)du$. A long lifetime therefore makes the modulation weak about an accumulated background. Intrinsic gene-expression noise makes successive peak heights and periods irregular; averaging cells with unrelated oscillation phases can hide the single-cell oscillations.