Past exam of the biological physics course of the University of Oxford / 2022 / 2 / d / Solution

Normalize each reporter by its mean intensity, after correcting background and unequal brightness, so equal expression lies on $y=x$. Write $x=1+e+i_C$, $y=1+e+i_Y$, where the common fluctuation $e$ is extrinsic gene-expression noise and the independent reporter-specific fluctuations $i_C,i_Y$ are intrinsic gene-expression noise. The extrinsic gene-expression noise includes shared variation in cell size, expression machinery and growth state; intrinsic gene-expression noise includes independent molecular production and removal events at each reporter.

If intrinsic gene-expression noise is much smaller, points form a narrow elongated cloud along $y=x$: a cell that is bright in cyan is usually bright in yellow. If the two contributions have comparable magnitude, the cloud is still positively correlated but has substantial width perpendicular to $y=x$, because the two reporters can differ within the same cell. With variance ${\sigma }_e^2$ in $e$ and equal variance ${\sigma }_i^2$ in each independent intrinsic term,Equal intrinsic and extrinsic variances therefore give correlation coefficient $1/2$, rather than an uncorrelated circular cloud. For mean-normalized reporters, ${\eta }_{\mathrm{int}}^2=⟨(x-y{)}^2⟩/2$ and ${\eta }_{\mathrm{ext}}^2=⟨xy⟩-1$. Measurement noise must be separately estimated before assigning perpendicular scatter to expression itself. These are the geometric diagnostics of the original dual-reporter experiment.^[ref]

The lower panels of c show these two cases. Repeated observations of one cell can move both along and across the diagonal; a static population cloud alone does not determine the time scales of the fluctuations.