Past exam of the biological physics course of the University of Oxford / 2019 / 1 / b

Treat the target as a stationary perfectly absorbing sphere of reaction radius $r_0$, surrounded by an effectively uniform reservoir of protein number density $C_A$. At steady state the radial diffusion equation givesThe inward capture rate per target is the integral of the Fick's first law:For dilute independent targets of number density $C_B$, the volume association rate is $J{C}_B$, soThis is the Smoluchowski association rate, in volume per time because the concentrations are number densities. If the target also diffuses, use the relative diffusion coefficient $D_A+D_B$. Orientation constraints, incomplete absorption and chromosome geometry can reduce the actual association rate relative to this ideal collision limit.

The given sphere has radius $a=5\mathrm{nm}$. The Stokes–Einstein relation at an illustrative $T=300\mathrm{K}$ givesFor $N=30$ well-mixed independent searchers in cell volume $V$, the target-arrival rate is $N{k}_a/V$. The corresponding mean first arrival time isThe source does not fix $V$, the reaction radius or temperature. For an order-of-magnitude estimate, a bacterial volume around $1\mu {\mathrm{m}}^3$ and reaction radius $5$–$10\mathrm{nm}$ give about $0.06$–$0.12\mathrm{s}$. A 20-base-pair B-form site is about $6.8\mathrm{nm}$ long, supporting a nanometre-scale capture radius but not uniquely defining its spherical value. The estimate is the ideal diffusion-limited time, not a measured intracellular search time: nonspecific binding, orientational matching, crowding and limited accessibility are excluded.

If the outcome is the first successful binding to any operator, ten well-separated functionally independent unoccupied targets give ten times the total capture rate. Henceor roughly $6$–$12\mathrm{ms}$ for the estimates above. This assumes nonoverlapping capture zones and no significant depletion before the first event. The first-arrival time at one designated operator is not divided by ten merely because other sites are added, and filling all ten sites is a different occupancy problem.