Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 1 / D7

Let ${\xi }_j$ be the displacement in step $j$, so the random walk position is $x(n)={\sum }_{j=1}^n{\xi }_j$. The single-step expected value and second moment areBy linearity of expected value, $\boxed{⟨x(n)⟩=0}$. The steps are independent random variables, so for $i\ne j$, $\mathbb{E}[{\xi }_i{\xi }_j]=\mathbb{E}[{\xi }_i]\mathbb{E}[{\xi }_j]=0$. Expanding the square gives the mean squared displacementThese are ensemble expected values; the empirical average over a finite number $N$ of independent cells fluctuates about them. The expected mean squared displacement itself does not depend on $N$.

For unbiased one-dimensional diffusion, the mean squared displacement is $⟨x^2(t)⟩=2Dt$. With $t=n\tau$, compare this to the mutant expression:Therefore the mutant diffusion coefficient is

On a moving step, the largest speed resolved by the discrete model is the distance per interval,This is a step-averaged speed; the model does not specify motion within the interval. Use the cell length $a=2r=20\mu \mathrm{m}=2.0\times 10^{-3}\mathrm{cm}$ in the stated Reynolds number formula. In consistent centimetre–gram–second units,Thus viscous effects dominate inertia overwhelmingly. The surrounding flow is in the Stokes flow regime: sustained movement needs applied forces, and inertial coasting is negligible. Using radius rather than the specified length as the characteristic size would halve the Reynolds number, without changing that physical conclusion.

Take the membrane particle flux density to be positive from cytoplasm into the cell nucleus. With membrane permeability $\alpha$,For fixed compartment volumes, conservation of mass and the absence of exterior flux giveSubtract the two concentration derivatives:This linear ordinary differential equation has solutionIts relaxation time is $\boxed{T=V_c{V}_n/[\alpha A_n(V_c+V_n)]}$. Here $\alpha$ has dimensions length/time, so $\alpha A_n/V$ has dimensions inverse time. A positive initial concentration difference decays monotonically toward zero; a negative one approaches zero from below.

The conserved total molecule amount fixes the final common concentration:Solve $c_c-c_n=\Delta c(t)$ together with $V_c{c}_c+V_n{c}_n=(V_c+V_n)c_{\infty }$ to obtainSubstituting the decaying concentration difference from part (d) supplies both explicit time dependences. The concentrations approach the volume-weighted initial average, which need not be the arithmetic average if the compartment volumes differ.