= Solution
\b[Diffusion and mobility.] For a dilute isolated sphere in a <Newtonian fluid>, <Stokes's law> gives the drag magnitude $F_{\mathrm{drag}}=6\pi\eta Rv=\xi v$, where $\eta$ is the <dynamic viscosity> and $R$ is the <hydrodynamic radius>. The <mechanical mobility> is therefore $1/\xi$; the translational <hydrodynamic mobility matrix> is $\xi^{-1}I$.
To relate mobility to <diffusion>, put the particles in a weak external <potential energy> $U(x)$. Their drift speed is $-U'(x)/\xi$. For <number density> $n(x)$, the sum of drift and <Fick's first law> fluxes is
$$
J=-\frac{nU'}\xi-D\frac{dn}{dx}.
$$
At <thermal equilibrium>, the <Boltzmann distribution> gives $n\propto e^{-U/(k_BT)}$, and the net flux is zero. Substituting $n'=-nU'/(k_BT)$ into $J=0$ yields $D=k_BT/\xi$. Thus the <Stokes–Einstein relation> is
$$
\boxed{D=\frac{k_BT}{6\pi\eta R}}.
$$
Here $k_B$ is the <Boltzmann constant> and $T$ is absolute <temperature>. The assumptions include a no-slip approximately spherical particle, a continuum fluid, a small particle <Reynolds number>, and negligible hydrodynamic interactions with other particles or boundaries.
\b[Gravitational sedimentation.] The <buoyant force> must be included, since the mass of displaced water is comparable to the protein mass. Using $1\,\mathrm{Da}=1.66054\times10^{-27}\,\mathrm{kg}$ gives
$$
m=6.1108\times10^{-23}\,\mathrm{kg},\qquad
V_p=\frac{4\pi R^3}{3}=4.2213\times10^{-26}\,\mathrm{m^3},
$$
so the buoyant mass in the stated spherical model is
$$
m_b=m-\rho_wV_p=1.8894\times10^{-23}\,\mathrm{kg}.
$$
The drag coefficient is $\xi=6\pi(10^{-3})(2.16\times10^{-9})=4.0715\times10^{-11}\,\mathrm{kg\,s^{-1}}$. At terminal speed, $\xi v_g=m_bg$, so
$$
\boxed{v_g=\frac{(m-\rho_wV_p)g}{6\pi\eta R}=4.55\times10^{-12}\,\mathrm{m\,s^{-1}}\quad\text{downwards}}.
$$
Ignoring <buoyancy> would give $1.47\times10^{-11}\,\mathrm{m\,s^{-1}}$ and would not use the supplied water density. This calculation treats the stated radius as both the hydrodynamic radius and the radius determining displaced volume; a real hydrated <protein> can require a separately measured partial specific volume.
\b[Electrical drift.] In the specified simple <electrophoresis> model, terminal force balance is $Ze\mathbf E=\xi\mathbf v$. Therefore
$$
\boxed{\mathbf v=\frac{Ze}{6\pi\eta R}\mathbf E},\qquad
\boxed{\mu=\frac{Ze}{6\pi\eta R}},
$$
where $\mu$ is the <electrophoretic mobility> and $e$ the elementary <electric charge>. For the given positive <beta-lactoglobulin> charge,
$$
\boxed{v_{20}=3.94\times10^{-5}\,\mathrm{m\,s^{-1}}=39.4\,\mu\mathrm{m\,s^{-1}}},
$$
in the direction of the <electric field>. This exceeds the gravitational <sedimentation> speed by about $8.6\times10^6$.
\b[Separation time.] The two populations have equal drag coefficients because their radii are equal. Their difference in drift speed is therefore set by one elementary charge:
$$
\Delta v=v_{21}-v_{20}=\frac{eE}{6\pi\eta R}=1.97\times10^{-6}\,\mathrm{m\,s^{-1}}.
$$
Starting with coincident band centres, their separation is $\Delta v\,t$. Hence
$$
\boxed{t=\frac{10^{-3}}{\Delta v}=5.08\times10^2\,\mathrm s\simeq8.47\,\mathrm{min}}.
$$
At that time the original band has travelled $20\,\mathrm{mm}$ and the mutant $21\,\mathrm{mm}$. Their masses do not enter this ideal electrical force balance: equal-radius particles with different masses but the same charge would have equal <electrophoretic mobility> in this model.
\b[Practical discrimination.] A one-charge difference is enough for the stipulated separation, so the method is sensitive to mutations that change effective charge or friction. It cannot distinguish every mutation: substitutions that preserve charge and hydrodynamic size need not separate. Buffer <pH> changes protein ionization, and realistic <electrophoresis> also involves the ionic atmosphere, <electro-osmosis>, walls or gel interactions, and finite band widths.
For example, the <Stokes–Einstein relation> at $293.15\,\mathrm K$ gives $D=9.94\times10^{-11}\,\mathrm{m^2\,s^{-1}}$. For an initially narrow band, <diffusive broadening of electrophoretic bands> gives
$$
\sigma_x=\sqrt{2Dt}=0.318\,\mathrm{mm}
$$
at the calculated separation time. Thus the centres are about $3.15$ diffusion standard deviations apart, rather than forming infinitely sharp bands. The stated $1\,\mathrm{mm}$ centre-separation criterion is achievable in the model, but actual resolution also depends on the loading width and other dispersion mechanisms.
\Image[/past-exam-of-the-soft-condensed-matter-course-of-the-university-of-cambridge/2019-scm-electrophoretic-bands.png]
{title=Predicted band centres and diffusion widths after 508 seconds}
{description=Original Gaussian drift-diffusion calculation for initially narrow bands, using the stipulated equal-radius model.}
\b[Equivalent centrifugal acceleration.] For the original protein to attain its electrical drift speed by <sedimentation>, its buoyant centrifugal force must equal $20eE$:
$$
m_ba_{\mathrm{eff}}=20eE.
$$
Consequently
$$
\boxed{a_{\mathrm{eff}}=\frac{20eE}{m_b}=8.48\times10^7\,\mathrm{m\,s^{-2}}\simeq8.64\times10^6g}.
$$
This is the acceleration at the protein's position in the rotating solution; no rotation rate can be assigned without a centrifuge radius. If the intended comparison is instead the relative separation speed of the two bands, use $\Delta m_b=600\,\mathrm{Da}$, since their displaced volumes are equal. Matching $|\Delta m_b|a_{\mathrm{eff}}=eE$ then gives $8.04\times10^7\,\mathrm{m\,s^{-2}}$. In that separation experiment the heavier original protein moves faster centrifugally, whereas the more highly charged mutant moves faster electrically.
Back to article page
