(a) A chemoreceptor exists in two states: ligand-free, with statistical weight , and…

PHYSICS · Subjective · Fermi University Physics · Condensed Matter, Materials & Soft Matter · Biophysics & Medical Physics · Macromolecules, membranes and cell architecture

Forces, Machines and Signals, Ch Free energy at work: states, weights and entropic forces

Problem

(a) A chemoreceptor exists in two states: ligand-free, with statistical weight 11, and ligand-bound, with statistical weight c/Kdc/K_d, where cc is the free-ligand concentration and KdK_d the dissociation constant, both in the same units. Derive the occupancy θ(c)\theta(c) (the probability the receptor is bound) from the ratio of weights. Then derive θ(c)\theta(c) a second way, starting from the mass-action expression Kd=[R][L]/[RL]K_d = [R][L]/[RL] for receptor RR and ligand LL, using conservation of total receptor number and taking the free-ligand concentration to remain cc throughout (the ligand is in large excess). Show the two results agree, then evaluate θ\theta for a chemotaxis receptor with Kd=2 μMK_d = 2\,\mathrm{\mu M} at a free attractant concentration c=8 μMc = 8\,\mathrm{\mu M}. (b) Two identical, rigid viral capsids of radius R=15 nmR = 15\,\mathrm{nm} sit in a solution of inert PEG crowders, modelled as impenetrable ideal coils of radius δ=3 nm\delta = 3\,\mathrm{nm} at volume fraction ϕ=0.20\phi = 0.20. The centre of a crowder is excluded from a shell of radius R+δR+\delta around each capsid's centre. Treating the overlap of the two exclusion shells as the lens formed by two spheres of radius a=R+δa = R+\delta with centres separated by dd, derive its volume as a function of dd, then set d=2Rd = 2R (capsids touching) and take R≫δR \gg \delta to show the overlap volume reduces to Vov≈2πRδ2V_{\rm ov} \approx 2\pi R\delta^2. Hence derive the contact depletion free energy Fdep≈−32ϕ(R/δ)kBTF_{\rm dep} \approx -\tfrac32\phi(R/\delta)k_BT and evaluate it in units of kBTk_BT and in pN nm\mathrm{pN\,nm} at 298 K298\,\mathrm{K} (take kBT=4.11 pN nmk_BT = 4.11\,\mathrm{pN\,nm}).

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