A silica bead of radius is held in a single-beam optical trap in water of viscosity at ( )

PHYSICS · Multi-part · Fermi University Physics · Biophysics & Medical Physics · Molecular machines, transport and bioenergetics · Molecular machines, transport and bioenergetics

Forces, Machines and Signals, Ch Measuring one molecule: traps, tweezers and single-molecule fluorescence

Problem

A silica bead of radius a=0.50 μma = 0.50\ \mu\mathrm{m} is held in a single-beam optical trap in water of viscosity η=1.0×10−3 Pa s\eta = 1.0\times10^{-3}\ \mathrm{Pa\,s} at T=298 KT = 298\ \mathrm{K} (kBT=4.11 pN nmk_BT = 4.11\ \mathrm{pN\,nm}), with a measured stiffness κ=0.050 pN nm−1\kappa = 0.050\ \mathrm{pN\,nm^{-1}}. Treat the bead as a sphere in an unbounded fluid, so its drag coefficient is ζ=6πηa\zeta = 6\pi\eta a, and treat the trap as a harmonic spring. ((a)) The bead's position x(t)x(t) obeys the overdamped Langevin equation ζx˙=−κx+ξ(t)\zeta\dot x = -\kappa x + \xi(t), with a random thermal force satisfying ⟨ξ(t)ξ(t′)⟩=2ζkBT δ(t−t′)\langle \xi(t)\xi(t')\rangle = 2\zeta k_BT\,\delta(t-t'). Starting from this equation, derive the one-sided power spectral density Sx(f)S_x(f) and identify the corner frequency fcf_c in terms of κ\kappa and ζ\zeta. Compute fcf_c for this bead, and show, by integrating Sx(f)S_x(f) over all frequencies, that the result recovers the equipartition value of ⟨x2⟩\langle x^2\rangle. ((b)) A single bond is pulled by a force ramp f=rtf = rt. Its instantaneous rupture rate is k(f)=k0exp⁡(fx‡/kBT)k(f) = k_0\exp(fx^\ddagger/k_BT), where k0k_0 is the unstressed off-rate and x‡x^\ddagger the distance to the transition state along the pulling direction. Starting from the survival probability S(t)S(t) obeying dS/dt=−k(rt)S\mathrm{d}S/\mathrm{d}t = -k(rt)S, derive the most probable rupture force f∗f^\ast (the mode of the rupture-force distribution) as a function of rr. Evaluate f∗f^\ast for a receptor-ligand bond with k0=0.010 s−1k_0 = 0.010\ \mathrm{s^{-1}}, x‡=0.30 nmx^\ddagger = 0.30\ \mathrm{nm}, pulled at r=1.0×103 pN s−1r = 1.0\times10^{3}\ \mathrm{pN\,s^{-1}}, at T=298 KT = 298\ \mathrm{K}.

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