Question

Suppose that a force of 1 pN is applied to a globular 100 kDa protein. In the absence of damping, how fast will the protein be moving after 1 ns? During this time, how far will the protein have moved? GIven the damping coefficient quoted in Table 2.2, what is the actual terminal velocity of the protein?

Table 2.2 Physical properties of a globular protein of molecular mass To0 Property Mass Density Volume Radius Drag coefficient 60 pN-s/m Diffusion coefficient 67 um2/s Average speed 8.6 m/s ass 100 kDa Value Comment 166 x 10- kg Mass of 1 mole/Avogadro constant 1.38 x 10 kg/m3 1.38 times the density of water 120 nm3 3 nm Mass/density Assuming it is spherical From Stokes' law (Chapter 3) From the Einstein relation (Chapter 4) From the Equipartition principle Chapter 4) Note: 1 nm- 10 m, but 1 nm-(1 nm)3 10-m3 In water at 20 C Root-mean-square (the square root of the average value of the square of the velocity)

Answer 1

Here F = 1 pN = 1 times 10^-12N t = 1ns = 1 times 10^-9 sec We know F = integral dp/dt F = m(v_f - v_i)/t (v_i = 0 starting at) v = Ft/m v = 10^-12 times 10^-9/166 times 10^-24 m = 166 times 10^-24 kg v = 6.02 m/s distance travelled by protein x = vt = 6.02 times 10^-9 = 6.02 times 10^-9 m Now, terminal velocity is given by v = 2r^2 (rho - sigma)g/9 eta rho = density of body sigma = density of medium eta = viscosity of medium r = radius of body (assuming medium is water) v = 2(3 time 10^-9)(1.38 times 10^3 -997) times 9.8/9 times 8.90 times 10^-4 v = 8.43 times 10^-12 m/s terminal velocity of protein in water