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Homework Help: Probability of finding a spherical cavity of a radius in proteins

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    The hydrophobic effect is important in stabilizing the folding of globular proteins. This is because of the drive to reduce interfaces between hydrophobic side chains and the solvent, usually polar water. One can estimate the contributions from hydrophobic forces to protein stability by using a model that assesses the work done to create hydrophobic cavities. According to this model, the likelihood of finding a spherical cavity of radius r is:

    [tex] p(r) = \frac{e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}}{\int_0^\infty{e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}} \,dr} [/tex]

    where k_B = 1.38 x 10-23 JK-1 and γ = 7.2 x 10-2 Nm-1 is the surface tension of water at T = 300 K.

    (a) (3 points) If r1 > r2, is it more or less likely that you will be able to create a cavity of radius r1 versus a cavity of radius r2. Explain.
    (b) (7 points) Compute the average radius of a cavity in water at T= 3000 K. The average radius is given by:

    [tex] <r> = \frac{\int_0^\infty{r e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}} \,dr}{\int_0^\infty{e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}} \,dr} [/tex]

    2. Relevant equations
    [tex] p(r) = \frac{e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}}{\int_0^\infty{e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}} \,dr} [/tex]

    [tex] <r> = \frac{\int_0^\infty{r e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}} \,dr}{\int_0^\infty{e^{\frac{-4*\pi*r^2*\gamma}{k_B T}}} \,dr} [/tex]

    Also, the Eötvös Rule, which was not given with the problem (this is a common occurrence in this class---we are given problems without all the equations necessary to solve it, and are expected to find those equations ourselves from other sources; I found the Eötvös Rule on Wikipedia):
    [tex] \gamma V^{2/3} = k(T_c - T) [/tex]

    where V is the molar volume, T_c is the critical temperature, and T is the temperature in question (and the independent variable).

    The Eötvos Rule in a form specific to water (I verified that this is correct by using the values for T and γ from the first part of the problem):
    [tex] \gamma = 0.07275\mathrm{N/m} \cdot (1-0.002 \cdot (T - 291 K)) [/tex]

    3. The attempt at a solution
    I was able to successfully determine the answer to part A (a cavity of larger radius is less likely to form) by using the general form of the integral for the Gaussian function, but I cannot figure out the answer to part (b).

    This is because I am getting a negative value for the surface tension (γ) when I attempt to solve for it using the Eötvös Rule:
    [tex] \gamma = 0.07275 (1-0.002 (3000 - 291)) = -.32140950 [/tex]

    Considering that the pressure is most likely below 225 atm, a temperature of 3000 K would turn the water into a superheated vapour (if the pressure were above 225 atm, the water would turn into a supercritical fluid.

    Moreover, the final form of <r> for part (b) is as follows (please let me know if you would like to see the work for this part---there is a lot of it, and I would rather not type it if it's not necessary):
    [tex] <r> = \frac{\sqrt{\frac{k_BT}{\gamma}}}{2\pi} [/tex]

    Therefore, I would get an imaginary value for <r> if I were to use this value of γ. My guess is that this means that means no spherical cavities will be formed, which subjectively implies that the average spherical cavity radius is 0 m.

    Any ideas?

    Last edited: Dec 9, 2008
  2. jcsd
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