Probability of finding a spherical cavity of a radius in proteins

In summary, the hydrophobic effect plays a crucial role in stabilizing the folding of globular proteins by reducing the interfaces between hydrophobic side chains and the polar solvent, water. A model can be used to estimate the contribution of hydrophobic forces to protein stability by assessing the work done to create hydrophobic cavities. The likelihood of finding a spherical cavity of radius r is given by the probability distribution function p(r), which is an exponential function. The average radius of a cavity in water at 300 K can be calculated using the general forms of integrals for the Gaussian or exponential functions. However, the Eötvös Rule, which predicts surface tension, may not accurately predict the surface tension at high temperatures far from
  • #1
vaizard
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Homework Statement


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]

Homework 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]

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?

TIA!
 
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  • #2




Thank you for bringing up this interesting problem! I can understand your confusion regarding the negative value for surface tension at such a high temperature. However, it is important to keep in mind that the Eötvös Rule is only applicable for temperatures close to the critical temperature of a substance. In the case of water, the critical temperature is 647 K, which is much lower than the 3000 K that you are considering. Therefore, the Eötvös Rule may not accurately predict the surface tension at such a high temperature.

In order to solve part (b) of the problem, you can try using the general form of the integral for the Gaussian function, as you did for part (a). This should give you a numerical value for the average radius of a cavity in water at 3000 K. Alternatively, you can also try using the general form of the integral for the exponential function, since the probability distribution function p(r) is an exponential function. This should also give you a numerical value for <r>.

I hope this helps! Let me know if you have any further questions.
 
  • #3


I would first verify the values and equations given in the problem to ensure accuracy. If the Eötvös Rule does not apply to water at such a high temperature, I would discuss this with my colleagues or consult other sources to find a suitable equation for the surface tension at 3000 K.

Once a suitable equation is found, I would then use it to calculate the surface tension and proceed with solving for <r> using the given equation. If the resulting value for <r> is still negative, it could indicate that the assumptions made in the model are not applicable at such extreme conditions. In this case, I would discuss with my colleagues and possibly re-evaluate the model or seek alternative methods for estimating the average spherical cavity radius in water at 3000 K.
 

1. What is the significance of studying the probability of finding a spherical cavity of a certain radius in proteins?

The probability of finding a spherical cavity in proteins is an important aspect of protein structure and function. It can provide insight into the stability, binding affinity, and enzymatic activity of proteins.

2. How is the probability of finding a spherical cavity calculated in proteins?

The probability of finding a spherical cavity in proteins is calculated using computational methods such as molecular dynamics simulations or homology modeling. These methods take into account the size and shape of the protein, as well as its surrounding environment.

3. What factors can influence the probability of finding a spherical cavity in proteins?

The probability of finding a spherical cavity in proteins can be influenced by various factors such as the amino acid composition, size and shape of the protein, and the surrounding solvent and ions. Mutations, post-translational modifications, and ligand binding can also affect the probability of finding a cavity.

4. How does the probability of finding a spherical cavity relate to protein function?

The probability of finding a spherical cavity in proteins is closely related to protein function. Cavity size and shape can determine the types of molecules that can bind to the protein, while the presence of cavities can also affect the stability and enzymatic activity of the protein.

5. Can the probability of finding a spherical cavity in proteins be experimentally determined?

While computational methods are commonly used to calculate the probability of finding a spherical cavity in proteins, it is also possible to experimentally determine this probability using techniques such as X-ray crystallography or NMR spectroscopy. However, these experimental methods can be challenging and time-consuming.

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