Biolphysm1

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Midterm I, Biological Physics, Fall 2009, PHY150/250

Problem 1)

a) What you need to do for each minimum is to complete the square; at the closed value this gives

E_{-}(R) = {1\over 2}k(R-R_{0-}-2\pi\Delta R\tau/k)^2+\epsilon_{0-} -2\pi\tau\Delta R R_{0-} - {2\pi^2 \tau^2\Delta R^2\over k} and

E_{+}(R) = {1\over 2}k(R-R_{0+}-2\pi\Delta R\tau/k)^2+\epsilon_{0+} -2\pi\tau\Delta R R_{0+} - {2\pi^2 \tau^2\Delta R^2\over k}

At the shifted minima then we can find the critical value of τc by demanding that E0 − ,min = E0 + ,min which gives

\tau_c = {(\epsilon_{0+}-\epsilon_{0-})\over 2\pi\Delta R(R_{0+}-R_{0-})}

b) Here we want to find where for τ = 0 the two parabolas intersect, at a position RB (shown in the figure) given by

{1\over 2} k(R_B-R_{0-})^2 + \epsilon_{0-} = {1\over 2} k(R_B-R_{0+})^2 + \epsilon_{0+}

We can solve this to obtain

R_B = {\Delta \epsilon\over k(R_{0+}-R_{0-})} + {1\over 2}k(R_{0+}+R_{0-})

c) This is the point at which the barrier goes away, ie, δε = λ Then the minimum disappears (replaced by a cusp at the lowest intersecting surfaces).

Then the barrier energy is the net spring energy of the + side at the barrier position:

\Delta E_B = E_+(R_B) -E_+(R_{0+}) = {1\over 2}k(R_B-R_{0+})^2 = {(\Delta\epsilon - {1\over 2}k(R_{0+}-R_{0-})^2)^2\over 2k (R_{0+}-R_{0-})^2} = {(\Delta \epsilon-\lambda)^2\over 4\lambda}

where the "reorganization energy" λ is given by

\lambda = {1\over 2} k(R_{0+}-R_{0+})^2

and is interpreted as the spring energy if you made a sudden leap from the equilibrium position of one surface to the other surface.

2) a) From Chap. 2, the number of proteins in yeast cell is about 2\times 10^8, and the volume of the cell, which we take to be all cytosol as a rough approximation, is 60 μ3 from Chap. 1. This gives an osmotic pressure

p_{protein} = {2\times 10^8 \over 60 x 10^9 nm^3} k_BT = .013 pN/nm^2

given that kBT = 4pN − nm

This is a pressure of 1.3 \times 10^4 Pascals \approx .13 Atm.

b) The idea here is that in equlibrium any slight change in energy from expansion is offset by surface tension.

So assuming a spherical cell -

pdV = σdA

p(4πR2dR) = σ(8πR)dR

R = {2\sigma \over p}

For the given threshold σ value and the estimated p value we get

R = {2 \times 1.8 \times 10^{-3} N/m \over 1.3\times 10^4 N/m} = 0.3 \times 10^{-6} m = 0.3\mu

So the question of course is why the cell does not rupture given a radius of 2.5 microns! There are other things inside giving osmotic pressure after all!

Well, one thing is this - there are inorganic constituents and proteins outside the cell which can help to render the osmotic pressure difference - which is what contributes to the net outward pressure - smaller. This is obviously important for red blood cells. Also, in yeast colonies, and in eukaryotic organisms, the internal cells are typically closely packed, so the outward pressure may be against other cells.

3) I will do this all at once. The idea is as follows - you first write down the chain of relations for equilibrium association of monomers - for convenience I will drop the concentration brackets in what follows:

K_1 X_1^2 = X_2, ~~K_1 = K

K_2 X_2 X_1 = X_3,~~K_2 = \phi K_1

K_3 X_3 X_1 = X_4,~~K_3 = \phi K_2 = \phi^2 K

Then, substituting in X_2 = KX_1^2, X_3 = \phi K^2 x_1^3 gives the general results

X_2 = KX_1^2, ~~X_3 = \phi K^2 X_1^3, ~~X_4 = \phi^3 K^3 X_1^4,~~ X_5 = \phi^6 K^4 X_1^5....

and we see that

X_n = \phi^{(n-1)(n-2)/2} K^{n-1} X_1^n

Formally then

p_n = {\phi^{(n-1)(n-2)/2} K^{n-1} X_1^n \over \sum_{j=1}^{\infty} \phi^{(j-1)(j-2)/2} K^{j-1} X_1^j}

and

<n> = {\sum_{j=1}^{\infty} j\phi^{(j-1)(j-2)/2} K^{j-1} X_1^j \over \sum_{j=1}^{\infty} \phi^{(j-1)(j-2)/2} K^{j-1} X_1^j}

However the problem is that the radius of convergence of the series as an expansion in X1 is zero, using the ratio test:

r = \lim_{n\to\infty} {\phi^{(n-1)(n-2)/2}K^{n-1}\over \phi^{n(n-1)/2} K^n} = \lim_{n\to\infty} {1\over \phi^{n-1}K} = 0~for~\phi>1

This can be fixed if we modify the cooperativity to the form (decaying cooperative effect with increased length)

K_n = \phi^{1\over (n-1)^a} K~~n\ge 2,~~a>1

This will give a radius of convergence

r = {1\over \phi^{\zeta(a)}K}

where

\zeta(a) = \sum_{p=1}^{\infty} p^{-a}

is the Riemann zeta function.

Then you would have a finite probability and mean length for X1 < r and the mean length would diverge at X1 = r.


4) Given a step of 10 nm and a force of 5 pN the work done is 50 pN-nm =12.5 kBT. Under physiological conditions, burning ATP releases 20 kBT, so the efficiency is 12.5/20 = .625 or 62.5%. This is pretty good considering that an internal combustion engine runs at 20-25% efficiency! The power delivered in a .05 second stroke is 50 pN-nm/.05 sec = 1000 pN-nm = 10 − 18 watts

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