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Thermal fluctuation of the arms

Introduction

In order to estimate how broad a broad spread of the data points around the mean angle in TEM angle measurements must be expected, we calculated the width of a Boltzmann distributed fluctuation of the arms. Hereby we assumed the DNA helices to be cylinders with a certain stretch modulus E, persistence length p and to fluctuate independently. Within these measurements we always quantified the angle between the end points of the arms end the other end (at the base).

Derivation

Fluctuation of one arm

The probability of finding a rod with a 10 helix bundle cross section at a certain angle [math]\displaystyle{ \phi }[/math] is given by a Boltzmann distribution

[math]\displaystyle{ p(\phi - \phi_0) = \frac{1}{Z} e^{- \frac{E(\phi - \phi_0)}{k_B T}} }[/math]

where the energy is given by

[math]\displaystyle{ E(r ) = \frac{1}{2} \frac{E_{DNA} I_{10 hb} L}{r^2} }[/math]

with r the radius of curvature, L the contour length of the arms and I10 hb the area moment of inertia of a 10 helix bundle. If we assume a constant curvature of the arms, the radius of curvature can be translated in an angle [math]\displaystyle{ \phi }[/math] (see Fig. 1).

[math]\displaystyle{ 2 \phi = 2π \frac{L}{2\pi r} }[/math]

[math]\displaystyle{ E(\phi- \phi_0) = \frac{1}{2} \frac{4 E_{DNA} I_{10 hb} (\phi - \phi_0)^2}{ L } }[/math]

The area moment of inertia of a rod is given by

[math]\displaystyle{ I = \frac{R^4}{4} \pi }[/math]

Now we are able to calculate the area moment of inertia of the 10 helix bundle with the parallel axis theorem (see Fig. 2).

[math]\displaystyle{ I = \sum_n (I_n + {z_n}^2 A_n) = I_{DNA} (10 + 320 sin (\frac{\pi}{6})) }[/math]

Fig. 2: Cross section of one of the 10 helix bundle arms. The distances for the parallel axis theorem and the direction of the fluctuation we are looking at are indicated in red.
Fig. 2: Cross section of one of the 10 helix bundle arms. The distances for the parallel axis theorem and the direction of the fluctuation we are looking at are indicated in red.

and therefore the energy needed to bend one arm by the angle [math]\displaystyle{ \phi }[/math] as well as the probability.

Fluctuation of the measured angles

Since the fluctuation of the arms is stochastically independent, we have to take the convolution of the two probabilities:

[math]\displaystyle{ p(\Delta\phi) = \int p_1 (\phi_1 ) p_2 (\Delta\phi - \phi_1) d\phi_1 }[/math]

with [math]\displaystyle{ \Delta\phi = \phi_2 - \phi_1 }[/math]

With all this follows

[math]\displaystyle{ p(\Delta\phi) = \frac{1}{Z^2} \sqrt{\frac{π k_B T}{E_{eff}}} e^{-\frac{E_{eff}}{2 k_B T}(\Delta \phi - \phi_0)^2} }[/math]

with

[math]\displaystyle{ E_{eff} = \frac{2 E_{DNA} I_{DNA} (10 + 320 sin(\frac{\pi}{6}))}{L} }[/math]

Therefore we get a standard deviation of the measured angles of

[math]\displaystyle{ \sigma = \sqrt{\frac{k_B T}{E_{eff}}} = \sqrt{\frac{L}{2 p (10 + 320 sin(\frac{\pi}{6}))}} }[/math]

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