Consider the following model for the electrostatic stretching of DNA. A spring w
ID: 1405501 • Letter: C
Question
Consider the following model for the electrostatic stretching of DNA. A spring with spring constant kDNA = 250 pN/m is tethered to an immobile surface on one end and a charge of qDNA = 350e on the other. A uniform electric eld of magnitude |E| = 2.5 N/C is applied pointing perpendicular to the surface.
a) How far does the spring stretch in response to the eld?
b) Now an transcription factor (TF), with charge qRNA = +15e, is introduced at the location specied in the gure. Calculate the electric eld due to the TF at the location of the DNA. Remember, biophysics experiments are done in water, which has a dielectric constant K = 80. (hint: the TF is very far from both the DNA and the surface, so you can ignore the string stretching distance in this calculation)
c) What is the angle that the spring makes with the x-axis in the presence of both the TF and the external eld?
Explanation / Answer
kDNA = 250 pN/m = 250 * 10^-12 N/m
Charge QDNA = -350 * 1.6 * 10^-19 C = 5.6 * 10^-17 C
E = 2.5 N/C
a)
Force due to Electric Field Fe = q*E
We also know, Force on Spring = k*x
where k is spring constant
x is spring Elongation.
250 * 10^-12 N/m * x = 5.6 * 10^-17 * 2.5
x = (5.6 * 10^-17 * 2.5)/(250 * 10^-12) m
x = 5.6 * 10^-7 m
spring stretch x = 5.6 * 10^-7 m
b)
charge qRNA = +15e
E = k * q / r^2
E = 8.9* 10^9 * 15*1.6 * 10^-19 / (10* 10^-6 )^2
E = 213.6 N/C
E due to TF at qDNA = 213.6/80
E due to TF at qDNA = 2.67 N/C
c)
Net Field in X direction = E due to TF + Ex due to External Field
Net Field in X direction = 2.5 + 2.67 * cos(45)
Net Field in X direction = 4.39 N/C
Net Field in Y direction = Ey due to External Field = E * cos(45)
Net Field in Y direction = 2.67 * cos(45)
Net Field in Y direction = 1.89 N/C
= tan^-1(1.89/4.39)
= 23.29