Download LS50 pset 9

LS50a Problem Set 9
Due November 13, 2015
Problem 0
Complete the mid-semester survey for completion points.
Problem 1 - Practice with vectors
Consider two vectors, ~x and ~y ; ~x = (3, 2) and ~y = (−5, 1). The angle between ~x and ~y is θ = 135◦ , as
indicated in the figure above. (The figure is not meant to depict these vectors in particular; it’s just meant
to show you what we mean by “the angle between vectors.”)
a) Draw a picture of these vector sums ~x + ~y and the vector ~x − ~y using the ‘tip to tail’ method. Then
calculate them. Confirm that your calculated vector sums match what you predicted from your drawing.
b)
Calculate the dot product ~x · ~y using the formula ~x · ~y = |~x||~y | cos θ.
c) Calculate the dot product ~x · ~y using the formula ~x · ~y = ~xx ∗ ~yx + ~xy ∗ ~yy . Confirm that this is equal to
your result from above.
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Problem 2 - Electric Fields and Potentials of Point Charges
The image above shows an arrangement of six fixed, charged particles, where a = 2.0cm and θ = 30◦ .
All six particles have the same magnitude of charge, q = 3.0 ∗ 106 C with signs indicated in the figure.
a) What is the net electrostatic force F~ acting on q1 due to all of the other charges? Show your work.
Hint: you’ll have to use vector addition.
b)
What is the potential (i.e., with V = 0 at infinity) at the position of q1 due to the other charges?
c) What is the potential difference between the position of q1 and the position where the y axis intersects
the line connecting q3 and q5 ? Explain how you calculated this and why.
*question and picture adapted from Halliday & Resnick
*picture from Halliday and Resnick
Problem 3 - Electric Fields and Potentials of a Charged Tube
Suppose that a charge is spread evenly on the surface of a very long tube of radius R, with a charge-perlength density λ along the tube.
a)
Far from the ends of the tube, what symmetry do you expect the electric field to have?
b)
In this region, what would be the direction of the field, inside and outsie of the cylinder?
c)
~ for this surface?
What is the appropriate Gaussian surface in this case? What is the direction of dA
d) Use Gauss law to calculate the magnitude of the electric field. Deal separately with the case r < R (the
field inside the tube) and r > R (the field outside the tube). Here r is the distance of the point from the
central axis of the tube.
e) Use the superposition principle to intuitively explain (without calculations) your result for r < R?
f)
What is the electric potential?
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Problem 4 - Gel Electrophoresis
In lab, you have used agarose gel electrophoresis to separate double-stranded DNA fragments by applying
an electric field. Consider a DNA fragment n basepairs long.
a)
What is the net charge q on this DNA fragment at pH 7? Express your answer in units of Coulombs.
b) Typically, the electric potential difference applied in gel electrophoresis is ∆Φ = +120 mV for a gel
of length 10 cm. (The potential is more positive at the end of the gel opposite from the wells where DNA
samples are added.) What accelerating force will the DNA fragment experience as a result of this electric
potential difference?
c) Calculate the acceleration experienced by the DNA fragment as a result of the force you found in part
(b). You may assume that the molecular weight of a basepair of DNA is approximately 650 Daltons.
The accelerating force associated with the electric field is countered by a drag force proportional the
DNA fragment’s velocity, but pointing in the opposite direction: F̂drag = −ζ v̂.
d) Find an expression for the terminal velocity of the DNA fragment in terms of the drag coefficient ζ and
the fragment’s length n.
e) What does this suggest about how quickly the drag coefficient ζ must increase with n during gel electrophoresis? Explain why this scaling is reasonable for migration through an agarose gel (though it would
not be reasonable for migration through water).
Problem 5 - Electrostatics of Phospholipids in the Bilayer
Phosphatydilserine is a phospholipid with a negatively charged headgroup (Q = −1.6 × 10−19 C) that
often finds itself on the cytosilic side of the lipid bilayer. Although most phospholipids in the bilayers are
not charged, we can use a charged headgroup as an upper bound of the effects of electrostatic interactions
between the phospholipids.
a) Let’s assume that most of the headgroup’s immediate environment is the hydrocarbon chains of the
surrounding phospholipids. Calculate the effective lengthscale of the headgroup’s electrostatic field for T =
300K. The dielectric constant of hydrocarbon chains is about 2.3, and the hydrocarbon chains are uncharged.
(Hint: Do you want to use the Bjerrum length or the Debye length κ−1 ? )
b) Now let’s consider the other possible environment: the cytosol. Estimate the effective lengthscale of the
headgroup’s electrostatic field, assuming it is mostly surrounded by the ionic solution of the cytosol. The
dielectric constant of water is about 80, and take the ionic density n = 0.1M for a representative ionic charge
with Z = 1, again at T = 300K. (Hint: Do you want to use the Bjerrum length or the Debye length κ−1 ? )
c) If an average phospholipid headgroup is treated like a square piece of the lipid bilayer, it has an area of
about 1nm2 . Give a rough estimate (order of magnitude) of the number of lipids affected by the electrostatic
field of a phosphatidylserine on the cytosylic side of the bilayer.
It turns out that phospholipids in general can flip from one side of the bilayer to the other, although
often fairly slowly. In the absence of any ions in the inner or outer solutions surrounding the bilayer,
phophatidylserine flips from one leaflet of the membrane to the other with some rate k. You can treat
this flipping as a simple reaction with a free energy barrier ∆G†† for moving the charged headgroup of a
phosphatidylserine through the bilayer from one side to the other. This means that the rate is given by
Arrhenius’ equation:
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k = Ae−∆G
††
/RT
For flipping in a non-ionic solution, the free energy of a lipid being on either side of the bilayer is the
same, and so the rates forward and backward are the same. However, for human cells, the typical ionic
content of either side creates a voltage of V = -70mV (the cytosylic side being the more negative side).
d) Draw a reaction-energy diagram for the flipping a phosphatidylserine from the extracellular side to
the cytoslyic side of the membrane, labeling the energy barrier with Q and V . Assume the electric field
is negligible in the middle of the bilayer, and that the bilayer still presents a significant barrier to flipping
despite the voltage difference.
e) How much quicker is the rate of phosphatidylserine flipping from cytosilic-to-extracellular than in the
other direction?
f ) Is your answer consistent with the empirical observation that phosphatidylserine is found almost exclusively on the cytosylic side? Regardless of your answer, ATP is needed to maintain this asymmetric
distribution. Give a short sentence describing a process that ATP might be facilitating.
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