LPS Assignment 6
MCB 65 Spring 2016
20 points
due Wednesday March 23th at 10 AM
Lab:
(1 point) Make a figure (and figure legend) depicting the standard curve from your Bradford
assay. Figure legends should contain enough information for the reader to understand the figure
but should not include extensive experimental detail (which belongs in the Materials and
Methods section). Refer to the “Lab report and scientific writing resources” section on the Lab
page of the course website for instructions on creating figures and writing figure legends.
Problem Set: As usual, show your work for calculations, justify your answers, and cite your
sources as appropriate!
1. Entropy and Temperature
a. (1 point) Define the temperature as a function of internal energy (U) and entropy (S).
𝑇=
𝜕𝑈
𝜕𝑆
b. (1.5 points) Suppose you have 5 distinguishable particles that can each have a spin of -1/2 or
+1/2, and the corresponding energy of these levels is 0 and 1 energy units (E.U.). This is
analogous to the spins of NMR samples. Construct a plot of multiplicity vs. energy from 0 to 5
E.U.
Total E.U.
0
1
2
3
4
5
Multiplicity
1
5
10
10
5
1
c. (1 point) According to equation 1, what property does the temperature have at E.U < 2? What
about for E.U. > 3?
At E.U < 2 the temperature is a positive number because the entropy increases as the multiplicity
does. Above E.U = 3, the temperature is negative.
d. (0.5 points) Suppose the system of 5 spins with E.U. of 5 is put in thermal contact with
another system of 5 spins with an E.U. of 2. How does the entropy of the two systems change
after equilibrium is reached?
The entropy of the high energy state increases as it loses energy. The energy of the low energy
state is constant.
e. (1 points) What can you say about the flow of energy in the system? What seems paradoxical
in this statement?
Energy flows from the high-energy system to the low-energy system. This is paradoxical because
the high-energy has negative temperature, meaning that negative temperature is even hotter than
infinite temperature. Also paradoxical because system 1 loses energy but gains entropy.
f. (2 points) To see whether this paradoxical behavior extends to large systems, suppose you have
a spin system of 10,000 spins that can either have an energy of 0 or 1. Calculate the multiplicity
of the system if there is a total energy of 3,000. Repeat calculation for a total energy of 5,000.
Probabilistic definition of entropy works best in this case (large numbers).
E1 = 7000 molecules
E2 = 3000 molecules
We first compute the entropy using the probabilistic definition (**note that the answer assumes
-1
kB is 1 J K for simplicity)
S = −Nt kB ∑ pi lnpi
P1 = 7000 / (10000) = .7
P2 = 3000/(10000) = .3
lnW = -NΣplnp = -10000 (0.7*ln(0.7) + 0.3*ln(0.3)) = 6109
W  e^6109
For the 5000 energy system:
lnW = -NΣplnp = -10000 (0.5*ln(0.5) + 0.5*ln(0.005)) = 6931
W = e^6931
g. (0.5 point) What is the multiplicity at an energy of 7,000 E. U.? Does this paradoxical
behavior hold?
e^6109, yes
h. (1 points) What property of spin systems allows for this paradoxical behavior? How would
you best define the requirements for a system that has this paradoxical behavior? (For more
reading on “negative temperature,” see: http://news.mit.edu/2013/its-a-negative-on-negativeabsolute-temperatures-1220)
A spin system has a finite number of energy levels. Negative temperature requires a system
where there are a finite number of energy levels such that when you put enough energy into the
system, the entropy actually decreases.
2. Protein Structure Determination and the Boltzmann Distribution
As we saw in lecture, the chemical shift of an NMR-active nucleus is highly dependent on its
local environment. Let’s explore how local secondary structure and solvent exposure affects the
chemical shift of residues in glutaredoxin, a small protein that uses the tripeptide glutathione to
regulate the cellular redox balance. For this question, please use the solution NMR structure of
glutaredoxin 3 (PDB ID: 1FOV).
a. (1 point) One simple and widely-employed method to analyze protein NMR data is the
chemical shift index (CSI). CSI is a graphical technique that plots the difference between the
observed chemical shift of an atom in the folded protein (δnative) and the chemical shift of the
corresponding atom in an unfolded random coil (δrandom), Δδ = δnative - δrandom. This technique is
mostly used to analyze protein backbone conformations, and proton on the alpha carbon of each
amino acid (1Hα) or the alpha carbon itself (13Cα) are the most commonly used atoms for CSI
plots. Is an atom with a positive Δδ more or less shielded from the external magnetic field than
its random coil counterpart?
A positive Δδ means that δnative is larger than δrandom. A larger δ means that the atom is less
shielded.
b. (1 point) The CSI plot for 1Hα of glutaredoxin 3 is provided below (Åslund, F., et al. (1996) J.
Biol. Chem. 271(12), 6736-6745.) Using the NMR structure of protein and the amino acid
sequence in the plot, label which residues correspond to α-helices and β-sheets.
c. (0.5 points) Which are more shielded, α-helices or β-sheets?
α-helices have a negative Δδ and are thus more shielded.
d. (2 points) Let’s now consider the effect of solvent exposure on the chemical shift index in
PyMol. First, separate each of the 20 states in the ensemble using the command “split_states
1FOV.” Next, delete all states except for 1FOV_0001. Display the contours of the solventaccessible surface of the protein using S → Mesh. Display the backbone as a cartoon, hide the
side chains, color the structure 50% grey, and change the background color to white. Now let’s
look at two residues, Q29 and Q54. Select both, display the residues as sticks, and color by
element. Label both residues. Right-click on the α-proton of each residue and display it as a
sphere. Please prepare two figures to highlight the surface exposure (or lack thereof) of the αproton of each residue. If you want to move the label, switch to editing mode by clicking on
Mouse Mode in the bottom right menu, and hold Ctrl to click and move the label. Resize the
label with label_size set to #, where # is the font size.
e. (0.5 points) In what secondary structure elements are Q29 and Q54 found?
Both are in β-sheets.
f. (0.5 points) Which 1Hα is more surface exposed? How can you tell?
Q54 is buried, while Q29 is exposed to solvent, as indicated by the VdW sphere aligning with
the solvent-accessible mesh for Q29 but not for Q54.
g. (0.5 points) Which factor appears to be more significant in determining Δδ for 1Hα, secondary
structure or solvent exposure? (Note: this result is not necessarily intuitive and is still being
investigated! For additional optional reading, see Avbelj, F., Kocjan, D., & Baldwin, R. L.
(2004). PNAS, 101(50), 17394-17397)
Based on the CSI plot from part b, Q29 and Q54 have similar, positive Δδ values. This suggests
that their residence in a β-sheet contributes to their Δδ value more than solvent exposure for 1Hα.
CSI plots for side-chain protons are much less useful for predicting secondary structure in part
because rotation about side-chain bonds results in different conformational isomers (rotamers)
that experience different chemical environments and will thus have variable chemical shifts. The
distribution of side-chain orientations in different rotamers states can be described by the
Boltzmann distribution. Consider the following two hypothetical distributions (image modified
Energy
Energy
B
A
from The Molecules of Life (© Garland Science 2008)):
h. (0.5 points) Which distribution represents the χ1 rotamers of alanine and isoleucine, assuming
all other factors are equivalent?
Isoleucine is β-branched and experiences more steric/VdW clash as it rotates, resulting in a
larger energy difference between rotamers, ie. a smaller partition function Q. A smaller partition
function means more molecules in the lower energy states, so A. isoleucine and B. alanine.
i. (0.5 points) Which distribution represents the rotamers of alanine at 20 °C and 37 °C?
Increasing the temperature means that more molecules can occupy higher energy states, so A. 20
°C and B. 37 °C.
j. (0.5 points) Which distribution represents the rotamers of glutamic acid in a structure where
there are no possible salt bridges and a structure where a salt bridge is possible?
Same logic as part h. A conformation with a salt bridge is assumed to be stabilizing, so the
energy difference between the salt-bridge conformation and all other conformations will be
large. Thus, A. salt bridge and B. no salt bridge.
3. Free Energy and Biological Processes
a. (1.5 points) A direct measurement of the standard free-energy change associated with the
hydrolysis of ATP is technically demanding due to difficulties in accurately measuring the
minute amount of ATP remaining at equilibrium. However, ΔG’° can be calculated indirectly
from the equilibrium constants of two other enzymatic reactions having less favorable
equilibrium constants:
ATP + glucose → ADP + glucose 6-phosphate
K’eq = 890
glucose + Pi → Glucose 6-phosphate + H2O
K’eq = 3.70x10-3
Calculate the standard free energy of hydrolysis of ATP at 25 °C in kJ/mol.
For the summed reactions to cancel to give the reaction of ATP hydrolysis, we need to reverse
the order of the bottom equation, and therefore invert the K’eq value, 1/3.70x10-3 = 270.
Reaction 1)
Reaction 2)
Sum)
ATP + glucose → ADP + glucose 6-phosphate
+ Glucose 6-phosphate + H2O → glucose + Pi
ATP + H2O → ADP + Pi
ΔG’°sum = ΔG’°1 + ΔG’°2 = -RTln(K’eq,1*K’eq,2)
= -(8.314x10-3kJ/mol*K)(298 K)ln(890*270)
= -30.7 kJ/mol
b. (1.5 points) The free energy released by the hydrolysis of ATP as calculated above assumes
standard conditions at pH 7.0. If ATP is hydrolyzed under standard conditions but at pH 8.0,
would more or less energy be released? Explain your reasoning. [Hint: Pi is shorthand notation
for H2PO4-, which has a pKa of 7.2] (For additional reading, see: Phillips, R. C., George, P., &
Rutman, R. J. (1969), J. Biol. Chem., 244(12), 3330-3342.)
The relevant equations can be written as follows:
ATP4- + H2O → ADP3- + H2PO41H2PO41- → H1PO42- + H+
pH 8 favors the dissociation of H2PO41- to H1PO42-, effectively depleting the concentration of the
products of ATP hydrolysis. Decreasing the concentration of the products shifts the equilibrium
to the right, so more energy would be released upon ATP hydrolysis.