Homework #4 Physics 498Bio Spring 2012 Prof. Paul Selvin

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1 Assigned Wednesday Feb. 22, 2012: Due Wednesday February 29, 10:30am. Hand in at start of class. Late homework is not accepted. (Solution sets will be posted shortly after deadline.) Note: Marco will give office hours on Friday from 7-8pm. Marco will also hold a Skype office hour Monday 9-10pm. His username is: tjioe2. (This will be only voice; no video.) There will be no class Monday Feb 27. (We both will be gone this weekend and early next week.) 0. Read handout on Protein Folding (on web). (Professor Maria Spies s write-up.) 1. Protein Folding: Energy, Entropy and Gibb s Free Energy: Consider the following two-dimensional lattice model for a helical 6-mer peptide. The configuration drawn below is the native state (correctly folded): Energy rule: any adjacent pair of amino acids not connected by a backbone bond counts as connected by a hydrogen bond (2 in the example). Assume each hydrogen bond (shown dotted) stabilizes the system by an enthalpy of -500 J (negative means stabilizing). (-500 J is for one mole of the molecules) Rotated versions of this peptide do not count as separate structures, and only 90 swing moves are allowed to go from one structure to another (as in lecture). a. What is the enthalpy of the helical ground state of this model? What is its configurational entropy? What is its Gibbs free energy at 275 K and 375 K? Hint: use the equation G = H TS. There is only one state for this configuration. (We assume H = E, which is true since we are dealing with solids and liquids.) b. Draw all structures that have an enthalpy of 500 J. Label each structure with a choice from this list: helix initiating state or completely unfolded state or offpathway trap or hairpin state or almost hairpin state. Hint: the total number of structures should be 6. c. Draw as many structures of unfolded states (H=0) as you can. What is the Gibbs free energy of your unfolded ensemble at 275 K and 375 K? Comparing to (a), will this peptide fold to a helix somewhere in that temperature range? Hint: the total number of structures should be 13. d. According to the energy rule, the ground state is actually degenerate! Draw the other low energy state, and label it with a choice from the list in (b). Is this peptide a good design for isolated unique secondary structure, according to the energy rule? e. Secondary structure is never isolated in proteins. Introducing three black circles as hydrophobic amino acids, draw a structure of a perfect amphipathic (having both a hydrophobic and hydrophilic) helix on the lattice. Let us say such a structure gets an extra score of 1000 J for having all non-polar amino acids on one side, -500 for having 2 on one side. Draw the structure of the low energy state in (d) with the same position of the hydrophobic residues as in the helix. Now what is the enthalpy of the 1

2 helix vs. the state in (d)? What secondary structure would this peptide form in the context of a protein, where hydrophobic residues can be buried if they all lie on one side of the peptide? 2

3 2. Big Wheel, Keep on Turning In class, we discussed F 0 F 1 -ATPase, a biological rotor that uses a proton gradient to create ATP 1. The α and β subunits are proteins that form a barrel, which holds the γ subunit, another protein. Protons traversing the membrane cause the γ subunit to rotate, which drives ATP synthesis. Left: The structure of F 0 -ATPase, Right: The barrel structure. a) In 1996, Kinosita, et al attached an actin filament to the γ subunit, and watched it rotate. The actin filament may be approximated as a rod of of length 2.6 m, and radius 5 nm. They observed that, at high concentrations of ATP, the rod rotated at a maximum speed of 0.5 revolutions per second. They calculated that the drag on a rigid rod held at one end and rotated in a solution is 4 3 L 3 /[ln( L / 2r) 0.447], where r is the radius of the rod, L is its length, is the rotation rate in radians per second, and is the viscosity of water, 10-3 N s m -2. Calculated the torque on the rod at this speed, in pn nm (piconewton nanometers.) b) Kinosita et al. found that the rod didn t rotate continuously, but would make a series of discrete jumps. Considering the symmetry of the structure of the barrel shown in the figure on the right above, what is a likely value for how big each jump is, in degrees? c) Each time the rod undergoes a jump from one state to the next, it synthesizes a single ATP. Combine your answers from part a and part b to find the energy required to synthesize a singe ATP. d) Considering the amount of energy in a single molecule of ATP, what is the efficiency of this process? If your answer is unreasonable, (e.g. significantly above 100%) try reconsidering your answer to part b. 1 Figure copied from Direct observation of the rotation of F 1 -ATPase, Noji, et al, Nature 386:300,

4 3. Molecules in a Room Imagine that there are four molecules in a room. (They are pictured as black dots in the figure, although you might want to picture each one as a dot of different color to help keep track of them.) If you look at a particular instant in time there is a chance that you will find four molecules on the left or right side of the room. There is also a chance that you will find one molecule on the right side and three on the left side, and so on. Left Side Right Side A Possible Configuration a) Draw a diagram showing all the possible configurations of the four molecules (i.e. one on the left and three on the right, etc.) along with the number of ways, i.e. states, this can be achieved for each configuration. What is the total number of ways the molecules can be arranged, i.e. the number of accessible states? The answer can be written as M N. What do M and N correspond to both numerically and physically in this problem? b) What is the probability of finding all four molecules on the left side of the room? What is the probability of finding two molecules on the left and two on the right side of the room? Remember that the fundamental postulate of statistical mechanics is that every accessible state is equally probable. c) An important result from statistical mechanics is that as the number of molecules increases, the most probable distribution dominates the properties of the system, i.e. is overwhelmingly more likely. There is a formula for counting the number of states for a given configuration of molecules. If N is the total number of molecules, L is the number of molecules on the left, and R is the number of molecules on the right, the number of accessible states for that configuration is N!/(R!L!). How many states are there with all molecules on the left if N=2, N=10, N=100? How many states are there with the molecules evenly divided between left and right if N=2, N=10, N=100? Use your results along with the fundamental postulate of statistical mechanics to explain the statement, As the number of molecules increases, the most probable distribution dominates the properties of the system, i.e. is overwhelmingly more likely. 4

5 4. Thermal motion and stiffness of an AFM Cantilever. In class, we discussed using the AFM for applying and measuring forces to proteins. Now we will calculate some of the relevant energy, length and force scales associated with the AFM cantilever (the arm used for measurement.) In all calculations for this problem, give spring constants in units of pn/nm. We model the cantilever as a spring with a certain spring constant. a. To measure the stiffness of a biomolecule, the cantilever is pulled up and the deflection of the cantilever measured (see lecture notes). In order to get a reasonably large (i.e. measurable) deflection, what should the relative spring constant of the cantilever be compared to the biomolecule, i.e. larger, smaller, about the same? Briefly explain why. b. Thermal forces acting on the cantilever will introduce noise into our measurements, i.e. will cause random deflections of the cantilever. If the spring constant of the cantilever is too small, we will be unable to detect the deflection of the cantilever due to the stiffness of the biomolecule we re pulling on because thermal noise will overwhelm the measurement. If we want to be able to measure a deflection of 10 nm, what is the minimum spring constant the cantilever can have? c. In the course lecture notes, the spring constant is given for a cantilever is 4 4 Er. 3 3 L The cantilever is typically made from glass and has a length L, a Young s Modulus of E, and a radius of r. Recall the E, the elastic modulus, also called the Young s Modulus, is just the k of Hook s law, F= -kx: F/A = E L/L, where you are pulling on a rod of length L by an amount of distance L, with a cross-sectional area A object. E of glass is 70 GPa. Given that we want to make a long, thin rod with a length to radius ratio L/r = 800, how long must our cantilever be for it to have a spring constant of 0.1 pn/nm? d. What is the minimum force that we can measure with our AFM in pn? e. What sets the upper limit on the force we can measure? (A qualitative answer is sufficient here.) 5

6 4. Springiness of an Alpha Helix Parts of proteins often fold in shapes called alpha helices. The alpha helix consists of a string of amino acids wrapped in a helix such that the C=O group in the main chain is hydrogen bonded to the NH group of the amino acid that is situated four residues ahead in the linear sequence (see pg 82, Secondary Structure of Campbell s text). There are 3.6 amino acids per turn of the helix. The helix has a diameter of 5Å and a pitch of 5.4Å (pitch is the distance between repeat units in the helix). Using the fact that the energy of a hydrogen bond is approximately 5kT and it will break after being stretched ~0.5Å, determine E, the Young s Modulus of an alpha helix. Remember that: F/A =EL/L Model the helix as a repeating unit with 3.6 springs in parallel. Express your answer in units of GPa. Compare this answer to the Young s Modulus calculated in the notes for a covalent bond and for a typical protein. Hydrogen Bond 5.4Å 5.0 Å 6

7 6. Replication Fork As soon as Watson and Crick announced their double-helix model of DNA structure, both physicists and chemists were quick to point out some troubling features of the model. For example, how are the two strands separated for replication, when they re wound around each other? Below is a simplified figure showing a replication complex (e.g. DNA polymerase and/or helicase) (triangle) unwinding DNA. If the replication machinery cannot rotate, because, e.g. it s tied down to cellular components, the DNA must rotate about its long axis 360 degrees every 10 bases unzipped. We will now calculate the frictional drag associated with replication (i.e. the drag caused by the DNA rotating about its axis) and compare the energy lost due to friction to the energy required to form the phosphate-sugar backbone bonds. Model a length of DNA as a cylinder, which is rotated with constant angular velocity (in rad/s) during replication. The frictional torque is given by, f = 4R 2 L, where is the viscosity of water (at 37 C), L is the length of DNA, and R is the radius of DNA. a) Write the equation for work done against the frictional torque for one helical turn. b) Consider a DNA piece 6000 helical turns long (still assuming it s a straight line which is not really true). What is L? What is R? c) If it takes 100 seconds to replicate this DNA, what is? d) Calculate the value for work done to overcome friction per helical turn. e) Let s assume the energy of each phosphate-sugar bond in the backbone is about 15kT. How does the energy required to synthesize these bonds per helical turn compare with the energy required to overcome the rotational drag? Is the rotational drag a significant fraction of the cost of replicating DNA? 7

8 7. Equipartition of energy Earlier in the class Professor Selvin stated without a rigorous proof that for each degree of freedom in which the energy of the molecule depends on either x 2 (position) or v 2 (velocity) there is ½k B T of energy on average. This is known as the Equipartition theorem. For example, a free particle has E = ½ mv 2, so <E> = ½k B T in one dimension, or 3/2 k B T in three dimensions. Another example: a particle on a spring has E = ½ kx 2 + ½ mv 2, so <E> = k B T. The goal of this problem is to derive the Equipartition theorem from the a Boltzmann distribution, i.e. the Equipartition theorem is a direct consequence of the Boltzmann distribution. We will determine the expectation value of the potential energy (which will be proportional to x 2 ) and show that the average energy is ½k B T. By analogy we will show the same for the kinetic energy (which is proportional to v 2 ). In general, the expectation value of a quantity x is shown below Expectation Value x n x n p( x) dx where p(x) is the probability of finding a particle at a position x. In our case the probability happens to be nothing more than the Boltzmann factor shown below normalized by the partition function. 1 E( x) / kbt p( x) e Z In this case we are going to write the partition function not as a sum of discrete states, as done in class, but rather as an integral (since x is continuous, not discrete). Z e E( x) / k T B dx Useful Integrals dx So now we want to find the expectation value of the potential energy, which is equal to ½kx 2. a) First, write down the partition function in terms of k (the spring constant) and x. Integrate the expression to determine Z. (See table of useful integrals, above.) b) Write down what p(x) is in terms of k and x. Now determine the expectation value of the potential energy, <½kx 2 >. Remember you already have been told that it is ½k B T, but please show all your work to receive full credit. c) Now lets look at kinetic energy term, ½mv 2. What happens if you substitute m for k and v for x into the expectation value calculation do integrals change? Will you get the same result, ½k B T? x e 2 e 2 ax 2 ax dx 3 / 2 2a 8