#### Details of this Paper

##### The energy of the folded state of a protein is 50 kJ/mol lower than the energy of the unfolded state

**Description**

solution

**Question**

PS4.pdf Download Attachment;Chemistry C130/MCB C100A Problem Set 4;due February 28, 2014 (15 pts);Please turn in your homework at the beginning of class on Friday.;Reading: Chapter 6;1. (2 pts);(a) The energy of the folded state of a protein is 50 kJ/mol lower than the energy of the;unfolded state. Calculate the fraction of proteins in the folded state at 300 K.;(b) How large of an energy dierence between the folded and unfolded states is required;to ensure that no more than 0.1% of proteins are unfolded at 300 K?;2. (3 pts);A system you are studying has three available states: A, B, and C. At 298 K, you observe;the system spending 90% of the time in state A, 9.98% of the time in state B, and 0.02%;of the time in state C. Relative to the energy of state A, what are the energies of states;B and C? What assumption must you make to do this calculation? (3 pts);3. (3 pts);One mole of an ideal monatomic gas is initially at 300 K and 1.01 kJ/L pressure inside a;cylinder with a frictionless piston. The gas expands until the pressure is 0.101 kJ/L.;(a) Calculate the change in energy U, the heat q transferred to the system, and the;work w done by the system when the expansion is isothermal and reversible.;(b) Calculate U, q, and w for an expansion that is adiabatic and reversible.;4. (2 pts);A salt bridge between arginine and glutamate has the bridging atoms 3.5 apart.;A;(a) What is the energy of this interaction on the protein surface?;(b) What is the energy of this interaction in the hydrophobic core (= 2)?;5. (2 pts);The van der Waals energy can be approximated by the Lennard-Jones potential;U (r) =;Rmin;r;12;2;Rmin;r;6;where r is the distance between two atoms. If = 1 kJ/mol and the energy of interaction;is 2.94 kJ/mol when the atoms are 1.5 apart, then what is the value of Rmin? Show;A;your work.;1;6. (3 pts);A particular side chain has a C C dihedral angle for which the energy is described;by the following equation;1;U () = K [1 + cos (3)];2;where = 10 kJ/mol and = 0.;(a) What is the energy of a side chain with an angle of 60? An angle of 90?;(b) What is the relative population of side chains with an angle of 60 versus 90 at 300;K?;(c) Draw a diagram illustrating the origin of the energetic dierence between the = 60;and = 90 conformations.;7. (3 pts extra credit) The partition function Q is dened as a sum of Boltzmann factors;eEi /kB T;Q=;(1);i;where the summation is over all states of the system, Ei is the energy of state i, and kB T;is the thermal energy at temperature T. For a system with continuous degrees of freedom;this sum turns into an integral. For example, a single monatomic, ideal gas particle has;a partition function that looks like;Q=;1;h;dp;dr e H(p,r).;Together, the position r and momentum p of the gas particle fully specify the state of the;system. The integral is then taken over all the allowed values of p and r, giving a sum;over states analogous to equation 1 above. For an ideal gas particle, the energy function;H is given by;p2.;H(p, r) =;2m;To simplify the problem and avoid dealing with vectors, well consider an ideal gas particle;in one dimension. In this case, the energy simplies to;H=;p2;2m;and the partition function becomes;Q=;1;h;dp;dx exp;p2;2mkB T.;Note that h is simply a constant factor that ensures the partition function remains unitless.;(a) Suppose our gas particle is conned to the region between x = 0 and x = L. Perform;the integration over x.;2;(b) The momentum can take on all real values. In other words, the integral over p ranges;from and. Perform the integration over p. You may need to look up how to;perform a Gaussian integral.;(c) In one dimension, the pressure can be obtained from the partition function via the;relation;ln Q.;p = kB T;L;Using this formula, nd an expression for pressure p in terms of the density n = 1/L;and the temperature T.;3

Paper#17950 | Written in 18-Jul-2015

Price :*$22*