III.) The unit cell of a crystal of the E. coli enzyme transaldolase is a rectan
ID: 3279113 • Letter: I
Question
III.) The unit cell of a crystal of the E. coli enzyme transaldolase is a rectangular solid with edge dimensions of 6.89 nm times 9.13 nm times 13.0 nm. There are eight enzyme molecules of molecular weight 35,072 Da in each unit cell. What is the protein concentration in the crystal in mg/mL? How does this value compare with the typical intracellular protein concentration in prokaryotic cells (to be looked up)? IV.) A spring is placed in a large thermostat at 27 degree C and stretched isothermally and reversibly from its equilibrium length, L_0, to 5L_0. During this process 1.0 J of heat is absorbed by the spring. The stretched spring, still in the thermostat, is then released without any restraining back tension and snaps irreversibly back to its equilibrium length L_0. During the contraction process 2.5 J of heat is lost from the spring to the thermostat. Calculate: a. The overall Delta E (stretching plus contraction) for the spring b. The work done on the spring in the stretching process c. The value of Delta S for the spring in the stretching step d. The value of Delta S for the spring in the contraction step e. The value of Delta S for the spring plus the surroundings for the overall process f. Does this process violate any laws of thermodynamics? Justify your answer.Explanation / Answer
The volume of the unit cell is (6.89*9.13*13)*10^(-9-9-9)m^3=825.95*10^(-27) m^3=.825.95 e-21 ml.
Mass of the unit cell is 8*35072 Da=280576 Da=280576*1.6605e-21 mg=4.65896 e-16 mg.
So the concentration is 4.65896 e-16 mg/825.95 e-21 ml=564.07 mg/ml.