Physics 200 Problem Set #2- Molecular Interactions, Viscosity and Mass Spectrometry Solutions 2. Which in the following pairs will produce the greater viscosity when suspended in water? a. Two spheres of identical mass but different radii (Larger) b. A solid sphere and a porous sphere of the same radius but through which solvent can flow (Solid sphere) c. A rigid rod and a flexible rod of the same dimensions (Rigid Rod) d. A sphere and a sphere of the same size but with a linear branch (a lollipop) (Lollipop) 3. Which will produce the greater relative viscosity, a molecule in water or the same molecule in a 20% glycerol solution? (Since rel = soln/solvent = 1 + , rel is independent of solvent and hence they should have the same value) 4. The following data describe the determination of the intrinsic viscosity of <a href="/keyword/bovine-serum-albumin/" >bovine serum albumin</a> (BSA) in a solvent containing 33.3 % dioxane and 0.03 M KCl at pH 2.0, T = 25oC. The protein concentration, c, is in grams/100 cm3; /o is the ratio of solution density to that of the solvent. c flow time readings (sec) /o 0.000 1.000 398.1 398.2 398.2 0.417 1.0011 439.3 439.1 439.1 0.685 1.0019 467.6 467.6 467.5 0.844 1.0023 485.8 485.5 485.7 a. Calculate []. c rel = (ts/to)(s/o) 0.000 0.417 1.104 0.685 1.177 0.844 1.223 Plot and fit gives [] = 23.4 0.1 ml/gm sp = rel - 1 0.104 0.177 0.223 sp/c (ml/gm) 24.9 25.8 26.4 b. What would you conclude about BSA under these conditions? Compare with results in Table presented in classnotes (see Powerpoint slide) (not globular under these conditions, probably indicates a highly charged asymmetric denatured molecule) 5. Derive Equation B1.3 from first principles (F=ma and energy conservation) explaining in words each step in the derivation. Newton's second law gives: Fnet = Fmagnetic = qvB = ma = mv2/r (centripetal acceleration, since motion is circular) Also since the charge is accelerated through a potential V, it gains KE given by KE = qV = mv2. From the first equation we have that v = qBr/m, or v2 = (qBr/m)2 and from the second equation, we have that v2 = 2qV/m. Equating these two expressions, we find that (qBr/m)2 = 2qV/m, or (qB2r2/m) = 2V. Then solving for m/q, we find m/q = B2r2/2V. 6. A supposedly homogeneous protein is studied by mass spectrometry. Rather than a single peak in the spectrum, the spectrograph records peaks arriving at times of 50 s, 35 s, 29 s, and 25 s in a TOF mass spectrometer. Show that these results are consistent with a protein that is homogeneous in mass. (Hint: examine Equation B1.7 to see how the time depends on the m/z ratio, and see what set of times are expected if only z changes in steps of 1 unit.) ( Can show that t2 ~ m/z, so that if we make the simplifying assumption that m is large and not affected by the addition of protons (m+1 ~ m), then we can show that m t12 z1 z2 = = . So, if the slowest peak (at 50 s) has a charge z = 1, then the second 2 t2 m z1 z2 slowest has z = 2 and so forth. Then we would expect the ratio t12 z3 = = 3 , etc. so we can construct the following chart: 2 t3 z1 Charge (z) time (s) t12 Predicted t z2 50 1 1 35 2 2 29 3 3 25 4 4 t12 z2 = = 2 , and 2 t2 z1 t12 Measured t z2 1 2.04 2.97 4.00 7. The following peptide fragments were generated after digestion of an 11 amino acid polypeptide with trypsin and separately with thermolysin. (Note that trypsin recognizes the basic amino acids lysine (K) and arginine (R) and...