1. Calculate the isoelectric point (pI) of arginine. Take the average of the two pKa which straddle the point at which the net charge on the amino acid straddles 0. At pH 6, the carboxylate has a -1 charge, but the others groups are protonated so the net charge is +1. At pH 10.0, the amino group would be deprotonated so the net charge is about 0. At 11 it would have a net negative charge. Average the two pKas which straddle pH 10-11 gives you approx (8+12)/2 or about 10
2. How would you prepare 2.00 L of 0.1 M glycine buffer, pH 9.0, from glycine and 1.00 M NaOH. You could do this in a variety of ways. The simplest would be to add 0.2 mol glycine, MW = 75, (or 15 g) to about 2 L of water. When you by glycine in several forms. One, the acid form, would have both amino group and the carboxy group in the protonated state. The other would consist of a zwitterion, in which the carboxy group is deprotonated (-1) and the amino group protonated (+1). The pKa of the amino group is about 9.6. At pH 7.6, it would be fully protonated. At pH 9.6 it would be half-protonated. At pH 9.0 it would be more protonated. If you used either the acid form or the zwitterion form of Gly, you would need to add some base (NaOH) to raise the pH of your glycine solution to 9.0. Then make to 2.0 L with water, and presto, you have a glycine buffer solution make without evening using the HH equation. You could use the HH equation: pH = pKa + log A/HA or 9 = 9.8 + log A-/HA. Let HA tot = A- + HA = 0.1 M. A- = x, HA = 0.1 -x. Solve for x, and add that much NaOH to the zwitterion form of glycine to convert that much of the protonated amino group to the A- or deprotonated form.
3. A polypeptide is subjected to the following degradative techniques resulting in polypeptide fragments with the indicated amino acid sequences. What is the amino acid sequence of the entire polypeptide?
CNBr treatment:
i. Asp-Ile-Lys-Gln-Met
ii. Lys
iii. Lys-Phe-Ala-Met
iv. Tyr-Arg-Gly-Met
Trypsin treatment:
v. Gln-Met-Lys
vi. Gly-Met-Asp-Ile-Lys
vii. Phe-Ala-Met-Lys
viii.Tyr-Arg
ANSWER: Tyr-Arg-Gly-Met-Asp-Ile-Lys-Gln-Met-Lys-Phe-Ala-Met-Lys
4. A protein is subjected to end group analysis by dansyl chloride. The liberated dansyl amino acids are found to be present with a molar ratio of two parts Ser to one part Ala. What conclusion can be drawn about the nature of the protein?
Several possible explanations exists. The protein might consist of three polypeptide chains held together by intermolecular forces with or without disulfides. It could consist of two identical subunits each with N-terminal Ser, and one different subunit with a N-terminal Ala. The protein displays quaternary structure. Alternatively, the protein might not be pure, and you actually have three separate, non-linked proteins, two of which happen to have an N-terminal Ser, and the other an Ala.
5. N-methylacetamide - this is a model compound to study H-bonding, since it has one H bond acceptor (carbonyl O) and 1 H bond donor (amide H), mimicking a peptide bond in a protein. By detemining the equilbrium constant, and hence delGo for making a dimer of 2 N-MAM in a nonpolar medium from isolated NMAM in aqueous solution, you may help determine if the analogous reactions in proteins is favorable. (i.e. two isolated donors and acceptors on a non-native protein in aqueous solution, with the donors and acceptors either H bonded to water or to each other but solvent exposed, forming a intrachain H bond buried inside the folded protein (in a more polar environment) would be favored and hence drive protein folding. See the web site for the explanation of how this could be done using thermodynamic cycles.
6. You have a mixture of 5 peptides, given below, which you wish to separate. You will separate them on a sheet of paper using two techniques, one paper chromatography, which like TLC, separates them on the basis of polarity, followed by electrophoresis, which separates them on the basis of charge. The peptides and their respective charges at pH 4.5 and 6.5 are:
CHARGE AT PH 4.5 6.5
a. Asn-Arg-Lys 2 2
b. Asn-His-Phe 1 ~0.3
c. Asp-Leu-Phe -0.8 -1
d. Asn-Leu-Phe 0 0
e. Val-Leu-Phe 0 0
The mixture is spotted on the paper as shown below, and chromatography performed in the direction shown, using a water/butanol/acetic acid solvent system at pH 4.5. At the end, electrophoresis is carried out at pH 6.5 with the paper placed in the solvent system with the electrodes placed as shown. After the electrophoresis, the paper is dried and stained to visualize the peptides as spots. Draw on the paper what the distribution of spots would look like and identify the spots.
ORDER OF DECREASING POLARITY AT PH 4.5:
A > B > C > D > E (A +2, B +1, C -1 but 2 hypo. aa, D has 1 polar aa, E all nonpolar
So they spread out from bottom to top in that direction. Then they separate by electrophoresis top to bottom. A and B moves down to the negative electrode, C moves up to the positive electrode, and D and E don't move. See diagram in envelop outside my door.
7. 2D-electrophoresis of peptides + perfomic acid treatement. The electrophoresis is conducted at pH 6.5 You need to calculate the net charge on each peptide. Please use the + and - 2 rule for determining the charge on the functional group given the pKa. At 6.5, all the carboxylates should be about fully deprotonated (-1 charge) and the lys and argininine side chains fully protonated (+1 charge). Please note that for all the peptides the charge on the N term (+) and the charge on the C term (-) cancel, so that total charge can be determined by adding the charges on the side chains.
a. glu-glu-glu-gly (net charge -3)
b. asp-glu-glu-arg (net charge -2)
c. ala-asp-glu-lys (net charge -1)
d. ala-cys-arg-gly covalently (net charge +1) attached through a disulfide bond to ala-cys-asp-ala. (net charge -1); Therefore net charge on disulfide-linked peptide is 0.
e. thr-tyr-lys-gly (net charge +1)
f. lys-arg-pro-ile (net charge +2)
Therefore, a-c are negative and will move to the anode (+ electrode), d will stay in place, and e-f, which are positive, will move to the cathode (- electrode). During the first electrophoresis, the peptides will move horizontally either right or left from the center spot. The peptides will be distributed in small spots from left to right (cathode to anode) of g,f,e,d (at initial spot since it will not move), c, b, and then a.
After you complete the electrophoresis (without runing the peptides off the paper), you treat the paper with performic acid fumes, rotate it 90 degrees, and electrophorses again. Performic acid will affect only the disulfide peptide, cleaving it to form:
ala-cys acid -arg-gly (net charge +0) and ala-cys acid -asp-ala. (net charge -2);
After the second electrophoresis, the spots will appear on a diagonal (since they will move the same distance in the horizontal direction in each electrophoresis) expect the peptide with the disulfide. If you knew the sequence each peptide, and also the sequence of the entire protein, then you could detemine which cys in the protein were linked by disulfides.