Exam 1, BICH 410 (TuTh 12:45-2), Tuesday, Sept. 24, 1996


Write your name on each page. Write concise answers to demonstrate effectively your mastery of the subject material. Show your work in order to receive partial credit where applicable.

1. (15 pts) (A) Draw the structure of the tripeptide (leucine-lysine-histidine) as it would exist at pH 5. You do not have to depict the proper stereochemistry in the structure.
Answer: See structures in book. Note that at pH 5, the N-terminal amino group, epsilon amino group on lysine, and imidazole group on histidine are all protonated with positive charges. The C-terminal carboxyl group is deprotonated with a negative charge.
(B) Calculate the isoelectric pH for this tripeptide given the following pKa values
amino-terminal amino 9.0
carboxy-terminal carboxyl 2.0
side-chain amino 10.5
imidazole 6.0
Answer: First, estimate net charge for the tripeptide at a range of pH values
pHN-termlys NhisC-termsum
pH 1+++0+3
pH 3+++-1+2
pH 5+++-1+2
pH 7++0-1+1
pH 100+0-10
pH 12000-1-1

Therefore net charge changes sign between pH7 and pH12. You must average the pKa values in this range.
pI = (9.0 + 10.5) / 2 = 9.75

2. (6 pts) Which compound do you expect to be more soluble in water, ethane (CH3CH3) or ethanol (CH3CH2OH)? Explain why.
Answer: Ethanol is polar whereas ethane is nonpolar. Furthermore, the hydroxyl group on ethanol can form hydrogen bonds with water molecules, whereas ethane cannot hydrogen-bond to water. Hence, formation of ordered arrays of H2O around the ethyl group can be partially compensated by the ability of the hydroxyl to form favorable interactions.

3. (15 pts) A 500 ml sample of a 0.1 M formate buffer, pH 3.75, is treated with 5 ml of 1.0 M KOH. What is the pH following this addition? The pKa of formic acid is 3.75.
Answer: The dissociation reaction for this weak acid is:
HCOOH (HA) <-> HCOO- (A-) + H+
At pH 3.75 = pKa, [HA] = [A-]
Hence, moles HA = moles A- = (0.5)(0.1 moles/liter)(0.5 liter) = 0.025 moles
Add 5 ml of 1M KOH, or (1 mole/liter)(0.005 liter) = 0.005 moles OH-
Then, you have 0.025 - 0.005 = 0.02 moles HA, and 0.025 + 0.005 = 0.03 moles A-.
From the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]),
so, pH = 3.75 + log(0.03/0.02) = 3.75 + 0.18 = 3.93

4. (6 pts) Differentiate between configuration and conformation. Give examples to help define each.
Answer: Configuration denotes the spatial arrangement of a molecule that is determined by the presence of either double bonds, around which there is no free rotation, or chiral centers, which give rise to stereoisomers. Examples are L- and D-amino acids. Conformation refers to the spatial arrangement of substituent groups that are free to assume different positions in space, without breaking any bonds, because of the freedom of bond rotation. Examples are the staggered and eclipsed conformations of ethane that result from rotation about the C-C bond.

5. (30 pts) Fill in the blanks with an appropriate answer.
a. Net charge on any protein when electrophoresed in an SDS polyacrylamide gel (Write positive, negative or uncharged) ___________
Answer: negative
b. Two phosphate species that act to buffer the cytoplasm of cells __________, ____________
Answer: H2PO4- and HPO4-2
c. If the absorbance at 280 nm of 0.1 mM tyrosine is 0.15, what is the absorbance of a 0.4 mM solution of tyrosine at this same wavelength? __________
Answer: 0.6; concentration is 4X so absorbance is 4X
d. Name a reagent that yields a colored or fluorescent compound when reacted with a free amino group of an amino acid __________________
Answer: ninhydrin or Sangers reagent (FDNB, fluorodinitrobenzene) or dansyl chloride or dabsyl chloride or fluorescamine
e. Approximate number of amino acids in a protein of molecular weight 110,000 ___________
Answer: 1000; 110,000 divided by 110 (ave. formula wt. of an amino acid in a protein)
f. Name two amino acids that contain sulfur _______________, ________________
Answer: cysteine (cys) and methionine (met)
g. Weakest type of noncovalent interaction ___________________
Answer: van der Waals interaction
h. Equation that defines pKa in terms of the equilibrium constant, Ka ___________________
Answer: pKa = -log Ka
i. What is the pOH of 0.01 M HCl? __________
Answer: 12; pH = -log[H+] = -log (10-2) = 2; pOH = 14 - pH
j. Name for a conjugated protein that contains carbohydrate or sugar groups ______________
Answer: glycoprotein
k. An amino acid containing an amido (amide) R-group ___________________
Answer: glutamine (gln) or asparagine (asn)
l. The ionization constant of water, Kw = ___________
Answer: 10-14 M2
m. Physical basis for separation of proteins by isoelectric focusing ______________________
Answer: pI (isoelectric point or isoelectric pH)
n. The aromatic amino acid with the least polar (actually nonpolar) R-group ______________
Answer: phenylalanine (phe)

6. (8 pts) A rigorously-purified protein sample is electrophoresed on an SDS-polyacrylamide gel both before and after treatment with beta-mercaptoethanol. Before treatment, the sample exhibited three bands with approximate molecular weights of 75,000, 50,000 and 40,000. After treatment with beta-mercaptoethanol, four bands of 75,000, 40,000, 30,000 and 20,000 were detected by electrophoresis. Explain (A): Why are there three bands in the purified protein sample, and (B): Why are there four bands after mercaptoethanol treatment vs. three bands before treatment?
Answer:
(A): The pure protein exhibits 3 bands upon electrophoresis means that it is a multisubunit protein.
(B): Beta-mercaptoethanol treatment reduces disulfide bonds. The protein contains a disulfide bond that links two polypeptide chains (the 20,000 and 30,000 molecular weight subunits).

7. (10pts) An unknown dipeptide was treated with fluorodinitrobenzene (Sangers reagent) and then hydrolyzed into individual amino acids by boiling in 6M HCl. Upon separation by HPLC, it was found that both amino acids were derivatized by dinitrobenzene. Furthermore, one of the amino acids was not optically-active. From this information, deduce the probable composition of the original dipeptide, including the amino to carboxyl polarity.
Answer: Both amino acids are derivatized with FDNB. Therefore one must be lysine with a free amino group on its sidechain. The other amino acid is modified on its alpha amino group. One of the amino acids is not optically-active, meaning that it is not present in a stereoisomeric form. It must by glycine. Glycine must be present on the N-terminus in order for both amino acids to have been derivatized. So the answer is:
N-terminus: Glycine (gly) - lysine (lys) : C-terminus

8. (10pts) You are planning to separate the amino acids, arginine, leucine, aspartic acid and glutamine by ion-exchange chromatography. Using an anion-exchange polystyrene resin at pH 11.0, predict the order of elution of these amino acids. The primary criterion for separation should be charge, and a secondary consideration should be hydrophobic character. It is important to justify your predictions in your answer.
Answer:Determine net charge at pH 11 for each amino acid:
arg: -1 from carboxyl, 0 from amino, +1 from guanidino (very high pKa); Sum = 0
leu: -1 from carboxyl, 0 from amino; Sum = -1
asp: -1 from alpha-carboxyl, -1 from side-chain carboxyl, 0 from amino; Sum = -2
gln: -1 from carboxyl, 0 from amino; Sum = -1
Therefore, asp is most negative, binds tightest to resin, and elutes last. Arg is relatively most positive, so it does not bind well to the resin and elutes first.
To differentiate leu and gln, need to consider hydrophobicity. Leu is less polar than gln and therefore more hydrophobic. Therefore, leu binds to polystyrene resin more than gln and elutes later.
Overall order of elution, from first to last: arg, gln, leu, asp