Exam 1, BICH 410 (MWF 3-3:50P), Monday September 27, 1999

Write your name on each page. Write concise answers to demonstrate effectively your mastery of the subject. Show your work in order to receive partial credit where applicable.
gas constant R 8.315 J/mol-K

1. (15 pts) Consider the following reaction:
glucose + ATP <-> glucose-6-phosphate + ADP, deltaGo' = -20.9 kJ/mole
A) What is the equilibrium constant for this reaction at 25oC?
rearrange eqn: Keq = e-deltaGo'/RT
Keq = e-(-20.9 kJ/mole)(1000 J/kJ) / (8.315 J/mole-K)(298 K) = e+8.435 = 4600

B) Under physiological conditions for the above reaction, deltaG = -27.2 kJ/mole. If the concentrations of glucose and glucose-6-phosphate are 2.0 mM and 1.58 mM, respectively, what is the ratio of [ATP] / [ADP] at 25oC?
rearrange eqn: [ATP]/[ADP] = [G6P]/[G] e-((deltaG - deltaGo') / (RT))
[ATP]/[ADP] = (1.58 mM / 2.0 mM) e-(-27.2-(-20.9) kJ/mole)(1000 J/mole) / (8.315 J/moleK)(298 K)

2. (12 pts) Titration and isoelectric point for arginine
A) Sketch the titration curve for the amino acid, arginine. Label the axes on your diagram. In addition, indicate the positions on your curve that correspond to the pKa's (alpha-carboxyl: 1.82; alpha-amino: 8.99; guanidino: 12.48).
B) Calculate the isoelectric point for arginine.
Answer: There are three ionizable groups: carboxyl, amino, and guanidino
need to average the two pKa's over which the net charge on arg goes from positive to negative
estimate net charge at several pH values using the following analysis:
grouppH 1pH 6pH 10pH 14
carboxyl0-1-1-1
amino+1+100
guanidino+1+1+10
net+2+10-1
Hence, pI = average of two high pKa's
pI = (8.99 + 12.48) / 2 = 10.74

3. (8 pts) Draw the structures of the complete sidechains (R groups) of tyrosine and glutamine engaged in an optimally-oriented hydrogen bond interaction in which the tyrosine sidechain is the donor.
Answer: see hard copy answer key; -OH of tyrosine is donor; acceptor can be either carbonyl O of gln sidechain or N of gln sidechain; three atoms in H-bond (-OH, and O or N) must be in straight line for strongest bond.

4. (15 pts) Starting with 1 liter of 0.03 M acetate buffer, pH 4.46, your goal is to adjust the pH to 5.06. Using your choice of either 1 M HCl or 1 M NaOH, what volume of which solution must be added to make this pH adjustment? The pKa for the ionization of acetic acid is 4.76.
Answer:Strategy: Use Henderson-Hasselbalch equation to calculate ratio of OAc- and HOAc at each pH. Then calculate moles of each form of acetate at each pH. Then calculate number of moles of NaOH needed to convert HOAc to OAc- (which is equal to difference in acetate between pH's).
HOAc <-> OAc- + H+
pH = pKa + log [OAc-]/[HOAc]
[OAc-]/[HOAc] = 10pH-pKa
One liter of 0.03 M acetate means 0.03 moles total HOAc + OAc-
At pH 4.46: [OAc-]/[HOAc] = 104.46-4.76 = 10-0.3 = 0.5
Hence, 0.01 moles OAc- and 0.02 moles HOAc
At pH 5.06: [OAc-]/[HOAc] = 105.06-4.76 = 10+0.3 = 2.0
Hence, 0.02 moles OAc- and 0.01 moles HOAc
To go from pH 4.46 to pH 5.06, need to convert 0.02 moles HOAc to 0.01 moles HOAc, which is a difference of 0.01 moles.
Therefore, need to add 0.01 moles NaOH or 0.01 moles / 1 mole/L = 0.01 L = 10 mL of 1 M NaOH

5. (6 pts) Sketch the pattern of mobilities when the following proteins are separated by electrophoresis on a single 2D gel (first dimension: isoelectric focusing; second dimension: SDS polyacrylamide gel). Be sure to label the gel dimensions and the identity of each protein spot.
protein pIsize
A 6.5 one subunit of 30,000
B 9.0 two subunits: 20,000 and 40,000
C 6.5 one subunit of 50,000
D 5.0 two subunits: 30,000 and 50,000

Answer:see hard copy answer key; remember that SDS separates the subunits and on an SDS gel, the bigger polypeptides run slower.

6. (12 pts) Consider the peptides labeled A-E, below. Choose one peptide that best answers each question. A given peptide can be used in more than one blank.
A. leucine-methionine-valine-tryptophan
B. glycine-histidine
C. proline-aspartic acid-glutamine
D. lysine-isoleucine-glycine-tyrosine-asparagine-arginine
E. phenylalanine-serine-cysteine-alanine

_____ has the lowest pI.
_____ contains an imidazole functional group.
_____ could form a disulfide bond with another molecule of itself.
_____ predicted to bind tightest to a hydrophobic interaction chromatography resin.
Answer:A; contains most hydrophobic amino acids with no polar sidechains
_____ could be split with cyanogen bromide.
_____ predicted to elute first from a gel filtration column.
Answer:D; largest peptide (6 amino acids)

A. Name two amino acids that are prevalent in silk fibroin: _________________, ________________
B. Approximate length in nm or angstroms of an alpha helix containing 5 turns:
Answer: 0.54 (or 0.56) nm/turn X 5 turns = 2.7 (or 2.8) nm, or 27 angstroms
C. Name two amino acids that are not likely to exist in an alpha helix: _________________, ________________
D. Thermodynamic value that can be obtained from the slope of a van't Hoff plot:
E. pH of 10-9 M NaOH:
Answer: Here is the answer I was looking for, but as pointed out by a student, the dissociation of water (Kw) would contribute more OH-, so the answer should be pH 7.
pOH = -log (10-9) = 9; pH = 14 - pOH = 14 - 9 = 5
F. Draw the structure of the sidechain of de-protonated cysteine:
Answer: see key; neg.-charged thiolate anion
G. Draw the structure of the sidechain of gamma-carboxy glutamic acid:
Answer: see key; contains two carboxlate groups on gamma carbon
H. Trypsin cleaves a polypeptide chain to the C-terminal side of these two amino acids: _________________, ________________
I. Name or structure of a reagent that can be used to modify (alkylate) a free thiol group of cysteine: _________________