Exam 1, BICH 410, Section 503 (MWF 3-3:50), Monday, September 28, 1998

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.
gas constant R 8.315 J/mol-K

1. (15 pts) (A) Draw the structure of the dipeptide valine-glutamic acid as it would exist at pH7. You do not need to depict a particular stereochemistry. Clearly identify the bonds that are described by a phi angle and psi angle.
(B) What is the net charge on this dipeptide at pH 1? pH 12?

Answer:(A) see book or answer key; (B) At pH 1, all the groups are protonated, so +1 from amino terminus and zero from carboxyl groups, net charge = +1; at pH 12, all groups de-protonated, so -1 from each carboxylate and zero from amino group, net charge = -2

2. (8 pts) Draw the backbone structure of a portion of 2 strands of an antiparallel beta sheet aligned to show three hydrogen bonds between the strands.

Answer: from your notes, or pg. 171 in book (except you need to identify the atoms)

3. (15 pts) You are in the process of preparing a 0.09 M glycine buffer for which the final pH should be 9.6. Unfortunately, after finding that the pH of 100 mL glycine solution is actually 9.9, the pH meter malfunctions. But you do not need a pH meter to complete your task. Choosing either 1 M HCl or 1 M NaOH, how much should you add to obtain a pH of 9.6? The pKas for glycine are 2.3 and 9.6. You must show your work.

Answer: At pH 9.9, pH = pKa + log [A-]/[HA]
9.9 = 9.6 + log [A-]/[HA]
[A-]/[HA] = 2
Since glycine molarity is 0.09 with 0.1 liter, total number of moles of glycine = 0.009 moles; HA + A- = 0.009 moles, so amount of HA = 0.003 moles and amount of A- = 0.006 moles
At pH 9.6, pH = pKa, so [A-]/[HA] = 1, or we have 0.0045 moles of each form of glycine
So, to go from pH 9.9 to 9.6, need to convert 0.006 moles A- to 0.0045 moles A-, which is a difference of 0.0015 moles
Need 0.0015 moles of HCl to lower pH, which is (0.0015 moles) / (1 mole/liter) = 0.0015 liter = 1.5 milliliter

4. (15 pts) Deduce the sequence of a peptide from the following information. You must show your reasoning to obtain full credit.
i) Amino acid analysis: alanine, arginine, 2 equivalents of cysteine, glutamic acid, histidine, leucine, methionine, valine
ii) Treatment with fluorodinitrobenzene, followed by hydrolysis, demonstrated two modified amino acids, 2,4-dinitrophenylalanine and 2,4-dinitrophenylglutamic acid.
iii) Treatment with beta-mercaptoethanol yielded two peptides, a 5-mer and a 4-mer.
iv) Treatment with carboxypeptidase indicated two C-terminal amino acids, histidine and valine.
v) Treatment with trypsin yielded a 7-mer and a dipeptide that contained leucine and histidine.
vi) Treatment with cyanogen bromide yielded a 7-mer and a dipeptide that contained methionine and glutamic acid.

Answer: two peptides joined by a disulfide bond via cysteine residues on each peptide: glu-met-cys-val and ala-cys-arg-leu-his
Reasoning:
amino acid analysis (i) implies 9 amino acids, of which 2 are cys
FDNB treatment indicates two amino-termini: ala and glu; this suggests that there are 2 peptide chains
Betamercaptoethanol treatment reduces disulfide bonds; since obtain two peptides after BME treatment, this means we have two peptides joined by a disulfide; one peptide contains 4 aas and the other contains 5 aas; each peptide must have a cys residue
carboxypeptidase treatment liberates C-terminal amino acids: 2 C-termini are his and val; this observation corroborates that there are two peptide chains
trypsin cleaves on C-terminal side of arg or lys residues (our sample contains only one arg); digestion gives a 7-mer (which must be joined by a S-S bond) and leu + his; sequence of last two must be leu-his since his is a C-terminus; furthermore, leu must be preceded by arg, since that is where trypsin cuts; this peptide must contain a cys, and it must be the 5-mer, because we have accounted for 4 amino acids which do not contain one of the possibilities for the amino terminus; so we know the 5-mer is (ala or glu)-cys-arg-leu-his
cyanogen bromide cleaves on C-terminal side of met; cleavage gives a 7-mer (which must contain the S-S bond) plus glu and met; glu is at the N-terminus, so must start with glu-met-; these aas must be part of the 4-mer, which must contain a cys and the "other" C-terminal aa, which is val; so the sequence of the 4-mer must be: glu-met-cys-val
By process of elimination, the sequence of the 5-mer must be ala-cys-arg-leu-his

5. (10 pts) Draw the structure of adenosine diphosphate (ADP), pH 7.

Answer: See notes or book; one negative charge on alpha phosphate and two neg. charges on beta phosphate

6. (15 pts) Use this information to answer parts (A) and (B) of this problem.
(1) fructose-6-phosphate + Pi <-> fructose-1,6-bisphosphate; deltaG°' = +17 kJ/mole
(2) ATP <-> ADP + Pi; deltaG°' = -31 kJ/mole
(A) What is the equilibrium constant for the top reaction (1)?
(B) In the glycolytic pathway, reactions (1) and (2) are coupled to give the reaction:
fructose-6-phosphate + ATP <-> fructose-1,6-bisphosphate + ADP
Calculate the actual free energy change (deltaG) at 37°C for this coupled reaction when the concentrations of each reactant and product are: [fructose-6-phosphate] = 0.4 mM, [fructose-1,6-bisphosphate] = 0.2 mM, [ATP] = 3 mM, [ADP] = 1 mM.

Answer: (A) deltaG°' = -RT ln Keq
Rearranged, Keq = e-deltaG°' / RT
Keq = e - (17 kJ/mole)(1000 J/kJ) / (8.315 J/mole-K)(310 K)
Keq = e -6.595 = 1.37 X 10-3

(B) For coupled reaction, deltaG°' = +17 kJ/mole + (-31 kJ/mole) = -14 kJ/mole
deltaG = deltaG°' + RT ln [F-1,6-BP][ADP] / [F-6-P][ATP]
deltaG = -14 kJ/mole + (8.315 J/mole-K)(310 K) / (1000 J/kJ) ln (.0002)(.001)/(.0004)(.003)
deltaG = -14 kJ/mole + (2.58)ln.167 = -14 +(-4.6) = -18.6 kJ/mole

7. (22 pts) Answer the following questions:
A) What is the relative elution from a gel filtration column for three proteins with molecular weights of 32,000, 58,000, and 98,000? 1st________ 2nd________ 3rd_________
Answer: 98,000 first, then 58,000, then 32,000
B) Which amino acid has the lowest isoelectric point (pI)?: methionine, tyrosine, or glutamic acid? _____________________
Answer: glutamic acid
(C) What is the specific activity of a protein preparation (10 ml) containing 50,000 Units of activity and a total protein concentration of 5 mg/ml? ____________________
Answer: 50,000 Units / (10 ml)(5 mg/ml) = 1000 U/mg
(D) Name an amino acid whose sidechain could form a salt-bridge with the sidechain of arginine ____________________.
Answer: glutamic acid or aspartic acid
(E) Predit the type of ion exchange column that would bind the dipeptide histidine-valine at pH 5 (cation-exchange or anion-exchange) _____________________
Answer: at pH 5, net charge is +1; so cation-exchange
(F) A conservative amino acid substitution for tryptophan is ________________.
Answer: best answers would be phenylalanine or tyrosine, but any non-polar side-chain amino acid would be OK
(G) A three amino-acid repeat that is likely to be found in the primary sequence of collagen is _______________________________________________________.
Answer: glycine-X-Y, where X=proline and Y=hydroxyproline
(H) The length (in nm or Å) of an alpha helix of 18 amino acid residues is ____________.
Answer: 18 amino acids / 3.6 amino acids per turn = 5 turns; (5 turns)(0.54 nm per turn) = 2.7 nm
(I) Name three chemical or physical treatments that are likely to denature a protein: _____________________________________________________________
Answer: heat, pH (low or high), aliphatic alcohols, urea, guanidine-HCl, SDS
(J) Write the mathematical expression that defines the energy charge:
Answer: energy charge = [ATP] + 1/2[ADP] / [ATP] + [ADP] + [AMP]