Answer the following.  Please avoid writing the Great American Novel – none of these questions require more than a paragraph or so.  Show all calculations for maximum credit.  Good luck!


1)                  A hemorrhagic fever of unknown origin has broken out in Waco, right after several hundred citizens received a scratch-and-sniff advertisement for a new perfume, postmarked from a post office box in Kabul.  An emergency response team from Fort Marlene has isolated the virus and is attempting to devise a means of control.  You are a part of the team; failure to get the correct answers to the following questions will result not only in loss of points but also in the destruction of Western Civilization.  No pressure…


a.      People infected with the virus begin to produce large amounts of a protein, which destroys collagen in the connective tissues.  You have isolated a large amount of the protein and are attempting to determine the amino acid sequence.  The protein is cleaved into two fragments by treatment with cyanogen bromide (CNBr).  You have isolated one of the fragments, which can itself be cleaved by the proteases trypsin and chymotrypsin into the following fragments:















For 8 points and the salvation of humanity, what is the amino acid sequence of the fragment?






For 1 extra point, would this fragment be at the amino (NH2) terminus or the carboxy (COOH) terminus of the intact protein?  For 2 extra points, why?


Amino terminus. The protein cleaved into two fragments; this fragment ends with a methionine indicating an internal cleavage site.  CNBr cleaves after Met.


2)                  Identify the following four amino acids by name(3 pts each).and one-letter code (2 pts each).



Name: _Threonine_____


Code:  _T________

Name: __Histidine____


Code:  __H_______




Name: __Leucine______


Code: ___L________


Name: __Lysine______


Code: ___K______







3)                  Define or describe the following terms.  One or a few sentences should suffice – do not write the Great American Novel. (5 pts each)


a.      Hydrogen Bond


Bond between a hydrogen atom covalently bound to an electronegative atom and a second electronegative atom.  Relatively weak; highly directional.



b.      Gel Filtration Chromatography


Separation based on molecular size.  Beads contain pores of restricted size, allowing smaller molecules to “see” a greater effective volume than larger molecules.

c.      Entropy


S = k*lnW, where k is Boltzmann’sconstant and W is the number of possible states the system can occupy.  A measure of the randomness – or randomizability – of a system.



d.      Beer’s Law (hint – ultraviolet)


I = 10-ecl


A = ecl

Where e is the extinction coefficient, c is the concentration, and l is the path length


e.      Peptide Bond


Condensation product between carboxyl group of one amino acid and the amino group of another.  Partially double bond in nature and cannot rotate



f.        Secondary Structure


Local folding of a protein into a set of characteristic stuctures (a-helix, b-sheet, etc.) based on hydrogen bonds formed between backbone atoms

(carbonyl oxygens and amide nitrogens).



g.      Edman Degradation (hint: Phenyl Isothiocyanate)


Method used in chemical sequencing of a protein.  PITC reacts quantitatively with the amino terminus of a peptide to cleave off the terminal amino acid.  Reaction can be repeated to cleave and identify one amino acid at a time.



4)                 Mustard gas, Cl-CH2-CH2-S-CH2-CH2-Cl (also known as HS or Yperite) is an inexpensive, easily manufactured and highly effective blistering agent.  Although fatalities are rare, victims frequently suffer blindness, agonizing burns, long term respiratory damage, permanent incapacitation, and drastically reduced lifespans.  The agent is also persistent - casualties can result from individuals taking shelter in contaminated soil.  You have discovered an enzyme from a rare species of Bolivian poison dart frog that inactivates mustard gas, converting it into an effective deodorant.  The battlefield uses are obvious.


Your initial extract contains several proteins with the following properties:



Molecular Weight




























a.      I apply the mixture to a chromatographic column containing an an anion exchanger (Diethylaminoethylcellulose) buffered at pH 6.5.  Based on your knowledge of pH and pI, and of ion-exchange chromatography, which proteins will stick to the column at this pH? (5 pts)


B, C and F have a negative charge at pH 6.5 and will stick to the column


b.      SNEAKY QUESTION ALERT!  We talked about gel electrophoresis of proteins in the presence and absence of sodium dodecyl sulfate.  If I electrophorese the mixture in the absence of SDS, I get 6 separate bands of protein on the gel.  If I electrophorese the mixture in the presence of SDS, I only see 4 protein bands.  Why would that be true?  If you get this one, you get a gold star. (and 4 pts)


A and F, and D and E, have the same molecular weights (83,000 and 33,000 respectively).  In the absence of SDS, they migrate according to their individual charge-to-mass ratios and separate.  In the presence of SDS, they migrate according to their molecular weights and do not separate.  One band is seen for both A dnd F, and one band for D and E.


c.      The purified inhibitor is electrophoresed on a gel.  Lane 1 contains Protein E plus Sodium Dodecyl Sulfate (SDS).  Lane 2 contains Protein E, SDS, and b-mercaptoethanol.




The gel is seen below:


         1            2





1: SDS, - b-ME

2. SDS, + b-ME











What does the fact that lane 2 contains 1 band of protein while lane 3 contains 3 bands signify?  (i.e. What is the effect of b-mercaptoethanol and what does it tell us about Protein E?) (Straight off the practice test – should be a freebie!) (4 pts)


Protein E contains three separate chains linked by disulfides.  These disulfides are reduced by BME, allowing the chains to separate.




5)                  For the conversion of phosphoenolpyruvate to pyruvate:

under normal cellular conditions, DG ≈ -42 kJ/mol. 


a.      Would you expect the reaction to spontaneously proceed to the right or to the left under cellular conditions?  Why?  (5 pts)



                        To the right; DG < 0



b.      What is DGo  Keq at 37oC  (5pts)  (TYPO ON TEST)




6)                  CAPS (3-(Cyclohexylamino)-1-propanesulfonic acid) is a common biochemical buffer that is frequently used when experimental conditions require a high pH.


Molecular Weight: 221.32 g/mol

pKa 10.4


I wish to make 3 liters of 30 mM (0.03 molar) CAPS buffer pH 11.0.  I have a jar of solid CAPS free acid, a stock solution of 10 M NaOH, a 4-liter graduated cylinder, and all the distilled water I can possibly want.  Tell me how to make the buffer.  (10 pts)


(a)   3l * .03 mol/l * 221.32 g/mol = 19.92 g CAPS

(b)   pH = pKa + log[base]/[acid]

11 = 10.4 +  log[base]/[acid]

0.6 = log[base]/[acid]

3.98 = [base]/[acid]

      [base]+[acid] = .03 M

      4.98 [acid] = .03M

      [acid] = 0.006M

      [base] = .03 - .006 = 0.024M


      Need to add .024M NaOH

(.024 M/ 10M) * 3l = .0072l = 7.2 ml of 10M NaOH

                        Water to 3l




7)                  The town of Langtry, Texas was originally named “Vinegaroon” after a common southwest Texas arachnid called Mastigoproctus giganteus.  Besides their rather alarming appearance, vinegaroons can defend themselves by squirting 450 mM acetic acid on their enemies.  (In real life, the concentration is much higher – 10 to 11 M - but that would have messed up the question.)



(about life size for adult female)

You have just been sprayed in the eyes.  Calculate the pH of the solution while you wait for the blinding pain to stop.  Assume that the pKa for CH3COOH is 4.76 . (7 pts)



Ka = [H+]{Ac-]/[HAc]

Ka = [H+]2/[HAc]


Ka = 10-pKa since pKa = log Ka

Ka = 10-4.76 = 1.74*10-5



[H+] = (Ka*[HAc])1/2

[H+] = .00246M

pH = -log[H+] = 2.61


Useful Physical Constants




Avagadro’s Number

6.02 * 1023

Boltzmann’s Constant

1.3807 * 10-23 JK-1

Charge on electron

-1.602 x 10-19 coulomb

Gas Constant (R)

8.314 JK-1mol-1

Faraday’s Constant

96,485 JV-1mol-1

k = 1/(4pe)

8.99 x 109 Nm2coulomb-2


2.303 log (x)

Dielectric constant of water