Part I of this midterm is to be completed using only two 8.5 x 11 inch pages of your hand-written notes. Closed book, closed internet etc. Complete all problems. Time Limit 1 hour in class

Open notebooks, textbook chapters 1-10, lab reports, handouts for laboratory and lecture. Closed library, internet. You may need to construct a graph or consult a periodic table. You will need to consult the two articles on lactalbumin. Time Limit 2 hours (or extra time as negociated). DUE Tuesday, March 6 at 5 PM in Room 202.

You will need to use a calculator and you may word process your answers. Time limits do not apply to time needed for printing answers. No consultations are permitted; if a question is unclear, please call me in my office (610-526-5107) or at home (610-527-8054) or restate the question in a clearer manner.

Practice precise use of scientific terminology. Get in the habit of drawing structures.

On numerical problems think about an intuitive approach to check your answer.

If the question calls for cause and effect reasoning make sure you have done more than restate the question.

Correct answers are in red.

CLOSED BOOK PORTION--You may consult ONLY 2 sheets of your handwritten notes.

Question #1-- -Closed Book--10 points

Multiple Choice-Circle the best answer

1 All tRNAs adopt more or less the same overall L-shaped structure because

A their codon must base pair to the messenger RNA.

B Charging with an amino acid is done by the same synthetase enzyme.

C All charged tRNAs must enter the ribosome, base pair with the mRNA, and allow formation of the new peptide bond.

D Different amino acids get attached to the free 5' end.

2 Enzymatic DNA synthesis occurs in the 5' to 3' direction because

A DNA is metastable so its hydrolysis is favored.

B The 5' end of DNA is less reactive.

C The 3' end of the growing chain attacks the innermost phosphate of the incoming nucleotide triphosphate.

D Many anticancer and antiviral drugs contain nucleotide triphosphates whose 3' OH groups have been replaced with other functional groups.

3 The standard state energies of hydrolysis of deoxyribo-ATP and ribo-ATP are the same because

A The sodium / potassium /ATPase pump uses the two forms interchangeably

C Whether the sugar is 2 'endo or 3' endo doesn't matter.

D ribo-ATP has an additional reactive group, the 2' OH.

4 Dinitrophenol (DNP) destroys proton gradients across membranes and thereby decouples ATP synthesis from respiration. At low pH, DNP is protonated and electrically neutral, whereas at higher pH it is deprotonated. In order to explain DNP's proton gradient destroying ability, what must be true?

A The protonated form is more soluble in hydrophobic substances than the deprotonated form.

B The deprotonated form is more soluble in hydrophobic substances than the protonated form.

C DNP "carries" protons from a basic to an acidic milieu.

D Lowering the temperature would increase the destruction of the proton concentration gradient.

5 X-ray crystallography has been helpful to those studying muscle contraction because

A it produces "molecular movies" that show the molecular details of muscle contraction.

B Its shows the light and dark banding patterns of the sarcomere.

C It helps measure the "power" of the myosin headpiece power stroke.

D It has solved the atomic level structure of small proteins such as actin-G and myosin S1 fragments.

Draw the elution curve (A₂₈₀ vs. fraction number) obtained by passing a mixture of the following proteins through a Sephadex G-100 column.

Catalase (222,000 Da.)

Lysozyme (14,314 Da.)

Serum albumin (68,500 Da.)

Chymotrypsinogen (23,240 Da.)

Protein elute in REVERSE order of size--Catalase is first, then Serum albumin, Chymotrypsinogen, and last lysosyme.

Explain the purpose of the following components in the different solutions (in one sentence or less):

a) Coomassie G-250 in Bradford reagent

Reacts with protein to produce a colored complex that has an absorbance at 590 nm.

This is used for protein quantitation.

b) CuSO₄ in IMAC preparation solution

Cu(II) will bind to the stationary phase to form a metal binding affinity matrix.

c) Imidazole in elution buffer from IMAC (immobilized metal affinity chromatography) during alpha-lactalbumin isolation

Causes elution of LA from the column by binding to the Cu(II) ions and disrupting the metal-protein complex.

d) SDS in SDS-PAGE running and sample buffers

Denatures proteins in the sample and provides them with a negative charge which is proportional to the protein mass (or size).

e) Beta-mercaptoethanol in SDS-PAGE sample buffer

Reduces disulfide bonds.

f) Bromophenol blue in SDS-PAGE sample buffer

Tracking dye--indicates the extent of migration during electrophoresis.

g) Glycerol in SDS-PAGE sample buffer

Due to its high density, it helps to maintain samples in the wells (and not floating away).

h) Coomassie blue in gel staining solution

Stains proteins and enables us to see the different protein bands in the gel.

The smooth muscle myosin from chicken contains protein contain a salt bridge or electrostatic contact between Arg 245 and Glu 268. A salt bridge must be correctly formed between positions 245 and 268 in order for ATP hydrolysis to occur.

a) Draw a chemical structure in which you clearly indicate how the salt bridge is formed.

ARG-----CH2-NH-C(NH2)=NH2⁺ …..^-OOC----Glu

At the very least draw the functional groups which have + and - charges and show electrostatic bond.

b) A mutant in which residue 245 is mutated to Glu is constructed (and the rest of the protein remains the same). Will ATP hydrolysis occur?

NO. The salt bridge cannot form between two GLU^-.

c) A mutant in which residue 268 is mutated to Arg is constructed (and the rest of the protein remains the same). Will ATP hydrolysis occur?

NO. The salt bridge cannot form between two ARG⁺.

This is known as a compensatory mutation. If only a salt bridge is required for activity and neither mutation causes additional structural changes, ATP hydrolysis will occur and the myosin will indeed function.

Your textbook (p. 131) lists the pKa of Tyrosine as 10.1 whereas no value is listed for either Serine or Threonine.

a) What percentage of tyrosine is ionized at pH = 10.1?

50% by definition of pKa

b) What percentage of tyrosine is ionized at pH = 7.0?

0.079 %.

Intuitively, we are 3 pH units away from the pKa so 1/1000 molecules will be ionized.

pH = pKa + log [Tyr-] / [Tyr] so 7.0 - 10.1 = log Tyr-/Tyr so 10 ^ -3.1 = Tyr-/Tyr = 7.9 x 10^-4. The fraction = 7.9 x 10^-4 / (1+ 7.9 x 10^-4)= 7.9 x 10 ^-4.

3. Using chemical structure for tyrosine and the tyrosinate ion, explain why tyrosine is much more easily deprotonated than either of the other alcoholic amino acids.

The anion is stabilized by the aromatic group. You should draw relevant resonance structures. Ser and Thr aren't aromatic, so the anions aren't as stable and the pKa are much higher.

The membrane potential across a cellular membrane is measured to be -70 milliVolts with the inside of the cell being more negative. Given that the concentration of sodium ions inside the cell is 10 mM what must be the concentration of sodium ions outside the cell in order to maintain equilibrium?

Intuitively--The membrane potential favors sodiums entering the cell so the concentration gradient must favor sodiums leaving if we are at equilibrium. Therefore the concentration of sodium outside the cell must be less than 10 mM and decidedly non-physiologic.

10 mM x e ^ -.07 x 96500 / 8.3 x 310 = 10 mM x 0.072 = 0.72 mM = [Na⁺ out]

Damaged base pairs that closely resemble Watson-Crick base pairs are among the most difficult for your DNA repair systems to detect and repair. Suppose a chemical mutagen reacts with a thymidine to produce a base is predominantly the enol tautomer.

a) Draw the enol form of T

The enol form has an OH on top (the bottom has the N bonded to the sugar) and an N adjacent to it.

b) Which base will most likely pair with the enol form of T to produce a base pair isosteric to a Watson-Crick pair? Draw this base pair.

This enol T will pair with a normal G and you should be able to draw 3 H bonds.

T OH to G keto

T N to G H-N

T keto to G amine

3. Explain how this leads to a mutation.

During replication a G will be incorporated instead of an A. In subsequent generations, DNAs derived from this strand will have a GC pair instead of an AT. The could ultimately cause proteins with point mutations.

Based on 2 articles on Lactalbumin

JBC 275, 2000, 37021-37029 and FEBS Letters 473 (2000) 269-74.

a)"LA is homologous with the c-type lysozymes and provides an example of extreme functional divergence in homologous protein with closely similar structures"

Briefly explain the importance of this sentence.

Homologous proteins usually share a common ancestor and have similar structures and functions. Here is an example of divergent evolution where related proteins have similar structures, but different functions.

Refute. The molten glubule is an intermediate in protein folding and in the all-or-none model there are no intermediates.

*Data are accurate to within ± 0.05% which corresponds to about ± 11 Da.

*Taking into account the uncertainty, peaks 1 and 2 have the same mass which is expected since iodoacetaminde reacts only with reduced thiol groups.

*The mass in 3 is increased by an amount consistent with 2 CH2CONH2 groups.

We expect an increase of 58 x 2 Da = 116 Da.

23209 - 23109 or 23099 = 100 Da or 110 Da so this can only be 1 -S--S- linkage.

Problem 9 continues…

Figure B--SDS gel with Molecular Weight Standards on the Left.

Aging the protein and adding Cu (II) convert the protein to the disease-associated beta sheet form. Production of the beta sheet form is typically associated with increased resistance to protease K (PK) hydrolysis. In Figure B below, samples were incubated for various times with or without Cu(II) then digested or not with protease K. All protease digestions were done for the same amount of time.

b) What is the approximate molar mass of the protease resistant fragment?

About 12 kD

c) Why do half the lanes labelled "+" have no bands visible?

Protease digestion hydrolyzes the protein completely and the small peptides which remain either run off the bottom of the gel or stain very poorly.

d) What may you conclude from this experiment?

Both Cu(II) and incubation for at least 6 hours are required for development of the protease-resistant form.

Asparagine residues from aged protein samples sometimes deaminate to form aspartic acid. To test whether this conversion occurred, the protein was digested with trypsin and then analyzed by mass spectroscopy. The authors maintain that the peaks at 475.7 and 476.7 represent tryptic fragments containing Asn and Asp respectively.

Problem 9 continues…

Figure C--Mass spectroscopy of trypsin fragments.

3. Do you agree that there should be a mass difference of 1 Dalton (or g/mol)? Support your argument with chemical structures and calculations.

If Asp is protonated (as it is under the acidic conditions used) it is 1 Da heavier than Asn.

f) Why was it necessary to digest the protein with trypsin?

Given the uncertainty in the measurement, in order to detect a difference of 1 Da, the MW needs to be quite small.

3. Is there a connection between the increased effect of Cu (II) and the conversion to Asp? Use your chemical intuition and a drawing.

Lamprey, a primitive vertebrate, has hemoglobin which differs significantly from the model discussed in class. Hemoglobin binds oxygen as a monomer and there is no Bohr effect for the oxygen binding to the monomer. However, the deoxygenated hemoglobin monomers bind two protons upon dimerization.

This mutant is not cooperative and is a monomer incapable of dimerization. It is always in the strong binding state.

The pKa of the interface Glutamic acids is 6.0. Explain why the pKa for Glu is abnormal.

Normally the pKa of Glu is around 4.5. Here the uptake of protons is facilitated by the cluster of negative charges.