Multiple Choice 1. Citric acid H+ + citrate- Citrate- H+ + citrate2- Citrate2- H+ + citrate3- Acids pKs at 25C Citric acid 2.15 Citrate- 4.76 Citrate2- 5.40

Given the above information, which statement(s) is/are most true if the pH of a citric acid solution is 5.08? a. [citrate-] > [citrate2-] b. [citrate-] = [citrate2-] c. [citrate-] < [citrate2-] d. [citrate2-] > [citrate3-] e. Both a and d

2. N-2-Hydroxyethylpiperazine-N-2-ethanesulfonic acid (HEPES) has a pK of 7.47 in water. What

would happen if HEPES was dissolved in formamide? Dielectric constant of water = 78.5 Dielectric constant of formamide = 110.0 a. The pK of HEPES decreases b. The pK of HEPES increases c. HEPES ions coalesce to form a salt d. Both a and c e. Both b and c 3. As climbers approach the summit of a mountain they usually increase their rate of breathing to

compensate for the thinner air due to the oxygen pressures at higher elevations. This increased ventilation rate results in a reduction in the levels of CO2 dissolved in the blood. Which of the following accurately describes the effect of lowering the [CO2]dissolved on blood pH?

a. It will result in an increase in the dissociation of H2CO3 H+ + HCO3- and a drop in the pH. b. It will result in an increase in the dissociation of H2CO3 H2O + CO2 and a drop in pH. c. It will result in an increase in the association of H+ + HCO3- H2CO3 and an increase in pH. d. It will result in a decrease in the association of H+ + HCO3- H2CO3 and a decrease in pH. e. Lowering [CO2]dissolved has no effect on blood pH.

4. Histidine in an aqueous solution with a pH of 8 a. has a net charge of -1 b. has a net charge of zero c. has a net charge of +1 d. is a zwitterions e. both b and d

5. In gel filtration, a. large molecules have a longer path through the column. b. excess ligand is added to free the protein from the column. c. proteins are separated based on charge. d. an electric field is applied that causes migration of ions. e. the size of a protein can be analyzed. 6. In two-dimensional gel electrophoresis, what properties of proteins are exploited? a. density b. ionic charge c. molecular size d. both a and b e. both b and c 7. DEAE-cellulose a. is a ligand b. polymerizes to form a gel c. is a detergent d. is an anion exchanger e. is a cation exchanger 8. CYTKAVMWEFRDGMQKPNS You exposed the polypeptide to cyanogen bromide. You also exposed the polypeptide to trypsin. What fragments would you expect to find?

a. AVMWEFR, CYTK, CYTKAVM, DGMQK, PNS, QKPNS, WEFRDGM b. AVMWEFR, CY, CYTK, DGMQK, DGMQKPNS, EF, R, PNS, TKAVMW c. CY, CYTKAVM, DGMQKPNS, EF, R, QKPNS, TKAVMW, WEFRDGM d. CY, CYTKAVM, EF, RDGMQKPNS, QKPNS, TKAVMW, WEFRDGM e. CY, DGMQKPNS, EF, R, RDGM, M, QKPNS, TKAV, TKAVMW, W

9. Most screens for finding protein crystallization conditions are done by the

____________ method, mixing about _______ microliters each of the protein and the precipitant solutions where the protein is at about ________ mg/mL. a. batch / 1/ 1 b. hanging drop / 200 / 1 c. dialysis / 20 / 20 d. hanging drop / 3 / 15 e. hanging drop / 5 / 50

10. The resolution of a typical X-ray structure reported in the literature is about

_____________, compared to the image obtained by negative stain electron microscopy that is about ____________ resolution. a. 1 nm / 10 nm b. 0.25 nm / 2.5 nm c. 0.1 nm / 1 nm d. 10 nm / 1 nm e. 0.20 nm / 0.20 nm

11. A peptide bond is all of these except: a. an ester bond b. a special type of amide bond c. a very stable type of amide bond d. formed when water is split out from an amino group and a carboxylic

acid e. a bond which displays resonance 12. All of the following statements concerning -sheets are true except: a. hydrogen bonds occur between neighboring polypeptide chains b. amino acid residues extend above and below -sheets c. -sheets exist at least numbers of two d. parallel -sheets are more stable than antiparallel -sheets e. -sheets have an inherent right-handed twist 13. Which of the following is a/are characteristic(s) of collagen? Collagen: a. is stretchy due to its helical structure b. exists as a right-handed triple helical cable c. contains crosslinked disulfide bonds between cysteine residues d. both a and b e. both b and c

14. What is the major determining factor for a proteins three-dimensional structure? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure e. hydrophobic interactions 15. The three-dimensional structure of macromolecules is formed and maintained

primarily through noncovalent interactions. Which of the following is not considered a noncovalent interaction? a. hydrogen bonds b. van der Waals interactions c. disulfide bonds d. ionic interactions e. hydrophobic interactions

16. Which of the following is not an example of a folding accessory protein? a. peptidyl prolyl cis-trans isomerase b. GroEL/ES c. lysyl oxidase d. protein disulfide isomerase e. heat shock protein 60 17. The sugar maltose contains the following simple sugars: a. galactose b. glucose c. fructose d. both a and b e. both b and c 18. In regard to the YO2 versus pO2 graph, the effect of high levels of BPG would

cause a shift to the ___________ of the ____________ curve. a. right / hyperbolic b. left / hyperbolic c. right / sigmoidal d. left / sigmoidal e. There would be no change.

19. The Levinthal golf course landscape theory of protein folding

a. shows that as a protein forms more intrachain contacts, internal energy of the protein decreases with its conformational freedom. b. shows that the protein randomly searches for its native conformation until it reaches a canyon in the energy surface and then the protein follows a particular pathway until it reaches it energy minimum. c. shows that there is a rugged energy surface containing local minima in which the folding protein can become trapped. d. shows that the protein chain randomly searches for its native conformation on a level energy surface. e. represents what is thought to be the folding funnel of real proteins.

20. Consider polysaccharides. Which statement(s) is/are true? a. Cellulase degrades the (14) linkages of glucose in cellulose. b. The (14) linkages in cellulose allow a straight chain formation

since each glucose residue can be flipped 180 relative to the previous residue.

c. Both starch and glycogen contain (14) linkages with (16) branch points. d. Both a and b e. All of the above

Written Answers 1. a) The Henderson-Hasselbalch equation is

Show how it proves that the pKa of a weak acid is equal to the pH of the solution at the midpoint of its titration. At the midpoint of a titration [proton acceptor] = [proton donor]. Therefore the Henderson-Hasselbalch equation reduces to pH = pKa + log 1 = pKa + 0 therefore pH = pKa at midpoint b) Draw a chemically accurate structure of the following peptide at pH 7.4. Gly-Pro-Gly-His

c) What % of the histidine side chain (pK = 6.04) will be protonated at this pH? Show your working. From the Henderson-Hasselbalch equation 7.4 = 6.04 + log A AH Therefore 1.36 = log A/AH A/AH = antilog 1.36 = 22.9 therefore protonated = 1/22.9 = 1/22.9 x 100% = 4.36% d) From the amino acid analysis of a single polypeptide protein you know that it contains 2 methionine residues. However, when the protein is treated with cyanogens bromide, which cleaves at Met residues, and the products are run on an SDS-Page gel only 2 fragments are observed. Provide an explanation as to why there are only 2 fragments instead of 3. Describe an experiment to determine whether your hypothesis is correct. There are a couple of possibilities that might give the result shown. a) Two of the fragments released by CNBr treatment are of the same size and thus would not separate on an SDS gel. ______________________Met__________Met__________ 10 a.a. long 10 a.a. long One way to detect whether this was the case would be to run the fragments on a native Page gel which separates fragments by charge well as size. b) The problem may contain a disulfide bridge which was not broken under the conditions of the CNBr treatment e.g., Met Met In this case three products would be obtained if the protein was subjected to disulfide bridge reduction and blocking (e.g. by iodoacetate) prior to CNBr treatment. Alternately the peptide mixture could be subjected to reduction and blocking of the disulfide after CNBr treatment. In either case SDS-Page could be used to see whether a total of 3 peptides was produced.

S S

c) Another possibility is that there is a Met residue at the C-terminus of the polypeptide In this case you would digest the polypeptide with carboxypeptidase to determine what residue is at the C-terminus. 2. a) Given that free heme binds CO 20,000 times better than it binds O2, why dont we all succumb to carbon monoxide poisoning? There are two aspects to this question. First, the normal concentration of CO in the air is very low compared with that of oxygen. Second, while free heme binds CO very well, the protein component of hemoglobin greatly reduces the relative affinity for CO compared with oxygen. b) Can you provide any reasoning for why it is important that erythrocytes are small cells. Because erythrocytes are small cells it is very much easier for oxygen to diffuse in and out of the cell to the enclosed hemoglobin molecules. If the red cells were much larger, the path for oxygen to get from blood plasma to deoxyhemoglobins in the middle of such a cell would be similarly large. Likewise for oxygen to diffuse out of the cell. A second reason for red cells to be small is that they can pass through narrow blood vessels (capillaries) deep into tissue. c) Could a myoglobin molecule substitute for one of the subunits in the hemoglobin tetramer? Why or why not? A myoglobin molecule could not effectively substitute for one of the subunits in the hemoglobin tetramer because the surface charge of myoglobin, with charged residues distributed all around the molecule, would not permit it to bind properly with the hemoglobin subunits. In hemoglobin a lot of the intersubunit interactions involve non-charged and quite hydrophobic regions of the subunits. Secondly, even if the myoglobin chain could bind it will not have the appropriate amino acid sequences at the alpha/beta contacts, which are crucial to the propagation of allosteric interactions governing appropriate oxygen binding and release. d) Draw oxygen dissociation curves for hemoglobin in the presence and absence of 2,3-bisphosphoglycerate. Be sure to label the axes.

e) Draw the structure of 2,3-bisphosphoglycerate.

3. a) Why are monosaccharides soluble in water but not in non-polar solvents? Monosaccharides typically contain many hydroxyl groups and these can form hydrogen bonds with water. The water molecules which then surround the sugars help

prevent the sugars from coming together (i.e. it keeps them in solution). By contrast, in non-polar solvents the lack of hydrophobic regions of the sugars, with which to interact (by hydrophobic interactions), is lacking b) Draw the structures of the sugars lactose and sucrose and indicate on each the glycosidic bond. and the anomeric carbon atoms. Anomeric carbons labeled a, glycosidic bonds labeled b

g

a

g

a

c) Suppose you had three polysaccharides, amylose, amylopectin, and glycogen each with the same number of monosaccharide subunits. Solutions of equal amounts of these polysaccharides are treated with the same amount of an enzyme which hydrolyzes maltose units from the non-reducing ends of the polymers. Which of the polysaccharides would be degraded most quickly? Explain why. The enzyme removes disaccharide units from non-reducing ends of the substrate. If the activity of the enzyme is at all limited by the amount of substrate available it will work fastest with glycogen, which is highly branched and has lots of non-reducing chain endings, next fastest with amylopectin, which is branched but less frequently than glycogen, and slowest with amylose which is linear and has only one non-reducing end per molecule. 4. You are given the task of purifying an important enzyme from the mitochondria of a rhinoceros liver. Give 4 different types of procedures that might be used during the purification of this protein and describe the physical basis for the protein separation by each technique. (8 points). Also indicate the approximate order in which these techniques might be used and why (2 points). Any four of the following techniques (and there may be others) is satisfactory (see pp. 75-100 Voet & Voet 3rd Edition) Salt fractionation early in purification because it is easily used with large volumes Basis = ionic strength plus ions size and hydration Organic solvents early, because it can be used with large volumes. Basis = change of dielectric constants pH early, easily used with large volumes Basis = proteins least soluble at isoelectric point Crystallization late, usually needs fairly pure proteins Basis = use of pH, salts, ionic strength which provide best conditions for crystals to form Ion exchange chromatography middle stages. It can handle a wide range of volumes and purities of protein solutions Basis = relative charge on different proteins compared with charge on chromatographic media (can include all sub-types mentioned in Table 5.2) Gel filtration (molecular size) middle to late stages, mainly because the media is expensive and is usually not very suitable for high pressures Basis = separation by size Affinity chromatography late stages because media needs to be custom made for specific protein Basis = exploits biological property of protein binding to a specific ligand. Ligand is bound to a support. Protein mixture passed over support and specific protein binds to immobilized ligand. Protein of interest later eluted from all affinity material.

Electrophoresis late stages because can only be used with small amounts of material Basis = separation by charge on protein when electric current is applied. Can have separation by size if electrophoresis carried out in presence of SDS. Isoelectric focusing late stages because only works with small volumes. Basis = separation by electric charge in an ampholyte medium. Movement of protein in the electric field ceases when protein reaches position in apparatus where pI of protein is equal to external pH. Ultracentrifugation At many stages. Centrifugation used for bulk separation of components and also as an analytical tool (for relatively pure samples). Basis = rate of sedimentation depends on mass of component and the applied centrifugal force. 5. The peptide bond in polypeptides can be described as having partial double bond character.

a) Describe and illustrate what is meant by this double bond character (5 points). Amino acids in a polypeptide are joined by a CO-NH link known as the peptide bond. The peptide group has a rigid planar structure that is a consequence of resonance interactions that give that peptide bond an approximately 40% double bond character. Whereas there is free rotation around single bonds, movement about double bonds is highly restricted.

The peptide bonds resonance energy is maximal when the O, C, N and H atoms are in a single plane.

b) How does this double bond character of the peptide bond affect the possible conformations of a polypeptide chain? (5 points)

Because of the planar character of the peptide bond, the only reasonably free movements in a polypeptide chain are around the C-N bond and the C-C bond (see Fig. 8-4). Even here rotation is restricted by steric interference between the carbonyl oxygen and the amide hydrogen (see Fig. 8-6). For these reasons only a very small amount of the possible conformations around the C-N and C-C are observed in polypeptides.

6. a) X-ray diffraction methods deal with structure factors that are characterized by two properties. One of these properties, the ___amplitude_________, can be obtained directly by experiment while the other, the __phase angle____________, needs to be calculated. b) In a typical protein structure determination using NMR methods, the final structural model is derived from distance measurements between _hydrogen____atoms separated by a distance of less than approximately _4 , 5 , 6 , or 4 or 6 ______Angstroms. c) What is the approximate length of a C-H single bond in Angstroms? In nanometers? 1.0 to 1.4 0.1 to 0.14 nm d) One of the most fundamental equations of NMR spectroscopy is the Larmor equation: = H / 2

i) Briefly define each of the terms in the Larmor equation. Frequency, gyromagnetic ratio (partial credit for a constant), magnetic field strength 6 ii) When a 1H nucleus is placed in a magnetic field, it can be in either of two allowed energy states. How much does the energy difference between the two allowed states change if the magnetic field is increased by 10 percent? The energy difference increases by 10%

iii) As a tool for determining protein structures, NMR is restricted to proteins with less than approximately how many amino acids? 200, or 250, or 300, or 400, or 200-400 (or somewhere in this range)

MetMet