Download Chemistry 695C Fall 2001 Exam 1 Key

Chemistry 695C
Fall 2001
Exam 1 Key
Multiple Choice
1. In an α-helix, the R groups on the amino acid residues:
a) are found on the outside of the helix spiral.
b) generate the hydrogen bonds that form the helix.
c) stack within the interior of the helix.
d) cause only right-handed helices to form.
e) alternate between the outside and the inside of the helix.
2. Kendrew’s studies of myoglobin structure demonstrated that:
a) the α helix predicted by Pauling and Corey was not found in myoglobin; β-pleated sheet
structure was found.
b) “corners” between α-helical regions invariably contained a glycine residue, which, because of
its unique properties, cannot fit into the helix.
c) highly polar or charged amino acid residues tended to be located at the interior of the protein
d) the structure was very compact, with virtually no internal space available for water molecules.
e) the structure of myoglobin proved to be completely different from that of hemoglobin, as
expected from their very different biological roles.
3. Experiments on denaturation and renaturation after the reduction and reoxidation of the –S-Sbonds in the enzyme ribonuclease (RNase) have shown that:
a) the primary sequence of RNase is sufficient to determine the formation of a specific secondary
and tertiary structure.
b) the enzyme, dissolved in water, is thermodynamically stable relative to the mixture of amino
acids whose residues are contained in RNase.
c) native ribonuclease does not have a unique secondary and tertiary structure.
d) the completely unfolded enzyme, with all –S-S- bonds broken, is still enzymatically active.
e) the folding of denatured RNase into the native, active conformation, requires the input of
energy in the form of heat.
4. A sequence of amino acids in a certain protein is found to be –Ser-Gly-Pro-Gly-. The
sequence is most probably part of a(n):
a) β turn
b) parallel β sheet
c) α helix
d) α sheet
e) antiparallel β sheet
5. Protein S will fold into its native conformation only when protein Q is also present in
solution. However, protein Q can fold into its native conformation without protein S. Protein Q,
therefore, may function as a ____________ for protein S.
a) molecular chaperone
b) protein precursor
c) ligand
d) supersecondary structural unit
e) structural motif
6. In the binding of oxygen to myoglobin, the relationship between the concentration of oxygen
and the fraction of binding sites occupied can best be described as:
a) hyperbolic
b) sigmoidal
c) linear with a positive slope
d) linear with a negative slope
e) random
7. An allosteric interaction between a ligand and a protein is one in which:
a) two different ligands can bind to the same binding site.
b) the binding of the ligand to the protein is covalent.
c) multiple molecules of the same ligand can bind to the same binding site.
d) the binding of a molecule to a binding site affects the binding of an additional molecule to the
same site.
e) the binding of a molecule to a binding site affects the binding properties of another site on the
protein.
8. Phosphoric acid is tribasic, with pKas of 2.14, 6.86, and 12.4. The ionic form that
predominates at pH 3.2 is:
a) H3PO4
b) H2PO4c) HPO42d) PO439. A nonapeptide was determined to have the following amino acid composition:
(Lys)2,(Gly)2,(Phe)2,His,Leu,Met. The native peptide was incubated with 1-fluoro-2,4dinitrobenzene (FDNB) and then hydrolyzed; 2,4-dinitrophenylhistidine was identified by
HPLC. When the native peptide was exposed to cyanogen bromide (CNBr), an octapeptide and
free glycine were recovered. Incubation of the native peptide with trypsin gave a pentapeptide, a
tripeptide, and free Lys. 2,4-Dinitrophenylhistidine was recovered from the pentapeptide, and
2,4-dinitrophenylphenyalanine was recovered from the tripeptide. Digestion with the enzyme
pepsin produced a dipeptide, a tripeptide, and a tetrapeptide. The tetrapeptide was composed of
(Lys)2, Phe, and Gly. The native sequence was determined to be:
a) Gly-Phe-Lys-Lys-Gly-Leu-Met-Phe-His
b) Met-Leu-Phe-Lys-Phe-Gly-Gly-Lys-His
c) His-Phe-Leu-Gly-Lys-Lys-Phe-Met-Gly
d) His-Leu-Phe-Gly-Lys-Lys-Phe-Met-Gly
e) His-Leu-Gly-Lys-Lys-Phe-Phe-Gly-Met
10. How would the pKa value of a glutamic acid residue be affected if it were moved from the
surface of a protein into an interior region containing primarily hydrophobic residues?
a) pKa would be increased
b) pKa would be decreased
c) pKa would not be changed
d) cannot predict the behavior
11. By adding SDS (sodium dodecyl sulfate) during the electrophoresis of proteins, it is possible
to:
a) determine the amino acid composition of the protein.
b) preserve a protein’s native structure and biological activity.
c) determine a protein’s isoelectric point.
d) separate proteins exclusively on the basis of molecular weight.
e) determine an enzyme’s specific activity.
12. Which of the following statements about sterols is true?
a) Cholesterol is found in the membranes of all living cells.
b) The principle sterol of animal cells is ergosterol.
c) All sterols share a fused-ring structure with four rings.
d) Sterols are very soluble in water, but less so in organic solvents such as chloroform.
13. Which of the following monosaccharides is not an aldose?
a) ribose
b) D-glucose
c) fructose
d) glyceraldehyde
e) L-glucose
14. The pH of a sample of blood is 7.4. The pH of a sample of gastric juice is 1.4. The blood
sample has:
a) 5.29 times lower [H+] than the gastric juice.
b) 6 times lower [H+] than the gastric juice.
c) 6,000 times lower [H+] than the gastric juice.
d) a million times lower [H+] than the gastric juice.
e) 0.189 times the [H+] as the gastric juice.
15. The Henderson-Hasselbach equation:
a) relates the pH of a solution to the pKa and the concentrations of acid and salt.
b) allows the graphic determination of the molecular weight of a weak acid from its pH alone.
c) is equally useful with solutions of acetic acid and of hydrochloric acid.
d) employs the same value for pKa for all weak acids.
16. From the abbreviated name of the compound Gal(β1→4)Glc, we know that:
a) the glucose residue is the β anomer.
b) the galactose residue is at the reducing end.
c) C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
d) the compound is a D-enantiomer.
e) the glucose is in its pyranose form.
17. Which of the following amino acid residues is not a point of oligosaccharide attachment in
glycoproteins?
a) Thr
b) Gly
c) Ser
d) Asn
18. The buffer system used in biological systems to keep the intracellular pH constant is
a) HPO4-2/H2PO4b) HCO3-/H2CO3
c HPO4-2/H2PO4- and histidine
d) HCO3-/H2CO3 and histidine
19. Which of the amino acids below is classified as a basic amino acid?
a) only A
b) only C
c) A and C
d) C and D
See original exam for the figures.
20. Hydrophobic interactions most likely occur between which of the following R groups in
amino acids?
a) tyrosine and glycine
b) arginine and histidine
c) phenylalanine and tryptophan
d) valine and asparagine
Written Answers
1. A. Name two nonstandard amino acids that occur in proteins. By what route do they get into
proteins?
_________________________
______________________
Explanation:
1. A. Some examples are 4-hydroxyproline, 5-hydroxylysine, γ-carboxyglutamate, Nmethyllysine, desmosine, and selenocysteine. Nonstandard amino acids in proteins (other than
selenocysteine) usually result from chemical modifications of standard amino acid R groups after
a protein has been synthesized.
B. Proteins that span biological membranes also often contain α helices. Given that the
insides of membranes are highly hydrophobic, predict which amino acids would be in such a
helix.
B. The amino acids would be hydrophobic in nature.
C. If a set of identical helical segments were to form a pore through a membrane, with a
hydrophilic interior to the pore, indicate which amino acids in the following sequence you would
expect to be hydrophobic (H) and which hydrophilic (P)
C. Alpha helices from flavodoxin and alcohol dehydrogenase show how this question can be
answered.
In an α-helix with 3.6 residues per turn, each residue is displaced around the helix by 100o.
Allowable answers are therefore as shown below or close to this.
2. Explain why the structure of myoglobin makes it function well as an oxygen-storage protein,
whereas the structure and properties of hemoglobin makes it function well as an oxygentransport protein. Take into account the different oxygen concentrations in the lungs and in
actively respiring tissues.
2. The hyperbolic binding of oxygen to the single binding site of myoglobin results in a high
affinity even at the relatively low partial pressures of O2 that occur in tissues. In contrast, the
cooperative (sigmoidal) binding of O2 to the multiple binding sites of hemoglobin results in high
affinity at high partial pressures such as occur in the lungs, but lower affinity in the tissues. This
permits hemoglobin to bind O2 in the lungs and release it in the tissues.
3. A. The following reagents are often used in protein chemistry:
CNBr
dansyl chloride
Urea
6 N HCl
β-mercaptoethanol phenyl isothiocyanate
performic acid
chymotrypsin
Which one is the best suited for accomplishing each of the following tasks?
a) Determination of the amino acid sequence of a small peptide.
a) phenyl isothiocyanate
b) Identification of the amino-terminal residue of a peptide (of which you have less than 0.1 µg)
b) dansyl chloride
c) Reversible denaturation of a protein devoid of disulfide bonds. Which additional reagent
would you need if disulfide bonds were present?
c) urea; {beta}-mercaptoethanol to reduce disulfides
d) Hydrolysis of peptide bonds on the carboxyl side of aromatic residues
d) chymotrypsin
e) Cleavage of peptide bonds on the carboxyl side of methionines
e) CNBr
f) cleavage of proteins into individual amino acids
f) 6 N HCl
B. The relative electrophoretic mobilities of a 30 kD protein and a 92-kD protein used as
standards on an SDS-polyacrylamide gel are 0.80 and 0.41, respectively. What is the apparent
mass of a protein having a mobility of 0.62 on this gel? Show your work. (Hint: use a graph)
B. 50 kD
You should use a plot of Log Mol Wt versus mobility to get this answer
C. The gene encoding a protein with a single disulfide bond undergoes a mutation that changes a
serine residue into a cysteine residue. You want to find out whether the disulfide pairing in this
mutant is the same as in the original protein. Propose an experiment to directly answer this
question.
C. The position of disulfide bonds can be determined by diagonal electrophoresis. The disulfide
pairing is inaltered by the mutation if the off-diagonal peptides formed from the native and
mutant proteins are the same.
4. A. Describe three of the important features of a β sheet polypeptide structure. Provide one or
two sentences for each feature.
. A. In the β sheet structure, several extended polypeptides, or two regions of the same
polypeptide, lie side by side and are stabilized by hydrogen bonding between adjacent chains.
Adjacent chains may be either parallel (with a repeat distance of about 6.5 A) or antiparallel (7 A
repeat). The R groups are often small, and protrude from either side of the β sheet.
B. Both amylose and cellulose are polymers of glucose. Explain how the manner in which the
glucose units are linked results in the very different physical properties of each polymer.
B. In amylose the glucose units are linked by α(1-4) glycosidic bonds. The α-configuration of
the linkages result in the formation of a loosely coiled structure for amylose (see Fig. 10-17
V&V). By contrast, the glucose units in cellulose are linked β(1-4) and this causes each
successive glucose residue to flip 180o resulting in an essentially straight chain polymer which
readily forms interstrand hydrogen bonds (see Fig. 10-15 V&V) and close packing of the strands
C. A mutation that changes an alanine residue in the interior of a protein to a valine is found to
lead to a loss of activity. However, activity is regained when a second mutation at a different
position changes an isoleucine residue to a glycine. How might this second mutation lead to a
restoration of activity?
C. The first mutation destroys activity because valine occupies more space than alanine, and so
the protein must take a different shape. The second mutation restores activity because of a
compensatory reduction of volume; glycine is smaller than isoleucine.
6. A. α-D-mannose is a sweet tasting sugar. β-D-mannose, on the other hand, tastes bitter. A
pure solution of α-D-mannose loses its sweet taste with time as it is converted into the βanomer. Draw the β-anomer and explain how it is formed from the α-anomer. The Fischer
projection of D-mannose is
See original exam for figure.
. A. The hemiketal linkage of the α-isomer is broken to form the open form. Rotation about the
C1 and C2 bond allows formation of the β-isomer, and a mixture of isomers results.
B . Describe and illustrate diagrammatically the difference in hydrogen bonding patterns of
parallel and antiparallel β-sheets.
B. In both parallel and antiparallel β-sheets the hydrogen bonding is between two polypeptide
strands (i.e. it is interstrand). Hydrogen bonds occur between the C=O and H-N groups of the
polypeptide backbone. Because of the arrangement of the strands the hydrogen bonding is more
linear in the antiparallel sheet compared with that in parallel pleated sheets. In the parallel sheets
the hydrogen bonds are bent significantly (see figure below).
C. The triacylglycerols are the major storage form of energy in animals. Contrast the structure of
these materials with the types of lipids found in membranes mentioning both similarities and
differences.
C. The triacyl glycerols have glycerol as a backbone, appended to three fatty acid moieties by
ester linkages. The molecule is neutral in charge and because of the long hydrocarbon chains is
highly hydrophobic. This results in clumping of the triglycerides in fat stores with exclusion of
water.
By contrast the membrane lipids are amphiphilic, with a charged head group and
appended fatty acids. The backbone can be either glycerol or sphingosine. The charged,
hydrophilic components include esterified phosphate groups plus additional components such as
amino acids, amines and sugars. The amphiphilic nature of these lipids allows for self assembly
into micelles and bilayers.