Lehninger Principles of Biochemistry 5th Edition by David L. Nelson -Test Bank
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Sample Test
Chapter
3 Amino Acids, Peptides, and Proteins
Multiple
Choice Questions
1. Amino
acids
Page: 72 Difficulty: 1
Ans: C
The chirality of an amino acid results from the fact that its a
carbon:
1. has
no net charge.
2. is a
carboxylic acid.
3. is
bonded to four different chemical groups.
4. is in
the l absolute configuration in naturally occurring proteins.
5. is
symmetric.
2. Amino
acids
Page: 72 Difficulty: 2
Ans: B
Of the 20 standard amino acids, only ___________ is not
optically active. The reason is that its side chain ___________.
1. alanine;
is a simple methyl group
2. glycine;
is a hydrogen atom
3. glycine;
is unbranched
4. lysine;
contains only nitrogen
5. proline;
forms a covalent bond with the amino group
3. Amino
acids
Page: 72 Difficulty: 1
Ans: C
Two amino acids of the standard 20 contain sulfur atoms.
They are:
1. cysteine
and serine.
2. cysteine
and threonine.
3. methionine
and cysteine
4. methionine
and serine
5. threonine
and serine.
4. Amino
acids
Page:
75 Difficulty: 1 Ans: A
All of the amino acids that are found in proteins, except for
proline, contain a(n):
1. amino
group.
2. carbonyl
group.
3. carboxyl
group.
4. ester
group.
5. thiol
group.
5. Amino
acids
Pages: 75–76
Difficulty: 3 Ans: C
Which of the following statements about aromatic amino acids is
correct?
1. All
are strongly hydrophilic.
2. Histidine’s
ring structure results in its being categorized as aromatic or basic, depending
on pH.
3. On a
molar basis, tryptophan absorbs more ultraviolet light than tyrosine.
4. The
major contribution to the characteristic absorption of light at 280 nm by
proteins is the phenylalanine R group.
5. The
presence of a ring structure in its R group determines whether or not an amino
acid is aromatic.
6. Amino
acids
Page: 77 Difficulty: 2
Ans: A
Which of the following statements about cystine is correct?
1. Cystine
forms when the —CH2—SH R group is oxidized to form a —CH2—S—S—CH2—
disulfide bridge between two cysteines.
2. Cystine
is an example of a nonstandard amino acid, derived by linking two standard
amino acids.
3. Cystine
is formed by the oxidation of the carboxylic acid group on cysteine.
4. Cystine
is formed through a peptide linkage between two cysteines.
5. Two
cystines are released when a —CH2—S—S—CH2—
disulfide bridge is reduced to —CH2—SH.
7. Amino
acids
Page: 77 Difficulty: 2
Ans: A
The uncommon amino acid selenocysteine has an R group with the
structure —CH2—SeH (pKa »
5). In an aqueous solution, pH = 7.0, selenocysteine would:
1. be a
fully ionized zwitterion with no net charge.
2. be
found in proteins as d-selenocysteine.
3. never
be found in a protein.
4. be
nonionic.
5. not
be optically active.
8. Amino
acids
Pages: 78–79
Difficulty: 1 Ans: A
Amino acids are ampholytes because they can function as either
a(n):
1. acid
or a base.
2. neutral
molecule or an ion.
3. polar
or a nonpolar molecule.
4. standard
or a nonstandard monomer in proteins.
5. transparent
or a light-absorbing compound.
9. Amino
acids
Pages: 79–80
Difficulty: 2 Ans: D
Titration of valine by a strong base, for example NaOH, reveals
two pK’s.
The titration reaction occurring at pK2 (pK2 =
9.62) is:
1. A)
—COOH + OH– ®
—COO– + H2
2. B)
—COOH + —NH2
® —COO– + —NH2+.
3. C)
—COO– + —NH2+
® —COOH + —NH2.
4. D)
—NH3+ + OH– ®
—NH2 + H2
5. E)
—NH2 + OH– ®
—NH– + H2
10. Amino
acids
Pages: 79–80
Difficulty: 1 Ans: C
In a highly basic solution, pH = 13, the dominant form of
glycine is:
1. NH2—CH2—COOH.
2. NH2—CH2—COO–.
3. NH2—CH3+—COO–.
4. NH3+—CH2—COOH.
5. NH3+—CH2—COO–.
11. Amino
acids
Pages: 80–81
Difficulty: 2 Ans: B
For amino acids with neutral R groups, at any pH below the pI of
the amino acid, the population of amino acids in solution will have:
1. a net
negative charge.
2. a net
positive charge.
3. no
charged groups.
4. no
net charge.
5. positive
and negative charges in equal concentration.
12. Amino
acids
Pages: 80–81 Difficulty:
2 Ans: A
At pH 7.0, converting a glutamic acid to g-carboxyglutamate,
will have what effect on the overall charge of the protein containing it?
1. it
will become more negative
2. it
will become more positive.
3. it
will stay the same.
4. there
is not enough information to answer the question.
5. the
answer depends on the salt concentration.
13. Amino
acids
Pages: 80–81
Difficulty: 2 Ans: C
At pH 7.0, converting a proline to hydroxyproline, will have
what effect on the overall charge of the protein containing it?
1. it
will become more negative
2. it
will become more positive.
3. it
will stay the same.
4. there
is not enough information to answer the question.
5. the
answer depends on the salt concentration.
14. Amino
acids
Pages: 80–81
Difficulty: 3 Ans: B
What
is the approximate charge difference between glutamic acid and a-ketoglutarate
at pH 9.5?
1. 0
2. ½
3. 1
4. 1½
5. 2
15. Peptides
and proteins
Page: 82 Difficulty: 1
Ans: B
The formation of a peptide bond between two amino acids is an example
of a(n) ______________ reaction.
1. cleavage
2. condensation
3. group
transfer
4. isomerization
5. oxidation
reduction
16. Peptides
and proteins
Page: 82 Difficulty: 1
Ans: C
The peptide alanylglutamylglycylalanylleucine has:
1. a
disulfide bridge.
2. five peptide
bonds.
3. four
peptide bonds.
4. no
free carboxyl group.
5. two
free amino groups.
17. Peptides
and proteins
Pages: 82–83
Difficulty: 1 Ans: C
An octapeptide composed of four repeating glycylalanyl units
has:
1. one
free amino group on an alanyl residue.
2. one
free amino group on an alanyl residue and one free carboxyl group on a glycyl
residue.
3. one
free amino group on a glycyl residue and one free carboxyl group on an alanyl
residue.
4. two
free amino and two free carboxyl groups.
5. two
free carboxyl groups, both on glycyl residues.
18. Peptides
and proteins
Page:
82–83 Difficulty:
1 Ans: C
At the isoelectric pH of a tetrapeptide:
1. only
the amino and carboxyl termini contribute charge.
2. the
amino and carboxyl termini are not charged.
3. the
total net charge is zero.
4. there
are four ionic charges.
5. two
internal amino acids of the tetrapeptide cannot have ionizable R groups.
19. Peptides
and proteins
Pages: 83–84
Difficulty: 2 Ans: C
Which of the following is correct with respect to the amino acid
composition of proteins?
1. Larger
proteins have a more uniform distribution of amino acids than smaller proteins.
2. Proteins
contain at least one each of the 20 different standard amino acids.
3. Proteins
with different functions usually differ significantly in their amino acid
composition.
4. Proteins
with the same molecular weight have the same amino acid composition.
5. The
average molecular weight of an amino acid in a protein increases with the size
of the protein.
20. Peptides
and proteins
Page: 83 Difficulty: 2
Ans: B
The average molecular weight of the 20 standard amino acids is
138, but biochemists use 110 when estimating the number of amino acids in a
protein of known molecular weight. Why?
1. The
number 110 is based on the fact that the average molecular weight of a protein
is 110,000 with an average of 1,000 amino acids.
2. The
number 110 reflects the higher proportion of small amino acids in proteins, as
well as the loss of water when the peptide bond forms.
3. The
number 110 reflects the number of amino acids found in the typical small
protein, and only small proteins have their molecular weight estimated this
way.
4. The
number 110 takes into account the relatively small size of nonstandard amino
acids.
5. The
number 138 represents the molecular weight of conjugated amino acids.
21. Peptides
and proteins
Page: 84 Difficulty: 1
Ans: C
In a conjugated protein, a prosthetic group is:
1. a
fibrous region of a globular protein.
2. a
nonidentical subunit of a protein with many identical subunits.
3. a
part of the protein that is not composed of amino acids.
4. a
subunit of an oligomeric protein.
5. synonymous
with “protomer.”
22. Peptides
and proteins
Pages: 84–85
Difficulty: 1 Ans: A
Prosthetic groups in the class of proteins known as
glycoproteins are composed of:
1.
2. flavin
nucleotides.
3.
4. metals
.
5.
23. Working
with proteins
Page: 85 Difficulty: 1
Ans: E
For the study of a protein in detail, an effort is usually made
to first:
1. conjugate
the protein to a known molecule.
2. determine
its amino acid composition.
3. determine
its amino acid sequence.
4. determine
its molecular weight.
5. purify
the protein.
24. Working
with proteins
Page: 87 Difficulty: 2
Ans: B
In a mixture of the five proteins listed below, which should
elute second in size-exclusion (gel- filtration) chromatography?
1. cytochrome c Mr =
13,000
2. immunoglobulin
G Mr =
145,000
3. ribonuclease
A Mr =
13,700
4. RNA
polymerase Mr =
450,000
5. serum
albumin Mr =
68,500
25. Working
with proteins
Page: 89 Difficulty: 2
Ans: E
By adding SDS (sodium dodecyl sulfate) during the
electrophoresis of proteins, it is possible to:
1. determine
a protein’s isoelectric point.
2. determine
an enzyme’s specific activity.
3. determine
the amino acid composition of the protein.
4. preserve
a protein’s native structure and biological activity.
5. separate
proteins exclusively on the basis of molecular weight.
26. Working
with proteins
Page: 90 Difficulty: 2
Ans: B
To determine the isoelectric point of a protein, first establish
that a gel:
1. contains
a denaturing detergent that can distribute uniform negative charges over the
protein’s surface.
2. exhibits
a stable pH gradient when ampholytes become distributed in an electric field.
3. is
washed with an antibody specific to the protein of interest.
4. neutralizes
all ionic groups on a protein by titrating them with strong bases.
5. relates
the unknown protein to a series of protein markers with known molecular
weights, Mr.
27. Working
with proteins
Pages: 90–91
Difficulty: 3 Ans: A
The first step in two-dimensional gel electrophoresis generates
a series of protein bands by isoelectric focusing. In a second step, a
strip of this gel is turned 90 degrees, placed on another gel containing SDS,
and electric current is again applied. In this second step:
1. proteins
with similar isoelectric points become further separated according to their
molecular weights.
2. the
individual bands become stained so that the isoelectric focus pattern can be
visualized.
3. the
individual bands become visualized by interacting with protein-specific
antibodies in the second gel.
4. the
individual bands undergo a second, more intense isoelectric focusing.
5. the
proteins in the bands separate more completely because the second electric
current is in the opposite polarity to the first current.
28. Working
with proteins
Page: 91 Difficulty: 1
Ans: B
The term specific activity differs
from the term activity in
that specific activity:
1. is
measured only under optimal conditions.
2. is
the activity (enzyme units) in a milligram of protein.
3. is
the activity (enzyme units) of a specific protein.
4. refers
only to a purified protein.
5. refers
to proteins other than enzymes.
29. Peptides
and proteins
Page: 92 Difficulty: 1
Ans: B
Which of the following refers to particularly stable
arrangements of amino acid residues in a protein that give rise to recurring
patterns?
1. Primary
structure
2. Secondary
structure
3. Tertiary
structure
4. Quaternary
structure
5. None
of the above
30. Peptides
and proteins
Page:
92 Difficulty: 1 Ans: D
Which of the following describes the overall three-dimensional
folding of a polypeptide?
1. Primary
structure
2. Secondary
structure
3. Tertiary
structure
4. Quaternary
structure
5. None
of the above
31. The
covalent structure of proteins
Page: 93 Difficulty: 1
Ans: B
The functional differences, as well as differences in
three-dimensional structures, between two different enzymes from E. coli result
directly from their different:
1. affinities
for ATP.
2. amino
acid sequences.
3. roles
in DNA metabolism.
4. roles
in the metabolism of coli.
5. secondary
structures.
32. The
covalent structure of proteins
Page: 95 Difficulty: 2
Ans: C
One method used to prevent disulfide bond interference with
protein sequencing procedures is:
1. cleaving
proteins with proteases that specifically recognize disulfide bonds.
2. protecting
the disulfide bridge against spontaneous reduction to cysteinyl sulfhydryl
groups.
3. reducing
disulfide bridges and preventing their re-formation by further modifying the
—SH groups.
4. removing
cystines from protein sequences by proteolytic cleavage.
5. sequencing
proteins that do not contain cysteinyl residues.
33. The
covalent structure of proteins
Pages: 96–97
Difficulty: 3 Ans: C
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-2,4-dinitrobenzene (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-Dinitrophenyl-histidine was
recovered from the pentapeptide, and 2,4-dinitrophenylphenylalanine 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:
1. Gly–Phe–Lys–Lys–Gly–Leu–Met–Phe–His.
2. His–Leu–Gly–Lys–Lys–Phe–Phe–Gly–Met.
3. His–Leu–Phe–Gly–Lys–Lys–Phe–Met–Gly.
4. His–Phe–Leu–Gly–Lys–Lys–Phe–Met–Gly.
5. Met–Leu–Phe–Lys–Phe–Gly–Gly–Lys–His.
34. The
covalent structure of proteins
Pages: 96–97
Difficulty: 1 Ans: C
Even when a gene is available and its sequence of nucleotides is
known, chemical studies of the protein are still required to determine:
1. molecular
weight of the protein.
2. the
amino-terminal amino acid.
3. the
location of disulfide bonds.
4. the
number of amino acids in the protein.
5. whether
the protein has the amino acid methionine in its sequence.
35. The
covalent structure of proteins
Page: 100 Difficulty: 1 Ans: C
The term “proteome” has been used to describe:
1. regions
(domains) within proteins.
2. regularities
in protein structures.
3. the
complement of proteins encoded by an organism’s DNA.
4. the
structure of a protein-synthesizing ribosome.
5. the
tertiary structure of a protein.
36. The
covalent structure of proteins
Pages: 98–100 Difficulty:
2 Ans: C
A major advance in the application of mass spectrometry to
macromolecules came with the development of techniques to overcome which of the
following problems?
1. Macromolecules
were insoluble in the solvents used in mass spectrometry.
2. Mass
spectrometric analyses of macromolecules were too complex to interpret.
3. Mass
spectrometric analysis involved molecules in the gas phase.
4. Most
macromolecules could not be purified to the degree required for mass
spectrometric analysis.
5. The
specialized instruments required were prohibitively expensive.
37. Protein
sequences and evolution
Pages: 102–106 Difficulty:
3 Ans: A
Compare the following sequences taken from four different
proteins, and select the answer that best characterizes their relationships.
A
B
C
1 DVEKGKKIDIMKCS
HTVEKGGKHKTGPNLH
GLFGRKTGQAPGYSYT
2 DVQRALKIDNNLGQ
HTVEKGAKHKTAPNVH
GLADRIAYQAKATNEE
3 LVTRPLYIFPNEGQ
HTLEKAAKHKTGPNLH
ALKSSKDLMFTVINDD
4 FFMNEDALVARSSN
HQFAASSIHKNAPQFH
NLKDSKTYLKPVISET
1. Based
only on sequences in column B, protein 4 reveals the greatest evolutionary
divergence.
2. Comparing
proteins 1 and 2 in column A reveals that these two proteins have diverged the
most throughout evolution.
3. Protein
4 is the protein that shows the greatest overall homology to protein 1.
4. Proteins
2 and 3 show a greater evolutionary distance than proteins 1 and 4.
5. The
portions of amino acid sequence shown suggest that these proteins are
completely unrelated.
Short
Answer Questions
38. Amino
acids
Page: 72 Difficulty: 1
What are the structural characteristics common to all amino
acids found in naturally occurring proteins?
Ans: All amino acids found in naturally
occurring proteins have an a carbon to which are attached a carboxylic acid, an
amine, a hydrogen, and a variable side chain. All the amino acids are
also in the l configuration.
39. Amino
acids
Page: 75 Difficulty: 1
Only one of the common amino acids has no free a-amino
group. Name this amino acid and draw its structure.
Ans: The amino acid l-proline has no
free a-amino group, but rather has an imino group formed by cyclization of the
R-group aliphatic chain with the amino group (see Fig. 3-5, p. 79).
40. Amino
acids
Pages: 74–77
Difficulty: 2
Briefly describe the five major groupings of amino acids.
Ans: Amino acids may be categorized by
the chemistry of their R groups: (1) nonpolar aliphatics; (2) polar, uncharged;
(3) aromatic; (4) positively charged; (5) negatively charged. (See Fig. 3-5, p.
79.)
41. Amino
acids
Pages: 73–75
Difficulty: 2
A
B
C
D
E
__________________________________________________________________
Tyr-Lys-Met
Gly-Pro-Arg
Asp-Trp-Tyr
Asp-His-Glu Leu-Val-Phe
Which one of the above tripeptides:
____(a) is most negatively charged at pH 7?
____(b) will yield DNP-tyrosine when reacted with
l-fluoro-2,4-dinitrobenzene and hydrolyzed in acid?
____(c) contains the largest number of nonpolar R groups?
____(d) contains sulfur?
____(e) will have the greatest light absorbance at 280 nm?
Ans: (a) D; (b) A; (c) E; (d) A; (e) C
42. Amino
acids
Pages: 73–75
Difficulty: 2
Draw the structures of the amino acids phenylalanine and
aspartate in the ionization state you would expect at pH 7.0. Why is aspartate
very soluble in water, whereas phenylalanine is much less soluble?
Ans: Aspartate has a polar (hydrophilic)
side chain, which forms hydrogen bonds with water. In contrast,
phenylalanine has a nonpolar (hydrophobic) side chain. (See Fig. 3-5, p. 79 for
structures.)
43. Amino
acids
Pages: 77–78
Difficulty: 3
Name two uncommon amino acids that occur in proteins. By
what route do they get into proteins?
Ans: Some examples are 4-hydroxyproline,
5-hydroxylysine, g-carboxyglutamate, N-methyllysine,
desmosine, and selenocysteine. Uncommon amino acids in proteins (other
than selenocysteine) usually result from chemical modifications of standard
amino acid R groups after a protein has been synthesized.
44. Amino
acids
Pages: 78–79
Difficulty: 1
Why do amino acids, when dissolved in water, become zwitterions?
Ans: Near pH = 7, the carboxylic acid
group (—COOH) will dissociate to become a negatively charged —COO– group,
and the —NH2 amino group will attract a proton to become a positively
charged —NH3+ group.
45. Amino
acids
Page: 79 Difficulty: 1
As more OH– equivalents
(base) are added to an amino acid solution, what titration reaction will occur
around pH = 9.5?
Ans: Around pH = 9.5, the —NH3+ group
will be titrated according to the reaction: —NH3+ +
OH– ® —NH2 +
H2O.
46. Amino
acids
Page: 80 Difficulty: 3
In the amino acid glycine, what effect does the positively
charged —NH3+ group have on
the pKa of
an amino acid’s —COOH group?
Ans: The positively charged amino group
stabilizes the negatively charged ionized form of the carboxyl group, —COO–, and
repels the departing H+ thereby
promoting deprotonation. The effect is to lower the pKa of
the carboxyl group (see Fig. 3-11, p. 80).
47. Amino
acids
Page:
79 Difficulty: 3
How does the shape of a titration curve confirm the fact that
the pH region of greatest buffering power for an amino acid solution is around
its pK’s?
Ans: In a certain range around the pKa’s of
an amino acid, the titration curve levels off. This indicates that for a solution
with pH » pK,
any given addition of base or acid equivalents will result in the smallest
change in pH—which is the definition of a buffer.
48. Amino
acids
Page: 79 Difficulty: 2
Leucine has two dissociable protons: one with a pKa of
2.3, the other with a pKa of
9.7. Sketch a properly labeled titration curve for leucine titrated with
NaOH; indicate where the pH = pK and
the region(s) in which buffering occurs.
Ans: See the titration curve for glycine
in Fig. 3-10, p. 79.
49. Amino
acids
Page: 80 Difficulty: 2
What is the pI, and how is it determined for amino acids that
have nonionizable R groups?
Ans: The pI is the isoelectric
point. It occurs at a characteristic pH when a molecule has an equal
number of positive and negative charges, or no net charge. For amino acids with
nonionizable R groups, pI is the arithmetic mean of a molecule’s two pKa values:
pI = 1/2 (pK1 +
pK2)
50. Amino
acids
Page: 80 Difficulty: 2
The amino acid histidine has a side chain for which the pKa is
6.0. Calculate what fraction of the histidine side chains will carry a
positive charge at pH 5.4. Be sure to show your work.
Ans: pH = pKa +
log
pKa –
pH = log
antilog (pKa – pH)
=
antilog (6.0 – 5.4) =
4 = [acid]/[conjugate base], or
4[conjugate base] = [acid]
Therefore, at pH 5.4, 4/5 (80%) of the histidine will be in the
protonated form.
51. Amino
acids
Page: 80 Difficulty: 2
The amino acid histidine has three ionizable groups, with pKa values
of 1.8, 6.0, and 9.2. (a) Which pKa corresponds
to the histidine side chain? (b) In a solution at pH 5.4, what percentage
of the histidine side chains will carry a positive charge?
Ans: (a) 6.0; (b) 80%. (See the
previous problem for expanded solution to this problem.)
52. Amino
acids
Page: 81 Difficulty: 2
What is the uniquely important acid-base characteristic of the
histidine R group?
Ans: Only the imidazole ring of the
histidine R group has a pKa near
physiological pH (pKa =
6.0), which suggests that histidine may provide buffering power in
intercellular and intracellular fluids.
53. Peptides
and proteins
Page: 82 Difficulty: 1
How can a polypeptide have only one free amino group and one
free carboxyl group?
Ans: This is possible only if the
peptide has no side chains with carboxyl or amino groups. Then, with the
exception of the single amino-terminal amino acid and the single
carboxyl-terminal amino acid, all the other a-amino and carboxyl groups are
covalently condensed into peptide bonds.
54. Peptides
and proteins
Page: 82 Difficulty: 1
Hydrolysis of peptide bonds is an exergonic reaction. Why,
then, are peptide bonds quite stable?
Ans: Peptide bonds are stable because
hydrolysis of peptide (or amide) bonds has a high activation energy and as a
result occurs very slowly.
55. Peptides
and proteins
Page: 82 Difficulty: 2
Draw the structure of Gly–Ala–Glu in the ionic form that
predominates at pH 7.
Ans: The peptide must have an
amino-terminal Gly residue, a carboxyl-terminal Glu residue, and ionized amino
and carboxyl groups.
56. Peptides
and proteins
Page:
82 Difficulty: 2
The artificial sweetener NutraSweet®, also called aspartame, is
a simple dipeptide, aspartylphenylalanine methyl ester, on which the free
carboxyl of the dipeptide is esterified to methyl alcohol. Draw the structure
of aspartame, showing the ionizable groups in the form they have at pH 7.
(The ionizable group in the side chain of aspartate has a pKa of
3.96.)
Ans: See the structure on p. 83.
57. Peptides
and proteins
Page: 84 Difficulty: 1
If the average molecular weight of the 20 standard amino acids
is 138, why do biochemists divide a protein’s molecular weight by 110 to
estimate its number of amino acid residues?
Ans: For each peptide bond formed, a
molecule of water is lost, bringing the average molecular weight down to
120. To reflect the preponderance of low-molecular-weight amino acids,
the average molecular weight is lowered further to 110.
58. Peptides
and proteins
Page: 84 Difficulty: 2
Lys residues make up 10.5% of the weight of ribonuclease.
The ribonuclease molecule contains 10 Lys residues. Calculate the
molecular weight of ribonuclease.
Ans: From the structure of lysine, we
can calculate its molecular weight (146); when it condenses (loses H2O, Mr =
18) to form a peptide bond, the resulting residue contributes 146 – 18 = 128 to
the protein’s molecular weight. If 10 Lys residues contribute 10.5% of
the protein’s molecular weight, each Lys residue is 1.05%. To calculate
the total molecular weight, divide 128 by 1.05% (0.0105); the result is
12,190. (The actual value is 13,700.)
59. Working
with proteins
Pages: 86-87
Difficulty: 2
Why do smaller molecules elute after large molecules when a
mixture of proteins is passed through a size-exclusion (gel filtration) column?
Ans: The column matrix is composed of
cross-linked polymers with pores of selected sizes. Smaller molecules can
enter pores in the polymer beads from which larger molecules would be
excluded. Smaller molecules therefore have a larger three-dimensional
space in which to diffuse, making their path through the column longer.
Larger molecules migrate faster because they pass directly through the column,
unhindered by the bead pores.
60. Working
with proteins
Pages: 86-87
Difficulty: 2
For each of these methods of separating proteins, describe the
principle of the method, and tell what property of proteins allows their
separation by this technique.
(a) ion-exchange chromatography
(b) size-exclusion (gel filtration) chromatography
·
affinity chromatography
Ans: (a) Ion-exchange chromatography
separates proteins on the basis of their charges. (b) Size-exclusion or
gel filtration chromatography separates on the basis of size (c) Affinity
chromatography separates proteins with specific, high affinity for some ligand
(attached to an inert support) from other proteins with no such affinity. (See
Fig. 3-17, p. 87.)
61. Working
with proteins
Pages: 86-88
Difficulty: 2
A biochemist is attempting to separate a DNA-binding protein
(protein X) from other proteins in a solution. Only three other proteins
(A, B, and C) are present. The proteins have the following properties:
pI
(isoelectric
Size
Bind to
point) Mr
DNA?
––––––––––––––––––––––––––––––––––––––––––
protein A
7.4
82,000
yes
protein B
3.8
21,500
yes
protein C
7.9
23,000
no
protein X
7.8
22,000
yes
––––––––––––––––––––––––––––––––––––––––––
What type of protein separation techniques might she use to
separate
(a) protein X from protein A?
(b) protein X from protein B?
(c) protein X from protein C?
Ans: (a) size-exclusion (gel filtration)
chromatography to separate on the basis of size; (b) ion-exchange
chromatography or isoelectric focusing to separate on the basis of charge; (c)
specific affinity chromatography, using immobilized DNA.
62. Working
with proteins
Pages: 88-89
Difficulty: 2
What factors would make it difficult to interpret the results of
a gel electrophoresis of proteins in the absence of sodium dodecyl sulfate
(SDS)?
Ans: Without SDS, protein migration
through a gel would be influenced by the protein’s intrinsic net charge—which
could be positive or negative—and its unique three-dimensional shape, in
addition to its molecular weight. Thus, it would be difficult to ascertain the
difference between proteins based upon a comparison of their mobilities in gel
electrophoresis.
63. Working
with proteins
Pages: 90-91
Difficulty: 2
How can isoelectric focusing be used in conjunction with SDS gel
electrophoresis?
Ans: Isoelectric focusing can separate
proteins of the same molecular weight on the basis of differing isoelectric
points. SDS gel electrophoresis can then separate proteins with the same
isoelectric points on the basis of differing molecular weights. When combined
in two-dimensional electrophoresis, a great resolution of large numbers of
proteins can be achieved.
64. Working
with proteins
Pages: 91-92
Difficulty: 3
You are given a solution containing an enzyme that converts B
into A. Describe what you would do to determine the specific activity of
this enzyme solution.
Ans: First, add a known volume of the
enzyme solution (say, 0.01 mL) to a solution of its substrate B and measure the
initial rate at which product A is formed, expressed as mmol/mL of enzyme
solution/min. Then measure the total protein concentration, expressed as
mg/mL. Finally, divide the enzyme activity (mmol/min/mL) by the protein
concentration (mg/mL); the quotient is the specific activity.
65. Working
with proteins
Pages: 91-92
Difficulty: 2
As a protein is purified, both the amount of total protein and
the activity of the purified protein decrease. Why, then, does the specific activity of the
purified protein increase?
Ans: Specific activity is the units of
enzyme activity (mmol of product/min) divided by the amount of protein (mg).
As the protein is purified, some of it is lost in each step, resulting in a
drop in activity. However, other contaminating proteins are lost to a
much greater extent. Therefore, with each purification step, the purified
protein constitutes a greater proportion of the total, resulting in an increase
in specific activity. (See also Table 3-5, p. 88.)
66. Peptides
and proteins
Page: 84 Difficulty: 1
Define the primary structure of a protein.
Ans: The primary structure of a protein
is its unique sequence of amino acids and any disulfide bridges present in the
native structure, that is, its covalent bond structure.
67. The
covalent structure of proteins
Pages: 94-100 Difficulty: 2
In one or two sentences, describe the usefulness of each of the following
reagents or reactions in the analysis of protein structure:
(a) Edman reagent (phenylisothiocyanate)
(b) Sanger reagent (1-fluoro-2,4-dinitrobenzene, FDNB)
·
trypsin
Ans: (a) used in determination of the
amino acid sequence of a peptide, starting at its amino terminus; (b)
used in determination of amino-terminal amino acid of a polypeptide; (c)
used to produce specific peptide fragments from a polypeptide.
68. The
covalent structure of proteins
Pages: 96-97
Difficulty: 2
A polypeptide is hydrolyzed, and it is determined that there are
3 Lys residues and 2 Arg residues (as well as other residues). How many peptide
fragments can be expected when the native polypeptide is incubated with the
proteolytic enzyme trypsin?
Ans: Six fragments would be expected,
unless the carboxyl-terminal residue is Lys or Arg; in which case there would
be five.
69. The
covalent structure of proteins
Pages: 94-95
Difficulty: 2
The following reagents are often used in protein chemistry.
Match the reagent with the purpose for which it is best suited. Some answers
may be used more than once or not at all; more than one reagent may be suitable
for a given purpose.
(a) CNBr (cyanogen
bromide)
(e) performic acid
(b) Edman reagent (phenylisothiocyanate) (f)
chymotrypsin
(c)
FDNB
(g) trypsin
(d) dithiothreitol
___ hydrolysis of peptide bonds on the carboxyl side of Lys and
Arg
___ cleavage of peptide bonds on the carboxyl side of Met
___ breakage of disulfide (—S—S—) bonds
___ determination of the amino acid sequence of a peptide
___ determining the amino-terminal amino acid in a polypeptide
Ans: g; a; d and e; b; c
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