Chapter 21–1
Chapter 21 Amino Acids, Proteins, and Enzymes
Solutions to In-Chapter Problems
21.1 Identify the other functional groups in each amino acid.
21.2 Compare the OH groups in each amino acid.
21.3 Draw the enantiomers as in Example 21.1.
21.4 Refer to Table 21.2 to determine which amino acids are naturally occurring. Naturally occurring
amino acids are L-amino acids.
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Chapter 21–2
21.5 Draw the structure of valine at each pH as in Example 21.2.
21.6 Draw the different forms of phenylalanine.
21.7
The amino acid with the free –NH3+ group on the  carbon is called the N-terminal amino acid.
The amino acid with the free –COO– group on the  carbon is called the C-terminal amino acid.
21.8 Peptides are named from left to right as substituents of the C-terminal amino acid.
21.9 Draw the structure of each dipeptide by joining the –COO– and –NH3+ groups of the two different
amino acids.
21.10 Draw the two dipeptides formed from leucine and asparagine.
21.11 Identify the amino acids in the dipeptides as in Example 21.3. The amide bonds joining the two
amino acids are drawn in bold.
21.12 Name the dipeptide and indicate the peptide bond as in Example 21.3.
21.13 Answer each question about met-enkephalin.
21.14 When cysteine is oxidized, a disulfide bond forms.
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Chapter 21–3
21.15 Yes, two different proteins can be composed of the same amino acids since the amino acids can be
ordered differently. For example: Ala–Gly–Ile–Trp and Gly–Trp–Ala–Ile.
21.16 Draw each pair of amino acids.
21.17 Glycine has no large side chain and this allows for the  sheets to stack well together.
21.18 Hemoglobin is more water soluble than keratin because it is a globular protein that folds with its
polar groups on its exterior, thus allowing for hydrogen bonding. Keratin is water insoluble
because it has more nonpolar groups on its exterior, and this does not allow for hydrogen bonding.
21.19 Draw the products formed by the hydrolysis of each tripeptide as in Example 21.4. The products of
hydrolysis are the individual amino acids.
21.20 Heating collagen disrupts the hydrogen bonding and other intermolecular forces. The superhelix
unwinds, making it less ordered, turning it into the jelly-like substance gelatin.
21.21 The names of most enzymes end in the suffix -ase. Sucrase (b), lactase (d), and
phosphofructokinase (e) are all enzymes.
21.22 Answer each question about fumarase.
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Chapter 21–4
b. The enzyme is a hydratase because it adds the elements of water to a double bond; that is, the
enzyme is needed to hydrate the double bond.
21.23 A competitive inhibitor has a shape and structure similar to the substrate, so it competes with the
substrate for binding to the active site.
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Chapter 21–5
21.24 Sarin is an irreversible inhibitor since it forms a covalent bond at the enzyme’s active site.
21.25 Fibrin and thrombin circulate as inactive zymogens (fibrinogen and prothrombin) so that the blood
does not clot unnecessarily. They are activated as required at a bleeding point to form a clot.
21.26 Both captopril and enalapril are derived from the amino acid proline and contain an amide and a
carboxylate anion. Captopril also contains a thiol (SH), while enalapril contains three additional
functional groups—an amine, a carboxylate anion, and a benzene ring.
Solutions to End-of-Chapter Problems
21.27
The L designation refers to the configuration at the chirality center. With a vertical carbon chain
in the Fischer projection, the L isomer has the –NH3+ drawn on the left side. The -amino acid
designation indicates that the amino group is bonded to the carbon adjacent to the carbonyl group.
21.28
Neutral amino acids exist as zwitterions with no net charge because all amino acids contain a base
(NH2 group) and an acid (COOH) and proton transfer occurs from the acid to the base to form the
zwitterion.
21.29
Alanine is an ionic salt with extremely strong electrostatic forces, leading to its high melting
point, and making it a solid at room temperature. Pyruvic acid is a neutral polar molecule with
weaker intermolecular forces, so it is a liquid at room temperature.
21.30
Phenylalanine can form a salt called a zwitterion that contains both a positive (–NH3+) and a
negative (–CO2–) charge. Zwitterions are water soluble. 4-Phenylbutanoic acid, on the other hand,
is not an amino acid and does not form a zwitterion. Instead, 4-phenylbutanoic acid consists of a
nonpolar phenyl group and a nonpolar four-carbon chain with a polar carboxylic acid group on
one end. Even if the carboxylic acid group hydrogen bonds with water, the nonpolar portion of
the 4-phenylbutanoic acid is so large that the compound is insoluble in water.
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Chapter 21–6
21.31
Draw an amino acid to fit each requirement.
21.32
Draw an amino acid to fit each requirement.
21.33
Both isoleucine and threonine contain two chirality centers.
21.34
Proline is different from all of the other amino acids in Table 21.2 because the amino group and
carboxyl group are in a ring not a chain.
21.35
Answer each question about the amino acids.
[1] L Enantiomer
[2] Classification
[3] Three-letter
Symbol
[4] One-letter
Symbol
neutral
Leu
L
neutral
Trp
W
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Chapter 21–7
21.36
basic
Lys
K
acidic
Asp
D
Answer each question about the amino acids.
[1] L Enantiomer
[2] Classification
[3] Three-letter
Symbol
[4] One-letter
Symbol
basic
Arg
R
neutral
Tyr
Y
acidic
Glu
E
neutral
21.37
Val
V
Draw and label the enantiomers of each amino acid.
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Chapter 21–8
21.38
Name each amino acid and designate it as a D or L isomer. The L isomer in part a is naturally
occurring.
21.39
Answer each question about the amino acids.
[1] Amino Acid
21.40
[3] One-letter
Symbol
[4] Classification
Gln
Q
neutral
Tyr
Y
neutral
[2] Three-letter
Symbol
[3] One-letter
Symbol
[4] Classification
Met
M
neutral
His
H
basic
Answer each question about the amino acids.
[1] Amino acid
21.41
[2] Three-letter
Symbol
Draw the structure of leucine at each pH as in Example 21.2.
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Chapter 21–9
21.42
Draw the structure of isoleucine at each pH as in Example 21.2.
21.43
Draw the structure of tyrosine at each pH as in Example 21.2.
21.44
Draw the structure of valine at each pH as in Example 21.2.
21.45
Locate the peptide bond and name the dipeptide as in Example 21.3.
21.46
Label the N-terminal and C-terminal amino acids and name the dipeptide.
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Chapter 21–10
21.47
The amino acid with the free –NH3+ group on the  carbon is called the N-terminal amino acid.
The amino acid with the free –COO– group on the  carbon is called the C-terminal amino acid.
21.48
The amino acid with the free –NH3+ group on the  carbon is called the N-terminal amino acid.
The amino acid with the free –COO– group on the  carbon is called the C-terminal amino acid.
21.49
Answer each question about the tripeptides.
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Chapter 21–11
21.50
Answer each question about the tripeptides.
21.51
Answer each question about the tripeptide.
21.52
Answer each question about the tripeptide.
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Chapter 21–12
21.53
Draw the structure of the three tripeptides.
21.54
Six different tripeptides can be formed from three different amino acids. The six tripeptides that
can be formed from methionine, histidine, and arginine are Met–His–Arg, His–Arg–Met, Arg–
Met–His, Met–Arg–His, His–Met–Arg and Arg–His–Met.
21.55
Draw the products formed by the hydrolysis of the tripeptide as in Example 21.4.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 21–13
21.56
Draw the products formed by the hydrolysis of the tripeptide as in Example 21.4.
21.57
Draw the products formed by the hydrolysis of each tripeptide as in Example 21.4.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 21–14
21.58
Draw the products formed by the hydrolysis of each tripeptide as in Example 21.4.
21.59
Draw the products of the hydrolysis of bradykinin.
21.60
Draw the products of the hydrolysis of aspartame.
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Chapter 21–15
21.61
Draw the products of the hydrolysis of the peptide with chymotrypsin.
21.62
Draw the products of the hydrolysis of the peptide with trypsin.
21.63
The primary structure of a protein is the order of its amino acids. The secondary structure refers
to the three-dimensional arrangement of regions within the protein.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 21–16
21.64
The tertiary structure is the three-dimensional shape adopted by the entire peptide chain. The
quaternary structure is the shape adopted when two or more folded polypeptide chains come
together into one protein complex.
21.65
Use Figure 21.10 to determine which types of intermolecular forces occur between each amino
acid pair.
a. isoleucine and valine: London dispersion forces of the nonpolar side chains
b. threonine and phenylalanine: London dispersion forces. Since the side chain of phenylalanine
has no O or N atom, no hydrogen bonding is possible.
c. Lys and Glu: electrostatic attraction of the charged side chains
d. Arg and Asp: electrostatic attraction of the charged side chains
21.66
Use Figure 21.10 to determine which types of intermolecular forces occur between each amino
acid pair.
a. Tyrosine has an OH group on its side chain, so hydrogen bonding is possible between two
tyrosine residues.
b. Serine and threonine both have OH groups on their side chains, so hydrogen bonding is
possible between the two residues.
c. Tyrosine has an OH group on its side chain, but phenylalanine has no O or N atom, so
hydrogen bonding is not possible.
d. Threonine has an OH group on its side chain, but alanine has no O or N atom, so hydrogen
bonding is not possible.
21.67
Draw the structure.
21.68
Draw the structure.
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Chapter 21–17
21.69
Use Figure 21.6 to label the regions of secondary structure.
The -helix is the corkscrew-shaped region.
The -pleated sheets are drawn as flat ribbons with gentle curves.
The areas of random coil are shown as thin string-like lines.
21.70
Use Figure 21.6 to label the regions of secondary structure.
The -helix is the corkscrew-shaped region.
The -pleated sheets are drawn as flat ribbons with gentle curves.
The areas of random coil are shown as thin, string-like lines.
21.71
Compare keratin and hemoglobin.
a. Secondary Structure
b. H2O Solubility
c. Function
d. Location
Hemoglobin
globular with much
-helix
soluble
carries oxygen to
tissues
blood
Keratin
-helix
insoluble
gives strength to
tissues
nails, hair
a. Secondary Structure
b. H2O Solubility
c. Function
d. Location
Collagen
triple helix
insoluble
connects tissues to
bones
connective
tissue
Myoglobin
globuluar with -helix
soluble
stores oxygen in
muscles
muscles
21.72
Compare collagen and myoglobin.
21.73
When a protein is heated and denatured, the primary structure is unaffected. The 2o, 3o, and 4o
structures may be altered by unfolding.
21.74
a. The functional groups NH2 and C=O hydrogen bond to stabilize secondary structure.
b. The functional groups OH, NH2, and COOH can hydrogen bond to stabilize tertiary structure.
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Chapter 21–18
21.75
a. Insulin is a hormone that controls glucose levels.
b. Myoglobin stores oxygen in muscle.
c. -Keratin forms hard tissues such as hair and nails.
d. Chymotrypsin is a protease that hydrolyzes peptide bonds.
e. Oxytocin is a hormone that stimulates uterine contractions and induces the release of breast
milk.
21.76
a. Collagen connects tissues to bones.
b. Hemoglobin transports oxygen throughout the body.
c. Vasopressin is a hormone that acts as an antidiuretic.
d. Pepsin is a digestive enzyme.
e. Met-enkephalin acts as a pain killer
21.77
Reversible enzyme inhibition occurs when an enzyme’s activity is restored when the inhibitor is
released. Irreversible inhibition permanently renders the enzyme incapable of further activity.
21.78
a. The zymogen angiotensinogen, when activated to angiotensin, acts to increase blood pressure.
b. Digestive enzymes, such as trypsin and chymotrypsin, are formed as zymogens and are
converted to the active form in the intestines.
21.79
Captopril inhibits the angiotensin-converting enzyme, blocking the conversion of
angiotensinogen to angiotensin. This reduces the concentration of angiotensin, which in turn
lowers blood pressure.
21.80
Enzyme inhibitors, such as HIV protease inhibitors, inhibit the action of the HIV protease enzyme
needed by HIV to make copies of itself that go on to infect other cells. Amprenavir is an example
of a protease inhibitor.
21.81
A noncompetitive inhibitor binds to the enzyme but does not bind at the active site.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 21–19
21.82
In the lock-and-key model, the shape of the active site is rigid. The three-dimensional shape of
the substrate must exactly match the shape of the active site for catalysis to occur.
In the induced-fit model, the shape of the active site is more flexible. The shape of the active site
can adjust to fit the shape of the substrate.
21.83
The -keratin in nails has more cysteine residues to form disulfide bonds. The larger the number
of disulfide bonds, the harder the substance. Therefore, fingernails are harder than skin.
21.84
The -keratin in hair contains many cysteine residues because cysteine residues can form
disulfide bonds between adjacent polypeptide strands, thus imparting strength and flexibility to
the hair.
21.85
Humans cannot synthesize the amino acids methionine and lysine, and they therefore must be
obtained in the diet. Diets that include animal products readily supply all of the needed amino
acids, but no one plant source has sufficient amounts of all the essential amino acids. Grains—
wheat, rice, and corn—are low in lysine, and legumes—beans, peas, and peanuts—are low in
methionine, but a combination of these foods provides all of the needed amino acids.
21.86
Cooking meat makes it easier to digest because cooking denatures the protein, breaking down its
secondary and tertiary structure.
21.87
Cauterization denatures the proteins in a wound.
21.88
Insulin is administered by injection instead of taken in tablet form because insulin is a small
protein consisting of two polypeptide chains held together by two disulfide bonds. The disulfide
bonds would be broken in the acidic environment of the stomach if taken in tablet form, thus
denaturing the protein and rendering it useless for treating diabetes.
21.89
Sickle cell disease results from abnormal hemoglobin. In sickle hemoglobin there is a substitution
of a single amino acid, valine for glutamic acid, which changes the shape of the hemoglobin.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 21–20
21.90
Silk produced by a silkworm is a protein with a high glycine and alanine content. This makes the
silk fiber strong because it has regions of -pleated sheet and regions of -helix. The -pleated
sheet regions are almost fully extended and their highly ordered three-dimensional structure
imparts strength to the silk.
21.91
Penicillin inhibits the formation of the bacterial cell wall by irreversibly binding to an enzyme
needed for its construction. It does not affect humans since human cells have a cell membrane,
instead of a rigid cell wall. Sulfanilamide inhibits the production of folic acid and therefore
reproduction in bacteria, but humans do not synthesize folic acid (they must ingest it instead), so
it does not affect humans.
21.92
Blood enzyme levels are used to diagnose certain diseases because certain enzymes are present in
higher concentration in particular cells. When the cells are damaged by disease or injury, the cells
rupture and die, releasing the enzymes into the bloodstream. By measuring the activity of the
enzymes in the blood, the presence of disease or injury in an organ can be diagnosed. If a patient
presents with chest pain, the doctor can measure the level of creatine phosphokinase to determine
whether the patient has had a heart attack.
21.93
Both aspartic acid and glutamic acid have two carboxylic acid groups. At low pH they have a +1
charge with both acid groups protonated, but at a high pH both acid groups are ionized, leading to
a net charge of –2.
21.94
Proteins like collagen that are high in proline do not contain an -helix in their secondary
structure because proline has a cyclic structure. The cyclic structure makes it difficult for a
hydrogen atom bonded to the amino nitrogen atom to hydrogen bond to the C=O group of another
amino acid further along the polypeptide chain, an interaction that is necessary to stabilize the helix secondary structure.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.