final exam Flashcards
(241 cards)
1
Q
Name different types of chemical bonds
A
Covalent, ionic, metallic, & hydrogen
2
Q
Which is the strongest chemical bond?
A
Ionic
3
Q
Draw Lewis structure of H2O
A
S+ H__O S-__H S+
4
Q
What is the molecular geometry of H2O?
A
Bent
5
Q
What is a key biochemical advantage of the use of weak bonds in biochemistry?
A
A key biochemical advantage of the use of weak bonds in biochemistry is to
carry out metabolic processes (Ex: How insulin released by the pancreas results
in the breakdown of sugars)
6
Q
Is water a good buffer?
A
No, water is not a good buffer because it has a neutral pH, so it has small but
equal concentrations of H+ and OH- which means it is unable to counteract
acids and bases.
7
Q
Carbon dioxide (CO2) is a product of energy metabolism. Exercise will increase
metabolism and thus increase the amount of CO2 in the blood. Following a bout of
intense exercise, the pH of the exerciser’s blood was found to be 7.1. If the HCO3 −
concentration is 8 mM and the pKa for HCO3 − is 6.1, what is the concentration of CO2 in
the blood? The normal concentration of CO2 in the blood is approximately 25 mM.
A
6). a)
8
Q
What is the hydrogen ion concentration of a solution with pH=3.75?
A
7.) a)
9
Q
A
8.) a)
10
Q
What is the chiral carbon center in molecule A and B? (q.13)
A
A does not have a chiral center; B has a chiral center @ carbon 3
11
Q
What is the orbital
hybridization of the molecule?;
A
sp 3
12
Q
What is the molecular geometry of chiral carbon?
A
Tetrahedral
13
Q
Identify the following amino acids as D or L enationmers: (q14.)
A
A= L-phenylalanine and B=D-serine
14
Q
Can an amino acid act as an acid and base? Please explain your answer.
A
Yes, it depends whether the amino acid donates a proton, making it an acid, or
accepts a proton, making it a base.
15
Q
Draw the structure at acidic and basic pH: a. pH=1; b. pH=7; c. pH=12
A
q.16
16
Q
Which structure represents a. Alanine at a pH above its pI?; b. Alanine at a pH below its pI?
A
a) (2)
b) (1)
q.)17
17
Q
Consider the amino acid leucine with a pH of 6.0. a. At a pH of 3.0, how does leucine
change?
A
+1. Because the pH of 3.0 is more acidic than the pI at 6.0, the –COO - group
gains an H + to give –COOH. The remaining –NH 3
+ gives leucine an overall
positive charge (+1).
18
Q
b. At a pH of 9.0, how does leucine change?
A
-1. Because the pH of 9.0 is more basic than the pI at 6.0, the –NH 3
+ group loses
an H + to give –NH 2
. The remaining –COO - gives leucine an overall negative
charge (-1).
19
Q
Draw the zwitterions for the amino acids serine and aspartate.
A
q.19)
20
Q
a. Name three amino acids that are positively charged at neutral pH
A
Lysine (Lys), Histidine (His), and Arginine (Arg)
21
Q
Name three amino
acids that contain hydroxyl groups.
A
Serine (Ser), Threonine (Thr), and Tyrosine (Tyr)
22
Q
Which statement is true of the chirality of amino acids found in proteins?
A. Only L amino acids are found in proteins
B. Only D amino acids are found in proteins
C. Proteins contain both D and L amino acids
D. D amino acids alternate with L amino acids in certain regions of protein
A
A. Only L amino acids are found in proteins
23
Q
Of the 20 amino acids that make up proteins, which is most likely to be present with its R
group in a mixture ionization states near physiological pH?
A. Tyr
B. His
C. Glu
D. Gly
A
B. His
24
Q
Which statement on the amino acids found in proteins is true?
A. Only some contain a central alpha (α) carbon.
B. All are dipolar ions, or zwitterions, regardless of the pH of their environment.
C. All have a side chain except for glycine
D. All contain a central alpha (α) carbon and the majority are dipolar ions, or zwitterions at
physiological pH.
A
D. All contain a central alpha (α) carbon and the majority are dipolar ions, or
zwitterions at physiological pH.
25
The protonated amino group of an amino acid will lose a proton when the pH approaches
A. 3.3.
B. 5.5.
C. 7.4.
D. 9.
D. 9.
26
Which amino acid has the simplest side chain?
A. Alanine
B. Glycine
C. Valine
D. Histidine
B. glycine
27
Which amino acid contains a guanidinium group?
A. G
B. V
C. R
D. Y
C. R
28
Why is it important to know amino acid sequences that make a protein?
Because it determines the protein’s 3-D structure which then determines its
function
29
Are amino acids attached by covalent or non-covalent bonds?
Covalent bonds, called peptide bonds, which form when the carboxyl group of
an amino acid reacts with the amino group of another amino acid.
30
Name the amino acids and their one and three letter codes. (q.30)
Valine (Val, V) + Alanine (Ala, A) → Val-Ala
31
Draw the peptide Ser–Thr
q.31
32
Draw the structure and give the name for the tripeptide, Gly–Ser–Met.
q.32
33
Which statement is false regarding peptide bonds?
A. They tend to be planar.
B. They are generally in the trans and rarely in the cis configuration.
C. They tend to have the amide nitrogen protonated to give a positive charge.
D. They contain an unusually long carbon–carbon bond
C. They tend to have the amide nitrogen protonated to give a positive charge.
34
What is the amino acid terminus of the tripeptide Gly–Ala–Asp?
) Glycine (Gly, G)
35
What is the
approximate molecular weight of a protein composed of 300 amino acids?
( 300 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑𝑠 )( 110 𝑔/𝑚𝑜𝑙__1 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑 ) = 33000 𝑔 𝑚𝑜𝑙
36
Approximately
how many amino acids are required to form a protein with a molecular weight of 110,000?
110000 𝑔/𝑚𝑜𝑙 __110 𝑔/𝑚𝑜𝑙 ___1 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑
= 100 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑𝑠
37
For the peptide Ala-Arg-Lys-Ala-Asn-Ser-Ala-Ser, what would be the
expected charges at pH 1, 7, and 13?
A. +3, +3, 0
B. +2, +2, -1
C. +3. +2, -1
D. +2, +1, -1
C. +3, +2, -1
38
What do α helices and β sheets have in common?
A. Both are stabilized by hydrogen bonding involving carbonyl oxygens and amide
nitrogens.
B. The same amino acids stabilize both forms of secondary structure.
C. The lengths of a 10 amino acid α helix and β sheet strand will be the same
D. Both are stabilized by glycine and proline residues.
Both are stabilized by hydrogen bonding involving carbonyl oxygens and
amide nitrogens.
39
Water-soluble proteins, such as myoglobin, tend to fold such that:
A. hydrophobic amino acid R groups are on the interior of the protein and hydrophilic
groups are on the outside.
B. hydrophilic amino acid R groups are on the interior of the protein and hydrophobic
groups are on the outside.
C. all peptides form hydrogen bonds with water.
D. hydrophilic and hydrophobic amino acid R groups form hydrogen bonds with each other.
A. hydrophobic amino acid R groups are on the interior of the protein and
hydrophilic groups are on the outside.
40
When sequencing a protein, it is necessary to break disulfide bonds within a polypeptide
chain. What reagent is used to break the disulfide bonds?
A. guanidine hydrochloride
B. Iodoacetate
C. β-mercaptoethanol
D. sodium hydroxide
β-mercaptoethanol
41
The major conclusion of C. Anfinsen involving ribonuclease was that
A. The information dictating how the protein should fold is contained in the amino acid
sequence.
B. A denatured protein loses its enzymatic activity
C. A protein can be refolded correctly only if the disulfide bonds are left intact.
D. Proteins unfold upon heating
E. Reducing agents are required to unfold proteins
A. The information dictating how the protein should fold is contained in the
amino acid sequence.
42
What physical differences among proteins allow for their purification?
Solubility, size, charge, and bonding specificity/affinity
43
The first step in protein purification is called ______.
A. Differential centrifugation
B. Passage through gel-filtration column
C. Electrophoresis
D. Determining the mass of the protein
A. Differential centrifugation
44
What can SDS polyacrylamide electrophoresis be used to do?
A. Determine the molecular weight of an oligomeric (multi-subunit) protein
B. Determine the molecular weights of subunits of an oligomeric protein
C. Purify a monomeric enzyme in its active form
D. Determine the folding state of a monomeric enzyme
B. Determine the molecular weights of subunits of an oligomeric protein
45
What is the method for ensuring that proteins are separated by size and not the charge
during gel electrophoresis?
A. a negatively charged detergent binds to amino acids and denatures the protein
B. SDS binds to amino acids and does not denature the protein
C. a positively charged detergent binds amino acids and denatures the protein
D. none of the other answer choices are correct
A. a negatively charged detergent binds to amino acids and denatures the
proteins
46
An assay is useful for what during protein purification?
A. isolating a protein from other proteins in a sample
B. detecting a protein in a sample
C. modifying a protein in a sample
D. determining the primary sequence of a protein in a sample
B. detecting a protein in the sample
47
Which statement is true of the proteome?
A. It includes the interactions of peptides that yield a functional unit.
B. It varies with cell type.
C. It varies with environmental conditions.
D. All three statements are true.
D. All three statements are true.
48
What is the best method to precisely determine the mass of a protein?
A. gel-filtration chromatography
B. SDS polyacrylamide gel electrophoresis
C. mass spectrometry
D. high-pressure liquid chromatography
C. mass spectrometry
49
Antibodies used as regents to quantify proteins are the basis of a lab method called
_____________.
A. Titration
B. enzyme-linked immunosorbent assay
C. 2-D gel electrophoresis
D. monoclonal antibody purification
B. enzyme-linked immunosorbent assay
50
Enzyme-linked immunosorbent assays (ELISA) are useful in biochemistry because:
A. They are based on the specific binding of a substrate to the enzyme.
B. They can be used to detect very small amounts of a specific material.
C. They involve a specific recognition between the antibody and enzyme used in the color
formation.
D. They require only small amounts of enzymes for color formation.
B. They can be used to detect very small amounts of a specific material
51
Match the type of reaction with an enzyme.
1) Aminase; 2) Dehydrogenase; 3) Isomerase; 4) Synthetase
a. Converts a cis fatty acid to a trans fatty acid; b. Removes 2 H atoms to form a double bond;
c. Combines 2 molecules to make a new compound; d. Adds NH3
a) 3
b) 2
c) 4
d) 1
52
Which of the following two reactions will take place spontaneously? What are the ΔG°’
values for the reverse reactions?
A → B ΔG°’ = –10 kJ*mol –1
C → D ΔG°’ = +10 kJ*mol –1
A → B ΔG°’ = –10 kJ mol–1 because it is negative
53
What are some non-covalent interactions that can happen in the enzyme-substrate
complex?
Hydrogen bonds, ionic bonds, hydrophobic interactions, and van der Waals forces
54
Does the Lock-and-Key Model or the Induced-fit Model of Enzyme–Substrate Binding better
explain the binding of enzymes with substrate?
The induced-fit model has the substrate and enzyme working together to acquire a
geometrical arrangement that lowers the activation energy.
55
What must be true of the free-energy change (Δ𝐺) for a reaction to be spontaneous?
A. It must be negative.
B. It must be greater than the change in entropy (Δ𝑆).
C. It is dominated by the enthalpy change (Δ𝐻).
D. It must be positive.
E. It must be driven by a large change in both entropy and enthalpy.
A. It must be negative.
56
A reaction has an equilibrium constant (𝐾eq) of 50. When performed in the presence of an
appropriate enzyme, the forward rate constant is increased 20-fold. What will happen to the
reverse rate constant?
A. It will be unaffected.
B. It will increase 20-fold.
C. It will decrease 20-fold.
D. It will increase 10-fold.
E. It will decrease 10-fold.
B. It will decrease 20-fold.
57
What role does an enzyme play in catalysis?
A. An enzyme increases the equilibrium constant.
B. An enzyme decreases the activation energy.
C. An enzyme increases the energy of the transition state so that it breaks down more
rapidly.
D. An enzyme increases the rate of the forward reaction.
B. An enzyme decreases the activation energy.
58
Which statement is true of carbonic anhydrase?
A. It catalyzes a reaction involving water.
B. It catalyzes a reaction involving CO2.
C. It is a very fast enzyme.
D. All of these answer choices are correct.
D. All of these answer choices are correct.
59
The alteration of enzyme structure on binding of a substrate to an active site is referred to
as
A. enzyme denaturation.
B. enzyme inhibition.
C. induced fit.
D. enzyme adaptation.
C. induced fit
60
What does a proteolytic enzyme cleave?
A. Peptide bonds
B. Disulfide bonds
C. Double bonds
D. None of these
A. Peptide bonds
61
Which statement is true of a holoenzyme, but not of an apoenzyme?
A. A holoenzyme contains its necessary cofactor.
B. A holoenzyme is catalytically inactive.
C. A holoenzyme contains a denatured active site.
D. A holoenzyme contains more than one active site.
A. A holoenzyme contains its necessary cofactor.
62
Which statement regarding a system at equilibrium is false?
A. There is no net charge in the concentrations of the products and reactants.
B. Δ𝐺 is zero.
C. Δ𝐺ʹ is zero.
D. The rate of the forward reaction equals the rate of the reverse reaction.
C. Δ𝐺ʹ is zero.
63
What value of [S], as a fraction of KM, is required to obtain 80% Vmax? Vo = 80% of Vmax;
assume, Vmax = 1
q.61
64
The maximum velocity (𝑉max) of an enzyme-catalyzed reaction is
A. the rate observed when all enzyme active sites are saturated with substrate.
B. independent of the amount of enzyme present.
C. the rate observed at the highest substrate concentration that can be experimentally
obtained.
D. the initial rate observed at very low substrate concentrations.
A. the rate observed when all enzyme active sites are saturated with substrate.
65
When the rate constant for dissociation of the enzyme–substrate complex (𝑘−1) is
greater than the rate constant for conversion to product (𝑘2), the 𝐾M is most analogous to
A. the 𝐾d.
B. the 𝐾a.
C. the 𝑘cat.
D. 1/𝑘cat.
A. the 𝐾d.
66
To obtain the turnover number of an enzyme (𝑘2), one must
A. divide 𝑉max by 2.
B. divide 𝑉 by 𝑉max.
C. divide 𝑉max by the total enzyme concentration.
D. divide 𝑉max by 𝑘cat.
C. divide 𝑉max by the total enzyme concentration.
67
How do you find the 𝐾M from a double-reciprocal, or Lineweaver–Burk, plot?
A. 𝑦-intercept=–1/𝐾M
B. 𝑥-intercept=–1/𝐾M
C. slope=𝐾M
D. 𝑦-intercept=–𝑉max/𝐾M
E. 𝑥-intercept=–𝑉max/𝐾M
B. 𝑥-intercept=–1/𝐾M
68
On what basis are enzymes and proteins with allosteric properties different from
those without allosteric properties? Allosteric enzymes have
A. more than one subunit.
B. different responses to non-substrate molecules such as inhibitors and activators.
C. different dependence on the substrate concentration.
D. more than one substrate-binding site.
C. different dependence on the substrate concentration.
69
The effects of molecules other than substrate on allosteric enzymes are called
A. heterotropic effects.
B. homotropic effects.
C. allosteric effects.
D. competitive effects.
A. heterotropic effects.
70
What effect will a heterotropic activator have on a sigmoidal kinetic plot of 𝑉 versus [S]?
A. The curve will shift to the right.
B. The curve will shift to the left.
C. The curve will become more sigmoidal, such that the first half will be decreased and the
second half increased.
D. The curve will become hyperbolic.
The curve will shift to the left
71
Which statement is true of studying the rates of enzyme-related reactions?
A. It is called enzyme kinetics.
B. It can involve determining how fast the substrate disappears as it is converted to
product.
C. It can involve following the appearance of the product formed over time.
D. All three statements are true.
All three statements are true.
72
In the following graph, identify the curve that corresponds to each of the following
conditions: no inhibition, competitive inhibition, noncompetitive inhibition, and uncompetitive
inhibition.
a) 1: no inhibition
b) 2: competitive inhibition
c) 3: noncompetitive inhibition
d) 4: uncompetitive inhibition
73
Which of the following binds to an enzyme at its active site?
A. irreversible inhibitor
B. reversible competitive inhibitor
C. reversible noncompetitive inhibitor
D. reversible uncompetitive inhibitor
B. reversible competitive inhibitor
74
An uncompetitive inhibitor binds to
A. E
B. ES
C. P
D. E and ES
B. ES
75
In a Lineweaver-Burk Plot, competitive inhibitor shows which of the following effect?
A. it moves the entire curve to right
B. it moves entire curve to left
C. t changes the x intercept
D. it has no effect on the shape
C. t changes the x intercept
76
A reversible inhibitor that can bind to E alone or ES complex is referred to as:
A. competitive inhibitor
B. non-competitive inhibitor
C. uncompetitive inhibitor
D. irreversible inhibitor
B. non-competitive inhibitor
77
TPCK is a molecule with large hydrophobic groups. Why does TPCK inactivate chymotrypsin
but not trypsin?
A. TPCK looks like the substrate for chymotrypsin, but not trypsin, and thus can bind in its
active site and modify His-57.
B. Trypsin does not have a His residue present in the active site to react with TPCK.
C. TPCK is a transition state analog for chymotrypsin, but not trypsin.
D. TPCK binds to trypsin but does not modify its active site.
A. TPCK looks like the substrate for chymotrypsin, but not trypsin, and thus can
bind in its active site and modify His-57.
78
Which statement is false of a competitive inhibitor?
A. It irreversibly inhibits the enzyme by chemically modifying a group at the active site.
B. It often resembles the substrate for the enzyme it inhibits.
C. Its effects can be overcome by increasing the substrate concentration.
D. It competes with substrates for binding to the active site.
It irreversibly inhibits the enzyme by chemically modifying a group at the active
site.
79
An enzyme inhibitor that decreases the apparent 𝐾M and reduces the 𝑉max can be
classified as
A. a competitive inhibitor.
B. a noncompetitive inhibitor.
C. an uncompetitive inhibitor.
D. an irreversible inhibitor.
C. an uncompetitive inhibitor.
80
Diisopropylphosphofluoridate (DIPF) inactivates chymotrypsin by covalently modifying
serine 195. Which statement is true of DIPF's inhibitory mechanism?
A. DIPF looks like the substrate for chymotrypsin and binds in the active site as a
competitive inhibitor.
B. Serine 195 is in an environment that gives it a higher than normal reactivity with respect
to DIPF.
C. DIPF randomly modifies all serine residues on the protein, and if enough is added, the
serine in the active site will eventually be modified.
D. DIPF approaches serine 195 more closely than other substrates.
Serine 195 is in an environment that gives it a higher than normal reactivity
with respect to DIPF.
81
The mechanism of chymotrypsin can be viewed as a two-step process, acylation of the
enzyme active site followed by a deacylation reaction. What accounts for the observed "burst"
in rapid kinetic studies of the hydrolysis of N-acetyl-L-phenylalanine p-nitrophenyl ester by
chymotrypsin?
A. The rate of the acylation reaction is slower than the rate of the deacylation reaction.
B. The rate of the acylation reaction is faster than the rate of the deacylation reaction.
C. The rates of the acylation and deacylation reactions are equal.
D. The rate of hydrolysis of the acyl-enzyme intermediate is faster than the rate of forming
the acyl-enzyme intermediate.
B. The rate of the acylation reaction is faster than the rate of the deacylation
reaction.
82
You have isolated a new protease that cleaves peptide bonds on the carboxyl side of Asp
and Glu. Based on the enzyme's inactivation by DIPF, you suspect that it may utilize a
mechanism similar to chymotrypsin. The difference in specificity might be explained by the
A. presence of a positively charged residue in the S1 binding pocket.
B. presence of a negatively charged residue in the S1 binding pocket.
C. replacement of serine 195 with a positively charged residue.
D. absence of the S1 binding pocket.
A. presence of a positively charged residue in the S1 binding pocket.
83
Many drugs are competitive inhibitors of specific enzymes. For drugs that act as competitive
inhibitors,
A. The amount of drug needed to achieve the same effect is increased as the substrate
concentration increases.
B. The amount of drug needed to achieve the same effect is increased as the substrate
concentration decreases.
C. The substrate must be bound to the enzyme before the drug can bind.
D. The enzymes must be allosterically regulated.
A. The amount of drug needed to achieve the same effect is increased as the
substrate concentration increases.
84
Which mechanism does NOT regulate the activity of an enzyme?
A. a competitive inhibitor binding to the active site
B. a competitive inhibitor binding to the ES complex
C. an uncompetitive inhibitor binding to the ES complex
D. All three mechanisms regulate enzyme activity.
B. a competitive inhibitor binding to the ES complex
85
a. What is the electronic configuration of Fe, Fe2+ and Fe3+?; b. Why can iron be in different
ionization state. e.g. Fe2+ and Fe3+?
a) [Ar]3d 6 4s 2 , [Ar]3d 6 , and [Ar]3d 5
b) Due to electrons being at different energy levels
86
What are the two models for allosteric enzymes?
Sequential and concerted
87
Name three factors that stabilize the deoxy form of hemoglobin.
Low pH, high [CO2], and low temperature; BPG binding, salt bridges between acidic
and basic amino acids, and salt bridges that include terminal carbamate
88
The binding of oxygen to myoglobin and hemoglobin has what effect on the heme iron?
A. It causes the iron to move out of the plane of the porphyrin ring.
B. It causes the iron to move closer to the plane of the porphyrin ring.
C. It has no effect because the iron is restricted to being within the plane of the porphyrin
ring.
D. It stabilizes the iron in its position out of the plane of the porphyrin ring.
It causes the iron to move close to the plane of the porphyrin ring.
89
Which statements correctly describe how 2,3-bisphosphoglycerate (2,3-BPG) reduces
hemoglobin's affinity for oxygen?
A. 2,3-BPG binds to the heme iron and prevents oxygen from binding.
B. 2,3-BPG binds to positively charged Lys and His residues in the center of the hemoglobin,
stabilizing the T (low-affinity) state.
C. 2,3-BPG binds to the amino termini, stabilizing the R (high-affinity) state.
D. 2,3-BPG binds to the interface between α1β1 and α2β2, stabilizing the R (high-affinity)
state.
B. 2,3-BPG binds to positively charged Lys and His residues in the center of the
hemoglobin, stabilizing the T (low-affinity) state.
90
Rapidly metabolizing tissues generate large amounts of protons and carbon dioxide. The
result of this increase in protons and carbon dioxide is that the oxygen-binding curve of
hemoglobin
A. shifts to the right, meaning lower saturation at higher O2O2 levels.
B. shifts to the left, meaning more binding of O2O2 at lower O2O2 levels.
C. changes shape from sigmoidal to hyperbolic.
D. remains the same.
A. shifts to the right, meaning lower saturation at higher O2O2 levels
91
Each chain of hemoglobin can be viewed as existing in either the R state or the T state. What
is the relationship between these two hemoglobin states and oxygen binding?
A. Oxygen binds with greater affinity to the R state, and oxygen binding converts
hemoglobin to the R state.
B. Oxygen binds with greater affinity to the T state, and oxygen binding converts
hemoglobin to the R state.
C. The conversion between the R and T states is not directly related to oxygen binding.
D. The conversion between the R and T states takes place only when 2,3-BPG is bound.
A. Oxygen binds with greater affinity to the R state, and oxygen binding
converts hemoglobin to the R state.
92
How many chiral centers?
q. 90
a.) 4
93
Define the following: a. Enantiomer; b. Diastereomer; c. Epimer; d. Anomer
a) Mirror image stereoisomers that are nonsuperimposable (all chiral centers are
different) and have the same molecular formula
b) Non-mirror image stereoisomers that are nonsuperimposable and have the same
molecular formula
c) Diastereomers that differ at only one chiral center
d) Epimers specific to carbohydrates, differing only at the anomeric carbon
94
Please label R and S on optically active stereocenters.
q. 92
95
Please label R and S on optically active stereocenters.
q. 93
96
Is glucose more stable in linear or in cyclic form? Why?
Cyclic because it is lower in energy, its arrangement minimizes strain, and the
formation of hemiacetals.
97
a. Are these D-sugars or L-sugars? How can you tell?; b. Name each form of galactose.; c. Do
the Haworth projections show these monosaccharides to be pyranoses or furanoses?; d. Are
these monosaccharides chiral?; e. Number the carbon atoms in the ring.; f. Which carbon atom
is anomeric carbon?
q.95
a) D, because the C6 carbon is pointing upward
b) D-α-galactose and D-β-galactose
c) Pyranose
d) Yes
e) 6
f) C1
98
Draw the structure of the maltose (Hint: glucose is a monomer and has α 1-4 linkage).
q.96
99
D-glucose and L-glucose are
A. Diastereomers
B. Enantiomers
C. Epimers
D. Anomers
B. enantiomers
100
The furanose form of fructose is generated by the formation of a hemiketal, involving the
attack of the hydroxyl group on
A. C-5 by C-2.
B. C-2 by C-6.
C. C-6 by C-1.
D. C-6 by C-2.
A. C-5 by C-2
101
The formation of pyranose and furanose forms of sugar generates a new asymmetric
carbon, giving rise to α- and β-forms. If the resulting anomeric alcohol group is above the plane
of the sugar, the structure is known as
A. the α-anomer.
B. the epimeric stereoisomer.
C. the β-anomer.
D. the D-enantiomer.
E. the L-enantiomer.
C. the β-anomer.
102
Once formed, the α- and β-forms of D-glucose are
A. not capable of interconversion.
B. interconvertible directly with no intermediate and are in equilibrium with one another.
C. interconvertible only through a linear, noncyclic intermediate with which they are both
in equilibrium.
D. found in an equal ratio.
C. interconvertible only through a linear, noncyclic intermediate with which
they are both in equilibrium.
103
Which of the two fatty acids will have a high melting point? Why?
q.101
The lack of tight packing limits van der Waals interactions between chains lowers the
melting point.
104
Describe the relationship between MP and fatty acid saturation.
As saturation increases, MP increases (direct positive relationship)
105
Describe the relationship between MP and fatty acid unsaturation.
As unsaturation increases, MP decreases (inverse relationship)
106
What factors determine the MP of fatty acids?
Chain length and saturation
106
What properties determine the melting point for the series of 18-carbon fatty acids?
As the number of cis double bonds increases, the melting point decreases. Longer fatty
acid chains allow for more van der Waals interactions between the chains.
Consequently, an increase in chain length will result in an increase in melting point.
107
Is glycogen or triacylglycerol better at energy storage?
Triacylglycerol
108
Some lipid molecules are said to be amphipathic, meaning that they have
A. asymmetric carbons and can exist in left- and right-handed forms.
B. have a dual nature with part of the molecule being hydrophobic and the other part
being hydrophilic.
C. are capable of moving rapidly from one side of a lipid bilayer to the other.
D. carry a positive charge on one end and a negative charge on the other end.
B. have a dual nature with part of the molecule being hydrophobic and the
other part being hydrophilic.
109
Unsaturated fatty acids have double bonds that are in the cis configuration. One of the
consequences of this configuration is _____.
A. an alteration in the charge of the molecule.
B. an alteration in the number of carbons in the molecule.
C. a bend in the molecule.
D. enhanced flexibility of the molecule.
a bend in the molecule.
110
In phosphoglycerides, fatty acids are esterified at
A. C-1 and C-2 of glycerol.
B. C-1 and C-3 of glycerol.
C. any two of the three glycerol carbons.
D. a glycerol carbon and the phosphate group.
A. C-1 and C-2 of glycerol.
111
The fatty acid oleate contains 18 carbons and a cis double bond after C-9. Which
designation describes the composition and structure of oleate?
A. 18:9
B. 9:18
C. 18:1
D. 0:18
C. 18:1
112
In biological systems, fatty acids usually contain an even number of carbon atoms. Which
fatty acids are MOST common in biological systems?
A. fatty acids with 16 and 18 carbons
B. fatty acids with 18 and 20 carbons
C. fatty acids with 14 and 16 carbons
D. fatty acids with 12 and 14 carbons
A. fatty acids with 16 and 18 carbons
113
A triacylglycerol consists of fatty acids attached to
A. three cholesterols.
B. one glycerol phosphate.
C. two glycerol phosphates.
D. one glycerol.
D. one glycerol.
114
What are the building blocks of proteins, lipids, and carbohydrates?
Carbon, hydrogen, and oxygen
i)
ii)
iii)
iv)
Proteins: amino acids
Lipids: fatty acids and glycerol
Carbohydrates: monosaccharides (simple sugars)
Nucleic acids (made up of nitrogen and phosphorus): nucleotides
115
Why is it necessary to break down food into smaller pieces?
To increase surface area by breaking down food into smaller pieces for enzymes to
digest food more quickly and efficiently into smaller
116
Why are enzymes packaged as zymogens (inactive enzymes)?
To prevent them from digesting the body’s healthy tissues
117
Enzymes that are activated by specific proteolytic cleavage are called
A. Isozymes.
B. Zymogens.
C. Allosteric.
D. Heterozymes.
B. Zymogens
118
The process of digestion in the stomach is carried out in two main ways. One way
involves the environment that exists in the stomach. What environmental condition within the
stomach promotes digestion?
A. Nonpolar
B. low pH
C. high pH
D. neutral pH
B. low pH
119
Increased levels of cholecystokinin (CCK) lead to a feeling of satiety. CCK is a family of
peptide hormones released from the
A. Pancreas.
B. Liver.
C. gall bladder.
D. small intestine.
D. small intestine
120
Saliva is an aqueous solution of which ion?
A. Na +
B. K +
C. HCO 3-
D. all three ions
C. HCO 3-
121
Chylomicrons are
A. Micelles.
B. lipoprotein transport particles.
C. comprised of disaccharides and protein.
D. transported from the intestinal lumen to mucosal cells.
B. lipoprotein transport particles
122
Bile salts, which aid in lipid adsorption, are synthesized from cholesterol in the liver and
released into the small intestine from the
A. Pancreas.
B. gall bladder.
C. Stomach.
D. large intestine.
B. gall bladder
123
What do micelles and chylomicrons have in common?
A. They both contain protein.
B. They both contain carbohydrate.
C. They both contain fatty acids.
D. They both contain triacylglycerol.
C. They both contain fatty acids
124
Name the following bonds in ATP q.123
Phosphoanhydride and phosphoester (left to right)
125
How many phosphoanhydride bonds in ATP, ADP, and AMP?
ATP has 2, ADP has 1, and AMP has 0.
126
How many negative charges on ATP?
4
127
Resonance structure of phosphoric acid.
q.126
Orthophosphate
b) Y-phosphate
128
What properties of ATP make it an especially effective phosphoryl-transfer-potential
compound?
ATP has a high phosphoryl-transfer-potential because of 4 key factors:
i)
Charge repulsion
ii)
iii)
iv)
Resonance stabilization
Increase in entropy
Stabilization by hydration
129
a. Draw the resonance structure of the following.; b. Which of the 2 structures are
unstable?
q.128
130
Which statement correctly describes metabolic reactions?
A. Anabolic processes break down material and transform fuels into cellular energy,
whereas catabolic processes require energy for biosynthesis.
B. Catabolic processes break down material and transform fuels into cellular energy,
whereas anabolic processes require energy for biosynthesis.
C. Intermediary metabolism breaks down material and transforms fuels into cellular
energy, whereas metabolism requires energy for biosynthesis.
D. Metabolism breaks down material and transforms fuels into cellular energy, whereas
intermediary metabolism requires energy for biosynthesis.
Catabolic processes break down material and transform fuels into cellular
energy, whereas anabolic processes require energy for biosynthesis.
131
In a metabolic pathway,
A. a reaction with a positive Δ𝐺 can occur if it is coupled to a reaction with a more negative
Δ𝐺.
B. a reaction with a positive Δ𝐺 cannot occur.
C. it is Δ𝐺°, not Δ𝐺, that determines whether a reaction can occur.
D. a reaction with a positive Δ𝐺 can occur if there is an increase in the concentration of the
reaction's products.
A. a reaction with a positive Δ𝐺 can occur if it is coupled to a reaction with a
more negative Δ𝐺.
132
Which factor does NOT contribute to the high phosphoryl-transfer-potential of ATP?
A. resonance stabilization
B. the adenine ring structure
C. charge repulsion
D. the ability of water to interact more favorably with the products of ATP hydrolysis than
with ATP itself
B. the adenine ring structure
133
The hydrolysis of ATP drives metabolism by
A. shifting the equilibrium of coupled reactions.
B. providing part of the activation energy for a key reaction.
C. providing energy in the form of heat.
D. altering the conformation of metabolic reactants.
A. shifting the equilibrium of coupled reactions.
134
What material in vertebrate muscle serves as a reservoir for high-energy phosphate
groups?
A. Glycogen
B. Phosphoenolpyruvate
C. creatine phosphate
D. glucose 6-phosphate
C. creatine phosphate
135
During catabolic processes, the oxidation of energy-rich molecules often results in
the reduction of NAD+ to NADH. What comparable molecule is the most commonly used
reductant for reductive steps in anabolic processes?
A. FAD
B. NADPH
C. coenzyme A
D. FMNH2
E. ATP
B. NADPH
136
Acyl groups generated during metabolic processes involving carbohydrates and fatty acids
are activated by attachment to
A. glyceraldehyde 3-phosphate.
B. pyruvate.
C. coenzyme A.
D. biotin.
C. coenzyme A
137
Please number C on glucose from 1 to 6.
q.136
138
a. Is glucose an aldose or ketose sugar?; b. Is fructose an aldose or ketose sugar?; c. Are
fructose and glucose isomers?
a) Aldose
b) Ketose
c) Yes
139
How is carbon oxidation captured in 1,3-BPG?
The aldehyde groups of the triose sugars are oxidised, and inorganic phosphate is
added to them, forming 1,3-BPG; by the formation of a high-energy phosphate bond
140
Why is phosphate in PEP high in energy?
Because it is attached to an enol form of a molecule, creating a high-energy bond that,
when broken, releases a significant amount of free energy
141
The gross yield of ATP from the metabolism of glucose to two molecules of pyruvate is four
molecules of ATP. However, the net yield is only two molecules of ATP. Why are the gross and
net values different?
2 molecules of ATP are produced per molecule of glyceraldehyde 3-phosphate, and
because 2 molecules of GAP are produced per molecule of glucose, the total ATP yield
is 4. However, 2 molecules of ATP are required to convert glucose into fructose
1,6-biphosphate. Thus, the net yield is only 2 molecules of ATP.
142
Lactic acid fermentation and alcoholic fermentation are oxidation–reduction reactions.
Identify the ultimate electron donor and acceptor.
In both cases, the electron donor is glyceraldehyde 3-phosphate. In lactic acid
fermentation, the electron acceptor is pyruvate, converting it into lactate. In alcoholic
fermentation, acetaldehyde is the electron acceptor, forming ethanol.
143
A suspension of yeast cells is being grown under anaerobic conditions such that glucose is
degraded to ethanol and carbon dioxide. To follow this process by monitoring the release of
14CO2, at which positions in the glucose molecule would the 14C label need to be
incorporated?
A. C-1 and C-6
B. C-3 and C-4
C. C-3 and C-6
D. C-1 and C-4
B. C-3 and C-6
144
A suspension of yeast cells is being grown under anaerobic conditions such that glucose is
degraded to ethanol and carbon dioxide. Why would adding an inhibitor of alcohol
dehydrogenase to the yeast cell suspension result in rapid cell death?
A. Pyruvate is no longer being produced for use in gluconeogenesis, so the cells quickly
deplete this energy source.
B. Loss of this enzyme during anaerobic glucose degradation leads to an increased
requirement for ATP, which cannot be met.
C. NADH generated during glycolysis cannot be recycled back to NAD+, thus anaerobic ATP
production through glycolysis cannot be maintained.
D. Without alcohol dehydrogenase, the acetaldehyde concentration increases to levels that
are toxic to the yeast cells.
NADH generated during glycolysis cannot be recycled back to NAD+, thus
anaerobic ATP production through glycolysis cannot be maintained.
145
Glyceraldehyde 3-phosphate is converted to 1,3-bisphosphoglycerate by the enzyme
glyceraldehyde 3-phosphate dehydrogenase. This reaction reduces which electron carrier?
A. NAD+
B. NADH
C. NADP+
D. NADPH
A. NAD+
146
Catalysis by the enzyme glyceraldehyde 3-phosphate dehydrogenase does NOT involve
A. phosphorylation of the substrate using ATP.
B. oxidation and phosphorylation of the substrate.
C. a covalent intermediate.
D. an active site histidine to serve as a proton acceptor.
A. phosphorylation of the substrate using ATP.
147
Although the glycolytic degradation of glucose is a major source of energy, other dietary
carbohydrates, such as galactose and fructose, are also important fuels. When being
metabolized, fructose and galactose
A. cannot be degraded by steps in the glycolytic pathway.
B. must first be converted to glucose to be used as fuel.
C. are metabolized by conversion to intermediates in the glycolytic pathway.
D. can be used to form pyruvate by metabolic pathways that run parallel to glycolysis.
C. are metabolized by conversion to intermediates in the glycolytic pathway.
148
The three major regulatory enzymes in the glycolytic pathway are
A. hexokinase, phosphofructokinase, and pyruvate kinase.
B. phosphofructokinase, glyceraldehyde 3-phosphate dehydrogenase, and pyruvate kinase.
C. hexokinase, phosphofructokinase, and phosphoglycerate kinase.
D. glyceraldehyde 3-phosphate dehydrogenase, pyruvate kinase, and hexokinase.
A. hexokinase, phosphofructokinase, and pyruvate kinase.
149
During sustained exercise, the ATP/AMP ratio in muscle cells decreases. What effect does
this decreased ATP/AMP ratio have on the activity of phosphofructokinase and pyruvate kinase?
A. The activity of both enzymes increases due to a lack of allosteric binding of ATP.
B. Phosphokinase activity increases and pyruvate kinase activity decreases.
C. The activity of both enzymes decreases due to a lack of allosteric binding of ATP.
D. The activity of both enzymes remains the same, maintaining normal glucose
metabolism.
A. The activity of both enzymes increases due to a lack of allosteric binding of
ATP.
150
The processes of gluconeogenesis and glycolysis are said to be reciprocally regulated.
Reciprocal regulation means that
A. molecules that activate or inhibit one process have the same effect on the other process.
B. molecules that activate or inhibit one process have the opposite effect on the other
process.
C. opposing sets of molecules, such as ATP and AMP, have opposite effects on the process.
D. one cell predominantly uses glycolysis, whereas another predominantly uses
gluconeogenesis.
B. molecules that activate or inhibit one process have the opposite effect on
the other process.
151
Please draw the structure of glucose and galactose in Haworth projection.
q.150
152
What are monomers of lactose?
q.151
153
Why is PFK the pacemaker of glycolysis?
Because the first irreversible reaction unique to the glycolytic pathway is the
phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate. If PFK is
inhibited, then hexokinase is also inhibited. If PFK is inactive, the concentration of
fructose 6-phosphate increases and, thus, the concentration of glucose 6-phosphate
increases due to equilibrium.
154
Distinguish between the role of glycolysis in muscle and liver.
Glycolysis in skeletal muscle provides ATP primarily to power contraction. The liver
uses glycolysis to synthesize a host of building blocks for other biomolecules.
155
Arsenate (AsO43– ) closely resembles Pi in structure and reactivity. In the reaction
catalyzed by glyceraldehyde 3-phosphate dehydrogenase, arsenate can replace
phosphate in attacking the energy-rich thioester intermediate. The product of this reaction,
1-arseno-3-phosphoglycerate, is unstable. It and other acyl arsenates are rapidly and
spontaneously hydrolyzed. What is the effect of arsenate on ATP generation in a cell?
Arsenate inhibits ATP generation
156
Although the glycolytic degradation of glucose is a major source of energy, other dietary
carbohydrates, such as galactose and fructose, are also important fuels. When being
metabolized, fructose and galactose
A. cannot be degraded by steps in the glycolytic pathway.
B. must first be converted to glucose to be used as fuel.
C. are metabolized by conversion to intermediates in the glycolytic pathway.
D. can be used to form pyruvate by metabolic pathways that run parallel to glycolysis.
C. are metabolized by conversion to intermediates in the glycolytic pathway.
157
During sustained exercise, the ATP/AMP ratio in muscle cells decreases. What effect does
this decreased ATP/AMP ratio have on the activity of phosphofructokinase and pyruvate kinase?
A. The activity of both enzymes increases due to a lack of allosteric binding of ATP.
B. Phosphokinase activity increases and pyruvate kinase activity decreases.
C. The activity of both enzymes decreases due to a lack of allosteric binding of ATP.
D. The activity of both enzymes remains the same, maintaining normal glucose
metabolism.
A. The activity of both enzymes increases due to a lack of allosteric binding of
ATP.
158
What barrier prevents glycolysis from simply running in reverse to synthesize glucose? How
is this barrier overcome in gluconeogenesis?
The reverse of glycolysis is highly endergonic under cellular conditions. The
expenditure of 6 NTP molecules in gluconeogenesis renders gluconeogenesis
exergonic.
159
What are the regulatory means that prevent high levels of activity in glycolysis and
gluconeogenesis simultaneously? What would be the result if both pathways functioned rapidly
at the same time?
Reciprocal regulation at the key allosteric enzymes in the two pathways. For instance,
PFK is stimulated by fructose 2,6-biphosphate and AMP. The effect of these signals is
opposite that of fructose 1,6-biphosphatase. If both pathways were operating
simultaneously, a futile cycle would result. ATP would be hydrolyzed, yielding only
heat.
160
What would be the effect on an organism’s ability to use glucose as an energy source if a
mutation inactivated glucose 6-phosphatase in the liver?
The blood glucose concentration would fall, resulting in energy deprivation for
glucose-dependent tissues.
161
The process of glycolysis results in the formation of two high-energy phosphate bonds in
the form of ATP. How many high-energy phosphate bonds are consumed during the process of
gluconeogenesis using pyruvate as a starting material?
A. Two
B. Four
C. Six
D. Three
C. six
162
What product of contracting muscle tissue is used by the liver as the starting material for
gluconeogenesis?
A. Pyruvate
B. Lactate
C. Glycerol
D. Oxaloacetate
B. Lactate
163
Although glycolysis can occur in most cells, gluconeogenesis is mostly limited to two
tissues. Which two tissues are the primary sites of gluconeogenesis?
A. liver and kidney
B. muscle and kidney
C. liver and muscle
D. muscle and brain
A. liver and kidney
164
The generation of glucose from glucose 6-phosphate occurs in which membrane-bound
organelle(s)?
A. Nucleus
B. Golgi apparatus
C. endoplasmic reticulum
D. mitochondria
C. endoplasmic reticulum
165
Which statement is true of fructose 1,6-bisphosphate but not of fructose 6-phosphate?
A. It is a six-carbon sugar.
B. It is a five-carbon sugar.
C. It contains two phosphate groups.
D. It contains one phosphate group.
C. It contains two phosphate groups.
166
If glycolysis and gluconeogenesis were simultaneously operating at the same rate, which
statement would be false?
A. Phosphofructokinase activity would inhibit phosphoenolpyruvate carboxykinase activity.
B. Two ATP and two GTP molecules would be consumed per reaction cycle.
C. Hypoglycemia would eventually occur unless food was consumed.
D. More high-energy phosphate bonds would be hydrolyzed than formed.
A. Phosphofructokinase activity would inhibit phosphoenolpyruvate
carboxykinase activity.
167
What is substrate level phosphorylation? What is oxidative phosphorylation?
Substrate-level phosphorylation directly transfers a high-energy phosphate group from
a substrate to ADP to form ATP, while oxidative phosphorylation uses energy from the
electron transport chain to generate an electrochemical gradient that drives ATP
synthase.
168
Why are the electrons carried by FADH2 not as energy-rich as those carried by NADH?
What is the consequence of this difference?
The reduction potential of FADH2 is less than that of NADH. As a result, when those
electrons are passed along to oxygen, less energy is released. The consequence of the
difference is that electron flow from FADH2 to O2 pumps fewer protons than does the
electron flow from NADH.
169
Please write down the complete breakdown of glucose to extract energy. How many ATPs
and NADH/FADH (high energy carriers) are produced during it?
q.167
170
For a given pair of reduced and oxidized molecules (X− and X), a negative reducing
potential suggests that
A. X− has a higher affinity for electrons than X does.
B. X has a higher affinity for electrons than H2 does.
C. X− has a lower affinity for electrons than X does.
D. X has a lower affinity for electrons than H2 does
X has a lower affinity for electrons than H2 does.
171
Consider a substance that can exist in an oxidized form X and a reduced form X-. Such a
pair is called a(n)
A. redox couple.
B. reduction couple.
C. electron-transfer potential.
D. half-cell.
A. redox couple
172
Amytal is a barbiturate sedative that inhibits electron flow through Complex I. How would
the addition of amytal to actively respiring mitochondria affect the relative oxidation–reduction
states of the components of the electron-transport chain and the citric acid cycle?
Complex I would be reduced, whereas Complexes II, III, and IV would be oxidized. The
citric acid cycle would become reduced because it has no way to oxidize NADH.
173
Four electron carriers—a, b, c, and d—are required for electron transport in a bacterial
electron transport chain. The first carrier receives electrons from NADH, and the last carrier
passes the electrons to O2. The oxidized and reduced forms of the carriers are easily
distinguishable. In the presence of NADH and O2, three different inhibitors block respiration
yielding the patterns of oxidation states shown in the table. From the data in the table,
determine the order of the carriers from NADH to O2.
q.172
NADH → c → b → a → d → O2
174
Of the electron transfer complexes associated with the inner mitochondrial membrane,
which does NOT directly produce a portion of the proton gradient?
A. cytochrome c oxidase (Complex IV)
B. Q-cytochrome c oxidoreductase (Complex III)
C. succinate-Q reductase (Complex II)
D. NADH-Q reductase (Complex I)
C. succinate-Q reductase (Complex II)
175
What is the source of electrons transferred by succinate-Q reductase (Complex II)?
A. NADH from the citric acid cycle and glycolysis
B. NAD+ from the conversion of pyruvate to lactate
C. FADH2 from the citric acid cycle
D. FAD from pyruvate dehydrogenase
FADH2 from the citric acid cycle
175
What is the product of beta oxidation?
5 acetyl CoA molecules, 4 NADH, and 4 FADH2.
176
Which statement describes the direction of proton movement relative to electron transfer
and phosphorylation?
A. Electron transfer pumps protons out of the matrix. As the protons reenter through the
ATP synthase, ATP is produced in the matrix.
B. Electron transfer pumps protons into the matrix. As the protons leave the matrix
through the ATP synthase, ATP is produced on the opposite side of the membrane.
C. Electron transfer pumps protons into the matrix. As the protons leave the matrix
through the ATP synthase, ATP is produced in the matrix.
D. Electron transfer pumps protons out of the matrix. As the protons reenter through the
ATP synthase, ATP is produced in the intermembrane space
A. Electron transfer pumps protons out of the matrix. As the protons reenter
through the ATP synthase, ATP is produced in the matrix.
177
Is below reaction a redox reaction? What is getting oxidized and what is getting reduced?
Why do you think Glucose 6-phosphate dehydrogenase used of NADP+ for oxidation and not
NAD?
q.176
The enzyme is highly specific for NADP+; the Km for NAD+ is 1000 greater than for
NADP+.
178
What functional group is this? What is the hydrolysis product?
Ester
179
Please name metabolic pathways Fructose 6-phosphate is part of.
Glycolysis, gluconeogenesis, and pentose phosphate pathways.
180
Describe how the pentose phosphate pathway and glycolysis are linked by transaldolase
and transketolase.
The enzymes catalyze the transformation of the five-carbon sugar formed by the
oxidative phase of the pentose phosphate pathway into fructose 6-phosphate and
glyceraldehyde 3-phosphate, intermediates in glycolysis (and gluconeogenesis).
180
Upon entry of ribose into the bacterium, ribose 5-phosphate is formed at the expense of
ATP. Describe a pathway that metabolizes the ribose yielding the products.
q.180
180
Glucose 6-phosphate can be utilized by either the glycolytic or the pentose phosphate
pathway. What is the major factor regulating the distribution of glucose 6-phosphate between
these two pathways?
A. the relative levels of NADP+ and NADPH
B. the different location of the two pathways within the cell
C. allosteric inhibition of the glucose 6-phosphate dehydrogenase enzyme by ATP
D. the ratio of ATP to AMP in the cell
A. the relative levels of NADP+ and NADPH
180
In addition to making NADPH, the pentose phosphate pathway generates key molecules
for the synthesis of
A. Lipids.
B. amino acids.
C. Nucleotides.
D. Heme.
C. Nucleotides
180
Glucose 6-phosphate produced in the liver during glycogen breakdown can
A. enter glycolysis.
B. enter the pentose phosphate pathway.
C. form glucose through the action of glucose 6-phosphatase.
D. All of these fates are possible.
D. All of these fates are possible.
181
How many acetyl groups are in Capric Acid? How many cycles of beta oxidation?
5 acetyl groups and 4 cycles of beta oxidation
181
What is the product of hydration?
3-Hydroxyacyl CoA
181
The first phase of the pentose phosphate pathway involves the
A. oxidative generation of NADH.
B. oxidative generation of NADPH.
C. oxidative generation of FADH2.
D. nonoxidative generation of NADH.
B. oxidative generation of NADPH.
182
What is the product of oxidation?
trans-Enoyl CoA
182
What is the product of the second oxidation?
β-Ketoacyl CoA
183
What is the product of thiolysis?
Fatty acyl CoA (2 C atoms shorter than original) and acetyl CoA
184
What is the product of beta oxidation for the: a. Second cycle; b. Third cycle; c. Fourth
cycle?
a) Fatty acyl CoA (2 C atoms shorter than previous) and acetyl CoA
b) Fatty acyl CoA (2 C atoms shorter than previous) and Acetyl CoA
c) 2 acetyl CoA molecules
185
What is the total ATP produced when one acetyl CoA enters the citric acid cycle?
Produces one acetyl CoA that enters the citric acid cycle, generating 12 ATP.
186
Why might D-3-hydroxybutyrate be considered a superior ketone body compared with
acetoacetate?
D-3-Hydroxybutyrate is more energy-rich because its oxidation potential is greater
than that of acetoacetate. After having been absorbed by a cell, D-3-hydroxybutyrate
is oxidized to acetoacetate, generating high-energy electrons in the form of NADH. The
acetoacetate is then cleaved to yield acetyl CoA.
186
Describe the repetitive steps of β oxidation. Why is the process called β oxidation?
The steps are (1) oxidation by FAD, (2) hydration, (3) oxidation by NAD+, and (4)
thiolysis to yield acetyl CoA. In symbolic notation, the beta-carbon atom is oxidized
187
How many rounds of beta oxidation before encountering double bonds?
3 rounds
188
The cleavage of triacylglycerols into glycerol and fatty acids is performed by lipases. The
signaling process that initiates the activation of lipases is
A. the same as that which signals the cell to break down glycogen.
B. the same as that which signals the cell to shut down the citric acid cycle.
C. the same as that which signals the initiation of gluconeogenesis.
D. insensitive to the energy status of the cell.
A. the same as that which signals the cell to break down glycogen.
189
The glycerol produced during the lipolysis of triacylglycerols
A. is a waste product that is excreted from cells.
B. is stored within the cell for future synthesis of triacylglycerols.
C. can be used as either a source of energy or for gluconeogenesis.
D. can immediately be used in glycolysis.
C. can be used as either a source of energy or for gluconeogenesis.
190
Fatty acids are activated by attachment to coenzyme A at the outer mitochondrial
membrane. However, the resulting fatty acyl-CoA is not degraded until it reaches the
mitochondrial matrix. The fatty acyl-CoA
A. freely crosses the inner mitochondrial membrane due to its hydrophobic character.
B. is reconverted to the free fatty acid on one side of the inner mitochondrial membrane,
then reattached to coenzyme A on the other side of the membrane.
C. is transferred to another molecule that is involved in the transport of the fatty acid into
the mitochondrial matrix.
D. binds to the inner mitochondrial membrane, which facilitates its degradation.
C. is transferred to another molecule that is involved in the transport of the
fatty acid into the mitochondrial matrix.
191
The degradation of an 18-carbon saturated fatty acid would be expected to yield how
many NADH and FADH2 per fatty acid molecule?
A. 7
B. 8
C. 9
D. 6
B. 8
192
During the degradation of unsaturated fatty acids, the double bonds must be reduced. The
reductant used to reduce the double bonds is
A. FADH2 formed during other steps of fatty acid degradation.
B. NADH formed during other steps of fatty acid degradation.
C. NADPH.
D. cytochrome Q.
C. NADPH.
193
Animals are not capable of converting most fatty acids into glucose because
A. the products of fatty acid degradation cannot leave the mitochondrion.
B. fatty acid degradation through the citric acid cycle does not lead to a net increase in
oxaloacetate.
C. fatty acid degradation does not produce sufficient energy in the form of ATP to drive
gluconeogenesis.
D. they cannot turn on the enzymes for gluconeogenesis if fatty acids are being broken
down.
B. fatty acid degradation through the citric acid cycle does not lead to a net
increase in oxaloacetate.
194
Triacylglycerol catabolism can convert glycerol for use in energy production. How many net
ATPs will glycerol metabolism directly generate through substrate-level phosphorylation via
glycolysis?
A. 0
B. 1
C. 2
D. 3
B. 1
195
For every double bond added in a fatty acid during a β-oxidation cycle, there is a
subsequent loss of one
A. ATP.
B. NADH.
C. FADH2.
D. acetyl CoA.
C. FADH2
196
Why are supercoiling and nucleosome formation important for the structure of DNA in a
cell?
Both supercoiling and nucleosome formation help to compact the DNA.
197
Bases in a DNA double helix absorb less ultraviolet light of 260 nm than do bases in
single-stranded DNA, an effect called hypochromism. Hypochromism allows us to determine the
melting point of a DNA double helix. As a solution of double-stranded DNA is heated, the
absorbance at 260 nm increases with temperature as the double stranded DNA is converted
into two separate strands. The DNA is said to be denatured upon complete conversion to single
strands. The temperature at which denaturation is half complete is called the melting point, or
Tm. Predict the effect on Tm as: a. The length of the double-stranded DNA increases.; b. The salt
concentration of the solution containing double-stranded DNA is decreased from 0.1 M NaCl to
0.02 M NaCl.
a) Tm increases
i)
More hydrogen bonds between strands requires more thermal energy to
separate the strands
b) Tm decreases
i)
ii)
iii)
iv)
To answer this part of the question, we need to consider another aspect of the
structure of double-stranded DNA. What molecules compose the backbone of
DNA? The backbone is composed of alternating deoxyribose and phosphates.
Under physiological conditions, the phosphates bear negative charges.
What is the effect of bringing like charges together in close proximity? Like
charges repel, as you recall from playing with magnets.
How can the negative charges on the phosphate be neutralized? Adding a salt
such as NaCl to the solution of DNA would neutralize the charges on the DNA.
Na+ would form an ionic bond (p. 20) with the negative charge on the
phosphates. If salts such as NaCl can neutralize the negative charges on the
phosphate, then
What would be the effect of lowering the salt concentration on the Tm of the
DNA? The Tm would decrease. Less thermal energy would be required to
denature the DNA because charge repulsion between the negative charges on
the phosphates of the backbone is already weakening the double helix.
198
The process by which two complementary nucleic acid strands base-pair to form a
double-stranded structure is called
A. Freezing.
B. Annealing.
C. Supercoiling.
D. Ligation.
E. Recombination.
B. annealing
199
As a sample of DNA is heated, it is said to melt. What is occurring?
A. As hydrogen bonds are broken, the two DNA strands separate from one another.
B. The DNA is converted from a solid to a liquid state.
C. The large polymeric DNA is broken down into smaller pieces as the backbone is partially
hydrolyzed.
D. The DNA becomes less soluble and precipitates out of solution.
E. Glycosidic linkages are broken, causing separation of the bases from the backbone.
A. As hydrogen bonds are broken, the two DNA strands separate from one
another.
200
a. Why is pyrophosphate released in DNA elongation?; b. Can NDP/NMP be used in DNA
chain elongation?
a) Because it is a byproduct of the chemical reaction where a new nucleotide is added to
the growing DNA chain, forming a phosphodiester bond; this reaction requires energy,
which comes from the breaking of a high-energy phosphate bond within the incoming
deoxynucleotide triphosphate (dNTP), releasing the two remaining phosphates as
pyrophosphate (PPi) to maintain the overall energy balance and drive the reaction
forward.
b) No, NDP/NMP cannot be used directly in DNA chain elongation because they lack the
necessary energy-providing phosphate groups.
201
What is an enzyme and what is their purpose?
A protein that increases the rate of a chemical reaction without being changed by the
reaction. Helicase allows room for the next enzyme, DNA polymerase, to read each
strand of DNA and put together the corresponding base pairs.
201
What are the three key enzymes required for DNA synthesis, and what biochemical
challenges to replication do they address?
DNA polymerases, helicases, and topoisomerases. Polymerases faithfully replicate the
DNA. Helicases unwind the double helix, allowing the polymerases access to the base
sequence. Topoisomerases relax the DNA by removing negative supercoils.
202
a. What is the direction of replication for the leading strand (top)?; b. What is the direction
of replication for the lagging strand (bottom)?
a) DNA polymerase moves in the 3’ to 5’ direction, catalyzing the formation of new
phosphodiester linkages.
b) The lagging strand is synthesized in the 5’ to 3’ direction in short segments called
Okazaki fragments.
203
204
Match the following to their definitions:
1) helicase
2) DNA polymerase
3) replication fork
4) Okazaki fragments
A. short segments formed on the lagging strand
B. the starting point for synthesis in unwound DNA sections
C. the enzyme that unwinds the DNA double helix
D. the enzyme that catalyzes the formation of phosphodiester bonds of complementary
bases
1-C, 2-D, 3-B, and 4-A
205
For DNA replication to occur, the two strands of the double helix must be physically
separated. This separation is accomplished through the use of
A. helicase enzymes
B. topoisomerase enzymes.
C. breaking and rejoining of one of the DNA strands.
D. single-stranded-binding (SSB) protein.
E. primase.
A. helicase enzymes
206
The primase enzyme in E. coli
A. synthesizes a short stretch of RNA that is complementary to a part of the template DNA
strand.
B. synthesizes a short stretch of DNA that is complementary to a part of the template DNA
strand.
C. acts as the recognition site to which DNA polymerases bind to start the replication
process.
D. replaces the RNA primers used to initiate DNA replication with DNA polymerase.
E. ligates the nicks formed by discontinuous replication along the lagging strand.
A. synthesizes a short stretch of RNA that is complementary to a part of the
template DNA strand
207
Which factor accounts for the increased rate of DNA synthesis necessary for replication in
eukaryotic cells as compared to that in prokaryotes?
A. the use of different chemistry by the DNA polymerase enzymes
B. the lack of need for a proofreading mechanism to correct errors
C. the presence of multiple origins at which replication can take place at the same time
D. the ability to replicate in both the 5ʹ to 3ʹ and 3ʹ to 5ʹ directions
E. the presence of telomere DNA
C. the presence of multiple origins at which replication can take place at the
same time
208
During the replication of DNA,
A. both strands are simultaneously synthesized in a continuous manner.
B. one strand is synthesized in a continuous manner and the other is synthesized by reverse
transcription.
C. both stands are simultaneously synthesized in short pieces that are later joined by DNA
ligase.
D. one strand is synthesized continuously and the other is synthesized in short segments.
E. both strands are synthesized continuously, but the lagging strand is synthesized only
after the leading strand has been completely replicated.
D. one strand is synthesized continuously and the other is synthesized in short
segments.
209
There are five DNA polymerases in E. coli. Which is NOT a function of E. coli DNA
polymerases?
A. primer formation
B. primer removal
C. DNA replication
D. DNA repair
E. exonuclease activity
A. primer formation
210
Which compound is released during the addition of deoxyribonucleotides to the nascent
DNA chain?
A. dNTP
B. pyrophosphate (PPi)
C. ATP
D. Okazaki fragments
E. AMP
B. pyrophosphate (PPi)
211
During replication in E. coli, the removal of Okazaki RNA primers in the lagging strand and
resolution of the resulting gaps in DNA are achieved by
A. only DNA polymerase I.
B. only DNA polymerase III.
C. only DNA ligase.
D. DNA polymerase I and DNA ligase.
E. DNA polymerase III and DNA ligase.
DNA polymerase I and DNA ligase.
212
Much of the original research on DNA polymerase was done using cell extracts from E. coli
because the enzyme had not yet been purified. The enzyme assay consisted of incubating the
extract, which contained both the enzymes required for DNA synthesis and the DNA template,
with the four deoxyribonucleotides. All of the nucleotides contained 32P in the α position of the
triphosphate moiety of the deoxyribonucleotides. The reaction was halted by acidifying the
reaction mixture with trichloroacetic acid, which precipitates the macromolecules but not the
radioactive deoxyribonucleotides. The amount of radioactivity in the precipitate was a measure
of the activity of DNA polymerase. a. How would the omission of one deoxyribonucleotide
affect the amount of radioactivity found in the precipitate?; b. Suppose that the
deoxyribonucleotides contained 32P in both the β and γ positions rather than just the α
position. How would this alteration affect the level of radioactivity found in the precipitate?; c. If
only one of the four deoxyribonucleotides contained 32P, would the precipitate still be
radioactive?
Assuming a normal DNA molecule with approximately equal amounts of the four
deoxyribonucleotides, the answer would be no. All four of the deoxyribonucleotides
are required for DNA synthesis.
b) The DNA would contain no 32P. As shown in the figure below, the β and γ phosphates
are released during the polymerization reaction.
c) Again, assuming that the DNA had approximately equal amounts of each nucleotide,
we would expect the level of radioactivity to be about one-quarter of the radioactivity
present when all deoxyribonucleotides are labeled.
213
a. What strand has been transcribed?; b. How many amino acids in eukaryotes?; c. How
many base sequences to read to code for all amino acids?
a) Template or non-coding strand
b) 20 amino acids
c) 3 base sequences
214
a. What are the codons for Ala?; b. What is meant by codon degeneracy (or redundancy)?;
c. What is the advantage of codon degeneracy?
a) GCU, GCC, GCA, and GCG.
b) Multiple codons (sequences of three nucleotides) can code for the same amino acid.
c) Provides a buffer against mutations and ensures protein function despite some genetic
errors.
215
Why do we have 1 start codon but 3 stop codons?
Redundancy, such as having three stop codons, increases the likelihood of terminating
translation, even if there's a frameshift or mutation that might lead to an incorrect
reading of the mRNA.
216
Write the amino acids coded for by a section of mRNA with the following base sequence:
5ʹ GCC GUA GAC 3ʹ
Some possible codons to use are the following:
GGC = Glycine
GAC = Aspartic acid
CUC = Leucine
GCC = Alanine
GUA = Valine
CGC = Arginine
a) Ala-Val-Asp
217
How can we get a high energy bond in the physiological position?
High-energy bonds in a physiological position, primarily found in ATP, are formed
through the transfer of a phosphate group from ATP to another molecule. This process,
known as phosphorylation, is catalyzed by enzymes called kinases, and it's a crucial
mechanism for driving energetically unfavorable reactions in cells.
218
How many tRNA and tRNA synthetase are there?
20 tRNA and 20 tRNA synthetase
219
Why is the synthesis of aminoacyl-tRNA the crucial step in protein synthesis?
Aminoacyl-tRNA molecules are the actual substrates for protein synthesis.
Furthermore, the synthesis of the aminoacyl-tRNA is the step at which the translation
of nucleic acid information into amino acid information takes place.
220
The genetic code is said to be degenerate because
A. more than one codon may code for a specific amino acid.
B. a codon may specify more than one amino acid.
C. a codon may code for no amino acid.
D. an equal number of codons code for each amino acid.
E. the third base of the codon is irrelevant in determining the amino acid it codes for
A. more than one codon may code for a specific amino acid.
221
One benefit of the degeneracy of the genetic code is that
A. it conserves material, as the same codon can be used multiple times across different
genes.
B. it lessens the chance of deleterious mutations.
C. it facilitates the use of a nonoverlapping, unpunctuated code.
D. it allows for the tRNA to proofread the mRNA sequence and correct any mistakes.
E. it ensures that methionine will be used only at the start of translation.
B. it lessens the chance of deleterious mutations.
222
The translation process reads the mRNA in the _____ direction and synthesizes the protein
in the _____ direction.
A. 5ʹ-to-3ʹ; amino-to-carboxyl
B. 3ʹ-to-5ʹ; carboxyl-to-amino
C. 5ʹ-to-3ʹ; carboxyl-to-amino
D. 3ʹ-to-5ʹ; amino-to-carboxyl
A. 5ʹ-to-3ʹ; amino-to-carboxyl
223
During attachment of an amino acid to its tRNA, how is ATP required?
A. The free energy of hydrolysis of ATP is used to provide the energy necessary to link the
free amino acid to the tRNA.
B. The ATP forms an adenylate-tRNA complex that is activated in such a way that the
adenylate can be displaced by the amino acid.
C. The ATP phosphorylates the 3ʹ −OH3ʹ −OH of the tRNA that is activated in such a way
that the phosphate can be displaced by the amino acid.
D. The ATP forms an aminoacyl adenylate that transfers the activated amino acid to the
tRNA.
E. The ATP phosphorylates the carboxyl terminal of the amino acid, which transfers the
activated amino acid to the tRNA.
D. The ATP forms an aminoacyl adenylate that transfers the activated amino
acid to the tRNA.
224
The codon AUG specifies either initiation of protein synthesis or an internal methionine.
What would be the sequence of the anticodon of a tRNA that binds to this codon?
A. 5ʹ-TAC-3ʹ
B. 5ʹ-UAC-3ʹ
C. 5ʹ-GUA-3ʹ
D. 5ʹ-CAU-3ʹ
E. 5ʹ-CAT-3
D. 5ʹ-CAU-3ʹ
225
Which statement is NOT true of tRNA?
A. tRNA contains an amino acid attachment site.
B. tRNA stands for transfer RNA.
C. tRNA contains a codon.
D. tRNA contains unusual bases.
E. tRNA has an L shape.
C. tRNA contains a codon.
226
What is the direction of the polypeptide?
N-terminus (amino terminus) to C-terminus (carboxyl terminus)
227
During ribosomal protein synthesis, the growing peptide chain moves from the tRNA in the
____ site to the aminoacyl tRNA in the ____ site.
A. P; A
B. E; P
C. A; P
D. P; E
E. A; E
A. P; A
228
Which statement is true of the tRNA-binding sites of the ribosome? The E site stands for
the entry site.
A. The P site stands for the polypeptide channel site.
B. The A site stands for the active site.
C. The active site is located between the E and P sites.
D. The A site is closer to the 3ʹ end of the mRNA molecule being translated than the P site.
D. The A site is closer to the 3ʹ end of the mRNA molecule being translated than
the P site.
229
During protein synthesis, which process is NOT an energy-requiring step?
A. formation of aminoacyl tRNAs
B. delivery of aminoacyl tRNAs to the ribosome
C. formation of the peptide bond
D. translocation of the mRNA and tRNA within the ribosome
E. dissociation of the 70S ribosome after translation termination
C. formation of the peptide bond
230
Termination of protein synthesis occurs when a stop or termination codon is in the A site.
Termination occurs because
A. stop codons recognize specific termination tRNAs that do not have amino acids attached
to them.
B. as the stop codon moves from the A site to the P site, the tRNA containing the newly
synthesized protein is released.
C. stop codons are recognized by specific proteins that release the newly synthesized
protein chain from the ribosome.
D. there are no tRNAs that base-pair with stop codons, so translation stalls and the
ribosome dissociates.
E. EF-G binds to the empty A site and pushes the P site tRNA with attached peptide out of
the ribosome.
C. stop codons are recognized by specific proteins that release the newly
synthesized protein chain from the ribosome.
