Name: __________________________ Date: _____________ NUID:
Questions from the first lecture:
1. How can just a few elements give rise to all biological diversity? At what level, if any,
are all biological organisms similar? Given this biochemical similarity, how is the
structural and functional diversity of living things possible?
Ans: Living things are composed primarily of macromolecules, polymers of simple
compounds of just a few different types. The properties of these polymers are
determined by their sequence of monomers and these can be combined in many
different ways. Diversity is thus achieved through the nearly limitless variety of
sequences that can exist when amino acids are linked to form proteins, nucleotides
are linked to form nucleic acids, and monosaccharides are linked to form
polysaccharides. Branching in the latter can contribute additional heterogeneity.
Each type of organism constructs a unique set of macromolecules from these
monomeric units, resulting in the structural and functional diversity among
species.
2. What are five periodic elements most frequently seen incorporated into compounds of
biological organisms? Name three occasionally seen elements too.
Ans: Hydrogen, Oxygen, Nitrogen, Carbon, Phosphorus and Sulfur are the most
frequently seen elements. Chloride, Sodium, Potassium, and Calcium are
frequently seen, but not incorporated into compounds. The three elements could
include iron, selenium, magnesium, vanadium, chromium, manganese,
molybdenum, cobalt, nickel, copper, or zinc.
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3. Draw the structures of the following functional groups in their un-ionized forms:
(a) hydroxyl, (b) carboxyl, (c) amino, (d) phosphoryl.
Ans:
4. (a) List the types of noncovalent interactions that are important in providing stability to
the three-dimensional structures of macromolecules. (b) Why is it important that these
interactions be noncovalent, rather than covalent, bonds?
Ans: (a) Noncovalent interactions include hydrogen bonds, ionic interactions between
charged groups, van der Waals interactions, and hydrophobic interactions. (b)
Because noncovalent interactions are weak, they can form, break, and re-form
more rapidly and with less energy input than can covalent bonds. This is
important to maintain the flexibility needed in macromolecules.
5. Why is an asymmetric carbon atom called a chiral center?
Ans: An asymmetric carbon has four different substituents attached, and cannot be
superimposed on its mirror image—as a right hand cannot fit into a left glove.
Thus, a molecule with one chiral carbon will have two stereoisomers, which may
be distinguishable from one another in a biological system.
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6. A chemist working in a pharmaceutical lab synthesized a new drug as a racemic
mixture. Why is it important that she separate the two enantiomers and test each for its
biological activity?
Ans: Biomolecules such as receptors for drugs are stereospecific, so each of the two
enantiomers of the drug may have very different effects on an organism. One
may be beneficial, the other toxic; or one enantiomer may be ineffective and its
presence could reduce the efficacy of the other enantiomer.
7. How is the genetic information encoded in DNA and how is a new copy of DNA
synthesized?
Ans: The genetic information is encoded in the linear sequence (order) of the four
different deoxyribonucleotides in the DNA. When a new copy of DNA is needed,
the two strands of the DNA unwind and each strand serves as a template on which
a new strand is synthesized.
8. Name two functions of (a) proteins, (b) nucleic acids, (c) polysaccharides, (d) lipids.
Ans: Many answers are possible including: (a) proteins function as enzymes, structural
elements, signal carriers, transporters; (b) nucleic acids store and transmit genetic
information and act as both structural and catalytic elements; (c) polysaccharides
serve as energy-yielding fuel stores and cellular and extracellular structural and
recognition elements; (d) lipids function as membrane components, fuel stores,
and cellular signals.
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Questions from the second lecture:
9. How does the electronegativity of each atom affect the polarity of a bond? Use
electronegativity to explain why water is a good solvent.
Ans: Electronegativity describes the tendency of an atom to attract electrons / electron
density towards itself. The electronegativity of hydrogen is 2.1, while that of
oxygen is 3.5. Thus, electron density is more towards the oxygen, polarizing the
bond between hydrogen and oxygen. This makes water a good solvent, especially
for ions, because the negative and positive dipoles of the water can partially
satisfy the charges on the ions.
10. Explain the fact that ethanol (CH3CH2OH) is more soluble in water than is ethane
(CH3CH3).
Ans: Ethanol can form hydrogen bonds with water molecules, but ethane cannot.
When ethanol dissolves, the decrease in the system's entropy that results from
formation of ordered arrays of water around the CH3CH2– group is partly
compensated by the favorable interactions (hydrogen bonds) of the hydroxyl
group of ethanol with water molecules. Ethane cannot form such hydrogen bonds.
11. Describe how van der Waal's interactions work. What types of molecules can participate
in van der Waal's interactions?
Ans: Van der Waal's interactions work because of mutually attractive induced dipoles.
They are highly dependent on the distance between two participating atoms. They
are very weak, but any atoms can participate in them.
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12. Phosphoric acid (H3PO4) has three dissociable protons, with the pKa's shown below.
Which form of phosphoric acid predominates in a solution at pH 4? Explain your
answer.
pKa
Acid
H3PO4
2.14
H2PO4–
6.86
2–
HPO4
12.4
Ans: At pH 4, the first dissociable proton (pKa = 2.14) has been titrated completely, and
the second (pKa = 6.86) has just started to be titrated. The dominant form at pH 4
is therefore H2PO4–, the form with one dissociated proton (see Fig. 2-15).
13. Define pKa for a weak acid in the following two ways: (1) in relation to its acid
dissociation constant, Ka, and (2) by reference to a titration curve for the weak acid.
Ans: (1) pKa = –log Ka. (2) See Fig. 2-17, p. 59; pKa is the value of pH at the inflection
point in a plot of pH vs. extent of titration of the weak acid. At the pKa, the
concentration of ionized acid equals the concentration of un-ionized acid.
14. Give the general Henderson-Hasselbalch equation and sketch the plot it describes (pH
against amount of NaOH added to a weak acid). On your curve, label the pKa for the
weak acid and indicate the region in which the buffering capacity of the system is
greatest.
Ans: The inflection point, which occurs when the weak acid has been exactly one half
titrated with NaOH, occurs at a pH equal to the pKa of the weak acid. The region
of greatest buffering capacity (where the titration curve is flattest) occurs at pH
values of pKa ±1. (See Fig. 2-17, p. 59.)
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15. You have just made a solution by combining 50 mL of a 0.1 M sodium acetate solution
with 150 mL of 1 M acetic acid (pKa = 4.7). What your solution's pH?
Ans: pH = pKa + log
[conjugatebase]
= 4.7 + log (5/150)
[acid]
= 4.7 – 1.48 = 3.22
16. 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  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.
17. Only one of the common amino acids has no free -amino group. Name this amino acid
and draw its structure.
Ans: The amino acid L-proline has no free -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).
18. 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.)
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19. 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%.
[conjugatebase]
[acid]
[acid]
[conjugate base]
pH = pKa + log
pKa – pH = log
antilog (pKa – pH) =
[acid]
[conjugate base]
antilog (6.0 – 5.4) =
[acid]
[conjugate base]
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.
Questions from the third lecture:
20. 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.
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21.
A
Tyr-Lys-Met
B
Gly-Pro-Arg
C
Asp-Trp-Tyr
D
Asp-His-Glu
E
Leu-Val-Phe
Which one of the above tripeptides:
____(a) is most negatively charged at pH 7?
____(b) will not form an alpha helix or beta sheet?
____(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) B; (c) E; (d) A; (e) C
22. Any given protein is characterized by a unique amino acid sequence (primary structure)
and three-dimensional (tertiary) structure. How are these related?
Ans: The three-dimensional structure is determined by the amino-acid sequence. This
means that the amino-acid sequence contains all of the information that is
required for the polypeptide chain to fold up into a discrete three-dimensional
shape.
23. When a polypeptide is in its native conformation, there are weak interactions between
its R groups. However, when it is denatured there are similar interactions between the
protein groups and water. What then accounts for the greater stability of the native
conformation?
Ans: In the unfolded polypeptide, there are ordered solvation shells of water around the
protein groups. The number of water molecules involved in such ordered shells is
reduced when the protein folds, resulting in higher entropy. Hence, the lower free
energy of the native conformation.
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24. Draw the resonance structure of a peptide bond, and explain why there is no rotation
around the C—N bond.
Ans: The intermediate resonance structure imparts a partial double bond characteristic
to the C—N bond, thereby prohibiting rotation. (See Fig. 4-2, p. 116.)
25. Draw the hydrogen bonding typically found between two residues in an  helix.
Ans: Hydrogen bonds occur between every carbonyl oxygen in the polypeptide
backbone and the peptide —NH of the fourth amino acid residue toward the
amino terminus of the chain. (See Fig. 4-2, p. 116.)
26. Describe three important features of an -helix structure. Provide one or two sentences
describing why each feature is important.
Ans: The -helical structure of a polypeptide is tightly wound around a long central
axis; each turn of the right-handed helix contains 3.6 residues and stretches 5.4 Å
along the axis. The peptide NH is hydrogen-bonded to the carbonyl oxygen of the
fourth amino acid along the sequence toward the amino terminus. The R groups
of the amino acid residues protrude outward from the helical backbone.
27. Describe three of the important features of a  sheet polypeptide structure. Provide one
or two sentences for each feature.
Ans: In the  sheet structure, several extended polypeptides, or two regions of the same
polypeptide, lie side by side and are stabilized by hydrogen bonding between
adjacent chains. Adjacent chains may be either parallel (with a repeat distance of
about 6.5 Å) or antiparallel (7 Å repeat). The R groups are often small and
alternately protrude from opposite faces of the  sheet.
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28. Why are glycine and proline often found within a  turn?
Ans: A  turn results in a tight 180° reversal in the direction of the polypeptide chain.
Glycine is the smallest and thus most flexible amino acid, and proline can readily
assume the cis configuration, which facilitates a tight turn.
29. Explain how circular dichroism spectroscopy could be used to measure the denaturation
(unfolding) of a protein.
Ans: Circular dichroism spectroscopy measures the amount of -helix in a given
protein. As the protein denatures, the amount of -helix should decrease as the
protein chain becomes disordered; this change would be detectable using CD
spectrography.
30. How can changes in pH alter the conformation of a protein?
Ans: Changes in pH can influence the extent to which certain amino acid side chains
(or the amino and carboxyl termini) are protonated. The result is a change in net
charge on the protein, which can lead to electrostatic attractions or repulsions
between different regions of the protein. The final effect is a change in the
protein's three-dimensional shape or even complete denaturation.
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Questions from the fourth lecture:
31. Name four factors (bonds or other forces) that contribute to stabilizing the native
structure of a protein, and describe one condition or reagent that interferes with each
type of stabilizing force.
Ans: Any of the following forces stabilize native protein structures and are disrupted by
the listed conditions or reagents: (a) disulfide bonds by reducing conditions or
mercaptoethanol or dithiothreitol, (b) hydrogen bonds by pH extremes, high salt
or heat, (c) hydrophobic interactions of non-polar groups in aqueous solvent by
detergents, urea or guanidine hydrochloride, or organic solvent, (d) ionic
interactions by changes in pH or ionic strength, and (e) van der Waals interactions
by any unfolding condition.
32. What important concepts regarding protein thermal denaturation can be inferred from
the egg white of a boiled egg?
Ans: 1) Denatured proteins often precipitate and/or aggregate. 2) Denaturation is often
not a reversible process.
33. Once a protein has been denatured, how can it be renatured? If renaturation does not
occur, what might be the explanation?
Ans: Because a protein may be denatured through the disruption of hydrogen bonds
and hydrophobic interactions by salts or organic solvents, removal of those
conditions will reestablish the original aqueous environment, often permitting the
protein to fold once again into its native conformation. If the protein does not
renature, it may be because the denaturing treatment removed a required
prosthetic group, or because the normal folding pathway requires the presence of
a polypeptide chain binding protein or molecular chaperone. The normal folding
pathway could also be mediated by a larger polypeptide, which is then cleaved
(e.g., insulin). Denatured insulin would not refold easily.
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34. Each of the following reagents or conditions will denature a protein. For each, describe
in one or two sentences what the reagent/condition does to destroy native protein
structure.
(a) urea / guanidine hydrochloride
(b) high temperature
(c) detergent
(d) low pH
Ans: (a) Urea or guanidine hydrochloride acts primarily by disrupting hydrophobic
interactions.
(b) High temperature provides thermal energy greater than the strength of the
weak interactions (hydrogen bonds, electrostatic interactions, hydrophobic
interactions, and van der Waals forces, breaking these interactions.
(c) Detergents bind to hydrophobic regions of the protein, preventing hydrophobic
interactions among several hydrophobic patches on the native protein.
(d) Low pH causes protonation of the side chains of Asp, Glu, and His, preventing
electrostatic interactions.
35. What is the pI, and how is it determined for a protein?
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.
Hypothetically, one can average all the pKa's of a proteins amino acids to
determine the pI. However in practice, some of these amino acids are satisfied by
the proteins tertiary or quaternary structure. Thus, to be most accurate, pI needs to
be determined empirically by isoelectric focusing gel.
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36. 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.
37. A biochemist has accidentally isolated a second protein with the protein they were
trying to isolate. The reason for its co-purification is unknown, as is its identity. How
might the biochemist have identified its presence and what should they do to determine
its identity?
Ans: The biochemist most likely identified the second protein in a PAGE gel, either
SDS or isoelectric focusing or both. Since the biochemist presumably knows
what species they are working with, they only need a small amount of sequence to
correctly identify the protein. In this circumstance, either Edman degradation or
mass spectrometry may be performed. However, mass spectrometry is the most
likely to be used, since it is more readily available.
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38. 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.
39. Describe a reservation about the use of x-ray crystallography in determining the threedimensional structures of biological molecules.
Ans: To obtain an x-ray picture of a biomolecule, the molecule must be purified and
crystallized under laboratory conditions far different from those encountered by
the native molecule. Biomolecules in the cell also have more flexibility and
freedom of motion than can be accommodated in a rigid crystal structure.
Therefore, the static picture obtained from an x-ray analysis of a crystal may not
provide a complete or accurate representation of the biomolecule in vivo.
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40. Name two ways to determine the precise (high-resolution) three-dimensional structure
of a protein complex. Describe why you think these methods are appropriate.
Ans: The protein complex could be crystallized, and its structure determined by x-ray
crystallography. The pattern of diffracted x-rays yields, by Fourier
transformation (black magic), the three-dimensional distribution of electron
density. By matching electron density with the known sequence of amino acids in
the protein, each region of electron density is identified as a single atom. This
would be an appropriate choice if the complex can be crystallized, but
inappropriate if not.
Sometimes, the three-dimensional structure of a small protein or peptide can be
determined in solution by sophisticated analysis of the NMR spectrum of the
polypeptide. However, this technique is rarely appropriate for complexes, which
usually exceed the complexitiy limit.
A three dimensional structure can also be determined by cryo-electron
tomography, which is an excellent method to use with complexes that don't
crystallize. If an advanced machine is available, molecular details may be
sufficient to determine high resolution structure. This is most appropriate for
larger complexes, or those with individually crystallized components, but may
also be used as the first approach.
Homology modeling is only an appropriate answer if the above three are likely to
fail, and a good structure of a highly similar protein is available.
Questions from the fifth lecture:
41. For the binding of a ligand to a protein, what is the relationship between the Ka
(association constant), the Kd (dissociation constant), and the affinity of the protein for
the ligand?
Ans: Ka = 1/Kd. The larger the Ka (and hence the smaller the Kd), the higher the affinity
of the protein for the ligand.
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42. Describe how you would determine the Ka (association constant) for a ligand and a
protein.
Ans: An experiment would be carried out in which a fixed amount of the protein is
incubated with varying amounts of ligand (long enough to reach equilibrium).
The fraction of protein molecules that have a molecule of ligand bound is then
determined. A plot of this fraction () vs. ligand concentration [L] should yield a
hyperbola. The value of [L] when  = 0.5 is equal to 1/Ka.
43. What fraction of ligand binding sites are occupied () when [ligand] = Kd? Show your
work.
Ans:
From equation (5-8) on page 156:
[Ligand]
=
[Ligand] + K d
thus, when [Ligand] = K d , the equation becomes:
=
Kd
1

Kd + Kd 2
44. Explain why most multicellular organisms use an iron-containing protein for oxygen
binding rather than free Fe2+. Your answer should include an explanation of (a) the role
of heme and (b) the role of the protein itself.
Ans: (a) Binding of free Fe2+ to oxygen would result in the formation of reactive
oxygen species that can damage biological structures. Heme-bound iron is less
reactive in this regard. (b) Binding of oxygen to free heme can result in
irreversible oxidation of the Fe2+ to Fe3+ that does not bind oxygen. The
environment of the heme group in proteins helps to prevent this from occurring.
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45. Explain why the structure of myoglobin makes it function well as an oxygen-storage
protein, whereas the structure of hemoglobin makes it function well as an oxygentransport protein.
Ans: The hyperbolic binding of oxygen to the single binding site of myoglobin results
in a high affinity even at the relatively low partial pressures of O2 that occur in
tissues. In contrast, the cooperative (sigmoidal) binding of O2 to the multiple
binding sites of hemoglobin results in high affinity at high partial pressures such
as occur in the lungs, but lower affinity in the tissues. This permits hemoglobin to
bind O2 in the lungs and release it in the tissues.
46. How does BPG binding to hemoglobin decrease its affinity for oxygen?
Ans: BPG binds to a cavity between the  subunits. It binds preferentially to molecules
in the low-affinity T state, thereby stabilizing that conformation.
47. Fetal hemoglobin binds BPG with lower affinity than adult hemoglobin. How does this
property facilitate tranfers of O2 from mother to fetus?
Ans: Lower affinity for BPG means that fetal hemoglobin will have less BPG bound
that the mother's hemoglobin. This shifts the fetus' fractional O2 saturation curve
to the left (i.e., lower p50) of the mother's. At low pO2, O2 will dissociate from the
maternal hemoglobin and can be bound by the fetal hemoglobin.
48. Why is carbon monoxide (CO) toxic to aerobic organisms?
Ans: It binds to heme with a higher affinity than oxygen, and thus prevents oxygen
from binding to hemoglobin.
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49. Describe briefly the structure of myosin.
Ans: Myosin contains two copies of a large polypeptide (heavy chain) and four copies
of a small polypeptide (light chain). The  helix contributes significantly to the
structure of the heavy chains. At their carboxyl termini, the heavy chains are
wrapped around each other in a fibrous left-handed coil. At their amino termini,
they each have a globular domain with which the light chains are associated.
50. What is the role of ATP and ATP hydrolysis in the cycle of actin-myosin association
and disassociation that leads to muscle contraction?
Ans: ATP binding to myosin results in a conformational change that causes
dissociation of actin from the myosin. ATP hydrolysis results in a change of
orientation of the myosin relative to the actin filament, which allows movement to
the next actin subunit. This is followed initially by release of the phosphate
hydrolysis product and weak binding of the myosin to this actin subunit, and,
subsequently, by tight binding and release of the ADP hydrolysis product.
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