1
REVIEW THE CONCEPTS
1. Less energy is required to form noncovalent bonds than covalent bonds, and the
bonds that stick the gecko’s feet to the smooth surface need to be formed and broken
many times as the animal moves. Since van der Waals interactions are so weak, there
must be many points of contact (a large surface area) yielding multiple van der Waals
interactions between the septae and the smooth surface.
2. a. These are likely to be hydrophilic amino acids, and in particular, negatively
charged amino acids (aspartate and glutamate), which would have an affinity
for K+ via ionic bonds.
b. Like the phospholipid bilayer itself, this portion of the protein is likely to be
amphipathic, with hydrophobic amino acids in contact with the fatty acyl chains
and hydrophilic amino acids in contact with the hydrophilic heads.
c, d. Since both the cytosol and extracellular space are aqueous environments,
hydrophilic amino acids would contact these fluids.
3. At pH = 7.0, the net charge is –1 because of the negative charge on the carboxyl resi-
due of glutamate (E). After phosphorylation by a tyrosine kinase, two additional nega-
tive charges (because of attachment of phosphate residues to tyrosines (Y)) would be
added. Thus, the net charge would be –3. The most likely source of phosphate is ATP
since the attachment of inorganic phosphate (Pi
) to tyrosine is energetically highly
unfavorable, but when coupled to the hydrolysis of the high-energy phosphoan-
hydride bond of ATP, the overall reaction is energetically favorable.
2
CHEMICAL FOUNDATIONS

2 CHAPTER 2: Chemical Foundations
4. Disulfide bonds are formed between two cysteine residue side chains. The forma-
tion of disulfide bonds increases the order and therefore decreases the entropy
(S becomes more negative).
5. Stereoisomers are compounds that have the same molecular formula but are
mirror images of each other. Many organic molecules can exist as stereoisomers
because of two different possible orientations around an asymmetric carbon
atom (e.g., amino acids). Because stereoisomers differ in their three-dimensional
orientation and because biological molecules interact with one another based on
precise molecular complementarity, stereoisomers often react with different
molecules, or react differently with the same molecules. Therefore, they may
have very distinct physiological effects in the cell.
6. The compound is guanosine triphosphate (GTP). Although the guanine base is
found in both DNA and RNA, the sugar is a ribose sugar because of the 2′
hydroxyl group. Therefore, GTP is a component of RNA only. GTP is an impor-
tant intracellular signaling molecule.
7. At least three properties contribute to this structural diversity. First, monosaccha-
rides can be joined to one another at any of several hydroxyl groups. Second, the
C-1 linkage can have either an α or a β configuration. Third, extensive branching
of carbohydrate chains is possible.
8. What is the pH of 1 L of water? In all aqueous solutions, water spontaneously
dissociates into hydrogen and hydroxide ions according to the equilibrium reac-
tion H2 O G H+ + OH –
. The ionization constant for aqueous solutions at 25°C
is Kw = [H+ ][OH –
] = 1 × 10–14 M 2 . In a solution of pure water, the production of
one H+ ion will always be accompanied by the production of one OH– ion. In
other words, [H+ ] = [OH–
].
Kw = [H + ][OH– ] = [H+ ]2 = 1 × 10 –14 M 2
[H + ] = 1 × 10 –7 M
pH = –log10
[H +
] = –log10 (1 × 10–7 ) = 7
What is the pH after 0.008 moles NaOH are added? NaOH (sodium hydroxide)
is a strong base. This means all the added NaOH ionizes to increase the [OH–
]
concentration to 0.008 M.
[H + ] = Kw
/[OH–
] = (1 × 10–14 M 2 )/(0.008 M) = 1.25 × 10–12 M
pH = –log10
[H+ ] = –log 10
(1.25 × 10–12 ) = 11.903
What is the pH of the solution of 50 mM MOPS? MOPS is a weak acid. As such,
upon dissolving in water it will undergo partial dissociation yielding equal
concentrations of hydrogen ions and MOPS conjugate base according to the
equilibrium reaction:
MOPS(weak acid form) = H+ + MOPS(conjugate base form)

CHAPTER 2: Chemical Foundations 3
The extent to which this reaction goes forward determines the relative strength
of the MOPS weak acid and is given by its acid dissociation equilibrium
constant:
Ka = ([H+ ][MOPS(conjugate base form)])/[MOPS(weak acid form)]
pKa = –log10Ka = 7.20
If the relative concentrations are known for the weak acid and conjugate base
forms of a dissolved weak acid at equilibrium in water, then the solution pH can
be determined according to the equation:
pH = pKa + log10
([conjugate base]/[weak acid])
Therefore,
pH = 7.20 + log10
(0.39/0.61) = 7.01
What is the pH after 0.008 moles NaOH are added to the MOPS buffer solution?
Rather than simply increase the total [OH–
] by 0.008 moles, addition of the strong
base shifts the equilibrium of the dissolved MOPS such that [MOPS(weak acid)]
decreases by 0.008 M and [MOPS(conjugate base)] increases by 0.008 M.
Before addition of 0.008 moles NaOH:
[MOPS(weak acid)] = 0.61(0.050 M) = 0.0305 M
[MOPS(conjugate base)] = 0.39(0.050 M) = 0.0195 M
After addition of 0.008 moles NaOH:
[MOPS(weak acid)] = 0.0305 M – 0.008 M = 0.0225 M
[MOPS(conjugate base)] = 0.0195 M + 0.008 M = 0.0275 M
The fi nal pH after addition of 0.008 moles NaOH to 50 mM MOPS at pH 7.01 is,
therefore:
pH = 7.20 + log10
(0.0275/0.0225) = 7.29
9. In the acidic pH of a lysosome, ammonia is converted to ammonium ion. Ammo-
nium ion is unable to traverse the membrane because of its positive charge and is
trapped within the lysosome. The accumulation of ammonium ion decreases the
concentration of protons within lysosomes and therefore elevates lysosomal pH.
At neutral pH, ammonia has little, if any, tendency to protonate to ammonium
ion and thus has no effect on cytosolic pH.
10. Keq = [LR]/[L][R]
Since 90% of L binds R, the concentration of LR at equilibrium is 0.9(1 × 10–3
M) =
9 × 10 –4 M. The concentration of free L at equilibrium is the 10% of L that remains

4 CHAPTER 2: Chemical Foundations
unbound, 1 × 10–4 M. The concentration of R at equilibrium is (5 × 10−2 M) –
(9 × 10 –4 M) = 4.91 × 10 –2 M. Therefore, [LR]/[L][R] = 9 × 10–4 M/ ((1 × 10–4 M)
(4.91 × 10 –2 M)) = 183.3 M –1
.
The equilibrium constant is unaffected by the presence of an enzyme.
Kd = 1/Keq = 5. 4 × 10−3 M.
11. Figure 2.28 shows the titration curve for phosphoric acid. In the range of pH
values between 6 and 8, phosphoric acid loses its second proton.
H2
PO −
4 GHPO 4
2− + H + .
The pKa for dissociation of the second proton is 7.2. Thus, at pH 7.2, the
approximate pH of the cytoplasm of cells, about 50 percent of cellular phosphate
is H2
PO 4
− and about 50 percent is HPO 4
2− accorcing to the Henderson–
Hasselbalch equation.
Phosphoric acid is physiologically important because it serves as the buffering
agent in the cytosol. In addition, phosphate is used to covalently modify proteins
(for example for regulation) and small molecules (such as glucose, intermediates
in many metabolic pathways, AMP, ADP, ATP, GTP, etc.).
12. ΔG = ΔGº’ + RTln [products]/[reactants]
For this reaction, ΔG = −1000 cal/mol + [1.987 cal/(degree · mol) × (298 degrees)
× ln (0.01 M/(0.01 M × 0.01 M))].
ΔG = −1000 cal/mol + 2727 cal/mol = 1727 cal/mol
To make this reaction energetically favorable, one could increase the concen-
tration of reactants relative to products such that the term RTln [products]/
[reactants] becomes smaller than 1000 cal/mol. One might also couple this
reaction to an energetically favorable reaction.
13. The presence of one or more carbon-carbon double bonds is indicative of an
unsaturated or polyunsaturated fatty acid. The term saturated refers to the fact
that all carbons, except the carbonyl carbon, have four single bonds. In a cis
unsaturated fatty acid, the carbon atoms flanking the double bond are on the
same side, thus introducing a kink in the otherwise flexible straight chain.
There is no such kink in a trans unsaturated fatty acyl chain.
14. Glutamate is the amino acid that undergoes γ-carboxylation, resulting in the
formation of a host of blood clotting factors. Warfarin inhibits γ-carboxylation of
glutamate. Thus, blood clotting is severely compromised. Patients prone to
forming clots (thrombi) in blood vessels might be prescribed warfarin in order
to prevent an embolism, which would result if the clot dislodged and blocked
another vessel elsewhere in the body. Patients at risk for heart disease due to
blockages in the coronary arteries are also often prescribed this drug.

5
REVIEW THE CONCEPTS
1. The primary structure of a protein is the linear arrangement or sequence of amino
acids. The secondary structure of a protein is the various spatial arrangements that
result from folding localized regions of the polypeptide chain. The tertiary structure of
a protein is the overall conformation of the polypeptide chain, its three-dimensional
structure. Secondary structures, which include the alpha (a) helix and the beta (b)
sheet, are held together by hydrogen bonds. In contrast, the tertiary structure is pri-
marily stabilized by hydrophobic interactions between non-polar side chains of the
amino acids and hydrogen bonds between polar side chains. (The quaternary structure
describes the number and relative positions of the subunits in a multimeric protein.)
2. Despite the fact that folded proteins adopt conformations that are energetically
favorable, the amount of time required for a particular protein to arrive at this confor-
mation on its own can vary significantly. This is especially true if there are other
“quasi-stable” conformations available to the polypeptide. Molecular chaperones func-
tion to protect an unfolded protein from participating in interactions that will take it
off the “pathway” to its native, functional conformation. Chaperonins provide similar
support to an unfolded protein. However, chaperonins can also use encapsulation and
ATPase activity to give energetic “kicks” to misfolded proteins and get them back on
the pathway toward their native folded state.
3. The active site of an enzyme is the region within which the substrate binds and is
converted into product. The turnover number (kcat ) is the rate constant that can be
used to calculate the rate at which a product is formed (V). The Michaelis constant
3
PROTEIN STRUCTURE
AND FUNCTION

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6 CHAPTER 3: Protein Structure and Function
(Km
) is equal to the substrate concentration at which an enzyme will generate
product at precisely one-half its potential maximal velocity. This maximal veloc-
ity (Vmax ) is the theoretical limit to the rate at which product can be formed by a
particular preparation of enzyme. A rate constant never changes for a particular
enzyme. However, the actual measured rate of the chemical reaction will change
in response to many variables including the concentration of enzyme, its affinity
for substrate, the concentration of substrate, the potential of product to inhibit
substrate binding, etc. To calculate a rate (V), one would multiply the turnover
number (kcat
) and [ES] (the concentration of enzyme-substrate). The rate (V) will
equal the maximal rate (Vmax ) when all of the enzyme in a reaction is bound with
substrate ([ES] = [E]Total
).
4. The addition of enzyme does not affect the free energy of either the substrate or
the product. Therefore, the difference in free energy (DG) for a chemical does not
change as a consequence of adding enzyme. The free energy of the enzyme-
substrate complex (ES) is lower for E2. Therefore, E2 binds substrate with greater
affinity than E1. The transition state (X‡
) is stabilized equally well by both E1
and E2. Although both E1 and E2 stabilize X‡ equally well, E2 binds more tightly
to S than E1. This results in a greater activation energy barrier for the reaction
catalysed by E2 than by E1 and, therefore, E1 is the better catalyst. In fact, the
reaction should proceed at the uncatalyzed rate in the presence of E2 because
the height of the barrier between the lowest energy state of the substrate and its
transition state is unchanged. By contrast, in the presence of enzyme E1 that
barrier is significantly smaller resulting in a more rapid transition between
substrate and product.
5. In order for an antibody to catalyze a chemical reaction it should show preferen-
tial binding affinity for the transition state of a chemical reaction. As transition
states are, by definition, high-energy intermediates of chemical reactions rather
than stable molecules, one would first have to synthesize a stable molecule with
chemical properties similar to the transition state and use this “transition state
analog” as an antigen to promote an adaptive immune response in a test animal.
6. Ubiquitin is a 76–amino acid protein that serves as a molecular tag for proteins
destined for degradation. Ubiquitination of a protein involves an enzyme-
catalyzed transfer of a single ubiquitin molecule to the lysine side chain of a tar-
get protein. This ubiquitination step is repeated many times, resulting in a long
chain of ubiquitin molecules. The resulting polyubiquitin chain is recognized by
the proteasome, which is a large, cylindrical, multisubunit complex that proteo-
lytically cleaves ubiquitin-tagged proteins into short peptides and free ubiquitin
molecules. Proteasome inhibitors would be useful to treat cancers if they blocked
the degradation of proteins (e.g., tumor suppressors) required to halt the progres-
sion of uncontrolled cell growth. In the case of the proteasome inhibitor Velcade,
which is used to treat patients with multiple myeloma, cells undergo apoptosis
(programmed cell death), and because a protein serving as a pro-survival factor
called NFkB cannot be activated when proteasome activity is blocked (reviewed
in A. Fribley and C. Y. Wang, Cancer Biol. Ther., 2006 July 1; 5(7):745–8).
7. Cooperativity, or allostery, refers to any change in the tertiary or quaternary
structure of a protein induced by the binding of a ligand that affects the binding

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CHAPTER 3: Protein Structure and Function 7
of subsequent ligand molecules. In this way, a multisubunit protein can respond
more efficiently to small changes in ligand concentration compared to a protein
that does not show cooperativity. The activity of many proteins is regulated by
the reversible addition/removal of phosphate groups to specific serine, threo-
nine, and tyrosine residues. Protein kinases catalyze phosphorylation (the addi-
tion of phosphate groups), while protein phosphatases catalyze dephosphoryla-
tion (the removal of phosphate groups). Phosphorylation/dephosphorylation
changes the charge on a protein, which typically leads to a conformational
change and a resulting increase or decrease in activity. Some proteins are synthe-
sized as inactive propeptides, which must be enzymatically cleaved to release an
active protein.
8. Proteins can be separated by mass by centrifuging them through a solution of
increasing density, called a density gradient. In this separation technique, known
as rate-zonal centrifugation, proteins of larger mass generally migrate faster than
proteins of smaller mass. However, this is not always true because the shape of
the protein also influences the migration rate. Gel electrophoresis can also sepa-
rate proteins based on their mass. In this technique, proteins are separated
through a polyacrylamide gel matrix in response to an electric field. Because the
migration of proteins through a polyacrylamide gel is also influenced by shape
of proteins, the ionic detergent sodium dodecyl sulfate is added to denature pro-
teins and force proteins into similar conformations. During rate-zonal centrifuga-
tion, a protein of larger mass (transferrin) will sediment faster during centrifu-
gation, whereas a protein of smaller mass (lysozyme) will migrate faster during
electrophoresis.
9. Gel filtration, ion exchange, and affinity chromatography typically involve the
use of a bead consisting of polyacrylamide, dextran or agarose packed into a col-
umn. In gel filtration chromatography, the protein solution flows around the
spherical beads and interacts with depressions that cover the surface of the
beads. Small proteins can penetrate these depressions more readily than larger
proteins and thus spend more time in the column and elute later from the col-
umn; larger proteins do not interact with these depressions and elute first from
the column. In ion-exchange chromatography, proteins are separated on the basis
of their charge. The beads in the column are covered with amino or carboxyl
groups that carry a positive or negative charge, respectively. Positively charged
proteins will bind to negatively charged beads, and negatively charged proteins
will bind to positively charged beads. In affinity chromatography, ligand mole-
cules that bind to the protein of interest are covalently attached to beads in a col-
umn. The protein solution is passed over the beads and only those proteins that
bind to the ligand attached to the beads will be retained, while other proteins are
washed out. The bound protein can later be eluted from the column using an
excess of ligand or by changing the salt concentration or pH.
10. Proteins can be made radioactive by the incorporation of radioactively labeled
amino acids during protein synthesis. Methionine or cysteine labeled with sul-
fur-35 are two commonly used radioactive amino acids, although many others
have also been used. The radioactively labeled proteins can be detected by auto-
radiography. In one example of this technique, cells are labeled with a radioac-
tive compound and then overlaid with a photographic emulsion sensitive to

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8 CHAPTER 3: Protein Structure and Function
radiation. The presence of radioactive proteins will be revealed as deposits of
silver grains after the emulsion is developed. A Western blot is a method for
detecting proteins that combines the resolving power of gel electrophoresis, the
specifi city of antibodies, and the sensitivity of enzyme assays. In this method,
proteins are first separated by size using gel electrophoresis. The proteins are
then transferred onto a nylon filter. A specific protein is then detected by use of
an antibody specific for the protein of interest (primary antibody) and an
enzyme-antibody conjugate (secondary antibody) that recognizes the primary
antibody. The presence of this protein-primary antibody-enzyme-conjugated
secondary antibody complex is detected using an assay specific for the conju-
gated enzyme.
11. X-ray crystallography can be used to determine the three-dimensional structure
of proteins. In this technique, x-rays are passed through a protein crystal. The
diffraction pattern generated when atoms in the protein scatter the x-rays is a
characteristic pattern that can be interpreted into defined structures. Cryoelec-
tron microscopy involves the rapid freezing of a protein sample and examina-
tion with a cryoelectron microscope. A low dose of electrons is used to generate
a scatter pattern that can be used to reconstruct the protein’s structure. In
nuclear magnetic resonance (NMR) spectroscopy, a protein solution is placed in
a magnetic field and the effects of different radio frequencies on the spin of
different atoms are measured. From the magnitude of the effect of one atom on
an adjacent atom, the distances between residues can be calculated to generate
a three-dimensional structure.
X-ray crystallography can provide extremely high-resolution structural
information on molecules and molecular complexes of any size. The principal
disadvantage of x-ray crystallography is the challenge of producing samples in
the form of single crystals suitable for diffraction experiments. NMR spectros-
copy gives high-resolution information on protein structures in solution. It also
is ideal for monitoring protein dynamics. However, NMR spectroscopy is lim-
ited in its ability to conclusively determine the structures of very large proteins
and symmetrical macromolecular assemblies. The principal advantage of elec-
tron microscopy is the relative ease of sample preparation. However, structural
resolution is generally not so high as with the other methods, especially for
asymmetric assemblies. NMR is better for small proteins. Electron microscopy
and x-ray crystallography, so long as a suitable crystal can be obtained, are ideal
for large proteins and macromolecular assemblies.
12. The four features of a mass spectrometer are 1) the ion source, 2) the mass
analyzer, 3) the detector, and 4) a computerized data system. Basically, the
investigator would collect protein samples from the cancerous cells and from
the normal healthy cells, the latter serving as a control. Samples would be
prepared for 2D PAGE and after electrophoresis the gels would be dyed and
the profi les compared. If a protein “spot” were present in the sample from the
cancer cell and not the control, it would be isolated out of the gel, protease-
digested using trypsin to generate peptides that are mixed with a matrix, and
applied to a metal target. A laser is used to ionize the peptides, which are
vaporized into singly charged ions. In the case of a time of flight (TOF) mass
analyzer, the time it takes the ions to pass through the analyzer before reaching

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CHAPTER 3: Protein Structure and Function 9
the detector is inversely proportional to its mass and directly proportional to the
charge they carry, generating a spectrum in which each molecule has a distinct
signal, allowing the investigator to calculate each ion’s mass. The fourth essential
component is a computerized data system that acquires and stores the data,
which are then compared to information in databases. The mass and charge
signature, or fingerprint, of the unknown is compared to that of peptides in a
database and the best match protein is identified.

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11
REVIEW THE CONCEPTS
1. Electron microscopy has better resolution than light microscopy, but many light-
microscopy techniques allow observation and manipulation of living cells.
2. The total magnification of an image is described as the product of the magnification
of the individual lenses, where the objective lens magnification immediately above the
specimen is multiplied to that of the projection or eyepiece lens. Being able to clearly
distinguish between two closely spaced points at even the highest total magnification
is the ultimate goal because if the two objects are already blurred and cannot be dis-
criminated at a lower magnification, simply increasing the magnification will have no
effect. In fact, the formula defining the resolution (D) of a lens does not take magnifi-
cation into account and is written as D = 0.61l N sin a, where l is the wavelength of
light used to illuminate the specimen, N is the refractive index of the medium (usually
air) between the front face of the objective lens and the specimen, and a is the half-
angle of the cone of light entering the face of the objective lens. N sin a is often referred
to as the lens’ numerical aperture, which is physically stamped on the barrel of the
objective lens. Since only three of the values can be altered to achieve the best resolu-
tion (the smallest D possible), one has to either decrease the wavelength of light or
increase the numerical aperture by gathering more light into the front face of the
objective lens. In most circumstances, therefore, the limitations include the use of
wavelengths in the visible spectrum and the ability to gather more light to increase the
numerical aperture. Increasing the numerical aperture is accomplished by placing a
drop of oil or water, which have greater refractive indices (1.5 and 1.3, respectively)
relative to that of air (1), between the specimen and the objective lens.
4
CULTURING AND
VISUALIZING CELLS

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12 CHAPTER 4: Culturing and Visualizing Cells
3. Chemical stains are required for visualizing cells and tissues with the basic light
microscope because most cellular material does not absorb visible light and
therefore cells are essentially invisible in a light microscope. The chemical stains
that may be used to absorb light and thereby generate a visible image usually
bind to a certain class of molecules rather than a specific molecule within that
class. For example, certain stains may reveal where proteins are in a cell but not
where a specific protein is located. This limitation can be overcome by fluores-
cence microscopy, in which a fluorescent molecule may be either directly or indi-
rectly attached to a molecule of interest which is then viewed by an appropriately
equipped microscope. Only light emitted by the sample will form an image, so
the location of the fluorescence indicates the location of the molecule of interest.
Confocal scanning microscopy and deconvolution microscopy build on the abil-
ity of fluorescence microscopy by using either optical (confocal scanning) or com-
putational (deconvolution) techniques to remove out-of-focus fluorescence and
thereby produce much sharper images. As a result, these techniques facilitate
optical sectioning of thick specimens as opposed to physical sectioning and asso-
ciated techniques that may alter the specimen.
4. Certain electron microscopy methods rely on the use of metal to coat the speci-
men. The metal coating acts as a replica of the specimen, and the replica rather
than the specimen itself is viewed in the electron microscope. Methods that use
this approach include metal shadowing, freeze fracturing, and freeze etching.
Metal shadowing allows visualization of viruses, cytoskeletal fibers, and even
individual proteins, while freeze fracturing and freeze etching allow visualiza-
tion of membrane leaflets and internal cellular structures.
5. A cell strain is a lineage of cells originating from a primary culture taken from an
organism. Since these cells are not transformed, they have a limited lifespan in
culture. In contrast, a cell line is made of transformed cells and therefore these
cells can divide indefinitely in culture. Such cells are said to be immortal. A clone
results when a single cell is cultured and gives rise to genetically identical prog-
eny cells.
6. Normal B lymphocyte cells can produce a single type of antibody molecule.
However, such cells have a finite lifespan in culture. Researchers use cell fusion
of B lymphocytes and immortalized myeloma cells to create immortalized,
antibody-secreting cells. Such cells, called hybridoma cells, retain characteristics
of both parent cells, allowing for production of a single-type, or monoclonal,
antibody.
7. Specific types of cells in suspension may be isolated by a fluorescence-activated
cell sorter (FACS) machine in which cells previously “tagged” with a fluorescent-
labeled antibody are separated from cells not recognized by the antibody. The
scientist selects an antibody specific for the cell type desired. Specific organelles
are generally separated by centrifugation of lysed cells. A series of centrifuga-
tions of successive supernatant fractions at increasingly higher speeds and corre-
sponding higher forces serves to separate cellular organelles from one another on
the basis of size and mass (larger, heavier cell components pellet at lower

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CHAPTER 4: Culturing and Visualizing Cells 13
speeds). This is often combined with density-gradient separations to purify spe-
cifi c organelles on the basis of their buoyant density.
8. FACS (see Figure 9-2), whereby labeled cells pass through a laser light beam and
the fluorescent intensity of light emitted is measured, allowing the computer to
assign each cell with an electric charge proportional to the fluorescence. Fibro-
blasts having twice the amount of DNA (G2 phase) compared to the normal dip-
loid cells will emit more fluorescence and therefore have a different electric
charge, which allows them to be separated and collected.

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15
REVIEW THE CONCEPTS
1. Watson-Crick base pairs are interactions between a larger purine and a smaller pyri-
midine base in DNA. These interactions result in primarily G-C and A-T base pairing
in DNA and A-U base pairs in double-stranded regions of RNA. They are important
because they allow one strand to function as the template for synthesis of a comple-
mentary, antiparallel strand of DNA or RNA.
2. At 90°C, the double-stranded DNA template will denature and the strands will
separate. As the temperature slowly drops below the Tm of the plasmid DNA, the
single-stranded oligonucleotide primer present at higher concentration than the
plasmid DNA strands hybridizes to its complementary sequence on the plasmid
template. The resulting molecules contain a short double-stranded stretch the length
of the primer with a free 3’ OH that can be used by DNA polymerase enzyme in
sequencing reactions.
3. RNA is less stable chemically than DNA because of the presence of a hydroxyl group
on C-2 in the ribose moieties in the backbone. Additionally, cytosine (found in both
RNA and DNA) may be deaminated to give uracil. If this occurs in DNA, which does
not normally contain uracil, the incorrect base is recognized and repaired by cellular
enzymes. In contrast, if this deamination occurs in RNA, which normally contains
uracil, the base substitution is not corrected. Thus, the presence of deoxyribose and
thymine make DNA more stable and less subject to spontaneous changes in nucleotide
sequence than RNA. These properties might explain the use of DNA as a long-term
information-storage molecule.
5
FUNDAMENTAL MOLECULAR
GENETIC MECHANISMS

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Solution Manual for Molecular Cell Biology , 8th Edition

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