BIO 892 Midterm 2

subject Type Homework Help
subject Pages 9
subject Words 3078
subject Authors Alexander D. Johnson, Bruce Alberts, Julian Lewis

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An amino acid residue that is not part of the active site of an enzyme and does not
interact with the ligand is nevertheless critical for ligand binding and is highly
conserved. How can this be explained?
A.This residue is critical for the correct folding and the placement of ligand-binding site
residues.
B.The residue helps with restricting the access of water to the ligand-binding site.
C.This residue can affect the chemical properties of the residues in the ligand-binding
site.
D.All of the above.
Which of the following represents an "activated" carrier molecule?
A.AMP
B.NADH
C.NAD+
D.NADP+
E.CoA
In humans, nearly 80% of proteins are acetylated on their N-terminal residue, a
modification known to be recognized by a specific E3 enzyme, which directs the
ubiquitylation of the protein for rapid degradation. Does this mean that all of these
proteins are actively degraded at any given time?
A.Yes; this high turnover rate ensures that their activity is under tight control.
B.Yes; but they are not fully degraded in this way and can still function as protein
fragments.
C.No; the destruction signal can be buried in the interior of the protein or bound to
other proteins.
D.No; the E3 enzyme recognizing this mark is inactive most of the time.
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In classical experiments done half a century ago, the cells of early frog embryos were
disaggregated and later reaggregated in desired combinations. The cells managed to
rearrange and sort themselves out into an overall arrangement similar to that of a
normal embryo. This effect is mainly due to ...
A.chemotactic cell motility.
B.asymmetric cell division.
C.cell adhesion.
D.convergent extension.
E.lateral inhibition.
The actin-nucleating protein formin has flexible "whiskers" containing binding sites
that help recruit actin subunits in order to enhance polymerization by this protein. What
protein would you expect to bind to these sites?
A.Thymosin
B.Profilin
C.Cofilin
D.Gelsolin
E.Tropomodulin
All cells ___
A.have membrane transport proteins.
B.synthesize proteins on the ribosome.
C.replicate their genome by DNA polymerization.
D.transcribe their genetic information by RNA polymerization.
E.All of the above.
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A physician had prescribed a differential white blood cell count to help diagnose a sick
child with sustained fever and weight loss. The results show a significant increase in the
number of eosinophils beyond the normal range for healthy individuals. This is most
likely due to ...
A.bacterial infection.
B.parasitic infection.
C.viral infection.
D.kidney failure.
E.internal bleeding.
"Headless" kinesin mutants only contain the stalk (middle) and tail domains and can
therefore dimerize with their wild-type kinesin partners. However, since they lack the
motor (head) domain, the resulting dimers are unable to carry out processive transport
of their cargoes and the mutation thus behaves as "dominant negative," meaning that the
mutant not only is nonfunctional, but can also interfere with the function of its
wild-type counterparts. If a headless mutant of a kinesin heavy chain involved in
melanosome movement is overexpressed in fish melanocytes, what would you predict
happens in these cells?
A.Pigment dispersion would be inhibited and there would be more tug-of-war between
the motors.
B.Pigment aggregation would be inhibited and there would be more tug-of-war between
the motors.
C.Pigment dispersion would be inhibited and there would be less tug-of-war between
the motors.
D.Pigment aggregation would be inhibited and there would be less tug-of-war between
the motors.
Imagine a prometaphase chromosome pair that is attached to one spindle pole. Which of
the following would happen if both arms of the chromosome are severed with a strong
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laser beam?
A.All chromosome fragments (the centromere-containing fragment and the arm
fragments) would be pushed away from the pole.
B.Only the two arm fragments would be pulled toward the pole. The
kinetochore-containing fragment would remain stationary.
C.The kinetochore-containing fragment would be pulled toward the pole. The two arm
fragments would move away from the pole.
D.The kinetochore-containing fragment would move away from the pole, but the two
arm fragments would be pulled toward the pole.
E.All chromosome fragments would be pulled toward the pole.
Consider the signaling pathway depicted below, which involves two different receptors.
Assume that a protein is active only when it is activated but not inhibited by its
upstream signaling molecules. Activation and inhibition are indicated by normal and
blunt arrows, respectively. Under which of the following conditions is target-gene
expression induced?
A.Only in the presence of both signals (1 and 2)
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B.In the presence of either or both signals (1 or 2)
C.In the presence of either signal but not both of them
D.Only in the absence of both signals
E.Only in the presence of signal 2 but not signal 1
The EcoRI restriction enzyme recognizes and cuts at the sequence GAATTC. Such an
enzyme is expected to cut a typical double-stranded DNA at 1 in every ___ sites.
A.about 400
B.about 4000
C.about 40,000
D.about 400,000
E.about 4 million
Which of the following is true regarding energy production and storage in plants and
animals?
A.Plant and animal cells make starch for long-term energy storage.
B.Most of the ATP in a plant cell has been generated in the chloroplast and transported
to other parts of the cell.
C.Oxidation of one gram of starch releases more energy than oxidation of fat, but since
starch absorbs a lot of water, it is not as efficient as fat in energy storage.
D.Animals, but not plants, can store fats in the form of triacylglycerol (triglyceride).
E.Plant seeds often contain large amounts of fats and starch.
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Transcription regulation has similarities and differences in bacteria and in eukaryotes.
Which of the following is correct in this regard?
A.Most bacterial genes are regulated individually, whereas most eukaryotic genes are
regulated in clusters.
B.The rate of transcription for a eukaryotic gene can vary in a much wider range than
for a bacterial gene (which is, at most, only about 1000-fold).
C.DNA looping for gene regulation is the rule in bacteria but the exception in
eukaryotes.
D.Transcription regulators in both bacteria and eukaryotes usually bind directly to RNA
polymerase.
E.The default state of both bacterial and eukaryotic genomes is transcriptionally active.
Cajal bodies in the eukaryotic cell nucleus___
A.are stockpiles of fully mature snRNPs and other RNA processing components.
B.can only be observed by electron microscopy.
C.are absolutely required in all cell types.
D.speed up the maturation and assembly of snRNPs and snoRNPs.
E.are the main sites of pre-mRNA splicing.
Arabidopsis thaliana seedlings show positive phototropism: they bend and grow toward
the source of (blue) light. However, phototropism is not observed if the seedlings are
treated with an auxin efflux inhibitor called NPA. Knowing that auxin stimulates
elongation of the cells in the growing shoot, these observations are consistent with a
model in which, in the presence of a lateral light, ...
A.auxin efflux transporter proteins become localized to the side of the cell that is facing
the light source.
B.auxin efflux transporter proteins become localized to the side of the cell that is away
from the light source.
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C.auxin influx transporter proteins become localized to the side of the cell that is facing
the light source.
D.auxin influx transporter proteins become localized to the side of the cell that is away
from the light source.
E.auxin efflux transporter proteins become uniformly distributed around the cell.
Which of the following repair pathways can accurately repair a double-strand break?
A.Base excision repair
B.Nucleotide excision repair
C.Direct chemical reversal
D.Homologous recombination
E.Nonhomologous end joining
Indicate whether each of the following descriptions better applies to enveloped (E) or
nonenveloped (N) viruses. Your answer would be a four-letter string composed of
letters E and N only, e.g. NNNN.
( )They include retroviruses.
( )They include adenoviruses.
( )To leave the cell, they normally lyse it.
( )They are more sensitive to treatment with detergents, heat, or drying.
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A certain GTP-binding protein can exist in two main states. When bound to GDP, it is
mostly cytosolic. In its GTP-bound form, however, it associates with the cytosolic face
of the endoplasmic reticulum (ER) membrane, where it hydrolyzes the bound GTP after
a short delay and is released again into the cytosol. You have created and expressed
green fluorescent protein (GFP) fusions of the wild-type protein, as well as that of a
mutant protein that does not bind GTP as readily as the wild type. You then perform a
fluorescence recovery after photobleaching (FRAP) experiment by photobleaching a
small area of the ER membrane and measuring GFP fluorescence recovery over time.
According to the results below, which curve (1 or 2) do you think corresponds to the
wild-type fusion protein? Write down 1 or 2 as your answer.
The Ames test is used to test the mutagenicity of a compound suspected to be a
carcinogen. In a simple form of the test, the carcinogen is first mixed with a rat liver
extract. A disc of filter paper is soaked with this mixture and placed on a culture of a
strain of Salmonella typhimurium that is defective in a gene involved in the synthesis of
histidine, an amino acid that is essential for cell growth and proliferation. The strain is
thus normally unable to grow into visible colonies when the histidine in the culture
medium is depleted. In the presence of a mutagen, however, mutations (often "reverse
mutations" in the same gene) can enable the bacteria to produce histidine on their own,
and therefore grow into colonies. The results of the Ames test for three compounds A,
B, and C€each used at the same concentration€are shown in the schematic diagram
below. Colonies are indicated with black dots, and the disc is indicated with a white
circle at the center of each plate. Which compound (A to C) appears to be a stronger
mutagen in this assay? Write down A, B, or C as your answer.
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In the following simplified diagram, three mechanisms for the horizontal transfer of
virulence genes to an avirulent bacterium are depicted. Indicate which mechanism (A to
C) corresponds to conjugation, transduction, and transformation, respectively. Your
answer would be a three-letter string composed of letters A to C only, e.g. CAB.
Indicate whether each of the following descriptions better applies to the plant cell wall
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(W) or the extracellular matrix of animal cells (M). Your answer would be a three-letter
string composed of letters W and M only, e.g. WMM.
( )It is generally thicker and stronger.
( )It is more rigid.
( )It has higher nitrogen content.
A protein is drawn in the following simplified diagram undergoing a set of covalent
modifications including the addition of a chain of ubiquitin protein monomers (U) to
one of its lysine side chains, a phosphate moiety (P) to its tyrosine side chain, and a
lipid farnesyl (F) to its cysteine side chain. Indicate the order of these modifications (1
to 3) in the diagram. Your answer would be a three-letter string composed of letters F,
U, and P only, e.g. PFU.
Atomic Force Microscopy
Atomic force microscopy (AFM) is used in an experiment to unfold a multidomain
protein by applying mechanical force. The protein contains several copies of an
immunoglobulin domain that are unfolded one by one as the two ends of the molecule
(one attached to a cover slip, and the other to the AFM tip) are being pulled apart,
resulting in the 'sawtooth" force-extension curves shown below. The same experiment is
done twice, once in the presence and once in the absence of a chaperone protein that
stabilizes the immunoglobulin domains. Answer the following questions based on this
graph.
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According to the force-extension graph, how many immunoglobulin domains are
unfolded in each of these experiments? Write down the number as your answer, e.g. 9

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