AAMC FL2 B/B
The information in the passage suggests that in mice CRY1 most likely affects XPA
by:
A.
activating XPA protein activity.
B.
activating translation of XPA-encoding transcripts.
C.
repressing replication of the XPA-encoding gene.
D.
repressing transcription of the XPA-encoding gene.
The answer to this question is D because Figure 1 shows that XPA levels decrease
as CRY1 levels increase.
It shows that as CRY1 increases, XPA activity decreases, indicating an inhibitory
effect of CRY1 on XPA. Since XPA is a protein that is repressed by CRY1, only
option D is viable
Which cells harvested from adult mice were most likely used as the highly
proliferative benchmark in the experiment that generated the data shown in Figure
3?
A.
Adipocytes
B.
Cardiac muscle cells
C.
Gastrointestinal epithelial cells
D.
Neurons
The answer to this question is C because the epithelial cells that line the
gastrointestinal tract are typically highly proliferative.
After a section of a DNA strand containing a UVR-induced lesion is removed and
resynthesized, the newly synthesized strand is rejoined to the remainder of the DNA
strand by what type of bond?
A.
Disulfide
B.
Hydrogen
,C.
Peptide
D.
Phosphodiester
The answer to this question is D because phosphodiester bonds link the 3ʹ carbon
atom of one deoxyribose and the 5ʹ carbon atom of another deoxyribose within the
DNA molecules.
AlP exposed to an aqueous solution in which pH range will result in the largest
amount of phosphine production?
A.
pH < 4
B.
4 < pH < 7
C.
7 < pH < 10
D.
pH > 10
The answer to this question is A because H+is a reactant, and the increase in the
concentration of H+ at low pH will favor product formation.
The lower the pH, the more H+ ions on the reactants side, which pushes the
equlibrium toward the product side due to Le Chatlier's principle
When researchers determined the total cellular concentration of ATP in AlP-exposed
rat liver cells, they found the concentration to be equal to the control value. Which
conclusion about the metabolic state of the cell is best supported by these data?
A.
Glycolytic flux is increased after AlP treatment.
B.
Glycolytic flux is decreased after AlP treatment.
C.
Citric acid cycle flux is increased after AlP treatment.
D.
Citric acid cycle flux is decreased after AlP treatment.
The answer to this question is A because ATP production is the same in both control
and AlP-exposed cells, and the data in the passage show that mitochondrial ATP
production is decreased. This indicates that the flux through glycolysis is increased,
because this would be the major pathway for ATP production once the electron
transport chain is shut down.
, If the rate of ATP levels were reduced by 65% after treatment with AIP as stated in
the passage, then were subsiquently found to be equal to the control value - the only
logical answer is that this was due to an increase of glycolytic flux which makes up
for the decreased ATP levels due to AIP inhibiting the electron transport chain.
Why was it necessary for the researchers to determine the activity of the complexes
independent of one another?
A.
Complex stability is lost if the complexes are able to interact structurally.
B.
The complexes have different cellular locations, and it is not feasible to isolate them
together.
C.
The complexes all use the same substrates, so their use must be monitored
separately.
D.
The reactions catalyzed by the complexes are coupled to one another.
The answer to this question is D because studying the complexes all together would
lead to erroneous results because inhibition of complexes I and II affects the activity
of Complex III, which affects the activity of Complex IV.
A large carbohydrate is tagged with a fluorescent marker and placed in the
extracellular environment around a macrophage. The macrophage ingests the
carbohydrate via phagocytosis. Which cellular structure is most likely to be
fluorescently labeled upon viewing with a light microscope soon after phagocytosis?
A.
Nucleus
B.
Golgi apparatus
C.
Lysosome
D.
Endoplasmic reticulum
The answer to this question is C because when a macrophage ingests foreign
material, the material initially becomes trapped in a phagosome. The phagosome
then fuses with a lysosome to form a phagolysosome. Inside the phagolysosome,
enzymes digest the foreign object. Of the cell structures listed, the labeled
carbohydrate is most likely to be microscopically visualized within a lysosome
(phagolysosome).
Inhibition of phosphofructokinase-1 by ATP is an example of:
allosteric regulation.
feedback inhibition.
competitive inhibition.
A.
I only
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