AAMC FL2 B/B
The information inside the passage shows that during mice CRY1 most possibly impacts XPA
with the aid of:
A.
Activating XPA protein hobby.
B.
Activating translation of XPA-encoding transcripts.
C.
Repressing replication of the XPA-encoding gene.
D.
Repressing transcription of the XPA-encoding gene. - ANSThe solution to this question is D
because Figure 1 shows that XPA tiers decrease as CRY1 degrees increase.
It suggests that as CRY1 will increase, XPA pastime decreases, indicating an inhibitory impact
of CRY1 on XPA. Since XPA is a protein this is repressed through CRY1, best choice D is
possible
Which cells harvested from grownup mice were most probable used as the extraordinarily
proliferative benchmark in the test that generated the information shown in Figure 3?
A.
Adipocytes
B.
Cardiac muscle cells
C.
Gastrointestinal epithelial cells
D.
Neurons - ANSThe answer to this query is C because the epithelial cells that line the
gastrointestinal tract are commonly particularly proliferative.
After a segment of a DNA strand containing a UVR-induced lesion is eliminated and
resynthesized, the newly synthesized strand is rejoined to the remainder of the DNA strand by
way of what kind of bond?
A.
Disulfide
B.
Hydrogen
C.
Peptide
D.
,Phosphodiester - ANSThe solution to this query is D because phosphodiester bonds hyperlink
the 3ʹ carbon atom of 1 deoxyribose and the 5ʹ carbon atom of another deoxyribose inside the
DNA molecules.
AlP uncovered to an aqueous answer wherein pH range will result in the largest amount of
phosphine production?
A.
PH < four
B.
4 < pH < 7
C.
7 < pH < 10
D.
PH > 10 - ANSThe solution to this question is A because H+is a reactant, and the boom inside
the attention of H+ at low pH will prefer product formation.
The decrease the pH, the more H+ ions on the reactants aspect, which pushes the equlibrium
towards the product side because of Le Chatlier's precept
When researchers determined the whole cell concentration of ATP in AlP-exposed rat liver cells,
they found the concentration to be identical to the manage price. Which conclusion about the
metabolic state of the cell is exceptional supported with the aid of those data?
A.
Glycolytic flux is improved after AlP treatment.
B.
Glycolytic flux is decreased after AlP remedy.
C.
Citric acid cycle flux is accelerated after AlP remedy.
D.
Citric acid cycle flux is reduced after AlP remedy. - ANSThe solution to this question is A
because ATP production is the same in both control and AlP-exposed cells, and the data inside
the passage show that mitochondrial ATP manufacturing is reduced. This shows that the flux
through glycolysis is extended, due to the fact this would be the major pathway for ATP
manufacturing once the electron transport chain is shut down.
If the fee of ATP ranges had been reduced through 65% after remedy with AIP as stated in the
passage, then were subsiquently located to be same to the control price - the only logical
solution is this turned into because of an boom of glycolytic flux which makes up for the
decreased ATP degrees due to AIP inhibiting the electron transport chain.
Why was it necessary for the researchers to decide the activity of the complexes independent of
one another?
A.
Complex balance is lost if the complexes are capable of have interaction structurally.
, B.
The complexes have exclusive mobile places, and it isn't always possible to isolate them
collectively.
C.
The complexes all use the same substrates, so their use should be monitored one at a time.
D.
The reactions catalyzed through the complexes are coupled to one another. - ANSThe answer
to this query is D because studying the complexes all together might cause inaccurate effects
due to the fact inhibition of complexes I and II influences the pastime of Complex III, which
impacts the pastime of Complex IV.
A massive carbohydrate is tagged with a fluorescent marker and located within the extracellular
surroundings round a macrophage. The macrophage ingests the carbohydrate through
phagocytosis. Which cellular structure is maximum probable to be fluorescently categorized
upon viewing with a light microscope quickly after phagocytosis?
A.
Nucleus
B.
Golgi apparatus
C.
Lysosome
D.
Endoplasmic reticulum - ANSThe solution to this query is C due to the fact when a macrophage
ingests overseas cloth, the material to start with turns into trapped in a phagosome. The
phagosome then fuses with a lysosome to shape a phagolysosome. Inside the phagolysosome,
enzymes digest the foreign item. Of the cellular structures indexed, the categorized
carbohydrate is maximum probable to be microscopically visualized inside a lysosome
(phagolysosome).
Inhibition of phosphofructokinase-1 with the aid of ATP is an example of:
allosteric regulation.
Comments inhibition.
Aggressive inhibition.
A.
I best
B.
III only
C.
I and II handiest
D.
II and III simplest - ANSThe solution to this question is C because ATP, the stop made of
glycolysis, downregulates through comments inhibition the pastime of phosphofructokinase-1 by
binding to a regulatory web page aside from the lively website of the enzyme (allosteric
The information inside the passage shows that during mice CRY1 most possibly impacts XPA
with the aid of:
A.
Activating XPA protein hobby.
B.
Activating translation of XPA-encoding transcripts.
C.
Repressing replication of the XPA-encoding gene.
D.
Repressing transcription of the XPA-encoding gene. - ANSThe solution to this question is D
because Figure 1 shows that XPA tiers decrease as CRY1 degrees increase.
It suggests that as CRY1 will increase, XPA pastime decreases, indicating an inhibitory impact
of CRY1 on XPA. Since XPA is a protein this is repressed through CRY1, best choice D is
possible
Which cells harvested from grownup mice were most probable used as the extraordinarily
proliferative benchmark in the test that generated the information shown in Figure 3?
A.
Adipocytes
B.
Cardiac muscle cells
C.
Gastrointestinal epithelial cells
D.
Neurons - ANSThe answer to this query is C because the epithelial cells that line the
gastrointestinal tract are commonly particularly proliferative.
After a segment of a DNA strand containing a UVR-induced lesion is eliminated and
resynthesized, the newly synthesized strand is rejoined to the remainder of the DNA strand by
way of what kind of bond?
A.
Disulfide
B.
Hydrogen
C.
Peptide
D.
,Phosphodiester - ANSThe solution to this query is D because phosphodiester bonds hyperlink
the 3ʹ carbon atom of 1 deoxyribose and the 5ʹ carbon atom of another deoxyribose inside the
DNA molecules.
AlP uncovered to an aqueous answer wherein pH range will result in the largest amount of
phosphine production?
A.
PH < four
B.
4 < pH < 7
C.
7 < pH < 10
D.
PH > 10 - ANSThe solution to this question is A because H+is a reactant, and the boom inside
the attention of H+ at low pH will prefer product formation.
The decrease the pH, the more H+ ions on the reactants aspect, which pushes the equlibrium
towards the product side because of Le Chatlier's precept
When researchers determined the whole cell concentration of ATP in AlP-exposed rat liver cells,
they found the concentration to be identical to the manage price. Which conclusion about the
metabolic state of the cell is exceptional supported with the aid of those data?
A.
Glycolytic flux is improved after AlP treatment.
B.
Glycolytic flux is decreased after AlP remedy.
C.
Citric acid cycle flux is accelerated after AlP remedy.
D.
Citric acid cycle flux is reduced after AlP remedy. - ANSThe solution to this question is A
because ATP production is the same in both control and AlP-exposed cells, and the data inside
the passage show that mitochondrial ATP manufacturing is reduced. This shows that the flux
through glycolysis is extended, due to the fact this would be the major pathway for ATP
manufacturing once the electron transport chain is shut down.
If the fee of ATP ranges had been reduced through 65% after remedy with AIP as stated in the
passage, then were subsiquently located to be same to the control price - the only logical
solution is this turned into because of an boom of glycolytic flux which makes up for the
decreased ATP degrees due to AIP inhibiting the electron transport chain.
Why was it necessary for the researchers to decide the activity of the complexes independent of
one another?
A.
Complex balance is lost if the complexes are capable of have interaction structurally.
, B.
The complexes have exclusive mobile places, and it isn't always possible to isolate them
collectively.
C.
The complexes all use the same substrates, so their use should be monitored one at a time.
D.
The reactions catalyzed through the complexes are coupled to one another. - ANSThe answer
to this query is D because studying the complexes all together might cause inaccurate effects
due to the fact inhibition of complexes I and II influences the pastime of Complex III, which
impacts the pastime of Complex IV.
A massive carbohydrate is tagged with a fluorescent marker and located within the extracellular
surroundings round a macrophage. The macrophage ingests the carbohydrate through
phagocytosis. Which cellular structure is maximum probable to be fluorescently categorized
upon viewing with a light microscope quickly after phagocytosis?
A.
Nucleus
B.
Golgi apparatus
C.
Lysosome
D.
Endoplasmic reticulum - ANSThe solution to this query is C due to the fact when a macrophage
ingests overseas cloth, the material to start with turns into trapped in a phagosome. The
phagosome then fuses with a lysosome to shape a phagolysosome. Inside the phagolysosome,
enzymes digest the foreign item. Of the cellular structures indexed, the categorized
carbohydrate is maximum probable to be microscopically visualized inside a lysosome
(phagolysosome).
Inhibition of phosphofructokinase-1 with the aid of ATP is an example of:
allosteric regulation.
Comments inhibition.
Aggressive inhibition.
A.
I best
B.
III only
C.
I and II handiest
D.
II and III simplest - ANSThe solution to this question is C because ATP, the stop made of
glycolysis, downregulates through comments inhibition the pastime of phosphofructokinase-1 by
binding to a regulatory web page aside from the lively website of the enzyme (allosteric
