AAMC Biological and Biochemical Foundations of Living
Systems || Already Passed.
Which steps involved in the contraction of a skeletal muscle require binding and/or hydrolysis of
ATP?
I Dissociation of myosin head from actin filament
II Attachment of myosin head to actin filament
III Conformational change that moves actin and myosin filaments relative to one another
IV Binding of troponin to actin filament
V Release of calcium from the sarcoplasmic reticulum
VI Reuptake of calcium into the sarcoplasm
A. I, II, and III only
B. II, III, and IV only
C. I, III, and VI only
D. III, IV, and VI only correct answers C. I, III, and VI. Dissociation of myosin head from actin
filament, conformational changes that moves actin and myosin filaments relative to one another,
Reuptake of calcium into sarcoplasm
What control experiment was necessary to ensure that the apparent subcellular locations of the
cat6xbs and lacY6xbs transcripts were NOT skewed by the location preference of the bound
MS2-GFP?
A. Determination of expression of MS2-GFP in cells that lacked the 6xbs insertion upstream of
the cat and lacY genes
B. Use of E. coli cells that expressed only MS2 instead of the MS2-GFP fusion protein
C. Insertion of the 6xbs region upstream of both the cat and lacY genes in the same cells
D. Determination of expression of both MS2 and GFP as separate proteins rather than as a fusion
protein correct answers A. Lack of 6xbs left the protein without a binding partner, thus location
was based on protein itself
Question: https://www.mcatofficialprep.org/app/online-practice-questions-official-guide#quiz/
review/categories/28183/832703
Chloramphenicol did NOT inhibit translation in E. coli cells containing the cat6xbs expression
plasmid. What experimental parameter could be changed in order to affect translation inhibition?
A. Increase the chloramphenicol concentration.
B. Increase the chloramphenicol incubation time.
C. Alter the incubation temperature by a few degrees.
D. Use an alternate antibiotic. correct answers D. the cells are resistant to chloramphenicol, thus
a different antibiotic should be tried.
,The protozoan that causes human malaria, Plasmodium falciparum, completes part of its life
cycle inside human mature red blood cells (RBCs). Infected RBCs adhere to platelets,
endothelial cells, and other mature RBCs. The protozoan genome contains approximately 60 var
genes encoding transmembrane protein PfEMP1 variants, which localize to the RBC plasma
membrane and bind to endothelial cells. Only one var gene is transcribed at a time over multiple
mitotic generations, but the var gene that is expressed successively switches during an infection.
The expressed var gene is located in a different place in the periphery of the nucleus than are
silent var genes. Histone H3 on the active var promoter is trimethylated on lysine 4 and
acetylated on lysine 9, whereas on silent var genes, lysine 9 of histone H3 is trimethylated.
Figure 1 shows the relative distances between expressed and silent var genes and either P. falcipa
correct answers PfEMP1 is a parasite protein that is present in the plasma membrane of the RBC
that the parasite inhabits. This suggests that there must be a mechanism for transporting PfEMP1
from the parasite to the RBC plasma membrane.
Lysine and amino acids with similar chemical characteristics in histones most likely promote the
interaction of histones with which DNA components?
A. Purines
B. Pyrimidines
C. Deoxyribose
D. Phosphate groups correct answers D. Lysine is a basic, positively-charged amino acid that
interacts with negatively-charged phosphate groups on DNA.
Which finding, when combined with the data in the passage, is most likely to lead researchers to
conclude that the 5-HT2A and 5-HT2B receptor subtypes mediate serotonin-dependent liver
regeneration?
A. Administration of 5-HT2A receptor agonist resulted in reduced Ki67 staining.
B. RNA for seven different receptor subtypes was detectible in naïve liver tissue.
C. Up-regulation of 5-HT2A and 5-HT2B was observed during periods of peak hepatocyte
proliferation.
D. Administration of 5-HT2C and 5-HT3 receptor antagonists reduced the number of Ki67-
positive cells. correct answers A. This is incorrect. A reduction in Ki67 staining after
administration of a 5-HT2A agonist is not consistent with the data presented in the passage.
B. This is incorrect. The presence of RNA encoding all the receptor types in the liver tissues
would not lead researchers to conclude that the 5-HT2A and 5-HT2B receptor subtypes directly
mediate serotonin-dependent liver regeneration.
C. An observed up-regulation of 5-HT2A and 5-HT2B during periods of peak hepatocyte
proliferation corroborates the data presented in the passage (that administration of antagonists of
5-HT2A and 5-HT2B lead to decreased hepatocyte proliferation) and, of the options presented, is
most likely to lead researchers to conclude that the 5-HT2A and 5-HT2B receptor subtypes
mediate serotonin-dependent liver regeneration.
,D. This is incorrect. That 5-HT2C and 5-HT3 receptor antagonists failed to reduce the number of
positive Ki67 cells is consistent with the conclusion that the 5-HT2A and 5-HT2B receptor
subtypes mediate serotonin-dependent liver regeneration, but this observation is not evidence of
a role of 5-HT2A and 5-HT2B in this process. Thus, this observation provides less support for
the role of the 5-HT2A and 5-HT2B receptor subtypes in serotonin-dependent liver regeneration
than does option C.
ADP-ribosylation factor I (Arf1) plays an essential role in vesicle formation and is responsible
for the recruitment of cytosolic coat protein complexes (COPs) and subsequent retrograde
transport from the Golgi apparatus.
Proteins that are encapsulated in Arf1-COP derived vesicles are bound for the:
A.endoplasmic reticulum.
B.cellular membrane.
C.nucleus.
D.cytosol. correct answers The answer to this question is A because the passage states that Arf1-
COP undergoes retrograde transport. Protein translation principally occurs in the endoplasmic
reticulum (ER). The proteins are then transported to the Golgi apparatus for post translational
modification, processing, and packaging for proper localization. Retrograde transport would
reverse the motion and return the protein to the ER.
The scientist wanted to use antibody B clinically (to treat humans), but this proposal was
rejected. Which of the following is the most logical reason for the rejection?
A.Because the antibody was generated in the mouse, it can never be used in humans.
B.Because the antibody was generated in the mouse, repeated usage in the same patient would
elicit the production of human anti-mouse antibodies.
C.Because the antibody was generated in the mouse, it will not recognize human antigens.
D.Because the antibody was generated in the mouse, it can only be used in vitro. correct answers
The answer to this question is B because the human immune system will recognize the mouse
antibody as a foreign substance (antigen) and generate an immune response towards it, which
may include a mild to severe allergic response. The immune response will generate antibodies
against the mouse antibody, limiting its usefulness as a treatment.
A scientist claimed that hemophilia is caused by a functional deficiency in Factor VIII. The
evidence in the passage that would best support this claim is that Factor VIII is:
A.a soluble blood protein.
B.produced by a gene on the X chromosome.
C.able to relieve hemophilia symptoms.
D.encoded by a gene that contains introns. correct answers he answer to this question is C
because the passage states that the symptoms of hemophilia can be at least partially alleviated by
administration of functional Factor VIII, which would help counter functional deficiencies in
Factor VIII in the affected individual. All of the other options can be characteristics of both
functional and nonfunctional proteins and, thus, do not provide the best support for the claim.
, The cutting of introns and the splicing associated with the expression of the Factor VIII gene
occur during which of the following steps in the protein-synthesis process?
A.DNA → mRNA
B.DNA → tRNA
C.mRNA → tRNA
D.tRNA → protein correct answers the answer to this question is A because the removal of
introns and splicing of the remaining pieces (exons) is a step in the production of mature mRNA
from the transcript produced from DNA (pre-mRNA).
Organic acids, denoted by HA, are only minimally deprotonated when added to pure water.
HA(aq) BB-8-1-MCAT Double Arrow.gif H+(aq) + A-(aq)
When dissolved in blood, however, HA fully dissociate. What factor can be used to explain this
discrepancy?
A.In blood, the concentration of H+(aq) is maintained at low levels by other equilibria.
B.In blood, the reaction is coupled to ATP hydrolysis to make it more favorable.
C.In blood, the ionic strength of the solvent medium is much higher than pure water.
D.In blood, enzymes are used to catalyze the dissociation reaction. correct answers the answer to
this question is A because the chemical equation shows that H+ is a product of the reaction. Due
to Le Châtelier's Principle, the equilibrium will favor products when the concentration of
products is kept low.
In oxidative phosphorylation, cytochrome c acts as:
A.a 1-electron carrier.
B.a 2-electron carrier.
C.a 3-electron carrier.
D.a 4-electron carrier. correct answers The answer to this question is A because cytochrome c is
a heme protein that only cycles between a ferrous and ferric state during oxidative
phosphorylation. Therefore, only single electron transfers are possible.
Which primer is most suitable for PCR?
A.
5BB-8-1-prime.gif-ATTACGTTAACATGAAG-3BB-8-1-prime.gif
B.
5BB-8-1-prime.gif-ATATCGTTAACAAATTG-3BB-8-1-prime.gif
C.
5BB-8-1-prime.gif-GCTATAAAGATTGCAAA-3BB-8-1-prime.gif
