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AAMC MCAT Practice Exam 2 questions and answers graded A+ 2024/2025
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AAMC MCAT Practice Exam 2 questions and answers graded A+ 2024/2025
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AAMC MCAT Practice Exam 2C/P: What expression gives the quantity of light strength (in J in keeping with photon) that is
transformed to other kinds among the fluorescence excitation and emission events?
"intensity of fluorescence emission at 440 nm excitation at 360 nm) changed into monitored for
20 mins"
A) (6.62 × 10-34) × (3.Zero × 108)
B) (6.Sixty two × 10-34) × (three.0 × 108) × (360 × 10-9)
C) (6.62 × 10-34) × (three.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)]
D) (6.Sixty two × 10-34) × (three.0 × 108) / (440 × 10-nine) - ANSC) (6.62 × 10-34) × (3.0 × 108)
× [1 / (360 × 10-9) - 1 / (440 × 10-9)]
The answer to this query is C due to the fact the equation of hobby is E = hf = hc/λ, where h =
6.62 × 10 −34 J ∙ s and c = 3 × 10 eight m/s. Excitation happens at λe = 360 nm, however
fluorescence is located at λf = 440 nm. This means that an strength of E = (6.Sixty two × 10
−34) × (three × 10 eight) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon is transformed to
other forms among the excitation and fluorescence occasions.
C/P: Compared to the awareness of the proteasome, the awareness of the substrate is larger by
using what factor?
"purified rabbit proteasome (2 nM) become incubated within the presence of porphyrin...The
reaction changed into initiated by means of addition of the peptide (one hundred uM)"
A) five × one hundred and one
B) five × 102
C) 5 × 103
D) 5 × 104 - ANSD) five × 104
The answer to this query is D. The proteasome become gift at a awareness of 2 × 10-nine M,
even as the substrate became present at one hundred × 10-6 M. The ratio of these numbers is
5 × 104.
Sp2 hybridized - ANSpossess precisely one doubly bonded atom
C/P: The awareness of enzyme for every experiment turned into 5.Zero μM. What is kcat for the
reaction at pH four.5 with NO chloride introduced when Compound 3 is the substrate?
Rate of reaction = 125 nM/s
,A) 2.Five × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × one zero five s-1 - ANSA) 2.Five × 10-2 s-1
The solution to this query is A. The reality that the rate of product formation did now not range
over the years for the primary 5 minutes means that the enzyme changed into saturated with
substrate. Under those situations, kcat = Vmax/[E] = (125 nM/s)/five.Zero μM = 2.Five × 10-2
s-1.
Kcat, Vmax, [E] - ANSkcat = Vmax/[E]
C/P: Absorption of ultraviolet mild by natural molecules always consequences in what manner?
A) Bond breaking
B) Excitation of bound electrons
C) Vibration of atoms in polar bonds
D) Ejection of certain electrons - ANSB) Excitation of sure electrons
The solution to this question is B. The absorption of ultraviolet light by way of natural molecules
continually consequences in digital excitation. Bond breaking can sooner or later end result, as
can ionization or bond vibration, but none of these techniques are guaranteed to result from the
absorption of ultraviolet mild.
C/P: Four natural compounds: 2-butanone, n-pentane, propanoic acid, and n-butanol, present
as a mixture, are separated with the aid of column chromatography the usage of silica gel with
benzene as the eluent. What is the predicted order of elution of these four natural compounds
from first to ultimate?
A) n-Pentane → 2-butanone → n-butanol → propanoic acid
B) n-Pentane → n-butanol → 2-butanone → propanoic acid
C) Propanoic acid → n-butanol → 2-butanone → n-pentane
D) Propanoic acid → 2-butanone → n-butanol → n-pentane - ANSA) n-Pentane → 2-butanone
→ n-butanol → propanoic acid
The answer to this question is A. The four compounds have similar molecular weights, so the
order of elution will depend on the polarity of the molecule. Since silica gel serves as the desk
bound phase for the test, growing the polarity of the eluting molecule will increase its affinity for
the desk bound phase and boom the elution time (reduced Rf).
C/P: The 1/2-life of a radioactive fabric is:
A) half of the time it takes for all the radioactive nuclei to decay into radioactive nuclei.
B) 1/2 the time it takes for all the radioactive nuclei to decay into their daughter nuclei.
C) the time it takes for half of of all of the radioactive nuclei to decay into radioactive nuclei.
,D) the time it takes for half of all of the radioactive nuclei to decay into their daughter nuclei. -
ANSD) the time it takes for 1/2 of all the radioactive nuclei to decay into their daughter nuclei.
The answer to this query is D because the half-existence of a radioactive fabric is defined
because the time it takes for 1/2 of all the radioactive nuclei to decay into their daughter nuclei,
which may additionally or might not also be radioactive.
C/P: A character is sitting in a chair. Why need to the person either lean ahead or slide their feet
underneath the chair so as to get up?
A) to boom the force required to rise up
B) to use the friction with the floor
C) to reduce the energy required to get up
D) to hold the body in equilibrium even as growing - ANSD) to hold the body in equilibrium even
as growing
The answer to this query is D due to the fact because the person is trying to stand, the handiest
guide comes from the ft on the floor. The man or woman is in equilibrium handiest when the
center of mass is immediately above their ft. Otherwise, if the individual did no longer lean
ahead or slide the feet beneath the chair, the man or woman might fall backward because of the
big torque created via the mixture of the load of the body (implemented on the individual's
middle of mass) and the gap alongside the horizontal between the middle of mass and the
assist factor.
C/P: The facet chain of tryptophan will give upward push to the largest CD signal in the near UV
location while:
A) present as a unfastened amino acid
B) a part of an a-helix
C) part of a B-sheet
D) a part of a fully folded protein - ANSD) a part of a totally folded protein
The answer to this question is D due to the fact tryptophan has an fragrant side chain with the
intention to supply rise to a large CD signal in the close to UV vicinity if it's far located in a
completely folded protein.
C/P: Which amino acid will contribute to the CD sign inside the some distance UV place, but
NOT the near UV region, when a part of a completely folded protein?
"Asymmetry because of tertiary structural functions causes the biggest growth in CD signal
intensity within the close to UV area of peptides. The side chains of amino acid residues soak
up in this vicinity.
, The peptide bond absorbs inside the some distance UV place (190-250 nm). The CD indicators
of those bonds are dramatically impacted by means of their proximity to secondary structural
elements."
A) Trp
B) Phe
C) Ala
D) Tyr - ANSC) Ala
C/P: Based at the relative power of the absorbed electromagnetic radiation, which absorber, a
peptide bond or an aromatic aspect chain, exhibits an digital excited country that is closer in
power to the ground nation?
"Asymmetry resulting from tertiary structural functions causes the most important growth in CD
sign depth within the close to UV place of peptides. The side chains of amino acid residues take
in on this location.
The peptide bond absorbs within the some distance UV area (one hundred ninety-250 nm). The
CD indicators of these bonds are dramatically impacted by means of their proximity to
secondary structural factors."
A) An fragrant side chain; the absorbed photon electricity is higher.
B) An aromatic facet chain; the absorbed photon strength is decrease.
C) A peptide bond; the absorbed photon strength is higher.
D) A peptide bond; the absorbed photon energy is lower. - ANSB) An aromatic aspect chain; the
absorbed photon energy is lower.
The answer to this question is B due to the fact aromatic facet chains take in inside the close to
UV area of the electromagnetic spectrum, which has longer wavelengths, and subsequently
decrease electricity, than peptide bonds. Because the power of the photon fits the energy hole
among the floor and the excited country, this means that the fragrant side chain has extra
intently spaced energy tiers.
C/P: What is the net rate of sT-loop at pH 7.2?
"A artificial peptide with the amino acid sequence KTFCGPEYLA became generated as a mimic
of the T-loop. This artificial T-loop (sT-loop) turned into incubated with 32P-classified ATP within
the presence of PDK1 for special time intervals at 37 ° C and pH 7.2, and the quantity of
radioactivity integrated into sT-loop become measured by means of detection of β- decay."
A) -2
B) -1
C) zero
D) +1 - ANSC) 0
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