AAMC MCAT Practice Exam 2
Questions and Answers 100%
Correct
C/P: What expression gives the amount of light energy (in J per photon) that is
converted to other forms between the fluorescence excitation and emission events?
"intensity of fluorescence emission at 440 nm excitation at 360 nm) was monitored
for 20 minutes"
A) (6.62 × 10-34) × (3.0 × 108)
B) (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9)
C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)]
D) (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9) - ANSWER - C) (6.62 × 10-34) × (3.0
× 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)]
The answer to this question is C because the equation of interest is E = hf = hc/λ,
where h = 6.62 × 10 −34 J ∙ s and c = 3 × 10 8 m/s. Excitation occurs at λe = 360
nm, but fluorescence is observed at λf = 440 nm. This implies that an energy of E =
(6.62 × 10 −34) × (3 × 10 8) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon
is converted to other forms between the excitation and fluorescence events.
C/P: Compared to the concentration of the proteasome, the concentration of the
substrate is larger by what factor?
"purified rabbit proteasome (2 nM) was incubated in the presence of porphyrin...the
reaction was initiated by addition of the peptide (100 uM)"
A) 5 × 101
B) 5 × 102
C) 5 × 103
,D) 5 × 104 - ANSWER - D) 5 × 104
The answer to this question is D. The proteasome was present at a concentration of
2 × 10-9 M, while the substrate was present at 100 × 10-6 M. The ratio of these two
numbers is 5 × 104.
sp2 hybridized - ANSWER - possess exactly one doubly bonded atom
C/P: The concentration of enzyme for each experiment was 5.0 μM. What is kcat for
the reaction at pH 4.5 with NO chloride added when Compound 3 is the substrate?
C/P: The side chain of tryptophan will give rise to the largest CD signal in the near
UV region when:
A) present as a free amino acid
B) part of an a-helix
C) part of a B-sheet
D) part of a fully folded protein - ANSWER - D) part of a fully folded protein
The answer to this question is D because tryptophan has an aromatic side chain
that will give rise to a significant CD signal in the near UV region if it is found in a
fully folded protein.
C/P: Which amino acid will contribute to the CD signal in the far UV region, but NOT
the near UV region, when part of a fully folded protein?
"Asymmetry resulting from tertiary structural features causes the largest increase in
CD signal intensity in the near UV region of peptides. The side chains of amino acid
residues absorb in this region.
The peptide bond absorbs in the far UV region (190-250 nm). The CD signals of
these bonds are dramatically impacted by their proximity to secondary structural
elements."
A) Trp
,B) Phe
C) Ala
D) Tyr - ANSWER - C) Ala
C/P: Based on the relative energy of the absorbed electromagnetic radiation, which
absorber, a peptide bond or an aromatic side chain, exhibits an electronic excited
state that is closer in energy to the ground state?
Rate of reaction = 125 nM/s
A) 2.5 × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × 105 s-1 - ANSWER - A) 2.5 × 10-2 s-1
The answer to this question is A. The fact that the rate of product formation did not
vary over time for the first 5 minutes implies that the enzyme was saturated with
substrate. Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-
2 s-1.
kcat, Vmax, [E] - ANSWER - kcat = Vmax/[E]
C/P: Absorption of ultraviolet light by organic molecules always results in what
process?
A) Bond breaking
B) Excitation of bound electrons
C) Vibration of atoms in polar bonds
D) Ejection of bound electrons - ANSWER - B) Excitation of bound electrons
The answer to this question is B. The absorption of ultraviolet light by organic
molecules always results in electronic excitation. Bond breaking can subsequently
result, as can ionization or bond vibration, but none of these processes are
guaranteed to result from the absorption of ultraviolet light.
, C/P: Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n-
butanol, present as a mixture, are separated by column chromatography using silica
gel with benzene as the eluent. What is the expected order of elution of these four
organic compounds from first to last?
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 - ANSWER - A) n-Pentane
→ 2-butanone → n-butanol → propanoic acid
The answer to this question is A. The four compounds have comparable molecular
weights, so the order of elution will depend on the polarity of the molecule. Since
silica gel serves as the stationary phase for the experiment, increasing the polarity
of the eluting molecule will increase its affinity for the stationary phase and increase
the elution time (decreased Rf).
C/P: The half-life of a radioactive material is:
A) half the time it takes for all of the radioactive nuclei to decay into radioactive
nuclei.
B) half the time it takes for all of the radioactive nuclei to decay into their daughter
nuclei.
C) the time it takes for half of all the radioactive nuclei to decay into radioactive
nuclei.
D) the time it takes for half of all the radioactive nuclei to decay into their daughter
nuclei. - ANSWER - D) the time it takes for half of all the radioactive nuclei to decay
into their daughter nuclei.
The answer to this question is D because the half-life of a radioactive material is
defined as the time it takes for half of all the radioactive nuclei to decay into their
daughter nuclei, which may or may not also be radioactive.
Questions and Answers 100%
Correct
C/P: What expression gives the amount of light energy (in J per photon) that is
converted to other forms between the fluorescence excitation and emission events?
"intensity of fluorescence emission at 440 nm excitation at 360 nm) was monitored
for 20 minutes"
A) (6.62 × 10-34) × (3.0 × 108)
B) (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9)
C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)]
D) (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9) - ANSWER - C) (6.62 × 10-34) × (3.0
× 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)]
The answer to this question is C because the equation of interest is E = hf = hc/λ,
where h = 6.62 × 10 −34 J ∙ s and c = 3 × 10 8 m/s. Excitation occurs at λe = 360
nm, but fluorescence is observed at λf = 440 nm. This implies that an energy of E =
(6.62 × 10 −34) × (3 × 10 8) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon
is converted to other forms between the excitation and fluorescence events.
C/P: Compared to the concentration of the proteasome, the concentration of the
substrate is larger by what factor?
"purified rabbit proteasome (2 nM) was incubated in the presence of porphyrin...the
reaction was initiated by addition of the peptide (100 uM)"
A) 5 × 101
B) 5 × 102
C) 5 × 103
,D) 5 × 104 - ANSWER - D) 5 × 104
The answer to this question is D. The proteasome was present at a concentration of
2 × 10-9 M, while the substrate was present at 100 × 10-6 M. The ratio of these two
numbers is 5 × 104.
sp2 hybridized - ANSWER - possess exactly one doubly bonded atom
C/P: The concentration of enzyme for each experiment was 5.0 μM. What is kcat for
the reaction at pH 4.5 with NO chloride added when Compound 3 is the substrate?
C/P: The side chain of tryptophan will give rise to the largest CD signal in the near
UV region when:
A) present as a free amino acid
B) part of an a-helix
C) part of a B-sheet
D) part of a fully folded protein - ANSWER - D) part of a fully folded protein
The answer to this question is D because tryptophan has an aromatic side chain
that will give rise to a significant CD signal in the near UV region if it is found in a
fully folded protein.
C/P: Which amino acid will contribute to the CD signal in the far UV region, but NOT
the near UV region, when part of a fully folded protein?
"Asymmetry resulting from tertiary structural features causes the largest increase in
CD signal intensity in the near UV region of peptides. The side chains of amino acid
residues absorb in this region.
The peptide bond absorbs in the far UV region (190-250 nm). The CD signals of
these bonds are dramatically impacted by their proximity to secondary structural
elements."
A) Trp
,B) Phe
C) Ala
D) Tyr - ANSWER - C) Ala
C/P: Based on the relative energy of the absorbed electromagnetic radiation, which
absorber, a peptide bond or an aromatic side chain, exhibits an electronic excited
state that is closer in energy to the ground state?
Rate of reaction = 125 nM/s
A) 2.5 × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × 105 s-1 - ANSWER - A) 2.5 × 10-2 s-1
The answer to this question is A. The fact that the rate of product formation did not
vary over time for the first 5 minutes implies that the enzyme was saturated with
substrate. Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-
2 s-1.
kcat, Vmax, [E] - ANSWER - kcat = Vmax/[E]
C/P: Absorption of ultraviolet light by organic molecules always results in what
process?
A) Bond breaking
B) Excitation of bound electrons
C) Vibration of atoms in polar bonds
D) Ejection of bound electrons - ANSWER - B) Excitation of bound electrons
The answer to this question is B. The absorption of ultraviolet light by organic
molecules always results in electronic excitation. Bond breaking can subsequently
result, as can ionization or bond vibration, but none of these processes are
guaranteed to result from the absorption of ultraviolet light.
, C/P: Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n-
butanol, present as a mixture, are separated by column chromatography using silica
gel with benzene as the eluent. What is the expected order of elution of these four
organic compounds from first to last?
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 - ANSWER - A) n-Pentane
→ 2-butanone → n-butanol → propanoic acid
The answer to this question is A. The four compounds have comparable molecular
weights, so the order of elution will depend on the polarity of the molecule. Since
silica gel serves as the stationary phase for the experiment, increasing the polarity
of the eluting molecule will increase its affinity for the stationary phase and increase
the elution time (decreased Rf).
C/P: The half-life of a radioactive material is:
A) half the time it takes for all of the radioactive nuclei to decay into radioactive
nuclei.
B) half the time it takes for all of the radioactive nuclei to decay into their daughter
nuclei.
C) the time it takes for half of all the radioactive nuclei to decay into radioactive
nuclei.
D) the time it takes for half of all the radioactive nuclei to decay into their daughter
nuclei. - ANSWER - D) the time it takes for half of all the radioactive nuclei to decay
into their daughter nuclei.
The answer to this question is D because the half-life of a radioactive material is
defined as the time it takes for half of all the radioactive nuclei to decay into their
daughter nuclei, which may or may not also be radioactive.
