AAMC MCAT PRACTICE EXAM 2
QUESTIONS AND ANSWERS GRADED A+
2025/2026
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) - ANS 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.
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,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 - ANS 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 - ANS 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?
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 - ANS A) 2.5 × 10-2 s-1
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,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] - ANS 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 - ANS 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 - ANS A) n-Pentane → 2-
butanone → n-butanol → propanoic acid
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, 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. -
ANS 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.
C/P: A person is sitting in a chair. Why must the person either lean forward or slide their feet
under the chair in order to stand up?
A) to increase the force required to stand up
B) to use the friction with the ground
C) to reduce the energy required to stand up
D) to keep the body in equilibrium while rising - ANS D) to keep the body in equilibrium while
rising
The answer to this question is D because as the person is attempting to stand, the only support
comes from the feet on the ground. The person is in equilibrium only when the center of mass
4 @COPYRIGHT 2025/2026 ALLRIGHTS RESERVED.
QUESTIONS AND ANSWERS GRADED A+
2025/2026
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) - ANS 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.
1 @COPYRIGHT 2025/2026 ALLRIGHTS RESERVED.
,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 - ANS 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 - ANS 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?
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 - ANS A) 2.5 × 10-2 s-1
2 @COPYRIGHT 2025/2026 ALLRIGHTS RESERVED.
,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] - ANS 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 - ANS 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 - ANS A) n-Pentane → 2-
butanone → n-butanol → propanoic acid
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, 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. -
ANS 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.
C/P: A person is sitting in a chair. Why must the person either lean forward or slide their feet
under the chair in order to stand up?
A) to increase the force required to stand up
B) to use the friction with the ground
C) to reduce the energy required to stand up
D) to keep the body in equilibrium while rising - ANS D) to keep the body in equilibrium while
rising
The answer to this question is D because as the person is attempting to stand, the only support
comes from the feet on the ground. The person is in equilibrium only when the center of mass
4 @COPYRIGHT 2025/2026 ALLRIGHTS RESERVED.
