MCAT AAMC Practice Exam 2 | 160 Questions
with Correct Answers and Rationale | Graded A+ |
New 2026-2027 Update
Summary of Exam Content
Chemical and Physical Foundations (C/P)
Critical Analysis and Reasoning Skills (CARS)
Biological and Biochemical Foundations (B/B)
Psychological, Social, and Biological Foundations (P/S)
Section 1: Chemical and Physical Foundations of Biological Systems (C/P)
Question 1
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)
Correct Answer: C. (6.62 × 10⁻³⁴) × (3.0 × 10⁸) × [1/(360 × 10⁻⁹) - 1/(440 × 10⁻⁹)]
Rationale: The equation of interest is E = hf = hc/λ, where h = 6.62 × 10⁻³⁴ J·s and c
= 3 × 10⁸ 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 × 10⁸) × [1/(360 ×
10⁻⁹) − 1/(440 × 10⁻⁹)] J per photon is converted to other forms between the
excitation and fluorescence events.
,Question 2
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 μM)"
A. 5 × 10¹
B. 5 × 10²
C. 5 × 10³
D. 5 × 10⁴
Correct Answer: D. 5 × 10⁴
Rationale: 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.
Question 3
The concentration of enzyme for each experiment was 5.0 μM. What is k_cat 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
Correct Answer: A. 2.5 × 10⁻² s⁻¹
Rationale: 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, k_cat = V_max/[E] = (125 nM/s)/5.0 μM = 2.5 × 10⁻² s⁻¹.
,Question 4
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
Correct Answer: B. Excitation of bound electrons
Rationale: 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.
Question 5
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
Correct Answer: A. n-Pentane → 2-butanone → n-butanol → propanoic acid
Rationale: 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).
Question 6
, 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.
Correct Answer: D. the time it takes for half of all the radioactive nuclei to
decay into their daughter nuclei.
Rationale: 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.
Question 7
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
Correct Answer: D. to keep the body in equilibrium while rising
Rationale: 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 is
directly above their feet. Otherwise, if the person did not lean forward or slide the
feet under the chair, the person would fall backward due to the large torque
created by the combination of the weight of the body (applied at the person's
center of mass) and the distance along the horizontal between the center of mass
and the support point.
with Correct Answers and Rationale | Graded A+ |
New 2026-2027 Update
Summary of Exam Content
Chemical and Physical Foundations (C/P)
Critical Analysis and Reasoning Skills (CARS)
Biological and Biochemical Foundations (B/B)
Psychological, Social, and Biological Foundations (P/S)
Section 1: Chemical and Physical Foundations of Biological Systems (C/P)
Question 1
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)
Correct Answer: C. (6.62 × 10⁻³⁴) × (3.0 × 10⁸) × [1/(360 × 10⁻⁹) - 1/(440 × 10⁻⁹)]
Rationale: The equation of interest is E = hf = hc/λ, where h = 6.62 × 10⁻³⁴ J·s and c
= 3 × 10⁸ 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 × 10⁸) × [1/(360 ×
10⁻⁹) − 1/(440 × 10⁻⁹)] J per photon is converted to other forms between the
excitation and fluorescence events.
,Question 2
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 μM)"
A. 5 × 10¹
B. 5 × 10²
C. 5 × 10³
D. 5 × 10⁴
Correct Answer: D. 5 × 10⁴
Rationale: 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.
Question 3
The concentration of enzyme for each experiment was 5.0 μM. What is k_cat 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
Correct Answer: A. 2.5 × 10⁻² s⁻¹
Rationale: 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, k_cat = V_max/[E] = (125 nM/s)/5.0 μM = 2.5 × 10⁻² s⁻¹.
,Question 4
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
Correct Answer: B. Excitation of bound electrons
Rationale: 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.
Question 5
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
Correct Answer: A. n-Pentane → 2-butanone → n-butanol → propanoic acid
Rationale: 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).
Question 6
, 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.
Correct Answer: D. the time it takes for half of all the radioactive nuclei to
decay into their daughter nuclei.
Rationale: 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.
Question 7
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
Correct Answer: D. to keep the body in equilibrium while rising
Rationale: 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 is
directly above their feet. Otherwise, if the person did not lean forward or slide the
feet under the chair, the person would fall backward due to the large torque
created by the combination of the weight of the body (applied at the person's
center of mass) and the distance along the horizontal between the center of mass
and the support point.
