AAMC MCAT Practice Exam 2 2025/2026
Exam Questions and Verified Answers |
Already Graded A+
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?
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
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STATEMENT. ALL RIGHTS RESERVED
3
, 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
Exam Questions and Verified Answers |
Already Graded A+
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?
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
COPYRIGHT©NINJANERD 2025/2026. YEAR PUBLISHED 2025. COMPANY REGISTRATION NUMBER: 619652435. TERMS OF USE. PRIVACY
STATEMENT. ALL RIGHTS RESERVED
3
, 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
