AAMC MCAT
C/P: What expression gives the amount of light energy (in J per
photon) that
converted to is
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
D) × 10-9)]
(6.62 × 10-34) × (3.0 × 108) / (440 × 10-9) correct answersC) (6.62 ×
10-34)
108) ××[1(3.0
/ (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 =fluorescence
but 360 nm, is observed at λf = 440 nm. This implies that an energy of
E =−34)
10 (6.62×× (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
substrate of the
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 ×
B) 5 ×
101
102
C) 5 ×
103
D) 5 × 104 correct answersD) 5
× 104
The answer to this question is D. The proteasome was present at a
concentration
10-9 M, while theof 2substrate
× was present at 100 × 10-6 M. The ratio of
these twois 5 ×
numbers
104.
sp2 hybridized correct answerspossess 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 correct answersA) 2.5 ×
10-2 s-1
The answer to this question is A. The fact that the rate of product formation
did not
over timevary
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] correct answerskcat =
Vmax/[E]
C/P: Absorption of ultraviolet light by organic molecules always results in
what
A) process?
Bond
breaking
B) Excitation of bound
electrons
C) Vibration of atoms in polar
bonds
D) Ejection of bound electrons correct answersB) Excitation of bound
electrons
The answer to this question is B. The absorption of ultraviolet light
by organicalways results in electronic excitation. Bond breaking can
molecules
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 as
benzene with
the eluent. What is the expected order of elution of these
four organic from first to
compounds
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
answersA)
Pentane → n-
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
gel silicaas the stationary phase for the experiment, increasing the
serves
polaritymolecule
eluting of the will increase its affinity for the stationary phase and
increasetime
elution the (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
B) nuclei.
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 answersD) the time it takes for half of all the radioactive
nuclei
into to decay
their daughter
nuclei.
The answer to this question is D because the half-life of a radioactive
materialasisthe time it takes for half of all the radioactive nuclei to
defined
decay intonuclei,
daughter their 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
slideunder
feet their the chair in order to
stand up?
A) to increase the force required to
stand
B) up the friction with the
to use
ground
C) to reduce the energy required to
stand
D) up the body in equilibrium while rising correct answersD) to keep
to keep
the body in while
equilibrium
rising
The answer to this question is D because as the person is attempting to
stand, the
support only from the feet on the ground. The person is in equilibrium
comes
only when
center the is directly above their feet. Otherwise, if the person did not
of mass
lean
or forward
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
mass) andcenter of
the distance along the horizontal between the center of
mass and the
support
point.
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 correct answersD) part of a fully
folded protein
The answer to this question is D because tryptophan has an aromatic side
chain
will that
give rise to a significant CD signal in the near UV region if it is found in a
fully folded
protei
n.
C/P: Which amino acid will contribute to the CD signal in the far UV region,
but NOT
near the
UV region, when part of a fully folded
protein?
"Asymmetry resulting from tertiary structural features causes the largest
increase
signal in CD in the near UV region of peptides. The side chains of
intensity
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)
C)
Phe
Ala
D) Tyr correct answersC)
Ala
C/P: Based on the relative energy of the absorbed electromagnetic
radiation, a
absorber, which
peptide bond or an aromatic side chain, exhibits an electronic
excited
that state in energy to the ground
is closer
state?
"Asymmetry resulting from tertiary structural features causes the largest
increase
signal in CD in the near UV region of peptides. The side chains of
intensity
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) An aromatic side chain; the absorbed photon energy
is higher.
B) An aromatic side chain; the absorbed photon energy
is lower.
C) A peptide bond; the absorbed photon energy is
higher.
D) A peptide bond; the absorbed photon energy is lower. correct
answersB)side
aromatic An chain; the absorbed photon energy
is lower.
The answer to this question is B because aromatic side chains absorb in
the near
region of UV
the electromagnetic spectrum, which has longer wavelengths,
and hence
lower energy, than peptide bonds. Because the energy of the photon
matchesgap
energy thebetween the ground and the excited state, this implies that the
aromatic
chain hasside
more closely spaced energy
levels.
C/P: What is the net charge of sT-loop at pH
7.2?
"A synthetic peptide with the amino acid sequence KTFCGPEYLA was
generated
mimic of theasT-loop.
a This synthetic T-loop (sT-loop) was incubated with
32P-labeled
ATP in the presence of PDK1 for different time periods at 37 ° C and pH
7.2, and of
amount theradioactivity incorporated into sT-loop was measured by
detection of β-
decay.
"
A) -
2
C)
B) -
0
D) 1+1 correct
answersC) 0
The answer to this question is C because at pH 7.2, the N-terminus will be
positively
charged and the C-terminus will be negatively charged. In addition, the
lysine
will sideone
carry chain
positive charge and the glutamic acid side chain will carry
charge
one negative
.
C/P: What expression gives the amount of light energy (in J per
photon) that
converted to is
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
D) × 10-9)]
(6.62 × 10-34) × (3.0 × 108) / (440 × 10-9) correct answersC) (6.62 ×
10-34)
108) ××[1(3.0
/ (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 =fluorescence
but 360 nm, is observed at λf = 440 nm. This implies that an energy of
E =−34)
10 (6.62×× (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
substrate of the
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 ×
B) 5 ×
101
102
C) 5 ×
103
D) 5 × 104 correct answersD) 5
× 104
The answer to this question is D. The proteasome was present at a
concentration
10-9 M, while theof 2substrate
× was present at 100 × 10-6 M. The ratio of
these twois 5 ×
numbers
104.
sp2 hybridized correct answerspossess 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 correct answersA) 2.5 ×
10-2 s-1
The answer to this question is A. The fact that the rate of product formation
did not
over timevary
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] correct answerskcat =
Vmax/[E]
C/P: Absorption of ultraviolet light by organic molecules always results in
what
A) process?
Bond
breaking
B) Excitation of bound
electrons
C) Vibration of atoms in polar
bonds
D) Ejection of bound electrons correct answersB) Excitation of bound
electrons
The answer to this question is B. The absorption of ultraviolet light
by organicalways results in electronic excitation. Bond breaking can
molecules
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 as
benzene with
the eluent. What is the expected order of elution of these
four organic from first to
compounds
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
answersA)
Pentane → n-
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
gel silicaas the stationary phase for the experiment, increasing the
serves
polaritymolecule
eluting of the will increase its affinity for the stationary phase and
increasetime
elution the (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
B) nuclei.
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 answersD) the time it takes for half of all the radioactive
nuclei
into to decay
their daughter
nuclei.
The answer to this question is D because the half-life of a radioactive
materialasisthe time it takes for half of all the radioactive nuclei to
defined
decay intonuclei,
daughter their 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
slideunder
feet their the chair in order to
stand up?
A) to increase the force required to
stand
B) up the friction with the
to use
ground
C) to reduce the energy required to
stand
D) up the body in equilibrium while rising correct answersD) to keep
to keep
the body in while
equilibrium
rising
The answer to this question is D because as the person is attempting to
stand, the
support only from the feet on the ground. The person is in equilibrium
comes
only when
center the is directly above their feet. Otherwise, if the person did not
of mass
lean
or forward
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
mass) andcenter of
the distance along the horizontal between the center of
mass and the
support
point.
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 correct answersD) part of a fully
folded protein
The answer to this question is D because tryptophan has an aromatic side
chain
will that
give rise to a significant CD signal in the near UV region if it is found in a
fully folded
protei
n.
C/P: Which amino acid will contribute to the CD signal in the far UV region,
but NOT
near the
UV region, when part of a fully folded
protein?
"Asymmetry resulting from tertiary structural features causes the largest
increase
signal in CD in the near UV region of peptides. The side chains of
intensity
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)
C)
Phe
Ala
D) Tyr correct answersC)
Ala
C/P: Based on the relative energy of the absorbed electromagnetic
radiation, a
absorber, which
peptide bond or an aromatic side chain, exhibits an electronic
excited
that state in energy to the ground
is closer
state?
"Asymmetry resulting from tertiary structural features causes the largest
increase
signal in CD in the near UV region of peptides. The side chains of
intensity
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) An aromatic side chain; the absorbed photon energy
is higher.
B) An aromatic side chain; the absorbed photon energy
is lower.
C) A peptide bond; the absorbed photon energy is
higher.
D) A peptide bond; the absorbed photon energy is lower. correct
answersB)side
aromatic An chain; the absorbed photon energy
is lower.
The answer to this question is B because aromatic side chains absorb in
the near
region of UV
the electromagnetic spectrum, which has longer wavelengths,
and hence
lower energy, than peptide bonds. Because the energy of the photon
matchesgap
energy thebetween the ground and the excited state, this implies that the
aromatic
chain hasside
more closely spaced energy
levels.
C/P: What is the net charge of sT-loop at pH
7.2?
"A synthetic peptide with the amino acid sequence KTFCGPEYLA was
generated
mimic of theasT-loop.
a This synthetic T-loop (sT-loop) was incubated with
32P-labeled
ATP in the presence of PDK1 for different time periods at 37 ° C and pH
7.2, and of
amount theradioactivity incorporated into sT-loop was measured by
detection of β-
decay.
"
A) -
2
C)
B) -
0
D) 1+1 correct
answersC) 0
The answer to this question is C because at pH 7.2, the N-terminus will be
positively
charged and the C-terminus will be negatively charged. In addition, the
lysine
will sideone
carry chain
positive charge and the glutamic acid side chain will carry
charge
one negative
.
