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AAMC FL2 C/P questions and answers graded A+ 2024/2025
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AAMC FL2 C/P questions and answers graded A+ 2024/2025
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AAMC FL2 C/PWhat atom is the website of covalent attachment of AMC to the version tetrapeptide used inside
the research?
A.
I
B.
II
C.
III
D.
IV - ANSThis query is A because AMC is attached to the peptide on the carboxyl side. This
suggests that an amide linkage involving the N atom in AMC is used to covalently connect the
fluorophore to the peptide.
The passage says AMC is eliminated from a peptide via peptide bond hydrolysis. Thus, AMC
should shape a peptide bond with the peptide. Peptide bonds are between amine and carboxylic
acid groups to form an amide. There is only one amine in AMC. It additionally facilitates to
understand that chymotrypsin cleaves on the carboxyl facet of an aromatic amino acid, which
means the purposeful group on AMC that binds is the amine.
What expression gives the quantity of light power (in J according to photon) that is transformed
to other forms among the fluorescence excitation and emission activities?
A.
(6.Sixty two × 10-34) × (three.Zero × 108)
B.
(6.62 × 10-34) × (3.Zero × 108) × (360 × 10-9)
C.
(6.62 × 10-34) × (three.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)]
D.
(6.Sixty two × 10-34) × (3.Zero × 108) / (440 × 10-nine) - ANSThe answer to this question is C
because the equation of hobby is E = hf = hc/λ, where h = 6.Sixty two × 10 −34 J ∙ s and c = 3 ×
10 8 m/s. Excitation takes place at λe = 360 nm, however fluorescence is determined at λf = 440
nm. This implies that an power of E = (6.62 × 10 −34) × (3 × 10 eight) × [1 / (360 × 10 −9) − 1 /
(440 × 10 −9)] J in step with photon is converted to other kinds between the excitation and
fluorescence events.
Compared to the awareness of the proteasome, the attention of the substrate is larger through
what element?
A.
5 × one hundred and one
B.
Five × 102
,C.
Five × 103
D.
Five × 104 - ANSThe solution to this question is D. The proteasome become gift at a
concentration of 2 × 10-nine M, whilst the substrate become present at one hundred × 10-6 M.
The ratio of these numbers is five × 104.
Put the given concentrations into a ratio. The query asks with the aid of how a great deal the
substrate concentration is bigger, so positioned the substrate attention on pinnacle. This is
[S]/[E] = (100 microM)/(2 nanoM).Simplify the ratio with the aid of dividing. Since math and
preserving track of gadgets is not my strong point, I for my part want to rewrite all devices in SI
devices with clinical notation, like this: (100 x 10-6 M)/(2 x 10-9 M). Then I divide the numbers
(one hundred/2 = 50), and then I divide the powers of ten (10--9 = 103). So the solution is
50 x 103 M, which equals one in all the answer alternatives, 5 x 104 M.
The concentration of enzyme for each test changed into five.0 μM. What is kcat for the response
at pH 4.Five with NO chloride delivered when Compound three is the substrate?
A.
2.Five × 10-2 s-1
B.
1.3 × 102 s-1
C.
5.Three × 103 s-1
D.
7.0 × one zero five s-1 - ANSThe solution to this query is A. The fact that the rate of product
formation did no longer range through the years for the primary five minutes means that the
enzyme become saturated with substrate. Under these conditions, kcat = Vmax/[E] = (a
hundred twenty five nM/s)/5.Zero μM = 2.5 × 10-2 s-1.
Absorption of ultraviolet light by organic molecules continually results in what procedure?
A.
Bond breaking
B.
Excitation of sure electrons
C.
Vibration of atoms in polar bonds
D.
Ejection of bound electrons - ANSThe answer to this question is B. The absorption of ultraviolet
mild by means of natural molecules continually consequences in digital excitation. Bond
breaking can eventually end result, as can ionization or bond vibration, however none of these
methods are assured to end result from the absorption of ultraviolet mild.
C. Different results dominate at different frequencies. With IR, there isn't sufficient power inside
the photon to excite electrons, so the electricity is going into vibrations. With UV, there may be
, continually sufficient energy in the photon to reason an excitation of the electron, and that is
where the energy is going, in place of vibration.
D. This is an instance of the photoelectric effect. You can absorb electricity with out inflicting
emission.
Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n-butanol, present as a
aggregate, are separated by using column chromatography the use of silica gel with benzene as
the eluent. What is the expected order of elution of these four organic compounds from first to
final?
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 - ANSThe answer to this query is A.
The 4 compounds have similar molecular weights, so the order of elution will rely upon the
polarity of the molecule. Since silica gel serves because the desk bound section for the
experiment, growing the polarity of the eluting molecule will increase its affinity for the desk
bound section and boom the elution time (reduced Rf).
The half of-lifestyles of a radioactive cloth is:
A.
Half of 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 of the radioactive nuclei to decay into radioactive nuclei.
D.
The time it takes for half of all of the radioactive nuclei to decay into their daughter nuclei. -
ANSThe answer to this query is D because the 1/2-life of a radioactive fabric is defined because
the time it takes for 1/2 of all of the radioactive nuclei to decay into their daughter nuclei, which
may additionally or won't also be radioactive.
A person is sitting on a chair as shown.
Why must the character either lean ahead or slide their toes underneath the chair so as to get
up?
A.
To increase the pressure required to arise
B.
To use the friction with the ground
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