100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
AAMC MCAT Practice Exam 2 Questions and Answers 100% Correct $17.99   Add to cart

Exam (elaborations)

AAMC MCAT Practice Exam 2 Questions and Answers 100% Correct

 14 views  0 purchase
  • Course
  • AAMC MCAT
  • Institution
  • AAMC MCAT

AAMC MCAT Practice Exam 2 Questions and Answers 100% CorrectAAMC MCAT Practice Exam 2 Questions and Answers 100% CorrectAAMC MCAT Practice Exam 2 Questions and Answers 100% CorrectAAMC MCAT Practice Exam 2 Questions and Answers 100% Correct C/P: What expression gives the amount of light energy (in...

[Show more]

Preview 4 out of 75  pages

  • September 15, 2024
  • 75
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers
  • AAMC MCAT
  • AAMC MCAT
avatar-seller
NursingTutor1
AAMC MCAT Practice Exam 2
Questions and Answers 100%
Correct
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?


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 - ANSWER - D) part of a fully folded protein


The answer to this question is D because tryptophan has an aromatic side chain
that will give rise to a significant CD signal in the near UV region if it is found in a
fully folded protein.


C/P: Which amino acid will contribute to the CD signal in the far UV region, but NOT
the near UV region, when part of a fully folded protein?


"Asymmetry resulting from tertiary structural features causes the largest increase in
CD signal intensity in the near UV region of peptides. The side chains of 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) Phe
C) Ala
D) Tyr - ANSWER - C) Ala


C/P: Based on the relative energy of the absorbed electromagnetic radiation, which
absorber, a peptide bond or an aromatic side chain, exhibits an electronic excited
state that is closer in energy to the ground state?


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 - 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 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 - ANSWER - A) n-Pentane
→ 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
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. - ANSWER - 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.

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller NursingTutor1. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $17.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

67096 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$17.99
  • (0)
  Add to cart