100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
AAMC FL 5 C.P Exam Questions and Answers. $17.99   Add to cart

Exam (elaborations)

AAMC FL 5 C.P Exam Questions and Answers.

 1 view  0 purchase
  • Course
  • AAMC FL 5 C.P
  • Institution
  • AAMC FL 5 C.P

AAMC FL 5 C.P Exam Questions and Answers.AAMC FL 5 C.P Exam Questions and Answers.AAMC FL 5 C.P Exam Questions and Answers.

Preview 2 out of 10  pages

  • September 14, 2024
  • 10
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers
  • AAMC FL 5 C.P
  • AAMC FL 5 C.P
avatar-seller
Lectjoe
AAMC FL 5 C.P Exam Questions and Answers.
*What quantity of Compound 1 must be provided to prepare 100.00 mL of solution with
a concentration equal to Ki?

A. 48.4 mg
B. 24.2 mg
C. 5.64 mg
D. 2.92 mg

"Compound 1 (molar mass: 483.5 g/mol) has been shown to inhibit HIV-1 protease with
Ki = 60.3 μM (Table 1). Ki is the dissociation constant for the enzyme-bound inhibitor,
which is either EI or ESI, depending on the type of inhibitor." - ANS D
In 100.00 mL solution, 60.3 μM Compound 1 contains 6.03 μmol, which when converted
to mol and multiplied by the molar mass, yields 0.00292 g or 2.92 mg.

What functional group transformation occurs in the product of the reaction catalyzed by
Na+-NQR?

A. RC(=O)R → RCH(OH)R
B. ROPO32- → ROH + Pi
C. RC(=O)NHR'→ RCOOH + R'NH2
D. RC(=O)OR'→ RCOOH + R'OH

"The electron transport pathway in Na+-NQR is composed of four flavins (FAD, FMNc,
FMNb, and riboflavin) and a [2Fe-2S] center, with electrons flowing in the direction:
NADH → FAD → [2Fe-2S] → FMNc → FMNb → riboflavin → ubiquinone. Two
electrons are transferred from NADH to FAD in the first step of the cycle, but all
subsequent steps are one-electron transfers." - ANS A
This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed
by Na+-NQR.

flavin - ANS

ubiquinone - ANS coenzymeQ
- Biologically active quinone (electron acceptor in photosynthesis and aerobic
respiration)
- Reduced to ubiquinol upon the acceptance of electrons.
- Long alkyl chain = lipid soluble = act as an electron carrier within the phospholipid
bilayer.

What is the ratio of cation to enzyme in the spectroelectrochemical experiments
described in the passage?

A. 1:2
B. 2:1

, C. 20:1
D. 200:1

"...the researchers used spectroelectrochemistry to investigate the chemical changes
that take place during electron transfer and how these changes are impacted by the
presence of various cations. Na+-NQR was diluted to a final concentration of 0.75 mM
in 0.150 M LiCl, NaCl, KCl, RbCl, or NH4Cl (each solution also contained redox active
mediators) and placed in a glass instrument cell with CaF2 windows."

*ratio of various cations (0.150 M) to Na+-NQR (0.75mM) - ANS D
The ratio can be found by noting that the enzyme concentration was 0.75 mM, while the
concentration of cations was 0.150 M = 150 mM. The ratio is therefore 200:1.

*units!

Boyle's Law - ANS P1V1=P2V2; P is inversely proportional to V

Charles' Law - ANS V1/T1=V2/T2; T is directly proportional to V for an ideal gas

What is the number of neutrons in the nucleus of the atom used to produce laser
radiations?

A. 48
B. 49
C. 50
D. 51

"Researchers performed in situ laser-induced fluorescence imaging and spectral
analysis of different skin areas of patients with acne vulgaris. They used the
fluorescence spectrometer depicted in Figure 2, which employs a 86Kr+ laser that
simultaneously emits radiations of wavelengths 407 nm and 605 nm." - ANS C
The 86/36Kr atom contains 36 electrons and 36 protons. Therefore, the number of
protons is equal to 86 - 36 = 50 neutrons.

*the question states atom (Kr), not the ion (Kr+) actually used in the experiment

pyrrole - ANS a five-membered aromatic heterocycle containing one nitrogen atom
(C4H5N)

*The radiation of wavelength 605 nm CANNOT be used to produce the fluorescence
radiations depicted in Figure 3 because:

A. the energy of the absorbed radiation must be larger than the energy of the
fluorescence radiation.
B. the energy of the absorbed radiation must be smaller than the energy of the
fluorescence radiation.

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 Lectjoe. 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)

81989 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