FL Exam #2 Flashcards
(35 cards)
The general experimental procedure was as follows: (1) purified rabbit proteasome (2 nM) was incubated in the presence of porphyrin at 37°C for 30 minutes; (2) the reaction was initiated by addition of the peptide (100 μM); and (3) intensity of fluorescence emission at 440 nm (excitation at 360 nm) was monitored for 20 minutes.
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?
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)
The answer to this question is C.
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.
Which amino acid is found in greatest abundance in the active site of a laccase?
A. Phe
B. Tyr
C. His
D. Cys

The answer to this question is C.
The five-membered imidazole rings depicted in Figure 1 are associated with histidine, an amino acid whose side chain is frequently encountered at the active site of metalloenzymes.
A person is sitting on a chair as shown.
Why must the person either lean forward or slide their feet under the chair in order to stand up?
A. To increase the force required to stand up
B. To use the friction with the ground
C. To reduce the energy required to stand up
D. To keep the body in equilibrium while rising

The answer to this question is D.
As the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet.
Otherwise, if the person did not lean forward or 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 center of mass) and the distance along the horizontal between the center of mass and the support point.
Asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region (250–290 nm) of peptides. The side chains of aromatic amino acid residues absorb in this region. The asymmetry of the α-carbon atom does not impact the CD signal of the aromatic side chain nor do elements of secondary structure.
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 α-helix.
C. part of a β-sheet.
D. part of a fully folded protein.
The answer to this question is D.
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.
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.
Asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region (250–290 nm) of peptides. The side chains of aromatic amino acid residues absorb in this region. The asymmetry of the α-carbon atom does not impact the CD signal of the aromatic side chain nor do elements of secondary structure.
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?
A. Trp
B. Phe
C. Ala
D. Tyr
The answer to this question is C.
All chiral nonaromatic amino acids will contribute solely to the CD signal in the far UV region.
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?
Asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region (250–290 nm) of peptides. The side chains of aromatic amino acid residues absorb in this region. The peptide bond absorbs in the far UV region (190–250 nm).
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.
The answer to this question is B.
Aromatic side chains absorb in the near UV region of the electromagnetic spectrum, which has longer wavelengths, and hence lower energy, than peptide bonds.
Because the energy of the photon matches the energy gap between the ground and the excited state, this implies that the aromatic side chain has more closely spaced energy levels.
A synthetic peptide with the amino acid sequence KTFCGPEYLA was generated as a mimic of the T-loop.
What is the net charge of sT-loop at pH 7.2?
A. –2
B. –1
C. 0
D. +1
The answer to this question is C.
At pH 7.2, the N-terminus will be positively charged and the C-terminus will be negatively charged.
In addition, the lysine side chain will carry one positive charge and the glutamic acid side chain will carry one negative charge.
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 the amount of radioactivity incorporated into sT-loop was measured by detection of β– decay.
In designing the experiment, the researchers used which type of 32P labeled ATP?
A. α32P-ATP
B. β32P-ATP
C. γ32P-ATP
D. δ32P-ATP
The answer to this question is C.
The phosphoryl transfer from kinases comes from the γ-phosphate of ATP.
Therefore, the experiment should require γ32P-ATP.
This experiment was repeated in the presence of a synthetic peptide that mimics the HM domain (sHM) of Ser/Thr kinases with the amino acid sequence FLGFTY. Phosphorylated sHM (spHM) was also used in place of sHM.
When used in place of spHM, which peptide would be most likely to achieve the same experimental results?
A. FLGFAY
B. FLGFQY
C. FLGFGY
D. FLGFEY
The answer to this question is D.
The phosphorylated threonine would most likely be mimicked by glutamic acid.
The curve of a cooperative process has which shape?
sigmoidal
Which of the following best describes the bonds between Cu2+ and the nitrogen atoms of the ammonia molecules in [Cu(NH3)4]2+?
A. Ionic
B. Covalent
C. Coordinate ionic
D. Coordinate covalent
The answer to this question is D.
The Lewis acid–base interaction between a metal cation and an electron pair donor is known as a coordinate covalent bond.
In [Cu(NH3)4]2+, the subscript 4 indicates which of the following?
A. The oxidation number of Cu only
B. The coordination number of Cu2+ only
C. Both the oxidation number of Cu and the coordination number of Cu2+
D. Neither the oxidation number of Cu nor the coordination number of Cu2+
The answer to this question is B.
Because ammonia is neutral, the number 4 reflects only the number of ammonia molecules that bind to the central Cu2+ cation and does not indicate anything about its oxidation number.
4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)
Reaction 1
At STP, the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin would be closest to which of the following?
A. 5 L
B. 10 L
C. 20 L
D. 30 L
The answer to this question is D.
Based on the balanced equation provided, 4 moles of nitroglycerin produces 6 moles of N2(g).
Therefore, 1 mole of nitroglycerin will produce 1.5 moles of N2(g). At STP 1.5 moles of N2(g) will occupy 33.6 L since the molar volume of an ideal gas at STP is 22.4 L/mol.
Which of the following energy conversions best describes what takes place in a battery-powered resistive circuit when the current is flowing?
A. Electric to thermal to chemical
B. Chemical to thermal to electric
C. Electric to chemical to thermal
D. Chemical to electric to thermal
The answer to this question is D.
The chemical energy of the battery elements is used as electrical energy to set the charge carriers in motion through the resistor, where they experience drag from the crystal lattice of the resistive conductor and dissipate their energy as heat from the resistor.
Protein secondary structure is characterized by the pattern of hydrogen bonds between:
A. backbone amide protons and carbonyl oxygens.
B. backbone amide protons and side chain carbonyl oxygens.
C. side chain hydroxyl groups and backbone carbonyl oxygens.
D. side chain amide protons and backbone carbonyl oxygens.
The answer to this question is A.
Secondary structure is represented by repeated patterns of hydrogen bonds between the backbone amide protons and carbonyl oxygen atoms.
What is the major pathway for ATP production once the electron transport chain is shut down?
glycolysis
Why was it necessary for the researchers to determine the activity of the complex I, II, and III of the ETC independent of one another?
A. Complex stability is lost if the complexes are able to interact structurally.
B. The complexes have different cellular locations, and it is not feasible to isolate them together.
C. The complexes all use the same substrates, so their use must be monitored separately.
D. The reactions catalyzed by the complexes are coupled to one another.
The answer to this question is D.
Studying the complexes all together would lead to erroneous results because inhibition of complexes I and II affects the activity of Complex III, which affects the activity of Complex IV.
In the absence of Frizzled activation, CK1 and GSK3 sequentially phosphorylate β-catenin, which targets it for ubiquitination.
In the absence of Frizzled activation, β-catenin is covalently modified and:
A. bound by a proteasome to initiate degradation into short peptides.
B. translocated into the Golgi body for secretion through exocytosis.
C. engulfed by a lysosome where it is hydrolyzed by proteases.
D. stored in vesicles until the signaling pathway is activated.
The answer to this question is A.
According to the passage, in the absence of Frizzled activation, β-catenin is phosphorylated and ubiquitinated. Ubiquitination targets a protein for degradation by a proteasome.
What is the best experimental method to analyze the effect of tdh2 gene deletion on the rate of histone acetylation? Comparing histone acetylation in wild-type and Δtdh2 cells by:
A. Western blot
B. Southern blot
C. Northern blot
D. RT-PCR
The answer to this question is A.
Posttranslational modification of proteins such as histone acetylation is analyzed by Western blotting.
Which experimental approach(es) can be used to analyze the effect of ROS on the lifespan of yeast? Comparing the lifespans of:
I. wild-type yeast versus yeast lacking antioxidant enzymes
II. wild-type yeast versus yeast overexpressing antioxidant enzymes
III. yeasts growing in the presence or absence of hydrogen peroxide
A. I only
B. II only
C. II and III only
D. I, II, and III
The answer to this question is D.
All listed options influence ROS levels in yeast and can be used to analyze the role of ROS in regulating the lifespan of yeast.
Vasopressin regulates the insertion of aquaporins into the apical membranes of the epithelial cells of which renal structure?
A. Collecting duct
B. Proximal tubule
C. Bowman’s capsule
D. Ascending loop of Henle
The answer to this question is A.
Vasopressin regulates the fusion of aquaporins with the apical membranes of the collecting duct epithelial cells.
G542X is another CFTR allele. If a female heterozygous for G542X bears a child fathered by a male heterozygous for the ΔF508 allele, what is the probability that the child would be homozygous for the G542X allele, given that neither parent has CF?
A. 0.00
B. 0.25
C. 0.75
D. 1.00
The answer to this question is A.
Both parents would need to carry the G542X allele in order for a child to be homozygous for the G542X allele.
Because only the mother carries the G542X allele, the probability that the child will be homozygous for the G542X allele is 0.
The infectivity of VSV-EGP, vesicular stomatitis virus (VSV) particles engineered to contain EGP instead of VSV glycoprotein in the viral envelope, was reduced more than 99-fold by inhibitors of the mammalian proteases CatB and CatL.
Based on the passage, CatB or CatL or both would be expected to have which of the following effects, if any, on EGP?
A. No effect
B. Reduction of enzyme activity
C. Formation of protein dimers
D. Digestion into smaller protein fragments
The answer to this question is D.
The passage states that CatB and CatL are proteases. Proteases function to digest proteins into smaller fragments.
The precursor of EGP is translated from a transcript that has had one nontemplated nucleotide added to the open reading frame. This change does not create or eliminate a stop codon. Compared with the protein sGP, which is produced from the unedited transcript, EGP most likely has the same primary:
A. amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence.
B. carboxy-terminal sequence as sGP, but a different primary amino-terminal sequence.
C. sequence as sGP except that EGP has one additional amino acid.
D. sequence as sGP except that EGP has one less amino acid.
The answer to this question is A.
The addition of one nucleotide to the open reading frame of EGP results in a frameshift mutation and an aberrant carboxy-terminal domain.