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
... [Show More] 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) - ANSWERSC) (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 - ANSWERSD) 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 - 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 - ANSWERSA) 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] - ANSWERSkcat = 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 - ANSWERSB) 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 - ANSWERSA) 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. - ANSWERSD) 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.
C/P: A person is sitting in a chair. 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 - ANSWERSD) to keep the body in equilibrium while rising
The answer to this question is D because 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.
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 - ANSWERSD) 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 - ANSWERSC) 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?
"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) 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. - ANSWERSB) An aromatic side chain; the absorbed photon energy is lower.
The answer to this question is B because 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.
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 as a mimic of the T-loop. 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."
A) -2
B) -1
C) 0
D) +1 - 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 side chain will carry one positive charge and the glutamic acid side chain will carry one negative charge.
C/P: In designing the experiment, the researchers used which type of P-32 labeled ATP?
A) aP32-ATP
B) BP32-ATP
C) γP32-ATP
D) δP-32 ATP - ANSWERSD) δP-32 ATP
The answer to this question is C because the phosphoryl transfer from kinases comes from the γ-phosphate of ATP. Therefore, the experiment should require γ32P-ATP.
C/P: When used in place of spHM, which peptide would be most likely to achieve the same experimental results?
"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."
A) FLGFAY
B) FLGFQY
C) FLGFGY
D) FLGFEY - ANSWERSD) FLGFEY
The answer to this question is D because the phosphorylated threonine would most likely be mimicked by glutamic acid in terms of size and charge.
C/P: Based on the information in the passage, PDK1 catalyzes the addition of phosphate to what functional group?
"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."
A) Hydroxyl
B) Amine
C) Carboxyl
D) Phenyl - ANSWERSA) Hydroxyl
The answer to this question is A because reactions involving either Ser or Thr would involve the hydroxyl group in the side chain of these amino acids.
C/P: Which statement about the cooperativity of RIα/C activation and RIα protein folding is supported by the data in figures 2 and 3?
A) Both activation and folding are cooperative.
B) Activation is cooperative, but folding is not.
C) Folding is cooperative, but activation is not.
D) Neither activation nor folding is cooperative - ANSWERSA) Both activation and folding are cooperative.
The answer to this question is A because both curves have a sigmoidal shape, which is indicative of cooperative processes.
C/P: A patient puts on a mask with lateral openings and inhales oxygen from a tank.
Which phenomenon causes static air to be drawn into the mask when oxygen flows?
A) Doppler effect
B) Venturi effect
C) Diffusion
D) Dispersion - ANSWERSB) Venturi effect
The answer to this question is B because oxygen pressure is the sum of the oxygen static pressure P and the oxygen flow pressure rv2/2. In the area of the mask openings, Pair = P + rv2/2, thus Pair > P. Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen in flow. This is the Venturi effect, and the mask is called the Venturi mask.
Doppler effect - ANSWERSan increase (or decrease) in the frequency of sound, light, or other waves as the source and observer move toward (or away from) each other
This effect causes the sudden change noticeable in a passing siren, as well as the redshift seen by astronomers.
Venturi effect - ANSWERSreduction in fluid pressure that results when a fluid flows through a constricted section of a pipe
diffusion - ANSWERSnet movement of molecules from a region of higher concentration to a region of lower concentration
dispersion - ANSWERSthe separation of light into colors by refraction or diffraction with formation of a spectrum
C/P: What causes duplex DNA with a certain (A + T):(G + C) ratio to melt at a higher temperature than comparable length duplex DNA with a greater (A + T):(G + C) ratio?
A) Stronger van der Waals forces of pyrimidines
B) Stronger van der Waals forces of purines
C) Increased number of hydrogen bonds
D) Reduced electrostatic repulsion of phosphates - ANSWERSC) Increased number of hydrogen bonds
The answer to this question is C. GC base pairs involve three hydrogen bonds, while AT base pairs involve only two. This disparity has often been used to explain the increased melting temperature of DNA rich in GC content.
C/P: Which property of a substance is best used to estimate its relative vapor pressure?
A) Melting point
B) Boiling point
C) Molecular weight
D) Dipole moment - ANSWERSB) Boiling point
The answer to this question is B because of the properties listed, the boiling point of a substance will give the best estimate of its relative vapor pressure.
C/P: What are the structural features possessed by storage lipids?
A) Two fatty acids ester-linked to a single glycerol plus a charged head group
B) Three fatty acids ester-linked to a single glycerol
C) Two fatty acids ester-linked to a single sphingosine plus a charged head group
D) Three fatty acids ester-linked to a single sphingosine - ANSWERSB) Three fatty acids ester-linked to a single glycerol
The answer to this question is B because triacylglycerols are neutral storage lipids. They consist of three fatty acids ester-linked to a single glycerol.
C/P: In the overall electrochemical reaction:
N2(g) + H2(g) --> NH3(g)
Half reactions: H2(g) → 2H+ + 2e-
N2(g) + 6H+ + 6e- → 2NH3(g)
A) nitrogen is oxidized at the anode, and hydrogen is reduced at the cathode.
B) nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode.
C) nitrogen is reduced at the anode, and hydrogen is oxidized at the cathode.
D) nitrogen is oxidized at the cathode, and hydrogen is reduced at the anode. - ANSWERSB) nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode.
The answer to this question is B because oxidation always occurs at the anode and reduction at the cathode of an electrochemical cell. Since nitrogen decreases in oxidation state during the reaction, it is reduced. Hydrogen, on the other hand, increases in oxidation state and is, therefore, oxidized.
ANOX REDCAT - ANSWERSIn an electrochemical cell:
oxidation occurs at the anode
reduction occurs at the cathode
C/P: In industrial use, ammonia is continuously removed from the reaction mixture. This serves to drive Reaction 1 because of:
N2(g) + H2(g) --> NH3(g)
A) Boyle's law
B) Charles's law
C) Heisenberg's principle
D) Le Châtelier's principle - ANSWERSD) Le Châtelier's principle
The answer to this question is D because removing a product as it forms causes a displacement from the equilibrium condition. The system will respond by shifting more reactants to the product side. This is an example of Le Châtelier's principle.
Boyle's law - ANSWERSa law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature
Charles' Law - ANSWERSa law stating that the volume of an ideal gas at constant pressure is directly proportional to the absolute temperature
Heisenberg principle - ANSWERSthe position and the velocity of an object cannot both be measured exactly, at the same time, even in theory
Le Châtelier's principle - ANSWERSa principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint
C/P: The lone pair of electrons in ammonia allows the molecule to:
A) assume a planar structure
B) acts as an oxidizing agent
C) act as a Lewis acid in water
D) act as a Lewis base in water - ANSWERSD) act as a Lewis base in water
The answer to this question is D because, by definition, a Lewis base is a substance that donates an electron pair in forming a covalent interaction. [Show Less]