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
... [Show More] 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?
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 [Show Less]