Chemical And Physical Foundation Of Biological Sys
Chemical and Physical Foundation of Biological Sys
Exam (elaborations)
Chemical and Physical Foundation of Biological Systems Section Bank || Already Graded A+.
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Chemical and Physical Foundation of Biological Sys
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Chemical And Physical Foundation Of Biological Sys
PASSAGE: Wavelength: 266 and 325
Power (mW): 1.5 and 2.2
Pulse duration (ms): 5 and 2
Which laser is suitable for the MALDI technique after its frequency is doubled?
A. Laser A: wavelength 826 nm, power 1.2 mW
B. Laser B: wavelength 714 nm, power 1.2 mW
C. Laser C: wavelength 650 nm, powe...
Chemical and Physical Foundation of Biological Systems
Section Bank || Already Graded A+.
PASSAGE: Wavelength: 266 and 325
Power (mW): 1.5 and 2.2
Pulse duration (ms): 5 and 2
Which laser is suitable for the MALDI technique after its frequency is doubled?
A. Laser A: wavelength 826 nm, power 1.2 mW
B. Laser B: wavelength 714 nm, power 1.2 mW
C. Laser C: wavelength 650 nm, power 1.5 mW
D. Laser D: wavelength 532 nm, power 1.5 mW correct answers The answer to this question is D
because the wavelength must be either 266 nm or 325 nm, and by doubling the frequency of the
laser whose wavelength is 532 nm, the resulting wavelength is 532 nm/2 = 266 nm, because
electromagnetic radiation wavelength and frequency are inversely proportional to each other.
According to the data in Table 1 (previous question), what is one of the values of the
electromagnetic energy delivered during one pulse by the ionizing radiation?
A. 2.0 uJ
B. 3.5 uJ
C. 7.5 uJ
D. 8.0 uJ correct answers C
P = work/time = energy/time
Pt = Energy
(1.5 mw)(5 mW) = E
(1.5 x 10^-3)(5 x 10^-3)=
(7.5 x 10^-6) J or 7.5 uJ
The monochromatic pulsed-laser beam is focused on the sample cell using a thin lens. Each
photon causes the cleavage of a chemical bond in the solute molecules. Because the photon
energy is designed to be larger than the energy needed to break the bond, the excess energy is
absorbed by the solution, causing a local temperature increase. This fast and localized heating
process creates a sound wave that reaches the microphone, which sends the sound to an
oscilloscope for recording. The amplitude of the sound signal is proportional to the amount of
heat produced ΔHnr. The energy meter, based on the photoelectric effect, uses a detector with a
work function of 3.4 eV.
What is the reading of the energy meter in Figure 1 when an appropriate laser is used in PAC to
dissociate a particular chemical bond?
A. Em
B. Delta Hu
,C. Delta Hnr
D. 0 correct answers D
When the laser/photon energy is just right, that means no excess heat is released from bond
breakages inside the cell, and therefore no sound waves are produced. Thus the energy meter will
read 0!
Work function is 3.4 eV
What is the kinetic energy of a photoelectron produced in the energy meter of the PAC device
when the frequency of an incident photon that is NOT absorbed in the solution is f = 5.0 × 1015
Hz? (Note: Use h = 4.1 × 10-15 eV•s.)
A. 23.9 eV
B. 20.5 eV
C. 17.1 eV
D. 12.3 eV correct answers kinetic energy of a photoelectron is equal to hf - 3.4 eV = 20.5 eV -
3.4 eV = 17.1 eV.
*work function*: minimum photon energy required to liberate an electron from a substance, in
the photoelectric effect.
because the work function is the energy required to initially dislodge the electron from the metal
surface. that energy is gone. the remaining energy (if any) will be in the form of KE.
Which alcohol will most likely undergo substitution by an SN1 mechanism in acidic conditions?
A. CH3-CH2OH
B. CH3-CHCH3-CH2OH
D. tert-butylOH correct answers D
The answer to this question is D because this alcohol produces the most stable carbocation
(tertiary) and consequently will most easily lose a water molecule upon protonation of the
hydroxyl group in acidic media.
SN1 rate law? SN1 big barrier? Alkyl halide (electrophile)? SN1 products? correct answers SN1
Rate law: substrate only
Big Barrier: carbocation stability
Electrophile/alkyl halide: 3 > 2 >> 1
Products: mixture of inversion and retention (i.e *RACEMIC MIXTURE*)
SN2 rate law? SN2 big barrier? Alkyl halide (electrophile)? SN2 product? correct answers SN2
rate law: substrate and nucleophile
, Big barrier: steric hinderance (1 sided, backside attack)
Electrophile/alkyl halide: 1 > 2 >> 3
Product: inverted (*STEREOSPECIFIC* bc configuration of reactans leads to SPECIFIC config
in PRODCUTS)
Which extraction procedure will completely separate an amide from the by-product of the
reaction between an amine and excess carboxylic acid anhydride?
A. Add 0.1M NaOH to quench unreacted anhydride. Then add diethyl ether to separate the
layers. The amide can be obtained from the ether layer by evaporating the solvent.
B. Add 0.1M HCl to quench unreacted anhydride. Then add diethyl ether to separate the layers.
The amide can be obtained from the ether layer by evaporating the solvent.
C. Add 0.1M NaOH to quench unreacted anhydride. Then add diethyl ether to separate the
layers. The amide can be obtained from the aqueous layer by neutralizing .with HCL.
D. Add 0.1M HCL to quench unreacted anhydride. Then add diethyl ether to separate the layers.
The amide can be obtained from the aqueuous layer by neutralizing with NaOH. correct answers
A
by-product of the reaction will be an acidic carboxylic acid and the excess unreacted starting
material will also be acidic. Extraction with aqueous base will hydrolyze and extract both of
these into the aqueous layer, leaving the neutral amide in the ether layer.
sound intensity correct answers I = P/A
sound intensity level correct answers sometines Bi will be zero
sometimes Ii will be 10^-12
triacylglycerol correct answers 3 fatty acids linked to glycerol
phosphatides correct answers 2 FA + phosphate + glycerol
*NOTE: the phosphate can be attached to choline which has nitrogens*
also a PHOSPHOLIPID
The researchers mixed liposomes of different sizes and observed that those formed from
Compound 1 were stable to mixing, but mixing those from Compound 2 formed new liposomes
with an average size expected for the effective final lipid concentration.
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