MLS Review Harr: Clinical Chemistry – Questions
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Terms in this set (620)
D. All of these options
D Absorbance is proportional to the inverse log of
transmittance.
Which formula correctly A = -log T = log 1/T
describes the relationship Multiplying the numerator and denominator
between absorbance and by 100 gives:
%T ? A = log (100/100 X T)
100 X T = %T,
A. A = 2 - log %T substituting %T for 100 X T gives:
B. A = log 1/T A = log 100/%T
C. A = -log T A = log 100 - log %T
D. All of these options A = 2.0 - log %T
For example, if %T = 10.0, then:
A = 2.0 - log 10.0
log 10.0 = 1.0
A = 2.0-1.0 = 1.0
A solution that has a B. 2.0
transmittance of 1.0 %T
would have an absorbance B
of: A = 2.0 - log %T
A = 2.0 - log 1.0
A. 1.0 The log of 1.0 = 0
B. 2.0 A = 2.0
C. 1%
D. 99%
,In absorption D. Absorbance is directly proportional to concentration
spectrophotometry:
D Beer's law states that A = a × b × c, where a is the
A. Absorbance is directly absorptivity coefficient (a constant), b is the path
proportional to length, and c is concentration. Absorbance is directly
transmittance proportional to both b and c. Doubling the path length
B. Percent transmittance is results in incident light contacting twice the number of
directly proportional to molecules in solution. This causes absorbance to
concentration double, the same effect as doubling the concentration
C. Percent transmittance is of molecules.
directly proportional to the
light path length
D. Absorbance is directly
proportional to
concentration
Which wavelength would A. 450 nm
be absorbed strongly by a
red-colored solution? A A solution transmits light corresponding in
wavelength to its color, and usually absorbs light of
A. 450 nm wavelengths complementary to its color. A red solution
B. 585 nm transmits light of 600-650 nm and strongly absorbs
C. 600 nm 400-500 nm light.
D. 650 nm
A green-colored solution B. 525 nm
would show highest
transmittance at: B Green light consists of wavelengths from 500-550 nm.
A green-colored solution with a transmittance maximum
A. 475 nm of 525 nm and a 50-nm bandpass transmits light of 525
B. 525 nm nm and absorbs light below 475 nm and above 575 nm.
C. 585 nm A solution that is green would be quantitated using a
D. 620 nm wavelength that it absorbs strongly, such as 450 nm.
,SITUATION: A technologist A. Replace the source lamp
is performing an enzyme
assay at 340 nm using a A Visible spectrophotometers are usually supplied with
visible-range a tungsten or quartz halogen source lamp. Tungsten
spectrophotometer. After lamps produce a continuous range of wavelengths from
setting the wavelength and about 320-2,000 nm. Output increases as wavelength
adjusting the readout to becomes longer peaking at around 1,000 nm, and is
zero %T with the light path poor below 400 nm. As the lamp envelope darkens with
blocked, a cuvette with age, the amount of light reaching the photodetector at
deionized water is inserted. 340 nm becomes insufficient to set the blank reading to
With the light path fully 100%T. Quartz halogen lamps produce light from 300
open and the 100%T control nm through the infrared region. Deuterium or hydrogen
at maximum, the instrument lamps produce ultraviolet-rich spectra optimal for
readout will not rise above ultraviolet (UV) work. Mercury vapor lamps produce a
90%T. What is the most discontinuous spectrum that includes a high output at
appropriate first course of around 365 nm that is useful for fluorescent
action? applications. Xenon lamps generate a continuous
spectrum of fairly uniform intensity from 300-2,000 nm,
A. Replace the source lamp making them useful for both visible and UV applications.
B. Insert a wider cuvette
into the light path
C. Measure the voltage
across the lamp terminals
D. Replace the instrument
fuse
Which type of D. A prism and a variable exit slit
monochromator produces
the purest monochromatic D Diffraction gratings and prisms both produce a
light in the UV range? continuous range of wavelengths. A diffraction grating
produces a uniform separation of wavelengths. A prism
A. A diffraction grating and produces much better separation of high-frequency
a fixed exit slit light because refraction is greater for higher-energy
B. A sharp cutoff filter and a wavelengths. Instruments using a prism and a variable
variable exit slit exit slit can produce UV light of a very narrow
C. Interference filters and a bandpass. The adjustable slit is required in order to
variable exit slit allow sufficient light to reach the detector to set 100%T.
D. A prism and a variable
exit slit
, Which monochromator D. 5-nm bandpass
specification is required in
order to measure the true D Bandpass refers to the range of wavelengths passing
absorbance of a through the sample. The narrower the bandpass, the
compound having a natural greater the photometric resolution. Bandpass can be
absorption bandwidth of made smaller by reducing the width of the exit slit.
30 nm? Accurate absorbance measurements require a
bandpass less than one-fifth the natural bandpass of the
A. 50-nm bandpass chromophore
B. 25-nm bandpass
C. 15-nm bandpass
D. 5-nm bandpass
D. Photomultiplier tube
Which photodetector is
most sensitive to low levels
D The photomultiplier tube uses dynodes of increasing
of light?
voltage to amplify the current produced by the
photosensitive cathode. It is 10,000 times as sensitive as
A. Barrier layer cell
a barrier layer cell, which has no amplification. A
B. Photodiode
photomultiplier tube requires a DC-regulated lamp
C. Diode array
because it responds to light fluctuations caused by the
D. Photomultiplier tube
AC cycle.
Which condition is a C. Dispersion from second-order spectra
common cause of stray
light? C Stray light is caused by the presence of any light
other than the wavelength of measurement reaching the
A. Unstable source lamp detector. It is most often caused by second-order
voltage spectra, deteriorated optics, light dispersed by a
B. Improper wavelength darkened lamp envelope, and extraneous room light.
calibration
C. Dispersion from second-
order spectra
D. Misaligned source lamp
With Actual Solutions
Save
Terms in this set (620)
D. All of these options
D Absorbance is proportional to the inverse log of
transmittance.
Which formula correctly A = -log T = log 1/T
describes the relationship Multiplying the numerator and denominator
between absorbance and by 100 gives:
%T ? A = log (100/100 X T)
100 X T = %T,
A. A = 2 - log %T substituting %T for 100 X T gives:
B. A = log 1/T A = log 100/%T
C. A = -log T A = log 100 - log %T
D. All of these options A = 2.0 - log %T
For example, if %T = 10.0, then:
A = 2.0 - log 10.0
log 10.0 = 1.0
A = 2.0-1.0 = 1.0
A solution that has a B. 2.0
transmittance of 1.0 %T
would have an absorbance B
of: A = 2.0 - log %T
A = 2.0 - log 1.0
A. 1.0 The log of 1.0 = 0
B. 2.0 A = 2.0
C. 1%
D. 99%
,In absorption D. Absorbance is directly proportional to concentration
spectrophotometry:
D Beer's law states that A = a × b × c, where a is the
A. Absorbance is directly absorptivity coefficient (a constant), b is the path
proportional to length, and c is concentration. Absorbance is directly
transmittance proportional to both b and c. Doubling the path length
B. Percent transmittance is results in incident light contacting twice the number of
directly proportional to molecules in solution. This causes absorbance to
concentration double, the same effect as doubling the concentration
C. Percent transmittance is of molecules.
directly proportional to the
light path length
D. Absorbance is directly
proportional to
concentration
Which wavelength would A. 450 nm
be absorbed strongly by a
red-colored solution? A A solution transmits light corresponding in
wavelength to its color, and usually absorbs light of
A. 450 nm wavelengths complementary to its color. A red solution
B. 585 nm transmits light of 600-650 nm and strongly absorbs
C. 600 nm 400-500 nm light.
D. 650 nm
A green-colored solution B. 525 nm
would show highest
transmittance at: B Green light consists of wavelengths from 500-550 nm.
A green-colored solution with a transmittance maximum
A. 475 nm of 525 nm and a 50-nm bandpass transmits light of 525
B. 525 nm nm and absorbs light below 475 nm and above 575 nm.
C. 585 nm A solution that is green would be quantitated using a
D. 620 nm wavelength that it absorbs strongly, such as 450 nm.
,SITUATION: A technologist A. Replace the source lamp
is performing an enzyme
assay at 340 nm using a A Visible spectrophotometers are usually supplied with
visible-range a tungsten or quartz halogen source lamp. Tungsten
spectrophotometer. After lamps produce a continuous range of wavelengths from
setting the wavelength and about 320-2,000 nm. Output increases as wavelength
adjusting the readout to becomes longer peaking at around 1,000 nm, and is
zero %T with the light path poor below 400 nm. As the lamp envelope darkens with
blocked, a cuvette with age, the amount of light reaching the photodetector at
deionized water is inserted. 340 nm becomes insufficient to set the blank reading to
With the light path fully 100%T. Quartz halogen lamps produce light from 300
open and the 100%T control nm through the infrared region. Deuterium or hydrogen
at maximum, the instrument lamps produce ultraviolet-rich spectra optimal for
readout will not rise above ultraviolet (UV) work. Mercury vapor lamps produce a
90%T. What is the most discontinuous spectrum that includes a high output at
appropriate first course of around 365 nm that is useful for fluorescent
action? applications. Xenon lamps generate a continuous
spectrum of fairly uniform intensity from 300-2,000 nm,
A. Replace the source lamp making them useful for both visible and UV applications.
B. Insert a wider cuvette
into the light path
C. Measure the voltage
across the lamp terminals
D. Replace the instrument
fuse
Which type of D. A prism and a variable exit slit
monochromator produces
the purest monochromatic D Diffraction gratings and prisms both produce a
light in the UV range? continuous range of wavelengths. A diffraction grating
produces a uniform separation of wavelengths. A prism
A. A diffraction grating and produces much better separation of high-frequency
a fixed exit slit light because refraction is greater for higher-energy
B. A sharp cutoff filter and a wavelengths. Instruments using a prism and a variable
variable exit slit exit slit can produce UV light of a very narrow
C. Interference filters and a bandpass. The adjustable slit is required in order to
variable exit slit allow sufficient light to reach the detector to set 100%T.
D. A prism and a variable
exit slit
, Which monochromator D. 5-nm bandpass
specification is required in
order to measure the true D Bandpass refers to the range of wavelengths passing
absorbance of a through the sample. The narrower the bandpass, the
compound having a natural greater the photometric resolution. Bandpass can be
absorption bandwidth of made smaller by reducing the width of the exit slit.
30 nm? Accurate absorbance measurements require a
bandpass less than one-fifth the natural bandpass of the
A. 50-nm bandpass chromophore
B. 25-nm bandpass
C. 15-nm bandpass
D. 5-nm bandpass
D. Photomultiplier tube
Which photodetector is
most sensitive to low levels
D The photomultiplier tube uses dynodes of increasing
of light?
voltage to amplify the current produced by the
photosensitive cathode. It is 10,000 times as sensitive as
A. Barrier layer cell
a barrier layer cell, which has no amplification. A
B. Photodiode
photomultiplier tube requires a DC-regulated lamp
C. Diode array
because it responds to light fluctuations caused by the
D. Photomultiplier tube
AC cycle.
Which condition is a C. Dispersion from second-order spectra
common cause of stray
light? C Stray light is caused by the presence of any light
other than the wavelength of measurement reaching the
A. Unstable source lamp detector. It is most often caused by second-order
voltage spectra, deteriorated optics, light dispersed by a
B. Improper wavelength darkened lamp envelope, and extraneous room light.
calibration
C. Dispersion from second-
order spectra
D. Misaligned source lamp
