Chem 791 J (Biological Mass Spectrometry) – Exam 1 – Take Home Exam Answer Key
1. (10 pts) If a compound with the molecular formula of C5H11NBr2 was ionized by electron ionization, what
would be the m/z ratios and relative ion abundances of the monoisotopic ions and any isotopic ions?
The easiest way to answer this question is to use the program IsoPro.
There is one monoisotopic peak and 5 main isotopic peaks. The m/z and relative ion abundances
are shown below. The tricky part of this question is the fact that a second Br atom causes the most
abundant ion to be the M+2 peak.
m/z
242.9253
243.9283
244.9232
245.9263
246.9212
247.9242
relative ion abundance (%)
51.3 - monoisotopic peak
3.12
100
6.08
48.8
2.96
2. (10 pts) Determine the mass of the protein whose electrospray ionization mass spectrum is shown to the right.
For example,
m z
 1342.1
z
1007.3
100
80
Relative Intensity
Choose two adjacent peaks. Set the higher m/z equal to
(m+z)/z and set the lower m/z equal to [m+(z+1)]/(z+1). Solve
for m and z. This could then be repeated for all adjacent peaks
and averaged to find the molecular weight of the protein,
although it would have been fine to just calculate this value for
one pair of peaks.
1151.2
895.3
60
40
806.1
1342.1
20
732.8
1610.9
0
400 600 800 1000 1200 1400 1600 1800 2000
m/z
m  ( z 1)
 1151.2
z 1
m ≈ 8050 Da
1
3. (10 pts) Explain why singly-charged protein ions are predominantly formed during MALDI, whereas
multiply-charged ions are typically formed during electrospray ionization.
Multiply-charged ions are formed in electrospray because the application of the voltage to the
spraying needle results in the formation of droplets that have an excess charge. Ions emerge from
these charged droplets, and this excess of charge allows multiply-charged ions to emerge. In
MALDI, no such voltage is applied to produce species with excess charge. Instead, in MALDI the
laser provides enough energy to enable charge separation, but there are an equal number of
positive and negative ions produced. Analyte ions emerge with a net charge as a result of this
charge separation and eventual proton transfer with the matrix. The lack of excess charge and the
short time scale available for proton transfer means that analyte ions are typically produced with
only a single charge.
4. (6 pts) You have a sample with one of the following peptides that have the amino acid sequences of
VACHETQ and VACHETK. The (M+H)+ ion of the VACHETQ peptide has a monoisotopic m/z of 787.340.
The (M+H)+ ion of the VACHETK peptide has a monoisotopic m/z of 787.377. What mass accuracy (in ppm)
would be needed to identify which peptide you have?
minimum mass accuracy = (Δm / m) x 106 = ([787.377 – 787.340]/[787.377]) x 106 = 47 ppm
A mass accuracy just below 47 ppm would be necessary to confirm the identity. A mass accuracy
of exactly 47 ppm would still lead to an ambiguous result because one could still measure an ion
that is exactly between the m/z of the two peptides, and one would still be unsure about the
identity.
5. (10 pts) You have detected a new antibacterial compound that has a m/z ratio of 386.231. From a friend, who
knows a lot about mass spectrometry, you learn that a mass accuracy of 1 mmu (millimass unit) would be
sufficient to narrow down the number of possible molecular formulas for this compound to 1. Approximately,
what is minimum mass resolving power you will need to be sure that the compound of interest is not
contaminated by an interfering ion so that you can make your 1 mmu mass accuracy measurement with
confidence?
The ion of interest would need to be resolved from a potential interfering ion that is 0.001 Da (or
greater) away. The approximate minimum resolving power needed would be about 386,000 (m/Δm
386.231
= 386/0.001).
Δm = 0.001
potential interfering ion
6. (15 pts) The three methods used to improve resolution in time-of-flight mass analyzers are (i) orthogonal
acceleration, (ii) delayed extraction, and (iii) a reflectron. Considering these three methods, explain which of
these methods can be combined together in the same instrument. Briefly justify your answers.

Orthogonal acceleration and delayed extraction cannot be combined together to further
improve resolution. Orthogonal acceleration improves resolution by “turning” the ion beam so
2


that the
t ion spreead in spacee and time iss no longer in the direcction of the d
detector. Deelayed
extra
action impro
oves resoluttion by acceelerating thee ions after they have spread out in
n the
direcction of the detector
d
im
mmediately after
a
ionizattion. If delayyed extractiion is used, subsequentt
ortho
ogonal accelleration would negate the
t advantaage obtained
d by delayed
d extraction
n.
Orth
hogonal acceeleration an
nd a reflectrron can be ccombined toogether to fu
urther imprrove
resollution. The reflectron
r
im
mproves ressolution by correcting any spread in the kinettic energies
of a given
g
m/z, regardless
r
of when this spread arisses.
Delay
yed extractiion and a reeflectron can
n be combin
ned togetheer to furtherr improve reesolution.
The reflectron
r
im
mproves ressolution by correcting aany spread in the kinettic energies of a given
m/z regardless
r
of
o when it arrises. Any correction in
n the spread
d of ion kineetic energiess not
achieeved by dela
ayed extracttion will be corrected b
by the reflecctron, thereby further iimproving
resollution.
7. (15 pts) Using
g the equatio
ons for ax and
d qx below, will
w an with a m/z of 10000 be successsfully transm
mitted
throuugh a quadru
upole mass analyzer
a
thatt is operated under the foollowing connditions
DC voltaage: 1200 V
frequency
y = 1.0 MHzz (recall thatt Ω = 2πf)
8
8zeU
4zeV
ax = qx = 2
2
mr0 Ω
mr02Ω2
RF voltage:: 7200 V
r0 = 1 cm
Whaat about an io
on at m/z 500, will it be transmitted through a quuadrupole m
mass analyzerr that is operrated under
the aabove condittions?
for m/z 1000
1
ax = 0.2345
0
qx = 0.7036
(tthe key to geetting thesee values corrrect is to reccognize thatt ax and qx aare unit-lesss, so the
units that you use for th
he different values mustt all cancel in the end; also you neeed to use kg
g
fo
or mass and
d m for dista
ance)
d
so m/z 1000 paass through
h the quadru
upole
these ax and qx values fall in the stability diagram,
for m/z 500
5
ax = 0.469
0
qx = 1.407
these ax and qx values do not fa
all in the sta
ability diagrram, so m/z 500 does noot pass through the
quadrup
pole
m analyzeers that we taalked about iin class, givee one exampple of a methhod that
8. (12 pts) Consiidering the mass
imprroves resoluttion without sacrificing sensitivity.
s
Briefly
B
explaain how the rresolution iss improved in this
methhod and why
y sensitivity is not lost.
3
There arre at least th
hree possibiilities here.
 The reflectron
r
im
mproves ressolution witthout sacrifi
ficing sensitiivity by refllecting all ioons. It
correects spreadss in ion kineetic energiess, while alsoo increasingg the path length of the time-offlightt analyzer. It
I does thesee things witthout losingg ions.
 Addiing helium to
t a quadru
upole ion tra
ap improvess resolution without saccrificing sen
nsitivity. It
does this by giving the ions a more con
nsistent startting point b
before their ejection froom the ion
trap to the detecctor. This minimizes
m
the spread off ions in timee, thereby im
mproving rresolution,
whilee also impro
oving the effficiency of ion
i ejection by collapsin
ng the ion ccloud to a sm
maller
radiu
us, enabling
g fewer ions to be lost.
 Using
g resonancee ejection in
n a quadrupole ion trap
p improves rresolution w
without sacrrificing
sensiitivity. It does this by moving
m
the io
ons closer in
n space to oon another, so that ionss of the
samee m/z are ejeected closerr in time. It does
d
this wiithout losingg ion signall.
9. (12 pts) Consiider the plot to the right that shows the
t relationshhip
betw
ween resolutiion (or resolv
ving power) and m/z forr an FTICR aand an
Orbiitrap. Briefly
y explain (i)) why the ressolving poweer decreasess for
bothh mass analyzzers as m/z ratio
r
increases and (ii) why
w the resollving
poweer drops morre quickly att higher m/z ratios in an FTICR.
(i) Resolving power decreases at
a higher m//z values beecause
ion oscilllation frequ
uencies decrrease at high
her m/z valu
ues.
Recall th
hat maximu
um resolving
g power is ft/2
ft
kly at higherr m/z valuess in an FTIC
CR becausee the
(ii) Resolving powerr decreases more quick
m/z)-1 in an ICR
I
but onlly (m/z)-0.5 in
n an Orbitrrap.
frequenccy is proporrtional to (m
4