Problem Set # 1 Answer Key
Chemistry 791J – Biological Mass Spectrometry 2016
1. (15 pts)
(a) [C7H10N]+  108.081
contributes to the m/z)
(b)
C613C1H1014N
C71H92H14N
12
C71H1015N
12
12
(note that there is an extra H+ from Chemical Ionization that
109.084
109.087
109.078
(c) The easiest way to answer the problem would be to use the IsoPro program that I posted
on the website. If you used that, then you would arrive at a relative abundance of 8.28% if
the monoisotopic peak is 100% relative abundance. Alternately, you could assume that the
13
C is the only isotope that would significantly contribute to the peak that is 1 mass unit
higher. By only considering this isotope, you would arrive at a +1 isotope peak that
contributes 7.49% (7 x 1.07%). The monoisotope peak would then be 92.51% (100% 7.49%), so the relative abundance of the +1 isotope peak would be 8.10% if the monoisotopic
peak were 100% relative abundance. Either way, the answer is about 8%.
2. (12 pts)
K.E. = ½ mv2
1 eV = 1.60 x 10-19 J
mass of an electron = 9.109 × 10-31 kg
1 J = kg m2/s2
The key to this problem is to get the units correct.
70 eV electron:
v = 4.96 x 106 m/s (1.6% of the speed of light)
200 eV electron:
v = 8.38 x 106 m/s (2.8% of the speed of light)
In case you are interested. . . The higher electron energy in CI is necessary for the ions to make it
through the relatively dense gas of reagent used in CI. Because the reagent gas pressure is
typically high (up to 0.1 Torr), one needs higher energies for the electrons to make it through and
still ionize reagent molecules after they lose a lot of their kinetic energy upon collisions with the
other reagent gas molecules.
3. (10 pts)
A nucleus has a mass less than the sum of its protons and neutrons because of the binding
energy necessary to keep the nucleus together. It is the mass of the binding energy of the
nucleus that explains the defect (E = mc2). Mass is lost as it is converted to energy to keep the
protons and neutrons together in the nucleus. The more energy required to keep the protons
and neutrons together, the greater the negative mass defect. Another way to look at this is to
say that the mass defect is the difference between the calculated mass of the unbound
protons and neutrons and the actual measured mass. Interestingly, uranium has a positive
mass defect because the binding energy of the nucleus is relatively weak compared to other
elements.
4. (12 pts)
You would need to choose dimethylamine ((CH3)2NH) because it has proton affinity higher
than glycine, so it won’t transfer a proton to ionize this molecule, but lower than histidine, so
it will transfer a proton to histidine.
5. (11 pts)
Binding of calcium to Calbindin D28K causes a structural change that makes the protein
more compact and thus causes it to accommodate less charge during the electrospray
ionization process.
6. (15 pts)
Use the following equation and make sure to get the units correct:
Von = 2 x 105 (γrc)1/2ln(4d/rc)
After rearranging the equation to solve for the surface tension (γ), one will find that a
surface tension below about 0.0413 N/m is needed to be able to do electrospray at a voltage
of 3000 V or less. At temperatures below 45 ºC, the surface tension of a 25/75
methanol/water mixture is too high and thus requires greater than 3000 V to generate an
electrospray voltage. So, in order to electrospray this protein with the equipment available
to you, you would need to operate the electrospray source at about 45 ºC or greater.
7. (15 pts)
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 can then be repeated for all adjacent peaks and
averaged to find the molecular weight of the protein
For example,
mz
 1395
z
m  ( z  1)
 1268.4
z 1
m ≈ 13940 Da
8. (10 pts)
This is not a trick question. The answer is 16,804 Da if the measurement was done in the
positive mode because MALDI produces [M+H]+ ions. It is 16,806 Da if the measurement
was done in the negative mode because MALDI would then produce [M-H]- ions.