Breakthroughs in molecular mass determinations of biopolymers such as proteins were recognized with — Physical Chemistry — Electrochemistry Chemistry Question
Mass spectrometry of hemoglobin
Breakthroughs in molecular mass determinations of biopolymers such as proteins were recognized with the 2002 Nobel Prize in chemistry awarded to Fenn for developing electrospray ionization mass spectrometry (ESI MS) and to Tanaka for pioneering work leading to matrix–assisted laser desorption/ionization time–of–flight mass spectrometry (MALDI–TOF MS). In MALDI–TOF MS, proteins are embedded in a crystal of UV–absorbing matrix molecules and desorbed/ionized upon irradiation with a UV laser pulse. Singly charged protonated protein ion, [M+H]+, is produced as a major species from a protein with mass M.
If you lived in the 19th century, what method would you use to determine the molecular mass of hemoglobin? Explain.
Model Answer
Osmotic pressure (freezing point depression is too small, mass spectrometry is not available)
Consider hemoglobin with a 67,434 Da molecular mass. After desorption / ionization, the [M+H]+ ion is accelerated by 20.000 kV. Calculate the energy of the protein ion in joule. (coulomb × volt = joule)(e = 1.60218×10–19 coulomb)
Model Answer
For a singly charged protein ion (67,435 Da)
Electrical energy = e V = 1.60218×10–19 C × 2.0000×104 V = 3.20436×10–15 J
The accelerated protein ion is then allowed to travel 1.0000 m in an evacuated flight tube to a detector. All electrical energy is converted to kinetic energy (mv2 / 2). If the flight time of the protein ion was determined to be 1.3219×10–4 s, what is the molar mass of hemoglobin calculated from the flight time measurement? What is the mass accuracy in ppm?
Model Answer
mv 2 / 2 = electrical energy
m = (2)(electrical energy) / v 2
= 2 (3.20436×10–15 J) / (1.0000 m / 1.3219×10–4 s)2 = 1.11987×10–22 kg
M [M+H]+ = 1.11987×10–22 kg × 6.0221×1023 = 67.440 kg mol–1
Molar mass M of haemoglobin = 67,440 – 1 = 67,439 kg mol–1
mass accuracy = 67,439 / 67,434 = 1.000074 74 ppm
The flight tube is maintained under a high vacuum at 25 °C. What is the residual pressure in the flight tube at which the mean free path of air molecules is the same as the length of the flight tube? See Problem 2 for definition of mean free path. Assume that all air molecules are spheres with a diameter of 2 angstroms.
Model Answer
Volume of collision cylinder = π d2 v
Number of molecules in unit volume: N / V = p N0 / RT
Collision / sec = (volume of collision cylinder)(molecules / unit volume) =
= (πd2 v) (p N0 / RT)
Time between collisions = 1 / [(πd2 v) (pN0 / RT)]
mean free path = speed / time between collisions
= v / [(πd2 v) × (pN0 / RT)] = 1 m
p = (RT / N0) / [(πd2)(1 m)]
= (8.314 J mol–1 K–1) (298 K) / [(6.02×1023 mol–1) (3.14) (2×10–10 m)2 (1 m)]
= 3.3×10–2 Pa = 3.2×10–7 atm