Isotopic Distribution and Tandem Mass Spectrometry Notes

Isotopic profiles are somewhat tricky to calculate, but the availability of characteristic polynomials makes computation fairly simple.
Input:
- $p_{iA}$ = abundance of +i isotope of atom A
- $n_A$ = number of atoms of type A in the molecule
- For amino acids: carbon, oxygen, nitrogen, hydrogen, sulfur
- Example:
 - Carbon: mono-isotropic mass is 12 atomic mass units
 - C13 as the only isotope
 - $p{0C}$ (C12 abundance), $p{1C}$ (C13 abundance), $p_{2C} = 0$ (no C14 or C15)
 - If molecule has two carbon atoms, $n_C = 2$ , and n of any other atom is zero.
Output:
- $p_i$ = height of the +i isotopic peak
- Characteristic Polynomial: another way of writing the same information in polynomial form.
  - The coefficient of $x^i$ is the ith isotropic peak height.
  - $x$ is a placeholder.
  - Characteristic polynomial is equivalent to the isotropic profile; it's a computational tool.

Example: Molecule with two carbon atoms
- Characteristic polynomial: $(p{0C} + p{1C}x)^2$
- Expanded: $p{0C}^2 + 2p{0C}p{1C}x + p{1C}^2x^2$
- Semantics:
 - Mono-isotropic peak
 - Plus one peak
 - Plus two peak (mass is 26, both carbon atoms are C13). The probability = $p_{1C}^2$ .
Height of the +2 Isotopic Peak:
- Both carbon atoms are C13, so square the probability ( $p_{1C}^2$ ).
Height of the +1 Isotopic Peak:
- One carbon atom is C12, the other is C13. There are two ways to pick this (either carbon atom), so $2 * p{0C} * p{1C}$ .
The values can be thought of as either probabilities or relative heights, proportional to counts for heights.
Molecule with $n_C$ carbon atoms
- Characteristic polynomial: $(p{0C} + p{1C}x)^{n_C}$
Height of the +10 Isotropic Peak:
- $n_C$ carbon atoms; 10 of those are C13, and the rest are C12.
- ${nC choose 10} * p{0C}^{nC - 10} * p{1C}^{10}$

General Polynomial:
- $\sum{i=0}^{nC} {nC \choose i} p{0C}^{(nC - i)} p{1C}^i x^i$
- The coefficient of the $x^i$ term is the height of the ith isotopic peak.
Isotopic profile of H2O:
- $(p{0H} + p{1H}x)^2 * (p{0O} + p{2O}x^2)$
- For oxygen, only O16 and O18 are considered, thus there are peaks at 0 and +2.

Height:
- $p{1H}^2 * p{0O} + 2 * p{0H} * p{1H} * p_{2O}$
- Either two deuterium atoms or one O18 atom.

$(p{0C} + p{1C}x) * (p{0O} + p{2O}x^2)^2$
Coefficient of +5 Isotopic Peak:
- $p{1C} * p{2O}^2$
Characteristic polynomial for SAM (residue composition):
- $(p{0H} + p{1H}x)^{11} * (p{0N} + p{1N}x)^{22} * (p{0O} + p{2O}x^2)^3 * (p{0S} + p{1S}x)^5$

Characteristic polynomial is the product of each atom type raised to the number of atoms:
- $\prod{\text{atom types}} (p{0A} + p{1A}x)^{nA}$

Problem: Given m/z = 501, determine the charge (z) to find the true mass (m).
Isotopic Profile:
- Monoisotopic peak at m/z = $\frac{m+z}{z}$
- Plus one isotopic peak at $\frac{m+1+z}{z}$
- Plus two isotopic peak at $\frac{m+2+z}{z}$
The difference between adjacent isotopic peaks is the reciprocal of the charge (1/z).
Compute the charge using the isotopic peaks and then find the true mass of the peptide.

Use isotopic profile calculations to determine the 3D structure of a protein.
Native form: hydrophobic residues are buried in the core, hydrophilic ones on the surface.
Denature the protein in deuterium (heavy water).
- $p_{1H}$ is the fraction of hydrogen atoms replaced by deuterium.
Peptides in the core are not exposed to heavy water.
Compare isotropic profiles before and after denaturation to determine which residues are in the core.

Time of flight mass spectrometry: time to travel a distance is proportional to the square root of m/z.
Cut big proteins into small peptides using trypsin.
Measure the mass accurately; identify isotopic profiles to get the charge and mass.

Two mass spectrometry experiments in tandem.
Peptides accelerate towards a field; a gate opens and closes to select one peptide.
Peptides collide with a neutral gas molecule, breaking into at most two pieces (prefix and suffix).
Measure the mass of these fragments to create a fingerprint for peptide identification.

Amino acid sequences are chains of amino acid residues.
The total mass of the peptide is the sum of the residue masses plus the mass of water plus the charge divided by charge.
Prefix and suffix ions: charge can stick on either the prefix or suffix.

If only prefix ions are seen with the masses shown, can you identify the peptide?
Sort the masses and take the differences. Look up the residue masses to identify each amino acid in sequence.

Prefix ions are called b ions; suffix ions are called y ions.
$b1$ is the mass of S; $y1$ is the mass of K.
$bn$ are measured starting from the original and going to the last residue and vice versa for $yn$ .

Intact peptide: residue masses + extra OH + extra H + H+ (proton, one unit of charge).
Prefix ions (b ions): sum of residue masses + charge / z.
Suffix ions (y ions): sum of residue masses + OH + H + charge.
Calculation Example: Y2 for a Peptide
- The sum of last two residues + Water + 1 unit of charge divided by one, gives ion mass.

Our genome is sequenced and the vast majority of proteins are known along with sequences.
Enzymes like trypsin act after K or R.
Instead of directly identifying a sequence from spectrum, say to which of these is the BEST match.
Generate theoretical B and Y ions and compare.

Chromatography column acts like a marathon race for peptides by placing an absorbent material and washing sample through it.
Single peaks broken down.
Tandem gives identifications.
Result is listing of proteins.
A 15 amino acid sequence allows for confident identification.