Peroxidation of lipids, in particular of those found in biomembranes and lipoproteins, is considered — Physical Chemistry — Kinetics Chemistry Question
Unusual pathways of fatty acid oxidation: peroxidation
Peroxidation of lipids, in particular of those found in biomembranes and lipoproteins, is considered as an important stage in the development of numerous diseases including atherosclerosis. Lipids containing residues of polyunsaturated fatty acids (PUFA) are most liable to oxidation of this type.
Write down the overall reaction of exhaustive ozonolysis of timnodonic acid with subsequent treatment of the reaction mixture with dimethyl sulfide.
[VISUAL]
timnodonic acid (without stereochemical information)
Model Answer
timnodonic acid + 5 O3 + 5 (CH3)2S → CH3CH2CHO + 4 CH2(CHO)2 + OHC(CH2)3COOH + 5 (CH3)2SO
Where:
- CH3CH2CHO is propanal
- CH2(CHO)2 is malondialdehyde (X)
- OHC(CH2)3COOH is 5-oxopentanoic acid
- (CH3)2SO is dimethyl sulfoxide (DMSO)
X is one of the final products of peroxidation of any polyunsaturated acids in mammals. X can by also obtained by reductive ozonolysis of PUFA.
X reveals high reaction ability towards various biomolecules including proteins. In particular, it interacts non-enzymatically with amino acid residues of albumin, an important transport protein of serum. As a result, side groups of two canonical amino acids are cross-linked. The linker formed in this reaction is depicted below (R1 and R2 are fragments of polypeptide chain of the protein):
[VISUAL]
Draw (with stereochemical details) the structures of X and canonical amino acids, side groups of which are involved in the cross-linking.
Model Answer
Since X is formed as a result of reductive ozonolysis of PUFA and contains only C, H, and O, all four nitrogen atoms in the linker must originate from the side chains of two amino acids. Among the six canonical amino acids with nitrogen-containing side chains (Asn, Gln, Lys, His, Arg, Trp), only arginine (providing the guanidine group) and lysine (providing the epsilon-amino group) fit the structure of the linker.
- X is malonic dialdehyde (malondialdehyde): OHC-CH2-CHO
- The two canonical amino acids are L-lysine and L-arginine.
Structures of the products in the solution:
- Malondialdehyde: OHC-CH2-CHO
- L-lysine: H2N-CH(COOH-(CH2)4-NH2 (with stereochemistry at the alpha-carbon)
- L-arginine: H2N-CH(COOH-(CH2)3-NH-C(=NH-NH2 (with stereochemistry at the alpha-carbon)
Suggest mechanism of the linker formation, if it is known that only water molecules are released during the cross-linking.
Model Answer
The guanidino group of the arginine residue (Arg) and the epsilon-amino group of the lysine residue (Lys) react with malondialdehyde (X) via nucleophilic additions to the carbonyl groups and subsequent cyclization and dehydrations, releasing two molecules of water to form the stable conjugated imidazole-like linker:
Arg-NH-C(=NH-NH2 + Lys-NH2 + OHC-CH2-CHO → + 2 H2O
Y is another product of peroxidation of lipids. It contains the same number of carbon atoms as X and interacts with both proteins and nucleic acids.
Interaction of Y with lysine residues present in a protein results in formation of residues of non-canonical amino acid Ne-(3-formyl-3,4-dehydropiperidino) lysine (FDP-lysine):
[VISUAL]
Draw the structure of Y, taking into account that equimolar amount of water is released upon FDP-lysine formation.
Model Answer
Since the adduct of lysine with Y contains six extra carbon atoms compared to the starting lysine, two molecules of Y (each containing 3 carbon atoms) must react with one lysine residue to form FDP-lysine.
Based on the mass balance and the release of an equimolar amount of water:
Formula of Y = (FDP-lysine - lysine + H2O) / 2 = (C12H20O3N2 - C6H14O2N2 + H2O) / 2 = C3H4O.
Since Y is a common product of lipid peroxidation containing a carbonyl group and reacts with lysine residues, it is acrolein (propenal):
CH2=CH-CHO
Suggest mechanism of formation of FDP-lysine residue if the starting lysine residue is a part of a protein. Note that Michael reaction is one of the steps of the pathway.
Model Answer
The mechanism proceeds through the following steps:
1. Nucleophilic conjugate addition (Michael-type addition) of the free epsilon-amino group of lysine (I) to the double bond (C-3) of the first acrolein molecule (Y) to form a secondary amine intermediate (II) with the carbonyl group intact.
2. Intermediate (II) undergoes a second Michael-type conjugate addition to the double bond of a second acrolein molecule to yield a tertiary amine intermediate (III) containing two 3-oxopropyl chains.
3. Intermediate (III) undergoes intramolecular aldol condensation to form a cyclic beta-hydroxy aldehyde intermediate (IV).
4. Subsequent dehydration (croton condensation) of intermediate (IV) with the elimination of one molecule of H2O yields the final FDP-lysine residue (V).
Interaction of Y with nucleoside Z found in nucleic acids results in an adduct, nucleoside Z1. Mass spectrum of Z1 obtained by using fast atom bombardment mass spectrometry (FAB-MS) contains two major peaks corresponding to monoprotonated fragments (M+H+), m/z values being equal to 191 and 307.
Draw the structure of Z, if its reaction with Y gives solely product Z1. Z1 contains a base, a fragment of which is given below:
[VISUAL]
Model Answer
The difference between the two FAB-MS peaks is 307 - 191 = 116, which corresponds to a deoxyribose residue (minus H2O, 134 - 18 = 116), indicating that Z is a deoxyribonucleoside found in DNA.
The mass of the protonated modified base is 191, so the mass of the modified base itself is 190. Knowing that the base is modified by N acrolein residues (molecular mass of each acrolein addition is 56):
Mass of unmodified base = 190 - N * 56
- For N = 1: 190 - 56 = 134, which matches the mass of adenine (molecular mass = 135, or 134 as a radical/fragment in the nucleoside).
Therefore, the original nucleoside Z is deoxyadenosine (deoxyadenine).
Structure of Z (deoxyadenosine):
- Base: Adenine attached at the N9 position to the 1'-carbon of deoxyribose.
Draw the structure of Z1.
Model Answer
Z1 is deoxyadenosine modified by one molecule of acrolein (Y) to form a tricyclic base. The structure consists of deoxyadenosine where the adenine base has a fused ring formed by reaction with acrolein, specifically forming a 1,N6-(2-hydroxypropane-1,3-diyl)deoxyadenosine adduct (or its regioisomer):
Structure of Z1 shows the deoxyadenosine nucleoside with the acrolein adduct bridging the N1 and N6 positions of the adenine ring to form an additional six-membered ring with a hydroxyl group.