ω-Oxidation is one of metabolic pathways of fatty acids, though less common than β-oxidation. This u — Physical Chemistry — Kinetics Chemistry Question
Unusual pathways of fatty acid oxidation: omega- and (omega-1)-oxidation
ω-Oxidation is one of metabolic pathways of fatty acids, though less common than β-oxidation. This unusual route starts with oxidation of the methyl group of a fatty acid to give new carboxyl group. The resulting dicarbonic acid is further involved into several β-oxidation cycles developing in the direction towards the carboxyl group initially present in the acid. All reactions of ω-oxidation are non-stereospecific.
Due to peculiar features of its structure, synthetic saturated fatty acid D can be involved in mammals into ω-oxidation only (neither in α- nor in β-oxidation). The resulting dicarbonic acid E is metabolized into corresponding acyl CoA, which is further subjected to seven consecutive cycles of β-oxidation to give seven acetyl CoA molecules. The formula of the remaining metabolite F1 of the pathway is C27H39N7P3SO19 5–. F1 exists as anion at physiological pH values. Its hydrolysis leads to two products, one of which, substance F2, does not contain chiral carbon atoms.
Reaction flow scheme:
D --(omega-oxidation--> E --(acyl CoA of E--> 7 beta-oxidation cycles ⟶ F1 --(+H2O--> F2 + ...
Draw the structures of compounds D, E, F2 and anion F1 at pH 7. Show evidence to prove that the answer is unambiguous.
Model Answer
Consideration of mechanisms of ω- and β-oxidation suggests that F1 is an acyl CoA of a dicarbonic acid. Actually, the first carboxyl group was initially present in D, whereas the second one is formed as a result of the final β-oxidation cycle. Taking into account the hydrolysis reaction:
R-HOOC-COSCoA + H2O → R-HOOC-COOH + CoA-SH
one can determine the formula of F2 from the following calculations:
Formula of F2 = Formula of anion F1 + H5 – Formula of non-ionized form of coenzyme A + H2O = C27H39N7P3SO19 + H5 – C21H36N7P3SO16 + H2O = C6H10O4.
Note that the second product of hydrolysis, coenzyme A, cannot be F2 because it contains chiral carbon atoms. All possible structures of dicarbonic acids free of chiral carbon atoms and described by the formula C6H10O4 are presented below, as well as fatty acids D corresponding to each variant of F2. Having in mind that D cannot be involved in either α- or β-oxidation, one can conclude that there is only one choice (highlighted in bold) meeting all above requirements.
[VISUAL]
Formulae of D and E are generated by addition of 14 carbon atoms (7 β-cycles) to the fourth carbon atom in F2. There is no branching in the molecules except one at the α-carbon atom, since only acetyl CoA (and not propionyl CoA, etc.) molecules are released after every β-oxidation cycle.
Thus,
[VISUAL]
Explain why fatty acid D cannot be involved in both α- and β-oxidation.
Model Answer
D cannot be involved in α- or β-oxidation because it does not contain hydrogen atoms bound to α-carbon atom. This makes impossible formation of hydroxyl group and double bond, which are necessary for α- and β-pathways, respectively.
Propose the structure (without stereochemical details) of synthetic fatty acid G, an isomer of compound D, which contains the same number of carbon atoms in the main chain and cannot be involved in both α- and β-oxidation for structural reasons.
Model Answer
Fatty acid D and its isomer G contain 18 carbon atoms in their main chains. Thus, for G only two variants of branching are possible: either two methyl groups or one ethyl group. Possible structures of G with the ethyl group are equivalent with respect to oxidation pathways to phytanic and pristanic acids containing methyl substituents (see problem 22). We have found in question 1 of this problem that α- and β-oxidation pathways are restricted for fatty acids containing two substituents at the α-carbon atom. At the same time, α-pathway is possible if two substituents are bound to β-carbon atom (see solution of question 1). Thus, only a fatty acid containing methyl groups at both α- and β-carbon atoms is left in consideration. In this case β-oxidation is not possible for the same reason as for phytanic acid, whereas α-oxidation is restricted due to formation of ketone instead of aldehyde as an intermediate (a ketone group can not be oxidized to a carboxyl one in vivo).
Thus, the structure of G is:
[VISUAL]
Draw the structures of H and I. Show evidence to prove that the answer is unambiguous.
Context:
(ω-1)-oxidation is another pathway of fatty acid degradation in mammals. It plays an important role in metabolism of prostaglandins and development of several genetic diseases. One (ω-1)-oxidation cycle includes five two-electron oxidation reactions of a fatty acid.
Fatty monocarbonic acid H that contains 75.97 % C, 12.78 % H, and 11.25 % O by mass is widespread in nature. It gives compound J as the final product of (ω-1)-oxidation cycle. Compound I (72.42 % C, 11.50 % H, 16.08 % O by mass) is one of intermediates of the pathway from H to J. 1H NMR spectrum of I contains two singlets with different integral intensities and a number of multiplets. Integral intensity of any multiplet differs from those of singlets. One of the singlets is characterized by the maximal integral intensity among all the signals in the spectrum.
Model Answer
Calculations to determine empirical formulae of compounds H and I:
H: n(C): n(H) : n(O) = 75.97/12.01 : 12.78/1.01 : 11.25/16.00 = 9 : 18 : 1;
I: n(C) : n(H) : n(O) = 72.42/12.01 : 11.50/1.01 : 16.08/16.00 = 6 : 11.33 : 1.
Empirical formula of I is C18H34O3. Fatty acid H cannot contain less carbon atoms than its metabolite. It should also contain two oxygen atoms (monocarbonic acid). Thus, the molecular formula of H is C18H36O2.
H is a saturated fatty acid. Formally, one oxygen atom substitutes two hydrogen atoms in H to give I. There are several options for such substitution, namely formation of: 1) carbonyl group; 2) epoxide; 3) unsaturated double bond and hydroxyl group at another carbon atom; 4) oxygen containing heterocycle. One of singlets corresponds to hydrogen atom of a carboxyl group (integral intensity is minimal). Thus, I is free of hydrogen atoms with the same intensity, and hydroxyl group, −CH−CH− fragment in epoxide cycle, and −CH− fragment in heterocycle are impossible. Carbonyl group is the only variant left, aldehyde group being impossible, since it produces a singlet with the same integral intensity as carboxyl group. Keto group is the final choice. This group should be located at [(ω)-1] carbon atom, because only in this case the methyl group would give a singlet with integral intensity three times higher than that of the carboxyl group. All multiplets give signals with integral intensity of 2 (higher than 1 and lower than 3). Thus, H is a linear fatty acid without branching (only nonequivalent CH2 groups are present between terminal carbon atoms).
Finally,
[VISUAL]
Determine how many steps of two-electron oxidation of H are required to produce I, if it is known that the entire ω-pathway is a part of (ω-1)-pathway.
Model Answer
All reactions of ω-1-pathway are two-electron oxidations of a fatty acid. Reverse analysis shows that I is formed from the corresponding secondary alcohol.
[VISUAL]
This alcohol is formed (do not forget two electrons) directly from stearic acid (H) by oxygenase reaction. Thus, H is converted into I in two steps.
Draw the structure of J.
Model Answer
Three steps are needed to metabolize I to the final product J, since (ω-1)-oxidation includes five consecutive steps. It is further needed to count the number of steps of ω-pathway, which allows formation of carboxyl group from the terminal methyl group. All steps of ω-pathway are also two-electron oxidations as it is a part of (ω-1)-pathway. At the first stage the fatty acid is transformed into primary alcohol by oxygenase reaction. The alcohol is then oxidized to aldehyde, and finally to carbonic acid (similar to (ω-1)-oxidation described above). Thus, ω-oxidation starts from I and includes the final product J. Finally,
[VISUAL]
Determine how many steps of oxidation pathways given below are needed to obtain C from A in organisms of patients with ARD, if it is known that malonyl CoA is not released at the first β-oxidation cycle.
β-oxidation _____
ω-oxidation _____
(ω-1)-oxidation _____
Context:
α-Oxidation is impossible for patients with hereditary pathology Adult Refsum Disease (ARD) due to genetically determined absence of an enzyme of this oxidation pathway. Metabolism of phytanic acid A (a mixture of two diastereomers enriched with R-epimer, i.e. R>S, see problem 21) in organisms of such patients leads to dicarbonic acid C (non-equivalent mixture of two enantiomers, R>S).
Model Answer
Structure of phytanic acid A (determined in problem 21):
[VISUAL]
In organisms of patients with ARD, oxidation of this fatty acid from carboxyl terminus is impossible by any of known pathways. Therefore, degradation should start from ω-terminus. Presence of the methyl group at ω-1 carbon atom does not allow (ω-1)-oxidation. So, the first step is ω-oxidation, which leads to the following intermediate:
[VISUAL]
Repetitive ω-oxidation of the intermediate would lead to tricarbonic acid. Subsequent β-oxidation of this acid would give malonyl CoA, which is in contradiction with the task conditions. Thus, β-oxidation is the only possible pathway of further metabolism of the above dicarbonic acid. The number of β-cycles can be found by analyzing data on compounds A and C. Being a mixture of two enantiomers, C contains one chiral carbon atom. Only two metabolites of β-oxidation pathway are in agreement with this condition:
[VISUAL]
β-Oxidation of metabolite (1) leads to metabolite (2). This transformation is accompanied by inversion of the stereochemical configuration due to changes of the substituents priority.
[VISUAL]
At the same time, five β-oxidation cycles of the dicarbonic acid (giving intermediate (1)) do not lead to inversion of the stereochemical configuration of the chiral carbon atom nearest to the initial carboxyl group. Since the R>S ratio is retained as a result of A metabolism to C, metabolite (1) is the final choice. Even an assumption that metabolite (2) is an AMCAR substrate will not allow treating this substance as appropriate (AMCAR will not alter the S>R ratio).
Thus, the number of steps needed on the way from A to C:
β-oxidation: 5 steps
ω-oxidation: 1 step
(ω-1)-oxidation: 0 steps (the pathway is impossible)
Draw formula(e) (with stereochemical details) of intermediate(s) of A oxidation in organisms of patients with ARD, that can be AMCAR substrates.
Model Answer
The enzyme catalyzing the first step of ω-oxidation is not stereospecific, thus a mixture of diastereomers will be obtained in the case of phytanic acid:
[VISUAL]
Therefore, acyl CoA formed by the product of ω-oxidation (15R-epimer) will be transformed by AMCAR into corresponding S-epimer.
As can be seen from the above scheme, ω-oxidation alters the absolute configuration of C-11 due to the changes in substituents priority, which makes AMCAR catalyzed reaction prior to the third β-oxidation cycle unnecessary. Similar considerations are true for C-7, the absolute configuration of which is changed after second β-oxidation cycle:
[VISUAL]
Thus, the only AMCAR substrate is:
[VISUAL]