Site of carbon removal in β-oxidation In mitochondrial β-oxidation of long-chain fatty acids, from which end of the fatty acid are the two-carbon acetyl units removed step-wise?

Introduction / Context:β-oxidation systematically shortens fatty acyl chains. Recognizing the directionality of carbon removal clarifies how acetyl-CoA units are generated and why carbon numbering matters in metabolic maps.

Given Data / Assumptions:

• We are considering standard mitochondrial β-oxidation of straight, even-chain fatty acids.
• Each round removes a two-carbon acetyl unit.
• Enzymatic reactions focus at the β-carbon relative to the carbonyl carbon.

Concept / Approach:In β-oxidation, oxidation occurs at the β-carbon (C-3 counting from the carboxyl carbon). Following dehydrogenation, hydration, and another dehydrogenation, thiolysis cleaves the bond between the α and β carbons, releasing acetyl-CoA from the carboxyl end.

Step-by-Step Solution:Number carbons from the carboxyl end (carbonyl carbon is C-1).Create a double bond between α and β carbons (dehydrogenation).Hydrate across the double bond; then oxidize the hydroxyl at C-3.Thiolysis between C-2 and C-3 liberates acetyl-CoA from the carboxyl end.

Verification / Alternative check:Labeling experiments (e.g., using 14C at the methyl end) show that the methyl carbon persists through multiple cycles, confirming removal from the carboxyl terminus.

Why Other Options Are Wrong:Methyl end removal would describe ω- or α-oxidation pathways, not classic β-oxidation.Random internal removal does not occur in β-oxidation; the pathway is highly ordered.“Both ends” is incorrect; directionality is carboxyl → methyl.

Common Pitfalls:Confusing β-oxidation with α-oxidation (phytanic acid) or peroxisomal shortening rules; misnumbering carbons.

Final Answer:Carboxyl end.