Human biochemistry – Through which metabolic entry routes is dietary fructose metabolized in human tissues (considering tissue-specific enzymes)?

Introduction / Context:
Fructose metabolism in humans follows tissue-dependent routes. The liver uses a specialized fructose 1-phosphate pathway that bypasses phosphofructokinase control, while extrahepatic tissues with sufficient hexokinase activity can phosphorylate fructose to fructose 6-phosphate. This question tests recognition of both physiologic entry points for fructose carbon into central metabolism.

Given Data / Assumptions:

• Liver expresses fructokinase (ketohexokinase) and aldolase B.
• Many extrahepatic tissues (e.g., muscle) possess hexokinase that can act on fructose at high enough concentrations.
• Downstream, triose phosphates merge into glycolysis/gluconeogenesis pathways.

Concept / Approach:
In hepatocytes, fructokinase converts fructose to fructose 1-phosphate; aldolase B cleaves it to dihydroxyacetone phosphate and glyceraldehyde, which is then phosphorylated to glyceraldehyde 3-phosphate. In extrahepatic tissues, hexokinase can phosphorylate fructose to fructose 6-phosphate, feeding directly into glycolysis. Therefore, both routes are valid in vivo; what differs is tissue distribution and regulatory context.

Step-by-Step Solution:

Verification / Alternative check:
Metabolic labeling studies trace fructose carbons entering triose phosphate pools via both hepatic and hexokinase-dependent routes; clinical genetics of hereditary fructose intolerance (aldolase B deficiency) selectively disrupts the fructose 1-phosphate pathway, confirming its liver specificity.

Why Other Options Are Wrong:

• “Glyceraldehyde 3-phosphate pathway” is not an independent entry route; it is a downstream product common to both routes.
• “Neither of the above” ignores well-established hepatic and extrahepatic metabolism.

Common Pitfalls:
Assuming fructose only uses the liver pathway; in fact, hexokinase can handle fructose where expressed, albeit with lower affinity than glucose.

Final Answer:
Both (a) and (b)