7. The synthesis of 1 mol palmitate from 1 mol acetyl-C0A requires 14 mol NADPH.

Question

7. The synthesis of 1 mol palmitate from 1 mol acetyl-C0A requires 14 mol NADPH. Some of this NADPH comes from the reaction of malic enzyme, and some from the pentose phosphate pathway. Assuming that the only source of malate is from the oxaloacetate produced by ATP-citrate lyase, what is the maximum amount of NADPH that malic enzyme can produce? If this is less than the NADPH required for palmitate synthesis, calculate how many moles of glucose will need to be used in the pentose phosphate pathway in order to produce the rest of the required NADPH.

Explanation / Answer

1mole oxaloacetate to 1 mole malate --- 1 mole NADPH

1 mol glucose -- pentose phosphate pathway --- 2 mole NADPH

Action of ATP-citrate lyase

citrate + ATP + Co enzyme A + H2O ---   oxaloacetate + Acetyl coenzyme A + ADP + Pi

assuming that only source of malate is oxaloacetate from ATP-citrate lyase,

oxaloacetate + ADP + pi ---- pyruvate + carbonic acid + ATP; oxaloacetate so produced forms pyruvate

pyruvate forms malate by the action of malic enzyme; ;

pyruvate + CO2 + NADH --- malate + NAD+ , thereby using 1 mole of NADH per mole of pyruvate.

hence to produce 1 mole of malate from oxaloacetate (from ATP-citrate lyase) 1 mole of NADH is required.

Dehydrogenation of this 1 mole of this malate will produce 1 mole of NADPH,

hence net production of NADPH is zero if the only source of Malate is  is oxaloacetate from ATP-citrate lyase.

now, to get 14 moles of NADPH 7 moles of glucose are required since 1 mole of glucose produces 2 moles of NADPH during petose phosphate pathway

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.