Question

Calculate the theoretical number of glucose molecules that can be synthesized from complete oxidation of tripalmitin. Assume that synthesis begins at pyruvate and that all ATP produced from tripalmitin oxidation is used directly for glucose synthesis.

Answer 1

Interestingly (fat synthesis from carbohydrate) synthesis of tripalmitin requires 14 moecues of glucose, but reverse reaction is not same.

This will be indirectly measured from oxidation reactions as this is not occurs biologically.

Glucose complete oxidation gives: C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O

tripalmitin complete oxidation gives: 2 C₅₁H₉₈O₆ + 145 O₂ --> 102 CO₂ + 98 H₂O

from above two balanced chemical reactions, we can conclude that 6 moles CO₂ & 6 moles H₂O are required to synthesize one molecule of glucose. In this way from 1 molecule of tripalmitin gives 102 / 2 = 51 moles of CO₂ and 98 / 2 = 49 moles of H₂O.

Dividing both above results with 6 gives how many glucose molecules can be prepared.

51 / 6 = 8.5 ; 49 / 6 = 8.17   ~ 8

Hence theoretically eight molecules of glucose can be synthesized from complete oxidation of tripalmitin.

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