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QCM 22 annales paces maraichers 2016


titoulou
Go to solution Solved by ElCassoulet,

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Bonjour je ne comprends pas cette correction: 

C) Vrai. L'acide laurique (C12:0) doit subir [(n/2)-1] hélices de Lynen, ici n=12 donc il doit subir 5 tours. On aura donc : - Activation de l'AG = -2 ATP - 6 Acétyl-CoA = 6 x 10 ATP = 60 ATP - 5 FADH2 = 5 x 1.5 ATP = 7.5ATP - 5 NADH = 5 x 2.5 ATP = 12.5 ATP => 78 ATP

 

Je ne comprends pas pourquoi ça fait 78 et pas 80 parce que on a 60+7,5+12,5=80?

Merci d'avance

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@titoulou

T'as oublier de compter l'étape d'activation dans ton calcul, on part d'un acide gras non estérifié et son activation coûte l'équivalent de 2ATP à la cellule.

Donc 80-2 = 78 ATP

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