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Exercice dirigé

Exercice dirigé. La panse des ruminants peut être assimilée à un réacteur microbiologique (chémostat) dans lequel se produisent les réactions microbiennes suivantes: 6 CH 2 O + 2 H 2 O  2 CH 3 COOH + 2 CO 2 + 4 H 2 (I) 4 H 2 + CO 2  CH 4 + 2 H 2 O (II)

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Exercice dirigé

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  1. Exercice dirigé La panse des ruminants peut être assimilée à un réacteur microbiologique (chémostat) dans lequel se produisent les réactions microbiennes suivantes: 6 CH2O + 2 H2O  2 CH3COOH + 2 CO2 + 4 H2 (I) 4 H2 + CO2 CH4 + 2 H2O(II) Les parois gastriques de l’animal absorbent au fur et à mesure tout l’acétate (CH3COOH ) produit; les gaz produits s’échappent par le gosier et la biomasse microbienne produite est transférée vers le second estomac de la chèvre qui s’en nourrit. Métabolismes microbiens(I) et (II)? (I) Métabolisme organotrophe: fermentation acétogénique (anaérobie) • Métabolisme chemolitotrophe: • méthanogénèse • (anaérobie)

  2. Quelle est, à l’état stationnaire, la production totale de biomasse microbienne dans la panse de Biquette (en gC/l.jour) ? Culture continue M(S) = S/(S+Ks) 2/6 Acetate 4/6 H2 CH 1/6 CH4 B2 2/6 CO2 B1 dB1/dt = µ1max . M(CH). B1 - dil. B1 CHo dB2/dt = µ2max . M(H2). B2 - dil. B2 Q dCH/dt = - µ1max/Y1. M(CH). B1 + dil. (CHo-CH) dH2/dt = µ1max/Y1. M(CH). B1 - µ2max/Y2 . M(H2). B2 - dil. H2 CH H2 B1 B2 V Etat stationnaire: dil = Q/V µ1max. M(CH) = dil µ2max. M(H2) = dil dil /Y1 . B1 = dil (CHo – CH) dil /Y1 . B1 = dil /Y2 . B2 - dil (H2)

  3. µ1max. M(CH) = dil µ2max. M(H2) = dil  CH= KCH / ( (µ1max / dil) -1) dil /Y1 . B1 = dil (CHo – CH)  B1 = Y1. (CHo – CH) dil /Y1 . B1 = dil /Y2 . B2 - dil (H2)  B2 = (Y2 / Y1) . B1 prod B1 = µ1max. M(CH). B1 = dil . B1 = dil . Y1. (CHo – CH) = dil.Y1.(CHo–KCH / ((µ1max/dil) -1)) prod B2 = µ2max. M(H2). B2 = dil . B2 = dil . (Y2 / Y1) . B1 = (Y2 / Y1) . prod B1

  4. dil = Q/V = 10 l/jour / 10 l = 1 jour-1 CHo = 720 gC/jour / 10 l = 72 gC/l prod B1 = dil.Y1.(CHo–KCH / ((µ1max/dil) -1)) = 1 j-1. 0.05. (72gC/l – 0.6gC/l/ ( (4 j-1 /1 j-1)-1) ) = 3.59 gC/l.j-1 prod B2 = (Y2 / Y1) . prod B1 = (1/6. 0.01 / 0.05) . 3.59 gC/l.j-1 = 0.12 gC/l.j-1 prod CH4 = prod B2 / Y2 = 0.12 gC/l.j-1 / 0.01 = 12 gCH4 /l.j-1 pour 10 l de panse: 10 l . (12 gC-CH4 /l.j-1 +12 gC-CO2 /l.j-1 ) = 240 gC/ j-1 = 240 gC/ j-1 / (12gC/mol) . 22.4 l/mol = 448 l j-1

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