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22-23 session 1


Go to solution Solved by MALOcclusionsphinctérienne,

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  • Ancien Responsable Matière
  • Solution
Posted

On repère pour la courbe A : 1/vmax = 0.2  *10^6 L.min/mol  (coupure avec l'axe des ordonnées)

donc vmax = 1/(0.2*10^6) = 5 * 10^-6 mol/min/L = 5micromol/min/L  =  donc une UI étant 1 quantité de substrat catalysant 1 micromol/min, là on a 5 UI/L

  • Responsable Matière
Posted
  On 12/17/2023 at 11:39 AM, ilhamoxiciIline said:

Ok merci ! et pour l'item A? je comprends pas nn plus

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faut tracer la courbe, tu vois que ça coupe les abscisses en -4, tu fais -4x104= -1/km donc km = 1/ x10^4

 

km=0.25 * 10^-4 mol/L = 0.025mmol/L

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