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Samenvatting Enzymologie ( Behaald Resultaat: 16/20): kinetische reacties
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  • Enzymologie
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  • Universiteit Antwerpen (UA)

Samenvatting Enzymologie ( Behaald Resultaat: 16/20): kinetische reacties: Dit document bevat een volledige samenvatting van het vak Enzymologie. Ik heb hiervoor notities van de les, de cursus van Sylvia Dewilde en de powerpoints van Ann Bracke gebruikt en dit bevat alles dat gekend moet zijn voor ...

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Document information

  • October 29, 2021
  • 9
  • 2021/2022
  • Summary

Subjects

  • cursus
  • enzymologie
  • ann bracke
  • sylvia dewilde
  • kinetische reacties
  • samenvatting

Written for

  • Universiteit Antwerpen (UA)
  • Biomedische Wetenschappen
  • Enzymologie
All documents for this subject (35)

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Effect van substraatconcentratie
Reactie:
k k
E+ S ⇌1 ES →2 E+ P
k −1
Voorwaarden:
d [ ES ]
 =0
dt
 [ E¿¿ 0]=[ E ] +[ ES]¿
 V =k 2 [ ES ]

In de stationaire fase is [ES] constant
k 1 [ E ] [ S ]=k 1 [ ES ] +k 2 [ ES ]

Rekening houdende met
[ E¿¿ 0]=[ E ] +[ ES]¿[ E ] =[ E 0 ]−[ES]
dan
k 1 ( [ E 0 ]−[ ES ] ) [ S ] =k −1 [ ES ] +k 2 [ ES ]k 1 [ E0 ] [ S ] −k 1 [ ES ][ S ] =k−1 [ ES ] + k 2 [ ES ]
k 1 [ E0 ] [ S ] =k −1 [ ES ] +k 2 [ ES ] + k 1 [ ES ] [ S ]k 1 [ E0 ] [ S ] =( k−1 +k 2 + k 1 [ S ] ) [ ES ]
k 1 [ E0 ] [ S ]
[ ES ] =
k −1+ k 2+ k 1 [ S ]

k [ E ][S ]
k1 k2 [ E 0 ] [ S] V = 2 0
Snelheid van de reactie V =k 2 [ ES ]V = k −1 +k 2
k −1+ k 2 +k 1 [ S ] +[ S ]
k1

k−1 +k 2
Stel V m =k 2 [ E0 ] en K m = , dan
k1
V m [S ]
V=
Km+ [ S ]

, Effect van de productconcentratie op de enzymatische reactie
Reactie:
k2
k 1 ES ⇌
E+ S ⇌ E+ P
k −1 k −2
Voorwaarden:
d [ ES ]
 =0
dt
 [ E¿¿ 0]=[ E ] +[ ES]¿
 V =k 2 [ ES ] −k −2 [ E][P]

In de stationaire fase is [ES] constant
k 1 [ E ] [ S ] +k −2 [ E ][ P ] =k −1 [ ES ] + k 2 [ ES ]k 1 [ E ] [ S ] +k −2 [ E ][ P ] =( k ¿ ¿−1+k 2 )[ ES ] ¿

Rekening houdende met
[ E¿¿ 0]=[ E ] +[ ES]¿[ E ] =[ E 0 ]−[ES]
dan k 1 ( [ E 0 ] −[ ES ] ) [ S ] +k −2( [ E0 ] −[ ES ] ) [ P ] =(k ¿ ¿−1+ k 2) [ ES ] ¿
k 1 [ E0 ] [ S ] −k 1 [ ES ][ S ] +k −2 [ E0 ] [ P ] −k −2 [ES ][P ]=(k ¿ ¿−1+k 2) [ ES ] ¿
k 1 [ E0 ] [ S ] + k−2 [ E0 ] [ P ]=k 1 [ ES ][ S ] + k−2 [ ES ][ P ] +( k ¿ ¿−1+k 2 ) [ ES ] ¿
( k ¿ ¿ 1 [ S ] +k −2 [ P ] ) [ E0 ]
(k ¿ ¿1 [ S ] +k −2 [ P ] ) [ E0 ]=(k ¿ ¿ 1 [ S ] + k−2 [ P ] + k−1 +k 2 ) [ ES ] ¿ ¿[ ES ] = ¿
k 1 [ S ] + k−2 [ P ] + k−1 +k 2

Snelheid van de reactie
V =k 2 [ ES ] −k −2 [ E ][ P ] V =k 2 [ ES ] −k −2 ( [ E 0 ] −[ ES ] ) [ P ]V =k 2 [ ES ] −k −2 [ E0 ] [ P ] + k−2 [ ES ][ P ]
V =( k 2 +k −2 [ P ] ) [ ES ] −k −2 [ E0 ] [ P ]
(k ¿ ¿ 1 [ S ] +k −2 [ P ] ) [ E0 ] ( k 1 k 2 [ S ] −k−1 k−2 [ P ] ) [E 0]
V =(k 2+ k−2 [P ]) −k −2 [ E 0 ] [ P ] ¿V =¿ ¿V =¿ ¿ ¿V =
k 1 [ S ] + k−2 [ P ] +k −1+ k 2 k 1 [ S ] +k −2 [ P ] + k −1 +k 2

Of
V k 1 k 2 [ S ]−k−1 k−2 [ P ]
=
[ E0 ] k 1 [ S ] +k −2 [ P ] +k −1 + k 2

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