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Samenvatting Enzymologie ( Behaald Resultaat: 16/20): kinetische reacties $7.06   Add to cart

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Samenvatting Enzymologie ( Behaald Resultaat: 16/20): kinetische reacties

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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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  • October 29, 2021
  • 9
  • 2021/2022
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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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