wyklad ch.f.10, Galeria, Muzyka, Dokumenty, ka, Materiały od Oli, fizyczna, Egzamin, fizyczna wyklady i zadania
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REAKCJE ENZYMATYCZNE
Mechanizm Michaelisa-Mentena
k
→
S
+
E
←
ES
1
k
ES
2
→
E
+
P
[ ]
d
P
[]
k
E
=
⋅
dt
[]
d
ES
[][] [] []
k
S
E
k
ES
k
ES
=
⋅
−
−
1
−
1
2
dt
[ ]
d
ES
=
0
dt
[]
[ ] [ ]
2
k
E
⋅
S
ES
=
1
k
k
+
−
1
[ ]
[ ]
)
[ ]
(
k
E
−
ES
⋅
S
[]
1
total
ES
=
k
+
k
−
1
2
[ ]
k
⋅
S
[]
[]
[]
total
ES
=
1
⋅
E
k
+
k
+
k
S
−
1
2
1
1
[ ]
S
[]
[]
total
ES
E
=
⋅
k
+
k
[]
−
1
2
+
S
k
1
[]
[ ]
S
[]
[]
total
ES
=
⋅
E
K
+
S
M
k
k
+
K
=
−
1
2
st
.
Michaelisa
M
k
1
[ ]
d
P
[]
=
k
⋅
ES
2
dt
[]
[ ]
d
P
k
S
[]
[]
total
2
=
⋅
E
dt
K
+
S
M
[ ]
d
P
[]
total
k
E
=
dt
[]
S
>>
K
M
[ ]
d
P
[]
total
=
k
⋅
E
2
dt
JEST TO SZYBKOŚĆ MAKSYMALNA
[]
d
P
[]
total
=
k
⋅
E
2
dt
MAKS
2
[ ]
[ ]
[]
d
P
S
[]
total
=
⋅
k
E
2
dt
K
+
S
M
[] [ ]
[]
[ ]
d
P
S
d
P
=
⋅
dt
K
+
S
dt
M
MAKS
[]
[ ]
d
P
d
P
r
r
=
=
max
dt
dt
MAKS
[ ]
[]
max
S
r
=
⋅
r
K
+
S
M
Metoda wyznaczenia stałej Mikaelisa – graficznie lub numerycznie
1
1
K
1
M
=
+
⋅
[]
r
r
r
S
max
max
1
1
=
f
[]
r
S
© wgrzybkowski 2003
3
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