作者elelith (trapped)
看板Chemistry
標題Re: [問題] enzyme kinetics
時間Thu May 18 09:05:23 2006
※ 引述《AppleAmily ()》之銘言:
: ※ 引述《elelith (trapped)》之銘言:
: : 這題是有關suicide substrate的
: : Thre reaction scheme is
: : K1 K2
: : E+S <----> ES ----> ED
: : K'
: : In this case [Et]-[ED] = [E] + [ES], where [Et] is the initail starting conc.
: : of enzyme, [ED] is the conc. of dead enzyme, and [E] and [ES] are the conc. of
: : viable enzyme that are repectively free and substrate-bound.
: : a) Using the steady state approximation for [ES], derive an expression for the
: : rate of creation of [ED]. (d[ED]/dt)
: : d[ES]/dt= K1[E][S] - K'[ES] - K2[ES] = 0
: : [ES] = K1[E][S]/(K'+K1)
: : d[ED]/dt = K2 [ES] 這樣不知道對不對 跟d[ES]/dt好像沒關係  ̄ー ̄;
: : 第二題跟第三題問在什麼時候 d[ED]/dt會是fist order and zeroth order in [S]
: : first order when [S] is at low conc. and zeroth when [S] > [E]?
: : when the rate of production of [ED] is zeroth in [S], what will be the rate?
: : K2[ES]?
: : 上面答案如果不對的還煩請指正Orz
: : 謝謝
: first of all, in part a) you need to eliminate the term [ES] by substituting
: the expression [ES] = K1[E][S]/(K'+K2) into d[ED]/dt = K2 [ES] since ES is
: an intermediate of which the actual concentration cannot be measured.
: the rate of formation for the product [ED] then becomes:
: d[ED]/dt = K1K2[E][S]/(K'+K2)
Well, for the first question, there was a hint I forgot to type out, it says
the expression will contain the term {[Et]-[ED]} instead of the usual [Et]
how and which term should I substitute with {[Et]-[ED]}?
: when the rate is zeroth order with respect to [S], the expression becomes
: d[ED]/dt = K1K2[E]/(K'+K2)
: for the second question....I am not really sure if your answer is right or
: not since you didn't provide with any justification, but, generally, I would
: assume the rate of the enzyme catalyzed reaction to be zeroth order with
: respect to [S] if the concentration of the substrate is large comparing to
: [E] (thus can be assumed constant throughout the reaction); the rate should
: be 1st order with respect to [S] in all other cases according to the kinetics.
thx for your reply
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