enzyme activity??

咩 enzyme 先.... 我估係 protease 啦...

好難答嫁喎.... 因為佢地既 optimum conditions 係唔同的... 因為佢地係三隻唔同既 enzyme...(同埋得5分...我唔會去幫你搵埋 data... 同埋不幸地我搵唔到 data...XD)(試下搵佢地既 optimum pH 同 temperature)
另外三種生果既 protease 都唔同既.

(以下 Scientific Name 應該用斜體)

Common Name: Kiwi fruit (with a "w")
Scientific Name: Actinidia deliciosa (various cultivars)
Protease in abundance: Actinidin (EC 3.4.22.14)


Common Name: Papaya
Scientific Name: Carica papaya
Protease in abundance: Papain (EC 3.4.22.2)

Common Name: Pineapple
Scientific Name: Ananas comosus
Protease in abundance: Bromelain (EC 3.4.22.32)

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Suggested expt. design:
*keep samples on ice!*
1. homogenize samples of the above fruits (攪爛佢地)
2. apply samples on a gelatin sample of known conc. & vol. at 37C(開魚膠粉, 之後變做 solid -> 之後用 sample digest 佢)
3. record the time needed to digest the samples completely.
4. repeat with different conc. of samples and gelatin
5a. compare activity by the time needed to complete the digestion
5b. (optional) compare activity of samples using method described below
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如果你係要做 project, 我會建議你去搵呢三種 enzyme 既 data.
搵下 佢地既 Km 同 Vmax.(提示: 這是大學本科課程，可能會有d深。)

如果你係中六以上既 science 人，你當 chem. equilibrium 去理解啦。

簡單D黎講 Vmax 係 enzyme 係 excess substrate 之下達到既最高 activity。不過我地係測量唔到呢個數既，因為 enzyme activity 同 conc. of substrate 唔係 linear relationship...
Km 係要幾多 substrate conc. 先會去到 1/2 Vmax.

我地會用一條方程式(Michaelis-Menten equation, 唔係度 derive 了 XD) 去 fit 返做到既 data... 即係我地會做好多 set 幾多 substrate 就會有幾多 activity 既 data, 之後 plot graph (1/v vs 1/[S]) 條直線既 intercept 就係 Km 同 Vmax... blah blah blah....