5 酶化学 生物化学习题汇编 sqh

5.0×10-3 5.0×10-2

不作图计算： (1)Vmax和Km。

0.25

0.25

(2)计算[S]=1.0×10-6 mol/L 和 [S]=1.0×10-1 mol/L 时的v。

(3)计算[S]＝2.0×10-3 mol/L 或[S]=2.0×10-6 mol/L 时最初 5min 内的产物总量。

(4)假如每一个反应体系中酶浓度增加至4倍时，Km，Vmax是多少?当[S]＝5.0×10-5 mol/L 时，v是多少? [1]

22、在很多酶的活性中心均有His残基参与，请解释。[2]

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αβεγδθλμπρζωφ°′→×÷±≈

参考文献

1、生物化学习题解析 第二版 陈钧辉等编 2001, 89-102

1

21、(1) Vmax = 0.25μmol/min, 因为, v = Vmax[S]/(Km + [S]), 所以，Km=0.25×1.0×10-5/0.20 - 5.0×10-5 = 1.23×10-5mol/L

(2) v1=Vmax[S]/(Km+[S]) = 0.25×1.0×10-6/(1.25×10-5 + 1.0×10-6) = 0.0185μmol/min, v2=Vmax=0.25μmol/min (3) v= 0.25μmol/min

p=0.25×5=1.25μmol < 2.0×10-3×0.01 = 20μmol

v= 0.25×2.0×10-6/(1.25×10-5 + 2.0×10-6) = 0.033μmol/min p=0.033×5=0.17μmol > 2.0×10-6×0.01 = 0.02 μmol 所以， p≤0.02 μmol

(4) 因为，Km与[E]无关，所以，Km不变。

因为，Vmax=k3[E]，[E]增大4倍，所以Vmax增大4倍=0.25×4=1.0μmol/min v= 1.0×5.0×10-5/(1.25×10-5 + 5.0×10-5) = 0.8μmol/min 2

22、酶蛋白分子中组氨酸的侧链咪唑基pK值为 6.0~7.0，在生理条件下，一半解离，一半不解离，因此既可以作为质子供体(不解离部分)，又可以作为质子受体(解离部分)，既是酸，又是碱，可以作为广义酸碱共同催化反应，因此常参与构成酶的活性中心。