生物化学工程基础(李强)生化作业9.docx

生化第13周作业—廖天琳_化02_2010011861
(1) Octyl Sepharose 是与一种多糖琼脂糖OH 基的一部分辛基 CHYCH^-结合的，市场上销售酶固定化用的凝胶颗粒。

100ml 溶 輕某种酶的缓冲液中(酶浓度15U/ml)加入10g 这种凝胶颗粒，搅
拌混合片刻后，过滤，得到100ml 滤液，滤液的酶活性为0.8U/ml o 且得到的固定化酶颗粒的活性为110U/go 回答下列问题。

列出三大类酶的固定化方法。

本题的实验属于其中哪一类固定化。

分别用百分率表示活性保持率及活性收率。

酶的固定化方法归纳起来大致可以分为三类，即载体结合法、交联法和包埋法。

(1) 载体结合法(carrier binding)载体结合法是指将酶固定到非水溶性载体上的方法。

根据固定方式的
不同，这种方法又可以分为物理吸附法(physical adsorption )＞离子 结合法(ionic bonding)和共价结合法(covalent bonding)。

(2) 交联法(cross-liking)采用双功能或多功能的交联剂，使酶与酶Z 间相互交联或酶 与载体之I'可交
联。

(3) 包埋法(entrapment)把溶液酶包埋于凝胶、屮空纤维或微囊内。

本题实验属于包埋法中的凝胶包埋法。

酶活收率：实际测定的固定化酶的活力与固定化时使用的全部游离酶的总活力之比。

110 U/g x 10g
15U/mL x 100 mL 酶活保持率：实际测定的固定化酶的活力与被固定化的酶在溶液状态时的活力Z 比。

s 十 E — + 110 U/a x 10a
酶活保持率=““ J 鳥爲八［仙 j = 77.5%
(15(//mL — 0.8〃/m 厶)x 100 mL
定化酶颗粒中，若以下变量增大，总有效系数是增大还是减
rjt - 1),随e 的增大而减小
外部传递速率_ ksR 内部传递速率一 De
酶活收率= =73.3%
2) 解释其理由。

酶浓
度, ,边界膜传质系数。

平均半径，米氏常数，底物浓度，分配 总有效系数加满足阻力串联关系
对球形颗粒，内扩散底物模数
V T 7 Bi
平均半径R t , 4）t , T]T I
酶浓度 f , ^max f，G f，丨
米氏常数心f,e丨，衍f
底物浓度t,外扩散抑制作用J ,衍t
边界膜传质系数人t , 77T t
由米氏方程
= ^max[s]_
v K m + [s]
分配系数p = SJS,其中式中£和S分别表示底物或其它效应物在微环境中的局部浓度和宏观体系中的总体（或平均）浓度。

从而
其中略=K m/p
当分配系数p t ,表观米氏常数I ,扩散的影响增强，衍I
⑶选做
Note6.1 ⑷(详见第49页)v 二k2@O-k.2PG 卜(少))=厂单
e 。

+k2 +k]S + k.2P Problem 6.8 Converting glucose to fructose using a glucose isomerase enzyme.
(a)In Note 6.1. (4). the rate of a first order reversible reaction S — P via the substrate-enzyme
complex ES is derived. If the substrate contains no product the feed concentration is
connected to s and p by
So = $ + p = £cq + “eq = Seq(呛)= $eq( 1 + Keq) (1)
In (1), Keq =殳=丹呂is the equilibrium constant which is a function of temperature only.
Insert these expressions in (4) of Note (6.1) and show that -r.=呉£二巩. where K m and k mtA serve the same purpose as in Michaelis-Mentcn kinetics, but the parameters Rg) are
complicated functions of the four rate constants (see Gram et al.(1990)).
k
For Km » (f - $eq): 一几竺(£ 一呵)=^0(^eq — X) ⑵
代m
In (2)t i is a first-order rate constant (unit: h *) and x is the conversion of S. X = 1 - $/$o・ and g =
(b)Gram et al. studied conversion of glucose to fructose using the Novo-lndustry (now
Novozymcs) immobilized glucose isomerase product.
They used a reaction temperaiure of 60°C, at which temperature the equilibrium conversion
8 = (1- s/%) = 0.51.
One aspect of the research was to find whether diffusion resistance was significant in catalyst pellets of different pellet radius R p. Specifically, it was desired to find the diffusion coefficient D cff for the pellet matrix and the rate constant 匕
which would be applicable if the glucose concentration inside the pellet was equal to
the glucose concentration in the bulk liquid outside the pellet. For this purpose,
steady-state experiments were conducted in a CSTR ・ The enzyme pellets were
homogenously dispersed in the reactor.
In two experiments at 60°C, the following results were obtained for $o = 540 g glucose L 1 and with dilution rate D = 0.25 h".
Experiment 1: /?p = 0.3 mm. s = ^emuen( = 289.0 gL
Experiment 2: /?p= 1.0 mm, $ = 365.2 gL L
Determine the rate of glucose consumption -(r,) for the two experiments・
(c) What is the significance of the ratio between r s(/?p = 0.3) and r s(/C p = 1.0)?
Determine an algebraic relation for the ratio between 6(0.3) and <I>( 1.0) (see Note
6.2 (15)).
Using this relation, you are required to find a numerical value for <l>(0.3).
Finally, determine k in (2) and the effective diffusion coefficient D cf f.
[E]t = [E] + [ES] 由准稳态假设,I*ES = 0 = k』E]⑸一k_】[ES] — k2〔ES] + k_2〔P][E]
=(& ⑸ + k_2[P])([E]t - [ES]) 一(J + k2)[ES]
得
r 1= [Eh(ki[S]) + k_2[P])
1」—k][S] + k_2[P] + k“ + k2
-r s = k][E][S] — k_dES] = k】([E]t — [ES])[S] — k.jES]二爲豐匚席k；同
[S]0 =⑸ + 旧=⑸eq + [P]eq =⑸eq ( + 霊)=⑸eq(l + K eq)
其中强
又平衡后，
邑二[P][E]也=[ES]
k^_"[EST，匚一WI
得[P]eq _ kjk?
⑸eq k_、k_2
[P]eq _ ^1^2
[S]eq k_\k 一2
在一「S 的表达式屮，用(⑸一 [S]eq) +⑸eq 代替⑸，用©q ⑸eq —(⑸一 [S]eq)代替[P] _ k]k2([S] — [S]eq) + kik2〔S]eq — k_ik_2Keq 卜扁 + k_ik_2([S] — [S]eq)旧 S k](⑸-⑸eq) + k 』S]eq + k-2K eq[S]eq 一 k_2([S] -⑸eq) + k_i + k 2 上
= ___________ (k]k2 + k_]k_2)(⑸—[S]eq) ______________ 二 伽]-⑸°q) k_! + k 2 4- (& + k_2K eq )[S]eq + (k x 一 k_2)([S] - [S]eq ) 1 K 办 + ([S] - [S]eq ) 其中
(k^ + Jk 一2)[E]t
ki - k-2
rz , _ k-i + k 2 4- (ki + k_2K eq )[S]eq
K M =
若» (⑸- [S]eq)‘ — r S =證(⑸一⑸eq) = k ⑸oUeg 一兀)
其中x = l - [S]/⑸o,
(b)由摩尔衡算
U ⑸o -讥S] + r s v = 0
u 为体积流量，V 为有效体积，稀释率D = u/V 得一rs = D([S]0 -⑸)
当Rp = 0.3 mm, - r s = 0.25/i"1 x (540 一 289・0)g/厶=62.75 g ・ IT — hr 1 当Rp = 1.0 mm. -r s = 0.25b-1 x (540 一 365.2)g/厶=43.7 g • L • h~r
(c)
乓(尺卩=0.3)_ rj (Rp = 0.3)
r s (/?p = 1.0)= z?(/?p = 1.0)
即乓(心=0.3)和心(心=1.0 )的比值等于两种粒径下的反应的有效系数之比。

3 H =77(0ico 址伽 一 1)
q (Rp = 0.3) (/>1(O.3)cot/i01(O.3) — 1 /0i(l.O)、2
r )(Rp = 1.0)= ei(l ・0)cot/u/)i(1.0) —1 (血(0.3)) 仞 Deff
平衡转化率为0.51,
-r = D ([S]0 一 [S]eq) = 0.25 x 540 x (1 - 0.51) = 66.16 g ・ L ・ h"1
/ 、 62.75
= °-3)= 66l6 = °-948 从而
x eq = 1 — [S]eq/[S]o = 1 — ⑸eq
[S]eq + [P]eq 1 + Keq。