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Biochemical Engineering Exam 1
Examination Instructions
The exam is open notes/open books/open resources
...

Your solutions should adhere to the following:
FINAL ANSWERS SHOULD BE BOXED FOR PROBLEM 1 and PROBLEM 2
SOLUTIONS/RESPONSES SHOULD BE ENCODED HERE IN MS WORD
...

Problem 1
Data on the enzymatic degradation of dichlorophenol at different concentrations are given in the
following table:
So (mg/L)
Rate (mg Phenol/L-hour)

30
2
...
8

90
5
...
4

150
8
...
1

250
4
...
5

You can assume that a substrate-inhibited kinetics facilitates this process and can be modelled
by:
𝑉=

𝑉𝑚
𝐾
𝑆
(1 + 𝑆𝑚 ) (1 + 𝐾 )
𝑆𝐼

What are the values (with correct units) for Vm, Km, and KSI?

Solution:
1
𝑆𝑜
1
𝑅𝑎𝑡𝑒

0
...
0166

0
...
0083

0
...
005

0
...
0033

0
...
2632

0
...
1351

0
...
1408

0
...
4

Biochemical Engineering Exam 1

1/S VS 1/RATE
0
...
5

0
...
4

0
...
3

0
...
2

0
...
1754

0
...
1351

0
...
1408

0
1

2

3

6

7

8

𝑦= 𝑚𝑥+ 𝑏
𝑦 = 8
...
151139
𝐴𝑡 𝑥 = 0 ; 𝑉 = 𝑉𝑚
𝑦 = 8
...
151139
1
= (8
...
151139
𝑉𝑚
1
𝑉𝑚

= 0
...
6164

𝐴𝑡 𝑦 = 0 ; 𝑥 =

𝑚𝑔 𝑃ℎ𝑒𝑛𝑜𝑙
𝐿−ℎ

−1
𝐾𝑚

𝑦 = 8
...
151139
0 = (8
...
151139
𝐾𝑚

(−0
...
203743
) (𝐾𝑚 )
𝐾𝑚

−0
...
203743
=
−0
...
151139
𝐾𝑚 = 54
...
2795
) (𝐾𝑆𝐼 )
𝐿
𝐿

(150)2

𝑚𝑔
𝑚𝑔
= (54
...
2795 𝐿

𝐾𝑆𝐼 = 414
...

The following data were obtained for enzymatic oxidation of phenylacetic acid at different
concentration of propionic acid:
S (mM)
V (mM/L-hour), I = 0
V (mM/L-hour), I = 1
mM
V (mM/L-hour), I = 2
mM

0
...
0
0
...
40
1
...
1

0
...
9
1
...
60
2
...
4

0
...
4
1
...
00
2
...
2

0
...
9

1
...
3

1
...
8

4

2
...
67

1
...
00

at I = 0 mM

1
...
588

0
...
476

0
...
4

at I = 1 mM

1
...
90

0
...
714

0
...
454

at I = 2 mM

1
...
1

1

0
...
714

0
...
For all the three conditions ([I] = 0, [I] = 1 mM, [I] = 2 mM)
...
5

1/V

1

0
...
5

2

3

4

5

1/S

Linear (1/v (I=0))

Linear (1/v (I=1))

Linear (1/v (I=2))

The figure 1 as shown above indicates that Vmax rises in the presence of an inhibitor, whereas
under all three circumstances [S] stays unchanged
...
Vmax is remains the same while in the
presence of inhibitors, Km is increasing
...

a
...
2035763 𝑥 + 0
...
2035763 𝑥 + 0
...
2035763)(0) + 0
...
1437161 ; 𝑉𝑚 = 6
...
2035763 𝑥 + 0
...
2035763) (

−1
) + 0
...
1437161)(𝐾𝑚 ) = (

−0
...
1437161 𝐾𝑚
−0
...
1437161
−0
...
4165

𝑚𝑔
𝐿

𝐴𝑡 𝐼 = 1:
𝑦= 𝑚𝑥+ 𝑏
𝑦 = 0
...
08477242
𝐴𝑡 𝑥 = 0 ; 𝑉 = 𝑉𝑚

𝑦 = 0
...
08477242
1
= (0
...
08477242
𝑉𝑚
1
𝑉𝑚

= 0
...
7963
𝐴𝑡 𝑦 = 0 ; 𝑥 =

𝑚𝑔 𝑃ℎ𝑒𝑛𝑜𝑙
𝐿−ℎ

−1
𝐾𝑚

𝑦 = 0
...
08477242
0 = (0
...
08477242
𝐾𝑚

(−0
...
3532294
) (𝐾𝑚 )
𝐾𝑚

Biochemical Engineering Exam 1

−0
...
3532294
=
−0
...
08477242
𝐾𝑚−𝑎𝑝𝑝 = 4
...
4165
𝛼=

𝐾𝐼 =

𝑚𝑔
𝐿

4
...
941616661
1
...
5150 𝑚𝑀
(2
...
4173847 𝑥 + 0
...
4173847 𝑥 + 0
...
4173847)(0) + 0
...
1251153 ; 𝑉𝑚 = 7
...
4173847 𝑥 + 0
...
4173847) (

−1
) + 0
...
1251153)(𝐾𝑚 ) = (

−0
...
1251153 𝐾𝑚
−0
...
1251153
−0
...
3360

𝑚𝑔
𝐿

𝑎𝑡 𝐼 = 0 ; 𝐾𝑚 = 1
...
3360 𝑚𝑀
= 2
...
4165 𝑚𝑀

2 𝑚𝑀
= 1
...
3551 − 1)𝑚𝑀

Problem 3
Michaelis-Menten kinetics are used to describe intracellular reactions
...

In in vitro batch reactors, the quasi-steady-state hypothesis does not hold for [Eo]  [So]
...
Rationalize why MichaelisMenten Kinetics and the quasi-steady-state approximation are still reasonable descriptions of
intracellular enzyme reactions
...
If your
answer exceeds three sentence, you will automatically be given a GRADE of ZERO in Problem
no
...


Biochemical Engineering Exam 1

Problem 4
You are working for company A and you join the research group working on immobilized
enzymes
...
His proof is that enzyme has a half-life of 10 days in free solution, but under identical
conditions of temperature, pH, and medium composition, the measured half-life of a packed
column is 30 days
...
Is Harry’s
reasoning right? Do you agree with him? Why or Why not? ANSWER IN NO MORE THAN
FOUR SENTENCES
...
4
Answer:
I agree with Harry, since the immobilization can extend the active lifespan of enzymes, but
it also can kill an enzyme with specific connections by changing the way the enzyme is related to
the enzyme's kinetics
...




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